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  "body": "\n## Definition\n\nLet $m \\ge 1$ be an integer, let $\\alpha \u003e 0$, and let $X$ be a complex $m \\times m$ [[symmetric matrix|0VR671PHE5]] with [[eigenvalues|0BQAA2BK3Q]] $x_1, \\dots, x_m$.\nWrite $\\kappa \\vdash k$ when $\\kappa = (\\kappa_1 \\ge \\kappa_2 \\ge \\dots)$ is an [[integer partition|0JV30XN73Z]] of $k$.\nLet $C_\\kappa^{(\\alpha)}(X)$ be the Jack polynomial of $\\kappa$ in $x_1, \\dots, x_m$, normalised so that $\\sum_{\\kappa \\vdash k} C_\\kappa^{(\\alpha)}(X) = (\\operatorname{tr} X)^k$.\nIt is $0$ when $\\kappa$ has more than $m$ nonzero parts.\nFor a complex number $a$, the *generalized Pochhammer symbol* is\n$$(a)_\\kappa^{(\\alpha)} = \\prod_{i \\ge 1} \\prod_{j = 1}^{\\kappa_i} \\left( a - \\frac{i - 1}{\\alpha} + j - 1 \\right).$$\nLet $a_1, \\dots, a_p$ and $b_1, \\dots, b_q$ be complex numbers such that $(b_l)_\\kappa^{(\\alpha)} \\ne 0$ for every $l$ and every $\\kappa$ with at most $m$ parts.\nThe *hypergeometric function of matrix argument* is the series\n$${}_pF_q^{(\\alpha)}(a_1, \\dots, a_p; b_1, \\dots, b_q; X) = \\sum_{k = 0}^{\\infty} \\sum_{\\kappa \\vdash k} \\frac{(a_1)_\\kappa^{(\\alpha)} \\cdots (a_p)_\\kappa^{(\\alpha)}}{(b_1)_\\kappa^{(\\alpha)} \\cdots (b_q)_\\kappa^{(\\alpha)}} \\frac{C_\\kappa^{(\\alpha)}(X)}{k!},$$\nwherever it converges.\n\n## Remarks\n\nFor $m = 1$ there is one partition $\\kappa = (k)$ of each $k$, $C_{(k)}^{(\\alpha)}(x) = x^k$ and $(a)_{(k)}^{(\\alpha)} = (a)_k$, so the series is the scalar [[generalized hypergeometric function|VDBAW9PN88]].\nThe value depends only on the eigenvalues of $X$.\nFor $\\alpha = 2$ the polynomials $C_\\kappa^{(2)}$ are the zonal polynomials of real symmetric matrices; for $\\alpha = 1$ they are multiples of [[Schur polynomials|43VFMJCDYJ]], suited to complex Hermitian matrices.\nExample: ${}_0F_0^{(\\alpha)}(X) = \\sum_k \\frac{1}{k!} \\sum_{\\kappa \\vdash k} C_\\kappa^{(\\alpha)}(X) = \\sum_k \\frac{(\\operatorname{tr} X)^k}{k!} = e^{\\operatorname{tr} X}$.\nNon-example: for $m = 2$ and $\\alpha = 1$, the choice $b_1 = 1$ is excluded, since $(1)_{(1, 1)}^{(1)} = 1 \\cdot (1 - 1) = 0$ appears in a denominator, although $b_1 = 1$ is allowed in the scalar case.\n",
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