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Implicit and inverse function theorem

  1. Nov 22, 2005 #1
    Lately, we've been going over these two theorems in class. I have a few questions to put forth.

    1) I know that in lower spaces, an inverse of a function exists locally (say around a point G) if it does not attain it's max/min at G (i.e. if f'(G) doesn't equal 0). Now, with the inverse function theorem, we have that det| Df | doesn't equal 0 (where Df is the matrix of partial derivatives?). I'm unsure how the determinant of the matrix relates to max/min at a point- So i can relate the two ideas.

    2)I know that the inverse function theorem implies the implicit function theorem, and have seen proofs of it, but how can we prove that the implicit function theorem implies the inverse function theorem? (any suggestions on how to go about this proof would be appreciated)

    3)In the implicit function theorem, where we consider a vector in R^m as [x y], x in R^(m-n), y in R^n, why is it that we only need the the determinant of DF/DY to not equal 0? Is this where the inverse function theorem comes in (to create a new function G s.t. G(x)=y?). What exactly is that matrix DF/DY? Also, why does there need exist a solution to the function?

    4) In both theorems, why is it required to have f continuously differentiable?

    Sorry if I'm asking too many questions, as I know some may flame me for this fact, but thanks in advance as any help is very appreciated.
     
  2. jcsd
  3. Nov 22, 2005 #2

    Hurkyl

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    For 1) and 3), when you have differentials, consider differential approximation!

    [tex]
    f(x) \approx f(x_0) + (Df)(x_0) \cdot (x - x_0)
    [/tex]

    Intuitively, this means that f looks like a linear function, if you zoom in far enough.

    Or, if you like, you could only approximate on some variables:

    [tex]
    f(x, y) \approx f(x_0, y) + L(x_0, y) \cdot (x - x_0)
    [/tex]

    What ought L be?


    So now, you have to apply your geometric insights! A very interesting number here is the rank of Df. (And you thought linear algebra didn't have any practical use!) If you knew the rank of Df at a point, (approximately) what can you say about the image of f near that point?

    (I'll stop with this hint, to give you a chance to work it out)

    In the case that Df is square, asking for a nonzero determinant is nothing more than asking for Df to be full rank.

    Also, you should note that, in one dimension, f'(x) = 0 does not mean that f attains a local extremum at x.


    For 2, I would see if I could write an implicit expression involving what ought to be the inverse of your function.


    For 4, I would brainstorm for interesting functions that aren't continuously differentiable, and see if I could come up with a counterexample. Looking for places in the proof that use the continuity of the derivative might suggest ways you want your counterexample to behave.
     
    Last edited: Nov 22, 2005
  4. Nov 24, 2005 #3
    Thanks for your help sir. You have cleared up a bit for me. However, I guess I'm not as well familiar with the topics as I should be, as most of what you talked about is confusing to me. Perhaps after I read up on both theorems a little more, I will get it. Nontheless, thanks a bunch, again.
     
  5. Nov 24, 2005 #4

    Hurkyl

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    I remember that when I came out of linear algebra, all "rank" meant to me was some useless number one could compute from a matrix!


    Anyways, the important thing is to try and figure out what the image of the function looks like near a point by considering what the image of the differential approximation looks like.
     
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