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F(f(x)) = exp(x)

  1. Aug 31, 2007 #1
    If f(f(x)) = exp(x), then what is f(x)?

    I don't think that f(x) can be an entire function. It would be nice to have f(x) be monotonically increasing for positive x. I tried to attack this problem by using a method in the book "Bypasses", and it works, but it involves numerical calculation of a power series and reversion of the series, and applying the two power series together, and seems very inelegant and I'm not even sure if it will converge.
     
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  3. Aug 31, 2007 #2
    if f(x) has inverse, then you can get it in high school, I think.
     
  4. Aug 31, 2007 #3

    Hurkyl

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    Isn't it enough to simply pick some a < 0, choose f to be a bijection from [-inf, a) to [a, 0), and then extend f by the functional equation?
     
    Last edited: Aug 31, 2007
  5. Aug 31, 2007 #4
    >if f(x) has inverse, then you can get it in high school, I think.

    I'm trying to find out a more direct way of expressing the unknown function f(x) -- not focusing on a particular value of x or known function f.

    >choose f to be a bijection

    Yes, the whole problem is how to "choose" f that satisfies the functional equation.

    Here is one idea on how to progress,

    f(f(a)) = exp(a) = a, at certain constants a; one such number being ln ln ln ln ... ln -1.

    Therefore

    (f^2n)(a) = a for n = 0, 1, 2...

    and it's reasonable to insist that

    (f^z)(a) = a for any complex number z

    Differentiating the function equation gives

    f(f(x)) = exp(x) = d/dx exp(x) = f'(x) f'(f(x))
    f(f(a)) = a f'(a) f'(f(a))

    So that f(a) = a, f'(a) = a^0.5

    I can see developing this into a power series expanded about a. Not sure what the radius of convergence is, or if it gives "nice" positive numbers for positive arguments.
     
    Last edited: Aug 31, 2007
  6. Aug 31, 2007 #5

    Hurkyl

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    I claim it's not a problem: any choice works. Even better, if you choose the bijection of [-inf, a) with [a, 0) to be monotone, then the resulting f is monotone.
     
  7. Aug 31, 2007 #6
    >I claim it's not a problem: any choice works.

    Um, f(x) = x + 3 is not a good choice: x + 3 + 3 = exp(x)?

    I'm trying to find a choice that satisfies the equation.
     
  8. Aug 31, 2007 #7

    Hurkyl

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    What a did you choose? And f(x) = x+3 is not a bijection from [-inf, a) to [a, 0) for any choice of a.
     
  9. Aug 31, 2007 #8
    >And f(x) = x+3 is not a bijection from [-inf, a) to [a, 0) for any choice of a.

    How about a = -1 and f(x) = -1 - 1/x? I don't see where that gets you.

    Going back to the power series idea (where the constant a i used is ln ln ln ... -1 )

    d^2/(dx)^2 f(f(x)) = f''(x) f'(f(x)) + f'(x)^2 f''(f(x))
    a = f''(a) f'(a) + f'(a)^2 f''(a) = f''(a)[ f'(a) + f'(a)^2] = f''(a)[ a^0.5 + a]
    f''(a) = 1/(1+a^-0.5)

    So

    f(x) = a + (x - a) a^-0.5 + (x - a)^2 / [(1 + a^-0.5) * 2!] + ...
     
    Last edited: Aug 31, 2007
  10. Aug 31, 2007 #9

    Hurkyl

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    Okay, so that's how f is defined on [-inf, -1). Now use the functional equation to figure out what f should be on [-1, 0)...
     
  11. Aug 31, 2007 #10
    >Okay, so that's the piece of f defined on [-inf, -1). Now use the functional equation to figure out what f should be on [-1, 0)...

    I guess, for y on [-1,0) you can say that f(y) = exp(-1/(1+y)):

    f(f(x)) = exp(x)
    f(y) = exp(f^-1(y))
    f(y) = exp(-1/(1+y))
     
  12. Aug 31, 2007 #11

    Hurkyl

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    I think that's right. And then you can figure out what the value of f should be on [0, e^a)...

    Basically, I'm trying to define f(x) piecewise. I know that f(f(-inf)) = 0, so I choose some value
    f(-inf) = a
    and then I immediately know that i have to choose
    f(a) = 0, f(0) = e^a, f(e^a) = 1, et cetera. Eyeballing this sequence, it appears plausible I can repeat this procedure if I start with any arbitrary increasing bijection [-inf, a) --> [a, 0), and maybe even with any bijection.
     
  13. Aug 31, 2007 #12
    I see now. Thanks Hurkyl!

    For [-inf,a), f(x) = f0(x) = a - aa/x
    For [a,0), f(x) = f1(x) = exp(f0^-1(x)) = exp(aa / (a - x))
    For [0,exp(a)), f(x) = f2(x) = exp(f1^-1(x)) = exp( a - aa / ln x )
    For [exp(a), exp(exp(a))), f(x) = f3(x) = exp(f2^-1(x)) = ...
    etc...

    The only problem is that there's a lot of arbitrariness and its piecewise, and I wonder if it can be a "nice" (differentiable) function. And also satisfy f(c) = c for constants c where exp(c) = c.

    On the other hand, my power series probably doesn't converge very well, and there are many constants c to choose to expand about. I wonder if it could help suggest a nice bijection.

    It seems like a rather unnatural problem overall.
     
    Last edited: Aug 31, 2007
  14. Aug 31, 2007 #13

    Hurkyl

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    Well, if you have additional constraints on f, you can attempt to translate those into conditions on your original choices. You could easily get f to be infintiely differentiable (even analytic!) on the interior of each interval, and it's easy enough to make it infintiely differentiable at a. Does that imply it's everywhere differentiable?

    exp doesn't have a fixed point, but if it did, I think the method I described would never reach c: it would define a function [-inf,c) to [-inf,c). You would have to make another choice to get an entire function.
     
  15. Sep 4, 2007 #14
    >exp doesn't have a fixed point

    There are complex, not real, fixed points. If you take -1 (or some other starting point) and apply the ln function repeatedly, it converges to some complex number c for which exp(c) = c. You can find infinitely many other fixed points by applying (2Npi + ln) instead of ln.
     
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