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2017 Paper 1 Q8
D: 1500.0 B: 1516.0
proof by induction sequences recurrence relation convergence inequality rational approximation quadratic identity sqrt(2)

Two sequences are defined by \(a_1 = 1\) and \(b_1 = 2\) and, for \(n \ge 1\), \begin{equation*} \begin{split} a_{n+1} & = a_n+ 2b_n \,, \\ b_{n+1} & = 2a_n + 5b_n \,. \end{split} \end{equation*} Prove by induction that, for all \(n \ge 1\), \[ a_n^2+2a_nb_n - b_n^2 = 1 \,. \tag{\(*\)}\]

  1. Let \(c_n = \dfrac{a_n}{b_n}\). Show that \(b_n \ge 2 \times 5^{n-1}\) and use \((*)\) to show that \[ c_n \to \sqrt 2 -1 \text{ as } n\to\infty\,. \]
  2. Show also that \(c_n > \sqrt2 -1\) and hence that \(\dfrac2 {c_n+1} < \sqrt2 < c_n+1\). Deduce that \(\dfrac{140}{99}< \sqrt{2} < \dfrac{99}{70 }\,\).

Solution: Claim \(a_n^2+2a_nb_n - b_n^2 = 1\) for all \(n \geq 1\) Proof: (By induction) Base case: (\(n = 1\)). When \(n = 1\) we have \(a_1^2 + 2a_1 b_1-b_1^2 = 1^2+2\cdot1\cdot2-2^2 = 1\) as required. (Inductive step). Now we assume our result is true for some \(n =k\), ie \(a_k^2+2a_kb_k - b_k^2 = 1\), now consider \(n = k+1\) \begin{align*} && a_{k+1}^2+2a_{k+1}b_{k+1} - b_{k+1}^2 &= (a_k+2b_k)^2+2(a_k+2b_k)(2a_k+5b_k) - (2a_k+5b_k)^2 \\ &&&= a_k^2+4a_kb_k+4b_k^2 +4a_k^2+18a_kb_k+20b_k^2 - 4a_k^2-20a_kb_k-25b_k^2 \\ &&&= (1+4-4)a_k^2+(4+18-20)a_kb_k +(4+20-25)b_k^2 \\ &&&= a_k^2+2a_kb_k -b_k^2 = 1 \end{align*} Therefore since our statement is true for \(n = 1\) and when it is true for \(n=k\) it is true for \(n=k+1\) by the POMI it is true for \(n \geq 1\)

  1. Notice that \(b_{n+1} \geq 5 b_n\) and therefore \(b_n \geq 5^{n-1} b_1 = 2\cdot 5^{n-1}\), so \begin{align*} && 1 &= a_n^2+2a_nb_n - b_n^2\\ \Rightarrow && \frac1{b_n^2} &= c_n^2 + 2c_n - 1 \\ \to && 0 &= c_n^2 + 2c_n - 1 \quad \text{ as } n\to \infty \\ \end{align*} This has roots \(c = -1 \pm \sqrt{2}\), and since \(c_n > 0\) it must tend to the positive value, ie \(c_n \to \sqrt{2}-1\)
  2. Notice that \(c_n^2 + 2c_n - 1 > 0\) so either \(c_n > \sqrt{2}-1\) or \(c_n < -1-\sqrt{2}\), but again, since \(c_n > 0\) we must have \(c_n > \sqrt{2}-1\). Therefore \(\sqrt{2} < c_n + 1\) and \(1+c_n > \sqrt{2} \Rightarrow \frac{1}{1+c_n} < \frac{\sqrt{2}}2 \Rightarrow \frac{2}{1+c_n} < \sqrt{2}\) \begin{array}{c|c|c} n & a_n & b_n \\ \hline 1 & 1 & 2 \\ 2 & 5 & 12 \\ 3 & 29 & 70 \end{array} Therefore \(c_3 = \frac{29}{70}\) and so \(\frac{2}{1 + \frac{29}{70}} = \frac{140}{99} < \sqrt{2} < \frac{29}{70} + 1 = \frac{99}{70}\)

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2001 Paper 2 Q1
D: 1600.0 B: 1500.0
generalised binomial theorem binomial expansion square root approximation polynomial approximation error estimation rational approximation small x expansion

Use the binomial expansion to obtain a polynomial of degree \(2\) which is a good approximation to \(\sqrt{1-x}\) when \(x\) is small.

  1. 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.)
  2. 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*}

  1. \(\,\) \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*}
  2. 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*}

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