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How can we reduce the summation \begin{equation} \sum_{k=0}^{N} \dfrac{\binom{2N}{2k}}{\binom{N}{k}} \lambda^k\, \Pi_{k}(x_1, \ldots, x_N), \end{equation} where ($\Pi_{k}(x_1, \ldots, x_N)$ is the $k$th elementary symmetric polynomial in $N$ variables) to a single expression? (Possibly even to a single integral in the real or complex plane?)

As a check, for $x_i = x_0$ for all $i$, we should find \begin{equation} \sum_{k=0}^{N} \binom{2N}{2k}(\lambda x_0)^k = \frac{1}{2}\Big[ \left(1 + \sqrt{\lambda x_0}\right)^{2N} + \left(1 - \sqrt{\lambda x_0}\right)^{2N}\Big]. \end{equation}

Background

The elementary symmetry polynomials $\Pi_k(x_1, \ldots, x_N)$ can be indirectly defined via $$ \prod_{\ell=1}^{N}(1+ \lambda x_{\ell}) = \sum_{k = 0}^{N} \lambda^k \, \Pi_{k}(x_1, \ldots, x_N). $$

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  • $\begingroup$ What does $\Pi_k$ mean? $\endgroup$ – saulspatz Apr 22 '18 at 11:12
  • $\begingroup$ It stands for the $k$th elementary symmetric polynomial. I have made a correction to the problem description to reflect this. $\endgroup$ – Olukayode Apr 22 '18 at 15:21

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