1

I'm trying to express $$_2F_1\left(\frac{1}{2} , 1;\,\frac{3}{2}+m;\,z\right), \quad m\in\mathbb{Z}\quad\text{and}\quad m\geq0$$ in terms of $\tanh^{-1}(\sqrt{z})$ and $\sqrt{z}$, basically generalizing these results.

Any ideas?

lel
  • 192

1 Answers1

2

Using Maple, I am getting $$ {}_2F_1\left(\frac12,1;\frac32+m;z\right) = A_m(z)\operatorname{atanh}\sqrt{z}+B_m(z) $$ where $$ A_m(z) = \frac{2(z-1)^m\Gamma(m+\frac32)}{z^{m+1/2}\sqrt{\pi}\;\Gamma(m+1)} $$ and $$ B_m(z) = \frac{\Gamma(m+\frac32)}{\Gamma(m+1)}\; \sum_{k=1}^m\frac{(z-1)^{m-k}\Gamma(k)}{z^{m+1-k}\Gamma(k+\frac12)} $$

For example, $$ A_{10}(z) = {\frac {969969\, \left( z-1 \right) ^{10}}{262144\,{z}^{21/2}}}, \\ B_{10}(z) = \frac{1}{{3932160\,{z}^{10}}}\; \big(68025825\,{z}^{9}-382331775\,{z}^{8}+1168982220{z}^{7}-2255541300\,{z}^{6}+2918514950{z}^{5}-2585198330{z}^{4}+1554721740\,{z}^{3}-609140532{z}^{2}+140645505z-14549535\big) $$

GEdgar
  • 111,679
  • That pesky sum is unfortunately hiding just another similar 2F1, so this basically amounts to a transformation. I also keep going in circles with this, which isn't really helping. But thanks for the contribution! – lel Mar 09 '21 at 23:35
  • 1
    But it’s a finite summation. – A rural reader Mar 10 '21 at 00:33
  • 1
    And something nice, @GEdgar, is you can write simple recurrence relations for $A_m(z)$ and $B_m(z)$. $A_m(z)/A_{m-1}(z) = \frac{m + 1/2}{m} \frac{z-1}{z}$ is very nice! – A rural reader Mar 10 '21 at 01:21
  • Hmmm, I'll give it another look, I was too frustrated last night x) – lel Mar 10 '21 at 12:02
  • Yup, this actually works, my bad! – lel Mar 10 '21 at 12:06