# What's the solution to this difference equation?

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.

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Originally posted by phoenixthoth
an+1=2an where n is a natural number and a0 is some fixed real number.
Well,

$$\begin{equation*} \begin{split} a_1 &= 2^{a_0}\\ a_2 &= 2^{a_1} = 2^{2^{a_0}}\\ a_3 &= 2^{a_2} = 2^{2^{2^{a_0}}}\\ &\cdots \end{split} \end{equation*}$$

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.

http://www-users.cs.york.ac.uk/~susan/cyc/b/big.htm

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.)

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Hurkyl
Staff Emeritus
Gold Member
You have the order of operations wrong.

Correct:

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

wrong

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