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Solution of e^x = x

  1. Nov 11, 2009 #1
    Okay, before you scream x = ∞, I'm finding the complex solution to the problem.

    I'll show you my working so far, maybe you'll see something I missed.

    First let x = a+bi

    e^(a+bi) = a+bi
    e^a * e^bi = a+bi

    Applying Euler's identity
    e^a*cos(b) + ie^a*sin(b) = a+bi

    e^a*cos(b) = a
    e^a*sin(b) = b

    Simple rearranging;
    [1] cos(b) = a/e^a
    [2] sin(b) = b/e^a
    [3] tan(b) = b/a

    Using the identity;
    cos^2(b) + sin^2(a) = 1

    It follows that..
    (a/e^a)^2 + (b/e^a)^2 = 1
    [4] a^2 + b^2 = e^2a

    Okay so I have these 4 equations and I still can't find solutions to any of them,
    I only need to find a or b and the solution to e^x = x will follow.
    Could you please help?

    +I'm only a college student and haven't done much uni level maths, so go easy on me if I've missed something blindingly obvious.

    Also, would analysis of the series of e^x help? (Just sprung into my mind as I was about to submit thread)



    EDIT: A solution can be found using Lamberts W-function, x =~ 0.318 + 1.337i, you can delete this thread if you want
     
    Last edited: Nov 11, 2009
  2. jcsd
  3. Feb 5, 2010 #2
    expand e^x as an infinite series and equate it to x:

    x=1+x+x^2/2!+x^3/3!+...

    or subtracting x from each side:

    0=1+x^2/2! +x^3/3! + x^4/4! ...

    This is an infinite degree equation and has an infinite number of solutions.
    I hope this helps.
     
  4. Feb 5, 2010 #3
    This is not true

    0=e^x=1+x+...

    Is too an "infinite degree" equation and has no solution at all.
     
  5. Feb 5, 2010 #4

    Mentallic

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    No real solutions. But there are infinite complex solutions.

    But again this is only because [itex]r.cis(\theta)=r.cis(\theta+2n\pi)[/itex] for all integers n.
     
  6. Feb 5, 2010 #5

    CRGreathouse

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    This is the only (finite) solution I find.

    0.3181315052047641353126542515876645172035176138713998669223786062294138715576... + 1.3372357014306894089011621431937106125395021384605124188763127819143505313612...i
     
  7. Feb 5, 2010 #6

    Mentallic

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    What the hell was the point of that post?
     
  8. Feb 5, 2010 #7
    Did you use the W-function to get that result? I've been hoping to find an alternate method to solve this which could maybe lead to a closed form expression.
    As of now, the expansion for e^x isn't getting me very far.

    Also, not that it's of much use, e^x=x also implies e^e^e^...e^x = x = logloglog...logx
     
  9. Feb 5, 2010 #8
    Well I don't think that infinite solutions are what the poster is aiming for.
     
  10. Feb 5, 2010 #9

    CRGreathouse

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    1. Verifying marcusmath's claim
    2. Providing more decimal places
    3. Noting that I could find no others

    That seems like two points more than the minimum (and one more point than this post).

    No, I used the secant method.
     
  11. Feb 5, 2010 #10

    CRGreathouse

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    Mentallic meant "infinitely many solutions", not "solutions involving infinities".
     
  12. Feb 5, 2010 #11

    Mentallic

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    Is a post necessary if the points containing it are redundant? Face it, marcusmath seemed to have no uncertainty in the finite solution he found and your post has no added value to this thread (except point 2, that was very helpful) :tongue:

    What CRGreathouse said.
     
  13. Feb 6, 2010 #12

    D H

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    You didn't look very hard. 0.318131505-1.337235701i is also a solution. In general, if z=x+iy is a solution to z-exp(z)=0, then so is z*.

    Here are a few more solutions:
    2.06227773±7.588631178i
    2.653191974±13.94920833i
    3.020239708±20.27245764i
    3.287768612±26.5804715i
     
  14. Feb 6, 2010 #13

    Hurkyl

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    Since ex - x has an essential singularity at infinity, we can invoke Picard's theorem:
    In any neighborhood of infinity, with at most one exception, ex-x takes on every complex value infinitely often​

    IMO, it's unlikely that 0 just happens to be the one exception, so we would expect infinitely many solutions.
     
  15. Feb 6, 2010 #14

    uart

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    Hi Marcus, did you want to use LambertW to find more solutions?

    I assume you've already worked out that the solution can be expressed as x=-W(-1) right. W() has multiple branches and most implementations of this function will allow you to include a second integer parameter to select the branch. Both Octave and MatLab for example use something like "lambertw(n,x)", where "n" is used to select different branches.

    For example :
    Code (Text):

    > lambertw(-1,-1)
    ans = -0.318131505204764 - 1.337235701430689i

    > lambertw(0,-1)
    ans = -0.318131505204764 + 1.337235701430690i

    > lambertw(1,-1)
    ans = -2.06227772959828 + 7.58863117847251i

    > lambertw(2,-1)
    ans =  -2.65319197403870 + 13.94920833453321i
     
    Hope that helps.
     
  16. Feb 6, 2010 #15
    Ah thanks, I never really knew what the first parameter was in matlabs lambertw function.
    Out of interest, is there some closed form expression for W(x)?
    I found the taylor expansion [tex]W_0 (x) = \sum_{n=1}^\infty \frac{(-n)^{n-1}}{n!}\ x^n[/tex]
    but it doesn't have an imaginary part so surely can't be valid for when [tex]W_0 (x)\notin\Re[/tex]?
     
  17. Feb 6, 2010 #16

    Hurkyl

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    The source where you found the Taylor series should have told you where it's valid -- or at least given you the radius of convergence. But... you can compute that yourself, can't you? I expect it to be a straightforward calculation.
     
  18. Feb 6, 2010 #17

    Mute

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    Hrm? e^z = 0 has no solutions, real or complex. The clause in Picard's theorem "...attains every value infinitely often with at most one exception" is usually demonstrated using e^z, which has an essentially singularity at infinity but you will never find it to be equal to zero anywhere near there (or anywhere else, of course).
     
  19. Feb 6, 2010 #18

    CRGreathouse

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    No, not for any reasonable definition of "closed form" anyway.
     
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