Mixing
Complex Analysis Extension of Bernoulli Number Generating function
$begingroup$
I have just recently started revising for my complex analysis module at university and come across and interesting exercise in a textbook while reading. I vaguely understand the concept but am struggling with extending a function. This is the question:
Consider the following complex differentiable function:
$$f(z)=frac{z}{e^z-1}, 0<|z|<2pi $$
(a) show that there exists a complex differentiable extension of $f$ onto the set $D={{zinmathbb{C} : |z|<2pi}}$. Denote this extension by $g$.
This is the part I am struggling with primarily as I feel I could do the other parts once I have this but really don't understand how to extend the function to include zero, all I can see is that $f(0)$ must be $1$.
(b) The power series expansion for $f$ around $0inmathbb{C}$ can be written as: $$f(z)=sum^infty_{n=0}frac{B_n}{n!}z^n $$
The coefficients $B_n$ are called Bernoulli numbers. Determine $B_0$ and $B_1$.
(c) Show that $$limsup_{ntoinfty}|B_n|^frac{1}{n}=infty$$
I have never come across Bernoulli numbers before this question and have looked at some more material on them and can see how to do part (b) once part (a) is complete. But I haven't found much material on any functions on how to extend them to include undefined points differentiable. Also I cannot really see how part (c) would follow from (b) or (a). Please help! Any hints or solutions are greatly appreciated.
complex-analysis complex-numbers power-series limsup-and-liminf bernoulli-numbers
asked Jan 23 at 15:02
Dani_5040Dani_5040
324
$endgroup$
add a comment |
$begingroup$
I have just recently started revising for my complex analysis module at university and come across and interesting exercise in a textbook while reading. I vaguely understand the concept but am struggling with extending a function. This is the question:
Consider the following complex differentiable function:
$$f(z)=frac{z}{e^z-1}, 0<|z|<2pi $$
(a) show that there exists a complex differentiable extension of $f$ onto the set $D={{zinmathbb{C} : |z|<2pi}}$. Denote this extension by $g$.
This is the part I am struggling with primarily as I feel I could do the other parts once I have this but really don't understand how to extend the function to include zero, all I can see is that $f(0)$ must be $1$.
(b) The power series expansion for $f$ around $0inmathbb{C}$ can be written as: $$f(z)=sum^infty_{n=0}frac{B_n}{n!}z^n $$
The coefficients $B_n$ are called Bernoulli numbers. Determine $B_0$ and $B_1$.
(c) Show that $$limsup_{ntoinfty}|B_n|^frac{1}{n}=infty$$
I have never come across Bernoulli numbers before this question and have looked at some more material on them and can see how to do part (b) once part (a) is complete. But I haven't found much material on any functions on how to extend them to include undefined points differentiable. Also I cannot really see how part (c) would follow from (b) or (a). Please help! Any hints or solutions are greatly appreciated.
complex-analysis complex-numbers power-series limsup-and-liminf bernoulli-numbers
asked Jan 23 at 15:02
Dani_5040Dani_5040
324
$endgroup$
$begingroup$
$frac{1}{f(z)}$ is obviously non-zero, analytic and complex-differentiable on $|z| < 2pi$ thus so is its inverse. Indeed the hardest part is (d) : going from "$f$ analytic for $|z|< 2pi$" to "the power series of $f$ at $0$ converges for $|z| < 2pi$". Then $lim sup_n |B_n|^{1/n} = r$ iff $sum_n B_n z^n$ converges for $|z| < ...$
$endgroup$
– reuns
Jan 23 at 20:23
$begingroup$
I still don't see how to extend f(z) to include z=0 as considering 1/f(z) this is still undefined at z=0???
$endgroup$
– Dani_5040
Jan 24 at 17:24
$begingroup$
Of course it is defined by continuity at $z=0$. Write down the power series of $e^z$ you'll see how obvious it is that $frac{e^z-1}{z}$ is analytic at $z=0$ and for $|z| < 2pi$. Then show it doesn't vanish so $frac{z}{e^z-1}$ is analytic too for $|z| < 2pi$.
$endgroup$
– reuns
Jan 25 at 3:14
add a comment |
$begingroup$
I have just recently started revising for my complex analysis module at university and come across and interesting exercise in a textbook while reading. I vaguely understand the concept but am struggling with extending a function. This is the question:
Consider the following complex differentiable function:
$$f(z)=frac{z}{e^z-1}, 0<|z|<2pi $$
(a) show that there exists a complex differentiable extension of $f$ onto the set $D={{zinmathbb{C} : |z|<2pi}}$. Denote this extension by $g$.
This is the part I am struggling with primarily as I feel I could do the other parts once I have this but really don't understand how to extend the function to include zero, all I can see is that $f(0)$ must be $1$.
(b) The power series expansion for $f$ around $0inmathbb{C}$ can be written as: $$f(z)=sum^infty_{n=0}frac{B_n}{n!}z^n $$
The coefficients $B_n$ are called Bernoulli numbers. Determine $B_0$ and $B_1$.
(c) Show that $$limsup_{ntoinfty}|B_n|^frac{1}{n}=infty$$
I have never come across Bernoulli numbers before this question and have looked at some more material on them and can see how to do part (b) once part (a) is complete. But I haven't found much material on any functions on how to extend them to include undefined points differentiable. Also I cannot really see how part (c) would follow from (b) or (a). Please help! Any hints or solutions are greatly appreciated.
complex-analysis complex-numbers power-series limsup-and-liminf bernoulli-numbers
asked Jan 23 at 15:02
Dani_5040Dani_5040
324
$endgroup$
I have just recently started revising for my complex analysis module at university and come across and interesting exercise in a textbook while reading. I vaguely understand the concept but am struggling with extending a function. This is the question:
Consider the following complex differentiable function:
$$f(z)=frac{z}{e^z-1}, 0<|z|<2pi $$
(a) show that there exists a complex differentiable extension of $f$ onto the set $D={{zinmathbb{C} : |z|<2pi}}$. Denote this extension by $g$.
This is the part I am struggling with primarily as I feel I could do the other parts once I have this but really don't understand how to extend the function to include zero, all I can see is that $f(0)$ must be $1$.
(b) The power series expansion for $f$ around $0inmathbb{C}$ can be written as: $$f(z)=sum^infty_{n=0}frac{B_n}{n!}z^n $$
The coefficients $B_n$ are called Bernoulli numbers. Determine $B_0$ and $B_1$.
(c) Show that $$limsup_{ntoinfty}|B_n|^frac{1}{n}=infty$$
I have never come across Bernoulli numbers before this question and have looked at some more material on them and can see how to do part (b) once part (a) is complete. But I haven't found much material on any functions on how to extend them to include undefined points differentiable. Also I cannot really see how part (c) would follow from (b) or (a). Please help! Any hints or solutions are greatly appreciated.
complex-analysis complex-numbers power-series limsup-and-liminf bernoulli-numbers
complex-analysis complex-numbers power-series limsup-and-liminf bernoulli-numbers
asked Jan 23 at 15:02
Dani_5040Dani_5040
324
asked Jan 23 at 15:02
Dani_5040Dani_5040
324
asked Jan 23 at 15:02
Dani_5040Dani_5040
324
asked Jan 23 at 15:02
Dani_5040Dani_5040
324
asked Jan 23 at 15:02
Dani_5040Dani_5040
324
324
$begingroup$
$frac{1}{f(z)}$ is obviously non-zero, analytic and complex-differentiable on $|z| < 2pi$ thus so is its inverse. Indeed the hardest part is (d) : going from "$f$ analytic for $|z|< 2pi$" to "the power series of $f$ at $0$ converges for $|z| < 2pi$". Then $lim sup_n |B_n|^{1/n} = r$ iff $sum_n B_n z^n$ converges for $|z| < ...$
$endgroup$
– reuns
Jan 23 at 20:23
$begingroup$
I still don't see how to extend f(z) to include z=0 as considering 1/f(z) this is still undefined at z=0???
$endgroup$
– Dani_5040
Jan 24 at 17:24
$begingroup$
Of course it is defined by continuity at $z=0$. Write down the power series of $e^z$ you'll see how obvious it is that $frac{e^z-1}{z}$ is analytic at $z=0$ and for $|z| < 2pi$. Then show it doesn't vanish so $frac{z}{e^z-1}$ is analytic too for $|z| < 2pi$.
$endgroup$
– reuns
Jan 25 at 3:14
add a comment |
$begingroup$
$frac{1}{f(z)}$ is obviously non-zero, analytic and complex-differentiable on $|z| < 2pi$ thus so is its inverse. Indeed the hardest part is (d) : going from "$f$ analytic for $|z|< 2pi$" to "the power series of $f$ at $0$ converges for $|z| < 2pi$". Then $lim sup_n |B_n|^{1/n} = r$ iff $sum_n B_n z^n$ converges for $|z| < ...$
$endgroup$
– reuns
Jan 23 at 20:23
$begingroup$
I still don't see how to extend f(z) to include z=0 as considering 1/f(z) this is still undefined at z=0???
$endgroup$
– Dani_5040
Jan 24 at 17:24
$begingroup$
Of course it is defined by continuity at $z=0$. Write down the power series of $e^z$ you'll see how obvious it is that $frac{e^z-1}{z}$ is analytic at $z=0$ and for $|z| < 2pi$. Then show it doesn't vanish so $frac{z}{e^z-1}$ is analytic too for $|z| < 2pi$.
$endgroup$
– reuns
Jan 25 at 3:14
$begingroup$
$frac{1}{f(z)}$ is obviously non-zero, analytic and complex-differentiable on $|z| < 2pi$ thus so is its inverse. Indeed the hardest part is (d) : going from "$f$ analytic for $|z|< 2pi$" to "the power series of $f$ at $0$ converges for $|z| < 2pi$". Then $lim sup_n |B_n|^{1/n} = r$ iff $sum_n B_n z^n$ converges for $|z| < ...$
$endgroup$
– reuns
Jan 23 at 20:23
$begingroup$
$frac{1}{f(z)}$ is obviously non-zero, analytic and complex-differentiable on $|z| < 2pi$ thus so is its inverse. Indeed the hardest part is (d) : going from "$f$ analytic for $|z|< 2pi$" to "the power series of $f$ at $0$ converges for $|z| < 2pi$". Then $lim sup_n |B_n|^{1/n} = r$ iff $sum_n B_n z^n$ converges for $|z| < ...$
$endgroup$
– reuns
Jan 23 at 20:23
$begingroup$
I still don't see how to extend f(z) to include z=0 as considering 1/f(z) this is still undefined at z=0???
$endgroup$
– Dani_5040
Jan 24 at 17:24
$begingroup$
I still don't see how to extend f(z) to include z=0 as considering 1/f(z) this is still undefined at z=0???
$endgroup$
– Dani_5040
Jan 24 at 17:24
$begingroup$
Of course it is defined by continuity at $z=0$. Write down the power series of $e^z$ you'll see how obvious it is that $frac{e^z-1}{z}$ is analytic at $z=0$ and for $|z| < 2pi$. Then show it doesn't vanish so $frac{z}{e^z-1}$ is analytic too for $|z| < 2pi$.
$endgroup$
– reuns
Jan 25 at 3:14
$begingroup$
Of course it is defined by continuity at $z=0$. Write down the power series of $e^z$ you'll see how obvious it is that $frac{e^z-1}{z}$ is analytic at $z=0$ and for $|z| < 2pi$. Then show it doesn't vanish so $frac{z}{e^z-1}$ is analytic too for $|z| < 2pi$.
$endgroup$
– reuns
Jan 25 at 3:14
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3084580%2fcomplex-analysis-extension-of-bernoulli-number-generating-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3084580%2fcomplex-analysis-extension-of-bernoulli-number-generating-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
$frac{1}{f(z)}$ is obviously non-zero, analytic and complex-differentiable on $|z| < 2pi$ thus so is its inverse. Indeed the hardest part is (d) : going from "$f$ analytic for $|z|< 2pi$" to "the power series of $f$ at $0$ converges for $|z| < 2pi$". Then $lim sup_n |B_n|^{1/n} = r$ iff $sum_n B_n z^n$ converges for $|z| < ...$
$endgroup$
– reuns
Jan 23 at 20:23
$begingroup$
I still don't see how to extend f(z) to include z=0 as considering 1/f(z) this is still undefined at z=0???
$endgroup$
– Dani_5040
Jan 24 at 17:24
$begingroup$
Of course it is defined by continuity at $z=0$. Write down the power series of $e^z$ you'll see how obvious it is that $frac{e^z-1}{z}$ is analytic at $z=0$ and for $|z| < 2pi$. Then show it doesn't vanish so $frac{z}{e^z-1}$ is analytic too for $|z| < 2pi$.
$endgroup$
– reuns
Jan 25 at 3:14