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A nonlinear difference equation

  1. Apr 4, 2005 #1
    I'm posting this under analysis because of the method I'm thinking about using to solve it.

    Here is the equation:
    [tex]\varepsilon \left( x\right) =e^{\varepsilon \left( x-1\right) }[/tex]
    and the initial condition is
    [tex]\varepsilon \left( 0\right) =1[/tex].

    My main goal is to consider x as a number in [0,1] but we can start with x being a natural number.

    Here's what I want to try now. I'm pretty sure [tex]\varepsilon[/tex] is not elementary so let's assume it's of the form
    [tex]\varepsilon \left( x\right) =\int_{0}^{\infty }f\left( t,x\right) dt[/tex].

    Now my question is how do i solve for f?
  2. jcsd
  3. Apr 4, 2005 #2

    matt grime

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    eps(x) is just repeated exponentiation, isn't it?

    eps(1) = e^{eps(0)} = e

    eps(2) =e^{eps(1)}= e^e

    eps(n) = e^{e^[e^...^{e}...} wit n e's in the expression.

    Note ine general the answer is

    eps(x) = e^{e^...e^{eps(floor(x)))...} with floor (x) e's in the expression.
  4. Apr 4, 2005 #3
    Thanks for the feedback but I was hoping to get a solution esp(x) where I can let x be a non-integer. My guess is the integral might be helpful in doing this, if it's like the gamma function.
  5. Apr 4, 2005 #4
    Matt's suggestion,

    [tex]\varepsilon (x) = e^{e^{\vdots^e}}[/tex]

    with [itex]\lfloor x \rfloor[/itex] [itex]e[/itex]'s in the expression works, doesn't it?

    For example,

    [tex]\varepsilon (0.5) = 1[/tex]

    [tex]\varepsilon (1.5) = e[/tex]

    [tex]\varepsilon (2.5) = e^e = e^{\varepsilon (1.5)}[/tex]

    [tex]\varepsilon (3.5) = e^{e^e} = e^{\varepsilon (2.5)}[/tex]
    Last edited: Apr 4, 2005
  6. Apr 4, 2005 #5
    yeah, it works but that's about as satisfying as defining 2.5! to be 2!.
  7. Apr 4, 2005 #6
    Alright, in that case try this:

    First define [itex]\langle x \rangle = n[/itex] where [itex]n[/itex] is the greatest integer strictly smaller than [itex]x[/itex] (ie. [itex]x \not {\in} \mathbb{N} \Longrightarrow \langle x \rangle = \lfloor x \rfloor[/itex], but [itex] x \in \mathbb{N} \Longrightarrow \langle x \rangle = \lfloor x \rfloor -1[/itex]). Then let

    [tex]\varepsilon (x) = e^{e^{\vdots^{e^{e^{x - \langle x \rangle}}}}}[/tex]

    where [itex]e[/itex] appears [itex]\langle x \rangle + 1[/itex] times. Thus

    [tex] \varepsilon(0) = 1[/tex]

    [tex]\varepsilon(0.2) = e^{\frac{1}{5}}[/tex]

    [tex] \varepsilon(0.5) = \sqrt{e}[/tex]

    [tex] \varepsilon(1) = e = e^{\varepsilon(0)}[/tex]

    [tex]\varepsilon(1.2) = e^{e^\frac{1}{5}} = e^{\varepsilon(0.2)}[/tex]

    [tex]\varepsilon(1.5) = e^{\sqrt{e}} = e^{\varepsilon (0.5)}[/tex]

    [tex]\varepsilon(2) = e^e = e^{\varepsilon (1)} [/tex]

    [tex]\varepsilon(2.5) = e^{e^{\sqrt{e}}} = e^{\varepsilon(1.5)}[/tex]


    I think you will find this function is actually pretty well-behaved... interesting!
    Last edited: Apr 4, 2005
  8. Apr 4, 2005 #7

    matt grime

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    sorry, i didn't mean floor(x) inside, I mean x - floor(x) ie the "bit after the decimal place"

    that is there is no remotely unique solution unless you specify eps(x) for all x in (0,1]
  9. Apr 6, 2005 #8
    I'm working on finding another way.

    Let f(x)=EXP(x) and note that p=W(-1) is a complex fixed point of f. Then I expanded a taylor series for the nth iterate of f about p and I got, where g is the nth iterate of f,
    [tex]g\left( x\right) =p+p^{n}\left( x-p\right) +\frac{p^{n}\left( p^{n}-1\right) }{2\left( p-1\right) }\left( x-p\right) ^{2}+O\left( x-p\right) ^{3}[/tex].

    Then, let n=a real number.

    1. I can't find a nice formula for the series
    2. I'm not sure the outputs for real x are real (they should be)
  10. Apr 7, 2005 #9
    Well...it's not even C^2 ...does somebody have any proof of the existence of an analytic solution ?
  11. Apr 7, 2005 #10

    matt grime

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    Any function from (0,1] to R yields a solution, which will be smooth on the domain R\Z if it is smooth on (0,1). If you wish to make it smooth at all points you'll need one such that it is infinitely differentiable at the the integers. Whether or not that can be done I do not wish to guess.

    If we think about it as a functional equation on C: exp(f(z)) =f(z+1), then I would be even more reluctant to state there was an analytic solution.
  12. Apr 16, 2005 #11
    Focusing on the half-iterate of e^x now...

    I rewrote the series above, which is centered at -W(-1), a fixed point of e^x, to be centered at 0. Lo and behold, the imaginary coefficients seem to be small on all coefficients (all of the five i calculated at least!).

    Here's the real part of the series for f, where ff(x)=e^x:

    f(x) = 0.50998 + 0.8876 x + 0.1720 x^2 + 0.0275 x^3 +0.0222 x^4 - 0.0018 x^5.

    Now the last term IS wrong because all derivatives should be positive. I think it came out negative to compensate for the truncation.

    Try graphing f(f(x)), f(x), x, and e^x on your CAS. You see that f(f(x)) is close to e^x for x in [-2,2] (at least) and the f(x) is between e^x and y=x which is to be expected.

    So is there a real analytic solution with a positive radius of convergence?

    And what is the formula for the nth derivative of f at 0?

    I'm pretty sure (:rolleyes:) that f(0)=1/2 exactly and that f'(0)=Sqrt(e)/2 but all I could get for the second derivative was this:
    f''(1/2)*e/4 + 2 f''(0) / Sqrt(e) = 1. I need f''(1/2) to get f''(0).

  13. Apr 21, 2005 #12
    Last edited by a moderator: Apr 21, 2017
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