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Proof/Disproof involving multi-variate functions

  1. Nov 22, 2009 #1
    1. The problem statement, all variables and given/known data

    Let some function, called f(u,v), be a differentiable function. Then, let its first partial with respect to u = 0 in some region, A. Then, for any u in the region A, f(u,v) will always equal itself for, let's say, f(ui, v) = f(uj, v).

    2. Relevant equations


    3. The attempt at a solution

    Alright, so it's obvious the function, f(u,v) really only depends on the variable v. The problem is then showing this, which I'm having a hard time doing. At first I tried to come up with a function f(u,v) that depends only on v but that didn't really fly.

    But if the first partial with respect to u equals zero, then isn't there no u-variable in the function? So wouldn't it be something like, f(u,v) = 2v^(2) or whatever? If u can't exist in the function, then a) How is it a multi-variable function? b) Does that prove or disprove that, say, f(ui, v) = f(uj, v)?

    Is there way to graph this?

    Obviously, I'm pretty confused.
     
  2. jcsd
  3. Nov 22, 2009 #2

    LCKurtz

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    You need (ui,v) and (uj,v) to be in A. For a fixed v, if you let

    g(u) = f(u,v), you have g'(u) = 0 on [ui,uj] and you are trying to show g is constant. You can apply your calculus theorem, or if that isn't allowed, suppose g(ui) isn't equal to g(uj), and apply the mean value theorem to get a contradiction.
     
    Last edited: Nov 22, 2009
  4. Nov 22, 2009 #3

    hth

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    How would you apply MVT? I don't understand that approach.
     
  5. Nov 22, 2009 #4

    LCKurtz

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    Try writing down what the mean value theorem says using ui, uj and assuming g(ui) isn't equal to g(uj), and see if it doesn't tell you something.
     
  6. Nov 22, 2009 #5

    hth

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    Here's my attempt.

    Assume g(ui) =/= g(uj).

    Let g'(x) exist on ui < x < uj.

    So, there exists some point, c, on ui < c < uj.

    Now,

    g(uj) - g(ui) = (uj - ui)g'(c)

    g'(c) = [g(uj) - g(ui)] / [(uj - ui)]

    Note that the LHS cannot equal zero because g(uj) =/= g(ui).

    So, g'(c) =/= 0....?? Man, I'm not really sure what to do. I know I'm supposed to show g(uj) = g(ui) to get the contradiction..right?
     
  7. Nov 22, 2009 #6

    LCKurtz

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    Instead of saying "Now" you should say "such that"
    You were given g'(u) = 0 and you are trying to show g is constant. By assuming g is not constant so you can find that ui and uj, you have come up with a point c where g'(c) =/= 0. Doesn't that contradict g'(u) = 0 for all u?
     
  8. Nov 22, 2009 #7

    hth

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    Yeah, it does. g'(u) =/= 0 for all u, so g isn't constant...
     
  9. Nov 22, 2009 #8

    LCKurtz

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    No, you still aren't quite getting it. You are given g'(u) = 0 for all u and you are trying to show g is constant. By supposing g is not constant you have found a c where g'(c) isn't 0. That contradicts g'(u) = 0 for all u. So supposing g is not constant has led to a contradiction. So g must be constant.
     
  10. Nov 22, 2009 #9

    hth

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    Alright, I see it now. Ty. Also, I'm sorry if I wasn't supposed to hijack the OP's thread like this.
     
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