Calculus – Sequences and series – Binomial series

Consider the function \(f(x)=(1+x)^{\alpha}\) with \(\alpha\in\mathbb{R}\), then we have:

\[f'(x)=\alpha(1+x)^{\alpha-1},\quad f''(x)=\alpha(\alpha-1)(1+x)^{\alpha-2}\,\quad f^{(3)}(x)=\alpha(\alpha-1)(\alpha-2)(1+x)^{\alpha-3},\quad\ldots.\]

This implies that \(f^{(n)}(0)=\alpha(\alpha-1)(\alpha-2)\cdots(\alpha-n+1)\) for \(n=1,2,3,\ldots\). So, if we define the binomial coefficients

\[\binom{\alpha}{0}=1,\quad\binom{\alpha}{n}=\frac{\alpha(\alpha-1)(\alpha-2)\cdots(\alpha-n+1)}{n!},\quad n=1,2,3,\ldots,\]

then we have:

Theorem: If \(\alpha\in\mathbb{R}\), then we have:

\[(1+x)^{\alpha}=\sum_{n=0}^{\infty}\binom{\alpha}{n}x^n=1+\frac{\alpha}{1!}x+\frac{\alpha(\alpha-1)}{2!}x^2 +\frac{\alpha(\alpha-1)(\alpha-2)}{3!}x^3+\cdots,\quad|x| < 1.\]

This series is called a binomial series.

In order to see that the radius of convergence equals \(1\) we apply the ratio test: for \(x\neq0\) let \(a_n=\displaystyle\binom{\alpha}{n}x^n\), then we have:

\[\lim\limits_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim\limits_{n\to\infty}\left|\binom{\alpha}{n+1}x^{n+1}\bigg/\binom{\alpha}{n}x^n\right| =\lim\limits_{n\to\infty}\left|\frac{\alpha(\alpha-1)(\alpha-2)\cdots(\alpha-n+1)(\alpha-n)}{\alpha(\alpha-1)(\alpha-2)\cdots(\alpha-n+1)}\frac{n!}{(n+1)!}x\right| =\lim\limits_{n\to\infty}\left|\frac{\alpha-n}{n+1}x\right|=|x|.\]

Examples:

1) \(\displaystyle\sqrt{1+x}=(1+x)^{\frac{1}{2}}=\sum_{n=0}^{\infty}\binom{\frac{1}{2}}{n}x^n=1+\frac{1}{2}x-\frac{1}{8}x^2+\frac{1}{16}x^3-\frac{5}{128}x^4+\cdots\) for \(|x| < 1\) since we have: \(\displaystyle\binom{\frac{1}{2}}{0}=1\), \(\displaystyle\binom{\frac{1}{2}}{1}=\frac{1}{2}\),

\[\binom{\frac{1}{2}}{2}=\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)}{2!}=-\frac{1}{8},\quad\binom{\frac{1}{2}}{3} =\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)}{3!}=\frac{1}{16},\quad\binom{\frac{1}{2}}{4} =\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)\left(\frac{1}{2}-3\right)}{4!}=-\frac{5}{128},\quad\ldots.\]

2) \(\displaystyle\frac{1}{\sqrt{1+x}}=(1+x)^{-\frac{1}{2}}=\sum_{n=0}^{\infty}\binom{-\frac{1}{2}}{n}x^n=1-\frac{1}{2}x+\frac{3}{8}x^2-\frac{5}{16}x^3+\cdots\) for \(|x| < 1\) since we have: \(\displaystyle\binom{-\frac{1}{2}}{0}=1\), \(\displaystyle\binom{-\frac{1}{2}}{1}=-\frac{1}{2}\),

\[\binom{-\frac{1}{2}}{2}=\frac{-\frac{1}{2}\left(-\frac{1}{2}-1\right)}{2!}=\frac{3}{8},\quad\binom{-\frac{1}{2}}{3} =\frac{-\frac{1}{2}\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right)}{3!}=\frac{5}{16},\quad\ldots.\]

Replacing \(x\) by \(-x^2\) this implies that \(\displaystyle\frac{1}{\sqrt{1-x^2}}=\sum_{n=0}^{\infty}(-1)^n\binom{-\frac{1}{2}}{n}x^{2n}=1+\frac{1}{2}x+\frac{3}{8}x^2+\frac{5}{16}x^3+\cdots\) for \(|-x^2| < 1\;\;\Longleftrightarrow\;\;|x| < 1\). Using integration we obtain

\[\arcsin(x)=\int\frac{1}{\sqrt{1-x^2}}\,dx=C+\sum_{n=0}^{\infty}(-1)^n\binom{-\frac{1}{2}}{n}\frac{x^{2n+1}}{2n+1},\quad|x| < 1\]

and since \(\arcsin(0)=0\) we conclude that \(C=0\). Similarly, we obtain

\[\arccos(x)=-\int\frac{1}{\sqrt{1-x^2}}\,dx=K-\sum_{n=0}^{\infty}(-1)^n\binom{-\frac{1}{2}}{n}\frac{x^{2n+1}}{2n+1},\quad|x| <1\]

and since \(\arccos(0)=\frac{1}{2}\pi\) we conclude that \(K=\frac{1}{2}\pi\). Hence we have

\[\arcsin(x)=\sum_{n=0}^{\infty}(-1)^n\binom{-\frac{1}{2}}{n}\frac{x^{2n+1}}{2n+1},\quad|x| < 1\quad\text{and}\quad \arccos(x)=\frac{1}{2}\pi-\sum_{n=0}^{\infty}(-1)^n\binom{-\frac{1}{2}}{n}\frac{x^{2n+1}}{2n+1},\quad|x| < 1.\]

Applications:

1) Using \(\displaystyle\sqrt{1+x^2}=\sum_{n=0}^{\infty}\binom{\frac{1}{2}}{n}x^{2n}=1+\frac{1}{2}x^2-\frac{1}{8}x^4+\cdots\) for \(|x| < 1\) we find that

\[\lim\limits_{x\to 0}\frac{\sqrt{1+x^2}-1-\frac{1}{2}x^2}{x^4}=\lim\limits_{x\to 0}\frac{1+\frac{1}{2}x^2-\frac{1}{8}x^4+\cdots-1-\frac{1}{2}x^2}{x^4} =-\frac{1}{8}.\]

2) Using \(\displaystyle\sqrt{1+x^3}=\sum_{n=0}^{\infty}\binom{\frac{1}{2}}{n}x^{3n}=1+\frac{1}{2}x^3-\frac{1}{8}x^6+\cdots\) for \(|x| < 1\) we find that

\[\int_0^{\frac{1}{2}}\sqrt{1+x^3}\,dx=\sum_{n=0}^{\infty}\binom{\frac{1}{2}}{n}\int_0^{\frac{1}{2}}x^{3n}\,dx =\sum_{n=0}^{\infty}\binom{\frac{1}{2}}{n}\frac{(\frac{1}{2})^{3n+1}}{3n+1}=\frac{1}{2}+\frac{1}{64}-\frac{1}{14680064}+\cdots\approx0.515625.\]

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Last modified on March 15, 2021