I'd like to solve the recurrence relation
$$\begin{align*} f(n+1) &= 2 f(n)^2 h(n) \\ h(n+1) &= h(n) f(n) \\ f(2) & = h(2) = 2 \end{align*}$$ for an explicit solution for $h$ and $f$. So far I found out that I can use RSolve
and came up with following code:
RSolve[ {f[n + 1] == 2 f[n]^2 h[n], h[n + 1] == h[n] f[n], h[2] == 2, f[2] == 2}, {h[n], f[n]}, n ]
Whenever I enter this code, I just get the same code back so I must be doing something wrong, but I cannot find out what. Strangely it immediately returns a result if I enter decoupled recurrence relations with the exact same syntax.
RSolve[ {f[n + 1] == 2 f[n], h[n + 1] == h[n]^2, h[2] == 2, f[2] == 2}, {h[n], f[n]}, n ]
Both are basically simple affine recurrence relations if you logarithmise them, which could be solved by linear algebra. Can anyone spot the mistake I'm making?
RSolve
expression. $\endgroup$ – mikado Jun 29 '16 at 21:41RSolve
works for me, returning terms like{f[n] -> 2^(-1 + 2^(-1 + n)), h[n] -> 2}
. I'm using Mathematica 10.4 $\endgroup$ – mikado Jun 29 '16 at 22:01RSolve
returns unevaluated very rapidly. I can't see any obvious mistakes in it, so I guess that Mathematica must recognise it as something it can't handle. $\endgroup$ – mikado Jun 29 '16 at 22:16