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A number raise to it self infinitely

  1. Jul 29, 2007 #1
    hi I have encountered this...

    .
    Code (Text):
                                         .
                                       X
                                    X
                                  X
                               X    =   3
     
    I tried to solve it using properties of logarithims however I always end up with the initial equation... how will I find the value of X? ( x is raised to it self endlessly ).

    :confused:
    .
     
    Last edited: Jul 29, 2007
  2. jcsd
  3. Jul 29, 2007 #2

    Dick

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    You have to parenthesize that exponential tower or it's ambiguous in meaning. 3^(3^3) is not equal to (3^3)^3. And as it's infinite, you need to express the meaning in terms of an infinite sequence. So start with a_0=x. Is a_n=x^(a_(n-1)) or is a_n=(a_(n-1))^x?
     
    Last edited: Jul 30, 2007
  4. Jul 30, 2007 #3

    Curious3141

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    The generally understood meaning of the left hand side of that equation is the infinite power tower function

    [tex]f(x) = x^{(x^{(x^{(x^...[/tex]

    or simply, work "downwards" from "infinity".

    The "trick" solution to f(x) = N is to see that this implies x^f(x) = N hence x^N = N when a solution exists, giving x = N^(1/N).

    The problem is that there is an upper bound on N for a solution to exist. I'll leave it to the OP to find that bound and hence deduce that f(x) = 3 has no solution.
     
  5. Jul 31, 2007 #4

    HallsofIvy

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    If [itex]x^{(x^{(x^{(x^...}}} = 3[/itex], Then
    x3= [tex]x^[{(x^{(x^{(x^...]}}}}= x^{(x^{(x^{(x^...}}}[/tex]. What is that equal to?

    Curios3141, I don't see why that "has no solution". Am I missing something?
     
  6. Jul 31, 2007 #5

    D H

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    Halls, try calculating [tex]f(x)=x^{x^{x^\cdots}}[/tex] when [tex]x=3^{1/3}[/tex]. It does not converge to 3. It does not converge, period. The upper bound on N is between 2 and 3 and is a very ubiquitous number.
     
  7. Aug 1, 2007 #6

    HallsofIvy

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    How are you calculating that? If, for example, you calculate 31/3, store it in "x" in a calculator, do x^x, then "ans"^x, repeatedly, it does not converge.

    But that is NOT [tex]f(x)=x^{x^{x^\cdots}}[/tex]! To calculate that you need to repeatedly do x^"ans" and that does converge, slowly, to 3.
     
  8. Aug 1, 2007 #7

    D H

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    I misspoke. It does converge, just not to the value you expect. Let

    [tex] f_1(x) = x[/tex]
    [tex] f_k(x) = x^{f_{k-1}(x)}[/tex]

    The desired function [itex]f(x)[/itex] is

    [tex] f(x) = x^{x^{x^{\cdots^x}} = \lim_{k\to\infty}f_k(x) [/tex]

    Now further define

    [tex]g(x) = x^{1/x}[/tex]

    THe question at hand is [itex]f(g(x))=x[/itex]? For [itex]x \in [ 1/e,e ] [/itex], [itex]f(g(x))=x[/itex]. The convergence is very slow as x approaches 1/e or e.

    Things get a bit interesting when x>e. For example, [itex]f(g(3)) = 2.478053\cdots[/itex] rather than 3. What's going on here? Note that [itex]g(3) = 1.44225\cdots[/itex]. However, [itex]g(2.478053\cdots) = 1.44225\cdots[/itex] also. The inverse of [itex]g(x)[/itex] has two branches, with a branch point at [tex]e^{1/e}[/tex]. [itex]f(g(x))[/itex] converges to [tex]g^{-1}(g(x))[/tex], where [tex]g^{-1}[/tex] refers to the lower branch.

    Things get even more interesting when x<1/e. The sequence [itex]\{f_k(x)\}[/itex] doesn't converge. Instead, the sequence toggles between a pair of values as [itex]k\to\infty[/itex], one close to zero and the other close to one.
     
  9. Aug 2, 2007 #8

    HallsofIvy

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    I'm a bit confused as to what converges, "converge, just not to the value you expect". The original problem was to solve
    [tex\ x^{x^x^{\cdot\cdot\cdot}}= 3[/tex].
    IF the sequence [itex]a_1= x^x[/itex], [itex]a_2= x^{a_1}, ... converges to 3, then x must be equal to 31/3.
     
  10. Aug 2, 2007 #9

    Dick

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    That's true. But the converse (if x=3^(1/3) then the sequence converges to 3) isn't true. The sequence does converge to a limit L that you can find by solving log(L)/L=log(3)/3 (not the L=3 root, the other root L=2.47805...).
     
  11. Aug 2, 2007 #10

    HallsofIvy

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    I am going to have to meditate upon this!
     
  12. Aug 2, 2007 #11

    D H

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    The problem is that no x exists such that the sequence [itex]\{a_k(x)\}[/itex] converges to 3.
     
  13. Aug 3, 2007 #12

    Curious3141

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    Halls, it's simple. The radius of convergence of the hyperpower function [tex]f(x) = ^{\infty}x[/tex] (using the tetration notation) is [tex]({(1/e)}^e, e^{(1/e)})[/tex]

    For x > e^(1/e), the function does not coverge.

    For a value like 3^(1/3), the function *does* converge. But it does not converge to 3.

    The function [tex]g(x) = x^{(\frac{1}{x})}[/tex] reaches a maximum at x = e. In other words, you cannot find a value of x that makes g(x) > e^(1/e), so g(f(x)) will always converge for any real positive x > e.

    So for N > e, f([N^(1/N)]) will never equal N, even though convergence will hold. Hence f(x) = N has no solution for x > e.

    Hope that clarifies things. I haven't explored the lower bound this carefully, so I shan't comment on it.
     
    Last edited: Aug 3, 2007
  14. Aug 3, 2007 #13

    HallsofIvy

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    Yes, thanks. I showed that "if the sequence of powers converges to 3, then x= 31/3. The converse, "if x= 31/3, then the sequence converges to 3" is not true.
     
  15. Aug 4, 2007 #14

    Curious3141

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    Of course, the first condition (convergence to 3) will never actually be met.
     
  16. May 14, 2008 #15
    D_H's f(g(x)) appears to provide solutions for x^y = y^x, x<>y, x>e.
    For instance y=f(g(3)) is the other solution for 3^y = y^3, besides the obvious y=3.
    When x=e, the only solution is y=e.
    However, such solutions also exist when 1 < x < e, eg 2^4 = 4^2, but f(g(x)) does not help then, eg f(g(2)) = 2.
     
    Last edited: May 14, 2008
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