How to extract coefficients from Dirichlet series.
$begingroup$
I know, there was a similar question like this:
Can the coefficients of a Dirichlet series be recovered?
But i can see that in case when given function(series) has only positive real numbers as a domain, then Perron's formula is useless here.
Could someone show me a other way to solve this problem?
Thanks in advance.
dirichlet-series
asked Dec 25 '18 at 10:38
mkultramkultra
1038
$endgroup$
add a comment |
$begingroup$
I know, there was a similar question like this:
Can the coefficients of a Dirichlet series be recovered?
But i can see that in case when given function(series) has only positive real numbers as a domain, then Perron's formula is useless here.
Could someone show me a other way to solve this problem?
Thanks in advance.
dirichlet-series
asked Dec 25 '18 at 10:38
mkultramkultra
1038
$endgroup$
$begingroup$
There are two ways to find the coefficients : either look at $lim_{s to infty} F(s)$ to find $f(1)$ then at $lim_{s to infty} n^s (F(s)-sum_{m=1}^n f(m) m^{-s})$ to find $f(n)$, or note that $f(n)n^{-sigma} = lim_{T to infty} frac{1}{2T} int_{-T}^T F(sigma+2ipi t) n^{it} dt$ for $sigma$ larger than the abscissa of absolute convergence.
$endgroup$
– reuns
Dec 25 '18 at 13:58
$begingroup$
But i don't understand the second way. I wrote that given function is determined only for pure real (and positive) numbers.
$endgroup$
– mkultra
Dec 25 '18 at 17:13
add a comment |
$begingroup$
I know, there was a similar question like this:
Can the coefficients of a Dirichlet series be recovered?
But i can see that in case when given function(series) has only positive real numbers as a domain, then Perron's formula is useless here.
Could someone show me a other way to solve this problem?
Thanks in advance.
dirichlet-series
asked Dec 25 '18 at 10:38
mkultramkultra
1038
$endgroup$
I know, there was a similar question like this:
Can the coefficients of a Dirichlet series be recovered?
But i can see that in case when given function(series) has only positive real numbers as a domain, then Perron's formula is useless here.
Could someone show me a other way to solve this problem?
Thanks in advance.
dirichlet-series
dirichlet-series
asked Dec 25 '18 at 10:38
mkultramkultra
1038
asked Dec 25 '18 at 10:38
mkultramkultra
1038
asked Dec 25 '18 at 10:38
mkultramkultra
1038
asked Dec 25 '18 at 10:38
mkultramkultra
1038
asked Dec 25 '18 at 10:38
mkultramkultra
1038
1038
$begingroup$
There are two ways to find the coefficients : either look at $lim_{s to infty} F(s)$ to find $f(1)$ then at $lim_{s to infty} n^s (F(s)-sum_{m=1}^n f(m) m^{-s})$ to find $f(n)$, or note that $f(n)n^{-sigma} = lim_{T to infty} frac{1}{2T} int_{-T}^T F(sigma+2ipi t) n^{it} dt$ for $sigma$ larger than the abscissa of absolute convergence.
$endgroup$
– reuns
Dec 25 '18 at 13:58
$begingroup$
But i don't understand the second way. I wrote that given function is determined only for pure real (and positive) numbers.
$endgroup$
– mkultra
Dec 25 '18 at 17:13
add a comment |
$begingroup$
There are two ways to find the coefficients : either look at $lim_{s to infty} F(s)$ to find $f(1)$ then at $lim_{s to infty} n^s (F(s)-sum_{m=1}^n f(m) m^{-s})$ to find $f(n)$, or note that $f(n)n^{-sigma} = lim_{T to infty} frac{1}{2T} int_{-T}^T F(sigma+2ipi t) n^{it} dt$ for $sigma$ larger than the abscissa of absolute convergence.
$endgroup$
– reuns
Dec 25 '18 at 13:58
$begingroup$
But i don't understand the second way. I wrote that given function is determined only for pure real (and positive) numbers.
$endgroup$
– mkultra
Dec 25 '18 at 17:13
$begingroup$
There are two ways to find the coefficients : either look at $lim_{s to infty} F(s)$ to find $f(1)$ then at $lim_{s to infty} n^s (F(s)-sum_{m=1}^n f(m) m^{-s})$ to find $f(n)$, or note that $f(n)n^{-sigma} = lim_{T to infty} frac{1}{2T} int_{-T}^T F(sigma+2ipi t) n^{it} dt$ for $sigma$ larger than the abscissa of absolute convergence.
$endgroup$
– reuns
Dec 25 '18 at 13:58
$begingroup$
There are two ways to find the coefficients : either look at $lim_{s to infty} F(s)$ to find $f(1)$ then at $lim_{s to infty} n^s (F(s)-sum_{m=1}^n f(m) m^{-s})$ to find $f(n)$, or note that $f(n)n^{-sigma} = lim_{T to infty} frac{1}{2T} int_{-T}^T F(sigma+2ipi t) n^{it} dt$ for $sigma$ larger than the abscissa of absolute convergence.
$endgroup$
– reuns
Dec 25 '18 at 13:58
$begingroup$
But i don't understand the second way. I wrote that given function is determined only for pure real (and positive) numbers.
$endgroup$
– mkultra
Dec 25 '18 at 17:13
$begingroup$
But i don't understand the second way. I wrote that given function is determined only for pure real (and positive) numbers.
$endgroup$
– mkultra
Dec 25 '18 at 17:13
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%2f3051998%2fhow-to-extract-coefficients-from-dirichlet-series%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%2f3051998%2fhow-to-extract-coefficients-from-dirichlet-series%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
$begingroup$
There are two ways to find the coefficients : either look at $lim_{s to infty} F(s)$ to find $f(1)$ then at $lim_{s to infty} n^s (F(s)-sum_{m=1}^n f(m) m^{-s})$ to find $f(n)$, or note that $f(n)n^{-sigma} = lim_{T to infty} frac{1}{2T} int_{-T}^T F(sigma+2ipi t) n^{it} dt$ for $sigma$ larger than the abscissa of absolute convergence.
$endgroup$
– reuns
Dec 25 '18 at 13:58
$begingroup$
But i don't understand the second way. I wrote that given function is determined only for pure real (and positive) numbers.
$endgroup$
– mkultra
Dec 25 '18 at 17:13