I didn't quite understand the expansion of, for instance $1 \over (1-x)^\alpha$, for $\alpha \in \mathbb Q$, for instance for $\alpha = {1\over 2}$ using the binomial coefficients. I know that for $\alpha \in \mathbb N$, ${1 \over (1-x)^\alpha}=\sum_{n=0}^∞ {n+\alpha-1 \choose \alpha-1}x^n$, but I don't know how to apply it when $\alpha$ isn't whole.
Thanks in advance for any explanation!
The fundamental formula is $$ (1+X)^\alpha = \sum_{k\geq0} \binom\alpha kX^k, $$ valid as an identity of formal power series in$~X$ for all $\alpha$, where by definition $$ \binom\alpha k=\frac{\alpha(\alpha-1)\ldots(\alpha-k+1)}{k!}. $$ The latter formula allows "negating the upper index": $$ (-1)^k\binom{-\alpha}k=\binom{\alpha+k-1}k $$ which is also valid for arbitrary$~\alpha$. This then allows the initial formula to be reformulated as $$ \frac1{(1-X)^\alpha}=\sum_{k\geq0}(-1)^k\binom{-\alpha}kX^k =\sum_{k\geq0}\binom{\alpha+k-1}kX^k. $$ So that is the correct formula, not the one given in the question.
What you appear to have done is apply symmetry: $\binom{\alpha+k-1}k=\binom{\alpha+k-1}{\alpha-1}$. But that is only valid when the upper index is a nonnegative integer, which requires (if it is to hold for all $k\geq0$) that $\alpha$ be a positive integer. So you cannot apply it when $\alpha=\frac12$, as you did (indeed the resulting binomial coefficient is not even defined by the above definition).
Consider the definition of $\displaystyle {m\choose k}\color{grey}{=\frac{m!}{(m-k!)k!}}$.
Now note that $\dfrac{m!}{(m-k!)k!}=\dfrac{m\cdot (m-1)\cdots(m-k+1)\cdot(m-k)!}{(m-k)!k!}=\dfrac{m\cdot (m-1)\cdots(m-k+1)}{k!}$.
Consider $\dfrac{m\cdot (m-1)\cdots(m-k+1)}{k!}$. What do you need $m$ to be for it to make sense?
