Writing a non integer power in terms of integer powers I would like to write $x^{2.5}$ in terms of $x$ to the power of integers, is there any way to do this. Taylor series etc. don't work when they depend on derivatives.
If it is not possible, do you have or know a proof.
Thanks
EDIT:
to clarify, I mean that I want to write $x^{2.5}$ in terms of a series of $x^{\mathrm{integer}}$'s, for example: $1 + x^2 + x^{3}$.
I tried to use Taylor series but since it depends on derivatives of  $x^{2.5}$ but they do not have integer powers...
 A: $x^{2.5}$ can't be represented by a series in integer powers of $x$, because it is not analytic (in fact not meromorphic) at $0$: instead, it has a branch point there. 
A: I'll quickly show in practice why Taylor series doesn't work in this case as one would think. Taylor's theorem says, that for a sufficiently smooth function we have
$$ f(x) = \sum_{k=0}^{n-1} \frac{\mathrm d^k}{dx^k}f(a)\frac{(x-a)^k}{k!} +R_n $$
for any $n\in \mathbb N$ and some remainder $R_n$. Surely we can apply this theorem to $x^{2.5}$. It's
$$ \frac{\mathrm d^k}{dx^k}x^{2.5} = 2.5^{\underline k} x^{2.5-k}$$where $2.5^{\underline k}$ denotes falling factorial. So for example evaluating the series for $a=1$ we get
$$x^{2.5} = \sum_{k=1}^{n-1} \frac{2.5^{\underline k}}{k!}(x-1)^k + R_k$$
However we get into trouble if the think, well let's do $n\to \infty$, forget about the Remainder and call 
$$(x+1)^{2.5} = \sum_{k=1}^{\infty} \frac{2.5^{\underline k}}{k!}x^k$$
the series we looked for. We have to be sure first that the series isn't all crap and the $R_n$ contains the useful information! For "nice" functions, we will have $R_n \to 0$, but not in this case.
By Lagrange form, there is some $\xi$ such that
$$R_n = \frac{2.5^{\underline n}}{n!}\xi^{2.5-n}(x-1)^n.$$
Now does this go to $0$? No, because $2.5^{\underline n}=2.5\cdot 1.5 \cdot \ldots \cdot (-998.5) \cdot (-999.5) \cdot ...$ behaves as bad as $n!$. So while we can use the series as above, it wont't probably converge to $x^{2.5}$ as we wished.
