3.6.2 Example: Taylor series of arctan function and \(\pi\)-calculation

\begin{eqnarray*} \arctan1&=&\frac{\pi}{4}\\ \arctan x&\stackrel{?}{=}&\sum_{k=0}^\infty a_kx^k\qquad\mbox{Taylor-Series}\end{eqnarray*}

We could calculate \(f^n(0)\) but it may be simpler in the following very instructive way: We assume that \(|x|\lt1\) \begin{eqnarray*} \arctan x&=&\int_0^x\frac{1}{1+t^2}dt\;\leftarrow\,\mbox{note: } [\arctan x]'=\frac{1}{1+x^2}\\ &=&\int_0^x\sum_{n=0}^\infty(-1)^n(t^2)^n dt\;\leftarrow\,\mbox{note:} \frac{1}{1-a}=\sum_{n=0}^\infty a^n,\;|a|\lt1\\ &=&\sum_{n=0}^\infty (-1)^n\int_0^x t^{2n} dt =\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n+1} \end{eqnarray*}

Thus: \begin{eqnarray*}\arctan x&=&x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\ldots\;\;;\mbox{also convergent for $x=1$ (can be shown)}\\ x&=&1\rightarrow\;\arctan1=\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\ldots\\ &\rightarrow&\mbox{very slow converging series: e.g.\ $10^{10}$ terms for {\bf10} precise digits of $\pi$!!}\end{eqnarray*}

Much better: To exploit the properties of the arctan function \begin{eqnarray*}\frac{\pi}{4}&=&4\arctan\frac{1}{5}-\arctan\frac{1}{239}\\ &=&\frac{4}{5}\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}\left(\frac{1}{5}\right)^{2k}-\frac{1}{239}\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}\left(\frac{1}{239}\right)^{2k}\\ &\sim&\mbox{10 terms for $\gt10$ precise digits of $\pi$!!}\;\;\left(\frac{1}{5}\right)^{20}\sim10^{-14} \end{eqnarray*}

It remains to prove that

This we show in three steps by each time applying the addition theorem for the arctan function

Combining these equation we easily get the desired result.

With frame

Taylor series and error estimation

© J. Carstensen (Math for MS)