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2023 Paper 2 Q5
D: 1500.0 B: 1500.0
recurrence relation sequences and series convergence inequality square root approximation newton-raphson method proof by induction

  1. The sequence \(x_n\) for \(n = 0, 1, 2, \ldots\) is defined by \(x_0 = 1\) and by \[x_{n+1} = \frac{x_n + 2}{x_n + 1}\] for \(n \geqslant 0\).
    1. Explain briefly why \(x_n \geqslant 1\) for all \(n\).
    2. Show that \(x_{n+1}^2 - 2\) and \(x_n^2 - 2\) have opposite sign, and that \[\left|x_{n+1}^2 - 2\right| \leqslant \tfrac{1}{4}\left|x_n^2 - 2\right|\,.\]
    3. Show that \[2 - 10^{-6} \leqslant x_{10}^2 \leqslant 2\,.\]
  2. The sequence \(y_n\) for \(n = 0, 1, 2, \ldots\) is defined by \(y_0 = 1\) and by \[y_{n+1} = \frac{y_n^2 + 2}{2y_n}\] for \(n \geqslant 0\).
    1. Show that, for \(n \geqslant 0\), \[y_{n+1} - \sqrt{2} = \frac{(y_n - \sqrt{2})^2}{2y_n}\] and deduce that \(y_n \geqslant 1\) for \(n \geqslant 0\).
    2. Show that \[y_n - \sqrt{2} \leqslant 2\left(\frac{\sqrt{2}-1}{2}\right)^{2^n}\] for \(n \geqslant 1\).
    3. Using the fact that \[\sqrt{2} - 1 < \tfrac{1}{2}\,,\] or otherwise, show that \[\sqrt{2} \leqslant y_{10} \leqslant \sqrt{2} + 10^{-600}\,.\]

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2007 Paper 2 Q1
D: 1600.0 B: 1516.0
generalised binomial theorem binomial expansion fractional exponent square root approximation cube root approximation numerical approximation series expansion

In this question, you are not required to justify the accuracy of the approximations.

  1. Write down the binomial expansion of \(\displaystyle \left( 1+\frac k {100} \right)^{\!\frac12}\)in ascending powers of \(k\), up to and including the \(k^3\) term.
    1. Use the value \(k=8\) to find an approximation to five decimal places for \(\sqrt{3}\,\).
    2. By choosing a suitable integer value of \(k\), find an approximation to five decimal places for \(\sqrt6\,\).
  2. By considering the first two terms of the binomial expansion of \(\displaystyle \left( 1+\frac k {1000} \right)^{\!\frac13}\), show that \(\dfrac{3029}{2100}\) is an approximation to \(\sqrt[3]{3}\).

Solution:

  1. Using the generalise binomial theorem \begin{align*} \left( 1+\frac k {100} \right)^{\frac12} &= 1 + \frac12 \frac{k}{100} + \frac{\tfrac12 \cdot \left ( -\tfrac12\right)}{2!} \left (\frac{k}{100} \right)^2 + \frac{\tfrac12 \cdot \left ( -\tfrac12\right)\cdot \left ( -\tfrac32\right)}{3!} \left (\frac{k}{100} \right)^3 + \cdots \\ &= 1 + \frac{1}{200}k - \frac{1}{80\,000}k^2 + \frac{1}{16\,000\,000}k^3 + \cdots \end{align*}
    1. If \(k = 8\), \begin{align*} && \left( 1+\frac 8 {100} \right)^{\frac12} &= 1 + \frac{1}{200}8 - \frac{1}{80\,000}8^2 + \frac{1}{16\,000\,000}8^3 + \cdots \\ \Rightarrow && \frac{6\sqrt{3}}{10} &\approx 1 + 0.04 - 0.0008 + 0.000032 \\ &&&= 1.039232\\ \Rightarrow && \sqrt{3} &\approx 1.73205 \, (5\, \text{d.p.}) \end{align*}
    2. If \(k = -4\), \begin{align*} && \left( 1-\frac 4 {100} \right)^{\frac12} &= 1 - \frac{1}{200}4 - \frac{1}{80\,000}4^2 - \frac{1}{16\,000\,000}4^3 + \cdots \\ \Rightarrow && \frac{4\sqrt{6}}{10} &\approx 1 -0.02-0.0002 -0.000004 \\ &&&= 0.979796\\ \Rightarrow && \sqrt{6} &\approx 2.44949\, (5\, \text{d.p.}) \end{align*}
  2. \(\,\) \begin{align*} && \left( 1+\frac k {1000} \right)^{\!\frac13} &= 1 + \frac13 \frac{k}{1000} + \cdots \\ &&&= 1 + \frac{k}{3\,000} + \cdots \\ && 3 \times 7^3 &= 1029 \\ \Rightarrow && \left( 1+\frac {29} {1000} \right)^{\!\frac13} &\approx 1 + \frac{29}{3\,000} \\ \Rightarrow && \frac{7\sqrt[3]{3}}{10} &\approx \frac{3\,029}{3000} \\ \Rightarrow && \sqrt[3]{3} &= \frac{3\,029}{2\,100} \end{align*}

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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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