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X with respect to Y?

  1. Apr 16, 2007 #1
    How do I find X with respect to Y if;

    Y = X + e^X?

    If I log it then I get Log Y = Log X + X, if this is the correct step how do I isolise X from here?
  2. jcsd
  3. Apr 16, 2007 #2
    Taking the log of the right-hand side is not log x + x, but instead log(x + e^x).
  4. Apr 16, 2007 #3
    ****, stupid mistake of mine :-).

    I still can't see that helping me at all though :-(.
  5. Apr 17, 2007 #4


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    It's not always possible to explicitely invert a function.
    Last edited: Apr 17, 2007
  6. Apr 17, 2007 #5
    So I can implicitly get one?

    y' = 1+e^x
    y'' = e^x

    y - y'' = x?
  7. Apr 17, 2007 #6


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    The equation that implicitly defines X as a function of Y is the original equation. Y = X + e^X. The fact that Y(X) is in bijection with its image assures you that X(Y) exists.

    It's not the case with say Y(X)=X² because X(Y) would be 2-valued: ±[itex]\sqrt{Y}[/itex].

    But your idea with differentiating bth sides was good. However, I think it's more interesting to acknowledge X(Y) exists, and then differentiate the equation wrt Y! I get



    This way you have a differential equation for X(Y), or inversely, an integral equation.


    I don't know if something can be done with those, but there's probably a way to at least extract some information or approximation. For instance, we know from Y(X) that when Y is around 1, Y is small, so you can keep only the first few terms in the series of exp(X(Y)) and integrate to get an approximation of the form of X(Y) in the nbhd of 1.
  8. Apr 17, 2007 #7

    matt grime

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    what on earth does that even mean?
  9. Apr 17, 2007 #8
    I hope this doesn't get into semantics.
  10. Apr 17, 2007 #9


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    May we know where this problem of yours comes from?

    P.S.This is nonsense, don't pay attention:

  11. Apr 17, 2007 #10
    Got to choose from 5 different questions to do animation type exercises in Maple. Chose number 5 but we had problems restricting the domain in the 2nd function to the time in the first function. The first function was in the form of y = t +e^t. We decided it was too hard and started on a different, easier, question. But it still bugged me that we couldn't get a function for t(y). Obviously we weren't supposed to do it like that, good idea we chose an easier one then eh? :-).
  12. Apr 17, 2007 #11

    Gib Z

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    y=x + e^x
    dx/dy = 1/(1+e^x)
  13. Apr 18, 2007 #12


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    1. Subt z=e^x to get : y = z + ln(z)

    2. Exponential each side to get : e^y = z e^z

    3. From the definition of the LambertW function : z = LambertW( e^y)

    4. So x = ln( LambertW( e^y ) )
    Last edited: Apr 18, 2007
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