- #1

Andrés85

- 10

- 0

## Homework Statement

Let G:R

^{2}[itex]\rightarrow[/itex]R be a C

^{2}function such that G(tx,ty)=t

^{2}G(x, y). Show that:

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

^{t}

## The Attempt at a Solution

G is C

^{2}, so its Taylor expansion is:

G(x,y) = G(0,0) + [itex]\nabla[/itex]G(0,0).(x,y) + [itex]\frac{1}{2}[/itex](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)=t

^{2}G(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).