# Expansion of a function with a negative fractional power

How would you expand a moment generating function with a negative fractional power as a power series??

What method would you be able to use?

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Have you just tried taking derivatives and setting them equal to zero? That's generally the first method to try. –  Qiaochu Yuan Dec 7 '10 at 21:17

See this, in particular equations (17)-(24). This is the moment-generating function of a gamma distribution, and so having the moments you are actually done.

EDIT:

Specifically, take $\theta = 2$ and $\alpha = n/2$. Then, the moments $\mu'_r$ are given by $$\mu'_r = \frac{{2^r \Gamma (n/2 + r)}}{{\Gamma (n/2)}} = n(n+2)(n+4) \cdots (n+2r-2),$$ where for the last equality we have used $\Gamma(p+1) = p \Gamma(p)$. Now, the moment-generating function can be expanded as $$M(t) = 1 + \mu' _1 t + \frac{1}{{2!}}\mu' _2 t^2 + \frac{1}{{3!}}\mu' _3 t^3 + \cdots,$$ giving us $$\frac{1}{{(1 - 2t)^{n/2} }} = 1 + nt + \frac{{n(n + 2)}}{{2!}}t^2 + \frac{{n(n + 2)(n + 4)}}{{3!}}t^3 + \cdots$$ (for $|t|<1/2$).

EDIT 2:

For completeness, let us show how to find the moments $\mu'_r$. Your moment-generating function corresponds to a gamma distribution with density $$f(x)= \frac{{x^{\alpha - 1} e^{ - x/\theta } }}{{\theta ^\alpha \Gamma (\alpha )}}, \;\; x>0,$$ where $\theta = 2$ and $\alpha = n/2$. So, $$\mu'_r = \frac{1}{{\theta ^\alpha \Gamma (\alpha )}}\int_0^\infty {x^{r + \alpha - 1} e^{ - x/\theta } \,{\rm d}x} = \frac{{\theta ^r }}{{\Gamma (\alpha )}}\int_0^\infty {x^{r + \alpha - 1} e^{ - x} \,{\rm d}x} = \theta ^r \frac{{\Gamma (r + \alpha )}}{{\Gamma (\alpha )}}.$$

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It is not necessary to assume that $n$ is an integer. –  Shai Covo Dec 8 '10 at 16:36
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Im trying to work out a similair question, but I am unsure as to how to expand it as a power series? I have taken the log of my function and differentiated it, but what is my final answer for the power series expansion?

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You should probably be asking your own question. :) –  Ｊ. Ｍ. Dec 8 '10 at 15:28
Sarah, that is what i'm looking for, i want to know how the final power series would turn out. –  R.M Dec 8 '10 at 15:37
I have asked it in a new question Rachel, if your needing any help have a look at my question and hopefully we can both find what we are looking for =). –  Sarah T Dec 8 '10 at 15:41