Difference between revisions of "Z2F1(1,1;2,-z) equals log(1+z)"

From specialfunctionswiki
Jump to: navigation, search
 
(2 intermediate revisions by the same user not shown)
Line 14: Line 14:
 
&= \log(1+z),
 
&= \log(1+z),
 
\end{array}$$
 
\end{array}$$
using a well-known formula for the [[logarithm|Taylor series of $\log(1+z)$]]. █  
+
by the [[Taylor series of log(1+z)|Taylor series of $\log(1+z)$]]. █  
  
 
==References==
 
==References==
 +
 +
[[Category:Theorem]]
 +
[[Category:Proven]]

Latest revision as of 08:33, 18 December 2016

Theorem

The following formula holds: $$\log(1+z)=z{}_2F_1(1,1;2;-z),$$ where $\log$ denotes the logarithm and ${}_2F_1$ denotes the hypergeometric pFq.

Proof

Calculate $$\begin{array}{ll} z{}_2F_1(1,1;2;-z) &= z\displaystyle\sum_{k=0}^{\infty} \dfrac{1^{\overline{k}}1^{\overline{k}}}{2^{\overline{k}}k!} (-z)^k \\ &= \displaystyle\sum_{k=0}^{\infty} \dfrac{\left( \frac{\Gamma(k+1)}{\Gamma(1)} \right)^2}{\left( \frac{\Gamma(2+k)}{\Gamma(2)} \right)k!}(-1)^k z^{k+1} \\ &= \displaystyle\sum_{k=0}^{\infty} \dfrac{(k!)^2(-1)^k}{(k+1)!k!} z^{k+1} \\ &= \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k}{k+1} z^{k+1} \\ &= -\displaystyle\sum_{k=1}^{\infty} \dfrac{(-1)^k z^k}{k} \\ &= \log(1+z), \end{array}$$ by the Taylor series of $\log(1+z)$. █

References