Exercise 8.22

Question

Exercise 22: 
    If $\alpha$ is real and $-1<x<1$, prove Newton's binomial theorem $$(1+x)^\alpha=1+\sum_{n=1}^\infty\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}x^n.$$ Show also that $$(1-x)^{-\alpha}=
    \sum_{n=0}^\infty\frac{\Gamma(n+\alpha)}{n!\,\Gamma(\alpha)}x^n$$ if $-1<x<1$ and $\alpha>0$.

analambanomenos · Accepted Answer

\def\eps{\varepsilon} We can show that the series on the right-hand side of the first equation converges by the Ratio Test, Theorem 3.34. Let $$a_n=\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}x^n.$$ Then for $n>\alpha$, $$\lim_{n ightarrow\infty}\bigg|\frac{a_{n+1}}{a_n}\bigg| = \lim_{n ightarrow\infty}\bigg(\frac{n-\alpha}{n+1}\bigg)|x| = \lim_{n ightarrow\infty} \bigg(1-\frac{\alpha+1}{n+1}\bigg)|x| = |x|<1.$$ Hence the right-hand side of the equation defines a function $f(x)$ for $|x|<1$. Following the hint, by Theorem 8.1 we can differentiate the series term-by-term to get $$f'(x) = \sum_{n=1}^\infty\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{(n-1)!}x^{n-1} = \sum_{n=0}^\infty \frac{\alpha(\alpha-1)\cdots(\alpha-n)}{n!}x^{n}.$$ Hence \begin{align*} (1+x)f'(x) &= \sum_{n=0}^\infty\frac{\alpha(\alpha-1)\cdots(\alpha-n)}{n!}x^{n}+\sum_{n=0}^\infty\frac{\alpha(\alpha-1)\cdots(\alpha-n)}{n!}x^{n+1} \ &= \alpha + \sum_{n=1}^\infty\bigg(\frac{\alpha(\alpha-1)\cdots(\alpha-n)}{n!}+\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{(n-1)!}\bigg)x^n \ &= \alpha + \sum_{n=1}^\infty\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)(\alpha-n+n)}{n!}x^n \ &= \alpha + \alpha\sum_{n=1}^\infty\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}x^n \ &= \alpha f(x) \end{align*} If we let $g(x)=(1+x)^\alpha$ be the left-hand side of the equation, then $(1+x)g'(x)=(1+x)\alpha(1+x)^{n-1}=\alpha g(x)$, so $g(x)$ also satisfies the differential equation $y'=\alpha y/(1+x)$. Also note that $g(0)=f(0)=1$. Since $$\bigg|\frac{\alpha y_1}{1+x}-\frac{\alpha y_2}{1+x}\bigg|=\bigg(\frac{\alpha}{1+x}\bigg)|y_1-y_2|$$ we can apply the uniqueness result of Exercise 5.27 to conclude that $g(x)=f(x)$ on any interval $[-1+\epsilon,1-\epsilon]$, and so on all of $(-1,1)$. Note that by Theorem 8.18(a) we have, for $\alpha>0$, \begin{align*} \Gamma(\alpha+n) &= (\alpha+n-1)\Gamma(\alpha+n-1) \ &= (\alpha+n-1)(\alpha+n-2)\Gamma(\alpha+n-2) \ &= \cdots = (\alpha+n-1)(\alpha+n-2)\cdots(\alpha+1)\alpha\Gamma(\alpha). \end{align*} Hence, for $-10$, \begin{align*} (1-x)^{-\alpha} &= 1+\sum_{n=1}^\infty\frac{-\alpha(-\alpha-1)\cdots(-\alpha-n+1)}{n!}(-1)^nx^n \ &= 1+\sum_{n=1}^\infty\frac{\alpha(\alpha+1)\cdots(\alpha+n-1)}{n!}x^n \ &= 1+\sum_{n=1}^\infty\frac{\Gamma(n+\alpha)}{n!\,\Gamma(\alpha)}x^n \ &= \sum_{n=0}^\infty\frac{\Gamma(n+\alpha)}{n!\,\Gamma(\alpha)}x^n \end{align*}

Exercise 8.22

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Exercise 8.22

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