# Is it true that all the “irrational power” functions are almost polynomial ?

Hello all, the $\Delta$ operator on functions $\mathcal{N} \to \mathbb{R}$ (where $\mathcal N$ denotes $\lbrace 1,2, \ldots , \rbrace$ )defined by $\Delta(f)(n)=f(n+1)-f(n)$ is well-known and it is not very hard to show by induction that $f$ is a polynomial of degree $\leq k$ iff $\Delta^{k+1}(f)$ is identically zero, where $\Delta^{k+1}$ denotes $\Delta$ iterated $k+1$ times. Now I say that a function $f : \mathcal{N} \to \mathbb{R}$ is "almost polynomial" iff $\Delta^{k}(f)$ is a bounded function for some $k$.

My question is : let $\lambda >0$ be a non-integer, and let $f(n)=n^{\lambda}$. Is it true that $f$ is almost polynomial ?

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Should that be $f(n)=n^\lambda$ in the question? –  Bruce Westbury Mar 13 '10 at 11:23
Strange. The $\lambda$ as exponent has now appeared. –  Bruce Westbury Mar 13 '10 at 11:25

One can also get this from a standard form of the remainder in Taylor's theorem. Namely, if $k>\lambda$, and $T_{k-1}(x)$ is the degree $k-1$ Taylor polynomial of $f(x)$ at $x=n$, then $$f(x) = T_{k-1}(x) + \frac{f^{(k)}(\xi_x)}{k!} (x-n)^k$$ for some $\xi_x \in [n,x]$. Applying $\Delta^k$ kills the Taylor polynomial term, and when we evaluate the result at $x=n$, the linear combination of the remainders is small when $n$ is large relative to $k$, because each $f^{(k)}(\xi_x)$ is $O(n^{\lambda-k})$ as $n \to \infty$.

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Thanks Bjorn. Your solution is simpler. –  Ewan Delanoy Mar 14 '10 at 5:27

There is an "iterated" version of the mean value theorem which states that for a smooth enough function $f$ on an interval that $\Delta^k f(x)=f^{(k)}(\xi)$ where $\xi$ is between $x$ and $x+k$. This should be enough for your purposes.

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