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.

 

[Ambitwistor]

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

Well,

[tex]
\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*}
[/tex]

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

 

[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:

[tex]
a^{b^c} = a^{\left( b^c\right)}
[/tex]

wrong

[tex]
a^{b^c} = \left( a^b \right) ^c
[/tex]