I am working on Special Functions. I have the following infinite series:

$f(z)=\left(z + \displaystyle\frac{(\gamma)_1}{\Gamma(4)}\frac{z^2}{1!}+\displaystyle\frac{(\gamma)_2}{\Gamma(6)}\frac{z^3}{2!}+\displaystyle\frac{(\gamma)_3}{\Gamma(8)}\frac{z^4}{3!}+\cdots\right)$,

where $(\gamma)_n=(\gamma)(\gamma+1)\cdots(\gamma+n-1)$ and is called as Pochhammer symbol.

I tried to bring this series to a closed form as follows: \begin{eqnarray} f(z)&=&\sqrt{z}\left(\sqrt{z} + \displaystyle\frac{(\gamma)_1}{1!}\frac{(\sqrt{z})^3}{3!}+\displaystyle\frac{(\gamma)_2}{2!}\frac{(\sqrt{z})^5}{5!}+\displaystyle\frac{(\gamma)_3}{3!}\frac{(\sqrt{z})^7}{7!}+\cdots\right)\\ &=&\sqrt{z}\sum_{k=0}^{\infty}\frac{(\gamma)_k}{(1)_k}\frac{(\sqrt{z})^{2k+1}}{(2k+1)!} \end{eqnarray} Can this be reduced to a Confluent Hypergeometric Function (or any other closed form)? Also, for $\gamma=1$, I obtain $f(z)=\sqrt{z}\sinh(\sqrt{z})$.

Can anyone provide me any hint?

  • 2
    $\begingroup$ Since $(2k+1)! = 4^k(1)_k(\frac32)_k$, the sum is the generalized hypergeometric function $z{}_1F_2\left(\gamma; 1,\frac32; \frac{z}{4}\right)$. $\endgroup$ – achille hui May 16 '18 at 9:50
  • $\begingroup$ Thanks a lot! This helped very much. $\endgroup$ – user374063 May 17 '18 at 6:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.