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Inverse Function (pt1)

  1. May 7, 2008 #1

    My 2 questions are underlined below. I have another problem but will first wait for an answer on this before posting the other question.

    Problem Statement

    Consider the function f:R2->R2 defined by
    f(x1,x2) = ( exp (x1-x2) + x[tex]^{2}_{1}[/tex]x2 + x1(x2-1)[tex]^{4}[/tex],
    1 + x[tex]^{2}_{1}[/tex] + x[tex]^{4}_{1}[/tex] + (x1x2)[tex]^{5}[/tex] )

    Q1- Find Df(x1,x2) at point (1,1) and show that there are open sets (1,1)[tex]\in[/tex] U and f(1,1)[tex]\in[/tex] V such that f:U->V has an inverse g

    Q2- Compute Dg(f(1,1))


    First, I can compute the matrix Df(x) by taking the partial derivatives wrt x1, x2 and then take its determinant evaluated at x= (1,1).
    If I evaluate it at (1,1) it comes out to A = [3 0 ; 11 5]

    Second, the determinant of A is 15 which is != 0. So an inverse exists at this point.

    My question for Q1 above: is it sufficient to prove that if det !=0 then an inverse exists. Is this all the question is askign. What does the question mean when asking to show there are open sets (1,1) [tex]\in[/tex] U ....

    My question for Q2 above: Dg(f(1,1)) = inv(A) evaluated at (1,1). Is this correct


  2. jcsd
  3. May 7, 2008 #2


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    Q1 is just the inverse function theorem.
    Q2 is a special case of the chain rule.
  4. May 7, 2008 #3
    For Q2:

    For a chain rule, don't I need a matrix for g. How do I get this?

    Is this g matrix, the inverse of f evaluated at (2,4) <-- I get this from f(1,1) = (2,4)
    And multiply this by Df(1,1)


  5. May 7, 2008 #4


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    Well, you can formulate the chain rule in terms of the jacobian evaluated at p.

    J(f○g)(p) = J(g)(f(p))*J(f)(p)
  6. May 7, 2008 #5
    Thanks... yes I think got the Jacobian part.
    However, what I am not getting is how do I get 'g'.

    BTW, I assume you are saying p=f(1,1)


  7. May 8, 2008 #6


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    No, my p in your case is (1,1).

    What happens on the left side when g is f's inverse?
    What can you then do to solve for J(g)(f(p))?
  8. May 8, 2008 #7
    When g is f's inverse, all I have to do is take inverse of f matrix evaluated at (1,1). This will give me the Jg(f(p)). Is this right?
    What I had missed was the Jf(p)


  9. May 8, 2008 #8


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    If g is f's inverse, then J(f○g)(p) = Id
    Id = J(g)(f(p))*J(f)(p)
    J^(-1)(f)(p) = J(g)(f(p))
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