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  "body": "\n## Proof\n\nWe take $k = 2$ in Taylor's theorem and rewrite the terms with matrices.\n\n1. Let $U$, $f$ and $a$ be as in the statement. Write $\\partial_i$ for the partial derivative in the direction $x_i$. Since $f$ is of [[class C^2|0TCBZMEYGN]], it is [[differentiable|0KXSK8TEEB]] at $a$, so $\\nabla f(a)$ is defined.\n2. By [[Taylor's theorem in several variables|0J8T4K4YTE]] with $k = 2$, we have $f(a + h) = \\sum_{\\abs{\\alpha} \\le 2} \\frac{D^\\alpha f(a)}{\\alpha!} h^\\alpha + R_2(h)$ with $R_2(h) / \\norm{h}^2 \\to 0$ as $h \\to 0$.\n3. It suffices to show that the sum in step 2 equals $f(a) + \\nabla f(a) \\cdot h + \\tfrac{1}{2} h^{\\mathsf{T}} H_f(a) h$. Then $R(h) = R_2(h)$ has the required limit.\n4. The only [[multi-index|0PJXBR6QA3]] with $\\abs{\\alpha} = 0$ is $\\alpha = 0$. Its term is $f(a)$.\n5. The multi-indices with $\\abs{\\alpha} = 1$ are the $e_i$, with $e_i! = 1$ and $h^{e_i} = h_i$. Their terms add up to $\\sum_i \\partial_i f(a) h_i$. This is $\\nabla f(a) \\cdot h$, since the [[gradient|0NS2QYVC4Y]] is the vector of partial derivatives.\n6. The multi-indices with $\\abs{\\alpha} = 2$ are $2 e_i$ for each $i$, and $e_i + e_j$ for each pair $i \u003c j$.\n7. For $\\alpha = 2 e_i$ we have $\\alpha! = 2$, so the term is $\\tfrac{1}{2} \\partial_i \\partial_i f(a) \\, h_i^2$. For $\\alpha = e_i + e_j$ with $i \u003c j$ we have $\\alpha! = 1$, so the term is $\\partial_i \\partial_j f(a) \\, h_i h_j$.\n8. Let $H = H_f(a)$, the [[Hessian matrix|0SCSQWXRA4]], with entries $H_{ij} = \\partial_i \\partial_j f(a)$. By [[Schwarz's theorem|0NNQY8WCW6]], $H_{ij} = H_{ji}$.\n9. Then $\\tfrac{1}{2} h^{\\mathsf{T}} H h = \\tfrac{1}{2} \\sum_{i, j} H_{ij} h_i h_j = \\tfrac{1}{2} \\sum_i H_{ii} h_i^2 + \\tfrac{1}{2} \\sum_{i \u003c j} (H_{ij} + H_{ji}) h_i h_j$.\n10. By step 8 the last sum is $\\sum_{i \u003c j} H_{ij} h_i h_j$. So $\\tfrac{1}{2} h^{\\mathsf{T}} H h$ equals the sum of the terms in step 7.\n11. Steps 4, 5, 7 and 10 give the identity of step 3, which proves the expansion.\n\n## Idea\n\nThe second-order expansion is the case $k = 2$ of Taylor's theorem. The only work is bookkeeping: the terms of order two in multi-index form match $\\tfrac{1}{2} h^{\\mathsf{T}} H h$, because the Hessian is symmetric and each mixed term appears twice in the matrix product.\n",
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