# Prove that a function is the quadratic form associated to

1. Feb 19, 2014

### Andrés85

1. The problem statement, all variables and given/known data

Let G:R2$\rightarrow$R be a C2 function such that G(tx,ty)=t2G(x, y). Show that:

2G(x,y)=(x,y).HG(0,0).(x,y)t

3. The attempt at a solution

G is C2, so its Taylor expansion is:

G(x,y) = G(0,0) + $\nabla$G(0,0).(x,y) + $\frac{1}{2}$(x,y).HG(c).(x,y)t,

where c lies on the line segment that goes from (0,0) to (x,y).

Using that G(tx,ty)=t2G(x,y) I get that G(0,0) and the linear term equals 0.

The problem is that I have HG(c) in the quadratic term, but I need HG(0,0).

2. Feb 20, 2014

### HallsofIvy

Staff Emeritus
If c is not (0,0) then what do you mean by "c" and "HG(c)"? You are taking the Taylor expansion at (0, 0), are you not?

3. Feb 20, 2014

### Andrés85

I can't increase the degree of the Taylor polynomial because G is C2, so the second degree term is the remainder written in matrix notation.

HG(c) is the Hessian matrix of G evaluated at c, where c lies on the segment that goes from (0,0) to (x,y).

4. Feb 20, 2014

### Andrés85

I solved the problem deriving two times the function f(t) = G(tx, ty). Thanks.