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What's the solution to this difference equation?

  1. Nov 18, 2003 #1
    an+1=2an where n is a natural number and a0 is some fixed real number. (in case it's not clear, what's on the right hand side is 2 to the an power.) thanks!

    i'm wondering if it will be possible to let n=1/2 and have the half iterate to 2^x i was looking for earlier, where by that i mean a function such that f(f(x))=2^x. i'm guessing that solving one problem would be roughly as difficult as the other...

    if you happen to know of a good article on nonlinear difference equations, i'd appreciate it. i'm also looking at the logistic.
    Last edited: Nov 18, 2003
  2. jcsd
  3. Nov 18, 2003 #2

    a_1 &= 2^{a_0}\\
    a_2 &= 2^{a_1} = 2^{2^{a_0}}\\
    a_3 &= 2^{a_2} = 2^{2^{2^{a_0}}}\\

    That's about as simple as you can express it. You could write it with Knuth's or Conway's shorthand notation, or maybe with Ackermann's function.

  4. Nov 18, 2003 #3
    an+1 = 2an = 22an-1 = ...

    And since: abc = abc

    an = 22a0(n - 1)

    I think. (And it seems to work too for a few calculations... the numbers do quickly get out of hand though.)
    Last edited: Nov 18, 2003
  5. Nov 18, 2003 #4


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    You have the order of operations wrong.


    a^{b^c} = a^{\left( b^c\right)}


    a^{b^c} = \left( a^b \right) ^c
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