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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.



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...

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What's wrong with Taylor series? Do you want a finite expression? – tomasz Aug 3 '12 at 17:41
What do you mean by "in terms of?" It's gonna be very hard to do this. In particular, integer powers are well-defined in the complex plane, but you get problems with non-integer powers, which need to be multi-valued. (The most obvious is the square root function, which can take +/- values, but if the exponent is irrational, you actually get an infinite number of possible values for exponentiation.) – Thomas Andrews Aug 3 '12 at 17:42
I think he means taylor series about $0$, since at some point, the derivatives become undefined at $x=0$. – Thomas Andrews Aug 3 '12 at 17:48
Yeah, I think that was the concern. A work-around to this is to use the Taylor series of $\sqrt{1+x}$ (or more generally, $\sqrt{c+x}$). – Hugh Denoncourt Aug 3 '12 at 18:08
What does Taylor series etc wont work when they depend on derivatives. mean? When doesn't Taylor depend on derivatives? Can you give an example to clarify that? – draks ... Aug 3 '12 at 18:11
up vote 6 down vote accepted

$x^{2.5}$ 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.

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Could you provide a link to some information on this? – John Echo Aug 3 '12 at 18:41
@EoinMurray In general, non-integer powers are "multivalued" functions in a complex sense. Branch points are points of discontinuity where the function "jumps" from one value to another. For example, $z^{1/2}$ is the basic example of a multivalued function. $z^{2.5} = z^2z^{.5}$ is also therefore multivalued. The term "analytic function" means a function that can be written with a power series expansion. Analyticity can be shown to not hold if certain conditions on the magnitude of the derivative do not hold. – Emily Aug 3 '12 at 18:49
Here are some helpful wiki links:… – Emily Aug 3 '12 at 18:50

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.

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