Difference between revisions of "Z2F1(1,1;2,-z) equals log(1+z)"
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| − | + | ==Theorem== | |
| − | + | 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]]. | ||
| − | + | ||
| − | + | ==Proof== | |
| + | Calculate | ||
$$\begin{array}{ll} | $$\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 \\ | 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 \\ | ||
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\end{array}$$ | \end{array}$$ | ||
using a well-known formula for the [[logarithm|Taylor series of $\log(1+z)$]]. █ | using a well-known formula for the [[logarithm|Taylor series of $\log(1+z)$]]. █ | ||
| − | + | ||
| − | + | ==References== | |
Revision as of 06:57, 4 June 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 $$\begin{array}{ll} z{}_2F_1(1,1;2;-z) &= z\displaystyle\sum_{k=0}^{\infty} \dfrac{1^{\overline{k}}1^{\overline{k}}}{2^{\overline{k}}k!} (-z)^k \\ &= \displaystyle\sum_{k=0}^{\infty} \dfrac{\left( \frac{\Gamma(k+1)}{\Gamma(1)} \right)^2}{\left( \frac{\Gamma(2+k)}{\Gamma(2)} \right)k!}(-1)^k z^{k+1} \\ &= \displaystyle\sum_{k=0}^{\infty} \dfrac{(k!)^2(-1)^k}{(k+1)!k!} z^{k+1} \\ &= \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k}{k+1} z^{k+1} \\ &= -\displaystyle\sum_{k=1}^{\infty} \dfrac{(-1)^k z^k}{k} \\ &= \log(1+z), \end{array}$$ using a well-known formula for the Taylor series of $\log(1+z)$. █
