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Thomchad wrote:
> Hi all,
>
> Take a function f(x,y,z,...)
>
> On the diagonal of the Hessian matrix, I understand we have a
> "standard" partial derivative. We write (using latex): \frac{\partial^2
> f}{\partial x^2}.
>
> Now, off diagonal, we write: \frac{partial^2 f}{\partial x \partial y}.
>
> There are two things I do not understand:
> 1. how to manipulate the "partial" signs. Are they "multiplied", which
> explains the ^2? I thought the partial sign was just some convention to
> say "partial"
> 2. What does it mean to have \partial x \partial y in the denominator?
It might be helpful to use a different notation: let D_1 denote the
derivative with respect to the first variable (x in this case) and D_2
the derivative wrt the second variable (y). Then g(x,y) = D_1 f(x,y)
is, in general, still a function of both x and y, and as such will have
derivatives with respect to x and y. Its y-derivative is D_2 g(x,y) =
D_2 D_1 f(x,y). There is a theorem that under certain broad conditions,
D_2 D_1 f = D_1 D_2 f, and so the off-diagonal elements of the Hessian
are equal; that is, the Hessian matrix is symmetric. (There was an
extensive recent thread dealing with this issue in this forum.)
> Could someone please give me an example? Do I take the partial with
> respect to x? to y? to both?
Example: f(x,y) = exp(2x)*y^3. Then D_1 f = 2*exp(2x)*y^3 and D_2 f =
3*exp(2x)*y^2.
Note that D_2D_1 f = D_2 (2*exp(2x)*y^3) = 3*2*exp(2x)*y^2,
while D_1D_2 f = D_1 (3*exp(2x)*y^2) = 2*3*exp(2x)*y^2 = D_2 D_1 f.
R.G. Vickson
>
> Hopefully I make some sense. Thank you for your help, which will be
> greatly appreciated.
> Tom
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