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For a multi-index $\alpha = \left (\alpha_1, \cdots, \alpha_n \right )$ we write $\left \lvert \alpha \right \rvert = \alpha_1 + \cdots + \alpha_n$ and $\alpha! = \alpha_1! \cdots \alpha_n!.$ For $z = (z_1, \cdots, z_n) \in \mathbb C^n$ and multi-index $\alpha = \left (\alpha_1, \cdots, \alpha_n \right )$ let $z^{\alpha} = z_1^{\alpha_1} \cdots z_n^{\alpha_n}.$ Define $e_{\alpha} : \mathbb D \longrightarrow \mathbb C$ by $$e_{\alpha} (z) = \left (\frac {\Gamma \left ( \left \lvert \alpha \right \rvert + n + \gamma + 1 \right )} {\Gamma \left (n + \gamma + 1 \right ) \alpha!} \right )^{\frac {1} {2}} z^{\alpha},$$

where $\gamma > - 1.$ Then find $\sum\limits_{\alpha} e_{\alpha} (z) \overline {e_{\alpha} (w)},$ where the sum is taken over all the multi-indices $\alpha = \left (\alpha_1, \cdots, \alpha_n \right ).$

This sum is actually the expression of the Bergman reproducing kernel of a certain weighted Bergman space. In the paper Compact Toeplitz Operators on Bergman Spaces by K.Stroethoff this sum is claimed to be $\frac {1} {\left (1 - \left \langle z, w \right \rangle \right )^{n + \gamma + 1}}.$

Could anyone shed some light on this fact as to how the sum is of the desired form? Any help would be much appreciated.

Thanks for your time.

Anacardium
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  • If you look at how the binomial theorem works for $(1-x)^{-s}$ you should be able to extend to a multinomial case which handles your series. – Lee Fisher May 21 '24 at 18:31
  • @LeeFisher$:$ Could you please explain in detail? I don't see how the coefficients are obtained in terms of Gamma function. – Anacardium May 21 '24 at 20:50
  • Try it with $\gamma\geq 0$ and an integer. You can replace the Gamma functions with factorials and write the coefficients explicitly as multinomial coefficients. – Lee Fisher May 21 '24 at 23:26

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