What's the solution to this difference equation?

  • #1
1,569
1

Main Question or Discussion Point

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:

Answers and Replies

  • #2
841
1
Originally posted by phoenixthoth
an+1=2an 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
 
  • #3
977
1
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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  • #4
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,916
17
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]
 
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