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Prove that a function is the quadratic form associated to

  • Thread starter Andrés85
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  • #1
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Homework Statement



Let G:R2[itex]\rightarrow[/itex]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

The Attempt at a Solution



G is C2, 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)=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).
 

Answers and Replies

  • #2
HallsofIvy
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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
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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
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I solved the problem deriving two times the function f(t) = G(tx, ty). Thanks.
 

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