Newton’s binomial theorem

By | October 22, 2015

Theorem (Newton’s binomial Theorem). If \alpha is real and -1<x<1, then

(1)   \begin{equation*} \left(1+x\right)^{\alpha}=1+\sum_{n=1}^{\infty}\frac{\alpha\left(\alpha-1\right)\cdots\left(\alpha-n+1\right)}{n!}x^{n}. \end{equation*}

Also we have

    \[ \left(1-x\right)^{-\alpha}=\sum_{n=0}^{\infty}\frac{\Gamma\left(\alpha+n\right)}{n!\Gamma\left(\alpha\right)}x^{n} \]

if -1<x<1 and \alpha>0.

Proof. Define f\left(x\right)=1+\sum_{n=1}^{\infty}\frac{\alpha\left(\alpha-1\right)\cdots\left(\alpha-n+1\right)}{n!}x^{n}. So if we define

    \[a_{n}=\frac{\alpha\left(\alpha-1\right)\cdots\left(\alpha-n+1\right)}{n!},\]

then

    \[ \rho=\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_{n}}\right|=1 \]

shows the radius of convergence of f is 1. So f converges pointwise on \left(-1,1\right).

Taking term by term differentiation, for -1<x<1, we get

    \begin{align*} f^{\prime}\left(x\right) & =\sum_{n=1}^{\infty}\frac{\alpha\left(\alpha-1\right)\cdots\left(\alpha-n+1\right)}{n!}nx^{n-1}\\ & =\sum_{n=0}^{\infty}\frac{\alpha\left(\alpha-1\right)\cdots\left(\alpha-n\right)}{n!}x^{n}. \end{align*}

So

    \begin{align*} \left(x+1\right)f^{\prime}\left(x\right) & =\left(x+1\right)\sum_{n=0}^{\infty}\frac{\alpha\left(\alpha-1\right)\cdots\left(\alpha-n\right)}{n!}x^{n}\\ & =\sum_{n=0}^{\infty}\frac{\alpha\left(\alpha-1\right)\cdots\left(\alpha-n\right)}{n!}x^{n+1}+\sum_{n=0}^{\infty}\frac{\alpha\left(\alpha-1\right)\cdots\left(\alpha-n\right)}{n!}x^{n}\\ & =\alpha+\sum_{n=0}^{\infty}\frac{\alpha\left(\alpha-1\right)\cdots\left(\alpha-n\right)}{n!}x^{n+1}+\sum_{n=1}^{\infty}\frac{\alpha\left(\alpha-1\right)\cdots\left(\alpha-n\right)}{n!}x^{n}\\ & =\alpha+\sum_{n=1}^{\infty}\frac{\alpha\left(\alpha-1\right)\cdots\left(\alpha-n+1\right)}{\left(n-1\right)!}x^{n}+\sum_{n=1}^{\infty}\frac{\alpha\left(\alpha-1\right)\cdots\left(\alpha-n\right)}{n!}x^{n}\\ & =\alpha+\alpha\sum_{n=1}^{\infty}\frac{\alpha\left(\alpha-1\right)\cdots\left(\alpha-n+1\right)}{n!}x^{n}\\ & =\alpha f\left(x\right) \end{align*}

for -1<x<1. So we get the functional equation

    \[ \frac{f^{\prime}\left(x\right)}{f\left(x\right)}=\frac{\alpha}{x+1}. \]

So

    \[ \ln f\left(x\right)=\ln\left(1+x\right)^{\alpha}+C \]

Since f\left(0\right)=1, C=0. Hence f\left(x\right)=\left(1+x\right)^{\alpha}. To prove the other relation, we observe that

    \begin{align*} \left(-\alpha\right)\left(-\alpha-1\right)\cdots\left(-\alpha-n+1\right) & =\left(-1\right)^{n}\alpha\left(\alpha+1\right)\cdots\left(\alpha+n-1\right)\\ & =\left(-1\right)^{n}\frac{\Gamma\left(n+\alpha\right)}{\Gamma\left(\alpha\right)}. \end{align*}

Then putting -x into x in (1) and -\alpha with \alpha>0 we get

    \begin{align*} \left(1-x\right)^{-\alpha} & =1+\sum_{n=1}^{\infty}\frac{\left(-\alpha\right)\left(-\alpha-1\right)\cdots\left(-\alpha-n+1\right)}{n!}\left(-1\right)^{n}x^{n}\\ & =1+\sum_{n=1}^{\infty}\left(-1\right)^{n}\frac{\Gamma\left(n+\alpha\right)}{\Gamma\left(\alpha\right)n!}\left(-1\right)^{n}x^{n}\\ & =1+\sum_{n=1}^{\infty}\frac{\Gamma\left(n+\alpha\right)}{\Gamma\left(\alpha\right)n!}x^{n}\\ & =\sum_{n=0}^{\infty}\frac{\Gamma\left(n+\alpha\right)}{\Gamma\left(\alpha\right)n!}x^{n} \end{align*}

so we are done.

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