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Implicit function Theorem

  1. Feb 7, 2012 #1
    Can anyone check if this argument works. I just made up this problem to check if I understand whats going on.

    Consider F = xy*e^x + y*e^y = K. I want to see if there is a unique solution.

    Fy = x*e^x + y*e^y + e^y.

    Since we are on the surface x*e^x + y*e^y = K.

    So if K >= 0 then Fy ≠ 0. So given any x we can find a unique y.

    Did I understand the theorem correctly?
     
  2. jcsd
  3. Feb 7, 2012 #2

    Char. Limit

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    You're missing a few things here. Firstly, if [itex]F = xy e^x + y e^y = K[/itex], then [itex]F_y = x e^x + (y+1) e^y = 0[/itex]. That's an implicit function. And secondly, given any x, there's not necessarily a unique y. For example, take x=0.05. For this point x, there are two solutions to y satisfying the derivative F_y (namely y=-4.06616... and y=-1.16923...).

    Is that what you mean?
     
  4. Feb 11, 2012 #3
    deluks, you are correct.

    The implicit function theorem implies that F=K is LOCALLY solvable for y as a function of x. That is, for each point there exists a neighbourhood of that point where y can be written as a function of x. In this case, F is only 2 variables, so "Locally solvable everywhere" is equivalent to "globally solvable" and so your conclusion is correct, but this would not be valid if F had more than 2 variables.

    To the other poster: He is referring to the Curve F=K having a unique x for each y, not anything about the derivative.
     
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