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I would like to show that for $s \in \mathbb{R}$ and a nonnegative integer $k$ $$ \triangle^k ((1+n)^s) \lesssim (1+|n|)^{s-k} $$

where $\triangle$ is the discrete derivative, i.e. $\triangle^1 ((1+n)^s) = (2+n)^s - (1+n)^s$.


This is easy when $s \in \mathbb{Z}$, and in the continuous analogue because $$ \partial_x^k (1+x)^s = (s)(s-1)\cdots (s-k+1) (1+x)^{s-k} $$


I think that you can use the generalized binomial theorem to prove this, but I was wondering if there was anything more straightforward, e.g. some kind of convexity argument to use the continuous case.


Note: I wasn't sure about the tags, feel free to re-tag as appropriate.

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up vote 2 down vote accepted

There's a generalization of the mean value theorem that can be applied here, namely that if $f$ is smooth enough then for each $x$ there is $y$ such that $$\Delta^kf(x)=f^{(k)}(y)$$ and $x < y < x+k$.

Added see Is it true that all the "irrational power" functions are almost polynomial ? .

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More like repeated application of MVT than a generalisation. :) –  Gjergji Zaimi Sep 12 '10 at 17:45
    
Thanks! This is what i needed. –  Otis Chodosh Sep 12 '10 at 17:51
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