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Very Awesome Nonlinear ODE

  1. Jul 9, 2011 #1
    [tex]y^2=y' \Rightarrow y=\frac{y'}{y} \Rightarrow \int y dx = ln \left( y \right) \Rightarrow y=e^{\int y dx}=e^{\int e^{\int y dx} dx}=e^{\int e^{\int e^{\int y dx} dx} dx}=\cdots[/tex]

    Is that correct?
     
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  3. Jul 9, 2011 #2

    hunt_mat

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    Why not just say:
    [tex]
    \frac{y'}{y^{2}}=1\Rightarrow -\frac{1}{y}=x+C
    [/tex]
     
  4. Jul 9, 2011 #3

    HallsofIvy

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    Which is the same as y= -1/(x+C) so that [itex]\int ydx= -ln(x+C)+ C_2[/itex] so that
    [itex]e^{\int ydx}=C_2 \frac{1}{x+ C}= \frac{-1}{x+ C}= y[/itex]
    so, yes, your chain of exponentials is correct- but a very complicated way of writing a very simple function.
     
    Last edited by a moderator: Jul 9, 2011
  5. Jul 9, 2011 #4
    This reminds me of a problem in a math contest I once saw. Solve:

    [tex]x^{x^{x^{x^\ldots}}} = 2[/tex]

    The answer is [itex]\sqrt{2}[/itex].
     
  6. Jul 9, 2011 #5
    Yeah, I found the real solution. But, it's the fact that something so contrived is equal to something so simple that makes it awesome.
     
  7. Jul 9, 2011 #6
    That was the first thing that came to mind. Btw, you [STRIKE]would[/STRIKE] could solve that "infinite power tower" using (4) here: http://mathworld.wolfram.com/LambertW-Function.html.
     
    Last edited: Jul 10, 2011
  8. Jul 10, 2011 #7

    HallsofIvy

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    If
    [tex]x^{x^{x^{x^\ldots}}}= 2[/tex]
    then
    [tex]x^{\left(x^{x^{x^\ldots}}\right)}= x^2=2[/tex]

    More generally, if
    [tex]x^{x^{x^{x^\ldots}}}= a> 0[/tex]
    then [itex]x=\sqrt{a}[/itex].
     
  9. Jul 10, 2011 #8

    mheslep

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    almost:
    [itex]x=\sqrt[a]{a}[/itex]
     
  10. Jul 10, 2011 #9
    Actually, it should be [itex]a^{1/a}[/itex], no?

    But this is true only if there is a solution. When I substitute [itex]\sqrt{2}[/itex] for x and iterate it, it does indeed converge to 2. But when I try it for cube root of 3, it doesn't converge to 3. It converges, but to 2.47805. What's more, when I ask Mathematica for [itex]-\frac{\text{ProductLog}[-\text{Log}[z]]}{\text{Log}[z]}[/itex], which is supposed to give the infinite power tower of z (see http://mathworld.wolfram.com/PowerTower.html), it does in fact come to 2.47805. MathWorld says it converges only up to [itex]e^{1/e}[/itex]. The interesting thing is, [itex]2.47805^{1/2.47805}[/itex] equals the cube root of 3.
     
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