Taylor Series for $(1-x)^p$

Question

Can anybody help me with the Taylor series for $(1-x)^p$? I have no idea how to do it. I know that:

$(1-x)^{-1}=1+x+x^2+x^3+...$

Any help would be much appreciated.

Answer 1

This was meant to be a reply but I am not sure how to properly render latex in comments so I apologise for that. $$(1-x)^{-3}=\sum^{\infty}_{n=0} \binom{-3}{n}(-x)^n=\sum^{\infty}_{n=0} \binom{n+2}{n}x^n=1+3x+6x^2+10x^3+15x^4+...$$ using $$\binom{-n}{k}=\binom{n+k-1}{k}(-1)^k$$

Answer 2

Apply the binomial expansion.

Since the Taylor series is unique, it is also the Taylor expansion.

The power can also be negative or fractional. In that case you need the generalised binomial theorem, that generalizes the binomium for negative and non-integer values. For instance $\binom{-3}{2} = \frac{-3 \cdot -2}{1 \cdot 2} = 3$.

Answer 3

Let $f(x)=(1-x)^p$. We see that $$f'(0)=-p$$ $$f''(0)=p(p-1)$$ and so on so $f^{(n)}(0)=(-1)^np(p-1)...(p-n+1)$. Hence $$(1-x)^p=\sum^{\infty}_{n=0}\frac{f^{(n)}(0)}{n!}x^n=\sum^{\infty}_{n=0}(-1)^n\frac{p(p-1)...(p-n+1)}{n!}x^n=\sum^{\infty}_{n=0}\binom{p}{n}(-1)^nx^n$$