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$f(n+1) = f(n) + f(n)^{a}$ 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 $m$?

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Standard comparison to the associated differential equation yields $y(n)=\Theta(n^b)$ with $b=1/(1-a)$ (and, with some more care, much more precise estimates) but this is not a research question. You might want to try instead. – Did May 6 '13 at 11:42
could you provide your derivation? – Turbo May 23 '13 at 10:32
up vote 2 down vote accepted

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: BSMF_1919_47/BSMF_1919_47_161_0/BSMF_1919_47_161_0.pdf

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I believe this is a good answer. However it is all in French. – Turbo May 6 '13 at 15:24

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