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System of complex functions

  1. Nov 14, 2012 #1
    Hi all,

    I posed this problem to my calculus teacher a few days ago and we have not been able to come close to solving it thus far. The problem is to find the intersection of the solids (complex functions require use of the 4d space, so I assume that the function would be a solid) f(z)=e^z and g(z)=z (where z is a complex number).

    What I managed to show was that this system is equivalent to the complex intersection between ln(z) and e^z. This is quite simple:

    e^z=z
    ln(e^z)=ln(z)
    z=ln(z)

    by basic substitution, a new system with equivalent solutions is born: e^z=ln(z).

    Furthermore, raising both sides of this equation to a power of e results in z=e^(e^z). Thus, this new function is equivalent to ln(z), and one can continue this process infinitely by replacing every z with e^z.

    Any ideas? Maybe there is a theorem regarding the intersections of inverse equations
     
  2. jcsd
  3. Nov 14, 2012 #2

    Mute

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    Homework Helper

    I'm sorry if this gives away more than you were hoping for, but there is a solution in terms of a special function. Read on if you are interested. I haven't given away all the details, I just tell you about the special function but leave it for you to try and solve the equation in terms of the special function.










    For complex numbers z, there is a solution in terms of a special function called the Lambert-W function, ##W_n(u)##. This special function is defined by solving ##u = w \exp(w)## for ##w = W_n(u)##. The index n indicates a "branch" of the Lambert-W function (similar to how when solving ##u = sin(w)## you have solutions ##w_n = \arcsin(u) + 2n\pi##).

    See if you can manipulate your equation ##z = e^z## into the form ##(\mbox{some number} = (\mbox{something} \times z) \exp((\mbox{something} \times z))##, in which case your solution will be ##z = W_n(\mbox{some number})/\mbox{something}##.

    In fact, the infinite tower

    $$z^{z^{z^\ldots}}$$

    can also be written in terms of the Lambert-W function (though for complex numbers z it is a definition). For real numbers, the infinite tower only converges to a non-infinite real number on a small range of z. The two approaches will agree for your case of z = e.
     
  4. Nov 16, 2012 #3

    Edit: Sorry for my ignorance. From the equation e^z=z, it is simple to show that this is an equivalent statement as -z*e^-z=-1.

    This means that z=-Wn(-1)... but what is this value? And what "branch" of the W function are we working with? Sorry that I don't really understand this; I am merely in Calculus.

    By the way, where might I go to learn how to type some basic math symbols? (e.g. integrals, exponents, fractions, etc)
     
    Last edited: Nov 16, 2012
  5. Nov 16, 2012 #4
    By the way, on the wikipedia article, it is noted that W(-1) is approximately -.31813-1.33723i, so my answer is just the negative of this. What is the n value in this case? Also, where do these values come from? Is there an exact value for w(z) when z is a real? How about w(a+bi)?
     
  6. Nov 16, 2012 #5

    Mute

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    Homework Helper

    If you go to www.wolfraalpha.com and enter "-ProductLog[n,-1]" in the search bar (without the quotation marks), where n is the "branch", it will evaluate the W function for the different branches, so you can see what some of the values are. In a problem like there, there is no one correct branch. There are simply infinitely many possible solutions to the equation z = exp(z).

    Here's a link for ##-W_0(-1)##: Link.

    There are some special values of z real for which W(z) has an exact value. Some are listed on the wikipedia page. In general, the solution is a complex number with no known exact expression.

    Only the branches n = 0 and n = -1 give real valued outputs for real valued inputs.

    You'll find that everything you try gives you a complex number. As you probably know from looking at plots of x and exp(x), there is no intersection so there is no real valued answer to z = exp(z). Also, the "power tower" that you derived, for real numbers z = x,

    $$x^{x^{x^\ldots}}$$
    only has a (finite-valued) solution on ##e^{-e} \leq x \leq e^{1/e}##. Since x = e is not in this range, there is no real solution.

    To learn how to type in ##\LaTeX## (the equation typesetting language used on this board), this pdf will teach you some basic commands. You can also see how people write their equations by quoting their post.

    Note that for inline equations, [ itex] stuff [ /itex] is equivalent to # # stuff # # (without spaces betweeen the hashtags or tags) and for full big equations [ tex] stuff [ /tex] is equivalent to $ $ stuff $ $ (without spaces between the dollar signs or tags).
     
    Last edited: Nov 16, 2012
  7. Nov 16, 2012 #6
    Thanks. So is it impossible to calculate exact values of Wn(z) for all numbers? By looking at the link you posted, it appears that the values of Wn(z) are calculated by an improper integral. Is this the case, similar to the gamma function?
     
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