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2012 Paper 2 Q1
D: 1600.0 B: 1500.0
generalised binomial theorem power series expansion coefficient extraction generating functions partial fractions combinatorics geometric series

Write down the general term in the expansion in powers of \(x\) of \((1-x^6)^{-2}\,\).

  1. Find the coefficient of \(x^{24}\) in the expansion in powers of \(x\) of \[ (1-x^6)^{-2} (1-x^3)^{-1}\,.\] Obtain also, and simplify, formulae for the coefficient of \(x^n\) in the different cases that arise.
  2. Show that the coefficient of \(x^{24}\) in the expansion in powers of \(x\) of \[ (1-x^6)^{-2} (1-x^3)^{-1} (1-x)^{-1}\,\] is \(55\), and find the coefficients of \(x^{25}\) and \(x^{66}\).

Solution: \(\displaystyle (1-x^6)^{-2} = \sum_{n=0}^{\infty} (n+1)x^{6n}\)

  1. \(\,\) \begin{align*} && f(x) &= (1-x^6)^{-2}(1-x^3)^{-1} \\ &&&= \left ( \sum_{n=0}^{\infty} (n+1)x^{6n} \right) \left ( \sum_{n=0}^{\infty} x^{3n} \right) \\ [x^{24}]: && c_{24} &= 1 + 2+ 3+4+5 = 15 \end{align*} Clearly \(n\) must be a multiple of \(3\). If \(n = 6k\) then we have \(1 + 2 + \cdots + (k+1) = \frac{(k+1)(k+2)}{2}\) If \(n = 6k+3\) then we have \(1 + 2 + \cdots + (k+1) = \frac{(k+1)(k+2)}{2}\) the same way, we just must always get one extra \(x^3\) term from the second expansion.
  2. We can obtain \(x^{24}\) from the product of \((1-x^6)^{-2}(1-x^3)^{-1}\) and \((1-x)^{-1}\) in the following ways: \begin{array}{c|c|c} (1-x^6)^{-2}(1-x^3)^{-1} & (1-x)^{-1} & \text{product} \\ \hline 15x^{24} & x^0 & 15x^{24} \\ 10x^{21} & x^3 & 10x^{24} \\ 10x^{18} & x^6 & 10x^{24} \\ 6x^{15} & x^9 & 6x^{24} \\ 6x^{12} & x^{12} & 6x^{24} \\ 3x^{9} & x^{15} & 3x^{24} \\ 3x^{6} & x^{18} & 3x^{24} \\ x^{3} & x^{21} & x^{24} \\ x^{0} & x^{24}& x^{24} \end{array} So the total is \(55\). Similarly for \(25\) we can only obtain this in the same ways but also taking an extra power of \(x\) from the geometric series, ie \(55\) For \(66\) we obtain by similar reasoning that it is: \(\frac{13\cdot12}{2} + 2 \left (1 + 3 + \cdots + \frac{13 \cdot 12}{2} \right) = \frac{13\cdot12}{2} + 2 \binom{14}{3} = \frac{13 \cdot 12}2 ( 1 + \frac{30}{3}) = 11 \cdot 6 \cdot 13 = 858\)

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1992 Paper 3 Q1
D: 1700.0 B: 1500.0
hyperbolic functions Maclaurin series logarithm differentiation power series expansion sinh cosh even functions Taylor expansion

  1. Given that \[ \mathrm{f}(x)=\ln(1+\mathrm{e}^{x}), \] prove that \(\ln[\mathrm{f}'(x)]=x-\mathrm{f}(x)\) and that \(\mathrm{f}''(x)=\mathrm{f}'(x)-[\mathrm{f}'(x)]^{2}.\) Hence, or otherwise, expand \(\mathrm{f}(x)\) as a series in powers of \(x\) up to the term in \(x^{4}.\)
  2. Given that \[ \mathrm{g}(x)=\frac{1}{\sinh x\cosh2x}, \] explain why \(\mathrm{g}(x)\) can not be expanded as a series of non-negative powers of \(x\) but that \(x\mathrm{g}(x)\) can be so expanded. Explain also why this latter expansion will consist of even powers of \(x\) only. Expand \(x\mathrm{g}(x)\) as a series as far as the term in \(x^{4}.\)

Solution:

  1. \begin{align*} && f(x) &= \ln (1+e^x) \\ && f'(x) &= \frac{1}{1+e^x} \cdot e^x \\ &&&= \frac{e^x}{1+e^x} \\ \Rightarrow && \ln [f'(x)] &= x - \ln (1+e^x) \\ &&&= x - f(x) \\ \\ \Rightarrow && \frac{f''(x)}{f'(x)} &= 1 - f'(x) \\ \Rightarrow && f''(x) &= f'(x) - [f'(x)]^2 \\ && f'''(x) &= f''(x) - 2f'(x) f''(x) \\ && f^{(4)}(x) &= f'''(x) - 2[f''(x)]^2-2f'(x)f'''(x) \end{align*} \begin{align*} f(0) &= \ln 2 \\ f'(0) &= \tfrac12 \\ f''(0) &= \tfrac12 -\tfrac14 \\ &= \tfrac 14 \\ f'''(0) &= \tfrac14 - 2 \tfrac12 \tfrac 14 \\ &= 0 \\ f^{(4)}(0) &= -2 \cdot \tfrac1{16} \\ &= -\frac18 \end{align*} Therefore \(f(x) = \ln 2 + \tfrac12 x + \tfrac18 x^2 - \frac1{8 \cdot 4!} x^4 + O(x^5)\)
  2. As \(x \to 0\), \(g(x) \to \infty\) therefore there can be no power series about \(0\). But as \(x \to 0, x g(x) \not \to \infty\) as \(\frac{x}{\sinh x}\) is well behaved. We can also notice that \(x g(x)\) is an even function, since \(\cosh x\) is even and \(\frac{x}{\sinh x}\) is even, therefore the power series will consist of even powers of \(x\) \begin{align*} \lim_{x \to 0} \frac{x}{\sinh x \cosh 2 x} &= \lim_{x \to 0} \frac{x}{\sinh x} \cdot \lim_{x \to 0} \frac{1}{\cosh2 x} \\ &= 1 \end{align*} Notice that \begin{align*} \frac{x}{\sinh x \cosh 2 x} &= \frac{4x}{(e^x - e^{-x})(e^{2x}+e^{-2x})} \\ &= \frac{4x}{(2x + \frac{x^3}{3} + \cdots)(2 + 4x^2 + \frac43 x^4 + \cdots )} \\ &= \frac{1}{1+\frac{x^2}{6}+\frac{x^4}{5!} + \cdots } \frac{1}{1 + 2x^2 + \frac23 x^4 + \cdots } \\ &= \left (1-(\frac{x^2}{6} + \frac{x^4}{5!})+ (\frac{x^2}{6} )^2 + O(x^6)\right) \left (1-(2x^2+\frac23 x^4)+ (2x^2)^2 + O(x^6)\right) \\ &= \left (1 - \frac16 x^2 + \frac{7}{360} x^4 + O(x^6) \right) \left (1 - 2x^2+ \frac{10}3x^4 + O(x^6) \right) \\ &= 1 - \frac{13}{6} x^2 + \frac{1327}{360}x^4 + O(x^6) \end{align*}

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1991 Paper 3 Q1
D: 1700.0 B: 1501.5
Taylor series partial fractions telescoping sum logarithmic series exponential series power series expansion geometric series

  1. Evaluate \[ \sum_{r=1}^{n}\frac{6}{r(r+1)(r+3)}. \]
  2. Expand \(\ln(1+x+x^{2}+x^{3})\) as a series in powers of \(x\), where \(\left|x\right|<1\), giving the first five non-zero terms and the general term.
  3. Expand \(\mathrm{e}^{x\ln(1+x)}\) as a series in powers of \(x\), where \(-1 < x\leqslant1\), as far as the term in \(x^{4}\).

Solution:

  1. \begin{align*} && \frac{6}{r(r+1)(r+3)} &= \frac{2}{r} - \frac{3}{r+1} + \frac{1}{r+3} \\ \Rightarrow && \sum_{r=1}^n \frac{6}{r(r+1)(r+3)} &= \sum_{r=1}^n \l \frac{2}{r} - \frac{3}{r+1} + \frac{1}{r+3} \r \\ &&&= \sum_{r=1}^n \frac{2}{r} - \sum_{r=1}^n \frac{3}{r+1} + \sum_{r=1}^n \frac{1}{r+3} \\ &&&= \sum_{r=1}^n \frac{2}{r} - \sum_{r=2}^{n+1} \frac{3}{r} + \sum_{r=3}^{n+2} \frac{1}{r} \\ &&& = \frac{2}{1} + \frac{2}{2} - \frac{3}{2} - \frac{3}{n+1} + \frac{1}{n+1} + \frac{1}{n+2} \\ &&& = \frac{3}{2} - \frac{2}{n+1} + \frac{1}{n+2} \end{align*}
  2. \begin{align*} && \ln (1 + x+ x^2 + x^3) &= \ln \l \frac{1-x^4}{1-x} \r \\ &&&= \ln (1-x^4) - \ln(1-x) \\ &&&= \sum_{k=1}^{\infty} -\frac{x^{4k}}{k} - \sum_{k=1}^{\infty} - \frac{x^k}{k} \\ &&&= x + \frac12x^2+\frac13x^3-\frac34x^4+\frac15x^5 + \cdots \\ &&&= \sum_{k=1}^{\infty}a_k x^k \end{align*} Where \(a_k = \frac{1}{k}\) if \(k \neq 0 \pmod{4}\) otherwise \(a_k = -\frac{3}{k}\) if \(k \equiv 0 \pmod{4}\)
  3. \begin{align*} \exp(x \ln (1+x) ) &= \exp\l x \l x-\frac12x^2+\frac13x^3-\cdots \r \r \\ &= \exp\l x^2-\frac12x^3+\frac13x^4 \r \\ &= 1 + \l x^2-\frac12x^3+\frac13x^4 \r + \frac12 \l x^2-\frac12x^3+\frac13x^4 \r^2 + \cdots \\ &= 1 + x^2-\frac12x^3+\frac13x^4 + \frac12x^4 + \cdots \\ &= 1 + x^2 -\frac12x^3+\frac56x^4+\cdots \end{align*}

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