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April 11th, 2007, 09:01 PM   #1
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Inner product product rule

Hi, how would one deduce the derivative product rule for the general inner product of functions? I was told to use chain rule, but I don't see how I could do that.
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April 11th, 2007, 11:45 PM   #2
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Do you mean you want to differentiate fg , or the scalar product of f and g, namely <f,g> ?
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May 4th, 2007, 08:03 AM   #3
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indeed I mean <f,g>.
Wow haven't been back for awhile!
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May 4th, 2007, 07:47 PM   #4
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Given a normed real vector space E and a scalar product <.,.>: E x E -> R (which is continuous for the euclidean norm by Cauchy-Schwarz inequality), by computing:

<x+h1,y+h2>=<x,y>+<x,h2>+<h1,y>+<h1,h2> you realize that:
<x+h1,y+h2>-<x,y>-f(h1,h2)=o((h1,h2)), where f(h=(h1,h2))=<x,h2>+<h1,y> is linear and continuous in h, and therefore is the derivative at (x,y) of our scalar product. We have that <h1,h2>=o(h) because |<h1,h2>| <= ||h1||.||h2|| by Cauchy-Schwarz.

This is a standard result for all continuous bilinear applications, and can be generalized to multinilear applications by induction.

If you are trying to compute the derivative of g: x-><x,x> at point h, then you find Dg(x)(h)=2<x,h>

If you are trying to compute the derivative of the norm: N: x->sqrt(<x,x>), then using the Chain rule, you find that DN(x)(h)=<x,h>/N(x).
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May 4th, 2007, 11:26 PM   #5
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Oh I see... so you mean f(h) is the derivative operator, so then f(h)(x) is the derivative at x?
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May 5th, 2007, 12:31 AM   #6
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I should have typed f_(x,y)(h1,h2). The derivative of f at point (x,y) (evaluated at (h1,h2)).

Usually the easiest way to evaluate the derivative of a function say g at point x is to compute limit( [g(x+a*h)-g(x)]/a, a->0 ). This gives you the candidate Dg(x)(h). You then have to verify that Dg(x) is indeed linear continuous (the continuity always holds in finite dimension) and that g(x+h)-g(x)-Dg(x)(h)=o(h) in order to conclude that Dg(x) is the derivative of g at point x. This method happens to be really useful for complicated expressions, for instance calculating the derivative of the Determinant.

For instance, what is the derivative of the application: Det: Mat(R^n)->R given by Det(M)= determinant of M for every matrix M ?
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May 6th, 2007, 04:27 PM   #7
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So... the Det: operation is a function... and you mean the derivative is taken with respect to these functions? I guess I could use chain rule??? Ummm.. (looking at the ceiling lol).
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