Question

Consider Gregory's expansion $$ \tan ^{-1} x=x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\cdots=\sum_{k=0}^{\infty} \frac{(-1)^{k}}{2 k+1} x^{2 k+1} $$ a. Derive Gregory's expansion using the definition $$ \tan ^{-1} x=\int_{0}^{x} \frac{d t}{1+t^{2}} $$ expanding the integrand in a Maclaurin series, and integrating the resulting series term by term. b. From this result, derive Gregory's series for \(\pi\) by inserting an appropriate value for \(x\) in the series expansion for \(\tan ^{-1} x\).

Short answer

a) Gregory's expansion for \(\tan^{-1}x\) can be derived by expanding the integrand \(\frac{1}{1+t^2}\) in a Maclaurin series and then integrating term by term leading to \(\tan^{-1} x = \sum_{k=0}^{\infty} \frac{(-1)^k}{2k+1} x^{2k+1}\). b) Gregory's series for \( \pi \) can be derived by inserting \( x = 1 \) into the series expansion that gives the series: \( \pi = 4\sum_{k=0}^{\infty} \frac{(-1)^k}{2k+1}\).

Step-by-step solution

Derive Gregory's expansion using the integral definition

Start by expanding the integrand \(\frac{1}{1+t^2}\) in a Maclaurin series around \(t=0\). This gives \(\frac{1}{1+t^2} = \sum_{k=0}^{\infty} (-1)^k t^{2k}\). Now integrate this series term by term, from 0 to \(x\), resulting in \(\int_{0}^{x} \frac{dt}{1+t^2} = \int_{0}^{x} \sum_{k=0}^{\infty} (-1)^k t^{2k} dt\). Finally, switch the integral and the sum (justification is required, but usually is accepted in undergraduate mathematics), resulting in: \(\sum_{k=0}^{\infty} (-1)^k \int_{0}^{x} t^{2k} dt = \sum_{k=0}^{\infty} \frac{(-1)^k}{2k+1} x^{2k+1}\), which is Gregory's expansion.

Derive Gregory's series for \( \pi \)

Inserting \( x = 1 \) in Gregory's expansion will yield the series expansion of \( \tan^{-1}(1) \), which is \( \frac{\pi}{4} \). So we can write: \( \sum_{k=0}^{\infty} \frac{(-1)^k}{2k+1} = \frac{\pi}{4} \). Then multiply through by 4 to isolate \( \pi \) on its own:\( 4\sum_{k=0}^{\infty} \frac{(-1)^k}{2k+1} = \pi \). This gives Gregory's series for \( \pi \)\.