Basic question about numerical hessian

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This discussion centers on computing the Hessian matrix for a 1-D signal represented as a vector v = [x_1, x_2, x_3, x_4...]. The numerical Hessian is defined as h = [0, (x_1+x_3 -2x_2), (x_2+x_4-2x_3), 0]. The user, Luca, seeks clarification on the transformation matrix H that allows the expression h = v * H * v' to hold true. The conversation highlights the distinction between the scalar output of the quadratic form v H v^T and the vector nature of h.

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pamparana
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Hello all,

Suppose I have a simple 1-D signal and I want to compute the hessian. In that case, it should generalise for second derivative for normal scalar functions.



So, I observe the signal as [itex]v = [x_1, x_2, x_3, x_4...][/itex]. Then, numerically the hessian is given as (assuming I am only conputing it at interior points):

[itex]h = [0, (x_1+x_3 -2x_2), (x_2+x_4-2x_3), 0][/itex].

Now, according to the document for example here (http://planetmath.org/HessianMatrix), I should be able to write this as:

[itex]h = v * H * v'[/itex]. Where H is some transformation. I am trying to figure out what this H should be for my simple 1-D case without any luck.

I would greatly appreciate any help anyone can give me with computing this Hessian operator.

Thanks,
Luca
 
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pamparana said:
Now, according to the document for example here (http://planetmath.org/HessianMatrix), I should be able to write this as:

[itex]h = v * H * v'[/itex]. Where H is some transformation.

The page at PlanetMath says [itex]v H v^T[/itex] is a quadratic form, so it is a scalar valued function. The thing you are calling [itex]h[/itex] looks like a vector.
 

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