Doubt in proof of Theorem 3.19 from Patrick Morandi’s *Field and Galois Theory*

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Relevant notations and results

  • Let $K / F$ be a field extension and let $alpha in K$ be algebraic over $F$. We denote the minimal polynomial of $alpha$ over $F$ by $min(F,alpha)$.
  • If $sigma : F to F’$ is a field homomorphism, then there is an induced ring homomorphism $F[x] to F'[x]$, which we also denote by $sigma$, given by $sigmaleft( sum a_i x^i right) = sum sigma(a_i) x^i$.
  • Lemma 3.17. Let $sigma : F to F’$ be a field isomorphism. Let $f(x) in F[x]$ be irreducible, let $alpha$ be a root of $f$ in some extension field $K$ of $F$, and let $alpha’$ be a root of $sigma(f)$ in some extension $K’$ of $F’$. Then there is an isomorphism $tau : F(alpha) to F'(alpha’)$ with $tau(alpha) = alpha’$ and $tau |_F = sigma$.

I am reading Patrick Morandi’s Field and Galois Theory, and on page 34, he states and proves the following theorem, which is a special case of the Isomorphism Extension Theorem.

Theorem 3.19. Let $sigma : F to F’$ be a field isomorphism, let $f(x) in F[x]$ and let $sigma(f) in F'[x]$ be the corresponding polynomial over $F’$. Let $K$ be a splitting field of $f$ over $F$, and let $K’$ be a splitting field of $sigma(f)$ over $F’$. Then there is an isomorphism $tau : K to K’$ with $tau|_F = sigma$. Furthermore, if $alpha in K$ and if $alpha’$ is any root of $sigma(min(F,alpha))$ in $K’$, then $tau$ can be chosen so that $tau(alpha) = alpha’$.

Proof: We prove this by induction on $n = [K:F]$. If $n = 1$, then $f$ splits over $F$, and the result is trivial in this case. So, suppose that $n > 1$ and that the result is true for splitting fields of degree less than $n$. $color{red}{text{If $f$ splits over $F$, then the result is clear.}}$ If not, let $p(x)$ be a nonlinear irreducible factor of $f(x)$, $color{blue}{text{let $alpha$ be a root of $p$,}}$ and let $alpha’$ be a root of $sigma(p)$. Set $L = F(alpha)$ and $L’ = F'(alpha’)$. Then $[L:F] > 1$, so $[K:L] < n$. By Lemma 3.17, there is a field isomorphism $rho : L to L’$ with $rho(alpha) = alpha’$. Since $K$ is a splitting field over $L$ for $f(x)$ and $K’$ is a splitting field over $L’$ for $sigma(f)$, by induction the isomorphism $rho$ extends to an isomorphism $tau : K to K’$. The isomorphism $tau$ is then an extension of $sigma$ (and $rho$), and $tau(alpha) = rho(alpha) = alpha’$. $$tag*{$blacksquare$}$$


My doubts are related to the highlighted parts in the proof:

  1. Upon assuming that $n = [K:F] > 1$, it is certain that $f$ cannot split over $F$, because then we would have $K = F implies [K:F] = 1$. So, isn’t the statement highlighted in red redundant?
  2. In the proof, $alpha$ is taken to be a root of (an irreducible factor of) $f$, and $alpha’$ a root of the corresponding polynomial. But in the statement of the theorem $alpha$ is an arbitrary element of $K$. It doesn’t seem to be obvious how to use the given proof to construct an isomorphism $K to K’$ which sends this arbitrary $alpha$ to an $alpha’$. What am I missing here?
share|cite|improve this question

    up vote
    4
    down vote

    favorite

    Relevant notations and results

    • Let $K / F$ be a field extension and let $alpha in K$ be algebraic over $F$. We denote the minimal polynomial of $alpha$ over $F$ by $min(F,alpha)$.
    • If $sigma : F to F’$ is a field homomorphism, then there is an induced ring homomorphism $F[x] to F'[x]$, which we also denote by $sigma$, given by $sigmaleft( sum a_i x^i right) = sum sigma(a_i) x^i$.
    • Lemma 3.17. Let $sigma : F to F’$ be a field isomorphism. Let $f(x) in F[x]$ be irreducible, let $alpha$ be a root of $f$ in some extension field $K$ of $F$, and let $alpha’$ be a root of $sigma(f)$ in some extension $K’$ of $F’$. Then there is an isomorphism $tau : F(alpha) to F'(alpha’)$ with $tau(alpha) = alpha’$ and $tau |_F = sigma$.

    I am reading Patrick Morandi’s Field and Galois Theory, and on page 34, he states and proves the following theorem, which is a special case of the Isomorphism Extension Theorem.

    Theorem 3.19. Let $sigma : F to F’$ be a field isomorphism, let $f(x) in F[x]$ and let $sigma(f) in F'[x]$ be the corresponding polynomial over $F’$. Let $K$ be a splitting field of $f$ over $F$, and let $K’$ be a splitting field of $sigma(f)$ over $F’$. Then there is an isomorphism $tau : K to K’$ with $tau|_F = sigma$. Furthermore, if $alpha in K$ and if $alpha’$ is any root of $sigma(min(F,alpha))$ in $K’$, then $tau$ can be chosen so that $tau(alpha) = alpha’$.

    Proof: We prove this by induction on $n = [K:F]$. If $n = 1$, then $f$ splits over $F$, and the result is trivial in this case. So, suppose that $n > 1$ and that the result is true for splitting fields of degree less than $n$. $color{red}{text{If $f$ splits over $F$, then the result is clear.}}$ If not, let $p(x)$ be a nonlinear irreducible factor of $f(x)$, $color{blue}{text{let $alpha$ be a root of $p$,}}$ and let $alpha’$ be a root of $sigma(p)$. Set $L = F(alpha)$ and $L’ = F'(alpha’)$. Then $[L:F] > 1$, so $[K:L] < n$. By Lemma 3.17, there is a field isomorphism $rho : L to L’$ with $rho(alpha) = alpha’$. Since $K$ is a splitting field over $L$ for $f(x)$ and $K’$ is a splitting field over $L’$ for $sigma(f)$, by induction the isomorphism $rho$ extends to an isomorphism $tau : K to K’$. The isomorphism $tau$ is then an extension of $sigma$ (and $rho$), and $tau(alpha) = rho(alpha) = alpha’$. $$tag*{$blacksquare$}$$


    My doubts are related to the highlighted parts in the proof:

    1. Upon assuming that $n = [K:F] > 1$, it is certain that $f$ cannot split over $F$, because then we would have $K = F implies [K:F] = 1$. So, isn’t the statement highlighted in red redundant?
    2. In the proof, $alpha$ is taken to be a root of (an irreducible factor of) $f$, and $alpha’$ a root of the corresponding polynomial. But in the statement of the theorem $alpha$ is an arbitrary element of $K$. It doesn’t seem to be obvious how to use the given proof to construct an isomorphism $K to K’$ which sends this arbitrary $alpha$ to an $alpha’$. What am I missing here?
    share|cite|improve this question

      up vote
      4
      down vote

      favorite

      up vote
      4
      down vote

      favorite

      Relevant notations and results

      • Let $K / F$ be a field extension and let $alpha in K$ be algebraic over $F$. We denote the minimal polynomial of $alpha$ over $F$ by $min(F,alpha)$.
      • If $sigma : F to F’$ is a field homomorphism, then there is an induced ring homomorphism $F[x] to F'[x]$, which we also denote by $sigma$, given by $sigmaleft( sum a_i x^i right) = sum sigma(a_i) x^i$.
      • Lemma 3.17. Let $sigma : F to F’$ be a field isomorphism. Let $f(x) in F[x]$ be irreducible, let $alpha$ be a root of $f$ in some extension field $K$ of $F$, and let $alpha’$ be a root of $sigma(f)$ in some extension $K’$ of $F’$. Then there is an isomorphism $tau : F(alpha) to F'(alpha’)$ with $tau(alpha) = alpha’$ and $tau |_F = sigma$.

      I am reading Patrick Morandi’s Field and Galois Theory, and on page 34, he states and proves the following theorem, which is a special case of the Isomorphism Extension Theorem.

      Theorem 3.19. Let $sigma : F to F’$ be a field isomorphism, let $f(x) in F[x]$ and let $sigma(f) in F'[x]$ be the corresponding polynomial over $F’$. Let $K$ be a splitting field of $f$ over $F$, and let $K’$ be a splitting field of $sigma(f)$ over $F’$. Then there is an isomorphism $tau : K to K’$ with $tau|_F = sigma$. Furthermore, if $alpha in K$ and if $alpha’$ is any root of $sigma(min(F,alpha))$ in $K’$, then $tau$ can be chosen so that $tau(alpha) = alpha’$.

      Proof: We prove this by induction on $n = [K:F]$. If $n = 1$, then $f$ splits over $F$, and the result is trivial in this case. So, suppose that $n > 1$ and that the result is true for splitting fields of degree less than $n$. $color{red}{text{If $f$ splits over $F$, then the result is clear.}}$ If not, let $p(x)$ be a nonlinear irreducible factor of $f(x)$, $color{blue}{text{let $alpha$ be a root of $p$,}}$ and let $alpha’$ be a root of $sigma(p)$. Set $L = F(alpha)$ and $L’ = F'(alpha’)$. Then $[L:F] > 1$, so $[K:L] < n$. By Lemma 3.17, there is a field isomorphism $rho : L to L’$ with $rho(alpha) = alpha’$. Since $K$ is a splitting field over $L$ for $f(x)$ and $K’$ is a splitting field over $L’$ for $sigma(f)$, by induction the isomorphism $rho$ extends to an isomorphism $tau : K to K’$. The isomorphism $tau$ is then an extension of $sigma$ (and $rho$), and $tau(alpha) = rho(alpha) = alpha’$. $$tag*{$blacksquare$}$$


      My doubts are related to the highlighted parts in the proof:

      1. Upon assuming that $n = [K:F] > 1$, it is certain that $f$ cannot split over $F$, because then we would have $K = F implies [K:F] = 1$. So, isn’t the statement highlighted in red redundant?
      2. In the proof, $alpha$ is taken to be a root of (an irreducible factor of) $f$, and $alpha’$ a root of the corresponding polynomial. But in the statement of the theorem $alpha$ is an arbitrary element of $K$. It doesn’t seem to be obvious how to use the given proof to construct an isomorphism $K to K’$ which sends this arbitrary $alpha$ to an $alpha’$. What am I missing here?
      share|cite|improve this question

      Relevant notations and results

      • Let $K / F$ be a field extension and let $alpha in K$ be algebraic over $F$. We denote the minimal polynomial of $alpha$ over $F$ by $min(F,alpha)$.
      • If $sigma : F to F’$ is a field homomorphism, then there is an induced ring homomorphism $F[x] to F'[x]$, which we also denote by $sigma$, given by $sigmaleft( sum a_i x^i right) = sum sigma(a_i) x^i$.
      • Lemma 3.17. Let $sigma : F to F’$ be a field isomorphism. Let $f(x) in F[x]$ be irreducible, let $alpha$ be a root of $f$ in some extension field $K$ of $F$, and let $alpha’$ be a root of $sigma(f)$ in some extension $K’$ of $F’$. Then there is an isomorphism $tau : F(alpha) to F'(alpha’)$ with $tau(alpha) = alpha’$ and $tau |_F = sigma$.

      I am reading Patrick Morandi’s Field and Galois Theory, and on page 34, he states and proves the following theorem, which is a special case of the Isomorphism Extension Theorem.

      Theorem 3.19. Let $sigma : F to F’$ be a field isomorphism, let $f(x) in F[x]$ and let $sigma(f) in F'[x]$ be the corresponding polynomial over $F’$. Let $K$ be a splitting field of $f$ over $F$, and let $K’$ be a splitting field of $sigma(f)$ over $F’$. Then there is an isomorphism $tau : K to K’$ with $tau|_F = sigma$. Furthermore, if $alpha in K$ and if $alpha’$ is any root of $sigma(min(F,alpha))$ in $K’$, then $tau$ can be chosen so that $tau(alpha) = alpha’$.

      Proof: We prove this by induction on $n = [K:F]$. If $n = 1$, then $f$ splits over $F$, and the result is trivial in this case. So, suppose that $n > 1$ and that the result is true for splitting fields of degree less than $n$. $color{red}{text{If $f$ splits over $F$, then the result is clear.}}$ If not, let $p(x)$ be a nonlinear irreducible factor of $f(x)$, $color{blue}{text{let $alpha$ be a root of $p$,}}$ and let $alpha’$ be a root of $sigma(p)$. Set $L = F(alpha)$ and $L’ = F'(alpha’)$. Then $[L:F] > 1$, so $[K:L] < n$. By Lemma 3.17, there is a field isomorphism $rho : L to L’$ with $rho(alpha) = alpha’$. Since $K$ is a splitting field over $L$ for $f(x)$ and $K’$ is a splitting field over $L’$ for $sigma(f)$, by induction the isomorphism $rho$ extends to an isomorphism $tau : K to K’$. The isomorphism $tau$ is then an extension of $sigma$ (and $rho$), and $tau(alpha) = rho(alpha) = alpha’$. $$tag*{$blacksquare$}$$


      My doubts are related to the highlighted parts in the proof:

      1. Upon assuming that $n = [K:F] > 1$, it is certain that $f$ cannot split over $F$, because then we would have $K = F implies [K:F] = 1$. So, isn’t the statement highlighted in red redundant?
      2. In the proof, $alpha$ is taken to be a root of (an irreducible factor of) $f$, and $alpha’$ a root of the corresponding polynomial. But in the statement of the theorem $alpha$ is an arbitrary element of $K$. It doesn’t seem to be obvious how to use the given proof to construct an isomorphism $K to K’$ which sends this arbitrary $alpha$ to an $alpha’$. What am I missing here?

      abstract-algebra field-theory galois-theory splitting-field

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      edited Sep 10 at 19:52

      asked Jun 17 at 18:52

      Brahadeesh

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          The proof is a little jumbled up. It is true that the sentence highlighted in red is redundant. Moreover, in the proof the author seems to be showing the existence of an isomorphism $tau : K to K’$ with $tau |_F = sigma$, which only incidentally maps a root of an irreducible factor $p$ of $f$ to a root of $sigma(p)$. The latter half of the theorem still needs proof.

          To correct/complete the proof, we can work in the following manner.


          If $n = 1$, then $f$ splits over $F$, and the result is trivial in this case. So, suppose that $n > 1$ and that the result is true for splitting fields of degree less than $n$.

          Let $alpha in K$ and $p = min(F,alpha)$. Suppose that $deg(p) > 1$. Let $alpha’ in K’$ be a root of $sigma(p)$. Set $L = F(alpha)$ and $L’ = F'(alpha’)$. Then, $[L:F] > 1$, so $[K:L] < n$. By Lemma 3.17, there is a field isomorphism $rho : L to L’$ with $rho(alpha) = alpha’$ and $rho |_F = sigma$. Since $K$ is a splitting field over $L$ for $f(x)$ and $K’$ is a splitting field over $L’$ for $sigma(f)$, by induction the isomorphism $rho$ extends to an isomorphism $tau : K to K’$. The isomorphism $tau$ is then an extension of $sigma$ (and $rho$), and $tau(alpha) = rho(alpha) = alpha’$.

          And, what if $deg(p) = 1$? Then, $alpha in F$, $p = x – alpha$ and $sigma(p) = x – sigma(alpha)$, so $alpha’ = sigma(alpha)$. So, any isomorphism $tau : K to K’$ with $tau |_F = sigma$ will satisfy $tau(alpha) = alpha’$. So, if $alpha in K$ is such that $deg(min(F,alpha)) = 1$, then we only need to show the existence of an isomorphism $tau : K to K’$ with $tau |_F = sigma$ to complete the proof. But this is the content of the proof as given in the textbook, so we are done.


          Incidentally, the proof of Theorem 3.20 (Isomorphism Extension Theorem), given in the textbook on pages 34-35, also suffers from the same flaw. There, too, the author constructs an isomorphism $tau : K to K’$ of splitting fields that extends the isomorphism $sigma : F to F’$ of the base fields, but neglects to prove that $tau$ can be chosen so that $tau(alpha) = alpha’$, where $alpha in K$ and $alpha’$ is any root of $sigma(min(F,alpha))$. A similar argument as above can complete that proof as well.

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            The proof is a little jumbled up. It is true that the sentence highlighted in red is redundant. Moreover, in the proof the author seems to be showing the existence of an isomorphism $tau : K to K’$ with $tau |_F = sigma$, which only incidentally maps a root of an irreducible factor $p$ of $f$ to a root of $sigma(p)$. The latter half of the theorem still needs proof.

            To correct/complete the proof, we can work in the following manner.


            If $n = 1$, then $f$ splits over $F$, and the result is trivial in this case. So, suppose that $n > 1$ and that the result is true for splitting fields of degree less than $n$.

            Let $alpha in K$ and $p = min(F,alpha)$. Suppose that $deg(p) > 1$. Let $alpha’ in K’$ be a root of $sigma(p)$. Set $L = F(alpha)$ and $L’ = F'(alpha’)$. Then, $[L:F] > 1$, so $[K:L] < n$. By Lemma 3.17, there is a field isomorphism $rho : L to L’$ with $rho(alpha) = alpha’$ and $rho |_F = sigma$. Since $K$ is a splitting field over $L$ for $f(x)$ and $K’$ is a splitting field over $L’$ for $sigma(f)$, by induction the isomorphism $rho$ extends to an isomorphism $tau : K to K’$. The isomorphism $tau$ is then an extension of $sigma$ (and $rho$), and $tau(alpha) = rho(alpha) = alpha’$.

            And, what if $deg(p) = 1$? Then, $alpha in F$, $p = x – alpha$ and $sigma(p) = x – sigma(alpha)$, so $alpha’ = sigma(alpha)$. So, any isomorphism $tau : K to K’$ with $tau |_F = sigma$ will satisfy $tau(alpha) = alpha’$. So, if $alpha in K$ is such that $deg(min(F,alpha)) = 1$, then we only need to show the existence of an isomorphism $tau : K to K’$ with $tau |_F = sigma$ to complete the proof. But this is the content of the proof as given in the textbook, so we are done.


            Incidentally, the proof of Theorem 3.20 (Isomorphism Extension Theorem), given in the textbook on pages 34-35, also suffers from the same flaw. There, too, the author constructs an isomorphism $tau : K to K’$ of splitting fields that extends the isomorphism $sigma : F to F’$ of the base fields, but neglects to prove that $tau$ can be chosen so that $tau(alpha) = alpha’$, where $alpha in K$ and $alpha’$ is any root of $sigma(min(F,alpha))$. A similar argument as above can complete that proof as well.

            share|cite|improve this answer

              up vote
              3
              down vote

              accepted

              The proof is a little jumbled up. It is true that the sentence highlighted in red is redundant. Moreover, in the proof the author seems to be showing the existence of an isomorphism $tau : K to K’$ with $tau |_F = sigma$, which only incidentally maps a root of an irreducible factor $p$ of $f$ to a root of $sigma(p)$. The latter half of the theorem still needs proof.

              To correct/complete the proof, we can work in the following manner.


              If $n = 1$, then $f$ splits over $F$, and the result is trivial in this case. So, suppose that $n > 1$ and that the result is true for splitting fields of degree less than $n$.

              Let $alpha in K$ and $p = min(F,alpha)$. Suppose that $deg(p) > 1$. Let $alpha’ in K’$ be a root of $sigma(p)$. Set $L = F(alpha)$ and $L’ = F'(alpha’)$. Then, $[L:F] > 1$, so $[K:L] < n$. By Lemma 3.17, there is a field isomorphism $rho : L to L’$ with $rho(alpha) = alpha’$ and $rho |_F = sigma$. Since $K$ is a splitting field over $L$ for $f(x)$ and $K’$ is a splitting field over $L’$ for $sigma(f)$, by induction the isomorphism $rho$ extends to an isomorphism $tau : K to K’$. The isomorphism $tau$ is then an extension of $sigma$ (and $rho$), and $tau(alpha) = rho(alpha) = alpha’$.

              And, what if $deg(p) = 1$? Then, $alpha in F$, $p = x – alpha$ and $sigma(p) = x – sigma(alpha)$, so $alpha’ = sigma(alpha)$. So, any isomorphism $tau : K to K’$ with $tau |_F = sigma$ will satisfy $tau(alpha) = alpha’$. So, if $alpha in K$ is such that $deg(min(F,alpha)) = 1$, then we only need to show the existence of an isomorphism $tau : K to K’$ with $tau |_F = sigma$ to complete the proof. But this is the content of the proof as given in the textbook, so we are done.


              Incidentally, the proof of Theorem 3.20 (Isomorphism Extension Theorem), given in the textbook on pages 34-35, also suffers from the same flaw. There, too, the author constructs an isomorphism $tau : K to K’$ of splitting fields that extends the isomorphism $sigma : F to F’$ of the base fields, but neglects to prove that $tau$ can be chosen so that $tau(alpha) = alpha’$, where $alpha in K$ and $alpha’$ is any root of $sigma(min(F,alpha))$. A similar argument as above can complete that proof as well.

              share|cite|improve this answer

                up vote
                3
                down vote

                accepted

                up vote
                3
                down vote

                accepted

                The proof is a little jumbled up. It is true that the sentence highlighted in red is redundant. Moreover, in the proof the author seems to be showing the existence of an isomorphism $tau : K to K’$ with $tau |_F = sigma$, which only incidentally maps a root of an irreducible factor $p$ of $f$ to a root of $sigma(p)$. The latter half of the theorem still needs proof.

                To correct/complete the proof, we can work in the following manner.


                If $n = 1$, then $f$ splits over $F$, and the result is trivial in this case. So, suppose that $n > 1$ and that the result is true for splitting fields of degree less than $n$.

                Let $alpha in K$ and $p = min(F,alpha)$. Suppose that $deg(p) > 1$. Let $alpha’ in K’$ be a root of $sigma(p)$. Set $L = F(alpha)$ and $L’ = F'(alpha’)$. Then, $[L:F] > 1$, so $[K:L] < n$. By Lemma 3.17, there is a field isomorphism $rho : L to L’$ with $rho(alpha) = alpha’$ and $rho |_F = sigma$. Since $K$ is a splitting field over $L$ for $f(x)$ and $K’$ is a splitting field over $L’$ for $sigma(f)$, by induction the isomorphism $rho$ extends to an isomorphism $tau : K to K’$. The isomorphism $tau$ is then an extension of $sigma$ (and $rho$), and $tau(alpha) = rho(alpha) = alpha’$.

                And, what if $deg(p) = 1$? Then, $alpha in F$, $p = x – alpha$ and $sigma(p) = x – sigma(alpha)$, so $alpha’ = sigma(alpha)$. So, any isomorphism $tau : K to K’$ with $tau |_F = sigma$ will satisfy $tau(alpha) = alpha’$. So, if $alpha in K$ is such that $deg(min(F,alpha)) = 1$, then we only need to show the existence of an isomorphism $tau : K to K’$ with $tau |_F = sigma$ to complete the proof. But this is the content of the proof as given in the textbook, so we are done.


                Incidentally, the proof of Theorem 3.20 (Isomorphism Extension Theorem), given in the textbook on pages 34-35, also suffers from the same flaw. There, too, the author constructs an isomorphism $tau : K to K’$ of splitting fields that extends the isomorphism $sigma : F to F’$ of the base fields, but neglects to prove that $tau$ can be chosen so that $tau(alpha) = alpha’$, where $alpha in K$ and $alpha’$ is any root of $sigma(min(F,alpha))$. A similar argument as above can complete that proof as well.

                share|cite|improve this answer

                The proof is a little jumbled up. It is true that the sentence highlighted in red is redundant. Moreover, in the proof the author seems to be showing the existence of an isomorphism $tau : K to K’$ with $tau |_F = sigma$, which only incidentally maps a root of an irreducible factor $p$ of $f$ to a root of $sigma(p)$. The latter half of the theorem still needs proof.

                To correct/complete the proof, we can work in the following manner.


                If $n = 1$, then $f$ splits over $F$, and the result is trivial in this case. So, suppose that $n > 1$ and that the result is true for splitting fields of degree less than $n$.

                Let $alpha in K$ and $p = min(F,alpha)$. Suppose that $deg(p) > 1$. Let $alpha’ in K’$ be a root of $sigma(p)$. Set $L = F(alpha)$ and $L’ = F'(alpha’)$. Then, $[L:F] > 1$, so $[K:L] < n$. By Lemma 3.17, there is a field isomorphism $rho : L to L’$ with $rho(alpha) = alpha’$ and $rho |_F = sigma$. Since $K$ is a splitting field over $L$ for $f(x)$ and $K’$ is a splitting field over $L’$ for $sigma(f)$, by induction the isomorphism $rho$ extends to an isomorphism $tau : K to K’$. The isomorphism $tau$ is then an extension of $sigma$ (and $rho$), and $tau(alpha) = rho(alpha) = alpha’$.

                And, what if $deg(p) = 1$? Then, $alpha in F$, $p = x – alpha$ and $sigma(p) = x – sigma(alpha)$, so $alpha’ = sigma(alpha)$. So, any isomorphism $tau : K to K’$ with $tau |_F = sigma$ will satisfy $tau(alpha) = alpha’$. So, if $alpha in K$ is such that $deg(min(F,alpha)) = 1$, then we only need to show the existence of an isomorphism $tau : K to K’$ with $tau |_F = sigma$ to complete the proof. But this is the content of the proof as given in the textbook, so we are done.


                Incidentally, the proof of Theorem 3.20 (Isomorphism Extension Theorem), given in the textbook on pages 34-35, also suffers from the same flaw. There, too, the author constructs an isomorphism $tau : K to K’$ of splitting fields that extends the isomorphism $sigma : F to F’$ of the base fields, but neglects to prove that $tau$ can be chosen so that $tau(alpha) = alpha’$, where $alpha in K$ and $alpha’$ is any root of $sigma(min(F,alpha))$. A similar argument as above can complete that proof as well.

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                edited Sep 10 at 19:56

                answered Jun 17 at 18:52

                Brahadeesh

                4,34131651

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                    Using induction to show associativity on $x_1+dots + x_n$

                    The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP

                    up vote
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                    favorite

                    I want to use induction to show that the sum $x_1 + dots + x_n$ of real numbers is defined independently of parentheses to specify order of addition.

                    I know how to apply induction(base, assumption, k+1 applying inductive hypothesis). Here I am not sure what the base would be. I have two ideas:

                    1) First case is $(x_1 + x_2)+x_3+dots+x_n$ and work through to $x_1+x_2+dots+x_{n-2} + (x_{n-1} + x_n)$

                    2) Start with $(x_1+x_2)+x_3=x_1+(x_2+x_3)$ and work up in number of elements to the full case.

                    Both seem wrong, I have no idea what to actually do.

                    I imagine above is sufficient effort, although I have shown no working. Before you downvote, please tell me why you are planning it, and I will edit.

                    share|cite|improve this question

                    • You cool with Peano’s axioms?
                      – Daniel W. Farlow
                      Jan 29 ’15 at 5:55

                    • @qqqqq Peano’s axioms are a definition of integers.
                      – Alice Ryhl
                      Jan 29 ’15 at 5:56

                    • @induktio Yeah from wiki it all seems familiar(except written full set theoretic). Equivalence relations, abelian group and ring construction, looks good
                      – qqqqq
                      Jan 29 ’15 at 6:00

                    up vote
                    1
                    down vote

                    favorite

                    I want to use induction to show that the sum $x_1 + dots + x_n$ of real numbers is defined independently of parentheses to specify order of addition.

                    I know how to apply induction(base, assumption, k+1 applying inductive hypothesis). Here I am not sure what the base would be. I have two ideas:

                    1) First case is $(x_1 + x_2)+x_3+dots+x_n$ and work through to $x_1+x_2+dots+x_{n-2} + (x_{n-1} + x_n)$

                    2) Start with $(x_1+x_2)+x_3=x_1+(x_2+x_3)$ and work up in number of elements to the full case.

                    Both seem wrong, I have no idea what to actually do.

                    I imagine above is sufficient effort, although I have shown no working. Before you downvote, please tell me why you are planning it, and I will edit.

                    share|cite|improve this question

                    • You cool with Peano’s axioms?
                      – Daniel W. Farlow
                      Jan 29 ’15 at 5:55

                    • @qqqqq Peano’s axioms are a definition of integers.
                      – Alice Ryhl
                      Jan 29 ’15 at 5:56

                    • @induktio Yeah from wiki it all seems familiar(except written full set theoretic). Equivalence relations, abelian group and ring construction, looks good
                      – qqqqq
                      Jan 29 ’15 at 6:00

                    up vote
                    1
                    down vote

                    favorite

                    up vote
                    1
                    down vote

                    favorite

                    I want to use induction to show that the sum $x_1 + dots + x_n$ of real numbers is defined independently of parentheses to specify order of addition.

                    I know how to apply induction(base, assumption, k+1 applying inductive hypothesis). Here I am not sure what the base would be. I have two ideas:

                    1) First case is $(x_1 + x_2)+x_3+dots+x_n$ and work through to $x_1+x_2+dots+x_{n-2} + (x_{n-1} + x_n)$

                    2) Start with $(x_1+x_2)+x_3=x_1+(x_2+x_3)$ and work up in number of elements to the full case.

                    Both seem wrong, I have no idea what to actually do.

                    I imagine above is sufficient effort, although I have shown no working. Before you downvote, please tell me why you are planning it, and I will edit.

                    share|cite|improve this question

                    I want to use induction to show that the sum $x_1 + dots + x_n$ of real numbers is defined independently of parentheses to specify order of addition.

                    I know how to apply induction(base, assumption, k+1 applying inductive hypothesis). Here I am not sure what the base would be. I have two ideas:

                    1) First case is $(x_1 + x_2)+x_3+dots+x_n$ and work through to $x_1+x_2+dots+x_{n-2} + (x_{n-1} + x_n)$

                    2) Start with $(x_1+x_2)+x_3=x_1+(x_2+x_3)$ and work up in number of elements to the full case.

                    Both seem wrong, I have no idea what to actually do.

                    I imagine above is sufficient effort, although I have shown no working. Before you downvote, please tell me why you are planning it, and I will edit.

                    induction

                    share|cite|improve this question

                    share|cite|improve this question

                    share|cite|improve this question

                    share|cite|improve this question

                    edited Jan 29 ’15 at 7:12

                    Asaf Karagila♦

                    295k32411738

                    295k32411738

                    asked Jan 29 ’15 at 5:47

                    qqqqq

                    61

                    61

                    • You cool with Peano’s axioms?
                      – Daniel W. Farlow
                      Jan 29 ’15 at 5:55

                    • @qqqqq Peano’s axioms are a definition of integers.
                      – Alice Ryhl
                      Jan 29 ’15 at 5:56

                    • @induktio Yeah from wiki it all seems familiar(except written full set theoretic). Equivalence relations, abelian group and ring construction, looks good
                      – qqqqq
                      Jan 29 ’15 at 6:00

                    • You cool with Peano’s axioms?
                      – Daniel W. Farlow
                      Jan 29 ’15 at 5:55

                    • @qqqqq Peano’s axioms are a definition of integers.
                      – Alice Ryhl
                      Jan 29 ’15 at 5:56

                    • @induktio Yeah from wiki it all seems familiar(except written full set theoretic). Equivalence relations, abelian group and ring construction, looks good
                      – qqqqq
                      Jan 29 ’15 at 6:00

                    You cool with Peano’s axioms?
                    – Daniel W. Farlow
                    Jan 29 ’15 at 5:55

                    You cool with Peano’s axioms?
                    – Daniel W. Farlow
                    Jan 29 ’15 at 5:55

                    @qqqqq Peano’s axioms are a definition of integers.
                    – Alice Ryhl
                    Jan 29 ’15 at 5:56

                    @qqqqq Peano’s axioms are a definition of integers.
                    – Alice Ryhl
                    Jan 29 ’15 at 5:56

                    @induktio Yeah from wiki it all seems familiar(except written full set theoretic). Equivalence relations, abelian group and ring construction, looks good
                    – qqqqq
                    Jan 29 ’15 at 6:00

                    @induktio Yeah from wiki it all seems familiar(except written full set theoretic). Equivalence relations, abelian group and ring construction, looks good
                    – qqqqq
                    Jan 29 ’15 at 6:00

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                    Your first idea is ambiguous, since the expressions aren’t fully parenthesized. Note that for $4$ terms, there are $5$ possible parenthesizations: $((x_1+x_2)+x_3)+x_4$, $(x_1+(x_2+x_3))+x_4$, $(x_1+x_2)+(x_3+x_4)$, $x_1+((x_2+x_3)+x_4)$, and $x_1+(x_2+(x_3+x_4))$. (In general, the number of parenthesizations of an expression with $n$ terms is the Catalan number $C_{n-1}$.)

                    I wouldn’t bother trying to go through all of the different parenthesizations in some order, but use your second idea instead. So the induction hypothesis is that every parenthesization of $k$ terms is equivalent to every other, and you want to prove the same for $k+1$ terms (for $kgeq3$). Actually, come to think of it, it will probably be easier to use strong induction: the induction hypothesis is that every parenthesization of $k$ terms is equivalent to every other whenever $k<n$, and prove the same for $n$ terms (for $n>3$). It may help to note that any given parenthesization has a position where the last operation (in the order that they are carried out) is applied.

                    share|cite|improve this answer

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                      Your first idea is ambiguous, since the expressions aren’t fully parenthesized. Note that for $4$ terms, there are $5$ possible parenthesizations: $((x_1+x_2)+x_3)+x_4$, $(x_1+(x_2+x_3))+x_4$, $(x_1+x_2)+(x_3+x_4)$, $x_1+((x_2+x_3)+x_4)$, and $x_1+(x_2+(x_3+x_4))$. (In general, the number of parenthesizations of an expression with $n$ terms is the Catalan number $C_{n-1}$.)

                      I wouldn’t bother trying to go through all of the different parenthesizations in some order, but use your second idea instead. So the induction hypothesis is that every parenthesization of $k$ terms is equivalent to every other, and you want to prove the same for $k+1$ terms (for $kgeq3$). Actually, come to think of it, it will probably be easier to use strong induction: the induction hypothesis is that every parenthesization of $k$ terms is equivalent to every other whenever $k<n$, and prove the same for $n$ terms (for $n>3$). It may help to note that any given parenthesization has a position where the last operation (in the order that they are carried out) is applied.

                      share|cite|improve this answer

                        up vote
                        1
                        down vote

                        Your first idea is ambiguous, since the expressions aren’t fully parenthesized. Note that for $4$ terms, there are $5$ possible parenthesizations: $((x_1+x_2)+x_3)+x_4$, $(x_1+(x_2+x_3))+x_4$, $(x_1+x_2)+(x_3+x_4)$, $x_1+((x_2+x_3)+x_4)$, and $x_1+(x_2+(x_3+x_4))$. (In general, the number of parenthesizations of an expression with $n$ terms is the Catalan number $C_{n-1}$.)

                        I wouldn’t bother trying to go through all of the different parenthesizations in some order, but use your second idea instead. So the induction hypothesis is that every parenthesization of $k$ terms is equivalent to every other, and you want to prove the same for $k+1$ terms (for $kgeq3$). Actually, come to think of it, it will probably be easier to use strong induction: the induction hypothesis is that every parenthesization of $k$ terms is equivalent to every other whenever $k<n$, and prove the same for $n$ terms (for $n>3$). It may help to note that any given parenthesization has a position where the last operation (in the order that they are carried out) is applied.

                        share|cite|improve this answer

                          up vote
                          1
                          down vote

                          up vote
                          1
                          down vote

                          Your first idea is ambiguous, since the expressions aren’t fully parenthesized. Note that for $4$ terms, there are $5$ possible parenthesizations: $((x_1+x_2)+x_3)+x_4$, $(x_1+(x_2+x_3))+x_4$, $(x_1+x_2)+(x_3+x_4)$, $x_1+((x_2+x_3)+x_4)$, and $x_1+(x_2+(x_3+x_4))$. (In general, the number of parenthesizations of an expression with $n$ terms is the Catalan number $C_{n-1}$.)

                          I wouldn’t bother trying to go through all of the different parenthesizations in some order, but use your second idea instead. So the induction hypothesis is that every parenthesization of $k$ terms is equivalent to every other, and you want to prove the same for $k+1$ terms (for $kgeq3$). Actually, come to think of it, it will probably be easier to use strong induction: the induction hypothesis is that every parenthesization of $k$ terms is equivalent to every other whenever $k<n$, and prove the same for $n$ terms (for $n>3$). It may help to note that any given parenthesization has a position where the last operation (in the order that they are carried out) is applied.

                          share|cite|improve this answer

                          Your first idea is ambiguous, since the expressions aren’t fully parenthesized. Note that for $4$ terms, there are $5$ possible parenthesizations: $((x_1+x_2)+x_3)+x_4$, $(x_1+(x_2+x_3))+x_4$, $(x_1+x_2)+(x_3+x_4)$, $x_1+((x_2+x_3)+x_4)$, and $x_1+(x_2+(x_3+x_4))$. (In general, the number of parenthesizations of an expression with $n$ terms is the Catalan number $C_{n-1}$.)

                          I wouldn’t bother trying to go through all of the different parenthesizations in some order, but use your second idea instead. So the induction hypothesis is that every parenthesization of $k$ terms is equivalent to every other, and you want to prove the same for $k+1$ terms (for $kgeq3$). Actually, come to think of it, it will probably be easier to use strong induction: the induction hypothesis is that every parenthesization of $k$ terms is equivalent to every other whenever $k<n$, and prove the same for $n$ terms (for $n>3$). It may help to note that any given parenthesization has a position where the last operation (in the order that they are carried out) is applied.

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                          share|cite|improve this answer

                          share|cite|improve this answer

                          answered Sep 10 at 21:25

                          Toby Bartels

                          528412

                          528412

                               
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                              Is this proof of $aleq b_1$ for every $b_1>b$, then $aleq b$ logically correct?

                              The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP

                              up vote
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                              favorite

                              Let $a,binBbb{R}$. Show if $aleq b_1$ for every $b_1>b$, then $aleq b$. Solve using proof-by-contradiction and write out using logical premises.

                              Proof: Let $Q: aleq b_1$ for every $b_1>b$ and $R:aleq b$ then $P:Qto R$. Suppose $neg P$, then we have $neg Piff Qlandneg R$. Thus suppose $a>b$ and let $b_1=dfrac{a+b}{2}$ and consider $Q$. We have $b_1>b$ but $a>b_1$ hence $neg Rtoneg Q$. Hence we have
                              $$(neg PRightarrow Qlandneg R) land(neg P Rightarrowneg Rtoneg Q)iffneg PRightarrow F$$
                              and thus $P$ must be true. $square$

                              I am uncertain about the last step where I have $neg P Rightarrowneg Rtoneg Q$ and use it to combine the contradiction $(Qlandneg R)land(neg Rtoneg Q)$.

                              share|cite|improve this question

                              • It would be a lot easier to understand what you’re doing if you don’t abbreviate your meaningful propositions as $P$, $Q$, $R$.
                                – Henning Makholm
                                Sep 10 at 21:45

                              • Also: It’s almost always a confusing detour to attempt to prove an implication by contradiction. If you want to prove $Qto R$, assume $Q$ and seek to derive $R$. Now you may choose to prove $R$ by contradiction if you must — but there’s no need to justify the initial assumption of $Q$ as proof-by-contradiction. Assuming $Q$ is simply how you prove an implication directly.
                                – Henning Makholm
                                Sep 10 at 21:48

                              • @HenningMakholm Yeah, sorry for that. I do it because I am practicing writing logical expressions.
                                – Bright
                                Sep 10 at 21:49

                              • @HenningMakholm My goal is to negate $P$ and show that $neg Pto F$ to prove $P$. I don’t understand what you mean by proving $R$.
                                – Bright
                                Sep 10 at 22:01

                              • 1

                                If you insist of writing your proof in that horribly confusing and convoluted way, suit yourself.
                                – Henning Makholm
                                Sep 10 at 23:35

                              up vote
                              1
                              down vote

                              favorite

                              Let $a,binBbb{R}$. Show if $aleq b_1$ for every $b_1>b$, then $aleq b$. Solve using proof-by-contradiction and write out using logical premises.

                              Proof: Let $Q: aleq b_1$ for every $b_1>b$ and $R:aleq b$ then $P:Qto R$. Suppose $neg P$, then we have $neg Piff Qlandneg R$. Thus suppose $a>b$ and let $b_1=dfrac{a+b}{2}$ and consider $Q$. We have $b_1>b$ but $a>b_1$ hence $neg Rtoneg Q$. Hence we have
                              $$(neg PRightarrow Qlandneg R) land(neg P Rightarrowneg Rtoneg Q)iffneg PRightarrow F$$
                              and thus $P$ must be true. $square$

                              I am uncertain about the last step where I have $neg P Rightarrowneg Rtoneg Q$ and use it to combine the contradiction $(Qlandneg R)land(neg Rtoneg Q)$.

                              share|cite|improve this question

                              • It would be a lot easier to understand what you’re doing if you don’t abbreviate your meaningful propositions as $P$, $Q$, $R$.
                                – Henning Makholm
                                Sep 10 at 21:45

                              • Also: It’s almost always a confusing detour to attempt to prove an implication by contradiction. If you want to prove $Qto R$, assume $Q$ and seek to derive $R$. Now you may choose to prove $R$ by contradiction if you must — but there’s no need to justify the initial assumption of $Q$ as proof-by-contradiction. Assuming $Q$ is simply how you prove an implication directly.
                                – Henning Makholm
                                Sep 10 at 21:48

                              • @HenningMakholm Yeah, sorry for that. I do it because I am practicing writing logical expressions.
                                – Bright
                                Sep 10 at 21:49

                              • @HenningMakholm My goal is to negate $P$ and show that $neg Pto F$ to prove $P$. I don’t understand what you mean by proving $R$.
                                – Bright
                                Sep 10 at 22:01

                              • 1

                                If you insist of writing your proof in that horribly confusing and convoluted way, suit yourself.
                                – Henning Makholm
                                Sep 10 at 23:35

                              up vote
                              1
                              down vote

                              favorite

                              up vote
                              1
                              down vote

                              favorite

                              Let $a,binBbb{R}$. Show if $aleq b_1$ for every $b_1>b$, then $aleq b$. Solve using proof-by-contradiction and write out using logical premises.

                              Proof: Let $Q: aleq b_1$ for every $b_1>b$ and $R:aleq b$ then $P:Qto R$. Suppose $neg P$, then we have $neg Piff Qlandneg R$. Thus suppose $a>b$ and let $b_1=dfrac{a+b}{2}$ and consider $Q$. We have $b_1>b$ but $a>b_1$ hence $neg Rtoneg Q$. Hence we have
                              $$(neg PRightarrow Qlandneg R) land(neg P Rightarrowneg Rtoneg Q)iffneg PRightarrow F$$
                              and thus $P$ must be true. $square$

                              I am uncertain about the last step where I have $neg P Rightarrowneg Rtoneg Q$ and use it to combine the contradiction $(Qlandneg R)land(neg Rtoneg Q)$.

                              share|cite|improve this question

                              Let $a,binBbb{R}$. Show if $aleq b_1$ for every $b_1>b$, then $aleq b$. Solve using proof-by-contradiction and write out using logical premises.

                              Proof: Let $Q: aleq b_1$ for every $b_1>b$ and $R:aleq b$ then $P:Qto R$. Suppose $neg P$, then we have $neg Piff Qlandneg R$. Thus suppose $a>b$ and let $b_1=dfrac{a+b}{2}$ and consider $Q$. We have $b_1>b$ but $a>b_1$ hence $neg Rtoneg Q$. Hence we have
                              $$(neg PRightarrow Qlandneg R) land(neg P Rightarrowneg Rtoneg Q)iffneg PRightarrow F$$
                              and thus $P$ must be true. $square$

                              I am uncertain about the last step where I have $neg P Rightarrowneg Rtoneg Q$ and use it to combine the contradiction $(Qlandneg R)land(neg Rtoneg Q)$.

                              calculus proof-verification logic proof-writing

                              share|cite|improve this question

                              share|cite|improve this question

                              share|cite|improve this question

                              share|cite|improve this question

                              edited Sep 10 at 22:23

                              asked Sep 10 at 21:41

                              Bright

                              18610

                              18610

                              • It would be a lot easier to understand what you’re doing if you don’t abbreviate your meaningful propositions as $P$, $Q$, $R$.
                                – Henning Makholm
                                Sep 10 at 21:45

                              • Also: It’s almost always a confusing detour to attempt to prove an implication by contradiction. If you want to prove $Qto R$, assume $Q$ and seek to derive $R$. Now you may choose to prove $R$ by contradiction if you must — but there’s no need to justify the initial assumption of $Q$ as proof-by-contradiction. Assuming $Q$ is simply how you prove an implication directly.
                                – Henning Makholm
                                Sep 10 at 21:48

                              • @HenningMakholm Yeah, sorry for that. I do it because I am practicing writing logical expressions.
                                – Bright
                                Sep 10 at 21:49

                              • @HenningMakholm My goal is to negate $P$ and show that $neg Pto F$ to prove $P$. I don’t understand what you mean by proving $R$.
                                – Bright
                                Sep 10 at 22:01

                              • 1

                                If you insist of writing your proof in that horribly confusing and convoluted way, suit yourself.
                                – Henning Makholm
                                Sep 10 at 23:35

                              • It would be a lot easier to understand what you’re doing if you don’t abbreviate your meaningful propositions as $P$, $Q$, $R$.
                                – Henning Makholm
                                Sep 10 at 21:45

                              • Also: It’s almost always a confusing detour to attempt to prove an implication by contradiction. If you want to prove $Qto R$, assume $Q$ and seek to derive $R$. Now you may choose to prove $R$ by contradiction if you must — but there’s no need to justify the initial assumption of $Q$ as proof-by-contradiction. Assuming $Q$ is simply how you prove an implication directly.
                                – Henning Makholm
                                Sep 10 at 21:48

                              • @HenningMakholm Yeah, sorry for that. I do it because I am practicing writing logical expressions.
                                – Bright
                                Sep 10 at 21:49

                              • @HenningMakholm My goal is to negate $P$ and show that $neg Pto F$ to prove $P$. I don’t understand what you mean by proving $R$.
                                – Bright
                                Sep 10 at 22:01

                              • 1

                                If you insist of writing your proof in that horribly confusing and convoluted way, suit yourself.
                                – Henning Makholm
                                Sep 10 at 23:35

                              It would be a lot easier to understand what you’re doing if you don’t abbreviate your meaningful propositions as $P$, $Q$, $R$.
                              – Henning Makholm
                              Sep 10 at 21:45

                              It would be a lot easier to understand what you’re doing if you don’t abbreviate your meaningful propositions as $P$, $Q$, $R$.
                              – Henning Makholm
                              Sep 10 at 21:45

                              Also: It’s almost always a confusing detour to attempt to prove an implication by contradiction. If you want to prove $Qto R$, assume $Q$ and seek to derive $R$. Now you may choose to prove $R$ by contradiction if you must — but there’s no need to justify the initial assumption of $Q$ as proof-by-contradiction. Assuming $Q$ is simply how you prove an implication directly.
                              – Henning Makholm
                              Sep 10 at 21:48

                              Also: It’s almost always a confusing detour to attempt to prove an implication by contradiction. If you want to prove $Qto R$, assume $Q$ and seek to derive $R$. Now you may choose to prove $R$ by contradiction if you must — but there’s no need to justify the initial assumption of $Q$ as proof-by-contradiction. Assuming $Q$ is simply how you prove an implication directly.
                              – Henning Makholm
                              Sep 10 at 21:48

                              @HenningMakholm Yeah, sorry for that. I do it because I am practicing writing logical expressions.
                              – Bright
                              Sep 10 at 21:49

                              @HenningMakholm Yeah, sorry for that. I do it because I am practicing writing logical expressions.
                              – Bright
                              Sep 10 at 21:49

                              @HenningMakholm My goal is to negate $P$ and show that $neg Pto F$ to prove $P$. I don’t understand what you mean by proving $R$.
                              – Bright
                              Sep 10 at 22:01

                              @HenningMakholm My goal is to negate $P$ and show that $neg Pto F$ to prove $P$. I don’t understand what you mean by proving $R$.
                              – Bright
                              Sep 10 at 22:01

                              1

                              1

                              If you insist of writing your proof in that horribly confusing and convoluted way, suit yourself.
                              – Henning Makholm
                              Sep 10 at 23:35

                              If you insist of writing your proof in that horribly confusing and convoluted way, suit yourself.
                              – Henning Makholm
                              Sep 10 at 23:35

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                              Strongly p-embedded subgroups and p-Sylow subgroups.

                              The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP

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                              Let $G$ be a finite group and $p$ a prime number. Recall that a subgroup $H$ of $G$ is strongly $p$-embedded if $p | |H|,$ and for each $x in G setminus H,$ $H cap {^x}H$ has order prime to $p,$ where ${^x}H$ denotes the group we get after conjugation by $x.$ I am reading a proof which claims that any strongly $p$-embedded group contains a $p$-Sylow group of $G$. I can prove this fact by other means, but I would like to understand the proof I am reading. This is the argument I am reading:

                              Assume $H$ is strongly $p$-embedded. Since $p | |H|,$ $p$ contains an element $g$ of order $p.$ Choose any $S in Syl_p(G)$ such that $g in S.$ For $x in Z(S)$ of order $p,$ $$g in H cap {^x}H$$ implies $x in H.$ Hence for all $y in S,$ $$x in H cap {^y}H $$ implies that $y in H.$ Thus $S subset H.$

                              Here is what I do not understand with the proof:

                              1. Why does it matter that $x in Z(S)$ has order $p?$ It seems to me that for any $x in Z(S),$ we have that $g in H cap {^x} H.$
                              2. Suppose I know that for $x in Z(S)$ of order $p,$ $g in H cap {^x}H$ implies $x in H,$ i.e. that the elements of order $p$ in $Z(S)$ is a subset of $H.$ Why does it follow then that if $y in S$ is arbitrary, that $x in H cap {^y}H?$
                              share|cite|improve this question

                              • I agree that for your first question the order of $x$ need not be equal to $p$. But the second question I find a bit confusing. Do you mean $Z(S)$ instead of $Z(P)$? If so, it seems to make sense.
                                – James
                                Sep 10 at 22:28

                              • @James Thanks for pointing that out. Yes, I meant $Z(S).$ Does it make sense in the sense that you understand the argument of the proof, or in the sense that you understand my second question?
                                – Dedalus
                                Sep 10 at 22:35

                              • Well, I now understand the question and, I think, also the argument. As per the first question, you need not restrict to elements of order $p$ so, in fact, we have $Z(S)leq H$ from the first part of the argument. And, there exists, therefore, $1neq tin Z(S)$ (so $t$ is also an element of $H$). So, if $y$ is now any element in $S$, we have $t = t^yin Hcap {}^{y}H$ and has $p$-power order, so $y$ must belong to $H$ by the assumption that $H$ is strongly $p$-embedded. This gives $Sleq H$.
                                – James
                                Sep 10 at 22:45

                              • Thanks, I misread the second statemnt and read it as $g in H cap {^y}H.$
                                – Dedalus
                                Sep 10 at 22:55

                              up vote
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                              down vote

                              favorite

                              Let $G$ be a finite group and $p$ a prime number. Recall that a subgroup $H$ of $G$ is strongly $p$-embedded if $p | |H|,$ and for each $x in G setminus H,$ $H cap {^x}H$ has order prime to $p,$ where ${^x}H$ denotes the group we get after conjugation by $x.$ I am reading a proof which claims that any strongly $p$-embedded group contains a $p$-Sylow group of $G$. I can prove this fact by other means, but I would like to understand the proof I am reading. This is the argument I am reading:

                              Assume $H$ is strongly $p$-embedded. Since $p | |H|,$ $p$ contains an element $g$ of order $p.$ Choose any $S in Syl_p(G)$ such that $g in S.$ For $x in Z(S)$ of order $p,$ $$g in H cap {^x}H$$ implies $x in H.$ Hence for all $y in S,$ $$x in H cap {^y}H $$ implies that $y in H.$ Thus $S subset H.$

                              Here is what I do not understand with the proof:

                              1. Why does it matter that $x in Z(S)$ has order $p?$ It seems to me that for any $x in Z(S),$ we have that $g in H cap {^x} H.$
                              2. Suppose I know that for $x in Z(S)$ of order $p,$ $g in H cap {^x}H$ implies $x in H,$ i.e. that the elements of order $p$ in $Z(S)$ is a subset of $H.$ Why does it follow then that if $y in S$ is arbitrary, that $x in H cap {^y}H?$
                              share|cite|improve this question

                              • I agree that for your first question the order of $x$ need not be equal to $p$. But the second question I find a bit confusing. Do you mean $Z(S)$ instead of $Z(P)$? If so, it seems to make sense.
                                – James
                                Sep 10 at 22:28

                              • @James Thanks for pointing that out. Yes, I meant $Z(S).$ Does it make sense in the sense that you understand the argument of the proof, or in the sense that you understand my second question?
                                – Dedalus
                                Sep 10 at 22:35

                              • Well, I now understand the question and, I think, also the argument. As per the first question, you need not restrict to elements of order $p$ so, in fact, we have $Z(S)leq H$ from the first part of the argument. And, there exists, therefore, $1neq tin Z(S)$ (so $t$ is also an element of $H$). So, if $y$ is now any element in $S$, we have $t = t^yin Hcap {}^{y}H$ and has $p$-power order, so $y$ must belong to $H$ by the assumption that $H$ is strongly $p$-embedded. This gives $Sleq H$.
                                – James
                                Sep 10 at 22:45

                              • Thanks, I misread the second statemnt and read it as $g in H cap {^y}H.$
                                – Dedalus
                                Sep 10 at 22:55

                              up vote
                              1
                              down vote

                              favorite

                              up vote
                              1
                              down vote

                              favorite

                              Let $G$ be a finite group and $p$ a prime number. Recall that a subgroup $H$ of $G$ is strongly $p$-embedded if $p | |H|,$ and for each $x in G setminus H,$ $H cap {^x}H$ has order prime to $p,$ where ${^x}H$ denotes the group we get after conjugation by $x.$ I am reading a proof which claims that any strongly $p$-embedded group contains a $p$-Sylow group of $G$. I can prove this fact by other means, but I would like to understand the proof I am reading. This is the argument I am reading:

                              Assume $H$ is strongly $p$-embedded. Since $p | |H|,$ $p$ contains an element $g$ of order $p.$ Choose any $S in Syl_p(G)$ such that $g in S.$ For $x in Z(S)$ of order $p,$ $$g in H cap {^x}H$$ implies $x in H.$ Hence for all $y in S,$ $$x in H cap {^y}H $$ implies that $y in H.$ Thus $S subset H.$

                              Here is what I do not understand with the proof:

                              1. Why does it matter that $x in Z(S)$ has order $p?$ It seems to me that for any $x in Z(S),$ we have that $g in H cap {^x} H.$
                              2. Suppose I know that for $x in Z(S)$ of order $p,$ $g in H cap {^x}H$ implies $x in H,$ i.e. that the elements of order $p$ in $Z(S)$ is a subset of $H.$ Why does it follow then that if $y in S$ is arbitrary, that $x in H cap {^y}H?$
                              share|cite|improve this question

                              Let $G$ be a finite group and $p$ a prime number. Recall that a subgroup $H$ of $G$ is strongly $p$-embedded if $p | |H|,$ and for each $x in G setminus H,$ $H cap {^x}H$ has order prime to $p,$ where ${^x}H$ denotes the group we get after conjugation by $x.$ I am reading a proof which claims that any strongly $p$-embedded group contains a $p$-Sylow group of $G$. I can prove this fact by other means, but I would like to understand the proof I am reading. This is the argument I am reading:

                              Assume $H$ is strongly $p$-embedded. Since $p | |H|,$ $p$ contains an element $g$ of order $p.$ Choose any $S in Syl_p(G)$ such that $g in S.$ For $x in Z(S)$ of order $p,$ $$g in H cap {^x}H$$ implies $x in H.$ Hence for all $y in S,$ $$x in H cap {^y}H $$ implies that $y in H.$ Thus $S subset H.$

                              Here is what I do not understand with the proof:

                              1. Why does it matter that $x in Z(S)$ has order $p?$ It seems to me that for any $x in Z(S),$ we have that $g in H cap {^x} H.$
                              2. Suppose I know that for $x in Z(S)$ of order $p,$ $g in H cap {^x}H$ implies $x in H,$ i.e. that the elements of order $p$ in $Z(S)$ is a subset of $H.$ Why does it follow then that if $y in S$ is arbitrary, that $x in H cap {^y}H?$

                              group-theory

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                              share|cite|improve this question

                              share|cite|improve this question

                              share|cite|improve this question

                              edited Sep 10 at 22:34

                              asked Sep 10 at 21:40

                              Dedalus

                              1,96211736

                              1,96211736

                              • I agree that for your first question the order of $x$ need not be equal to $p$. But the second question I find a bit confusing. Do you mean $Z(S)$ instead of $Z(P)$? If so, it seems to make sense.
                                – James
                                Sep 10 at 22:28

                              • @James Thanks for pointing that out. Yes, I meant $Z(S).$ Does it make sense in the sense that you understand the argument of the proof, or in the sense that you understand my second question?
                                – Dedalus
                                Sep 10 at 22:35

                              • Well, I now understand the question and, I think, also the argument. As per the first question, you need not restrict to elements of order $p$ so, in fact, we have $Z(S)leq H$ from the first part of the argument. And, there exists, therefore, $1neq tin Z(S)$ (so $t$ is also an element of $H$). So, if $y$ is now any element in $S$, we have $t = t^yin Hcap {}^{y}H$ and has $p$-power order, so $y$ must belong to $H$ by the assumption that $H$ is strongly $p$-embedded. This gives $Sleq H$.
                                – James
                                Sep 10 at 22:45

                              • Thanks, I misread the second statemnt and read it as $g in H cap {^y}H.$
                                – Dedalus
                                Sep 10 at 22:55

                              • I agree that for your first question the order of $x$ need not be equal to $p$. But the second question I find a bit confusing. Do you mean $Z(S)$ instead of $Z(P)$? If so, it seems to make sense.
                                – James
                                Sep 10 at 22:28

                              • @James Thanks for pointing that out. Yes, I meant $Z(S).$ Does it make sense in the sense that you understand the argument of the proof, or in the sense that you understand my second question?
                                – Dedalus
                                Sep 10 at 22:35

                              • Well, I now understand the question and, I think, also the argument. As per the first question, you need not restrict to elements of order $p$ so, in fact, we have $Z(S)leq H$ from the first part of the argument. And, there exists, therefore, $1neq tin Z(S)$ (so $t$ is also an element of $H$). So, if $y$ is now any element in $S$, we have $t = t^yin Hcap {}^{y}H$ and has $p$-power order, so $y$ must belong to $H$ by the assumption that $H$ is strongly $p$-embedded. This gives $Sleq H$.
                                – James
                                Sep 10 at 22:45

                              • Thanks, I misread the second statemnt and read it as $g in H cap {^y}H.$
                                – Dedalus
                                Sep 10 at 22:55

                              I agree that for your first question the order of $x$ need not be equal to $p$. But the second question I find a bit confusing. Do you mean $Z(S)$ instead of $Z(P)$? If so, it seems to make sense.
                              – James
                              Sep 10 at 22:28

                              I agree that for your first question the order of $x$ need not be equal to $p$. But the second question I find a bit confusing. Do you mean $Z(S)$ instead of $Z(P)$? If so, it seems to make sense.
                              – James
                              Sep 10 at 22:28

                              @James Thanks for pointing that out. Yes, I meant $Z(S).$ Does it make sense in the sense that you understand the argument of the proof, or in the sense that you understand my second question?
                              – Dedalus
                              Sep 10 at 22:35

                              @James Thanks for pointing that out. Yes, I meant $Z(S).$ Does it make sense in the sense that you understand the argument of the proof, or in the sense that you understand my second question?
                              – Dedalus
                              Sep 10 at 22:35

                              Well, I now understand the question and, I think, also the argument. As per the first question, you need not restrict to elements of order $p$ so, in fact, we have $Z(S)leq H$ from the first part of the argument. And, there exists, therefore, $1neq tin Z(S)$ (so $t$ is also an element of $H$). So, if $y$ is now any element in $S$, we have $t = t^yin Hcap {}^{y}H$ and has $p$-power order, so $y$ must belong to $H$ by the assumption that $H$ is strongly $p$-embedded. This gives $Sleq H$.
                              – James
                              Sep 10 at 22:45

                              Well, I now understand the question and, I think, also the argument. As per the first question, you need not restrict to elements of order $p$ so, in fact, we have $Z(S)leq H$ from the first part of the argument. And, there exists, therefore, $1neq tin Z(S)$ (so $t$ is also an element of $H$). So, if $y$ is now any element in $S$, we have $t = t^yin Hcap {}^{y}H$ and has $p$-power order, so $y$ must belong to $H$ by the assumption that $H$ is strongly $p$-embedded. This gives $Sleq H$.
                              – James
                              Sep 10 at 22:45

                              Thanks, I misread the second statemnt and read it as $g in H cap {^y}H.$
                              – Dedalus
                              Sep 10 at 22:55

                              Thanks, I misread the second statemnt and read it as $g in H cap {^y}H.$
                              – Dedalus
                              Sep 10 at 22:55

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                              Image of $f$ comes arbitrarily close to every $c$ in $mathbb{C}$.

                              The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP

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                              Let $p_1, p_2,ldots,p_n$ are points on the compact Riemann surface $X$ and $Y:=X-{p_1,…..,p_n}$. Suppose
                              $$f:Y→mathbb{C}$$
                              is a non-constant holomorphic function. Show that the image of $f$ comes arbitrarily close to every $c$ in $mathbb{C}$.$//$ Is this mean that, we have to show $f(Y)$ is dense in $mathbb{C}$? Then how can we proceed? Please help me.

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                                Let $p_1, p_2,ldots,p_n$ are points on the compact Riemann surface $X$ and $Y:=X-{p_1,…..,p_n}$. Suppose
                                $$f:Y→mathbb{C}$$
                                is a non-constant holomorphic function. Show that the image of $f$ comes arbitrarily close to every $c$ in $mathbb{C}$.$//$ Is this mean that, we have to show $f(Y)$ is dense in $mathbb{C}$? Then how can we proceed? Please help me.

                                share|cite|improve this question

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                                  Let $p_1, p_2,ldots,p_n$ are points on the compact Riemann surface $X$ and $Y:=X-{p_1,…..,p_n}$. Suppose
                                  $$f:Y→mathbb{C}$$
                                  is a non-constant holomorphic function. Show that the image of $f$ comes arbitrarily close to every $c$ in $mathbb{C}$.$//$ Is this mean that, we have to show $f(Y)$ is dense in $mathbb{C}$? Then how can we proceed? Please help me.

                                  share|cite|improve this question

                                  Let $p_1, p_2,ldots,p_n$ are points on the compact Riemann surface $X$ and $Y:=X-{p_1,…..,p_n}$. Suppose
                                  $$f:Y→mathbb{C}$$
                                  is a non-constant holomorphic function. Show that the image of $f$ comes arbitrarily close to every $c$ in $mathbb{C}$.$//$ Is this mean that, we have to show $f(Y)$ is dense in $mathbb{C}$? Then how can we proceed? Please help me.

                                  complex-analysis riemann-surfaces

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                                  share|cite|improve this question

                                  share|cite|improve this question

                                  share|cite|improve this question

                                  edited Sep 10 at 21:56

                                  aidangallagher4

                                  6571312

                                  6571312

                                  asked Sep 10 at 21:41

                                  abcdmath

                                  301110

                                  301110

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                                      Let’s say the image misses some ball $B(z,r)$. Then we consider the function $g(z) = frac{1}{f(z) – c}$. Then $g$ is non-constant and bounded by $frac{1}{r}$, so we may extend it to obtain a non-constant holomorphic function on $X$, which is impossible.

                                      share|cite|improve this answer

                                      • Sir why this is impossible to extend on $X$? There is no given condition that $p_k>’s$ are some type of singularities of $f$.
                                        – abcdmath
                                        Sep 10 at 22:10

                                      • 1

                                        There are no non-constant holomorphic functions on a compact Riemann surface. And what I’m extending is $g$ instead of $f$, the point is that since $g$ is bounded on $Y$, the $p_k$ are removable singularities for it.
                                        – Daminark
                                        Sep 10 at 22:16

                                      • Thank you sir..
                                        – abcdmath
                                        Sep 10 at 22:22

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                                      Let’s say the image misses some ball $B(z,r)$. Then we consider the function $g(z) = frac{1}{f(z) – c}$. Then $g$ is non-constant and bounded by $frac{1}{r}$, so we may extend it to obtain a non-constant holomorphic function on $X$, which is impossible.

                                      share|cite|improve this answer

                                      • Sir why this is impossible to extend on $X$? There is no given condition that $p_k>’s$ are some type of singularities of $f$.
                                        – abcdmath
                                        Sep 10 at 22:10

                                      • 1

                                        There are no non-constant holomorphic functions on a compact Riemann surface. And what I’m extending is $g$ instead of $f$, the point is that since $g$ is bounded on $Y$, the $p_k$ are removable singularities for it.
                                        – Daminark
                                        Sep 10 at 22:16

                                      • Thank you sir..
                                        – abcdmath
                                        Sep 10 at 22:22

                                      up vote
                                      1
                                      down vote

                                      accepted

                                      Let’s say the image misses some ball $B(z,r)$. Then we consider the function $g(z) = frac{1}{f(z) – c}$. Then $g$ is non-constant and bounded by $frac{1}{r}$, so we may extend it to obtain a non-constant holomorphic function on $X$, which is impossible.

                                      share|cite|improve this answer

                                      • Sir why this is impossible to extend on $X$? There is no given condition that $p_k>’s$ are some type of singularities of $f$.
                                        – abcdmath
                                        Sep 10 at 22:10

                                      • 1

                                        There are no non-constant holomorphic functions on a compact Riemann surface. And what I’m extending is $g$ instead of $f$, the point is that since $g$ is bounded on $Y$, the $p_k$ are removable singularities for it.
                                        – Daminark
                                        Sep 10 at 22:16

                                      • Thank you sir..
                                        – abcdmath
                                        Sep 10 at 22:22

                                      up vote
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                                      accepted

                                      up vote
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                                      down vote

                                      accepted

                                      Let’s say the image misses some ball $B(z,r)$. Then we consider the function $g(z) = frac{1}{f(z) – c}$. Then $g$ is non-constant and bounded by $frac{1}{r}$, so we may extend it to obtain a non-constant holomorphic function on $X$, which is impossible.

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                                      Let’s say the image misses some ball $B(z,r)$. Then we consider the function $g(z) = frac{1}{f(z) – c}$. Then $g$ is non-constant and bounded by $frac{1}{r}$, so we may extend it to obtain a non-constant holomorphic function on $X$, which is impossible.

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                                      answered Sep 10 at 21:58

                                      Daminark

                                      1,359610

                                      1,359610

                                      • Sir why this is impossible to extend on $X$? There is no given condition that $p_k>’s$ are some type of singularities of $f$.
                                        – abcdmath
                                        Sep 10 at 22:10

                                      • 1

                                        There are no non-constant holomorphic functions on a compact Riemann surface. And what I’m extending is $g$ instead of $f$, the point is that since $g$ is bounded on $Y$, the $p_k$ are removable singularities for it.
                                        – Daminark
                                        Sep 10 at 22:16

                                      • Thank you sir..
                                        – abcdmath
                                        Sep 10 at 22:22

                                      • Sir why this is impossible to extend on $X$? There is no given condition that $p_k>’s$ are some type of singularities of $f$.
                                        – abcdmath
                                        Sep 10 at 22:10

                                      • 1

                                        There are no non-constant holomorphic functions on a compact Riemann surface. And what I’m extending is $g$ instead of $f$, the point is that since $g$ is bounded on $Y$, the $p_k$ are removable singularities for it.
                                        – Daminark
                                        Sep 10 at 22:16

                                      • Thank you sir..
                                        – abcdmath
                                        Sep 10 at 22:22

                                      Sir why this is impossible to extend on $X$? There is no given condition that $p_k>’s$ are some type of singularities of $f$.
                                      – abcdmath
                                      Sep 10 at 22:10

                                      Sir why this is impossible to extend on $X$? There is no given condition that $p_k>’s$ are some type of singularities of $f$.
                                      – abcdmath
                                      Sep 10 at 22:10

                                      1

                                      1

                                      There are no non-constant holomorphic functions on a compact Riemann surface. And what I’m extending is $g$ instead of $f$, the point is that since $g$ is bounded on $Y$, the $p_k$ are removable singularities for it.
                                      – Daminark
                                      Sep 10 at 22:16

                                      There are no non-constant holomorphic functions on a compact Riemann surface. And what I’m extending is $g$ instead of $f$, the point is that since $g$ is bounded on $Y$, the $p_k$ are removable singularities for it.
                                      – Daminark
                                      Sep 10 at 22:16

                                      Thank you sir..
                                      – abcdmath
                                      Sep 10 at 22:22

                                      Thank you sir..
                                      – abcdmath
                                      Sep 10 at 22:22

                                       
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                                      Question on derivation of probability proportion in Deep Learning Book

                                      The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP

                                      up vote
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                                      I am working through the Deep Learning Book, I am currently on the regularization chapter (https://www.deeplearningbook.org/contents/regularization.html).

                                      My question concerns the third step in the following derivation:

                                      $$begin{align*}
                                      tilde{P}_{text{ensemble}}(y=yvert v)&propto sqrt[2^n]{prod_{din {0,1}^n} exp(W_{y,:}^T(dodot v) + b_y})\
                                      &=expleft(frac{1}{2^n} sum_{d in {0,1}^n} W_{y,:}^T(dodot v) + b_yright)\
                                      &=exp(frac{1}{2} W_{y,:}^Tv + b_y).
                                      end{align*}$$

                                      I’d like to clarify that $odot$ represents the operation of component-wise multiplication and $d$ and $v$ are vectors while $W$ is an $m times n$ matrix.

                                      The reason I am confused is because it seems that in the second to last step we are summing over all $n$ bit binary strings. So to me the simplification should be

                                      $$begin{align*}
                                      tilde{P}_{text{ensemble}}(y=yvert v)&propto sqrt[2^n]{prod_{din {0,1}^n} exp(W_{y,:}^T(dodot v) + b_y})\
                                      &=expleft(frac{1}{2^n} sum_{d in {0,1}^n} W_{y,:}^T(dodot v) + b_yright)\
                                      &=exp(frac{1}{2^n} 2^n(W_{y,:}^Tv + b_y))\
                                      &=exp(W_{y,:}^Tv + b_y)
                                      end{align*}$$

                                      Where does the extra factor $frac{1}{2}$ come from?

                                      share|cite|improve this question

                                        up vote
                                        1
                                        down vote

                                        favorite

                                        I am working through the Deep Learning Book, I am currently on the regularization chapter (https://www.deeplearningbook.org/contents/regularization.html).

                                        My question concerns the third step in the following derivation:

                                        $$begin{align*}
                                        tilde{P}_{text{ensemble}}(y=yvert v)&propto sqrt[2^n]{prod_{din {0,1}^n} exp(W_{y,:}^T(dodot v) + b_y})\
                                        &=expleft(frac{1}{2^n} sum_{d in {0,1}^n} W_{y,:}^T(dodot v) + b_yright)\
                                        &=exp(frac{1}{2} W_{y,:}^Tv + b_y).
                                        end{align*}$$

                                        I’d like to clarify that $odot$ represents the operation of component-wise multiplication and $d$ and $v$ are vectors while $W$ is an $m times n$ matrix.

                                        The reason I am confused is because it seems that in the second to last step we are summing over all $n$ bit binary strings. So to me the simplification should be

                                        $$begin{align*}
                                        tilde{P}_{text{ensemble}}(y=yvert v)&propto sqrt[2^n]{prod_{din {0,1}^n} exp(W_{y,:}^T(dodot v) + b_y})\
                                        &=expleft(frac{1}{2^n} sum_{d in {0,1}^n} W_{y,:}^T(dodot v) + b_yright)\
                                        &=exp(frac{1}{2^n} 2^n(W_{y,:}^Tv + b_y))\
                                        &=exp(W_{y,:}^Tv + b_y)
                                        end{align*}$$

                                        Where does the extra factor $frac{1}{2}$ come from?

                                        share|cite|improve this question

                                          up vote
                                          1
                                          down vote

                                          favorite

                                          up vote
                                          1
                                          down vote

                                          favorite

                                          I am working through the Deep Learning Book, I am currently on the regularization chapter (https://www.deeplearningbook.org/contents/regularization.html).

                                          My question concerns the third step in the following derivation:

                                          $$begin{align*}
                                          tilde{P}_{text{ensemble}}(y=yvert v)&propto sqrt[2^n]{prod_{din {0,1}^n} exp(W_{y,:}^T(dodot v) + b_y})\
                                          &=expleft(frac{1}{2^n} sum_{d in {0,1}^n} W_{y,:}^T(dodot v) + b_yright)\
                                          &=exp(frac{1}{2} W_{y,:}^Tv + b_y).
                                          end{align*}$$

                                          I’d like to clarify that $odot$ represents the operation of component-wise multiplication and $d$ and $v$ are vectors while $W$ is an $m times n$ matrix.

                                          The reason I am confused is because it seems that in the second to last step we are summing over all $n$ bit binary strings. So to me the simplification should be

                                          $$begin{align*}
                                          tilde{P}_{text{ensemble}}(y=yvert v)&propto sqrt[2^n]{prod_{din {0,1}^n} exp(W_{y,:}^T(dodot v) + b_y})\
                                          &=expleft(frac{1}{2^n} sum_{d in {0,1}^n} W_{y,:}^T(dodot v) + b_yright)\
                                          &=exp(frac{1}{2^n} 2^n(W_{y,:}^Tv + b_y))\
                                          &=exp(W_{y,:}^Tv + b_y)
                                          end{align*}$$

                                          Where does the extra factor $frac{1}{2}$ come from?

                                          share|cite|improve this question

                                          I am working through the Deep Learning Book, I am currently on the regularization chapter (https://www.deeplearningbook.org/contents/regularization.html).

                                          My question concerns the third step in the following derivation:

                                          $$begin{align*}
                                          tilde{P}_{text{ensemble}}(y=yvert v)&propto sqrt[2^n]{prod_{din {0,1}^n} exp(W_{y,:}^T(dodot v) + b_y})\
                                          &=expleft(frac{1}{2^n} sum_{d in {0,1}^n} W_{y,:}^T(dodot v) + b_yright)\
                                          &=exp(frac{1}{2} W_{y,:}^Tv + b_y).
                                          end{align*}$$

                                          I’d like to clarify that $odot$ represents the operation of component-wise multiplication and $d$ and $v$ are vectors while $W$ is an $m times n$ matrix.

                                          The reason I am confused is because it seems that in the second to last step we are summing over all $n$ bit binary strings. So to me the simplification should be

                                          $$begin{align*}
                                          tilde{P}_{text{ensemble}}(y=yvert v)&propto sqrt[2^n]{prod_{din {0,1}^n} exp(W_{y,:}^T(dodot v) + b_y})\
                                          &=expleft(frac{1}{2^n} sum_{d in {0,1}^n} W_{y,:}^T(dodot v) + b_yright)\
                                          &=exp(frac{1}{2^n} 2^n(W_{y,:}^Tv + b_y))\
                                          &=exp(W_{y,:}^Tv + b_y)
                                          end{align*}$$

                                          Where does the extra factor $frac{1}{2}$ come from?

                                          probability machine-learning regularization

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                                          asked Sep 10 at 21:42

                                          ClownInTheMoon

                                          1,119418

                                          1,119418

                                              1 Answer
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                                              accepted

                                              By linearity,

                                              $$
                                              sum_{din{0,1}^n}W_{y,:}^T(dodot v)=W_{y,:}^Tleft(left(sum_{din{0,1}^n}dright)odot vright);.
                                              $$

                                              So indeed you’re summing over all $2^n$ binary vectors $d$. In each component, half of them have a $0$ and half of them have a $1$. Thus every component of $v$ is multiplied by $frac12cdot2^n$.

                                              share|cite|improve this answer

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                                                1 Answer
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                                                active

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                                                accepted

                                                By linearity,

                                                $$
                                                sum_{din{0,1}^n}W_{y,:}^T(dodot v)=W_{y,:}^Tleft(left(sum_{din{0,1}^n}dright)odot vright);.
                                                $$

                                                So indeed you’re summing over all $2^n$ binary vectors $d$. In each component, half of them have a $0$ and half of them have a $1$. Thus every component of $v$ is multiplied by $frac12cdot2^n$.

                                                share|cite|improve this answer

                                                  up vote
                                                  2
                                                  down vote

                                                  accepted

                                                  By linearity,

                                                  $$
                                                  sum_{din{0,1}^n}W_{y,:}^T(dodot v)=W_{y,:}^Tleft(left(sum_{din{0,1}^n}dright)odot vright);.
                                                  $$

                                                  So indeed you’re summing over all $2^n$ binary vectors $d$. In each component, half of them have a $0$ and half of them have a $1$. Thus every component of $v$ is multiplied by $frac12cdot2^n$.

                                                  share|cite|improve this answer

                                                    up vote
                                                    2
                                                    down vote

                                                    accepted

                                                    up vote
                                                    2
                                                    down vote

                                                    accepted

                                                    By linearity,

                                                    $$
                                                    sum_{din{0,1}^n}W_{y,:}^T(dodot v)=W_{y,:}^Tleft(left(sum_{din{0,1}^n}dright)odot vright);.
                                                    $$

                                                    So indeed you’re summing over all $2^n$ binary vectors $d$. In each component, half of them have a $0$ and half of them have a $1$. Thus every component of $v$ is multiplied by $frac12cdot2^n$.

                                                    share|cite|improve this answer

                                                    By linearity,

                                                    $$
                                                    sum_{din{0,1}^n}W_{y,:}^T(dodot v)=W_{y,:}^Tleft(left(sum_{din{0,1}^n}dright)odot vright);.
                                                    $$

                                                    So indeed you’re summing over all $2^n$ binary vectors $d$. In each component, half of them have a $0$ and half of them have a $1$. Thus every component of $v$ is multiplied by $frac12cdot2^n$.

                                                    share|cite|improve this answer

                                                    share|cite|improve this answer

                                                    share|cite|improve this answer

                                                    answered Sep 11 at 5:33

                                                    joriki

                                                    169k10181337

                                                    169k10181337

                                                         
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                                                        Amount of Number Combinations to Reach a Sum of 10 With Integers 1-9 Using 2 or More Integers [closed]

                                                        The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP

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                                                        I received a word problem that goes like this.

                                                        A local kindergarten is thinking of making posters that show all the different ways of adding two or more integers from 1 to 9 to get a sum of 10. If there is enough space on each poster for up to 50 possible solutions, how many posters will the school need to make?

                                                        (Note: sums that contain the same number but in a different order are considered to be different; for example, 1 + 9 and 9 + 1 are two different solutions.)

                                                        What is the answer to this problem, and more importantly, how do I solve it?

                                                        share|cite|improve this question

                                                        closed as off-topic by Theoretical Economist, Adrian Keister, user99914, Xander Henderson, Deepesh Meena Sep 11 at 3:26

                                                        This question appears to be off-topic. The users who voted to close gave this specific reason:

                                                        • This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level.” – Theoretical Economist, Adrian Keister, Community, Xander Henderson, Deepesh Meena

                                                        If this question can be reworded to fit the rules in the help center, please edit the question.

                                                          up vote
                                                          0
                                                          down vote

                                                          favorite

                                                          I received a word problem that goes like this.

                                                          A local kindergarten is thinking of making posters that show all the different ways of adding two or more integers from 1 to 9 to get a sum of 10. If there is enough space on each poster for up to 50 possible solutions, how many posters will the school need to make?

                                                          (Note: sums that contain the same number but in a different order are considered to be different; for example, 1 + 9 and 9 + 1 are two different solutions.)

                                                          What is the answer to this problem, and more importantly, how do I solve it?

                                                          share|cite|improve this question

                                                          closed as off-topic by Theoretical Economist, Adrian Keister, user99914, Xander Henderson, Deepesh Meena Sep 11 at 3:26

                                                          This question appears to be off-topic. The users who voted to close gave this specific reason:

                                                          • This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level.” – Theoretical Economist, Adrian Keister, Community, Xander Henderson, Deepesh Meena

                                                          If this question can be reworded to fit the rules in the help center, please edit the question.

                                                            up vote
                                                            0
                                                            down vote

                                                            favorite

                                                            up vote
                                                            0
                                                            down vote

                                                            favorite

                                                            I received a word problem that goes like this.

                                                            A local kindergarten is thinking of making posters that show all the different ways of adding two or more integers from 1 to 9 to get a sum of 10. If there is enough space on each poster for up to 50 possible solutions, how many posters will the school need to make?

                                                            (Note: sums that contain the same number but in a different order are considered to be different; for example, 1 + 9 and 9 + 1 are two different solutions.)

                                                            What is the answer to this problem, and more importantly, how do I solve it?

                                                            share|cite|improve this question

                                                            I received a word problem that goes like this.

                                                            A local kindergarten is thinking of making posters that show all the different ways of adding two or more integers from 1 to 9 to get a sum of 10. If there is enough space on each poster for up to 50 possible solutions, how many posters will the school need to make?

                                                            (Note: sums that contain the same number but in a different order are considered to be different; for example, 1 + 9 and 9 + 1 are two different solutions.)

                                                            What is the answer to this problem, and more importantly, how do I solve it?

                                                            combinatorics combinations

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                                                            share|cite|improve this question

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                                                            share|cite|improve this question

                                                            edited Sep 10 at 23:54

                                                            N. F. Taussig

                                                            39.9k93253

                                                            39.9k93253

                                                            asked Sep 10 at 21:43

                                                            Jolly

                                                            31

                                                            31

                                                            closed as off-topic by Theoretical Economist, Adrian Keister, user99914, Xander Henderson, Deepesh Meena Sep 11 at 3:26

                                                            This question appears to be off-topic. The users who voted to close gave this specific reason:

                                                            • This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level.” – Theoretical Economist, Adrian Keister, Community, Xander Henderson, Deepesh Meena

                                                            If this question can be reworded to fit the rules in the help center, please edit the question.

                                                            closed as off-topic by Theoretical Economist, Adrian Keister, user99914, Xander Henderson, Deepesh Meena Sep 11 at 3:26

                                                            This question appears to be off-topic. The users who voted to close gave this specific reason:

                                                            • This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level.” – Theoretical Economist, Adrian Keister, Community, Xander Henderson, Deepesh Meena

                                                            If this question can be reworded to fit the rules in the help center, please edit the question.

                                                                2 Answers
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                                                                Imagine $10$ ones in a row, and insert plus signs between them; for instance, $|||+||+|||||$ stands for the sum $3+2+5$. So the sums you want to count correspond to the ways of inserting plus signs. There are $9$ slots between the $10$ ones, and you can either insert a plus sign or not in each of them independently, so the number of different ways of doing this is $2^9$. However, one of these corresponds to inserting no plus signs at all, which doesn’t yield a sum of at least two numbers, so we have to subtract that and the result is $2^9-1=512-1=511$.

                                                                share|cite|improve this answer

                                                                • Almost. Remember that there must be at least 1 + sign.
                                                                  – Rushabh Mehta
                                                                  Sep 10 at 21:57

                                                                • @RushabhMehta: Thanks, I realized that when I saw your answer and corrected mine accordingly.
                                                                  – joriki
                                                                  Sep 10 at 21:58

                                                                • Nice solution +1
                                                                  – Rushabh Mehta
                                                                  Sep 10 at 21:59

                                                                • Why is the number of different ways of doing it 2^9?
                                                                  – Jolly
                                                                  Sep 10 at 22:33

                                                                • @Jolly This is an application of the multiplication principle. If you have $a$ ways of choosing $A$ and $b$ ways of choosing $B$, and you choose $A$ and $B$ independently, then you have a total of $acdot b$ possible selections. In the present case, you have $9$ independent choices to make, with $2$ options for each of the $9$ choices: For each slot, you decide whether to insert a plus or not. That’s $2cdot2cdot2cdot2cdot2cdot2cdot2cdot2cdot2=2^9$ different ways to choose. Maybe try it with $3$ instead of $9$ to get a better feel for it.
                                                                  – joriki
                                                                  Sep 11 at 5:12

                                                                up vote
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                                                                down vote

                                                                We realize that the requirement of each number being at least $1$ and the sum being $10$ removes the need to cap the number at $9$, so we shall ignore that condition.

                                                                Let’s first start with the case of two integers. Realize that we can visualize this problem as us having 10 coins and two buckets, and we have to drop coins in each bucket. Each way we can drop coins is a different sum, i.e., if I drop 6 coins in the first bucket, but 4 in the second, then the equivalent sum is $6+4$.

                                                                But, we must make sure every bucket has a positive number of coins, so we put one coin in each bucket to start. So, we have 8 coins left to distribute across 2 buckets. So, we can represent this as

                                                                ********|
                                                                

                                                                Where the bar marks the divider between the buckets. So, we are essentially counting the number of ways to arrange those symbols, which is ${9choose1}=9$.

                                                                We can apply the same logic with three buckets, to get the following symbol list:

                                                                *******||
                                                                

                                                                So, the number of ways to arrange the symbols is ${9choose2}=36$.

                                                                We do this all the way to ${9choose9}=1$. So our answer is $$sumlimits_{n=1}^9{9choose n}=511$$

                                                                share|cite|improve this answer

                                                                • Does this answer account for a solution such as, “1+1+1+1+1+1+1+1+1+1”?
                                                                  – Jolly
                                                                  Sep 10 at 23:04

                                                                • @Jolly Yes indeed. That is the case when there are 10 numbers, which will be represented by $9choose9$
                                                                  – Rushabh Mehta
                                                                  Sep 11 at 0:02

                                                                2 Answers
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                                                                2 Answers
                                                                2

                                                                active

                                                                oldest

                                                                votes

                                                                active

                                                                oldest

                                                                votes

                                                                active

                                                                oldest

                                                                votes

                                                                up vote
                                                                2
                                                                down vote

                                                                accepted

                                                                Imagine $10$ ones in a row, and insert plus signs between them; for instance, $|||+||+|||||$ stands for the sum $3+2+5$. So the sums you want to count correspond to the ways of inserting plus signs. There are $9$ slots between the $10$ ones, and you can either insert a plus sign or not in each of them independently, so the number of different ways of doing this is $2^9$. However, one of these corresponds to inserting no plus signs at all, which doesn’t yield a sum of at least two numbers, so we have to subtract that and the result is $2^9-1=512-1=511$.

                                                                share|cite|improve this answer

                                                                • Almost. Remember that there must be at least 1 + sign.
                                                                  – Rushabh Mehta
                                                                  Sep 10 at 21:57

                                                                • @RushabhMehta: Thanks, I realized that when I saw your answer and corrected mine accordingly.
                                                                  – joriki
                                                                  Sep 10 at 21:58

                                                                • Nice solution +1
                                                                  – Rushabh Mehta
                                                                  Sep 10 at 21:59

                                                                • Why is the number of different ways of doing it 2^9?
                                                                  – Jolly
                                                                  Sep 10 at 22:33

                                                                • @Jolly This is an application of the multiplication principle. If you have $a$ ways of choosing $A$ and $b$ ways of choosing $B$, and you choose $A$ and $B$ independently, then you have a total of $acdot b$ possible selections. In the present case, you have $9$ independent choices to make, with $2$ options for each of the $9$ choices: For each slot, you decide whether to insert a plus or not. That’s $2cdot2cdot2cdot2cdot2cdot2cdot2cdot2cdot2=2^9$ different ways to choose. Maybe try it with $3$ instead of $9$ to get a better feel for it.
                                                                  – joriki
                                                                  Sep 11 at 5:12

                                                                up vote
                                                                2
                                                                down vote

                                                                accepted

                                                                Imagine $10$ ones in a row, and insert plus signs between them; for instance, $|||+||+|||||$ stands for the sum $3+2+5$. So the sums you want to count correspond to the ways of inserting plus signs. There are $9$ slots between the $10$ ones, and you can either insert a plus sign or not in each of them independently, so the number of different ways of doing this is $2^9$. However, one of these corresponds to inserting no plus signs at all, which doesn’t yield a sum of at least two numbers, so we have to subtract that and the result is $2^9-1=512-1=511$.

                                                                share|cite|improve this answer

                                                                • Almost. Remember that there must be at least 1 + sign.
                                                                  – Rushabh Mehta
                                                                  Sep 10 at 21:57

                                                                • @RushabhMehta: Thanks, I realized that when I saw your answer and corrected mine accordingly.
                                                                  – joriki
                                                                  Sep 10 at 21:58

                                                                • Nice solution +1
                                                                  – Rushabh Mehta
                                                                  Sep 10 at 21:59

                                                                • Why is the number of different ways of doing it 2^9?
                                                                  – Jolly
                                                                  Sep 10 at 22:33

                                                                • @Jolly This is an application of the multiplication principle. If you have $a$ ways of choosing $A$ and $b$ ways of choosing $B$, and you choose $A$ and $B$ independently, then you have a total of $acdot b$ possible selections. In the present case, you have $9$ independent choices to make, with $2$ options for each of the $9$ choices: For each slot, you decide whether to insert a plus or not. That’s $2cdot2cdot2cdot2cdot2cdot2cdot2cdot2cdot2=2^9$ different ways to choose. Maybe try it with $3$ instead of $9$ to get a better feel for it.
                                                                  – joriki
                                                                  Sep 11 at 5:12

                                                                up vote
                                                                2
                                                                down vote

                                                                accepted

                                                                up vote
                                                                2
                                                                down vote

                                                                accepted

                                                                Imagine $10$ ones in a row, and insert plus signs between them; for instance, $|||+||+|||||$ stands for the sum $3+2+5$. So the sums you want to count correspond to the ways of inserting plus signs. There are $9$ slots between the $10$ ones, and you can either insert a plus sign or not in each of them independently, so the number of different ways of doing this is $2^9$. However, one of these corresponds to inserting no plus signs at all, which doesn’t yield a sum of at least two numbers, so we have to subtract that and the result is $2^9-1=512-1=511$.

                                                                share|cite|improve this answer

                                                                Imagine $10$ ones in a row, and insert plus signs between them; for instance, $|||+||+|||||$ stands for the sum $3+2+5$. So the sums you want to count correspond to the ways of inserting plus signs. There are $9$ slots between the $10$ ones, and you can either insert a plus sign or not in each of them independently, so the number of different ways of doing this is $2^9$. However, one of these corresponds to inserting no plus signs at all, which doesn’t yield a sum of at least two numbers, so we have to subtract that and the result is $2^9-1=512-1=511$.

                                                                share|cite|improve this answer

                                                                share|cite|improve this answer

                                                                share|cite|improve this answer

                                                                edited Sep 10 at 22:02

                                                                answered Sep 10 at 21:55

                                                                joriki

                                                                169k10181337

                                                                169k10181337

                                                                • Almost. Remember that there must be at least 1 + sign.
                                                                  – Rushabh Mehta
                                                                  Sep 10 at 21:57

                                                                • @RushabhMehta: Thanks, I realized that when I saw your answer and corrected mine accordingly.
                                                                  – joriki
                                                                  Sep 10 at 21:58

                                                                • Nice solution +1
                                                                  – Rushabh Mehta
                                                                  Sep 10 at 21:59

                                                                • Why is the number of different ways of doing it 2^9?
                                                                  – Jolly
                                                                  Sep 10 at 22:33

                                                                • @Jolly This is an application of the multiplication principle. If you have $a$ ways of choosing $A$ and $b$ ways of choosing $B$, and you choose $A$ and $B$ independently, then you have a total of $acdot b$ possible selections. In the present case, you have $9$ independent choices to make, with $2$ options for each of the $9$ choices: For each slot, you decide whether to insert a plus or not. That’s $2cdot2cdot2cdot2cdot2cdot2cdot2cdot2cdot2=2^9$ different ways to choose. Maybe try it with $3$ instead of $9$ to get a better feel for it.
                                                                  – joriki
                                                                  Sep 11 at 5:12

                                                                • Almost. Remember that there must be at least 1 + sign.
                                                                  – Rushabh Mehta
                                                                  Sep 10 at 21:57

                                                                • @RushabhMehta: Thanks, I realized that when I saw your answer and corrected mine accordingly.
                                                                  – joriki
                                                                  Sep 10 at 21:58

                                                                • Nice solution +1
                                                                  – Rushabh Mehta
                                                                  Sep 10 at 21:59

                                                                • Why is the number of different ways of doing it 2^9?
                                                                  – Jolly
                                                                  Sep 10 at 22:33

                                                                • @Jolly This is an application of the multiplication principle. If you have $a$ ways of choosing $A$ and $b$ ways of choosing $B$, and you choose $A$ and $B$ independently, then you have a total of $acdot b$ possible selections. In the present case, you have $9$ independent choices to make, with $2$ options for each of the $9$ choices: For each slot, you decide whether to insert a plus or not. That’s $2cdot2cdot2cdot2cdot2cdot2cdot2cdot2cdot2=2^9$ different ways to choose. Maybe try it with $3$ instead of $9$ to get a better feel for it.
                                                                  – joriki
                                                                  Sep 11 at 5:12

                                                                Almost. Remember that there must be at least 1 + sign.
                                                                – Rushabh Mehta
                                                                Sep 10 at 21:57

                                                                Almost. Remember that there must be at least 1 + sign.
                                                                – Rushabh Mehta
                                                                Sep 10 at 21:57

                                                                @RushabhMehta: Thanks, I realized that when I saw your answer and corrected mine accordingly.
                                                                – joriki
                                                                Sep 10 at 21:58

                                                                @RushabhMehta: Thanks, I realized that when I saw your answer and corrected mine accordingly.
                                                                – joriki
                                                                Sep 10 at 21:58

                                                                Nice solution +1
                                                                – Rushabh Mehta
                                                                Sep 10 at 21:59

                                                                Nice solution +1
                                                                – Rushabh Mehta
                                                                Sep 10 at 21:59

                                                                Why is the number of different ways of doing it 2^9?
                                                                – Jolly
                                                                Sep 10 at 22:33

                                                                Why is the number of different ways of doing it 2^9?
                                                                – Jolly
                                                                Sep 10 at 22:33

                                                                @Jolly This is an application of the multiplication principle. If you have $a$ ways of choosing $A$ and $b$ ways of choosing $B$, and you choose $A$ and $B$ independently, then you have a total of $acdot b$ possible selections. In the present case, you have $9$ independent choices to make, with $2$ options for each of the $9$ choices: For each slot, you decide whether to insert a plus or not. That’s $2cdot2cdot2cdot2cdot2cdot2cdot2cdot2cdot2=2^9$ different ways to choose. Maybe try it with $3$ instead of $9$ to get a better feel for it.
                                                                – joriki
                                                                Sep 11 at 5:12

                                                                @Jolly This is an application of the multiplication principle. If you have $a$ ways of choosing $A$ and $b$ ways of choosing $B$, and you choose $A$ and $B$ independently, then you have a total of $acdot b$ possible selections. In the present case, you have $9$ independent choices to make, with $2$ options for each of the $9$ choices: For each slot, you decide whether to insert a plus or not. That’s $2cdot2cdot2cdot2cdot2cdot2cdot2cdot2cdot2=2^9$ different ways to choose. Maybe try it with $3$ instead of $9$ to get a better feel for it.
                                                                – joriki
                                                                Sep 11 at 5:12

                                                                up vote
                                                                1
                                                                down vote

                                                                We realize that the requirement of each number being at least $1$ and the sum being $10$ removes the need to cap the number at $9$, so we shall ignore that condition.

                                                                Let’s first start with the case of two integers. Realize that we can visualize this problem as us having 10 coins and two buckets, and we have to drop coins in each bucket. Each way we can drop coins is a different sum, i.e., if I drop 6 coins in the first bucket, but 4 in the second, then the equivalent sum is $6+4$.

                                                                But, we must make sure every bucket has a positive number of coins, so we put one coin in each bucket to start. So, we have 8 coins left to distribute across 2 buckets. So, we can represent this as

                                                                ********|
                                                                

                                                                Where the bar marks the divider between the buckets. So, we are essentially counting the number of ways to arrange those symbols, which is ${9choose1}=9$.

                                                                We can apply the same logic with three buckets, to get the following symbol list:

                                                                *******||
                                                                

                                                                So, the number of ways to arrange the symbols is ${9choose2}=36$.

                                                                We do this all the way to ${9choose9}=1$. So our answer is $$sumlimits_{n=1}^9{9choose n}=511$$

                                                                share|cite|improve this answer

                                                                • Does this answer account for a solution such as, “1+1+1+1+1+1+1+1+1+1”?
                                                                  – Jolly
                                                                  Sep 10 at 23:04

                                                                • @Jolly Yes indeed. That is the case when there are 10 numbers, which will be represented by $9choose9$
                                                                  – Rushabh Mehta
                                                                  Sep 11 at 0:02

                                                                up vote
                                                                1
                                                                down vote

                                                                We realize that the requirement of each number being at least $1$ and the sum being $10$ removes the need to cap the number at $9$, so we shall ignore that condition.

                                                                Let’s first start with the case of two integers. Realize that we can visualize this problem as us having 10 coins and two buckets, and we have to drop coins in each bucket. Each way we can drop coins is a different sum, i.e., if I drop 6 coins in the first bucket, but 4 in the second, then the equivalent sum is $6+4$.

                                                                But, we must make sure every bucket has a positive number of coins, so we put one coin in each bucket to start. So, we have 8 coins left to distribute across 2 buckets. So, we can represent this as

                                                                ********|
                                                                

                                                                Where the bar marks the divider between the buckets. So, we are essentially counting the number of ways to arrange those symbols, which is ${9choose1}=9$.

                                                                We can apply the same logic with three buckets, to get the following symbol list:

                                                                *******||
                                                                

                                                                So, the number of ways to arrange the symbols is ${9choose2}=36$.

                                                                We do this all the way to ${9choose9}=1$. So our answer is $$sumlimits_{n=1}^9{9choose n}=511$$

                                                                share|cite|improve this answer

                                                                • Does this answer account for a solution such as, “1+1+1+1+1+1+1+1+1+1”?
                                                                  – Jolly
                                                                  Sep 10 at 23:04

                                                                • @Jolly Yes indeed. That is the case when there are 10 numbers, which will be represented by $9choose9$
                                                                  – Rushabh Mehta
                                                                  Sep 11 at 0:02

                                                                up vote
                                                                1
                                                                down vote

                                                                up vote
                                                                1
                                                                down vote

                                                                We realize that the requirement of each number being at least $1$ and the sum being $10$ removes the need to cap the number at $9$, so we shall ignore that condition.

                                                                Let’s first start with the case of two integers. Realize that we can visualize this problem as us having 10 coins and two buckets, and we have to drop coins in each bucket. Each way we can drop coins is a different sum, i.e., if I drop 6 coins in the first bucket, but 4 in the second, then the equivalent sum is $6+4$.

                                                                But, we must make sure every bucket has a positive number of coins, so we put one coin in each bucket to start. So, we have 8 coins left to distribute across 2 buckets. So, we can represent this as

                                                                ********|
                                                                

                                                                Where the bar marks the divider between the buckets. So, we are essentially counting the number of ways to arrange those symbols, which is ${9choose1}=9$.

                                                                We can apply the same logic with three buckets, to get the following symbol list:

                                                                *******||
                                                                

                                                                So, the number of ways to arrange the symbols is ${9choose2}=36$.

                                                                We do this all the way to ${9choose9}=1$. So our answer is $$sumlimits_{n=1}^9{9choose n}=511$$

                                                                share|cite|improve this answer

                                                                We realize that the requirement of each number being at least $1$ and the sum being $10$ removes the need to cap the number at $9$, so we shall ignore that condition.

                                                                Let’s first start with the case of two integers. Realize that we can visualize this problem as us having 10 coins and two buckets, and we have to drop coins in each bucket. Each way we can drop coins is a different sum, i.e., if I drop 6 coins in the first bucket, but 4 in the second, then the equivalent sum is $6+4$.

                                                                But, we must make sure every bucket has a positive number of coins, so we put one coin in each bucket to start. So, we have 8 coins left to distribute across 2 buckets. So, we can represent this as

                                                                ********|
                                                                

                                                                Where the bar marks the divider between the buckets. So, we are essentially counting the number of ways to arrange those symbols, which is ${9choose1}=9$.

                                                                We can apply the same logic with three buckets, to get the following symbol list:

                                                                *******||
                                                                

                                                                So, the number of ways to arrange the symbols is ${9choose2}=36$.

                                                                We do this all the way to ${9choose9}=1$. So our answer is $$sumlimits_{n=1}^9{9choose n}=511$$

                                                                share|cite|improve this answer

                                                                share|cite|improve this answer

                                                                share|cite|improve this answer

                                                                answered Sep 10 at 21:56

                                                                Rushabh Mehta

                                                                2,588222

                                                                2,588222

                                                                • Does this answer account for a solution such as, “1+1+1+1+1+1+1+1+1+1”?
                                                                  – Jolly
                                                                  Sep 10 at 23:04

                                                                • @Jolly Yes indeed. That is the case when there are 10 numbers, which will be represented by $9choose9$
                                                                  – Rushabh Mehta
                                                                  Sep 11 at 0:02

                                                                • Does this answer account for a solution such as, “1+1+1+1+1+1+1+1+1+1”?
                                                                  – Jolly
                                                                  Sep 10 at 23:04

                                                                • @Jolly Yes indeed. That is the case when there are 10 numbers, which will be represented by $9choose9$
                                                                  – Rushabh Mehta
                                                                  Sep 11 at 0:02

                                                                Does this answer account for a solution such as, “1+1+1+1+1+1+1+1+1+1”?
                                                                – Jolly
                                                                Sep 10 at 23:04

                                                                Does this answer account for a solution such as, “1+1+1+1+1+1+1+1+1+1”?
                                                                – Jolly
                                                                Sep 10 at 23:04

                                                                @Jolly Yes indeed. That is the case when there are 10 numbers, which will be represented by $9choose9$
                                                                – Rushabh Mehta
                                                                Sep 11 at 0:02

                                                                @Jolly Yes indeed. That is the case when there are 10 numbers, which will be represented by $9choose9$
                                                                – Rushabh Mehta
                                                                Sep 11 at 0:02

                                                                Proving that a sequence is not Cauchy

                                                                The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP

                                                                up vote
                                                                0
                                                                down vote

                                                                favorite

                                                                1

                                                                I’m trying to prove that the sequence $left(frac{1}{2},frac{1}{3},frac{2}{3},frac{1}{4},frac{2}{4},frac{3}{4},frac{1}{5},frac{2}{5},frac{3}{5},frac{4}{5},cdotsright)$ is not a Cauchy sequence.

                                                                I know that a sequence of real numbers is not Cauchy if there exists an $epsilon>0$ such that, for all $Ninmathbb{N}$, there exists $m,n>N$ such that $|x_{m}-x_{n}|geqepsilon$. It intuitively makes sense to me that the sequence cannot be Cauchy, as the distance between points where the denominator changes (like $cdotsfrac{99}{101},frac{100}{101},frac{1}{102},cdots$) keeps growing larger. However, I’m not sure how to find indices $m$ and $n$ in general with $|x_{m}-x_{n}|geqepsilon$. Thanks in advance for any help!

                                                                share|cite|improve this question

                                                                  up vote
                                                                  0
                                                                  down vote

                                                                  favorite

                                                                  1

                                                                  I’m trying to prove that the sequence $left(frac{1}{2},frac{1}{3},frac{2}{3},frac{1}{4},frac{2}{4},frac{3}{4},frac{1}{5},frac{2}{5},frac{3}{5},frac{4}{5},cdotsright)$ is not a Cauchy sequence.

                                                                  I know that a sequence of real numbers is not Cauchy if there exists an $epsilon>0$ such that, for all $Ninmathbb{N}$, there exists $m,n>N$ such that $|x_{m}-x_{n}|geqepsilon$. It intuitively makes sense to me that the sequence cannot be Cauchy, as the distance between points where the denominator changes (like $cdotsfrac{99}{101},frac{100}{101},frac{1}{102},cdots$) keeps growing larger. However, I’m not sure how to find indices $m$ and $n$ in general with $|x_{m}-x_{n}|geqepsilon$. Thanks in advance for any help!

                                                                  share|cite|improve this question

                                                                    up vote
                                                                    0
                                                                    down vote

                                                                    favorite

                                                                    1

                                                                    up vote
                                                                    0
                                                                    down vote

                                                                    favorite

                                                                    1
                                                                    1

                                                                    I’m trying to prove that the sequence $left(frac{1}{2},frac{1}{3},frac{2}{3},frac{1}{4},frac{2}{4},frac{3}{4},frac{1}{5},frac{2}{5},frac{3}{5},frac{4}{5},cdotsright)$ is not a Cauchy sequence.

                                                                    I know that a sequence of real numbers is not Cauchy if there exists an $epsilon>0$ such that, for all $Ninmathbb{N}$, there exists $m,n>N$ such that $|x_{m}-x_{n}|geqepsilon$. It intuitively makes sense to me that the sequence cannot be Cauchy, as the distance between points where the denominator changes (like $cdotsfrac{99}{101},frac{100}{101},frac{1}{102},cdots$) keeps growing larger. However, I’m not sure how to find indices $m$ and $n$ in general with $|x_{m}-x_{n}|geqepsilon$. Thanks in advance for any help!

                                                                    share|cite|improve this question

                                                                    I’m trying to prove that the sequence $left(frac{1}{2},frac{1}{3},frac{2}{3},frac{1}{4},frac{2}{4},frac{3}{4},frac{1}{5},frac{2}{5},frac{3}{5},frac{4}{5},cdotsright)$ is not a Cauchy sequence.

                                                                    I know that a sequence of real numbers is not Cauchy if there exists an $epsilon>0$ such that, for all $Ninmathbb{N}$, there exists $m,n>N$ such that $|x_{m}-x_{n}|geqepsilon$. It intuitively makes sense to me that the sequence cannot be Cauchy, as the distance between points where the denominator changes (like $cdotsfrac{99}{101},frac{100}{101},frac{1}{102},cdots$) keeps growing larger. However, I’m not sure how to find indices $m$ and $n$ in general with $|x_{m}-x_{n}|geqepsilon$. Thanks in advance for any help!

                                                                    real-analysis sequences-and-series cauchy-sequences

                                                                    share|cite|improve this question

                                                                    share|cite|improve this question

                                                                    share|cite|improve this question

                                                                    share|cite|improve this question

                                                                    asked Sep 10 at 21:38

                                                                    Sir_Math_Cat

                                                                    792618

                                                                    792618

                                                                        3 Answers
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                                                                        up vote
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                                                                        accepted

                                                                        Let $(a_n)_{ninmathbb N}$ be your sequence. Take $varepsilon=frac12$. Given $Ninmathbb N$, take $ngeqslant N$ such that $a_n$ is of the form $frac k{k+1}$ for some $kinmathbb N$ and let $m=n+1$. Then $a_m=frac1{k+2}$. Therefore,$$leftlvert a_m-a_nrightrvert=frac k{k+1}-frac1{k+2}geqslantfrac12=varepsilon.$$


                                                                        Or you can say that your sequence diverges, since the subsequence$$frac12,frac13,frac14,ldots$$converges to $0$, whereas the subsequence$$frac12,frac23,frac34,ldots$$converges to $1$. So, the sequence diverges and therefore it is not a Cauchy sequence.
                                                                        share|cite|improve this answer

                                                                          up vote
                                                                          0
                                                                          down vote

                                                                          You only need such an $epsilon$ to exist, so you can choose a convenient value.

                                                                          Then you need to show that there are gaps bigger than $epsilon$ between elements of the sequence as far a=out as you care to go – that gaps as big as $epsilon$ don’t fade out and disappear in the tail of the sequence.

                                                                          Well you have identified some chunky gaps which persist (you don’t need every gap to be big). What $epsilon$ would work for the gaps you have identified?

                                                                          share|cite|improve this answer

                                                                            up vote
                                                                            0
                                                                            down vote

                                                                            Cauchy sequence: $$forallvarepsilon>0exists Nforall n,mquad n,m>Nimplies|a_n-a_m|<varepsilon$$

                                                                            We want to show the negative:$$existsvarepsilon>0forall Nexists n,mquad n,m>Nland |a_n-a_m|gevarepsilon$$

                                                                            Take $varepsilon=1/10$, and then, for all $N$ take $n=sum_{i=1}^{N+1} i=frac{(N+1)^2+N+1}{2}, m=n+1$ and you have $a_n$ an element in the form of $frac{k}{k+1}$ and $a_m$ in the form of $frac{1}{k+2}$

                                                                            share|cite|improve this answer

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                                                                              3 Answers
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                                                                              3 Answers
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                                                                              active

                                                                              oldest

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                                                                              up vote
                                                                              3
                                                                              down vote

                                                                              accepted

                                                                              Let $(a_n)_{ninmathbb N}$ be your sequence. Take $varepsilon=frac12$. Given $Ninmathbb N$, take $ngeqslant N$ such that $a_n$ is of the form $frac k{k+1}$ for some $kinmathbb N$ and let $m=n+1$. Then $a_m=frac1{k+2}$. Therefore,$$leftlvert a_m-a_nrightrvert=frac k{k+1}-frac1{k+2}geqslantfrac12=varepsilon.$$


                                                                              Or you can say that your sequence diverges, since the subsequence$$frac12,frac13,frac14,ldots$$converges to $0$, whereas the subsequence$$frac12,frac23,frac34,ldots$$converges to $1$. So, the sequence diverges and therefore it is not a Cauchy sequence.
                                                                              share|cite|improve this answer

                                                                                up vote
                                                                                3
                                                                                down vote

                                                                                accepted

                                                                                Let $(a_n)_{ninmathbb N}$ be your sequence. Take $varepsilon=frac12$. Given $Ninmathbb N$, take $ngeqslant N$ such that $a_n$ is of the form $frac k{k+1}$ for some $kinmathbb N$ and let $m=n+1$. Then $a_m=frac1{k+2}$. Therefore,$$leftlvert a_m-a_nrightrvert=frac k{k+1}-frac1{k+2}geqslantfrac12=varepsilon.$$


                                                                                Or you can say that your sequence diverges, since the subsequence$$frac12,frac13,frac14,ldots$$converges to $0$, whereas the subsequence$$frac12,frac23,frac34,ldots$$converges to $1$. So, the sequence diverges and therefore it is not a Cauchy sequence.
                                                                                share|cite|improve this answer

                                                                                  up vote
                                                                                  3
                                                                                  down vote

                                                                                  accepted

                                                                                  up vote
                                                                                  3
                                                                                  down vote

                                                                                  accepted

                                                                                  Let $(a_n)_{ninmathbb N}$ be your sequence. Take $varepsilon=frac12$. Given $Ninmathbb N$, take $ngeqslant N$ such that $a_n$ is of the form $frac k{k+1}$ for some $kinmathbb N$ and let $m=n+1$. Then $a_m=frac1{k+2}$. Therefore,$$leftlvert a_m-a_nrightrvert=frac k{k+1}-frac1{k+2}geqslantfrac12=varepsilon.$$


                                                                                  Or you can say that your sequence diverges, since the subsequence$$frac12,frac13,frac14,ldots$$converges to $0$, whereas the subsequence$$frac12,frac23,frac34,ldots$$converges to $1$. So, the sequence diverges and therefore it is not a Cauchy sequence.
                                                                                  share|cite|improve this answer

                                                                                  Let $(a_n)_{ninmathbb N}$ be your sequence. Take $varepsilon=frac12$. Given $Ninmathbb N$, take $ngeqslant N$ such that $a_n$ is of the form $frac k{k+1}$ for some $kinmathbb N$ and let $m=n+1$. Then $a_m=frac1{k+2}$. Therefore,$$leftlvert a_m-a_nrightrvert=frac k{k+1}-frac1{k+2}geqslantfrac12=varepsilon.$$


                                                                                  Or you can say that your sequence diverges, since the subsequence$$frac12,frac13,frac14,ldots$$converges to $0$, whereas the subsequence$$frac12,frac23,frac34,ldots$$converges to $1$. So, the sequence diverges and therefore it is not a Cauchy sequence.

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                                                                                  answered Sep 10 at 21:44

                                                                                  José Carlos Santos

                                                                                  124k17101186

                                                                                  124k17101186

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                                                                                      You only need such an $epsilon$ to exist, so you can choose a convenient value.

                                                                                      Then you need to show that there are gaps bigger than $epsilon$ between elements of the sequence as far a=out as you care to go – that gaps as big as $epsilon$ don’t fade out and disappear in the tail of the sequence.

                                                                                      Well you have identified some chunky gaps which persist (you don’t need every gap to be big). What $epsilon$ would work for the gaps you have identified?

                                                                                      share|cite|improve this answer

                                                                                        up vote
                                                                                        0
                                                                                        down vote

                                                                                        You only need such an $epsilon$ to exist, so you can choose a convenient value.

                                                                                        Then you need to show that there are gaps bigger than $epsilon$ between elements of the sequence as far a=out as you care to go – that gaps as big as $epsilon$ don’t fade out and disappear in the tail of the sequence.

                                                                                        Well you have identified some chunky gaps which persist (you don’t need every gap to be big). What $epsilon$ would work for the gaps you have identified?

                                                                                        share|cite|improve this answer

                                                                                          up vote
                                                                                          0
                                                                                          down vote

                                                                                          up vote
                                                                                          0
                                                                                          down vote

                                                                                          You only need such an $epsilon$ to exist, so you can choose a convenient value.

                                                                                          Then you need to show that there are gaps bigger than $epsilon$ between elements of the sequence as far a=out as you care to go – that gaps as big as $epsilon$ don’t fade out and disappear in the tail of the sequence.

                                                                                          Well you have identified some chunky gaps which persist (you don’t need every gap to be big). What $epsilon$ would work for the gaps you have identified?

                                                                                          share|cite|improve this answer

                                                                                          You only need such an $epsilon$ to exist, so you can choose a convenient value.

                                                                                          Then you need to show that there are gaps bigger than $epsilon$ between elements of the sequence as far a=out as you care to go – that gaps as big as $epsilon$ don’t fade out and disappear in the tail of the sequence.

                                                                                          Well you have identified some chunky gaps which persist (you don’t need every gap to be big). What $epsilon$ would work for the gaps you have identified?

                                                                                          share|cite|improve this answer

                                                                                          share|cite|improve this answer

                                                                                          share|cite|improve this answer

                                                                                          answered Sep 10 at 21:46

                                                                                          Mark Bennet

                                                                                          77.5k775175

                                                                                          77.5k775175

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                                                                                              Cauchy sequence: $$forallvarepsilon>0exists Nforall n,mquad n,m>Nimplies|a_n-a_m|<varepsilon$$

                                                                                              We want to show the negative:$$existsvarepsilon>0forall Nexists n,mquad n,m>Nland |a_n-a_m|gevarepsilon$$

                                                                                              Take $varepsilon=1/10$, and then, for all $N$ take $n=sum_{i=1}^{N+1} i=frac{(N+1)^2+N+1}{2}, m=n+1$ and you have $a_n$ an element in the form of $frac{k}{k+1}$ and $a_m$ in the form of $frac{1}{k+2}$

                                                                                              share|cite|improve this answer

                                                                                                up vote
                                                                                                0
                                                                                                down vote

                                                                                                Cauchy sequence: $$forallvarepsilon>0exists Nforall n,mquad n,m>Nimplies|a_n-a_m|<varepsilon$$

                                                                                                We want to show the negative:$$existsvarepsilon>0forall Nexists n,mquad n,m>Nland |a_n-a_m|gevarepsilon$$

                                                                                                Take $varepsilon=1/10$, and then, for all $N$ take $n=sum_{i=1}^{N+1} i=frac{(N+1)^2+N+1}{2}, m=n+1$ and you have $a_n$ an element in the form of $frac{k}{k+1}$ and $a_m$ in the form of $frac{1}{k+2}$

                                                                                                share|cite|improve this answer

                                                                                                  up vote
                                                                                                  0
                                                                                                  down vote

                                                                                                  up vote
                                                                                                  0
                                                                                                  down vote

                                                                                                  Cauchy sequence: $$forallvarepsilon>0exists Nforall n,mquad n,m>Nimplies|a_n-a_m|<varepsilon$$

                                                                                                  We want to show the negative:$$existsvarepsilon>0forall Nexists n,mquad n,m>Nland |a_n-a_m|gevarepsilon$$

                                                                                                  Take $varepsilon=1/10$, and then, for all $N$ take $n=sum_{i=1}^{N+1} i=frac{(N+1)^2+N+1}{2}, m=n+1$ and you have $a_n$ an element in the form of $frac{k}{k+1}$ and $a_m$ in the form of $frac{1}{k+2}$

                                                                                                  share|cite|improve this answer

                                                                                                  Cauchy sequence: $$forallvarepsilon>0exists Nforall n,mquad n,m>Nimplies|a_n-a_m|<varepsilon$$

                                                                                                  We want to show the negative:$$existsvarepsilon>0forall Nexists n,mquad n,m>Nland |a_n-a_m|gevarepsilon$$

                                                                                                  Take $varepsilon=1/10$, and then, for all $N$ take $n=sum_{i=1}^{N+1} i=frac{(N+1)^2+N+1}{2}, m=n+1$ and you have $a_n$ an element in the form of $frac{k}{k+1}$ and $a_m$ in the form of $frac{1}{k+2}$

                                                                                                  share|cite|improve this answer

                                                                                                  share|cite|improve this answer

                                                                                                  share|cite|improve this answer

                                                                                                  answered Sep 10 at 21:49

                                                                                                  Holo

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                                                                                                      Question involving Nested Intervals Theorem

                                                                                                      The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP

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                                                                                                      Let ${[a_n,b_n]}_{ninmathbb{N}}$ be a sequence of closed bounded intervals in $mathbb{R}$ such that $(forall ninmathbb{N})([a_{n+1},b_{n+1}]subset [a_n,b_n])$ e $lim_{ntoinfty}(b_n-a_n)=0$. Let $Ssubset [a_0,b_0]$ a set such that $(forall ninmathbb{N})(Scap [a_n,b_n]neq emptyset )$.

                                                                                                      We know from the Nested Intervals Theorem that exists $sigmainmathbb{R}$ such that $bigcap _{n=1}^infty [a_n,b_n]={sigma}$. My question: the element $sigma$ belongs to the set $S$? That is, is it true that $sigmain S$?

                                                                                                      I have not been able to prove either that it is false or that it is true. I tried to prove that $sigmain S$ by contradiction:

                                                                                                      Assume that $sigmanotin S$, then $Scap left(bigcap _{n=1}^infty [a_n,b_n]right)=emptyset $. This implies that $(forall xin S)left(xnotin bigcap _{n=1}^infty [a_n,b_n]right)$.

                                                                                                      Given $xin S$, we have $xnotin bigcap _{n=1}^infty [a_n,b_n]Rightarrow (exists ninmathbb{N})(xnotin [a_n,b_n])$.

                                                                                                      Therefore,

                                                                                                      $(forall xin S)left(xnotin bigcap _{n=1}^infty [a_n,b_n]right)Rightarrow (forall xin S)(exists ninmathbb{N})(xnotin [a_n,b_n])$

                                                                                                      But I could think of nothing else. Since $(forall xin S)left(xnotin bigcap _{n=1}^infty [a_n,b_n]right)$ is true, we have that $ (forall xin S)(exists ninmathbb{N})(xnotin [a_n,b_n])$ is also true. But I can not infer from this proposition that $sigmain S$.

                                                                                                      Can someone help me please?

                                                                                                      share|cite|improve this question

                                                                                                      • Is there a material difference between $a_n$ and $a^{(n)}$?
                                                                                                        – Saucy O’Path
                                                                                                        Sep 10 at 21:48

                                                                                                      • No, I just forgot to change the notation. I will correct.
                                                                                                        – rfloc
                                                                                                        Sep 10 at 21:55

                                                                                                      up vote
                                                                                                      0
                                                                                                      down vote

                                                                                                      favorite

                                                                                                      Let ${[a_n,b_n]}_{ninmathbb{N}}$ be a sequence of closed bounded intervals in $mathbb{R}$ such that $(forall ninmathbb{N})([a_{n+1},b_{n+1}]subset [a_n,b_n])$ e $lim_{ntoinfty}(b_n-a_n)=0$. Let $Ssubset [a_0,b_0]$ a set such that $(forall ninmathbb{N})(Scap [a_n,b_n]neq emptyset )$.

                                                                                                      We know from the Nested Intervals Theorem that exists $sigmainmathbb{R}$ such that $bigcap _{n=1}^infty [a_n,b_n]={sigma}$. My question: the element $sigma$ belongs to the set $S$? That is, is it true that $sigmain S$?

                                                                                                      I have not been able to prove either that it is false or that it is true. I tried to prove that $sigmain S$ by contradiction:

                                                                                                      Assume that $sigmanotin S$, then $Scap left(bigcap _{n=1}^infty [a_n,b_n]right)=emptyset $. This implies that $(forall xin S)left(xnotin bigcap _{n=1}^infty [a_n,b_n]right)$.

                                                                                                      Given $xin S$, we have $xnotin bigcap _{n=1}^infty [a_n,b_n]Rightarrow (exists ninmathbb{N})(xnotin [a_n,b_n])$.

                                                                                                      Therefore,

                                                                                                      $(forall xin S)left(xnotin bigcap _{n=1}^infty [a_n,b_n]right)Rightarrow (forall xin S)(exists ninmathbb{N})(xnotin [a_n,b_n])$

                                                                                                      But I could think of nothing else. Since $(forall xin S)left(xnotin bigcap _{n=1}^infty [a_n,b_n]right)$ is true, we have that $ (forall xin S)(exists ninmathbb{N})(xnotin [a_n,b_n])$ is also true. But I can not infer from this proposition that $sigmain S$.

                                                                                                      Can someone help me please?

                                                                                                      share|cite|improve this question

                                                                                                      • Is there a material difference between $a_n$ and $a^{(n)}$?
                                                                                                        – Saucy O’Path
                                                                                                        Sep 10 at 21:48

                                                                                                      • No, I just forgot to change the notation. I will correct.
                                                                                                        – rfloc
                                                                                                        Sep 10 at 21:55

                                                                                                      up vote
                                                                                                      0
                                                                                                      down vote

                                                                                                      favorite

                                                                                                      up vote
                                                                                                      0
                                                                                                      down vote

                                                                                                      favorite

                                                                                                      Let ${[a_n,b_n]}_{ninmathbb{N}}$ be a sequence of closed bounded intervals in $mathbb{R}$ such that $(forall ninmathbb{N})([a_{n+1},b_{n+1}]subset [a_n,b_n])$ e $lim_{ntoinfty}(b_n-a_n)=0$. Let $Ssubset [a_0,b_0]$ a set such that $(forall ninmathbb{N})(Scap [a_n,b_n]neq emptyset )$.

                                                                                                      We know from the Nested Intervals Theorem that exists $sigmainmathbb{R}$ such that $bigcap _{n=1}^infty [a_n,b_n]={sigma}$. My question: the element $sigma$ belongs to the set $S$? That is, is it true that $sigmain S$?

                                                                                                      I have not been able to prove either that it is false or that it is true. I tried to prove that $sigmain S$ by contradiction:

                                                                                                      Assume that $sigmanotin S$, then $Scap left(bigcap _{n=1}^infty [a_n,b_n]right)=emptyset $. This implies that $(forall xin S)left(xnotin bigcap _{n=1}^infty [a_n,b_n]right)$.

                                                                                                      Given $xin S$, we have $xnotin bigcap _{n=1}^infty [a_n,b_n]Rightarrow (exists ninmathbb{N})(xnotin [a_n,b_n])$.

                                                                                                      Therefore,

                                                                                                      $(forall xin S)left(xnotin bigcap _{n=1}^infty [a_n,b_n]right)Rightarrow (forall xin S)(exists ninmathbb{N})(xnotin [a_n,b_n])$

                                                                                                      But I could think of nothing else. Since $(forall xin S)left(xnotin bigcap _{n=1}^infty [a_n,b_n]right)$ is true, we have that $ (forall xin S)(exists ninmathbb{N})(xnotin [a_n,b_n])$ is also true. But I can not infer from this proposition that $sigmain S$.

                                                                                                      Can someone help me please?

                                                                                                      share|cite|improve this question

                                                                                                      Let ${[a_n,b_n]}_{ninmathbb{N}}$ be a sequence of closed bounded intervals in $mathbb{R}$ such that $(forall ninmathbb{N})([a_{n+1},b_{n+1}]subset [a_n,b_n])$ e $lim_{ntoinfty}(b_n-a_n)=0$. Let $Ssubset [a_0,b_0]$ a set such that $(forall ninmathbb{N})(Scap [a_n,b_n]neq emptyset )$.

                                                                                                      We know from the Nested Intervals Theorem that exists $sigmainmathbb{R}$ such that $bigcap _{n=1}^infty [a_n,b_n]={sigma}$. My question: the element $sigma$ belongs to the set $S$? That is, is it true that $sigmain S$?

                                                                                                      I have not been able to prove either that it is false or that it is true. I tried to prove that $sigmain S$ by contradiction:

                                                                                                      Assume that $sigmanotin S$, then $Scap left(bigcap _{n=1}^infty [a_n,b_n]right)=emptyset $. This implies that $(forall xin S)left(xnotin bigcap _{n=1}^infty [a_n,b_n]right)$.

                                                                                                      Given $xin S$, we have $xnotin bigcap _{n=1}^infty [a_n,b_n]Rightarrow (exists ninmathbb{N})(xnotin [a_n,b_n])$.

                                                                                                      Therefore,

                                                                                                      $(forall xin S)left(xnotin bigcap _{n=1}^infty [a_n,b_n]right)Rightarrow (forall xin S)(exists ninmathbb{N})(xnotin [a_n,b_n])$

                                                                                                      But I could think of nothing else. Since $(forall xin S)left(xnotin bigcap _{n=1}^infty [a_n,b_n]right)$ is true, we have that $ (forall xin S)(exists ninmathbb{N})(xnotin [a_n,b_n])$ is also true. But I can not infer from this proposition that $sigmain S$.

                                                                                                      Can someone help me please?

                                                                                                      real-analysis elementary-set-theory

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                                                                                                      edited Sep 10 at 21:57

                                                                                                      asked Sep 10 at 21:38

                                                                                                      rfloc

                                                                                                      12518

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                                                                                                      • Is there a material difference between $a_n$ and $a^{(n)}$?
                                                                                                        – Saucy O’Path
                                                                                                        Sep 10 at 21:48

                                                                                                      • No, I just forgot to change the notation. I will correct.
                                                                                                        – rfloc
                                                                                                        Sep 10 at 21:55

                                                                                                      • Is there a material difference between $a_n$ and $a^{(n)}$?
                                                                                                        – Saucy O’Path
                                                                                                        Sep 10 at 21:48

                                                                                                      • No, I just forgot to change the notation. I will correct.
                                                                                                        – rfloc
                                                                                                        Sep 10 at 21:55

                                                                                                      Is there a material difference between $a_n$ and $a^{(n)}$?
                                                                                                      – Saucy O’Path
                                                                                                      Sep 10 at 21:48

                                                                                                      Is there a material difference between $a_n$ and $a^{(n)}$?
                                                                                                      – Saucy O’Path
                                                                                                      Sep 10 at 21:48

                                                                                                      No, I just forgot to change the notation. I will correct.
                                                                                                      – rfloc
                                                                                                      Sep 10 at 21:55

                                                                                                      No, I just forgot to change the notation. I will correct.
                                                                                                      – rfloc
                                                                                                      Sep 10 at 21:55

                                                                                                      1 Answer
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                                                                                                      By the limit condition, at least one of the sequences $(a_n)$ and $(b_n)$ is not eventually constant, say $(a_n)$. Then if you take $S={a_nmid ninmathbb N}$, $sigma$ will not belong to $S$. So, the answer is: $sigma$ doesn’t have to belong to $S$.

                                                                                                      share|cite|improve this answer

                                                                                                      • I have noticed that in fact $sigmain S$ in the case where $S$ is a closed set.
                                                                                                        – rfloc
                                                                                                        Sep 10 at 22:41

                                                                                                      • To the proposer: Suppose $a_n<a_{n+1}<b_{n+1}<b_n$ for every $n$ and $S_1=[a_0,b_0] $ while $S_2=[a_0,b_0]backslash {sigma}.$
                                                                                                        – DanielWainfleet
                                                                                                        Sep 11 at 6:53

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                                                                                                      By the limit condition, at least one of the sequences $(a_n)$ and $(b_n)$ is not eventually constant, say $(a_n)$. Then if you take $S={a_nmid ninmathbb N}$, $sigma$ will not belong to $S$. So, the answer is: $sigma$ doesn’t have to belong to $S$.

                                                                                                      share|cite|improve this answer

                                                                                                      • I have noticed that in fact $sigmain S$ in the case where $S$ is a closed set.
                                                                                                        – rfloc
                                                                                                        Sep 10 at 22:41

                                                                                                      • To the proposer: Suppose $a_n<a_{n+1}<b_{n+1}<b_n$ for every $n$ and $S_1=[a_0,b_0] $ while $S_2=[a_0,b_0]backslash {sigma}.$
                                                                                                        – DanielWainfleet
                                                                                                        Sep 11 at 6:53

                                                                                                      up vote
                                                                                                      1
                                                                                                      down vote

                                                                                                      By the limit condition, at least one of the sequences $(a_n)$ and $(b_n)$ is not eventually constant, say $(a_n)$. Then if you take $S={a_nmid ninmathbb N}$, $sigma$ will not belong to $S$. So, the answer is: $sigma$ doesn’t have to belong to $S$.

                                                                                                      share|cite|improve this answer

                                                                                                      • I have noticed that in fact $sigmain S$ in the case where $S$ is a closed set.
                                                                                                        – rfloc
                                                                                                        Sep 10 at 22:41

                                                                                                      • To the proposer: Suppose $a_n<a_{n+1}<b_{n+1}<b_n$ for every $n$ and $S_1=[a_0,b_0] $ while $S_2=[a_0,b_0]backslash {sigma}.$
                                                                                                        – DanielWainfleet
                                                                                                        Sep 11 at 6:53

                                                                                                      up vote
                                                                                                      1
                                                                                                      down vote

                                                                                                      up vote
                                                                                                      1
                                                                                                      down vote

                                                                                                      By the limit condition, at least one of the sequences $(a_n)$ and $(b_n)$ is not eventually constant, say $(a_n)$. Then if you take $S={a_nmid ninmathbb N}$, $sigma$ will not belong to $S$. So, the answer is: $sigma$ doesn’t have to belong to $S$.

                                                                                                      share|cite|improve this answer

                                                                                                      By the limit condition, at least one of the sequences $(a_n)$ and $(b_n)$ is not eventually constant, say $(a_n)$. Then if you take $S={a_nmid ninmathbb N}$, $sigma$ will not belong to $S$. So, the answer is: $sigma$ doesn’t have to belong to $S$.

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                                                                                                      answered Sep 10 at 21:48

                                                                                                      SMM

                                                                                                      2,03049

                                                                                                      2,03049

                                                                                                      • I have noticed that in fact $sigmain S$ in the case where $S$ is a closed set.
                                                                                                        – rfloc
                                                                                                        Sep 10 at 22:41

                                                                                                      • To the proposer: Suppose $a_n<a_{n+1}<b_{n+1}<b_n$ for every $n$ and $S_1=[a_0,b_0] $ while $S_2=[a_0,b_0]backslash {sigma}.$
                                                                                                        – DanielWainfleet
                                                                                                        Sep 11 at 6:53

                                                                                                      • I have noticed that in fact $sigmain S$ in the case where $S$ is a closed set.
                                                                                                        – rfloc
                                                                                                        Sep 10 at 22:41

                                                                                                      • To the proposer: Suppose $a_n<a_{n+1}<b_{n+1}<b_n$ for every $n$ and $S_1=[a_0,b_0] $ while $S_2=[a_0,b_0]backslash {sigma}.$
                                                                                                        – DanielWainfleet
                                                                                                        Sep 11 at 6:53

                                                                                                      I have noticed that in fact $sigmain S$ in the case where $S$ is a closed set.
                                                                                                      – rfloc
                                                                                                      Sep 10 at 22:41

                                                                                                      I have noticed that in fact $sigmain S$ in the case where $S$ is a closed set.
                                                                                                      – rfloc
                                                                                                      Sep 10 at 22:41

                                                                                                      To the proposer: Suppose $a_n<a_{n+1}<b_{n+1}<b_n$ for every $n$ and $S_1=[a_0,b_0] $ while $S_2=[a_0,b_0]backslash {sigma}.$
                                                                                                      – DanielWainfleet
                                                                                                      Sep 11 at 6:53

                                                                                                      To the proposer: Suppose $a_n<a_{n+1}<b_{n+1}<b_n$ for every $n$ and $S_1=[a_0,b_0] $ while $S_2=[a_0,b_0]backslash {sigma}.$
                                                                                                      – DanielWainfleet
                                                                                                      Sep 11 at 6:53

                                                                                                       
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                                                                                                      Proving that $xf'(x) = frac{x}{x+1}(x+1)f(x+1) – xf(x) – frac{1}{2}xf”(ξ)$

                                                                                                      The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP

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                                                                                                      Exercise :

                                                                                                      Let $f:(0,infty) to mathbb R$ be a function such that $f in C^2$, $lim_{x to infty} xf(x) = 2$ and $lim_{x to infty} xf”(x) = 0$. If $x>0$, prove that for some $ξ in (x,x+1)$ it is
                                                                                                      $$xf'(x) = frac{x}{x+1}(x+1)f(x+1) – xf(x) – frac{1}{2}xf”(ξ)$$
                                                                                                      and then calculate the limit $lim_{x to infty} xf'(x)$.

                                                                                                      Question/Request : I need some help, hints or tips on how to prove the formula in the center of the question. The final limit asked is simple as you just use the formula proven, so no help needed on that.

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                                                                                                      • Is the first term in the right-hand side what you meant to write? Because then $x+1$ would cancel out.
                                                                                                        – Îµ-δ
                                                                                                        Sep 10 at 21:29

                                                                                                      • @ε-δ Was weird enough on when I was copying it from my textbook, but yes, according to it, this is the form of the expression.
                                                                                                        – Rebellos
                                                                                                        Sep 10 at 21:30

                                                                                                      • why do you write $frac{x}{x+1}(x+1)f(x+1)$ ?
                                                                                                        – Ahmad Bazzi
                                                                                                        Sep 10 at 21:33

                                                                                                      up vote
                                                                                                      0
                                                                                                      down vote

                                                                                                      favorite

                                                                                                      2

                                                                                                      Exercise :

                                                                                                      Let $f:(0,infty) to mathbb R$ be a function such that $f in C^2$, $lim_{x to infty} xf(x) = 2$ and $lim_{x to infty} xf”(x) = 0$. If $x>0$, prove that for some $ξ in (x,x+1)$ it is
                                                                                                      $$xf'(x) = frac{x}{x+1}(x+1)f(x+1) – xf(x) – frac{1}{2}xf”(ξ)$$
                                                                                                      and then calculate the limit $lim_{x to infty} xf'(x)$.

                                                                                                      Question/Request : I need some help, hints or tips on how to prove the formula in the center of the question. The final limit asked is simple as you just use the formula proven, so no help needed on that.

                                                                                                      share|cite|improve this question

                                                                                                      • Is the first term in the right-hand side what you meant to write? Because then $x+1$ would cancel out.
                                                                                                        – Îµ-δ
                                                                                                        Sep 10 at 21:29

                                                                                                      • @ε-δ Was weird enough on when I was copying it from my textbook, but yes, according to it, this is the form of the expression.
                                                                                                        – Rebellos
                                                                                                        Sep 10 at 21:30

                                                                                                      • why do you write $frac{x}{x+1}(x+1)f(x+1)$ ?
                                                                                                        – Ahmad Bazzi
                                                                                                        Sep 10 at 21:33

                                                                                                      up vote
                                                                                                      0
                                                                                                      down vote

                                                                                                      favorite

                                                                                                      2

                                                                                                      up vote
                                                                                                      0
                                                                                                      down vote

                                                                                                      favorite

                                                                                                      2
                                                                                                      2

                                                                                                      Exercise :

                                                                                                      Let $f:(0,infty) to mathbb R$ be a function such that $f in C^2$, $lim_{x to infty} xf(x) = 2$ and $lim_{x to infty} xf”(x) = 0$. If $x>0$, prove that for some $ξ in (x,x+1)$ it is
                                                                                                      $$xf'(x) = frac{x}{x+1}(x+1)f(x+1) – xf(x) – frac{1}{2}xf”(ξ)$$
                                                                                                      and then calculate the limit $lim_{x to infty} xf'(x)$.

                                                                                                      Question/Request : I need some help, hints or tips on how to prove the formula in the center of the question. The final limit asked is simple as you just use the formula proven, so no help needed on that.

                                                                                                      share|cite|improve this question

                                                                                                      Exercise :

                                                                                                      Let $f:(0,infty) to mathbb R$ be a function such that $f in C^2$, $lim_{x to infty} xf(x) = 2$ and $lim_{x to infty} xf”(x) = 0$. If $x>0$, prove that for some $ξ in (x,x+1)$ it is
                                                                                                      $$xf'(x) = frac{x}{x+1}(x+1)f(x+1) – xf(x) – frac{1}{2}xf”(ξ)$$
                                                                                                      and then calculate the limit $lim_{x to infty} xf'(x)$.

                                                                                                      Question/Request : I need some help, hints or tips on how to prove the formula in the center of the question. The final limit asked is simple as you just use the formula proven, so no help needed on that.

                                                                                                      calculus differential-equations derivatives

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                                                                                                      asked Sep 10 at 21:25

                                                                                                      Rebellos

                                                                                                      10.2k21039

                                                                                                      10.2k21039

                                                                                                      • Is the first term in the right-hand side what you meant to write? Because then $x+1$ would cancel out.
                                                                                                        – Îµ-δ
                                                                                                        Sep 10 at 21:29

                                                                                                      • @ε-δ Was weird enough on when I was copying it from my textbook, but yes, according to it, this is the form of the expression.
                                                                                                        – Rebellos
                                                                                                        Sep 10 at 21:30

                                                                                                      • why do you write $frac{x}{x+1}(x+1)f(x+1)$ ?
                                                                                                        – Ahmad Bazzi
                                                                                                        Sep 10 at 21:33

                                                                                                      • Is the first term in the right-hand side what you meant to write? Because then $x+1$ would cancel out.
                                                                                                        – Îµ-δ
                                                                                                        Sep 10 at 21:29

                                                                                                      • @ε-δ Was weird enough on when I was copying it from my textbook, but yes, according to it, this is the form of the expression.
                                                                                                        – Rebellos
                                                                                                        Sep 10 at 21:30

                                                                                                      • why do you write $frac{x}{x+1}(x+1)f(x+1)$ ?
                                                                                                        – Ahmad Bazzi
                                                                                                        Sep 10 at 21:33

                                                                                                      Is the first term in the right-hand side what you meant to write? Because then $x+1$ would cancel out.
                                                                                                      – Îµ-δ
                                                                                                      Sep 10 at 21:29

                                                                                                      Is the first term in the right-hand side what you meant to write? Because then $x+1$ would cancel out.
                                                                                                      – Îµ-δ
                                                                                                      Sep 10 at 21:29

                                                                                                      @ε-δ Was weird enough on when I was copying it from my textbook, but yes, according to it, this is the form of the expression.
                                                                                                      – Rebellos
                                                                                                      Sep 10 at 21:30

                                                                                                      @ε-δ Was weird enough on when I was copying it from my textbook, but yes, according to it, this is the form of the expression.
                                                                                                      – Rebellos
                                                                                                      Sep 10 at 21:30

                                                                                                      why do you write $frac{x}{x+1}(x+1)f(x+1)$ ?
                                                                                                      – Ahmad Bazzi
                                                                                                      Sep 10 at 21:33

                                                                                                      why do you write $frac{x}{x+1}(x+1)f(x+1)$ ?
                                                                                                      – Ahmad Bazzi
                                                                                                      Sep 10 at 21:33

                                                                                                      1 Answer
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                                                                                                      I don’t know why you would write the identity in this peculiar why, but this is just Taylor’s theorem on the interval $[x,x+1]$. Since $fin C^2$, the theorem guarantees the existence of $xiin(x,x+1)$, so that $f(x+1)=f(x)+f'(x)+frac12f”(xi)$. Multiply by $x$ and you’re done.

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                                                                                                        I don’t know why you would write the identity in this peculiar why, but this is just Taylor’s theorem on the interval $[x,x+1]$. Since $fin C^2$, the theorem guarantees the existence of $xiin(x,x+1)$, so that $f(x+1)=f(x)+f'(x)+frac12f”(xi)$. Multiply by $x$ and you’re done.

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                                                                                                          up vote
                                                                                                          2
                                                                                                          down vote

                                                                                                          I don’t know why you would write the identity in this peculiar why, but this is just Taylor’s theorem on the interval $[x,x+1]$. Since $fin C^2$, the theorem guarantees the existence of $xiin(x,x+1)$, so that $f(x+1)=f(x)+f'(x)+frac12f”(xi)$. Multiply by $x$ and you’re done.

                                                                                                          share|cite|improve this answer

                                                                                                            up vote
                                                                                                            2
                                                                                                            down vote

                                                                                                            up vote
                                                                                                            2
                                                                                                            down vote

                                                                                                            I don’t know why you would write the identity in this peculiar why, but this is just Taylor’s theorem on the interval $[x,x+1]$. Since $fin C^2$, the theorem guarantees the existence of $xiin(x,x+1)$, so that $f(x+1)=f(x)+f'(x)+frac12f”(xi)$. Multiply by $x$ and you’re done.

                                                                                                            share|cite|improve this answer

                                                                                                            I don’t know why you would write the identity in this peculiar why, but this is just Taylor’s theorem on the interval $[x,x+1]$. Since $fin C^2$, the theorem guarantees the existence of $xiin(x,x+1)$, so that $f(x+1)=f(x)+f'(x)+frac12f”(xi)$. Multiply by $x$ and you’re done.

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                                                                                                            answered Sep 10 at 22:12

                                                                                                            ε-δ

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