Extracting Dirichlet series coefficientsIs the maximum domain to which a Dirichlet series can be continued...



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













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$begingroup$


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







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$endgroup$








  • 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 $c_n = frac{F^{(n)}(0)}{n!} = lim_{z to 0} frac{sum_{k=0}^n {k choose n} (-1)^{k-n} F(nz)}{z^n n!}$
    $endgroup$
    – reuns
    1 hour ago


















1












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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 $c_n = frac{F^{(n)}(0)}{n!} = lim_{z to 0} frac{sum_{k=0}^n {k choose n} (-1)^{k-n} F(nz)}{z^n n!}$
    $endgroup$
    – reuns
    1 hour ago
















1












1








1


1



$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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 $c_n = frac{F^{(n)}(0)}{n!} = lim_{z to 0} frac{sum_{k=0}^n {k choose n} (-1)^{k-n} F(nz)}{z^n n!}$
    $endgroup$
    – reuns
    1 hour ago
















  • 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 $c_n = frac{F^{(n)}(0)}{n!} = lim_{z to 0} frac{sum_{k=0}^n {k choose n} (-1)^{k-n} F(nz)}{z^n n!}$
    $endgroup$
    – 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 $c_n = frac{F^{(n)}(0)}{n!} = lim_{z to 0} frac{sum_{k=0}^n {k choose n} (-1)^{k-n} F(nz)}{z^n n!}$
$endgroup$
– 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 $c_n = frac{F^{(n)}(0)}{n!} = lim_{z to 0} frac{sum_{k=0}^n {k choose n} (-1)^{k-n} F(nz)}{z^n n!}$
$endgroup$
– reuns
1 hour ago












2 Answers
2






active

oldest

votes


















3












$begingroup$

Yes. If $f(s)$ has a finite abscissa of absolute convergence $sigma_a$, then $forall sigma > sigma_a$:
$$
lim_{Ttoinfty} frac{1}{2T} int_{-T}^{T} f(sigma+ it)n^{it} mathrm{d}t = frac{a_n}{n^{sigma}}.
$$
IIRC, the proof can be found in Apostol's book on Analytic Number Theory.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    I think it holds for $sigma > sigma_c$ the absicssa of simple convergence because $f(s) = (s-s_0)int_1^infty (sum_{m le x} a_m m^{-s_0}) x^{-s+s_0-1}dx = O(s)$ so we can replace $f(sigma+it)$ by $f(sigma+i.) ast frac{ k}{sqrt{2pi}} e^{-t^2 k^2/2} = sum_m a_m m^{-sigma-it} e^{- frac{ln^2m}{2 k^2}}$ to make it absolutely convergent
    $endgroup$
    – reuns
    1 hour ago



















2












$begingroup$

Even for more general Dirichlet series
$$f(z)=sum_{0}^infty a_n e^{-lambda_nz}$$
there is the formula
$$a_ne^{-lambda_nsigma}=lim_{Ttoinfty}frac{1}{T}int_{t_0}^Tf(sigma+it)e^{lambda_n it}dt,$$
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.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Yes. If $f(s)$ has a finite abscissa of absolute convergence $sigma_a$, then $forall sigma > sigma_a$:
    $$
    lim_{Ttoinfty} frac{1}{2T} int_{-T}^{T} f(sigma+ it)n^{it} mathrm{d}t = frac{a_n}{n^{sigma}}.
    $$
    IIRC, the proof can be found in Apostol's book on Analytic Number Theory.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      I think it holds for $sigma > sigma_c$ the absicssa of simple convergence because $f(s) = (s-s_0)int_1^infty (sum_{m le x} a_m m^{-s_0}) x^{-s+s_0-1}dx = O(s)$ so we can replace $f(sigma+it)$ by $f(sigma+i.) ast frac{ k}{sqrt{2pi}} e^{-t^2 k^2/2} = sum_m a_m m^{-sigma-it} e^{- frac{ln^2m}{2 k^2}}$ to make it absolutely convergent
      $endgroup$
      – reuns
      1 hour ago
















    3












    $begingroup$

    Yes. If $f(s)$ has a finite abscissa of absolute convergence $sigma_a$, then $forall sigma > sigma_a$:
    $$
    lim_{Ttoinfty} frac{1}{2T} int_{-T}^{T} f(sigma+ it)n^{it} mathrm{d}t = frac{a_n}{n^{sigma}}.
    $$
    IIRC, the proof can be found in Apostol's book on Analytic Number Theory.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      I think it holds for $sigma > sigma_c$ the absicssa of simple convergence because $f(s) = (s-s_0)int_1^infty (sum_{m le x} a_m m^{-s_0}) x^{-s+s_0-1}dx = O(s)$ so we can replace $f(sigma+it)$ by $f(sigma+i.) ast frac{ k}{sqrt{2pi}} e^{-t^2 k^2/2} = sum_m a_m m^{-sigma-it} e^{- frac{ln^2m}{2 k^2}}$ to make it absolutely convergent
      $endgroup$
      – reuns
      1 hour ago














    3












    3








    3





    $begingroup$

    Yes. If $f(s)$ has a finite abscissa of absolute convergence $sigma_a$, then $forall sigma > sigma_a$:
    $$
    lim_{Ttoinfty} frac{1}{2T} int_{-T}^{T} f(sigma+ it)n^{it} mathrm{d}t = frac{a_n}{n^{sigma}}.
    $$
    IIRC, the proof can be found in Apostol's book on Analytic Number Theory.






    share|cite|improve this answer









    $endgroup$



    Yes. If $f(s)$ has a finite abscissa of absolute convergence $sigma_a$, then $forall sigma > sigma_a$:
    $$
    lim_{Ttoinfty} frac{1}{2T} int_{-T}^{T} f(sigma+ it)n^{it} mathrm{d}t = frac{a_n}{n^{sigma}}.
    $$
    IIRC, the proof can be found in Apostol's book on Analytic Number Theory.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 2 hours ago









    M.G.M.G.

    3,00022740




    3,00022740








    • 1




      $begingroup$
      I think it holds for $sigma > sigma_c$ the absicssa of simple convergence because $f(s) = (s-s_0)int_1^infty (sum_{m le x} a_m m^{-s_0}) x^{-s+s_0-1}dx = O(s)$ so we can replace $f(sigma+it)$ by $f(sigma+i.) ast frac{ k}{sqrt{2pi}} e^{-t^2 k^2/2} = sum_m a_m m^{-sigma-it} e^{- frac{ln^2m}{2 k^2}}$ to make it absolutely convergent
      $endgroup$
      – reuns
      1 hour ago














    • 1




      $begingroup$
      I think it holds for $sigma > sigma_c$ the absicssa of simple convergence because $f(s) = (s-s_0)int_1^infty (sum_{m le x} a_m m^{-s_0}) x^{-s+s_0-1}dx = O(s)$ so we can replace $f(sigma+it)$ by $f(sigma+i.) ast frac{ k}{sqrt{2pi}} e^{-t^2 k^2/2} = sum_m a_m m^{-sigma-it} e^{- frac{ln^2m}{2 k^2}}$ 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 $f(s) = (s-s_0)int_1^infty (sum_{m le x} a_m m^{-s_0}) x^{-s+s_0-1}dx = O(s)$ so we can replace $f(sigma+it)$ by $f(sigma+i.) ast frac{ k}{sqrt{2pi}} e^{-t^2 k^2/2} = sum_m a_m m^{-sigma-it} e^{- frac{ln^2m}{2 k^2}}$ 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 $f(s) = (s-s_0)int_1^infty (sum_{m le x} a_m m^{-s_0}) x^{-s+s_0-1}dx = O(s)$ so we can replace $f(sigma+it)$ by $f(sigma+i.) ast frac{ k}{sqrt{2pi}} e^{-t^2 k^2/2} = sum_m a_m m^{-sigma-it} e^{- frac{ln^2m}{2 k^2}}$ to make it absolutely convergent
    $endgroup$
    – reuns
    1 hour ago











    2












    $begingroup$

    Even for more general Dirichlet series
    $$f(z)=sum_{0}^infty a_n e^{-lambda_nz}$$
    there is the formula
    $$a_ne^{-lambda_nsigma}=lim_{Ttoinfty}frac{1}{T}int_{t_0}^Tf(sigma+it)e^{lambda_n it}dt,$$
    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.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Even for more general Dirichlet series
      $$f(z)=sum_{0}^infty a_n e^{-lambda_nz}$$
      there is the formula
      $$a_ne^{-lambda_nsigma}=lim_{Ttoinfty}frac{1}{T}int_{t_0}^Tf(sigma+it)e^{lambda_n it}dt,$$
      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.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Even for more general Dirichlet series
        $$f(z)=sum_{0}^infty a_n e^{-lambda_nz}$$
        there is the formula
        $$a_ne^{-lambda_nsigma}=lim_{Ttoinfty}frac{1}{T}int_{t_0}^Tf(sigma+it)e^{lambda_n it}dt,$$
        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.






        share|cite|improve this answer









        $endgroup$



        Even for more general Dirichlet series
        $$f(z)=sum_{0}^infty a_n e^{-lambda_nz}$$
        there is the formula
        $$a_ne^{-lambda_nsigma}=lim_{Ttoinfty}frac{1}{T}int_{t_0}^Tf(sigma+it)e^{lambda_n it}dt,$$
        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 1 hour ago









        Alexandre EremenkoAlexandre Eremenko

        51.9k6145265




        51.9k6145265






























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