Binomial Expansion of $(1-x)^{1/n}$.

Question

Please tell me the binomial expansion of $(1-x)^{1/n}$ where $n\ge 0$.

Answer 1

$$(1+x)^\alpha = \sum_{n=0}^\infty {\alpha \choose n} x^n\quad\text{ for all }|x| < 1 \text{ and all complex } \alpha\! $$ where $${\alpha\choose n} = \prod_{k=1}^n \frac{\alpha-k+1}k = \frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}.$$

Answer 2

Well, using the formula that DiMath suggested occurs that \begin{equation*} (1-x)^{\frac{1}{n}}=\sum_{k \geq 0} \binom{\frac{1}{n}}{k}(-x)^{k}=\sum_{k \geq 0} (-1)^{k}\binom{\frac{1}{n}}{k}x^{k} \end{equation*} Now we have to compute $(-1)^{k}\binom{\frac{1}{n}}{k}$. So, \begin{eqnarray*} (-1)^{k}\binom{\frac{1}{n}}{k} &= (-1)^{k} \dfrac{(-\frac{1}{n})(-\frac{1}{n}-1)(-\frac{1}{n}-2) \cdots (-\frac{1}{n}-k+1)}{k!} \\ &= (-1)^{2k}\dfrac{(n+1)(2n+1)\cdots (1+n(k-1))}{n^{k} k!} \\ &= \dfrac{n^{k}(n+1)(2+\frac{1}{n})^{(k)}}{n^{k}k!} \\ &= (n+1) \dfrac{(2+\frac{1}{n})^{(k)}}{k!} \\ \end{eqnarray*} We used the Pochhammer symbol (or rising factorial) $$x^{(n)}=x(x+1)(x+2) \cdots (x+n-1)$$ for the formulation $(2+\frac{1}{n})^{(k)}$. If we combine them, we get the binomial expansion of $(1-x)^{\frac{1}{n}}$ \begin{equation*} (1-x)^{\frac{1}{n}} = \sum_{k \geq o} (n+1) \dfrac{(2+\frac{1}{n})^{(k)}}{k!} x^{k} \end{equation*} There are certain relations for the Pochhammer symbol. I believe that you can simplify the last expression. I hope that I helped you.

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EDIT: I made a mistake in some calculation and I did it all over.

Answer 3

We have $$ (x+a)^n = \sum_{k=0}^{\infty} \binom{n}{k}x^ka^{n-k}$$ where $\binom{n}{k}$ is the binomial coefficient $n$ is a real number. This series converges for integer $n$ or when $|x/a| < 1$.

Specifically we have \begin{align}\displaystyle (1-x)^{1/n} &= \sum_{k=0}^{\infty} \dfrac{(n)_k}{k!}x^k \\ \newline &= 1 + nx + \frac{n(n+1)}{2!}x^2 + \frac{n(n+1)(n+2)}{3!}x^3 + \ldots\end{align}

Where $(n)_k$ is the Pochhammer symbol defined as $$(n)_k = \frac{\Gamma(n+k)}{\Gamma(n)} = n(n+1)\cdots(n+k-1)$$