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2023 Paper 3 Q6
D: 1500.0 B: 1500.0
hyperbolic functions maclaurin series inequality curve sketching differentiation inverse trigonometric functions stationary points monotonic functions

  1. By considering the Maclaurin series for \(\mathrm{e}^x\), show that for all real \(x\), \[\cosh^2 x \geqslant 1 + x^2.\] Hence show that the function \(\mathrm{f}\), defined for all real \(x\) by \(\mathrm{f}(x) = \tan^{-1} x - \tanh x\), is an increasing function. Sketch the graph \(y = \mathrm{f}(x)\).
  2. Function \(\mathrm{g}\) is defined for all real \(x\) by \(\mathrm{g}(x) = \tan^{-1} x - \frac{1}{2}\pi \tanh x\).
    1. Show that \(\mathrm{g}\) has at least two stationary points.
    2. Show, by considering its derivative, that \((1+x^2)\sinh x - x\cosh x\) is non-negative for \(x \geqslant 0\).
    3. Show that \(\dfrac{\cosh^2 x}{1+x^2}\) is an increasing function for \(x \geqslant 0\).
    4. Hence or otherwise show that \(\mathrm{g}\) has exactly two stationary points.
    5. Sketch the graph \(y = \mathrm{g}(x)\).

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2022 Paper 3 Q4
D: 1500.0 B: 1500.0
hyperbolic functions proof by induction infinite product maclaurin series limits logarithms trigonometric identities

You may assume that all infinite sums and products in this question converge.

  1. Prove by induction that for all positive integers \(n\), \[ \sinh x = 2^n \cosh\!\left(\frac{x}{2}\right) \cosh\!\left(\frac{x}{4}\right) \cdots \cosh\!\left(\frac{x}{2^n}\right) \sinh\!\left(\frac{x}{2^n}\right) \] and deduce that, for \(x \neq 0\), \[ \frac{\sinh x}{x} \cdot \frac{\dfrac{x}{2^n}}{\sinh\!\left(\dfrac{x}{2^n}\right)} = \cosh\!\left(\frac{x}{2}\right) \cosh\!\left(\frac{x}{4}\right) \cdots \cosh\!\left(\frac{x}{2^n}\right)\,. \]
  2. You are given that the Maclaurin series for \(\sinh x\) is \[ \sinh x = \sum_{r=0}^{\infty} \frac{x^{2r+1}}{(2r+1)!}\,. \] Use this result to show that, as \(y\) tends to \(0\), \(\dfrac{y}{\sinh y}\) tends to \(1\). Deduce that, for \(x \neq 0\), \[ \frac{\sinh x}{x} = \cosh\!\left(\frac{x}{2}\right) \cosh\!\left(\frac{x}{4}\right) \cdots \cosh\!\left(\frac{x}{2^n}\right) \cdots\,. \]
  3. Let \(x = \ln 2\). Evaluate \(\cosh\!\left(\dfrac{x}{2}\right)\) and show that \[ \cosh\!\left(\frac{x}{4}\right) = \frac{1 + 2^{\frac{1}{2}}}{2 \times 2^{\frac{1}{4}}}\,. \] Use part (ii) to show that \[ \frac{1}{\ln 2} = \frac{1 + 2^{\frac{1}{2}}}{2} \times \frac{1 + 2^{\frac{1}{4}}}{2} \times \frac{1 + 2^{\frac{1}{8}}}{2} \cdots\,. \]
  4. Show that \[ \frac{2}{\pi} = \frac{\sqrt{2}}{2} \times \frac{\sqrt{2+\sqrt{2}}}{2} \times \frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2} \cdots\,. \]

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2013 Paper 3 Q2
D: 1700.0 B: 1516.0
Taylor series Maclaurin series Leibniz rule differential equation arcsin recurrence relation infinite series evaluation repeated differentiation

In this question, you may ignore questions of convergence. Let \(y= \dfrac {\arcsin x}{\sqrt{1-x^2}}\,\). Show that \[ (1-x^2)\frac {\d y}{\d x} -xy -1 =0 \] and prove that, for any positive integer \(n\), \[ (1-x^2) \frac{\d^{n+2}y}{\d x^{n+2}} - (2n+3)x \frac{\d ^{n+1}y}{\d x ^{n+1}} -(n+1)^2 \frac{\d^ny}{\d x^n}=0\, . \] Hence obtain the Maclaurin series for \( \dfrac {\arcsin x}{\sqrt{1-x^2}}\,\), giving the general term for odd and for even powers of \(x\). Evaluate the infinite sum \[ 1 + \frac 1 {3!} + \frac{2^2}{5!} + \frac {2^2\times 3^2}{7!}+\cdots + \frac {2^2\times 3^2\times \cdots \times n^2}{(2n+1)!} + \cdots\,. \]

Solution: \begin{align*} && y &= \frac{\arcsin x}{\sqrt{1-x^2}} \\ && \frac{\d y}{\d x} &= \frac{(1-x^2)^{-1/2} \cdot (1-x^2)^{1/2}-\arcsin x \cdot (-x)(1-x^2)^{-1/2}}{1-x^2} \\ &&&= \frac{1+ xy}{1-x^2} \\ \Rightarrow && 0 &= (1-x^2) \frac{\d y}{\d x} -xy-1\\ \\ \frac{\d^n}{\d x^{n+1}}: && 0 &= \left ( (1-x^2) y' \right)^{(n+1)} - (xy)^{(n+1)} \\ \Rightarrow && 0 &= (1-x^2)y^{(n+2)} + \binom{n+1}{1}(1-x^2)^{(1)}y^{(n+1)}+\binom{n+1}{2} (1-x^2)^{(2)}y^{(n)} - (xy^{(n+1)} +\binom{n+1}{1} y^{(n)} ) \\ &&&= (1-x^2)y^{(n+2)}+\left ( (n+1)\cdot(-2x)-x \right)y^{(n+1)} + \left ( \frac{(n+1)n}{2} \cdot (-2)-(n+1) \right)y^{(n)} \\ &&&= (1-x^2)y^{(n+2)}-\left ( 2n+3 \right)xy^{(n+1)} - \left ( (n+1)n+(n+1)\right)y^{(n)} \\ &&&= (1-x^2)y^{(n+2)}-\left ( 2n+3 \right)xy^{(n+1)} - \left ( n+1\right)^2y^{(n)} \\ \end{align*} Since \(y(0) = 0, y'(0) = 1\) we can look at the recursion: \(y^{(n+2)} - (n+1)^2y^{(n)}\) for larger terms, ie \(y^{(2k)}(0) = 0\) \(y^{(1)}(0) = 1, y^{(3)}(0) = (1+1)^2 \cdot 1 = 2^2, y^{(5)}(0) = (3+1)^2 y^{(3)} = 4^2 \cdot 2^2\) and \(y^{(2k+1)}(0) = (2k)^2 \cdot (2k-2)^2 \cdots 2^2 \cdot 1^2 = 2^{2k} \cdot (k!)^2\). Therefore \begin{align*} && \frac{\arcsin x}{\sqrt{1-x^2}} &= \sum_{k=0}^{\infty} \frac{2^{2k} \cdot (k!)^2}{(2k+1)!} x^{2k+1} \\ \\ \Rightarrow && \frac{\arcsin \frac12}{\sqrt{1-\left (\frac12 \right)^2}} &= \sum_{k=0}^{\infty} \frac{2^{2k} \cdot (k!)^2}{(2k+1)!} 2^{-2k-1}\\ &&&= \frac12 \sum_{k=0}^{\infty} \frac{ (k!)^2}{(2k+1)!} \\ &&&= \frac12 \left ( 1 + \frac1{3!} + \frac{2^2}{5!} + \cdots+ \right) \\ \Rightarrow&& S &= 2 \frac{2\frac{\pi}{6}}{\sqrt{3}} = \frac{2\pi}{3\sqrt{6}} \end{align*}

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2010 Paper 3 Q7
D: 1700.0 B: 1516.0
differential equation Maclaurin series proof by induction Leibniz rule recurrence relation Chebyshev polynomial trigonometric higher derivatives

Given that \(y = \cos(m \arcsin x)\), for \(\vert x \vert <1\), prove that \[ (1-x^2) \frac {\d^2 y}{\d x^2} -x \frac {\d y}{\d x} +m^2y=0\,. \] Obtain a similar equation relating \(\dfrac{\d^3y}{\d x^3}\,\), \(\dfrac{\d^2y}{\d x^2}\, \) and \(\, \dfrac{\d y}{\d x}\,\), and a similar equation relating \(\dfrac{\d^4y}{\d x^4}\,\), \(\dfrac{\d^3y}{\d x^3}\,\) and \(\,\dfrac{\d^2 y}{\d x^2}\,\). Conjecture and prove a relation between \(\dfrac{\d^{n+2}y}{\d x^{n+2}}\,\), \(\dfrac{\d^{n+1}y}{\d x^{n+1}}\;\) and \(\;\dfrac{\d^n y}{\d x^n}\,\). Obtain the first three non-zero terms of the Maclaurin series for \(y\). Show that, if \(m\) is an even integer, \(\cos m\theta\) may be written as a polynomial in \(\sin\theta\) beginning \[ 1 - \frac{m^2\sin^2\theta}{2!}+ \frac{m^2(m^2-2^2)\sin^4\theta}{4!} -\cdots \,. \, \tag{\(\vert\theta\vert < \tfrac12 \pi\)} \] State the degree of the polynomial.

Solution: \begin{align*} && y &= \cos(m \arcsin x) \\ && y' &= -m \sin (m \arcsin x) \cdot (1-x^2)^{-\frac12} \\ && y'' &= -m^2 \cos(m \arcsin x) \cdot (1-x^2)^{-1} -m \sin(m \arcsin x) \cdot (1-x^2)^{-\frac32} \cdot (-x) \\ &&&= -m^2 y (1-x^2)^{-1} + x(1-x^2)^{-1} y' \\ \Rightarrow && 0 &= (1-x^2)y'' - x y' + m^2y \\ \\ && 0 &= (1-x^2)y^{(3)} -2xy'' - xy''-y' + m^2y' \\ &&&= (1-x^2)y^{(3)} - 3xy'' + (m^2-1)y' \\ \\ && 0 &= (1-x^2)y^{(4)} - 2xy^{(3)} - 3xy^{(3)} - 3y^{(2)} + (m^2-1)y^{(2)} \\ &&&= (1-x^2)y^{(4)}- 5xy^{(3)} - (m^2-4)y^{(2)} \end{align*} Claim: \(0 = (1-x^2)y^{(n+2)} - (2n+1)y^{(n+1)} + (m^2-n^2)y^{(n)}\) Proof: (By induction) Clearly the first few base cases are true. Suppose it is true for some \(n\), then \begin{align*} && 0 &= (1-x^2)y^{(n+2)} - (2n+1)xy^{(n+1)} + (m^2-n^2)y^{(n)} \\ \Rightarrow && 0 &= (1-x^2)y^{(n+3)} - 2xy^{(n+2)} - (2n+1)xy^{(n+2)} - (2n+1)y^{(n+1)} + (m^2-n^2)y^{(n+1)} \\ &&&= (1-x^2)y^{(n+3)} - (2n+3)xy^{(n+2)} + (m^2-n^2-2n-1)y^{(n+1)} \\ &&&= (1-x^2)y^{(n+1+2)} - (2(n+1)+1)xy^{(n+1+1)} +(m^2-(n+1)^2)y^{(n)} \end{align*} And so we can conclude the result by induction. Notice that \begin{align*} && y(0) &= \cos(m 0) = 1 \\ && y'(0) &= -m\sin(m 0) = 0 \\ && y''(0) &= -m^2 y(0) = -m^2\\ \end{align*} Notice that \(y^{(n+2)}(0) + (m^2-n^2)y^{(n)} = 0\) so in particular all the odd terms will be \(0\) and the even terms will be \(1, -m^2, m^2(m^2-2^2), \cdots\), therefore \begin{align*} && \cos (m \arcsin x) &= 1 -\frac{m^2}{2!} x^2 + \frac{m^2(m^2-2^2)}{4!}x^4 - \cdots \\ \Rightarrow && \cos(m \theta) &= 1 - \frac{m^2}{2!} \sin^2 \theta + \frac{m^2(m^2-2^2)}{4!} \sin^4 \theta \end{align*} Notice that if \(m\) is even, then at some point we will have \(m^2-m^2\) appearing in our expansion and all remaining terms will be zero. Therefore we will end up with a polynomial series, of degree \(m\) in \(\sin \theta\)

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2009 Paper 3 Q2
D: 1700.0 B: 1484.0
second order differential equation power series series solution recurrence relation comparing coefficients trigonometric Maclaurin series

  1. Let \(\displaystyle y= \sum_{n=0}^\infty a_n x^n\,\), where the coefficients \(a_n\) are independent of \(x\) and are such that this series and all others in this question converge. Show that \[ \displaystyle y'= \sum_{n=1}^\infty na_n x^{n-1}\,, \] and write down a similar expression for \(y''\). Write out explicitly each of the three series as far as the term containing \(a_3\).
  2. It is given that \(y\) satisfies the differential equation \[ xy''-y'+4x^3y =0\,. \] By substituting the series of part (i) into the differential equation and comparing coefficients, show that \(a_1=0\). Show that, for \(n\ge4\), \[ a_n =- \frac{4}{n(n-2)}\, a_{n-4}\,, \] and that, if \(a_0=1\) and \(a_2=0\), then \( y=\cos (x^2)\,\). Find the corresponding result when \(a_0=0\) and \(a_2=1\).

Solution:

  1. Let \(\displaystyle y= \sum_{n=0}^\infty a_n x^n\,\) then \begin{align*} y' &= \frac{\d}{\d x} \l \sum_{n=0}^\infty a_n x^n \r \\ &= \sum_{n=0}^\infty \frac{\d}{\d x} \l a_n x^n \r \\ &= \sum_{n=0}^\infty n a_n x^{n-1} \\ &= \sum_{n=1}^\infty n a_n x^{n-1} \\ \\ y'' &= \frac{\d}{\d x} \l\sum_{n=1}^\infty n a_n x^{n-1} \r \\ &= \sum_{n=1}^\infty \frac{\d}{\d x} \l n a_n x^{n-1} \r \\ &= \sum_{n=1}^\infty n(n-1) a_n x^{n-2} \\ &= \sum_{n=2}^\infty n(n-1) a_n x^{n-2} \\ \\ y &= a_0 + a_1 x+ a_2x^2 + a_3x^3 + \cdots \\ y'&= a_1 + 2a_2x+3a_3x^2 + \cdots \\ y'' &= 2a_2 + 6a_3x + \cdots \end{align*}
  2. \begin{align*} && 0 &= xy''-y'+4x^3y \\ &&&= x\sum_{n=2}^\infty n(n-1) a_n x^{n-2} - \sum_{n=1}^\infty n a_n x^{n-1} + 4x^3 \sum_{n=0}^\infty a_n x^n \\ &&&= \sum_{n=2}^\infty n(n-1) a_n x^{n-1} - \sum_{n=1}^\infty n a_n x^{n-1} + \sum_{n=0}^\infty 4a_n x^{n+3} \\ &&&= \sum_{n=2}^\infty n(n-1) a_n x^{n-1} - \sum_{n=1}^\infty n a_n x^{n-1} + \sum_{n=4}^\infty 4a_{n-4} x^{n-1} \\ &&&= \sum_{n=4}^{\infty} \l n(n-1) a_n- n a_n +4a_{n-4} \r x^{n-1} + 2a_2x + 6a_3x^2-a_1-2a_2x-3a_3x^2 \\ &&&= \sum_{n=4}^{\infty} \l n(n-2) a_n +4a_{n-4} \r x^{n-1}+ 3a_3x^2-a_1 \\ \end{align*} Therefore since all coefficients are \(0\), \(a_1 = 0\), \(a_3 = 0\) and \(\displaystyle a_n = -\frac{4}{n(n-2)}a_{n-4}\). If \(a_0 = 1, a_2 = 0\), and since \(a_1 = 0, a_3 = 0\) the only values which will take non-zero value are \(a_{4k}\). We can compute these values as: \(a_{4k} = -\frac{4}{(4k)(4k-2)} a_{4k-4} = \frac{1}{2k(2k-1)}a_{4k-r}\) so \(a_{4k} = \frac{(-1)^k}{(2k)!}\), which are precisely the coefficients in the expansion \(\cos x^2\). If \(a_0 = 0, a_2 = 1\) then since \(a_1 = 0, a_3 = 0\) the only values which take non-zero values are \(a_{4k+2}\) we can compute these values as: \(a_{4k+2} = -\frac{4}{(4k+2)(4k)}a_{4k-2} = -\frac{1}{(2k+1)2k}a_{4k-2}\) so we can see that \(a_{4k+2}= \frac{(-1)^k}{(2k+1)!}\) precisely the coefficients of \(\sin x^2\)

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2006 Paper 3 Q4
D: 1700.0 B: 1516.0
Taylor series Maclaurin series functional equation proof logarithm exponential function inverse trigonometric functions t-substitution

The function \(f\) satisfies the identity \begin{equation} f(x) +f(y) \equiv f(x+y) \tag{\(*\)} \end{equation} for all \(x\) and \(y\). Show that \(2\f(x)\equiv \f(2x)\) and deduce that \(f''(0)=0\). By considering the Maclaurin series for \(\f(x)\), find the most general function that satisfies \((*)\). [{\it Do not consider issues of existence or convergence of Maclaurin series in this question.}]

  1. By considering the function \(\G\), defined by \(\ln\big(\g(x)\big) =\G(x)\), find the most general function that, for all \(x\) and \(y\), satisfies the identity \[ \g(x) \g(y) \equiv \g(x+y)\,. \]
  2. By considering the function \(H\), defined by \(\h(\e^u) =H(u)\), find the most general function that satisfies, for all positive \(x\) and \(y\), the identity \[ \h(x) +\h(y) \equiv \h(xy) \,. \]
  3. Find the most general function \(t\) that, for all \(x\) and \(y\), satisfies the identity \begin{equation*} t(x) + t(y) \equiv t(z)\,, \end{equation*} where \(z= \dfrac{x+y}{1-xy}\,\).

Solution: \begin{align*} &&2f(x) &\equiv f(x) + f(x) \\ &&&\equiv f(x+x) \\ &&&\equiv f(2x) \\ \\ \Rightarrow && 2f(0) &= f(0) \\ \Rightarrow && f(0) &= 0 \\ && f''(0) &= \lim_{h \to 0} \frac{f(2h)-2f(0)+f(-2h)}{h^2} \\ &&&= \lim_{h \to 0} \frac{f(2h)+f(-2h)}{h^2} \\ &&&= \lim_{h \to 0} \frac{f(0)}{h^2} \\ &&&= 0 \\ \Rightarrow && f''(0) &= 0 \end{align*} If \(f(x)\) satisfies the equation, then \(f'(x)\) satisfies the equation. In particular this means that \(f^{(n)}(0) = 0\) for all \(n \geq 2\). Therefore the only non-zero term in the Maclaurin series is \(x^1\). Therefore \(f(x) = cx\)

  1. Suppose \(g(x)g(y) \equiv g(x+y)\), then if \(G(x) = \ln g(x)\) we must have \(G(x)+G(y) \equiv G(x+y)\), ie \(G(x) = cx \Rightarrow g(x) = e^{cx}\)
  2. Suppose \(h(x)+h(y) \equiv h(xy)\), then if \(h(e^u) = H(u)\) we must have that \(H(u)+H(v) \equiv h(e^u) + h(e^v) \equiv h(e^{u+v}) \equiv H(u+v)\).Therefore \(H(u) = cu\), ie \(h(e^u) = cu\) or \(h(x) = h(e^{\ln x}) = c \ln x\).
  3. Finally if \(t(x) + t(y) \equiv t(z)\), the considering \(T(w) = t(\tan w)\) then \(T(x) + T(y) \equiv t(\tan x) + t(\tan y) \equiv t( \frac{\tan x + \tan y}{1- \tan x \tan y}) \equiv t (\tan (x+y)) \equiv T(x+y)\). Therefore \(T(x) = cx\) Therefore \(t(\tan w) = c w \Rightarrow t(x) = c \tan^{-1} x\)

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2001 Paper 3 Q1
D: 1700.0 B: 1500.0
Taylor series Maclaurin series proof by induction implicit differentiation Leibniz rule recurrence relation logarithmic function inverse hyperbolic function

Given that \(y = \ln ( x + \sqrt{x^2 + 1})\), show that \( \displaystyle \frac{\d y}{\d x} = \frac1 {\sqrt{x^2 + 1} }\;\). Prove by induction that, for \(n \ge 0\,\), \[ \l x^2 + 1 \r y^{\l n + 2 \r} + \l 2n + 1 \r x y^{\l n + 1 \r} + n^2 y^{\l n \r} = 0\;, \] where \(\displaystyle y^{(n)} = \frac{\d^n y}{\d x^n}\) and \(y^{(0)} =y\,\). Using this result in the case \(x = 0\,\), or otherwise, show that the Maclaurin series for \(y\) begins \[ x - {x^3 \over 6} +{3 x^5 \over 40} \] and find the next non-zero term.

Solution: \begin{align*} && y & = \ln ( x + \sqrt{x^2+1}) \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{1}{x+\sqrt{x^2+1}} \cdot \frac{\d }{\d x} \left ( x + \sqrt{x^2+1} \right) \\ &&&= \frac{1}{x+\sqrt{x^2+1}} \left (1 + \frac12 \frac{2x}{\sqrt{x^2+1}} \right) \\ &&&= \frac{1}{x+\sqrt{x^2+1}} \left ( \frac{\sqrt{x^2+1} + x}{\sqrt{x^2+1}}\right) \\ &&&= \frac{1}{\sqrt{x^2+1}} \end{align*} Note that \(\displaystyle y^{(2)} = - \frac12 \frac{2x}{(x^2+1)^{3/2}} = - \frac{x}{(x^2+1)^{3/2}}\), and in particular \((x^2+1)y^{(2)} + xy^{(1)} = 0\). Now applying Leibnitz formula: \begin{align*} 0 &= \left ( (x^2+1)y^{(2)} + xy^{(1)} \right )^{(n)} \\ &= \left ( (x^2+1)y^{(2)}\right )^{(n)} + \left (xy^{(1)} \right )^{(n)} \\ &= (x^2+1)y^{(n+2)} +n2xy^{(n+1)} + \binom{n}{2}2y^{(n)} + xy^{(n+1)} + n y^{(n)} \\ &= (x^2+1)y^{(n+2)} + (2n+1)xy^{(n+1)} + (n^2-n+n)y^{(n)} \\ &= (x^2+1)y^{(n+2)} + (2n+1)xy^{(n+1)} + n^2y^{(n)} \end{align*} as required. When \(x = 0\): \begin{align*} && y(0) &= \ln(0 + \sqrt{0^2+1}) \\ &&&= \ln 1 = 0 \\ && y'(0) &= \frac{1}{\sqrt{0^2+1}} = 1 \\ && y^{(n+2)} &= -n^2 y^{(n)} \\ && y^{(2k)} &= 0 \\ && y^{(3)} &= -1 \\ && y^{(5)} &= 3^2 \\ && y^{(7)} &= - 5^2 \cdot 3^2 \\ \end{align*} Therefore the Maclaurin series about \(x = 0\) is \begin{align*} y &= x - \frac{1}{3!} x^3 + \frac{3^2}{5!} x^5 - \frac{3^2 \cdot 5^2}{7!} x^7 + \cdots \\ &= x - \frac{1}{6} x^3 + \frac{3}{1 \cdot 2 \cdot 4 \cdot 5} x^5 - \frac{5}{1 \cdot 2 \cdot 4 \cdot 2 \cdot 7} x^7 + \cdots \\ &= x - \frac{1}{6}x^3 + \frac{3}{40} x^5 - \frac{5}{56} x^7 + \cdots \end{align*}

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1997 Paper 3 Q1
D: 1700.0 B: 1500.0
Taylor series exponential function series expansion polynomial multiplication summation identity Maclaurin series

  1. By considering the series expansion of \((x^2+5x+4){\rm \; e}^x\) show that \[10{\rm\, e}=4+\frac{3^2}{1!}+\frac{4^2}{2!}+\frac{5^2}{3!}+\cdots\;.\]
  2. Show that \[5{\rm\, e}=1+\frac{2^2}{1!}+\frac{3^2}{2!}+\frac{4^2}{3!}+\cdots\;.\]
  3. Evaluate \[1+\frac{2^3}{1!}+\frac{3^3}{2!}+\frac{4^3}{3!}+\cdots\;.\]

Solution:

  1. \begin{align*} (x^2+5x+4)e^x &= \sum_{k=0}^\infty \frac{1}{k!} x^{k+2}+\sum_{k=0}^\infty \frac{5}{k!} x^{k+1}+\sum_{k=0}^\infty \frac{4}{k!} x^{k} \\ &= \sum_{k=0}^{\infty} \l \frac{1}{k!}+\frac{5}{(k+1)!}+\frac{4}{(k+2)!} \r x^{k+2} + 5x+4+4x \\ &= 4 + 9x + \sum_{k=0}^{\infty} \l \frac{(k+2)(k+1)}{(k+2)!}+\frac{5(k+2)}{(k+2)!}+\frac{4}{(k+2)!} \r x^{k+2}\\ &= 4 + 9x + \sum_{k=0}^{\infty} \l \frac{k^2+3k+2+5k+10+4}{(k+2)!} \r x^{k+2}\\ &= 4 + 9x + \sum_{k=0}^{\infty} \frac{(k+4)^2}{(k+2)!} x^{k+2}\\ &= 4 + 9x + \sum_{k=2}^{\infty} \frac{(k+2)^2}{k!} x^{k}\\ \end{align*} So when \(x = 1\) we have \[10e = 4 + \frac{3^2}{1!} + \frac{4^2}{2!} + \frac{5^2}{3!} + \cdots \]
  2. \begin{align*} (x^2+3x+1)e^x &= \sum_{k=0}^\infty \frac{1}{k!}x^{k+2}+\sum_{k=0}^\infty 3\frac{1}{k!}x^{k+1} + \sum_{k=0}^{\infty} \frac{1}{k!} x^k \\ &= 1+3x+\sum_{k=1}^{\infty} \l \frac1{(k-1)!}+\frac{3}{k!} + \frac{1}{(k+1)!} \r x^{k+1} \\ &= 1+3x+\sum_{k=1}^{\infty} \frac{(k+1)k + 3(k+1)+1}{(k+1)!}x^{k+1} \\ &= 1+3x+\sum_{k=1}^{\infty} \frac{k^2+4k+4}{(k+1)!}x^{k+1} \\ &= 1+3x+\sum_{k=0}^{\infty} \frac{(k+2)^2}{(k+1)!}x^{k+1} \\ &=1+3x+ \sum_{k=1}^{\infty} \frac{(k+1)^2}{k!}x^k \end{align*} Plugging in \(x=1\) we get the desired result.
  3. \begin{align*} && xe^x &= \sum_{k=0}^{\infty} \frac{x^{k+1}}{k!} \\ x\frac{\d}{\d x} : && x(1+x)e^x &= \sum_{k=0}^{\infty} \frac{(k+1)x^{k+1}}{k!} \\ x\frac{\d}{\d x} : && x(x(1+x)+1+2x)e^x &= \sum_{k=0}^{\infty} \frac{(k+1)^2x^{k+1}}{k!} \\ &&(x^3+3x^2+x)e^x &= \sum_{k=0}^{\infty} \frac{(k+1)^2x^{k+1}}{k!} \\ \frac{\d}{\d x} : && e^x(x^3+3x^2+x+3x^2+6x+1) &=\sum_{k=0}^{\infty} \frac{(k+1)^3x^{k}}{k!} \\ \Rightarrow && 15e &= 1 + \frac{2^3}{1!} + \frac{3^3}{2!} + \cdots \end{align*}

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1995 Paper 3 Q5
D: 1700.0 B: 1500.0
differential equation second order Maclaurin series Leibniz rule proof by induction inverse trigonometric functions repeated differentiation general term

Show that \(y=\sin^{2}(m\sin^{-1}x)\) satisfies the differential equation \[ (1-x^{2})y^{(2)}=xy^{(1)}+2m^{2}(1-2y), \] and deduce that, for all \(n\geqslant1,\) \[ (1-x^{2})y^{(n+2)}=(2n+1)xy^{(n+1)}+(n^{2}-4m^{2})y^{(n)}, \] where \(y^{(n)}\) denotes the \(n\)th derivative of \(y\). Derive the Maclaurin series for \(y\), making it clear what the general term is.

Solution: \begin{align*} && y &= \sin^2 (m \sin^{-1} x) \\ \Rightarrow && y' &= 2 \sin (m \sin^{-1} x) \cdot \cos (m \sin^{-1} x) \cdot m \cdot \frac1{\sqrt{1-x^2}} \\ \Rightarrow && y'' &= 2 \cos^2(m \sin^{-1} x) \cdot m^2 \cdot \frac{1}{1-x^2} + \\ &&&\quad\quad-2\sin^2(m \sin^{-1} x) m^2 \frac{1}{1-x^2} + \\ &&&\quad\quad\quad-\sin(m \sin^{-1} x) \cdot \cos(m \sin^{-1} x) \cdot m \cdot (1-x^2)^{-\frac32} \cdot (-2x) \\ \Rightarrow && (1-x^2)y^{(2)} &= 2m^2-4m^2y+xy' \\ &&&= xy^{(1)} + 2m^2(1-2y) \\ \\ \Rightarrow && (1-x^2)y^{(n+2)}-2nxy^{(n+1)}-2\binom{n}{2}y^{(n)} &= xy^{(n+1)}+ny^{(n)} -4m^2y^{(n)} \\ \Rightarrow && (1-x^2)y^{(n+2)} &= (2n+1)xy^{(n+1)}+(n(n-1)+n-4m^2)y^{(n)} \\ &&&= (2n+1)xy^{(n+1)}+(n^2-4m^2)y^{(n)} \\ \end{align*} \begin{align*} && y(0) &= \sin^2(m \sin^{-1} 0) \\ &&&= \sin^2 0 = 0 \\ \\ && y'(0) &= 0 \\ && (1-0^2)y^{(2)}(0) &= 2m^2(1-2y(0)) \\ \Rightarrow && y^{(2)}(0) &= 2m^2 \\ \\ && y^{(n+2)} (0) &= (2n+1) \cdot 0 \cdot y^{(n+1)} +(n^2-4m^2)y^{(n)}(0) \\ &&&= (n^2-4m^2)y^{(n)}(0) \\ \\ && y^{(2)}(0) &= 2m^2 \\ && y^{(4)}(0) &= (4-4m^2) \cdot 2m^2 \\ &&&= -8m(m+1)m(m-1) \\ && y^{(6)}(0) &= 32m(m+2)(m+1)m(m-1)(m-2) \\ && y^{(2k)}(0) &= (-1)^{k+1}2^{2k-1}m (m+k)\cdots(m-k) \text{ if }k < m \\ \\ && y &= m^2x^2 -2m\binom{m+1}{3} x^4 + \frac{16}{3}m\binom{m+2}{5}x^6 - \cdots \\ &&&+ (-1)^{k}\frac{2^{2k}}{k+1} m \binom{m+k}{2k+1}x^{2k+2}+\cdots \\ &&&= mx^2\sum_{k=0}^{m-1} \frac{(-1)^k2^{2k}}{k+1}\binom{m+k}{2k+1}x^{2k} \end{align*}

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1994 Paper 3 Q5
D: 1700.0 B: 1486.8
Taylor series Maclaurin series inverse trigonometric functions arcsin Leibniz rule differential equation logarithmic function inequality coefficient comparison proof by induction

The function \(\mathrm{f}\) is given by \(\mathrm{f}(x)=\sin^{-1}x\) for \(-1 < x < 1.\) Prove that \[ (1-x^{2})\mathrm{f}''(x)-x\mathrm{f}'(x)=0. \] Prove also that \[ (1-x^{2})\mathrm{f}^{(n+2)}(x)-(2n+1)x\mathrm{f}^{(n+1)}(x)-n^{2}\mathrm{f}^{(n)}(x)=0, \] for all \(n>0\), where \(\mathrm{f}^{(n)}\) denotes the \(n\)th derivative of \(\mathrm{f}\). Hence express \(\mathrm{f}(x)\) as a Maclaurin series. The function \(\mathrm{g}\) is given by \[ \mathrm{g}(x)=\ln\sqrt{\frac{1+x}{1-x}}, \] for \(-1 < x < 1.\) Write down a power series expression for \(\mathrm{g}(x),\) and show that the coefficient of \(x^{2n+1}\) is greater than that in the expansion of \(\mathrm{f},\) for each \(n > 0\).

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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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