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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!

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closed as off-topic by gen-z ready to perish, Martin R, Cesareo, YuiTo Cheng, cmk Jul 12 at 15:20

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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.

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  • $\begingroup$ Thank you so much! I am doing research related to Beta Function and you really saved my life! $\endgroup$ – Dingbang Chen Jul 12 at 7:26
  • $\begingroup$ That is not a Maclaurin series for non-integral $a$. $\endgroup$ – Martin R Jul 12 at 7:43
  • $\begingroup$ @MartinR there does not exist a Maclaurin series expansion for non-integral $a$ because $f^{(\lceil a\rceil)}(0)$ is undefined. $\endgroup$ – Peter Foreman Jul 12 at 7:54
  • $\begingroup$ Is there a way to take a limit? That can be helpful! $\endgroup$ – Dingbang Chen Jul 12 at 7:56
  • $\begingroup$ Puiseux series is amazing! Thank you $\endgroup$ – Dingbang Chen Jul 12 at 7:57

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