# Maclaurin expansion for $~x^a~(1-x)^b~$ [closed]

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!

## closed as off-topic by gen-z ready to perish, Martin R, Cesareo, YuiTo Cheng, cmkJul 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.

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