How to use a Maclaurin Series Expansion on $~x^a~(1-x)^b~$?
There is a singularity at $~x = 0~$ when derivatives are taken.
Thank you so much!
2026-03-25 21:01:32.1774472492
Maclaurin expansion for $~x^a~(1-x)^b~$
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$$(1-x)^b=\sum_{k=0}^\infty \binom{b}k (-x)^k$$ $$\therefore x^a(1-x)^b=\sum_{k=0}^\infty\binom{b}k (-1)^k x^{a+k}$$ This expansion is valid for $-1\le x\le1$ where $x,a,b\in\mathbb{R}$. If $a\not\in\mathbb{N}^0$ then this becomes a Puiseux series.