Find the function for this Taylor series

[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?

 

[HallsofIvy]

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

 

[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.

 

[g_edgar]

FIRST ANSWER...
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.

 

[g_edgar]

SECOND ANSWER...
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 &lt; 0

 

[daudaudaudau]

I think it is pretty clever to construct a differential equation like that. Well done.