How to compute the Laurent series of $(1+D/x)^{6/5}$?

Question

$(1+D/x)^{6/5}$

How do we compute the Laurent seires when we have fractional power like this and the iside is negative power? Where should I expand the seires?

Answer 1

Expanding Lubin's answer:

$(1+D/x)^{6/5} =\sum_{n=0}^{\infty} \binom{6/5}{n}\frac{D^n}{x^n} $ and

$\begin{array}\\ \binom{6/5}{n} &=\dfrac{\prod_{k=0}^{n-1}(\frac65-k)}{n!}\\ &=\dfrac{\prod_{k=0}^{n-1}(6-5k)}{5^nn!}\\ &=\dfrac{(-1)^n\prod_{k=0}^{n-1}(5k-6)}{5^nn!}\\ \end{array} $

You can play around with this to try to get a nicer looking result like you get for $\binom{1/2}{n}$.

Answer 2

Presumably you’re looking for a series in nonpositive powers of $x$. Assuming that, I say: write out the series for $(1+Dx)^{6/5}$ — you can use the Binomial expansion — and then just change the sign on all exponents of $x$.

Answer 3

This is the same as finding, around $y=0$, the Taylor series of $$f(y)=(1+y)^{\frac 65}\qquad \text{with} \qquad y=\frac Dx$$ and the binomial series is the power series $$(1+y)^\alpha= \sum_{n=0}^\infty\binom{\alpha}{n}y^n$$ where $$\binom{\alpha}{n} = \prod_{k=1}^n \frac{\alpha-k+1}k = \frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}$$