Difference between revisions of "Z2F1(1,1;2,-z) equals log(1+z)"
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&= \log(1+z), | &= \log(1+z), | ||
\end{array}$$ | \end{array}$$ | ||
| − | + | 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 $\log(1+z)$. █
