Difference between revisions of "Z2F1(1,1;2,-z) equals log(1+z)"
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| − | <strong>[[Z2F1(1,1;2,-z) equals log(1+z) | + | <strong>[[Z2F1(1,1;2,-z) equals log(1+z)|Theorem]]:</strong> The following formula holds: |
| − | |Theorem]]:</strong> The following formula holds: | ||
$$\log(1+z)=z{}_2F_1(1,1;2;-z),$$ | $$\log(1+z)=z{}_2F_1(1,1;2;-z),$$ | ||
where $\log$ denotes the [[logarithm]] and ${}_2F_1$ denotes the [[hypergeometric pFq]]. | where $\log$ denotes the [[logarithm]] and ${}_2F_1$ denotes the [[hypergeometric pFq]]. | ||
Revision as of 07:31, 11 April 2015
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 $$\begin{array}{ll} z{}_2F_1(1,1;2;-z) &= z\displaystyle\sum_{k=0}^{\infty} \dfrac{1^{\overline{k}}1^{\overline{k}}}{2^{\overline{k}}k!} (-z)^k \\ &= \displaystyle\sum_{k=0}^{\infty} \dfrac{\left( \frac{\Gamma(k+1)}{\Gamma(1)} \right)^2}{\left( \frac{\Gamma(2+k)}{\Gamma(2)} \right)k!}(-1)^k z^{k+1} \\ &= \displaystyle\sum_{k=0}^{\infty} \dfrac{(k!)^2(-1)^k}{(k+1)!k!} z^{k+1} \\ &= \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k}{k+1} z^{k+1} \\ &= -\displaystyle\sum_{k=1}^{\infty} \dfrac{(-1)^k z^k}{k} \\ &= \log(1+z), \end{array}$$ using a well-known formula for the Taylor series of $\log(1+z)$. █
