# Find the function for this Taylor series

1. Jul 13, 2009

### rman144

Find the function that has the following Taylor series representation:

$$\sum$$$$^{\infty}_________{m=0}$$$$\frac{(m+s)^{-1}x^{m}}{m!}$$

Where s is a constant such that 0<Re(s)<1.

Any ideas?

2. Jul 14, 2009

### HallsofIvy

Do you have any reason to believe it is an elementary function?

3. Jul 14, 2009

### Gib Z

Well the first thing to do would be to consider what a Taylor series itself looks like of a function f, assuming it has one;

$$f(x) = \sum_{m=0}^{\infty} f^m (0) \frac{x^m}{m!}$$ where f^m indicates the m-th derivative, not exponent.

So if we use the theorem that two power series are equal if and only iff their coefficients are identical, we get;

$$f^m (0) = \frac{1}{m+s}$$

I have tried for a while at seeing what that tells us exactly, and my intution leads me to believe this derivatives look like something from an exponential function, so if I were you, I would now try letting $$f(x) = e^{g(x)}$$ and seeing what I can find out about g(x) .

EDIT: Whoops I didn't see Halls post.

4. Jul 14, 2009

### g_edgar

We can get a differential equation for this function, call it $$f(x)$$. That denominator $$m+s$$ looks scary , let's get rid of it. Multiply by $$x^s$$ so that we have $$\sum x^{m+s}/((m+s)m!)$$, differentiate to get rid of the $$m+s$$. Then multiply by $$x^{1-s}$$ and we end up with the series $$\sum x^m/m!$$ which can be recognized as $$e^x$$. So the result is...
$$sf \left( x \right) +x{\frac {d}{dx}}f \left( x \right) ={{\rm e}^{x}}$$
Is this helpful? The solution is not elementary, so maybe or maybe not.

5. Jul 14, 2009

### g_edgar

This function is a hypergeometric function $${{}_1\mathrm{F}_1(s;\,1+s;\,x)}$$, that's just the definition... Maple evaluates this in terms of the incomplete Gamma function, so we conclude it is:
$$s \left( -x \right) ^{-s}\int _{0}^{-x}\!{{\rm e}^{-t}}{t}^{s-1}{dt}$$
at least for $$x < 0$$