Extracting Dirichlet series coefficientsIs the maximum domain to which a Dirichlet series can be continued always a halfplane?Dirichlet series expansion of an analytic functionMultiplicative functions whose Dirichlet series have essential singularitiesWhat is known about the polynomial factorization of power series?Smoothing Dirichlet Series partial sums by means of Pontifex Path Bending FunctionsDirichlet series decomposition of arbitrary functionFormal theory of (some) generating functions in $t$ and $t^-1$?Reaching Hecke eigenvalues from a trace formulaAsymptotic growth of the of Taylor coefficients of the inverse of a functionDirichlet series associated with polynomials
Extracting Dirichlet series coefficients
Is the maximum domain to which a Dirichlet series can be continued always a halfplane?Dirichlet series expansion of an analytic functionMultiplicative functions whose Dirichlet series have essential singularitiesWhat is known about the polynomial factorization of power series?Smoothing Dirichlet Series partial sums by means of Pontifex Path Bending FunctionsDirichlet series decomposition of arbitrary functionFormal theory of (some) generating functions in $t$ and $t^-1$?Reaching Hecke eigenvalues from a trace formulaAsymptotic growth of the of Taylor coefficients of the inverse of a functionDirichlet series associated with polynomials
$begingroup$
Cauchy's integral formula is a powerful method to extract the $n$'th power series coefficient of an analytic function by evaluating a single complex integral. Is there any such analytic method to extract (ordinary) Dirichlet series coefficients? That is, assuming that a given function $f(s)$ admits a Dirichlet series expansion $$sum_n a_n n^-s,$$ is there any known method to compute a desired coefficient $a_n$?
analytic-number-theory power-series dirichlet-series
asked 2 hours ago
MCHMCH
29819
$endgroup$
add a comment |
$begingroup$
Cauchy's integral formula is a powerful method to extract the $n$'th power series coefficient of an analytic function by evaluating a single complex integral. Is there any such analytic method to extract (ordinary) Dirichlet series coefficients? That is, assuming that a given function $f(s)$ admits a Dirichlet series expansion $$sum_n a_n n^-s,$$ is there any known method to compute a desired coefficient $a_n$?
analytic-number-theory power-series dirichlet-series
asked 2 hours ago
MCHMCH
29819
$endgroup$
2
$begingroup$
Otherwise you can find the $a_n$ one by one from the asymptotic as $s to +infty$. I wonder if there is an analog of $c_n = fracF^(n)(0)n! = lim_z to 0 fracsum_k=0^n k choose n (-1)^k-n F(nz)z^n n!$
$endgroup$
– reuns
1 hour ago
add a comment |
$begingroup$
Cauchy's integral formula is a powerful method to extract the $n$'th power series coefficient of an analytic function by evaluating a single complex integral. Is there any such analytic method to extract (ordinary) Dirichlet series coefficients? That is, assuming that a given function $f(s)$ admits a Dirichlet series expansion $$sum_n a_n n^-s,$$ is there any known method to compute a desired coefficient $a_n$?
analytic-number-theory power-series dirichlet-series
asked 2 hours ago
MCHMCH
29819
$endgroup$
Cauchy's integral formula is a powerful method to extract the $n$'th power series coefficient of an analytic function by evaluating a single complex integral. Is there any such analytic method to extract (ordinary) Dirichlet series coefficients? That is, assuming that a given function $f(s)$ admits a Dirichlet series expansion $$sum_n a_n n^-s,$$ is there any known method to compute a desired coefficient $a_n$?
analytic-number-theory power-series dirichlet-series
analytic-number-theory power-series dirichlet-series
asked 2 hours ago
MCHMCH
29819
asked 2 hours ago
MCHMCH
29819
asked 2 hours ago
MCHMCH
29819
asked 2 hours ago
MCHMCH
29819
asked 2 hours ago
MCHMCH
29819
29819
2
$begingroup$
Otherwise you can find the $a_n$ one by one from the asymptotic as $s to +infty$. I wonder if there is an analog of $c_n = fracF^(n)(0)n! = lim_z to 0 fracsum_k=0^n k choose n (-1)^k-n F(nz)z^n n!$
$endgroup$
– reuns
1 hour ago
add a comment |
2
$begingroup$
Otherwise you can find the $a_n$ one by one from the asymptotic as $s to +infty$. I wonder if there is an analog of $c_n = fracF^(n)(0)n! = lim_z to 0 fracsum_k=0^n k choose n (-1)^k-n F(nz)z^n n!$
$endgroup$
– reuns
1 hour ago
2
2
$begingroup$
Otherwise you can find the $a_n$ one by one from the asymptotic as $s to +infty$. I wonder if there is an analog of $c_n = fracF^(n)(0)n! = lim_z to 0 fracsum_k=0^n k choose n (-1)^k-n F(nz)z^n n!$
$endgroup$
– reuns
1 hour ago
$begingroup$
Otherwise you can find the $a_n$ one by one from the asymptotic as $s to +infty$. I wonder if there is an analog of $c_n = fracF^(n)(0)n! = lim_z to 0 fracsum_k=0^n k choose n (-1)^k-n F(nz)z^n n!$
$endgroup$
– reuns
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes. If $f(s)$ has a finite abscissa of absolute convergence $sigma_a$, then $forall sigma > sigma_a$:
$$
lim_Ttoinfty frac12T int_-T^T f(sigma+ it)n^it mathrmdt = fraca_nn^sigma.
$$
IIRC, the proof can be found in Apostol's book on Analytic Number Theory.
answered 2 hours ago
M.G.M.G.
3,01022740
$endgroup$
1
$begingroup$
I think it holds for $sigma > sigma_c$ the absicssa of simple convergence because $f(s) = (s-s_0)int_1^infty (sum_m le x a_m m^-s_0) x^-s+s_0-1dx = O(s)$ so we can replace $f(sigma+it)$ by $f(sigma+i.) ast frac ksqrt2pi e^-t^2 k^2/2 = sum_m a_m m^-sigma-it e^- fracln^2m2 k^2$ to make it absolutely convergent
$endgroup$
– reuns
1 hour ago
add a comment |
$begingroup$
Even for more general Dirichlet series
$$f(z)=sum_0^infty a_n e^-lambda_nz$$
there is the formula
$$a_ne^-lambda_nsigma=lim_Ttoinftyfrac1Tint_t_0^Tf(sigma+it)e^lambda_n itdt,$$
where $t_0$ is arbitrary (real) and $sigma>sigma_u$, the abscissa of uniform convergence.
Ref. S. Mandelbrojt, Series de Dirichlet, Paris, Gauthier-Villars, 1969.
answered 1 hour ago
Alexandre EremenkoAlexandre Eremenko
51.9k6145265
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "504"
;
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%2fmathoverflow.net%2fquestions%2f329980%2fextracting-dirichlet-series-coefficients%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes. If $f(s)$ has a finite abscissa of absolute convergence $sigma_a$, then $forall sigma > sigma_a$:
$$
lim_Ttoinfty frac12T int_-T^T f(sigma+ it)n^it mathrmdt = fraca_nn^sigma.
$$
IIRC, the proof can be found in Apostol's book on Analytic Number Theory.
answered 2 hours ago
M.G.M.G.
3,01022740
$endgroup$
1
$begingroup$
I think it holds for $sigma > sigma_c$ the absicssa of simple convergence because $f(s) = (s-s_0)int_1^infty (sum_m le x a_m m^-s_0) x^-s+s_0-1dx = O(s)$ so we can replace $f(sigma+it)$ by $f(sigma+i.) ast frac ksqrt2pi e^-t^2 k^2/2 = sum_m a_m m^-sigma-it e^- fracln^2m2 k^2$ to make it absolutely convergent
$endgroup$
– reuns
1 hour ago
add a comment |
$begingroup$
Yes. If $f(s)$ has a finite abscissa of absolute convergence $sigma_a$, then $forall sigma > sigma_a$:
$$
lim_Ttoinfty frac12T int_-T^T f(sigma+ it)n^it mathrmdt = fraca_nn^sigma.
$$
IIRC, the proof can be found in Apostol's book on Analytic Number Theory.
answered 2 hours ago
M.G.M.G.
3,01022740
$endgroup$
1
$begingroup$
I think it holds for $sigma > sigma_c$ the absicssa of simple convergence because $f(s) = (s-s_0)int_1^infty (sum_m le x a_m m^-s_0) x^-s+s_0-1dx = O(s)$ so we can replace $f(sigma+it)$ by $f(sigma+i.) ast frac ksqrt2pi e^-t^2 k^2/2 = sum_m a_m m^-sigma-it e^- fracln^2m2 k^2$ to make it absolutely convergent
$endgroup$
– reuns
1 hour ago
add a comment |
$begingroup$
Yes. If $f(s)$ has a finite abscissa of absolute convergence $sigma_a$, then $forall sigma > sigma_a$:
$$
lim_Ttoinfty frac12T int_-T^T f(sigma+ it)n^it mathrmdt = fraca_nn^sigma.
$$
IIRC, the proof can be found in Apostol's book on Analytic Number Theory.
answered 2 hours ago
M.G.M.G.
3,01022740
$endgroup$
Yes. If $f(s)$ has a finite abscissa of absolute convergence $sigma_a$, then $forall sigma > sigma_a$:
$$
lim_Ttoinfty frac12T int_-T^T f(sigma+ it)n^it mathrmdt = fraca_nn^sigma.
$$
IIRC, the proof can be found in Apostol's book on Analytic Number Theory.
answered 2 hours ago
M.G.M.G.
3,01022740
answered 2 hours ago
M.G.M.G.
3,01022740
answered 2 hours ago
M.G.M.G.
3,01022740
answered 2 hours ago
M.G.M.G.
3,01022740
3,01022740
1
$begingroup$
I think it holds for $sigma > sigma_c$ the absicssa of simple convergence because $f(s) = (s-s_0)int_1^infty (sum_m le x a_m m^-s_0) x^-s+s_0-1dx = O(s)$ so we can replace $f(sigma+it)$ by $f(sigma+i.) ast frac ksqrt2pi e^-t^2 k^2/2 = sum_m a_m m^-sigma-it e^- fracln^2m2 k^2$ to make it absolutely convergent
$endgroup$
– reuns
1 hour ago
add a comment |
1
$begingroup$
I think it holds for $sigma > sigma_c$ the absicssa of simple convergence because $f(s) = (s-s_0)int_1^infty (sum_m le x a_m m^-s_0) x^-s+s_0-1dx = O(s)$ so we can replace $f(sigma+it)$ by $f(sigma+i.) ast frac ksqrt2pi e^-t^2 k^2/2 = sum_m a_m m^-sigma-it e^- fracln^2m2 k^2$ to make it absolutely convergent
$endgroup$
– reuns
1 hour ago
1
1
$begingroup$
I think it holds for $sigma > sigma_c$ the absicssa of simple convergence because $f(s) = (s-s_0)int_1^infty (sum_m le x a_m m^-s_0) x^-s+s_0-1dx = O(s)$ so we can replace $f(sigma+it)$ by $f(sigma+i.) ast frac ksqrt2pi e^-t^2 k^2/2 = sum_m a_m m^-sigma-it e^- fracln^2m2 k^2$ to make it absolutely convergent
$endgroup$
– reuns
1 hour ago
$begingroup$
I think it holds for $sigma > sigma_c$ the absicssa of simple convergence because $f(s) = (s-s_0)int_1^infty (sum_m le x a_m m^-s_0) x^-s+s_0-1dx = O(s)$ so we can replace $f(sigma+it)$ by $f(sigma+i.) ast frac ksqrt2pi e^-t^2 k^2/2 = sum_m a_m m^-sigma-it e^- fracln^2m2 k^2$ to make it absolutely convergent
$endgroup$
– reuns
1 hour ago
add a comment |
$begingroup$
Even for more general Dirichlet series
$$f(z)=sum_0^infty a_n e^-lambda_nz$$
there is the formula
$$a_ne^-lambda_nsigma=lim_Ttoinftyfrac1Tint_t_0^Tf(sigma+it)e^lambda_n itdt,$$
where $t_0$ is arbitrary (real) and $sigma>sigma_u$, the abscissa of uniform convergence.
Ref. S. Mandelbrojt, Series de Dirichlet, Paris, Gauthier-Villars, 1969.
answered 1 hour ago
Alexandre EremenkoAlexandre Eremenko
51.9k6145265
$endgroup$
add a comment |
$begingroup$
Even for more general Dirichlet series
$$f(z)=sum_0^infty a_n e^-lambda_nz$$
there is the formula
$$a_ne^-lambda_nsigma=lim_Ttoinftyfrac1Tint_t_0^Tf(sigma+it)e^lambda_n itdt,$$
where $t_0$ is arbitrary (real) and $sigma>sigma_u$, the abscissa of uniform convergence.
Ref. S. Mandelbrojt, Series de Dirichlet, Paris, Gauthier-Villars, 1969.
answered 1 hour ago
Alexandre EremenkoAlexandre Eremenko
51.9k6145265
$endgroup$
add a comment |
$begingroup$
Even for more general Dirichlet series
$$f(z)=sum_0^infty a_n e^-lambda_nz$$
there is the formula
$$a_ne^-lambda_nsigma=lim_Ttoinftyfrac1Tint_t_0^Tf(sigma+it)e^lambda_n itdt,$$
where $t_0$ is arbitrary (real) and $sigma>sigma_u$, the abscissa of uniform convergence.
Ref. S. Mandelbrojt, Series de Dirichlet, Paris, Gauthier-Villars, 1969.
answered 1 hour ago
Alexandre EremenkoAlexandre Eremenko
51.9k6145265
$endgroup$
Even for more general Dirichlet series
$$f(z)=sum_0^infty a_n e^-lambda_nz$$
there is the formula
$$a_ne^-lambda_nsigma=lim_Ttoinftyfrac1Tint_t_0^Tf(sigma+it)e^lambda_n itdt,$$
where $t_0$ is arbitrary (real) and $sigma>sigma_u$, the abscissa of uniform convergence.
Ref. S. Mandelbrojt, Series de Dirichlet, Paris, Gauthier-Villars, 1969.
answered 1 hour ago
Alexandre EremenkoAlexandre Eremenko
51.9k6145265
answered 1 hour ago
Alexandre EremenkoAlexandre Eremenko
51.9k6145265
answered 1 hour ago
Alexandre EremenkoAlexandre Eremenko
51.9k6145265
answered 1 hour ago
Alexandre EremenkoAlexandre Eremenko
51.9k6145265
51.9k6145265
add a comment |
add a comment |
Thanks for contributing an answer to MathOverflow!
- 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%2fmathoverflow.net%2fquestions%2f329980%2fextracting-dirichlet-series-coefficients%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
2
$begingroup$
Otherwise you can find the $a_n$ one by one from the asymptotic as $s to +infty$. I wonder if there is an analog of $c_n = fracF^(n)(0)n! = lim_z to 0 fracsum_k=0^n k choose n (-1)^k-n F(nz)z^n n!$
$endgroup$
– reuns
1 hour ago