Extracting Dirichlet series coefficientsDirichlet series expansion of an analytic functionIs the maximum...
Extracting Dirichlet series coefficients
Dirichlet series expansion of an analytic functionIs 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 days ago
MCHMCH
32019
$endgroup$
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 days ago
MCHMCH
32019
$endgroup$
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
2 days ago
$begingroup$
Possibly relevant: mathoverflow.net/questions/30975/…
$endgroup$
– M.G.
yesterday
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 days ago
MCHMCH
32019
$endgroup$
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 days ago
MCHMCH
32019
asked 2 days ago
MCHMCH
32019
asked 2 days ago
MCHMCH
32019
asked 2 days ago
MCHMCH
32019
asked 2 days ago
MCHMCH
32019
32019
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
2 days ago
$begingroup$
Possibly relevant: mathoverflow.net/questions/30975/…
$endgroup$
– M.G.
yesterday
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 $c_n = frac{F^{(n)}(0)}{n!} = lim_{z to 0} frac{sum_{k=0}^n {k choose n} (-1)^{k-n} F(nz)}{z^n n!}$
$endgroup$
– reuns
2 days ago
$begingroup$
Possibly relevant: mathoverflow.net/questions/30975/…
$endgroup$
– M.G.
yesterday
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
2 days 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
2 days ago
$begingroup$
Possibly relevant: mathoverflow.net/questions/30975/…
$endgroup$
– M.G.
yesterday
$begingroup$
Possibly relevant: mathoverflow.net/questions/30975/…
$endgroup$
– M.G.
yesterday
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
This formula determines both $lambda_n$ and $a_n$: the RHS=0 when we integrate against $e^{ilambda t}$ with $lambdaneq lambda_n$.
The class of functions which can be represented by such a series is called (analytic) almost periodic functions (on a vertical line ${s=sigma+it:tin R}$). The "number-theoretic case" corresponds to $lambda_n=n$.
Ref. S. Mandelbrojt, Series de Dirichlet, Paris, Gauthier-Villars, 1969.
Dirichlet series with complex $lambda_n$ have been also studied (by A. F. Leont'ev and his school).
answered 2 days ago
Alexandre EremenkoAlexandre Eremenko
52k6146265
$endgroup$
$begingroup$
If we don't know the $lambda_n$ coefficients, is there a way to obtain them as well, or the problem becomes way too undetermined?
$endgroup$
– M.G.
yesterday
$begingroup$
@M.G. Number-theoretic case is $lambda_n=log n$. In general, they are determined by the same formula I wrote: apply it with arbitrary $lambda$, the RHS will be $0$ for all $lambda$ except countably many $lambda_n$. For the discussion of the class of functions $f$ which admit such a representation, see the book of Mandelbrojt that I mentioned.
$endgroup$
– Alexandre Eremenko
yesterday
add a comment |
$begingroup$
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.
answered 2 days ago
M.G.M.G.
3,04022740
$endgroup$
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
2 days ago
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
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$begingroup$
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.
This formula determines both $lambda_n$ and $a_n$: the RHS=0 when we integrate against $e^{ilambda t}$ with $lambdaneq lambda_n$.
The class of functions which can be represented by such a series is called (analytic) almost periodic functions (on a vertical line ${s=sigma+it:tin R}$). The "number-theoretic case" corresponds to $lambda_n=n$.
Ref. S. Mandelbrojt, Series de Dirichlet, Paris, Gauthier-Villars, 1969.
Dirichlet series with complex $lambda_n$ have been also studied (by A. F. Leont'ev and his school).
answered 2 days ago
Alexandre EremenkoAlexandre Eremenko
52k6146265
$endgroup$
$begingroup$
If we don't know the $lambda_n$ coefficients, is there a way to obtain them as well, or the problem becomes way too undetermined?
$endgroup$
– M.G.
yesterday
$begingroup$
@M.G. Number-theoretic case is $lambda_n=log n$. In general, they are determined by the same formula I wrote: apply it with arbitrary $lambda$, the RHS will be $0$ for all $lambda$ except countably many $lambda_n$. For the discussion of the class of functions $f$ which admit such a representation, see the book of Mandelbrojt that I mentioned.
$endgroup$
– Alexandre Eremenko
yesterday
add a comment |
$begingroup$
Even for more general Dirichlet series
$$f(z)=sum_{0}^infty a_n e^{-lambda_nz}$$
there is the formula
$$a_ne^{-lambda_nsigma}=lim_{Ttoinfty}frac{1}{T}int_{t_0}^Tf(sigma+it)e^{lambda_n it}dt,$$
where $t_0$ is arbitrary (real) and $sigma>sigma_u$, the abscissa of uniform convergence.
This formula determines both $lambda_n$ and $a_n$: the RHS=0 when we integrate against $e^{ilambda t}$ with $lambdaneq lambda_n$.
The class of functions which can be represented by such a series is called (analytic) almost periodic functions (on a vertical line ${s=sigma+it:tin R}$). The "number-theoretic case" corresponds to $lambda_n=n$.
Ref. S. Mandelbrojt, Series de Dirichlet, Paris, Gauthier-Villars, 1969.
Dirichlet series with complex $lambda_n$ have been also studied (by A. F. Leont'ev and his school).
answered 2 days ago
Alexandre EremenkoAlexandre Eremenko
52k6146265
$endgroup$
$begingroup$
If we don't know the $lambda_n$ coefficients, is there a way to obtain them as well, or the problem becomes way too undetermined?
$endgroup$
– M.G.
yesterday
$begingroup$
@M.G. Number-theoretic case is $lambda_n=log n$. In general, they are determined by the same formula I wrote: apply it with arbitrary $lambda$, the RHS will be $0$ for all $lambda$ except countably many $lambda_n$. For the discussion of the class of functions $f$ which admit such a representation, see the book of Mandelbrojt that I mentioned.
$endgroup$
– Alexandre Eremenko
yesterday
add a comment |
$begingroup$
Even for more general Dirichlet series
$$f(z)=sum_{0}^infty a_n e^{-lambda_nz}$$
there is the formula
$$a_ne^{-lambda_nsigma}=lim_{Ttoinfty}frac{1}{T}int_{t_0}^Tf(sigma+it)e^{lambda_n it}dt,$$
where $t_0$ is arbitrary (real) and $sigma>sigma_u$, the abscissa of uniform convergence.
This formula determines both $lambda_n$ and $a_n$: the RHS=0 when we integrate against $e^{ilambda t}$ with $lambdaneq lambda_n$.
The class of functions which can be represented by such a series is called (analytic) almost periodic functions (on a vertical line ${s=sigma+it:tin R}$). The "number-theoretic case" corresponds to $lambda_n=n$.
Ref. S. Mandelbrojt, Series de Dirichlet, Paris, Gauthier-Villars, 1969.
Dirichlet series with complex $lambda_n$ have been also studied (by A. F. Leont'ev and his school).
answered 2 days ago
Alexandre EremenkoAlexandre Eremenko
52k6146265
$endgroup$
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.
This formula determines both $lambda_n$ and $a_n$: the RHS=0 when we integrate against $e^{ilambda t}$ with $lambdaneq lambda_n$.
The class of functions which can be represented by such a series is called (analytic) almost periodic functions (on a vertical line ${s=sigma+it:tin R}$). The "number-theoretic case" corresponds to $lambda_n=n$.
Ref. S. Mandelbrojt, Series de Dirichlet, Paris, Gauthier-Villars, 1969.
Dirichlet series with complex $lambda_n$ have been also studied (by A. F. Leont'ev and his school).
answered 2 days ago
Alexandre EremenkoAlexandre Eremenko
52k6146265
edited yesterday
answered 2 days ago
Alexandre EremenkoAlexandre Eremenko
52k6146265
answered 2 days ago
Alexandre EremenkoAlexandre Eremenko
52k6146265
answered 2 days ago
Alexandre EremenkoAlexandre Eremenko
52k6146265
52k6146265
$begingroup$
If we don't know the $lambda_n$ coefficients, is there a way to obtain them as well, or the problem becomes way too undetermined?
$endgroup$
– M.G.
yesterday
$begingroup$
@M.G. Number-theoretic case is $lambda_n=log n$. In general, they are determined by the same formula I wrote: apply it with arbitrary $lambda$, the RHS will be $0$ for all $lambda$ except countably many $lambda_n$. For the discussion of the class of functions $f$ which admit such a representation, see the book of Mandelbrojt that I mentioned.
$endgroup$
– Alexandre Eremenko
yesterday
add a comment |
$begingroup$
If we don't know the $lambda_n$ coefficients, is there a way to obtain them as well, or the problem becomes way too undetermined?
$endgroup$
– M.G.
yesterday
$begingroup$
@M.G. Number-theoretic case is $lambda_n=log n$. In general, they are determined by the same formula I wrote: apply it with arbitrary $lambda$, the RHS will be $0$ for all $lambda$ except countably many $lambda_n$. For the discussion of the class of functions $f$ which admit such a representation, see the book of Mandelbrojt that I mentioned.
$endgroup$
– Alexandre Eremenko
yesterday
$begingroup$
If we don't know the $lambda_n$ coefficients, is there a way to obtain them as well, or the problem becomes way too undetermined?
$endgroup$
– M.G.
yesterday
$begingroup$
If we don't know the $lambda_n$ coefficients, is there a way to obtain them as well, or the problem becomes way too undetermined?
$endgroup$
– M.G.
yesterday
$begingroup$
@M.G. Number-theoretic case is $lambda_n=log n$. In general, they are determined by the same formula I wrote: apply it with arbitrary $lambda$, the RHS will be $0$ for all $lambda$ except countably many $lambda_n$. For the discussion of the class of functions $f$ which admit such a representation, see the book of Mandelbrojt that I mentioned.
$endgroup$
– Alexandre Eremenko
yesterday
$begingroup$
@M.G. Number-theoretic case is $lambda_n=log n$. In general, they are determined by the same formula I wrote: apply it with arbitrary $lambda$, the RHS will be $0$ for all $lambda$ except countably many $lambda_n$. For the discussion of the class of functions $f$ which admit such a representation, see the book of Mandelbrojt that I mentioned.
$endgroup$
– Alexandre Eremenko
yesterday
add a comment |
$begingroup$
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.
answered 2 days ago
M.G.M.G.
3,04022740
$endgroup$
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
2 days ago
add a comment |
$begingroup$
Yes. If $f(s)$ has a finite abscissa of absolute convergence $sigma_a$, then $forall sigma > sigma_a$:
$$
lim_{Ttoinfty} frac{1}{2T} int_{-T}^{T} f(sigma+ it)n^{it} mathrm{d}t = frac{a_n}{n^{sigma}}.
$$
IIRC, the proof can be found in Apostol's book on Analytic Number Theory.
answered 2 days ago
M.G.M.G.
3,04022740
$endgroup$
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
2 days ago
add a comment |
$begingroup$
Yes. If $f(s)$ has a finite abscissa of absolute convergence $sigma_a$, then $forall sigma > sigma_a$:
$$
lim_{Ttoinfty} frac{1}{2T} int_{-T}^{T} f(sigma+ it)n^{it} mathrm{d}t = frac{a_n}{n^{sigma}}.
$$
IIRC, the proof can be found in Apostol's book on Analytic Number Theory.
answered 2 days ago
M.G.M.G.
3,04022740
$endgroup$
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.
answered 2 days ago
M.G.M.G.
3,04022740
answered 2 days ago
M.G.M.G.
3,04022740
answered 2 days ago
M.G.M.G.
3,04022740
answered 2 days ago
M.G.M.G.
3,04022740
3,04022740
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
2 days ago
add a comment |
1
$begingroup$
I think it holds for $sigma > sigma_c$ the absicssa of simple convergence because $f(s) = (s-s_0)int_1^infty (sum_{m le x} a_m m^{-s_0}) x^{-s+s_0-1}dx = O(s)$ 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
2 days 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
2 days 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
2 days ago
add a comment |
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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
2 days ago
$begingroup$
Possibly relevant: mathoverflow.net/questions/30975/…
$endgroup$
– M.G.
yesterday