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Let \begin{eqnarray*} P_{k}(z)={_2F_1}(-k,\frac{1}{2}-k; -2k; z), \ \label{e0} Q_{k}(z)={_2F_1}(-k,-\frac{1}{2}-k; -2k; z) , \end{eqnarray*} where $k\ge 1$ is an integer.

How to show \begin{eqnarray*}\label{e1} \frac{Q_{k}(1-z^2)-zP_{k}(1-z^2)}{Q_{k}(1-z^2)+zP_{k}(1-z^2)}=\left(\frac{1-z}{1+z}\right)^{2k+1} \end{eqnarray*} analytically?

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Seems to me your $P$ and $Q$ are related to Legendre functions. Have you checked Abramowitz and Stegun? – Raskolnikov May 28 '11 at 13:00
up vote 6 down vote accepted

Those hypergeometric functions are essentially Chebyshev polynomials... $$ {}_2F_1 \Biggl(\biggl[-k,\frac{1}{2} - k\biggr],[-2k],1 - z^{2}\Biggr) = \frac{\bigl(1 - z^{2}\bigr)^{k} U_{2k} \biggl(\bigl(1 - z^{2}\bigr)^{-1/2}\biggr)}{2^{2 k}} $$ and the other one in terms of $T_{2k+1}$. Using the relation between the Chebyshev polynomials and the multiple angle formulas suggests the substitution $z=\tanh \theta$. Then the left-hand side becomes $$ \Bigl(\operatorname{cosh} \bigl((1 + 2 k) \theta\bigr) - \operatorname{sinh} \bigl((1 + 2 k) \theta\bigr)\Bigr)^{2} $$ and the right-hand side becomes $$ \Biggl(\frac{1 - \operatorname{tanh} (\theta)}{1 + \operatorname{tanh} (\theta)}\Biggr)^{1 + 2 k} $$ Converting to exponentials finishes.

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