Second differential of a function

  • Thread starter conquest
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  • #1
conquest
133
4
Hey everybody,

I was reading in Singularities of differentiable maps by Arnold, Gusein-Zade and Varchenko and I didn't even get through the first page before being confused.

My confusion concerns the sentence about the second differential of a function. This is supposed to be a quadratic form. So in what way should I view this since this should also just work for single variable functions.

I keep going back to either d(df)=0. Somehow I must be looking for the second derivative in some form.

I guess my question is just, what exactly is the second differential of a function (in what way is it a quadratic form)?

Cheers
 

Answers and Replies

  • #2
Muphrid
834
2
While it may not be as general, a scalar field [itex]\psi[/itex] obeys

[tex]\nabla \wedge \nabla \psi = 0[/tex]

In 2d, this reduces to

[tex]\left( e^x \partial_x + e^y \partial_y\right) \wedge \left(e^x \partial_x \psi + e^y \partial_y \psi \right) = e^x \wedge e^y \left( \partial_x \partial_y \psi - \partial_y \partial_x \psi \right) = 0[/tex]

This is trivially satisfied by the equality of mixed partial derivatives. In one dimension, you don't have two basis vectors, and there is no wedge product to speak of at all--I've heard of the wedge product referred to as an antisymmetric quadratic form, but that's started to get out of my area of familiarity.
 
  • #3
Vargo
350
1
I dont' know this book, but they are probably referring to the second order part of the Taylor expansion. This is a quadratic form that is composed of second derivatives, and it is the next best local information about a function once you know its value and its derivatives at a point. In other words, they are talking about the Hessian matrix.
 

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