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1991 Paper 2 Q5
D: 1600.0 B: 1516.0
reduction formulae integration curve sketching tan powers series alternating series ln 2 Leibniz formula for pi

Give a rough sketch of the function \(\tan^{k}\theta\) for \(0\leqslant\theta\leqslant\frac{1}{4}\pi\) in the two cases \(k=1\) and \(k\gg1\) (i.e. \(k\) is much greater than 1). Show that for any positive integer \(n\) \[ \int_{0}^{\frac{1}{4}\pi}\tan^{2n+1}\theta\,\mathrm{d}\theta=(-1)^{n}\left(\tfrac{1}{2}\ln2+\sum_{m=1}^{n}\frac{(-1)^{m}}{2m}\right), \] and deduce that \[ \sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{2m}=\tfrac{1}{2}\ln2. \] Show similarly that \[ \sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{2m-1}=\frac{\pi}{4}. \]

Solution:

TikZ diagram
Let \(\displaystyle I_n = \int_0^{\pi/4} \tan^{n} \theta \, \d \theta\), then \begin{align*} I_0 &= \int_0^{\pi/4} \tan \theta \d \theta \\ &= \left [ -\ln \cos \theta \right]_0^{\pi/4} \\ &= -\ln \frac{1}{\sqrt{2}} - 0 \\ &= \frac12 \ln 2 \\ \\ \\ I_{2n+1} &= \int_0^{\pi/4} \tan^{2n+1} \theta \, \d \theta \\&= \int_0^{\pi/4} \tan^{2n-1} \theta \tan ^2 \theta \, \d \theta \\ &= \int_0^{\pi/4} \tan^{2n-1} \theta (\sec^2 \theta - 1) \, \d \theta \\ &= \int_0^{\pi/4} \tan^{2n-1} \theta \sec^2 \theta - \tan^{2n-1} \theta \, \d \theta \\ &= \left[ \frac{1}{2n} \tan^{2n} \theta \right]_0^{\pi/4} - I_{2n-1} \\ &= \frac1{2n} - I_{2n-1} \end{align*} Therefore we can see that \(\displaystyle I_{2n+1} = (-1)^{n}\left(\tfrac{1}{2}\ln2+\sum_{m=1}^{n}\frac{(-1)^{m}}{2m}\right)\). As we can see as \(n \to \infty\), \(I_n \to 0\) Therefore \begin{align*} && 0 &= \tfrac{1}{2}\ln2+\lim_{n \to \infty} \sum_{m=1}^{n}\frac{(-1)^{m}}{2m} \\ \Rightarrow && \tfrac{1}{2}\ln2 &= \sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{2m} \end{align*} \begin{align*} && I_{-1} &= \int_0^{\pi/4} 1 \d \theta \\ &&&= \frac{\pi}{4} \end{align*} Therefore \(\displaystyle I_{2n} = (-1)^n \left ( \frac{\pi}{4} + \sum_{m=1}^n \frac{(-1)^m}{2m-1} \right)\) and since \(I_{2m} \to 0\) the same result follows.

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1987 Paper 3 Q7
D: 1500.0 B: 1500.0
sequences series convergence Taylor series integration inequality arctan remainder estimation Leibniz formula for pi proof

Prove that \[ \tan^{-1}t=t-\frac{t^{3}}{3}+\frac{t^{5}}{5}-\cdots+\frac{(-1)^{n}t^{2n+1}}{2n+1}+(-1)^{n+1}\int_{0}^{t}\frac{x^{2n+2}}{1+x^{2}}\,\mathrm{d}x. \] Hence show that, if \(0\leqslant t\leqslant1,\) then \[ \frac{t^{2n+3}}{2(2n+3)}\leqslant\left|\tan^{-1}t-\sum_{r=0}^{n}\frac{(-1)^{r}t^{2r+1}}{2r+1}\right|\leqslant\frac{t^{2n+3}}{2n+3}. \] Show that, as \(n\rightarrow\infty,\) \[ 4\sum_{r=0}^{n}\frac{(-1)^{r}}{(2r+1)}\rightarrow\pi, \] but that the error in approximating \(\pi\) by \({\displaystyle 4\sum_{r=0}^{n}\frac{(-1)^{r}}{(2r+1)}}\) is at least \(10^{-2}\) if \(n\) is less than or equal to \(98\).

Solution: We start by noticing that \(\displaystyle \tan^{-1} t = \int_0^t \frac{1}{1+x^2} \d x\). Consider the geometric series \(1-x^2+(-x^2)^2+ \cdots + (-x^2)^n = \frac{1-(-x^2)^{n+1}}{1+x^2}\). Therefore, \((1+x^2)(1-x^2+(-x^2)^2+ \cdots + (-x^2)^n) = 1-(-x^2)^{n+1}\) or \(1 = (1+x^2)(1-x^2+x^4-\cdots+(-1)^nx^{2n}) +(-1)^{n+1}x^{2n+2}\) \begin{align*} \tan^{-1} t &= \int_0^t \frac{1}{1+x^2} \d x \\ &= \int_0^t \frac{(1+x^2)(1-x^2+x^4-\cdots+(-1)^nx^{2n}) +(-1)^{n+1}x^{2n+2}}{x^2+1} \d x \\ &= \int_0^t (1-x^2+x^4-\cdots+(-1)^nx^{2n})\d x + \int_0^t \frac{(-1)^{n+1}x^{2n+2}}{x^2+1} \d x \\ &= t - \frac{t^3}{3}+\frac{t^5}{5}-\cdots + (-1)^n \frac{t^{2n+1}}{2n+1}+\int_0^t \frac{(-1)^{n+1}x^{2n+2}}{x^2+1} \d x \\ &= \sum_{r=0}^n \frac{(-1)^r t^{2r+1}}{2r+1} + \int_0^t \frac{(-1)^{n+1}x^{2n+2}}{x^2+1} \d x \\ \end{align*} Therefore we can say (for \(0 \leq t \leq 1\)) \begin{align*} \left | \tan^{-1} t - \sum_{r=0}^n \frac{(-1)^r t^{2r+1}}{2r+1} \right | &= \left | \int_0^t \frac{(-1)^{n+1}x^{2n+2}}{x^2+1} \d x \right | \\ &\leq \left | \int_0^t x^{2n+2} \d x \right | \\ &= \frac{t^{2n+3}}{2n+3} \\ \\ \left | \tan^{-1} t - \sum_{r=0}^n \frac{(-1)^r t^{2r+1}}{2r+1} \right | &= \left | \int_0^t \frac{(-1)^{n+1}x^{2n+2}}{x^2+1} \d x \right | \\ &\geq \left | \int_0^t \frac{(-1)^{n+1}x^{2n+2}}{1+1} \d x \right | \\ &= \frac{t^{2n+3}}{2(2n+3)} \\ \end{align*} Since \(\tan^{-1} 1 = \frac{\pi}{4}\) we must have that: \begin{align*} \lim_{n \to \infty} \left | \frac{\pi}{4} - \sum_{r=0}^{n}\frac{(-1)^{r}}{(2r+1)} \right | \to 0 \Rightarrow \lim_{n \to \infty} 4\sum_{r=0}^{n}\frac{(-1)^{r}}{(2r+1)} \to \pi \end{align*} However, \begin{align*} && \left | 4\sum_{r=0}^{n}\frac{(-1)^{r}}{(2r+1)} - \pi \right | &\geq 4 \frac{1}{2(2n+3)} \\ && &= \frac{2}{2n+3} \\ \\ && \frac{2}{2n+3} \geq 10^{-2} \\ \Leftrightarrow && 200 \geq 2n+3 \\ \Leftrightarrow && 197 \geq 2n \\ \Leftrightarrow && 98.5 \geq n \\ \end{align*} Therefore we need more than \(98\) terms to get two decimal places of accuracy. Not great!

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