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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

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

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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3  
The symbol is called a "nabla" or "del"; see en.wikipedia.org/wiki/Nabla_operator –  Arturo Magidin Sep 20 '10 at 18:23
    
Ok, so in this case the problem is a matter of taking the derivative with respect to x of the equation given? –  GBa Sep 20 '10 at 18:28
    
Vote to close, this question can be easily looked up. –  Noldorin Sep 20 '10 at 18:28
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@Noldorin: Ehr... fine, but "can be easily looked up" is not really the same thing as "not a real question", is it? You might have a case with "Off-topic" as too simple. –  Arturo Magidin Sep 20 '10 at 18:33
    
@Greg: Is it $\nabla _{x}f(x)=Ax+b$ or $\nabla f(x)=Ax+b$? –  Américo Tavares Sep 20 '10 at 19:36

2 Answers 2

up vote 9 down vote accepted

$\nabla f = (\partial f/\partial x_1, \ldots, \partial f/\partial x_n)^t$ 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, $\partial^2 f/\partial x_1^2 + \ldots + \partial^2 f/\partial x_n^2$.)

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Not that unusual; some optimization books use $\nabla^2$ as shorthand for the Hessian (in the sense of $\nabla$ of $\nabla f$ a.k.a. the gradient). –  J. M. Sep 20 '10 at 21:18
    
I am used to the following notation: a) $\nabla ^{2}f$ and $\nabla ^{2}\overrightarrow{F}$ as the laplacian of $f$ (scalar function) or $\overrightarrow{F}$ (vector field), in Physics b) $\nabla ^{2}A$ as the Hessian of the matrix $A$, in optimization. –  Américo Tavares Sep 20 '10 at 21:40
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Better notation, in my opinion: $\nabla ^{2}\cdot\overrightarrow{F}$, since here the operator $\nabla$ "is" a vector, while in $\nabla^2 f$ "is" a scalar. –  Américo Tavares Sep 20 '10 at 21:46

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

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