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Integration function homework question

  1. Dec 25, 2007 #1

    Typing correction: g maps to R^n, and id_R should be id_R^n
    Here id_R^n means the identity function on R^n. So the condition is saying g(x,b(x))=x.

    Any ideas how to do this one? Should I begin with the substitution y=g(x,t)?
    Last edited: Dec 26, 2007
  2. jcsd
  3. Dec 25, 2007 #2

    Gib Z

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    I don't really know what an identity map on R actually is, but otherwise I can get it down to [tex]f(x) = f( g(x, \beta (x))[/tex].

    EDIT: Am I right in assuming that it is sort of like the identity function h(x) = x, but for only the first variable? Basically I'm asking, if that condition the same as; [tex]g(x,\beta (a) ) = x[/tex] as long as [tex] x,a \in \mathbb_{R}[/tex]. If so, then my answer reduces down to f(x).
    Last edited: Dec 25, 2007
  4. Dec 25, 2007 #3
    I've retyped the condition g above.

    So if the substitution y=g(x,t) is made, then when t = b(x), we have
    y=g(x,b(x))=x by the condition on g.

    Now f(g(x,t))=f(y) is now a function of y only, but y itself is a function of n+1 variables x_1,...,x_n, t. So now the chain rule when we take df(y)/dt will be ......
    Last edited: Dec 25, 2007
  5. Dec 25, 2007 #4

    Gib Z

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  6. Dec 25, 2007 #5
    But Leibnit'z rule involves taking the derivative wrt, say, x, with the integral is wrt, say, t. That's not what's happening here in this question. The integral is wrt to t and the partial derivative also wrt to t. Leibnitz's rule does not apply here.

    Also, I don't think we can just cancel out the two dt 's in the integrand because they are not the same dt 's. I think more rigour is required to PROVE the result. I'm attempting to apply the chain rule here:

    f(g(x,t))= (fog)(x,t) so by the chain rule, (fog)'(x,t) = f'(g(x,t)*g'(x,t) yields a 1x(n+1) matrix and we are interested in the component that gives df(g(x,t))/dt only. ......
    Last edited: Dec 25, 2007
  7. Dec 25, 2007 #6
    Using the chain rule, I get df(g(x,t))/dt to be a messy sum of n terms.

    I need to know: Is [Df(g(x,t))/Dt]dt = df(g(x,t)), if the first D is partial derivative and the second d ordinary derivative ? If so, how do I prove that?
    Last edited: Dec 25, 2007
  8. Dec 26, 2007 #7
    I believe the answer is yes because in this case [Df(g(x,t))/Dt]dt is the integrand, and we are integrating with respect to t only. Thus we can view x as a constant (within the integral) and thus the partial derivative D/Dt can be treated as an ordinary derivative d/dt. Am I right?
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