Finding the Power Series Representation for \(\tan^{-1} {\bf x}\)
Show that the power series representation for \(\tan ^{-1}x\) is \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \tan ^{-1}x=x-\dfrac{x^{3}}{3}+\dfrac{x^{5}}{5}-\dfrac{ x^{7}}{7}+\cdots +(-1)^{n}\dfrac{x^{2n+1}}{2n+1}+\cdots =\sum\limits_{k\,=\,0}^{\infty }\dfrac{(-1) ^{k}x^{2k+1}}{2k+1} }} \]
The power series representation for \(\tan^{-1}x\) is called Gregory’s series (or Gregory–Leibniz series or Madhava–Gregory series). James Gregory (1638–1675) was a Scottish mathematician. His mother, Janet Anderson, was his teacher and taught him geometry. Gregory, like many other mathematicians of his time, was searching for a good way to approximate \(\pi\), a result of which was Gregory’s series. Gregory was also a major contributor to the theory of optics, and he is credited with inventing the reflective telescope.
Find the radius of convergence and the interval of convergence.
This series converges when \(\vert x^{2}\vert < 1\), or equivalently for \(-1 < x < 1\). We use the integration property of power series to integrate \( y^\prime =\dfrac{1}{1+x^{2}}\) and obtain a series for \(y=\tan ^{-1}x\). \[ \begin{eqnarray*} \int_{0}^{x}\frac{dt}{1+t^{2}} &=&\int_{0}^{x}(1-t^{2}+t^{4}-\cdots)\, dt \nonumber \\ \tan ^{-1}x &=& x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+\cdots +(-1)^{n}\frac{x^{2n+1}}{2n+1}+\cdots = \sum\limits_{k\,=\,0}^{\infty }(-1) ^{k}\dfrac{x^{2k+1}}{2k+1} \end{eqnarray*} \]
The radius of convergence is \(1\). To find the interval of convergence, we check \(x=-1\) and \(x=1\). For \(x=-1\), \[ -1+\frac{1}{3}-\frac{1}{5}+\frac{1}{7}-\cdots \]
For \(x=1\), we get \[ 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots \]
Both of these series satisfy the two conditions of the Alternating Series Test, and so each one converges. The interval of convergence is the closed interval \([-1,\,1] \).