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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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  • $\begingroup$ 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 math.stackexchange.com instead. $\endgroup$
    – Did
    May 6, 2013 at 11:42
  • $\begingroup$ could you provide your derivation? $\endgroup$
    – Turbo
    May 23, 2013 at 10:32

1 Answer 1

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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

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  • $\begingroup$ I believe this is a good answer. However it is all in French. $\endgroup$
    – Turbo
    May 6, 2013 at 15:24

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