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2016 Paper 2 Q8
Evaluate the integral \[ \hphantom{ \ \ \ \ \ \ \ \ \ (m> \tfrac12)\,.} \int_{m-\frac12} ^\infty \frac 1{x^2}\, \d x { \ \ \ \ \ \ \ \ \ (m > \tfrac12)\,.} \] Show by means of a sketch that \[ \sum_{r=m}^n \frac 1 {r^2} \approx \int_{m-\frac12}^{n+\frac12} \frac1 {x^2} \, \d x \,, \tag{\(*\)} \] where \(m\) and \(n\) are positive integers with \(m < n\).
- You are given that the infinite series \(\displaystyle \sum_{r=1}^\infty \frac 1 {r^2}\) converges to a value denoted by \(E\). Use \((*)\) to obtain the following approximations for \(E\): \[ E\approx 2\,; \ \ \ \ E\approx \frac53\,; \ \ \ \ E\approx \frac{33}{20} \,.\]
- Show that, when \(r\) is large, the error in approximating \(\dfrac 1{r^2}\) by \(\displaystyle \int_{r-\frac12}^{r+\frac12} \frac 1 {x^2} \, \d x\) is approximately \(\dfrac 1{4r^4}\,\). Given that \(E \approx 1.645\), show that \(\displaystyle \sum_{r=1}^\infty \frac1{r^4} \approx 1.08\, \).
Solution: \begin{align*} && \int_{m-\frac12}^\infty \frac{1}{x^2} \d x &= \lim_{K \to \infty} \left [ -x^{-1} \right]_{m-\frac12}^K \\ &&&= \frac{1}{m-\frac12} - \lim_{K \to \infty }\frac{1}K \\ &&&= \frac{1}{m-\frac12} \end{align*}
- \(\,\) \begin{align*} E &\approx \int_{1-\frac12}^\infty \frac1{x^2} \d x = \frac{1}{1-\frac12} = 2 \\ E &\approx 1 + \int_{2-\frac12}^\infty \frac1{x^2} \d x= 1 + \frac{1}{2 - \frac12} = \frac53 \\ E &\approx 1 +\frac14 + \int_{3-\frac12}^\infty \frac1{x^2} \d x= \frac54 + \frac{1}{3-\frac12} \\ &= \frac54+\frac{2}{5} = \frac{33}{20} \end{align*}
- The error is \begin{align*} && \epsilon &= \int_{r-\frac12}^{r+\frac12} \frac 1 {x^2} \, \d x - \frac1{r^2} \\ &&&= \frac{1}{r-\frac12} - \frac{1}{r + \frac12} - \frac1{r^2} \\ &&&= \frac{1}{r^2 - \frac14} - \frac1{r^2} \\ &&&= \frac{\frac14}{r^2(r^2-\frac14)} \\ &&&\approx \frac{1}{4r^4} \end{align*} Therefore \begin{align*} && \sum_{n=1}^\infty \frac1{r^4} &\approx 4 \left ( 1 +\frac14 + \int_{3-\frac12}^\infty \frac1{x^2} \d x-\sum_{r=1}^\infty \frac{1}{r^2} \right) + 1 + \frac{1}{2^4}\\ &&&= 4 \left ( \frac{33}{20}-1.645 \right) + 1 + \frac{1}{2^4} \\ &&&= 4 \left ( 1.65-1.645 \right) + 1 + \frac{1}{2^4} \\ &&&= 1.0825 \approx 1.08 \end{align*}
2001 Paper 2 Q1
Use the binomial expansion to obtain a polynomial of degree \(2\) which is a good approximation to \(\sqrt{1-x}\) when \(x\) is small.
- By taking \(x=1/100\), show that \(\sqrt{11}\approx79599/24000\), and estimate, correct to 1 significant figure, the error in this approximation. (You may assume that the error is given approximately by the first neglected term in the binomial expansion.)
- Find a rational number which approximates \(\sqrt{1111}\) with an error of about \(2 \times {10}^{-12}\).
Solution: \begin{align*} && \sqrt{1-x} &= (1-x)^{\frac12} \\ &&&= 1 -\frac12x+\frac{\frac12 \cdot \left (-\frac12 \right)}{2!}x^2 + \frac{\frac12 \cdot \left (-\frac12 \right) \cdot \left (-\frac32 \right)}{3!} x^3\cdots \\ &&&\approx 1-\frac12x - \frac18x^2 \end{align*}
- \(\,\) \begin{align*} && \frac{3\sqrt{11}}{10} &= \sqrt{1-1/100} \\ &&&\approx 1 - \frac{1}2 \frac{1}{100} - \frac{1}{8} \frac{1}{100^2} \\ &&&= \frac{80000-400-1}{80000} \\ &&&= \frac{79599}{80000}\\ \Rightarrow && \sqrt{11} &\approx \frac{79599}{24000} \\ \\ &&\text{error} &\approx \frac{1}{16} \frac{10}3 \frac{1}{100^3} \\ &&&= \frac{1}{48} 10^{-5} \\ &&&\approx 2 \times 10^{-7} \end{align*}
- Taking \(x = 1/10^4\) we have \begin{align*} && \frac{3 \sqrt{1111}}{100} &= \sqrt{1-1/10^4} \\ &&&\approx 1 - \frac12 \frac1{10^4} - \frac18 \frac{1}{10^8} \\ &&&= \frac{799959999}{800000000} \\ \Rightarrow && \sqrt{1111} & \approx \frac{266653333}{8000000} \\ \\ && \text{error} &\approx \frac{100}{3} \frac{1}{16} \frac{1}{10^{12}} \\ &&&= \frac{1}{48} \frac{1}{10^{10}} \\ &&&\approx 2 \times 10^{-12} \end{align*}