What's the solution to this difference equation?

[phoenixthoth]

a_n+1=2^a_n 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 a_n 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.

 

[Chen]

a_n+1 = 2^a_n = 2^2a_n-1 = ...

And since: a^bc = a^bc

a_n = 2^{2a₀(n - 1)}

I think. (And it seems to work too for a few calculations... the numbers do quickly get out of hand though.)

 

[Hurkyl]

You have the order of operations wrong.

Correct:

<br /> a^{b^c} = a^{\left( b^c\right)}<br />

wrong

<br /> a^{b^c} = \left( a^b \right) ^c<br />