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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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 $c_n = frac{F^{(n)}(0)}{n!} = lim_{z to 0} frac{sum_{k=0}^n {k choose n} (-1)^{k-n} F(nz)}{z^n n!}$
$endgroup$
– reuns
1 hour ago
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 $c_n = frac{F^{(n)}(0)}{n!} = lim_{z to 0} frac{sum_{k=0}^n {k choose n} (-1)^{k-n} F(nz)}{z^n n!}$
$endgroup$
– reuns
1 hour ago
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
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes. If $f(s)$ has a finite abscissa of absolute convergence $sigma_a$, then $forall sigma > sigma_a$:
$$
lim_{Ttoinfty} frac{1}{2T} int_{-T}^{T} f(sigma+ it)n^{it} mathrm{d}t = frac{a_n}{n^{sigma}}.
$$
IIRC, the proof can be found in Apostol's book on Analytic Number Theory.
answered 2 hours ago
M.G.M.G.
3,00022740
$endgroup$
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
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.
Ref. S. Mandelbrojt, Series de Dirichlet, Paris, Gauthier-Villars, 1969.
answered 1 hour ago
Alexandre EremenkoAlexandre Eremenko
51.9k6145265
$endgroup$
add a comment |
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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 $forall sigma > sigma_a$:
$$
lim_{Ttoinfty} frac{1}{2T} int_{-T}^{T} f(sigma+ it)n^{it} mathrm{d}t = frac{a_n}{n^{sigma}}.
$$
IIRC, the proof can be found in Apostol's book on Analytic Number Theory.
answered 2 hours ago
M.G.M.G.
3,00022740
$endgroup$
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
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 hours ago
M.G.M.G.
3,00022740
$endgroup$
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
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 hours ago
M.G.M.G.
3,00022740
$endgroup$
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 hours ago
M.G.M.G.
3,00022740
answered 2 hours ago
M.G.M.G.
3,00022740
answered 2 hours ago
M.G.M.G.
3,00022740
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
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
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
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.
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
$$a_ne^{-lambda_nsigma}=lim_{Ttoinfty}frac{1}{T}int_{t_0}^Tf(sigma+it)e^{lambda_n it}dt,$$
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.
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
$$a_ne^{-lambda_nsigma}=lim_{Ttoinfty}frac{1}{T}int_{t_0}^Tf(sigma+it)e^{lambda_n it}dt,$$
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.
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
$$a_ne^{-lambda_nsigma}=lim_{Ttoinfty}frac{1}{T}int_{t_0}^Tf(sigma+it)e^{lambda_n it}dt,$$
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.
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 |
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$begingroup$
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