$f(n+1) = f(n) + f(n)^{a}$ where $a \in (0,1)$ and $n \ge 1$ with $f(1) = m$. 


If $a=0$, we see $f(n) = m + n - 1$ and if $a=1$, we see $f(n) = 2^{n-1}m$. So the recursion seems to interpolate between linear and exponential forms.


Is there a closed form for $f(n)$ in terms of $n$, $a$ and $m$?



Closed form solution to an iterative equation.
There is no closed form except for the cases $a=0,1$. But you can find the asymptotic behavior.
See, for example Fatou, Sur les equations fonctionnelles, Bull Soc. Math. France, 47 (1919),
section 8 and further. Available here:


<http://archive.numdam.org/ARCHIVE/BSMF/>
BSMF\_1919\_*47*/BSMF\_1919\_*47*\_161\_0/BSMF\_1919\_*47*\_161\_0.pdf


