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

Clash Royale CLAN TAG#URR8PPP

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?

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?

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?

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

edited Sep 10 at 19:52

4,34131651

4,34131651

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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.

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.

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.

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.

edited Sep 10 at 19:56

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

Clash Royale CLAN TAG#URR8PPP

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.

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

• @qqqqq Peano’s axioms are a definition of integers.
âÂ 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
âÂ qqqqq
Jan 29 ’15 at 6:00

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.

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

• @qqqqq Peano’s axioms are a definition of integers.
âÂ 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
âÂ qqqqq
Jan 29 ’15 at 6:00

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.

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

edited Jan 29 ’15 at 7:12

Asaf Karagilaâ¦

295k32411738

295k32411738

asked Jan 29 ’15 at 5:47

qqqqq

61

61

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

• @qqqqq Peano’s axioms are a definition of integers.
âÂ 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
âÂ qqqqq
Jan 29 ’15 at 6:00

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

• @qqqqq Peano’s axioms are a definition of integers.
âÂ 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
âÂ qqqqq
Jan 29 ’15 at 6:00

You cool with Peano’s axioms?
âÂ Daniel W. Farlow
Jan 29 ’15 at 5:55

You cool with Peano’s axioms?
âÂ Daniel W. Farlow
Jan 29 ’15 at 5:55

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

@qqqqq Peano’s axioms are a definition of integers.
âÂ 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
âÂ 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
âÂ qqqqq
Jan 29 ’15 at 6:00

active

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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.

active

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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.

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.

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.

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.

Toby Bartels

528412

528412

Â

draft saved

function () {
}
);

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

Clash Royale CLAN TAG#URR8PPP

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 $(Qlandneg R)land(neg Rtoneg Q)$.

• It would be a lot easier to understand what you’re doing if you don’t abbreviate your meaningful propositions as $P$, $Q$, $R$.
âÂ 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.
âÂ Henning Makholm
Sep 10 at 21:48

• @HenningMakholm Yeah, sorry for that. I do it because I am practicing writing logical expressions.
âÂ 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 $R$.
âÂ Bright
Sep 10 at 22:01

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

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 $(Qlandneg R)land(neg Rtoneg Q)$.

• It would be a lot easier to understand what you’re doing if you don’t abbreviate your meaningful propositions as $P$, $Q$, $R$.
âÂ 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.
âÂ Henning Makholm
Sep 10 at 21:48

• @HenningMakholm Yeah, sorry for that. I do it because I am practicing writing logical expressions.
âÂ 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 $R$.
âÂ Bright
Sep 10 at 22:01

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

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 $(Qlandneg R)land(neg Rtoneg Q)$.

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

edited Sep 10 at 22:23

Bright

18610

18610

• It would be a lot easier to understand what you’re doing if you don’t abbreviate your meaningful propositions as $P$, $Q$, $R$.
âÂ 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.
âÂ Henning Makholm
Sep 10 at 21:48

• @HenningMakholm Yeah, sorry for that. I do it because I am practicing writing logical expressions.
âÂ 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 $R$.
âÂ Bright
Sep 10 at 22:01

• If you insist of writing your proof in that horribly confusing and convoluted way, suit yourself.
âÂ 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 $P$, $Q$, $R$.
âÂ 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.
âÂ Henning Makholm
Sep 10 at 21:48

• @HenningMakholm Yeah, sorry for that. I do it because I am practicing writing logical expressions.
âÂ 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 $R$.
âÂ Bright
Sep 10 at 22:01

• If you insist of writing your proof in that horribly confusing and convoluted way, suit yourself.
âÂ 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 $P$, $Q$, $R$.
âÂ 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 $P$, $Q$, $R$.
âÂ 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.
âÂ 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.
âÂ Henning Makholm
Sep 10 at 21:48

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

@HenningMakholm Yeah, sorry for that. I do it because I am practicing writing logical expressions.
âÂ 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 $R$.
âÂ 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 $R$.
âÂ Bright
Sep 10 at 22:01

1

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

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

active

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function () {
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);

## Strongly p-embedded subgroups and p-Sylow subgroups.

Clash Royale CLAN TAG#URR8PPP

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 $x in H cap {^y}H?$

• 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.
âÂ 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?
âÂ 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 $Sleq H$.
âÂ James
Sep 10 at 22:45

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

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 $x in H cap {^y}H?$

• 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.
âÂ 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?
âÂ 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 $Sleq H$.
âÂ James
Sep 10 at 22:45

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

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 $x in H cap {^y}H?$

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

edited Sep 10 at 22:34

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.
âÂ 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?
âÂ 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 $Sleq H$.
âÂ James
Sep 10 at 22:45

• Thanks, I misread the second statemnt and read it as $g in H cap {^y}H.$
âÂ 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.
âÂ 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?
âÂ 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 $Sleq H$.
âÂ James
Sep 10 at 22:45

• Thanks, I misread the second statemnt and read it as $g in H cap {^y}H.$
âÂ 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.
âÂ 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.
âÂ 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?
âÂ 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?
âÂ 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 $Sleq H$.
âÂ 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 $Sleq H$.
âÂ James
Sep 10 at 22:45

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

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

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

Clash Royale CLAN TAG#URR8PPP

Let $p_1, p_2,ldots,p_n$ are points on the compact Riemann surface $X$ and $Y:=X-{p_1,…..,p_n}$. Suppose
$$f:YÃ¢ÂÂmathbb{C}$$
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.

Let $p_1, p_2,ldots,p_n$ are points on the compact Riemann surface $X$ and $Y:=X-{p_1,…..,p_n}$. Suppose
$$f:YÃ¢ÂÂmathbb{C}$$
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.

Let $p_1, p_2,ldots,p_n$ are points on the compact Riemann surface $X$ and $Y:=X-{p_1,…..,p_n}$. Suppose
$$f:YÃ¢ÂÂmathbb{C}$$
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.

Let $p_1, p_2,ldots,p_n$ are points on the compact Riemann surface $X$ and $Y:=X-{p_1,…..,p_n}$. Suppose
$$f:YÃ¢ÂÂmathbb{C}$$
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

edited Sep 10 at 21:56

aidangallagher4

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6571312

abcdmath

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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.

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

• 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.
âÂ Daminark
Sep 10 at 22:16

• Thank you sir..
âÂ 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.

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

• 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.
âÂ Daminark
Sep 10 at 22:16

• Thank you sir..
âÂ abcdmath
Sep 10 at 22:22

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.

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

• 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.
âÂ Daminark
Sep 10 at 22:16

• Thank you sir..
âÂ abcdmath
Sep 10 at 22:22

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.

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.

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 $f$.
âÂ abcdmath
Sep 10 at 22:10

• 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.
âÂ Daminark
Sep 10 at 22:16

• Thank you sir..
âÂ 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 $f$.
âÂ abcdmath
Sep 10 at 22:10

• 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.
âÂ Daminark
Sep 10 at 22:16

• Thank you sir..
âÂ 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 $f$.
âÂ 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 $f$.
âÂ 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.
âÂ 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.
âÂ Daminark
Sep 10 at 22:16

Thank you sir..
âÂ abcdmath
Sep 10 at 22:22

Thank you sir..
âÂ abcdmath
Sep 10 at 22:22

Â

draft saved

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## Question on derivation of probability proportion in Deep Learning Book

Clash Royale CLAN TAG#URR8PPP

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?

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?

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?

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

ClownInTheMoon

1,119418

1,119418

active

oldest

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 $frac12cdot2^n$.

active

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oldest

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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 $frac12cdot2^n$.

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 $frac12cdot2^n$.

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 $frac12cdot2^n$.

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 $frac12cdot2^n$.

joriki

169k10181337

169k10181337

Â

draft saved

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

Clash Royale CLAN TAG#URR8PPP

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?

## 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.” â 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.

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?

## 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.” â 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.

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?

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

edited Sep 10 at 23:54

N. F. Taussig

39.9k93253

39.9k93253

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.” â 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.” â 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.

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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 $2^9-1=512-1=511$.

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

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

• Nice solution +1
âÂ Rushabh Mehta
Sep 10 at 21:59

• Why is the number of different ways of doing it 2^9?
âÂ 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.
âÂ joriki
Sep 11 at 5:12

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 $$sumlimits_{n=1}^9{9choose n}=511$$

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

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

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active

oldest

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 $2^9-1=512-1=511$.

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

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

• Nice solution +1
âÂ Rushabh Mehta
Sep 10 at 21:59

• Why is the number of different ways of doing it 2^9?
âÂ 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.
âÂ joriki
Sep 11 at 5:12

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 $2^9-1=512-1=511$.

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

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

• Nice solution +1
âÂ Rushabh Mehta
Sep 10 at 21:59

• Why is the number of different ways of doing it 2^9?
âÂ 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.
âÂ joriki
Sep 11 at 5:12

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 $2^9-1=512-1=511$.

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 $2^9-1=512-1=511$.

edited Sep 10 at 22:02

joriki

169k10181337

169k10181337

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

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

• Nice solution +1
âÂ Rushabh Mehta
Sep 10 at 21:59

• Why is the number of different ways of doing it 2^9?
âÂ 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.
âÂ joriki
Sep 11 at 5:12

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

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

• Nice solution +1
âÂ Rushabh Mehta
Sep 10 at 21:59

• Why is the number of different ways of doing it 2^9?
âÂ 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.
âÂ joriki
Sep 11 at 5:12

Almost. Remember that there must be at least 1 + sign.
âÂ Rushabh Mehta
Sep 10 at 21:57

Almost. Remember that there must be at least 1 + sign.
âÂ Rushabh Mehta
Sep 10 at 21:57

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

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

Nice solution +1
âÂ Rushabh Mehta
Sep 10 at 21:59

Nice solution +1
âÂ Rushabh Mehta
Sep 10 at 21:59

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

Why is the number of different ways of doing it 2^9?
âÂ 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.
âÂ 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.
âÂ joriki
Sep 11 at 5:12

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 $$sumlimits_{n=1}^9{9choose n}=511$$

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

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

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 $$sumlimits_{n=1}^9{9choose n}=511$$

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

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

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 $$sumlimits_{n=1}^9{9choose n}=511$$

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 $$sumlimits_{n=1}^9{9choose n}=511$$

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”?
âÂ Jolly
Sep 10 at 23:04

• @Jolly Yes indeed. That is the case when there are 10 numbers, which will be represented by $9choose9$
âÂ 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”?
âÂ Jolly
Sep 10 at 23:04

• @Jolly Yes indeed. That is the case when there are 10 numbers, which will be represented by $9choose9$
âÂ 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”?
âÂ 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”?
âÂ Jolly
Sep 10 at 23:04

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

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

## Proving that a sequence is not Cauchy

Clash Royale CLAN TAG#URR8PPP

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!

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!

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!

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

Sir_Math_Cat

792618

792618

active

oldest

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.

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?

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 $frac{1}{k+2}$

active

oldest

active

oldest

active

oldest

active

oldest

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.

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.

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.

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.

124k17101186

124k17101186

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?

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?

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?

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?

Mark Bennet

77.5k775175

77.5k775175

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 $frac{1}{k+2}$

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 $frac{1}{k+2}$

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 $frac{1}{k+2}$

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 $frac{1}{k+2}$

Holo

4,7312729

4,7312729

Â

draft saved

function () {
}
);

## Question involving Nested Intervals Theorem

Clash Royale CLAN TAG#URR8PPP

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 $sigmain S$.

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

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

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 $sigmain S$.

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

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

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 $sigmain S$.

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 $sigmain S$.

real-analysis elementary-set-theory

edited Sep 10 at 21:57

rfloc

12518

12518

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

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

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

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

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

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

No, I just forgot to change the notation. I will correct.
âÂ rfloc
Sep 10 at 21:55

No, I just forgot to change the notation. I will correct.
âÂ rfloc
Sep 10 at 21:55

active

oldest

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 $S$.

• I have noticed that in fact $sigmain S$ in the case where $S$ is a closed set.
âÂ 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 $S_2=[a_0,b_0]backslash {sigma}.$
âÂ DanielWainfleet
Sep 11 at 6:53

active

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active

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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 $S$.

• I have noticed that in fact $sigmain S$ in the case where $S$ is a closed set.
âÂ 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 $S_2=[a_0,b_0]backslash {sigma}.$
âÂ DanielWainfleet
Sep 11 at 6:53

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 $S$.

• I have noticed that in fact $sigmain S$ in the case where $S$ is a closed set.
âÂ 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 $S_2=[a_0,b_0]backslash {sigma}.$
âÂ DanielWainfleet
Sep 11 at 6:53

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 $S$.

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 $S$.

SMM

2,03049

2,03049

• I have noticed that in fact $sigmain S$ in the case where $S$ is a closed set.
âÂ 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 $S_2=[a_0,b_0]backslash {sigma}.$
âÂ DanielWainfleet
Sep 11 at 6:53

• I have noticed that in fact $sigmain S$ in the case where $S$ is a closed set.
âÂ 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 $S_2=[a_0,b_0]backslash {sigma}.$
âÂ DanielWainfleet
Sep 11 at 6:53

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

I have noticed that in fact $sigmain S$ in the case where $S$ is a closed set.
âÂ 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 $S_2=[a_0,b_0]backslash {sigma}.$
âÂ 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 $S_2=[a_0,b_0]backslash {sigma}.$
âÂ DanielWainfleet
Sep 11 at 6:53

Â

draft saved

function () {
}
);

## Proving that $xf'(x) = frac{x}{x+1}(x+1)f(x+1) – xf(x) – frac{1}{2}xf”(ÃÂ¾)$

Clash Royale CLAN TAG#URR8PPP

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 $ÃÂ¾ in (x,x+1)$ it is
$$xf'(x) = frac{x}{x+1}(x+1)f(x+1) – xf(x) – frac{1}{2}xf”(ÃÂ¾)$$
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.

• Is the first term in the right-hand side what you meant to write? Because then $x+1$ would cancel out.
âÂ ÃÂµ-ÃÂ´
Sep 10 at 21:29

• @ÃÂµ-ÃÂ´ Was weird enough on when I was copying it from my textbook, but yes, according to it, this is the form of the expression.
âÂ Rebellos
Sep 10 at 21:30

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

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 $ÃÂ¾ in (x,x+1)$ it is
$$xf'(x) = frac{x}{x+1}(x+1)f(x+1) – xf(x) – frac{1}{2}xf”(ÃÂ¾)$$
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.

• Is the first term in the right-hand side what you meant to write? Because then $x+1$ would cancel out.
âÂ ÃÂµ-ÃÂ´
Sep 10 at 21:29

• @ÃÂµ-ÃÂ´ Was weird enough on when I was copying it from my textbook, but yes, according to it, this is the form of the expression.
âÂ Rebellos
Sep 10 at 21:30

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

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 $ÃÂ¾ in (x,x+1)$ it is
$$xf'(x) = frac{x}{x+1}(x+1)f(x+1) – xf(x) – frac{1}{2}xf”(ÃÂ¾)$$
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.

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 $ÃÂ¾ in (x,x+1)$ it is
$$xf'(x) = frac{x}{x+1}(x+1)f(x+1) – xf(x) – frac{1}{2}xf”(ÃÂ¾)$$
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

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.
âÂ ÃÂµ-ÃÂ´
Sep 10 at 21:29

• @ÃÂµ-ÃÂ´ Was weird enough on when I was copying it from my textbook, but yes, according to it, this is the form of the expression.
âÂ Rebellos
Sep 10 at 21:30

• why do you write $frac{x}{x+1}(x+1)f(x+1)$ ?
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.
âÂ ÃÂµ-ÃÂ´
Sep 10 at 21:29

• @ÃÂµ-ÃÂ´ Was weird enough on when I was copying it from my textbook, but yes, according to it, this is the form of the expression.
âÂ Rebellos
Sep 10 at 21:30

• why do you write $frac{x}{x+1}(x+1)f(x+1)$ ?
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.
âÂ ÃÂµ-ÃÂ´
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.
âÂ ÃÂµ-ÃÂ´
Sep 10 at 21:29

@ÃÂµ-ÃÂ´ Was weird enough on when I was copying it from my textbook, but yes, according to it, this is the form of the expression.
âÂ Rebellos
Sep 10 at 21:30

@ÃÂµ-ÃÂ´ Was weird enough on when I was copying it from my textbook, but yes, according to it, this is the form of the expression.
âÂ Rebellos
Sep 10 at 21:30

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

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

active

oldest

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.

active

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oldest

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oldest

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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.

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.

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.

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.

ÃÂµ-ÃÂ´

1285

1285

Â

draft saved