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Given $f(x) = \frac{1}{2}x^TAx + b^Tx + \alpha $

where A is an nxn symmetric matrix, b is an n-dimensional vector, and alpha a scalar. Show that

$\bigtriangledown _{x}f(x) = Ax + b$

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and

$H = \bigtriangledown ^{2}_{x}f(x) = A$

Is this simply a matter of taking a derivative with respect to X, how would you attack this one?

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$\nabla f = (\partial f/\partial x_1, \ldots, \partial f/\partial x_n)^t$ denotes the vector of partial derivatives of $f$ and is a completely standard notation.

On the other hand, $\nabla^2 f$ seems to be used here in an unusual way, namely to denote the Hessian (the matrix of all second order partial derivatives), $(\partial^2 f/\partial x_i \partial x_j)_{i,j=1}^n$.

(The usual meaning of $\nabla^2 f$ is the Laplacian, $\partial^2 f/\partial x_1^2 + \ldots + \partial^2 f/\partial x_n^2$.)

$\bigtriangledown f$ finds the direction of maximal change in f.

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