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2016 Paper 3 Q4
D: 1700.0 B: 1484.0
sequences telescoping series partial fractions convergence hyperbolic functions sech cosech infinite series substitution

  1. By considering \(\displaystyle \frac1{1+ x^r} - \frac1{1+ x^{r +1}}\) for \(\vert x \vert \ne 1\), simplify \[ \sum_{r=1}^N \frac{x^r}{(1+x^r)(1+x^{r+1})} \] Show that, for \(\vert x \vert <1\), \[ \sum_{r=1}^\infty \frac{x^r}{(1+x^r)(1+x^{r+1})} = \frac x {1-x^2} \]
  2. Deduce that \[ \sum_{r=1}^\infty \textrm{sech}(ry)\textrm{sech}((r + 1)y) = 2\e^{-y} \textrm{cosech}(2 y) \] for \(y > 0\). Hence simplify \[ \sum_{r=-\infty}^\infty \textrm{sech}(ry) \textrm{sech}((r + 1)y) \] for \(y>0\).

Solution:

  1. \(\,\) \begin{align*} && \frac{1}{1+x^r} - \frac{1}{1+x^{r+1}} &= \frac{1+x^{r+1}-1-x^r}{(1+x^r)(1+x^{r+1})} \\ &&&= \frac{x^r(x-1)}{(1+x^r)(1+x^{r+1})} \\ \\ && \sum_{r=1}^N \frac{x^r}{(1+x^r)(1+x^{r+1})} &= \sum_{r=1}^N \frac{1}{x-1} \left ( \frac{1}{1+x^r} - \frac{1}{1+x^{r+1}}\right) \\ &&&= \frac{1}{x-1} \Bigg ( \frac{1}{1+x} + \cdots \\ &&& \qquad \qquad \quad - \frac{1}{1+x^2} + \frac{1}{1+x^2} + \cdots \\ &&& \qquad \qquad \quad - \frac{1}{1+x^3} + \frac{1}{1+x^3} + \cdots \\ &&& \qquad \qquad \quad - \cdots \\ &&& \qquad \qquad \quad - \frac{1}{1+x^{N+1}} \Bigg ) \\ &&&= \frac{1}{x-1} \left (\frac{1}{1+x} - \frac{1}{1+x^{N+1}} \right) \\ \\ && \sum_{r=1}^{\infty} \frac{x^r}{(1+x^r)(1+x^{r+1})} &= \lim_{N\to \infty} \frac{1}{x-1} \left (\frac{1}{1+x} - \frac{1}{1+x^{N+1}} \right) \\ &&&= \frac{1}{x-1} \left ( \frac{1}{1+x} - 1\right) \\ &&&= \frac{1}{x-1} \left ( \frac{-x}{1+x} \right) \\ &&&= \frac{x}{1-x^2} \end{align*}
  2. \(\,\) \begin{align*} && \sum_{r=1}^\infty \textrm{sech}(ry)\textrm{sech}((r + 1)y) &= \sum_{r=1}^\infty \frac{4}{(e^{ry}+e^{-ry})(e^{(r+1)y}+e^{-(r+1)y})} \\ &&&=\sum_{r=1}^\infty \frac{4e^{-(2r+1)y}}{(1+e^{-2ry})(1+e^{-2(r+1)y})} \\ x = e^{-2y}: &&&= \frac{4e^{-y}e^{-2y}}{1-e^{-4y}} \\ &&&= \frac{4e^{-y}e^{-2y}}{e^{-2y}(e^{2y}-e^{-2y})} \\ &&&=2e^{-y}\textrm{cosech}(2y) \end{align*} \begin{align*} && \sum_{r=-\infty}^\infty \textrm{sech}(ry) \textrm{sech}((r + 1)y) &= \sum_{r=1}^\infty \textrm{sech}(ry) \textrm{sech}((r + 1)y) + \sum_{r=-\infty}^0 \textrm{sech}(ry) \textrm{sech}((r + 1)y) \\ &&&= 2e^{-y}\textrm{cosech}(2y) + \sum_{r=0}^\infty \textrm{sech}(-ry) \textrm{sech}(-(r-1)y) \\ &&&= 2e^{-y}\textrm{cosech}(2y) + \sum_{r=0}^\infty \textrm{sech}((r-1)y) \textrm{sech}(ry) \\ &&&= 4e^{-y}\textrm{cosech}(2y) + \textrm{sech}(y) + \textrm{sech}(-y) \\ &&&= 4e^{-y}\textrm{cosech}(2y)+2\textrm{sech}(y) \\ &&&= 4e^{-y} \frac12 \textrm{sech}(y) \textrm{cosech}(y) + 2 \textrm{sech}(y) \\ &&&= 2\textrm{sech}(y) \left ( e^{-y} \textrm{cosech}(y)+1 \right) \\ &&&= 2\textrm{sech}(y) \left ( \frac{2}{e^{2y}-1} + 1 \right) \\ &&&= 2\textrm{sech}(y) \left ( \frac{e^{2y}+1}{e^{2y}-1} \right) \\ &&&= 2 \textrm{cosech}(y) \end{align*}

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2016 Paper 3 Q6
D: 1700.0 B: 1484.0
hyperbolic functions sech tanh curve intersection arcosh artanh identity manipulation necessary and sufficient conditions tangency optimisation

Show, by finding \(R\) and \(\gamma\), that \(A \sinh x + B\cosh x \) can be written in the form \(R\cosh (x+\gamma)\) if \(B>A>0\). Determine the corresponding forms in the other cases that arise, for \(A>0\), according to the value of \(B\). Two curves have equations \(y = \textrm{sech} x\) and \(y = a\tanh x + b\,\), where \(a>0\).

  1. In the case \(b>a\), show that if the curves intersect then the \(x\)-coordinates of the points of intersection can be written in the form \[ \pm\textrm{arcosh} \left( \frac 1 {\sqrt{b^2-a^2}}\right) - {\rm artanh \,} \frac a b .\]
  2. Find the corresponding result in the case \(a>b>0\,\).
  3. Find necessary and sufficient conditions on \(a\) and \(b\) for the curves to intersect at two distinct points.
  4. Find necessary and sufficient conditions on \(a\) and \(b\) for the curves to touch and, given that they touch, express the \(y\)-coordinate of the point of contact in terms of \(a\).

Solution: \begin{align*} && R\cosh(x + \gamma) &=R \cosh x \cosh \gamma + R \sinh x \sinh \gamma \\ \Rightarrow && R \cosh \gamma &= B \\ && R \sinh \gamma &= A \\ \Rightarrow && R^2 &= B^2 - A^2 \\ \Rightarrow && \tanh \gamma &= \frac{A}{B} \\ \end{align*} Therefore it is possible, by writing \(R = \sqrt{B^2-A^2}\) and \(\gamma = \textrm{artanh} \left ( \frac{A}{B} \right)\). This works as long as \(|B| > A > 0\). Supposing \(A >|B| \), try \(S \sinh (x + \delta) = S \sinh x \cosh \delta +S \cosh x \sinh \delta\) \begin{align*} && S \cosh \delta &= A \\ && -S \sinh \delta &= B \\ \Rightarrow && S^2 &= A^2 - B^2 \\ \Rightarrow && \tanh \delta &= \frac{B}{A} \\ \end{align*} Therefore in this case we can write \(\sqrt{A^2-B^2} \sinh \left (x + \tanh^{-1} \left ( \frac{B}{A} \right) \right)\) If \(A = \pm B > 0\) we can we have \(A \sinh x + B \cosh x = \pm Ae^{\pm x}\)

  1. Suppose \(y \cosh x = 1\) and \(y \cosh x = a \sinh x +b \cosh x\) so \begin{align*} && 1 & = a \sinh x + b \cosh x \\ &&&= \sqrt{b^2-a^2} \cosh(x + \textrm{artanh} \frac{a}{b} ) \\ \Rightarrow && x + \textrm{artanh} \frac{a}{b} &= \pm \textrm{arcosh} \left ( \frac{1}{\sqrt{b^2-a^2}} \right) \\ \Rightarrow && x &= \pm \textrm{arcosh} \left ( \frac{1}{\sqrt{b^2-a^2}} \right) -\textrm{artanh} \frac{a}{b} \end{align*}
  2. If \( a > b > 0\) we have \begin{align*} && 1 & = \sqrt{a^2-b^2} \sinh \left ( x - \textrm{artanh} \frac{b}{a} \right) \\ \Rightarrow && x &= \textrm{arsinh} \left ( \frac{1}{\sqrt{a^2-b^2}} \right) + \textrm{artanh} \left ( \frac{b}{a} \right) \end{align*}
  3. To intersect at distinct points we must have \(b > a\) and \(\textrm{arcosh} \left ( \frac{1}{\sqrt{b^2-a^2}} \right) \neq 0\) which is always true.
  4. For the curves to touch, we need them to intersect and have matching derivatives, ie \begin{align*} && -\tanh x \cdot \textrm{sech}x &= a\textrm{sech}^2 x \\ \Rightarrow && 0 &= \textrm{sech}^2 x (a + \sinh x) \\ \Rightarrow && x &= -\textrm{arsinh} \, a\\ \Rightarrow && \sinh x &= - a\\ \Rightarrow &&\cosh x &= \sqrt{1 + a^2} \\ \end{align*} So if the curves touch, we must have \(1 = -a^2+b\sqrt{1+a^2} \Rightarrow b = \sqrt{1+a^2}\) and since this does work it is a necessary and sufficient condition. We will also have the \(y\) coordinate is \(\frac{1}{\sqrt{1+a^2}}\)

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1994 Paper 3 Q1
D: 1700.0 B: 1516.0
integration reduction formula hyperbolic functions sech integration by parts definite integral

Calculate \[ \int_{0}^{x}\mathrm{sech}\, t\,\mathrm{d}t. \] Find the reduction formula involving \(I_{n}\) and \(I_{n-2}\), where \[ I_{n}=\int_{0}^{x}\mathrm{sech}^{n}t\,\mathrm{d}t \] and, hence or otherwise, find \(I_{5}\) and \(I_{6}.\)

Solution: \begin{align*} && \int_0^x \mathrm{sech}\, t \d t &= \int_0^x \frac{2}{e^t+e^{-t}} \d t \\ &&&= \int_0^x \frac{2e^t}{e^{2t}+1} \d t \\ &&&= \left [2 \arctan e^t \right ]_0^x \\ &&&= 2\tan^{-1}e^x- \frac{\pi}{2} \end{align*} \begin{align*} && I_n &= \int_0^x \mathrm{sech}^n\, t \d t \\ &&&= \int_0^x \mathrm{sech}^{n-2}\, t \mathrm{sech}^2\, t \d t \\ &&&= \left [ \mathrm{sech}^{n-2}\, t \cdot \tanh t\right ]_0^x - \int_0^x (n-2) \mathrm{sech}^{n-3} \, t \cdot (- \tanh t \mathrm{sech}\, t) \tanh t \d t \\ &&&= \mathrm{sech}^{n-2}\, x \cdot \tanh x+(n-2)\int_0^x \mathrm{sech}^{n-2} t \tanh^2 t \d t \\ &&&= \mathrm{sech}^{n-2}\, x \cdot \tanh x+(n-2)\int_0^x \mathrm{sech}^{n-2} t (1-\mathrm{sech}^2 \, t) \d t \\ &&&= \mathrm{sech}^{n-2}\, x \cdot \tanh x+(n-2)I_{n-2}-(n-2)I_n \\ \Rightarrow && (n-1)I_n &= \mathrm{sech}^{n-2}\, x \cdot \tanh x+(n-2)I_{n-2} \\ \Rightarrow && I_n &= \frac{1}{n-1} \left ( \mathrm{sech}^{n-2}\, x \cdot \tanh x+(n-2)I_{n-2} \right) \\ \end{align*} \begin{align*} I_1 &= 2\tan^{-1}e^x- \frac{\pi}{2} \\ I_3 &= \frac12 \left ( \mathrm{sech}\, x \cdot \tanh x+ 2\tan^{-1}e^x- \frac{\pi}{2}\right) \\ &= \frac12 \mathrm{sech}\, x \cdot \tanh x+ \tan^{-1}e^x- \frac{\pi}{4} \\ I_5 &= \frac14 \left (\mathrm{sech}^3\, x \cdot \tanh x + 3 \left ( \frac12 \mathrm{sech}\, x \cdot \tanh x+ \tan^{-1}e^x- \frac{\pi}{4} \right) \right) \\ &= \frac14 \mathrm{sech}^3\, x \cdot \tanh x +\frac38 \mathrm{sech}\, x \cdot \tanh x+\frac34 \tan^{-1}e^x- \frac{3\pi}{16} \\ \\ I_2 &= \tanh x \\ I_4 &= \frac13 \left ( \mathrm{sech}^2 x\tanh x +2\tanh x\right) \\ &= \frac13 \mathrm{sech}^2 x\tanh x +\frac23 \tanh x \\ I_6 &= \frac15 \left ( \mathrm{sech}^4 x \tanh x+4\left ( \frac13 \mathrm{sech}^2 x\tanh x +\frac23 \tanh x \right) \right) \\ &= \frac15 \mathrm{sech}^4 x \tanh x + \frac4{15}\mathrm{sech}^2 x\tanh x + \frac{8}{15} \tanh x \end{align*}

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1990 Paper 3 Q9
D: 1700.0 B: 1484.7
hyperbolic functions trigonometric chain rule implicit differentiation second derivative proof sech relationship between trig and hyperbolic

The real variables \(\theta\) and \(u\) are related by the equation \(\tan\theta=\sinh u\) and \(0\leqslant\theta<\frac{1}{2}\pi.\) Let \(v=\mathrm{sech}u.\) Prove that

  1. \(v=\cos\theta;\)
  2. \(\dfrac{\mathrm{d}\theta}{\mathrm{d}u}=v;\)
  3. \(\sin2\theta=-2\dfrac{\mathrm{d}v}{\mathrm{d}u}\quad\) and \(\quad\cos2\theta=-\cosh u\dfrac{\mathrm{d}^{2}v}{\mathrm{d}u^{2}};\)
  4. \({\displaystyle \frac{\mathrm{d}u}{\mathrm{d}\theta}\frac{\mathrm{d}^{2}v}{\mathrm{d}\theta^{2}}+\frac{\mathrm{d}v}{\mathrm{d}\theta}\frac{\mathrm{d}^{2}u}{\mathrm{d}\theta^{2}}+\left(\frac{\mathrm{d}u}{\mathrm{d}\theta}\right)^{2}=0.}\)

Solution:

  1. \begin{align*} v &= \mathrm{sech} u \\ &= \frac{1}{\mathrm{cosh } u} \\ &= \frac{1}{\sqrt{1+\mathrm{sinh}^2 u}} \tag{\(u > 0\)} \\ &= \frac{1}{\sqrt{1+\tan^2 \theta}} \\ &= \frac{1}{\sqrt{\mathrm{sec}^2 \theta}} \\ &= \cos \theta \tag{\(0 < \theta < \tfrac{\pi}{2}\)} \end{align*}
  2. \begin{align*} && \tan \theta &= \textrm{sinh} u \\ \underbrace{\Rightarrow}_{\frac{\d}{\d u}} && \sec^2 \theta \cdot \frac{\d \theta}{\d u} &= \cosh u \\ \Rightarrow && \frac{\d \theta}{\d u} &=\cosh u \cdot \cos^2 \theta \\ &&&= \frac{1}{v} \cdot v^2 \\ &&&=v \end{align*}
  3. \begin{align*} \sin 2 \theta &= 2 \sin \theta \cos \theta \\ &= 2 \sin \theta \cdot \frac{\d \theta}{\d u} \\ &= -2 \frac{\d v}{\d \theta} \cdot \frac{\d \theta}{\d u} \tag{\(\cos \theta = v\)} \\ &= -2 \frac{\d v}{\d u} \end{align*} \begin{align*} && \sin 2 \theta &= -2 \frac{\d v}{\d u} \\ \underbrace{\Rightarrow}_{\frac{\d}{\d u}} && 2 \cos 2 \theta \cdot \frac{\d \theta}{\d u} &= -2 \frac{\d^2 v}{\d u^2} \\ \Rightarrow && \cos 2 \theta &= - \frac{\d^2 v}{\d u^2} \frac{1}{v} \\ &&&= -\frac{\d ^2v}{\d u^2} \cosh u \end{align*}
  4. \begin{align*} && \frac{\d u}{\d \theta} &= \frac{1}{v} \\ \Rightarrow && \frac{\d^2 u}{\d \theta^2} &= -\frac{1}{v^2} \frac{\d v}{\d \theta} \\ &&&= \frac{1}{v^2} \sin \theta \\ && \frac{\d v}{\d \theta} &= -\sin \theta \\ \Rightarrow && \frac{\d^2 v}{\d \theta^2} &= -\cos \theta \\ &&&= - v \\ \end{align*} Therefore \begin{align*} \frac{\mathrm{d}u}{\mathrm{d}\theta}\frac{\mathrm{d}^{2}v}{\mathrm{d}\theta^{2}}+\frac{\mathrm{d}v}{\mathrm{d}\theta}\frac{\mathrm{d}^{2}u}{\mathrm{d}\theta^{2}}+\left(\frac{\mathrm{d}u}{\mathrm{d}\theta}\right)^{2} &= \frac{1}{v} \cdot \left (-v\right) + \left ( - \sin \theta \right ) \cdot \left (\frac{1}{v^2} \sin \theta \right) + \frac{1}{v^2} \\ &= -1 + \frac{1-\sin^2 \theta}{v^2} \\ &= -1 + \frac{\cos^2 \theta}{v^2} \\ &= -1 + 1 \\ &= 0 \end{align*}

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