Taylor Series of Gamma Function












1












$begingroup$


I have always wondered whether the Taylor Series of Gamma Function exists or not. I tried to find it, but in vain. I googled for it, but couldn't find it. Has anyone ever found its Taylor Series?






calculus soft-question taylor-expansion gamma-function







share|cite|improve this question









$endgroup$












  • $begingroup$
    It certainly has Taylor series (say around $1$), but it does not have a Maclaurin series. See How to obtain the Laurent Expansion of Gamma Function Around $z=0$ for its Laurent series.
    $endgroup$
    – Weijun Zhou
    Jan 24 '18 at 1:16












  • $begingroup$
    But Laurent series has negative exponents.
    $endgroup$
    – Shashank
    Jan 24 '18 at 1:20












  • $begingroup$
    If $f(z)$ has a Laurent series with negative exponent terms around $z=a$, it cannot have a Taylor series around $z=a$.
    $endgroup$
    – Weijun Zhou
    Jan 24 '18 at 1:21












  • $begingroup$
    Really!!? Ok, thank you soo much. What about its Taylor Series around 1? What are the coefficients?
    $endgroup$
    – Shashank
    Jan 24 '18 at 1:22












  • $begingroup$
    Not a simple job either, see section 6.1 of csie.ntu.edu.tw/~b89089/link/gammaFunction.pdf.
    $endgroup$
    – Weijun Zhou
    Jan 24 '18 at 1:45


















1












$begingroup$


I have always wondered whether the Taylor Series of Gamma Function exists or not. I tried to find it, but in vain. I googled for it, but couldn't find it. Has anyone ever found its Taylor Series?






calculus soft-question taylor-expansion gamma-function







share|cite|improve this question









$endgroup$












  • $begingroup$
    It certainly has Taylor series (say around $1$), but it does not have a Maclaurin series. See How to obtain the Laurent Expansion of Gamma Function Around $z=0$ for its Laurent series.
    $endgroup$
    – Weijun Zhou
    Jan 24 '18 at 1:16












  • $begingroup$
    But Laurent series has negative exponents.
    $endgroup$
    – Shashank
    Jan 24 '18 at 1:20












  • $begingroup$
    If $f(z)$ has a Laurent series with negative exponent terms around $z=a$, it cannot have a Taylor series around $z=a$.
    $endgroup$
    – Weijun Zhou
    Jan 24 '18 at 1:21












  • $begingroup$
    Really!!? Ok, thank you soo much. What about its Taylor Series around 1? What are the coefficients?
    $endgroup$
    – Shashank
    Jan 24 '18 at 1:22












  • $begingroup$
    Not a simple job either, see section 6.1 of csie.ntu.edu.tw/~b89089/link/gammaFunction.pdf.
    $endgroup$
    – Weijun Zhou
    Jan 24 '18 at 1:45
















1












1








1





$begingroup$


I have always wondered whether the Taylor Series of Gamma Function exists or not. I tried to find it, but in vain. I googled for it, but couldn't find it. Has anyone ever found its Taylor Series?






calculus soft-question taylor-expansion gamma-function







share|cite|improve this question









$endgroup$




I have always wondered whether the Taylor Series of Gamma Function exists or not. I tried to find it, but in vain. I googled for it, but couldn't find it. Has anyone ever found its Taylor Series?






calculus soft-question taylor-expansion gamma-function




calculus soft-question taylor-expansion gamma-function






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 24 '18 at 1:10









ShashankShashank

658




658












  • $begingroup$
    It certainly has Taylor series (say around $1$), but it does not have a Maclaurin series. See How to obtain the Laurent Expansion of Gamma Function Around $z=0$ for its Laurent series.
    $endgroup$
    – Weijun Zhou
    Jan 24 '18 at 1:16












  • $begingroup$
    But Laurent series has negative exponents.
    $endgroup$
    – Shashank
    Jan 24 '18 at 1:20












  • $begingroup$
    If $f(z)$ has a Laurent series with negative exponent terms around $z=a$, it cannot have a Taylor series around $z=a$.
    $endgroup$
    – Weijun Zhou
    Jan 24 '18 at 1:21












  • $begingroup$
    Really!!? Ok, thank you soo much. What about its Taylor Series around 1? What are the coefficients?
    $endgroup$
    – Shashank
    Jan 24 '18 at 1:22












  • $begingroup$
    Not a simple job either, see section 6.1 of csie.ntu.edu.tw/~b89089/link/gammaFunction.pdf.
    $endgroup$
    – Weijun Zhou
    Jan 24 '18 at 1:45




















  • $begingroup$
    It certainly has Taylor series (say around $1$), but it does not have a Maclaurin series. See How to obtain the Laurent Expansion of Gamma Function Around $z=0$ for its Laurent series.
    $endgroup$
    – Weijun Zhou
    Jan 24 '18 at 1:16












  • $begingroup$
    But Laurent series has negative exponents.
    $endgroup$
    – Shashank
    Jan 24 '18 at 1:20












  • $begingroup$
    If $f(z)$ has a Laurent series with negative exponent terms around $z=a$, it cannot have a Taylor series around $z=a$.
    $endgroup$
    – Weijun Zhou
    Jan 24 '18 at 1:21












  • $begingroup$
    Really!!? Ok, thank you soo much. What about its Taylor Series around 1? What are the coefficients?
    $endgroup$
    – Shashank
    Jan 24 '18 at 1:22












  • $begingroup$
    Not a simple job either, see section 6.1 of csie.ntu.edu.tw/~b89089/link/gammaFunction.pdf.
    $endgroup$
    – Weijun Zhou
    Jan 24 '18 at 1:45


















$begingroup$
It certainly has Taylor series (say around $1$), but it does not have a Maclaurin series. See How to obtain the Laurent Expansion of Gamma Function Around $z=0$ for its Laurent series.
$endgroup$
– Weijun Zhou
Jan 24 '18 at 1:16






$begingroup$
It certainly has Taylor series (say around $1$), but it does not have a Maclaurin series. See How to obtain the Laurent Expansion of Gamma Function Around $z=0$ for its Laurent series.
$endgroup$
– Weijun Zhou
Jan 24 '18 at 1:16














$begingroup$
But Laurent series has negative exponents.
$endgroup$
– Shashank
Jan 24 '18 at 1:20






$begingroup$
But Laurent series has negative exponents.
$endgroup$
– Shashank
Jan 24 '18 at 1:20














$begingroup$
If $f(z)$ has a Laurent series with negative exponent terms around $z=a$, it cannot have a Taylor series around $z=a$.
$endgroup$
– Weijun Zhou
Jan 24 '18 at 1:21






$begingroup$
If $f(z)$ has a Laurent series with negative exponent terms around $z=a$, it cannot have a Taylor series around $z=a$.
$endgroup$
– Weijun Zhou
Jan 24 '18 at 1:21














$begingroup$
Really!!? Ok, thank you soo much. What about its Taylor Series around 1? What are the coefficients?
$endgroup$
– Shashank
Jan 24 '18 at 1:22






$begingroup$
Really!!? Ok, thank you soo much. What about its Taylor Series around 1? What are the coefficients?
$endgroup$
– Shashank
Jan 24 '18 at 1:22














$begingroup$
Not a simple job either, see section 6.1 of csie.ntu.edu.tw/~b89089/link/gammaFunction.pdf.
$endgroup$
– Weijun Zhou
Jan 24 '18 at 1:45






$begingroup$
Not a simple job either, see section 6.1 of csie.ntu.edu.tw/~b89089/link/gammaFunction.pdf.
$endgroup$
– Weijun Zhou
Jan 24 '18 at 1:45












1 Answer
1






active

oldest

votes


















0












$begingroup$

If you want the Taylor series, you basically need the $n^{th}$ derivative of $Gamma(x)$. These express in terms of the polygamma function. Considering



$$d_n=frac{left[Gamma(x)right]^{(n)}}{Gamma(x)}$$ the first terms are
$$d_1=psi ^{(0)}(x)$$
$$d_2=psi ^{(0)}(x)^2+psi ^{(1)}(x)$$
$$d_3=psi ^{(0)}(x)^3+3 psi ^{(1)}(x) psi ^{(0)}(x)+psi ^{(2)}(x)$$
$$d_4=psi ^{(0)}(x)^4+6 psi ^{(1)}(x) psi ^{(0)}(x)^2+4 psi ^{(2)}(x) psi
^{(0)}(x)+3 psi ^{(1)}(x)^2+psi ^{(3)}(x)$$ which "simplify" (a little !) when you perform the expansion around $x=a$, $a$ being a positive integer.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    No, I am not looking for function evaluations as coefficients, but for the 'raw' numbers, like that in the Taylor Series of $sin(x)$.
    $endgroup$
    – Shashank
    Jan 24 '18 at 9:21










  • $begingroup$
    @Shashank. Then, good luck ... and have fun ! Cheers.
    $endgroup$
    – Claude Leibovici
    Jan 24 '18 at 9:34










  • $begingroup$
    I guess it is ridiculously hard. Anyways, thank you @Claude Leibovici
    $endgroup$
    – Shashank
    Jan 24 '18 at 9:48










  • $begingroup$
    @Shashank. Be sure that it is ridiculously hard and that you are very welcome !
    $endgroup$
    – Claude Leibovici
    Jan 24 '18 at 10:05











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

If you want the Taylor series, you basically need the $n^{th}$ derivative of $Gamma(x)$. These express in terms of the polygamma function. Considering



$$d_n=frac{left[Gamma(x)right]^{(n)}}{Gamma(x)}$$ the first terms are
$$d_1=psi ^{(0)}(x)$$
$$d_2=psi ^{(0)}(x)^2+psi ^{(1)}(x)$$
$$d_3=psi ^{(0)}(x)^3+3 psi ^{(1)}(x) psi ^{(0)}(x)+psi ^{(2)}(x)$$
$$d_4=psi ^{(0)}(x)^4+6 psi ^{(1)}(x) psi ^{(0)}(x)^2+4 psi ^{(2)}(x) psi
^{(0)}(x)+3 psi ^{(1)}(x)^2+psi ^{(3)}(x)$$ which "simplify" (a little !) when you perform the expansion around $x=a$, $a$ being a positive integer.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    No, I am not looking for function evaluations as coefficients, but for the 'raw' numbers, like that in the Taylor Series of $sin(x)$.
    $endgroup$
    – Shashank
    Jan 24 '18 at 9:21










  • $begingroup$
    @Shashank. Then, good luck ... and have fun ! Cheers.
    $endgroup$
    – Claude Leibovici
    Jan 24 '18 at 9:34










  • $begingroup$
    I guess it is ridiculously hard. Anyways, thank you @Claude Leibovici
    $endgroup$
    – Shashank
    Jan 24 '18 at 9:48










  • $begingroup$
    @Shashank. Be sure that it is ridiculously hard and that you are very welcome !
    $endgroup$
    – Claude Leibovici
    Jan 24 '18 at 10:05
















0












$begingroup$

If you want the Taylor series, you basically need the $n^{th}$ derivative of $Gamma(x)$. These express in terms of the polygamma function. Considering



$$d_n=frac{left[Gamma(x)right]^{(n)}}{Gamma(x)}$$ the first terms are
$$d_1=psi ^{(0)}(x)$$
$$d_2=psi ^{(0)}(x)^2+psi ^{(1)}(x)$$
$$d_3=psi ^{(0)}(x)^3+3 psi ^{(1)}(x) psi ^{(0)}(x)+psi ^{(2)}(x)$$
$$d_4=psi ^{(0)}(x)^4+6 psi ^{(1)}(x) psi ^{(0)}(x)^2+4 psi ^{(2)}(x) psi
^{(0)}(x)+3 psi ^{(1)}(x)^2+psi ^{(3)}(x)$$ which "simplify" (a little !) when you perform the expansion around $x=a$, $a$ being a positive integer.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    No, I am not looking for function evaluations as coefficients, but for the 'raw' numbers, like that in the Taylor Series of $sin(x)$.
    $endgroup$
    – Shashank
    Jan 24 '18 at 9:21










  • $begingroup$
    @Shashank. Then, good luck ... and have fun ! Cheers.
    $endgroup$
    – Claude Leibovici
    Jan 24 '18 at 9:34










  • $begingroup$
    I guess it is ridiculously hard. Anyways, thank you @Claude Leibovici
    $endgroup$
    – Shashank
    Jan 24 '18 at 9:48










  • $begingroup$
    @Shashank. Be sure that it is ridiculously hard and that you are very welcome !
    $endgroup$
    – Claude Leibovici
    Jan 24 '18 at 10:05














0












0








0





$begingroup$

If you want the Taylor series, you basically need the $n^{th}$ derivative of $Gamma(x)$. These express in terms of the polygamma function. Considering



$$d_n=frac{left[Gamma(x)right]^{(n)}}{Gamma(x)}$$ the first terms are
$$d_1=psi ^{(0)}(x)$$
$$d_2=psi ^{(0)}(x)^2+psi ^{(1)}(x)$$
$$d_3=psi ^{(0)}(x)^3+3 psi ^{(1)}(x) psi ^{(0)}(x)+psi ^{(2)}(x)$$
$$d_4=psi ^{(0)}(x)^4+6 psi ^{(1)}(x) psi ^{(0)}(x)^2+4 psi ^{(2)}(x) psi
^{(0)}(x)+3 psi ^{(1)}(x)^2+psi ^{(3)}(x)$$ which "simplify" (a little !) when you perform the expansion around $x=a$, $a$ being a positive integer.






share|cite|improve this answer









$endgroup$



If you want the Taylor series, you basically need the $n^{th}$ derivative of $Gamma(x)$. These express in terms of the polygamma function. Considering



$$d_n=frac{left[Gamma(x)right]^{(n)}}{Gamma(x)}$$ the first terms are
$$d_1=psi ^{(0)}(x)$$
$$d_2=psi ^{(0)}(x)^2+psi ^{(1)}(x)$$
$$d_3=psi ^{(0)}(x)^3+3 psi ^{(1)}(x) psi ^{(0)}(x)+psi ^{(2)}(x)$$
$$d_4=psi ^{(0)}(x)^4+6 psi ^{(1)}(x) psi ^{(0)}(x)^2+4 psi ^{(2)}(x) psi
^{(0)}(x)+3 psi ^{(1)}(x)^2+psi ^{(3)}(x)$$ which "simplify" (a little !) when you perform the expansion around $x=a$, $a$ being a positive integer.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 24 '18 at 8:16









Claude LeiboviciClaude Leibovici

123k1157134




123k1157134












  • $begingroup$
    No, I am not looking for function evaluations as coefficients, but for the 'raw' numbers, like that in the Taylor Series of $sin(x)$.
    $endgroup$
    – Shashank
    Jan 24 '18 at 9:21










  • $begingroup$
    @Shashank. Then, good luck ... and have fun ! Cheers.
    $endgroup$
    – Claude Leibovici
    Jan 24 '18 at 9:34










  • $begingroup$
    I guess it is ridiculously hard. Anyways, thank you @Claude Leibovici
    $endgroup$
    – Shashank
    Jan 24 '18 at 9:48










  • $begingroup$
    @Shashank. Be sure that it is ridiculously hard and that you are very welcome !
    $endgroup$
    – Claude Leibovici
    Jan 24 '18 at 10:05


















  • $begingroup$
    No, I am not looking for function evaluations as coefficients, but for the 'raw' numbers, like that in the Taylor Series of $sin(x)$.
    $endgroup$
    – Shashank
    Jan 24 '18 at 9:21










  • $begingroup$
    @Shashank. Then, good luck ... and have fun ! Cheers.
    $endgroup$
    – Claude Leibovici
    Jan 24 '18 at 9:34










  • $begingroup$
    I guess it is ridiculously hard. Anyways, thank you @Claude Leibovici
    $endgroup$
    – Shashank
    Jan 24 '18 at 9:48










  • $begingroup$
    @Shashank. Be sure that it is ridiculously hard and that you are very welcome !
    $endgroup$
    – Claude Leibovici
    Jan 24 '18 at 10:05
















$begingroup$
No, I am not looking for function evaluations as coefficients, but for the 'raw' numbers, like that in the Taylor Series of $sin(x)$.
$endgroup$
– Shashank
Jan 24 '18 at 9:21




$begingroup$
No, I am not looking for function evaluations as coefficients, but for the 'raw' numbers, like that in the Taylor Series of $sin(x)$.
$endgroup$
– Shashank
Jan 24 '18 at 9:21












$begingroup$
@Shashank. Then, good luck ... and have fun ! Cheers.
$endgroup$
– Claude Leibovici
Jan 24 '18 at 9:34




$begingroup$
@Shashank. Then, good luck ... and have fun ! Cheers.
$endgroup$
– Claude Leibovici
Jan 24 '18 at 9:34












$begingroup$
I guess it is ridiculously hard. Anyways, thank you @Claude Leibovici
$endgroup$
– Shashank
Jan 24 '18 at 9:48




$begingroup$
I guess it is ridiculously hard. Anyways, thank you @Claude Leibovici
$endgroup$
– Shashank
Jan 24 '18 at 9:48












$begingroup$
@Shashank. Be sure that it is ridiculously hard and that you are very welcome !
$endgroup$
– Claude Leibovici
Jan 24 '18 at 10:05




$begingroup$
@Shashank. Be sure that it is ridiculously hard and that you are very welcome !
$endgroup$
– Claude Leibovici
Jan 24 '18 at 10:05


















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