Polygamma expression for $frac{Gamma^{(k)}(z)}{Gamma(z)}$?











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1
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I'm trying to simplify



$$frac{Gamma^{(k)}(z)}{Gamma(z)}$$



for $k=1,2,cdots$, using polygamma notation




Try



I've calculated a few, using



$$Gamma^{(k)}(z) = int_0^infty (log x)^k x^{z-1} e^{-x} dx$$



but I'm not sure if there is any way to generalize.



$$
begin{aligned}
frac{Gamma^{(1)}(z)}{Gamma(z)} &= psi^{(0)}(z) \
frac{Gamma^{(2)}(z)}{Gamma(z)} &= psi^{(1)}(z) +left(psi^{(0)}(z)right)^2 \
frac{Gamma^{(3)}(z)}{Gamma(z)} &= psi^{(2)}(z) + 3 psi^{(1)}(z) psi^{(0)}(z)+left(psi^{(0)}(z)right)^3 \
frac{Gamma^{(4)}(z)}{Gamma(z)} &= psi^{(3)}(z) + 4 psi^{(2)}(z) psi^{(0)}(z)+ 6 psi^{(1)}(z) left(psi^{(0)}(z)right)^2+ 3 psi^{(1)}(z)^2 +left(psi^{(0)}(z)right)^4 \
end{aligned}
$$






gamma-function polygamma







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  • 1




    I think you've mixed up $x$ and $z$ in your integral formula.
    – Robert Israel
    Nov 21 at 15:29










  • @RobertIsrael True, a lot to edit. Thnx
    – Moreblue
    Nov 21 at 15:31















up vote
1
down vote

favorite












I'm trying to simplify



$$frac{Gamma^{(k)}(z)}{Gamma(z)}$$



for $k=1,2,cdots$, using polygamma notation




Try



I've calculated a few, using



$$Gamma^{(k)}(z) = int_0^infty (log x)^k x^{z-1} e^{-x} dx$$



but I'm not sure if there is any way to generalize.



$$
begin{aligned}
frac{Gamma^{(1)}(z)}{Gamma(z)} &= psi^{(0)}(z) \
frac{Gamma^{(2)}(z)}{Gamma(z)} &= psi^{(1)}(z) +left(psi^{(0)}(z)right)^2 \
frac{Gamma^{(3)}(z)}{Gamma(z)} &= psi^{(2)}(z) + 3 psi^{(1)}(z) psi^{(0)}(z)+left(psi^{(0)}(z)right)^3 \
frac{Gamma^{(4)}(z)}{Gamma(z)} &= psi^{(3)}(z) + 4 psi^{(2)}(z) psi^{(0)}(z)+ 6 psi^{(1)}(z) left(psi^{(0)}(z)right)^2+ 3 psi^{(1)}(z)^2 +left(psi^{(0)}(z)right)^4 \
end{aligned}
$$






gamma-function polygamma







share|cite|improve this question




















  • 1




    I think you've mixed up $x$ and $z$ in your integral formula.
    – Robert Israel
    Nov 21 at 15:29










  • @RobertIsrael True, a lot to edit. Thnx
    – Moreblue
    Nov 21 at 15:31













up vote
1
down vote

favorite









up vote
1
down vote

favorite











I'm trying to simplify



$$frac{Gamma^{(k)}(z)}{Gamma(z)}$$



for $k=1,2,cdots$, using polygamma notation




Try



I've calculated a few, using



$$Gamma^{(k)}(z) = int_0^infty (log x)^k x^{z-1} e^{-x} dx$$



but I'm not sure if there is any way to generalize.



$$
begin{aligned}
frac{Gamma^{(1)}(z)}{Gamma(z)} &= psi^{(0)}(z) \
frac{Gamma^{(2)}(z)}{Gamma(z)} &= psi^{(1)}(z) +left(psi^{(0)}(z)right)^2 \
frac{Gamma^{(3)}(z)}{Gamma(z)} &= psi^{(2)}(z) + 3 psi^{(1)}(z) psi^{(0)}(z)+left(psi^{(0)}(z)right)^3 \
frac{Gamma^{(4)}(z)}{Gamma(z)} &= psi^{(3)}(z) + 4 psi^{(2)}(z) psi^{(0)}(z)+ 6 psi^{(1)}(z) left(psi^{(0)}(z)right)^2+ 3 psi^{(1)}(z)^2 +left(psi^{(0)}(z)right)^4 \
end{aligned}
$$






gamma-function polygamma







share|cite|improve this question















I'm trying to simplify



$$frac{Gamma^{(k)}(z)}{Gamma(z)}$$



for $k=1,2,cdots$, using polygamma notation




Try



I've calculated a few, using



$$Gamma^{(k)}(z) = int_0^infty (log x)^k x^{z-1} e^{-x} dx$$



but I'm not sure if there is any way to generalize.



$$
begin{aligned}
frac{Gamma^{(1)}(z)}{Gamma(z)} &= psi^{(0)}(z) \
frac{Gamma^{(2)}(z)}{Gamma(z)} &= psi^{(1)}(z) +left(psi^{(0)}(z)right)^2 \
frac{Gamma^{(3)}(z)}{Gamma(z)} &= psi^{(2)}(z) + 3 psi^{(1)}(z) psi^{(0)}(z)+left(psi^{(0)}(z)right)^3 \
frac{Gamma^{(4)}(z)}{Gamma(z)} &= psi^{(3)}(z) + 4 psi^{(2)}(z) psi^{(0)}(z)+ 6 psi^{(1)}(z) left(psi^{(0)}(z)right)^2+ 3 psi^{(1)}(z)^2 +left(psi^{(0)}(z)right)^4 \
end{aligned}
$$






gamma-function polygamma




gamma-function polygamma






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edited Nov 21 at 15:34

























asked Nov 21 at 15:27









Moreblue

851216




851216








  • 1




    I think you've mixed up $x$ and $z$ in your integral formula.
    – Robert Israel
    Nov 21 at 15:29










  • @RobertIsrael True, a lot to edit. Thnx
    – Moreblue
    Nov 21 at 15:31














  • 1




    I think you've mixed up $x$ and $z$ in your integral formula.
    – Robert Israel
    Nov 21 at 15:29










  • @RobertIsrael True, a lot to edit. Thnx
    – Moreblue
    Nov 21 at 15:31








1




1




I think you've mixed up $x$ and $z$ in your integral formula.
– Robert Israel
Nov 21 at 15:29




I think you've mixed up $x$ and $z$ in your integral formula.
– Robert Israel
Nov 21 at 15:29












@RobertIsrael True, a lot to edit. Thnx
– Moreblue
Nov 21 at 15:31




@RobertIsrael True, a lot to edit. Thnx
– Moreblue
Nov 21 at 15:31










1 Answer
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Of course the exponential generating function is



$$ sum_{n=0}^infty frac{s^n}{n!} frac{Gamma^{(n)}(z)}{Gamma(z)} = frac{Gamma(z+s)}{Gamma(z)} $$
so basically you want the Taylor coefficients of $Gamma$ around $z$.



Now $ln(Gamma)$ has a nice series:



$$ ln(Gamma(z+s)) = ln(Gamma(z)) + sum_{k=1}^{infty} frac{Psi^{(k-1)}(z)}{k!} s^k $$



so



$$ frac{Gamma(z+s)}{Gamma(z)} = exp left(sum_{k=1}^infty frac{Psi^{(k-1)}(z)}{k!} s^k right) = prod_{k=1}^infty expleft(frac{Psi^{(k-1)}(z)}{k!} s^kright) $$



and the coefficient of $s^n$ here is



$$ sum_{sum_k k m_k = n} prod_{k=1}^infty frac{(Psi^{(k-1)}(z))^{m_k}}{(k!)^{m_k} m_k!}$$



the sum being over all sequences $m = (m_1, m_2, ldots)$ of nonnegative integers with $sum_k k m_k = n$. These correspond to partitions of $n$, where $m_k$ is the number of occurences of $k$ in the partition. Multiply by $n!$
to get $Gamma^{(n)}(z)/Gamma(z)$. Thus for $n=3$, the partitions of $3$ are $1+1+1$, $1+2$ and $3$, corresponding to the terms
$Psi^{(0)}(z)^3$, $3 Psi^{(0)}(z) Psi^{(1)}(z)$ and $Psi^{(2)}(z)$ respectively.






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    1 Answer
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    up vote
    1
    down vote



    accepted










    Of course the exponential generating function is



    $$ sum_{n=0}^infty frac{s^n}{n!} frac{Gamma^{(n)}(z)}{Gamma(z)} = frac{Gamma(z+s)}{Gamma(z)} $$
    so basically you want the Taylor coefficients of $Gamma$ around $z$.



    Now $ln(Gamma)$ has a nice series:



    $$ ln(Gamma(z+s)) = ln(Gamma(z)) + sum_{k=1}^{infty} frac{Psi^{(k-1)}(z)}{k!} s^k $$



    so



    $$ frac{Gamma(z+s)}{Gamma(z)} = exp left(sum_{k=1}^infty frac{Psi^{(k-1)}(z)}{k!} s^k right) = prod_{k=1}^infty expleft(frac{Psi^{(k-1)}(z)}{k!} s^kright) $$



    and the coefficient of $s^n$ here is



    $$ sum_{sum_k k m_k = n} prod_{k=1}^infty frac{(Psi^{(k-1)}(z))^{m_k}}{(k!)^{m_k} m_k!}$$



    the sum being over all sequences $m = (m_1, m_2, ldots)$ of nonnegative integers with $sum_k k m_k = n$. These correspond to partitions of $n$, where $m_k$ is the number of occurences of $k$ in the partition. Multiply by $n!$
    to get $Gamma^{(n)}(z)/Gamma(z)$. Thus for $n=3$, the partitions of $3$ are $1+1+1$, $1+2$ and $3$, corresponding to the terms
    $Psi^{(0)}(z)^3$, $3 Psi^{(0)}(z) Psi^{(1)}(z)$ and $Psi^{(2)}(z)$ respectively.






    share|cite|improve this answer



























      up vote
      1
      down vote



      accepted










      Of course the exponential generating function is



      $$ sum_{n=0}^infty frac{s^n}{n!} frac{Gamma^{(n)}(z)}{Gamma(z)} = frac{Gamma(z+s)}{Gamma(z)} $$
      so basically you want the Taylor coefficients of $Gamma$ around $z$.



      Now $ln(Gamma)$ has a nice series:



      $$ ln(Gamma(z+s)) = ln(Gamma(z)) + sum_{k=1}^{infty} frac{Psi^{(k-1)}(z)}{k!} s^k $$



      so



      $$ frac{Gamma(z+s)}{Gamma(z)} = exp left(sum_{k=1}^infty frac{Psi^{(k-1)}(z)}{k!} s^k right) = prod_{k=1}^infty expleft(frac{Psi^{(k-1)}(z)}{k!} s^kright) $$



      and the coefficient of $s^n$ here is



      $$ sum_{sum_k k m_k = n} prod_{k=1}^infty frac{(Psi^{(k-1)}(z))^{m_k}}{(k!)^{m_k} m_k!}$$



      the sum being over all sequences $m = (m_1, m_2, ldots)$ of nonnegative integers with $sum_k k m_k = n$. These correspond to partitions of $n$, where $m_k$ is the number of occurences of $k$ in the partition. Multiply by $n!$
      to get $Gamma^{(n)}(z)/Gamma(z)$. Thus for $n=3$, the partitions of $3$ are $1+1+1$, $1+2$ and $3$, corresponding to the terms
      $Psi^{(0)}(z)^3$, $3 Psi^{(0)}(z) Psi^{(1)}(z)$ and $Psi^{(2)}(z)$ respectively.






      share|cite|improve this answer

























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        Of course the exponential generating function is



        $$ sum_{n=0}^infty frac{s^n}{n!} frac{Gamma^{(n)}(z)}{Gamma(z)} = frac{Gamma(z+s)}{Gamma(z)} $$
        so basically you want the Taylor coefficients of $Gamma$ around $z$.



        Now $ln(Gamma)$ has a nice series:



        $$ ln(Gamma(z+s)) = ln(Gamma(z)) + sum_{k=1}^{infty} frac{Psi^{(k-1)}(z)}{k!} s^k $$



        so



        $$ frac{Gamma(z+s)}{Gamma(z)} = exp left(sum_{k=1}^infty frac{Psi^{(k-1)}(z)}{k!} s^k right) = prod_{k=1}^infty expleft(frac{Psi^{(k-1)}(z)}{k!} s^kright) $$



        and the coefficient of $s^n$ here is



        $$ sum_{sum_k k m_k = n} prod_{k=1}^infty frac{(Psi^{(k-1)}(z))^{m_k}}{(k!)^{m_k} m_k!}$$



        the sum being over all sequences $m = (m_1, m_2, ldots)$ of nonnegative integers with $sum_k k m_k = n$. These correspond to partitions of $n$, where $m_k$ is the number of occurences of $k$ in the partition. Multiply by $n!$
        to get $Gamma^{(n)}(z)/Gamma(z)$. Thus for $n=3$, the partitions of $3$ are $1+1+1$, $1+2$ and $3$, corresponding to the terms
        $Psi^{(0)}(z)^3$, $3 Psi^{(0)}(z) Psi^{(1)}(z)$ and $Psi^{(2)}(z)$ respectively.






        share|cite|improve this answer














        Of course the exponential generating function is



        $$ sum_{n=0}^infty frac{s^n}{n!} frac{Gamma^{(n)}(z)}{Gamma(z)} = frac{Gamma(z+s)}{Gamma(z)} $$
        so basically you want the Taylor coefficients of $Gamma$ around $z$.



        Now $ln(Gamma)$ has a nice series:



        $$ ln(Gamma(z+s)) = ln(Gamma(z)) + sum_{k=1}^{infty} frac{Psi^{(k-1)}(z)}{k!} s^k $$



        so



        $$ frac{Gamma(z+s)}{Gamma(z)} = exp left(sum_{k=1}^infty frac{Psi^{(k-1)}(z)}{k!} s^k right) = prod_{k=1}^infty expleft(frac{Psi^{(k-1)}(z)}{k!} s^kright) $$



        and the coefficient of $s^n$ here is



        $$ sum_{sum_k k m_k = n} prod_{k=1}^infty frac{(Psi^{(k-1)}(z))^{m_k}}{(k!)^{m_k} m_k!}$$



        the sum being over all sequences $m = (m_1, m_2, ldots)$ of nonnegative integers with $sum_k k m_k = n$. These correspond to partitions of $n$, where $m_k$ is the number of occurences of $k$ in the partition. Multiply by $n!$
        to get $Gamma^{(n)}(z)/Gamma(z)$. Thus for $n=3$, the partitions of $3$ are $1+1+1$, $1+2$ and $3$, corresponding to the terms
        $Psi^{(0)}(z)^3$, $3 Psi^{(0)}(z) Psi^{(1)}(z)$ and $Psi^{(2)}(z)$ respectively.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 21 at 16:17

























        answered Nov 21 at 16:07









        Robert Israel

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