72 problems found
A path is made up in the Argand diagram of a series of straight line segments \(P_{1}P_{2},\) \(P_{2}P_{3},\) \(P_{3}P_{4},\ldots\) such that each segment is \(d\) times as long as the previous one, \((d\neq1)\), and the angle between one segment and the next is always \(\theta\) (where the segments are directed from \(P_{j}\) towards \(P_{j+1}\), and all angles are measured in the anticlockwise direction). If \(P_{j}\) represents the complex number \(z_{j},\) express \[ \frac{z_{n+1}-z_{n}}{z_{n}-z_{n-1}} \] as a complex number (for each \(n\geqslant2\)), briefly justifying your answer. If \(z_{1}=0\) and \(z_{2}=1\), obtain an expression for \(z_{n+1}\) when \(n\geqslant2\). By considering its imaginary part, or otherwise, show that if \(\theta=\frac{1}{3}\pi\) and \(d=2\), then the path crosses the real axis infinitely often.
Solution: \begin{align*} && | \frac{z_{n+1}-z_{n}}{z_{n}-z_{n-1}} | &= d \\ && \arg \left ( \frac{z_{n+1}-z_{n}}{z_{n}-z_{n-1}} \right) &= \arg (z_{n+1}-z_{n}) - \arg(z_{n}-z_{n-1}) \\ &&&= \theta \\ \Rightarrow && \frac{z_{n+1}-z_{n}}{z_{n}-z_{n-1}} &= d e^{i \theta} \end{align*} \begin{align*} && z_1 &= 0 \\ && z_2 &= 1 \\ && \frac{z_3-z_2}{z_2-z_1} &= de^{i \theta} \\ \Rightarrow && z_3 &= de^{i \theta} + 1 \\ && \frac{z_4-z_3}{z_3-z_2} &= de^{i \theta} \\ \Rightarrow && z_4 &= (d e^{i \theta})^2 + d e^{i \theta} + 1\\ \Rightarrow && z_{n+1} &= \frac{(de^{i \theta})^{n}-1}{de^{i \theta}-1} \end{align*} If \(d = 2, \theta = \tfrac13 \pi\), then, \(2e^{i \tfrac13 \pi} = 1 + \sqrt{3}i\) \begin{align*} \textrm{Im}(z_{n+1})) &= \textrm{Im} \left ( \frac{(2e^{i \tfrac13 \pi})^{n}-1}{2e^{i \tfrac13 \pi}-1}\right) \\ &= \textrm{Im} \left ( \frac{(2e^{i \tfrac13 \pi})^{n}-1}{\sqrt{3}i}\right) \\ &= -\frac{1}{\sqrt{3}}\textrm{Re} \left (2^n e^{i \frac{n}{3} \pi} \right) + \frac{1}{\sqrt{3}} \end{align*} Which clearly changes sign infinitely many times, ie crosses the origin infinitely many times.
Let \(y=\cos\phi+\cos2\phi\), where \(\phi=\dfrac{2\pi}{5}.\) Verify by direct substitution that \(y\) satisfies the quadratic equation \(2y^{2}=3y+2\) and deduce that the value of \(y\) is \(-\frac{1}{2}.\) Let \(\theta=\dfrac{2\pi}{17}.\) Show that \[ \sum_{k=0}^{16}\cos k\theta=0. \] If \(z=\cos\theta+\cos2\theta+\cos4\theta+\cos8\theta,\) show that the value of \(z\) is \(-(1-\sqrt{17})/4\).
Solution: Note that \(\cos 4 \phi = \cos \phi, \cos 3 \phi = \cos 2 \phi\) \begin{align*} && LHS & = 2y^2 \\ &&&= 2 \left ( \cos \phi + \cos 2 \phi \right)^2 \\ &&&= 2 \cos ^2 \phi + 2 \cos^2 2 \phi + 4 \cos \phi \cos 2 \phi \\ &&&= \cos 2 \phi+1+ \cos4 \phi+1+2 \left ( \cos \phi + \cos 3 \phi \right) \\ &&&= \cos 2 \phi + 2 + \cos \phi + 2 \cos \phi + 2 \cos 2 \phi \\ &&&= 3(\cos \phi + \cos 2 \phi) + 2 \\ &&&= 3 y + 2 \\ &&&= RHS \end{align*} Therefore \(y\) satisfies \(2y^2 = 3y+2\), which we can solve: \begin{align*} && 0 &= 2y^2-3y-2 \\ &&&= (2y+1)(y-2) \\ \Rightarrow && y &= -\frac12,2 \end{align*} Since \(\cos \phi \neq 1\), \(y \neq 2\), therefore \(y = -\frac12\). \begin{align*} && \sum_{k=0}^{16} \cos k \theta &= \sum_{k=0}^{17} \textrm{Re} \left ( e^{ k \theta i} \right ) \\ &&&= \textrm{Re} \left ( \sum_{k=0}^{16}e^{ k \theta i} \right ) \\ &&&= \textrm{Re} \left ( \frac{1-e^{17 \theta i}}{1-e^{i \theta}} \right ) \\ &&&= 0 \end{align*} Suppose \(z = \cos \theta + \cos 2 \theta + \cos 4 \theta + \cos 8 \theta\) \begin{align*} z^2 &= \left (\cos \theta + \cos 2 \theta + \cos 4 \theta + \cos 8 \theta \right)^2 \\ &= \cos^2 \theta + \cos^2 2 \theta + \cos^2 4 \theta + \cos^2 8 \theta \\ & \quad \quad 2( \cos \theta \cos 2 \theta + \cos \theta \cos 4 \theta + \cos \theta \cos 8 \theta + \\ & \quad \quad \quad \cos 2 \theta \cos 4 \theta + \cos 2 \theta \cos 8 \theta + \cos 4 \theta \cos 8 \theta) \\ &= \frac12 \left (\cos 2 \theta + 1+ \cos 4 \theta + 1 + \cos 8 \theta + 1 + \cos 16 \theta + 1 \right ) + \\ &\quad \quad ( \cos \theta + \cos 3 \theta + \cos 3 \theta + \cos 5 \theta + \cos 7 \theta + \cos 9 \theta + \\ & \quad \quad \quad \cos 2 \theta + \cos 6 \theta + \cos 6 \theta + \cos 10 \theta +\cos 4 \theta + \cos 12 \theta ) \\ &= \frac12 z + 2 + \\ & \quad \quad ( \cos 3 \theta + \cos 6 \theta - \cos 8 \theta - \cos 11 \theta \\ & \quad \quad \quad - \cos 13 \theta - \cos 14 \theta - \cos 15 \theta - \cos 16 \theta - 1) \\ &= \frac12 z + 1 - z \\ &= -\frac12 z +1 \end{align*} Therefore \(z\) satisfies \(z^2=-\frac12 z+1 \Rightarrow z = \frac{-\frac12 \pm \sqrt{\frac14+4}}{2} = \frac{-1 \pm \sqrt{17}}{4}\) Therefore \(z = \frac{\sqrt{17}-1}{4}\) since \(z > 0\)
Solution:
A straight staircase consists of \(N\) smooth horizontal stairs each of height \(h\). A particle slides over the top stair at speed \(U\), with velocity perpendicular to the edge of the stair, and then falls down the staircase, bouncing once on every stair. The coefficient of restitution between the particle and each stair is \(e\), where \(e<1\). Show that the horizontal distance \(d_{n}\) travelled between the \(n\)th and \((n+1)\)th bounces is given by \[ d_{n}=U\left(\frac{2h}{g}\right)^{\frac{1}{2}}\left(e\alpha_{n}+\alpha_{n+1}\right), \] where \({\displaystyle \alpha_{n}=\left(\frac{1-e^{2n}}{1-e^{2}}\right)^{\frac{1}{2}}}\). If \(N\) is very large, show that \(U\) must satisfy \[ U=\left(\frac{L^{2}g}{2h}\right)^{\frac{1}{2}}\left(\frac{1-e}{1+e}\right)^{\frac{1}{2}}, \] where \(L\) is the horizontal distance between the edges of successive stairs.
Given that \(\sin\beta\neq0,\) sum the series \[ \cos\alpha+\cos(\alpha+2\beta)+\cdots+\cos(\alpha+2r\beta)+\cdots+\cos(\alpha+2n\beta) \] and \[ \cos\alpha+\binom{n}{1}\cos(\alpha+2\beta)+\cdots+\binom{n}{r}\cos(\alpha+2r\beta)+\cdots+\cos(\alpha+2n\beta). \] Given that \(\sin\theta\neq0,\) prove that \[ 1+\cos\theta\sec\theta+\cos2\theta\sec^{2}\theta+\cdots+\cos r\theta\sec^{r}\theta+\cdots+\cos n\theta\sec^{n}\theta=\frac{\sin(n+1)\theta\sec^{n}\theta}{\sin\theta}. \]
Solution: \begin{align*} \sum_{r = 0}^n \cos (\alpha + 2r \beta) &= \sum_{r = 0}^n \textrm{Re} \left ( \exp(i(\alpha + 2r \beta)) \right) \\ &= \textrm{Re} \left (\sum_{r = 0}^n \exp(i(\alpha + 2r \beta)) \right) \\ &= \textrm{Re} \left (e^{i \alpha} \sum_{r = 0}^n \ (e^{i 2 \beta})^r\right) \\ &= \textrm{Re} \left (e^{i \alpha} \frac{e^{2(n+1)\beta i}-1}{e^{2\beta i}-1} \right) \\ &= \textrm{Re} \left (e^{i \alpha} \frac{e^{(n+1)\beta i} (e^{(n+1)\beta i}-e^{-(n+1)\beta i})}{e^{\beta i}(e^{\beta i}-e^{-\beta i})} \right) \\ &= \textrm{Re} \left (\frac{e^{i \alpha} e^{(n+1)\beta i}}{e^{\beta i}} \frac{\sin (n+1) \beta}{\sin \beta} \right) \\ &= \textrm{Re} \left ( e^{i(\alpha + n \beta)}\frac{\sin (n+1) \beta}{\sin \beta} \right) \\ &= \frac{\cos (\alpha + n \beta)\sin (n+1) \beta}{\sin \beta} \end{align*} \begin{align*} \sum_{r = 0}^n \binom{n}{r} \cos (\alpha + 2r \beta) &= \sum_{r = 0}^n \textrm{Re} \left ( \binom{n}{r}\exp(i(\alpha + 2r \beta)) \right) \\ &= \textrm{Re} \left (\sum_{r = 0}^n \binom{n}{r} \exp(i(\alpha + 2r \beta)) \right) \\ &= \textrm{Re} \left (e^{i \alpha}(e^{2\beta i}+1)^n \right) \\ &= \textrm{Re} \left (e^{i \alpha}e^{n\beta i}(e^{\beta i}+e^{-\beta i})^n \right) \\ &= \textrm{Re} \left (e^{i \alpha}e^{n\beta i}2^n \cos^n \beta \right) \\ &= 2^n \cos(\alpha + n \beta) \cos ^n \beta \end{align*} \begin{align*} \sum_{r = 0}^n \cos r \theta \sec^r \theta &= \sum_{r = 0}^n \textrm{Re} ( e^{i r \theta})\sec^r \theta \\ &= \textrm{Re} \left ( \sum_{r=0}^n e^{i r \theta} \sec^r \theta\right) \\ &= \textrm{Re} \left ( \frac{e^{i (n+1) \theta}\sec^{n+1} \theta -1}{e^{i \theta}\sec \theta -1} \right) \\ &= \textrm{Re} \left ( \frac{e^{i (n+1) \theta}\sec^{n} \theta -\cos \theta}{e^{i \theta} -\cos \theta} \right) \\ &= \textrm{Re} \left ( \frac{e^{i (n+1) \theta}\sec^{n} \theta -\cos \theta}{i \sin \theta} \right) \\ &= \frac{1}{\sin \theta} \textrm{Im} \left ( e^{i (n+1) \theta}\sec^{n} \theta -\cos \theta \right) \\ &= \frac{\sin(n+1) \theta \sec^{n} \theta}{\sin \theta} \end{align*}
The matrix \(\mathbf{M}\) is given by \[ \mathbf{M}=\begin{pmatrix}\cos(2\pi/m) & -\sin(2\pi/m)\\ \sin(2\pi/m) & \cos(2\pi/m) \end{pmatrix}, \] where \(m\) is an integer greater than \(1.\) Prove that \[ \mathbf{M}^{m-1}+\mathbf{M}^{m-2}+\cdots+\mathbf{M}^{2}+\mathbf{M}+\mathbf{I}=\mathbf{O}, \] where $\mathbf{I}=\begin{pmatrix}1 & 0\\ 0 & 1 \end{pmatrix}\( and \)\mathbf{O}=\begin{pmatrix}0 & 0\\ 0 & 0 \end{pmatrix}.$ The sequence \(\mathbf{X}_{0},\mathbf{X}_{1},\mathbf{X}_{2},\ldots\) is defined by \[ \mathbf{X}_{k+1}=\mathbf{PX}_{k}+\mathbf{Q}, \] where \(\mathbf{P,Q}\) and \(\mathbf{X}_{0}\) are given \(2\times2\) matrices. Suggest a suitable expression for \(\mathbf{X}_{k}\) in terms of \(\mathbf{P},\) \(\mathbf{Q}\) and \(\mathbf{X}_{0},\) and justify it by induction. The binary operation \(*\) is defined as follows: \[ \mathbf{X}_{i}*\mathbf{X}_{j}\mbox{ is the result of substituting \ensuremath{\mathbf{X}_{j}}for \ensuremath{\mathbf{X}_{0}}in the expression for \ensuremath{\mathbf{X}_{i}}. } \] Show that if \(\mathbf{P=M},\) the set \(\{\mathbf{X}_{1},\mathbf{X}_{2},\mathbf{X}_{3},\ldots\}\) forms a finite group under the operation \(*\).
Solution: \(\mathbf{M}^m = \mathbf{I}\), we also have \(\mathrm{det}(\mathbf{M - I}) = \cos^2(2\pi/m) - 2\cos(2\pi/m) + 1 + \sin^2(2\pi/m) = 2(1-\cos(2\pi/m))\) therefore \(\mathbf{M-I}\) is invertible. Therefore since \(\mathbf{(M-I)(M^{m-1} + M^{m-1} + \cdots + M^2 + M + I)= M^m-I = 0}\) we can cancel the \(\mathbf{M-I}\) to obtain the desired result. \(\mathbf{X_0 = X_0}\) \(\mathbf{X_1 = PX_0+Q}\) \(\mathbf{X_2 = P(PX_0+Q)+Q = P^2X_0 + PQ + Q}\) Claim: \(\mathbf{X_k = P^k X_0 + (P^{k-1} + P^{k-2} + \cdots + I)Q}\) Proof: (By induction on \(k\)). Base case \(k = 0\) is true. Assume it's true for some \(k = l\), then consider \(k = l+1\) \(\mathbf{X_{l+1} = PX_l + Q = P( P^l X_0 + (P^{l-1} + P^{l-2} + \cdots + I)Q) + Q = P^{l+1}X_0 + (P^l + P^{l-1} + \cdots + P)Q + Q = P^{l+1}X_0 + (P^l + P^{l-1} + \cdots + P + I)Q}\) Suppose \(\mathbf{P} = \mathbf{M}\), then consider the set \(\{\mathbf{X_1, X_2}, \ldots\}\) with the operation \(*\) as defined. \(\mathbf{X_i * X_j} = M^{i}(X_j) + (M^{i-1} + M^{i-2} + \cdots + M + I)Q = M^{i}(M^jX_0 + (M^{j-1} + M^{j-2} + \cdots + M + I)Q) + (M^{i-1} + M^{i-2} + \cdots + M + I)Q = M^{i+j}X_0 + (M^{i+j-1}+\cdots + M + I)Q = X_{i+j}\) Since \(X_m = X_0\) we can check all the requirements of the group, but this is going to be isomorphic to the cyclic group with \(m\) elements.
Given that \(b>a>0\), find, by using the binomial theorem, coefficients \(c_{m}\) (\(m=0,1,2,\ldots\)) such that \[ \frac{1}{\left(1-ax\right)\left(1-bx\right)}=c_{0}+c_{1}x+c_{2}x^{2}+\ldots+c_{m}x^{m}+\cdots \] for \(b\left|x\right|<1\). Show that \[ c_{m}^{2}=\frac{a^{2m+2}-2(ab)^{m+1}+b^{2m+2}}{(a-b)^{2}}\,. \] Hence, or otherwise, show that \[ c_{0}^{2}+c_{1}^{2}x+c_{2}^{2}x^{2}+\cdots+c_{m}^{2}x^{m}+\cdots=\frac{1+abx}{\left(1-abx\right)\left(1-a^{2}x\right)\left(1-b^{2}x\right)}\,, \] for \(x\) in a suitable interval which you should determine.
Solution: \begin{align*} \frac{1}{(1-ax)(1-bx)} &=\frac{1}{b-a} \l \frac{b}{1-bx}-\frac{a}{1-ax}\r \\ &= \frac{1}{b-a} \l \sum_{k=0}^{\infty} b(bx)^k-\sum_{k=0}^{\infty} a(ax)^k \r \\ &= \frac{1}{b-a} \sum_{k=0}^{\infty} \l b^{k+1} - a^{k+1} \r x^k \end{align*} Therefore \(\displaystyle c_m = \frac{b^{k+1}-a^{k+1}}{b-a}\). \begin{align*} c_m^2 &= \frac{(b^{m+1}-a^{m+1})^2}{(b-a)^2} \\ &= \frac{a^{2m+2} - 2(ab)^{m+1} + b^{2m+2}}{(b-a)^2} \end{align*} \begin{align*} \sum_{m=0}^{\infty} c_m x^m &= \sum_{m=0}^{\infty} \l \frac{a^{2m+2} - 2(ab)^{m+1} + b^{2m+2}}{(b-a)^2} \r x^m \\ &= \frac{1}{(b-a)^2} \l \sum_{m=0}^{\infty} a^{2m+2} x^m-2\sum_{m=0}^{\infty} (ab)^{m+1} x^m+\sum_{m=0}^{\infty} b^{2m+2} x^m \r \\ &= \frac{1}{(b-a)^2} \l a^2\sum_{m=0}^{\infty} a^{2m} x^m-2ab\sum_{m=0}^{\infty} (ab)^{m} x^m+b^2\sum_{m=0}^{\infty} b^{2m} x^m \r \\ &= \frac{1}{(b-a)^2} \l \frac{a^2}{1-a^2x^2} - \frac{2ab}{1-abx} + \frac{b^2}{1-b^2x^2}\r \\ &= \frac{1+ab}{(1-a^2x)(1-abx)(1-b^2x)} \end{align*} Where geometric series will converge if \(|a^2x| < 1, |b^2x| < 1, |abx| < 1\), ie \(|x| < \min (\frac{1}{a^2}, \frac{1}{b^2} )\)
The function \(\mathrm{f}\) is defined, for \(x\neq1\) and \(x\neq2\) by \[ \mathrm{f}(x)=\frac{1}{\left(x-1\right)\left(x-2\right)} \] Show that for \(\left|x\right|<1\) \[ \mathrm{f}(x)=\sum_{n=0}^{\infty}x^{n}-\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{x}{2}\right)^{n} \] and that for \(1<\left|x\right|<2\) \[ \mathrm{f}(x)=-\sum_{n=1}^{\infty}x^{-n}-\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{x}{2}\right)^{n} \] Find an expression for \(\mbox{f}(x)\) which is valid for \(\left|x\right|>2\).
Solution: \begin{align*} && \f(x) &= \frac1{(x-1)(x-2)} \\ &&&=\frac{1}{x-2} -\frac{1}{x-1} \\ \end{align*} Therefore, for \(|x| < 1\) \begin{align*} && \f(x) &=\frac{1}{x-2} -\frac{1}{x-1} \\ &&&= -\frac12 \frac{1}{1-\frac{x}{2}} + \frac{1}{1-x} \\ &&&= \sum_{n=0}^{\infty} x^n - \frac12 \sum_{n=0}^\infty \l \frac{x}2 \r^2 \end{align*} where both geometric series converge since \(|x| < 1\) and \(|\frac{x}{2}| < 1\) When \(1 < |x|< 2 \Rightarrow |\frac{1}{x}| < 1\), we must have: \begin{align*} && \f(x) &=\frac{1}{x-2} -\frac{1}{x-1} \\ &&&= -\frac12 \frac{1}{1-\frac{x}{2}} + \frac1{x}\frac{1}{1-\frac{1}{x}} \\ &&&= - \frac12 \sum_{n=0}^\infty \l \frac{x}2 \r^2 - \frac{1}{x} \sum_{n=0}^{\infty} x^{-n} \\ &&&= - \frac12 \sum_{n=0}^\infty \l \frac{x}2 \r^2 - \sum_{n=1}^{\infty} x^{-n} \\ \end{align*} Finally, when \(|x| > 2\), ie \(|\frac{2}{x}| < 1\) we have \begin{align*} && \f(x) &=\frac{1}{x-2} -\frac{1}{x-1} \\ &&& =\frac1{x} \frac{1}{1-\frac{2}{x}} - \frac{1}{x}\frac{1}{1-\frac{1}{x}} \\ &&&= \frac1{x} \sum_{n=0}^{\infty} \l \frac{2}{x} \r^n - \sum_{n=1}^{\infty}x^{-n} \\ &&&= \sum_{n=1}^{\infty} 2^{n-1} x^{-n} - \sum_{n=1}^{\infty}x^{-n} \\ \end{align*}
A firm of engineers obtains the right to dig and exploit an undersea tunnel. Each day the firm borrows enough money to pay for the day's digging, which costs £\(c,\) and to pay the daily interest of \(100k\%\) on the sum already borrowed. The tunnel takes \(T\) days to build, and, once finished, earns £\(d\) a day, all of which goes to pay the daily interest and repay the debt until it is fully paid. The financial transactions take place at the end of each day's work. Show that \(S_{n},\) the total amount borrowed by the end of day \(n\), is given by \[ S_{n}=\frac{c[(1+k)^{n}-1]}{k} \] for \(n\leqslant T\). Given that \(S_{T+m}>0,\) where \(m>0,\) express \(S_{T+m}\) in terms of \(c,d,k,T\) and \(m.\) Show that, if \(d/c>(1+k)^{T}-1,\) the firm will eventually pay off the debt.
Solution: After \(n\) days they will have borrowed \(c\) for \(n-1\) days, \(c\) for \(n-2\) days, etc until \(c\) for no days. Therefore the outstanding balance will be: \begin{align*} c + (1+k)\cdot c+ (1+k)^2 \cdot c + \cdots + (1+k)^{n-1} \cdot c &= c\frac{(1+k)^n-1}{(1+k)-1} \\ &= \frac{c[(1+k)^n-1]}{k} \end{align*} At the end of \(T\) days the outstanding balance will be \(S_T = \frac{c[(1+k)^T-1]}{k}\). We can think of each payment of \(d\) during the subsequent period as being equivalent of a payment of \(d (1+k)^{m-1}\) \(m\) days later (as otherwise they would have accrued the equivalent amount in interest. Therefore after \(m\) days the amount paid back (equivalent) is: \begin{align*} (1+k)^{m-1} \cdot d + (1+k)^{m-2} \cdot d + \cdots + d &= \frac{d[(1+k)^m-1]}{k} \end{align*} Therefore the net position, \(S_{T+m}\) will be: \begin{align*} S_{T+m} &= \frac{c[(1+k)^T-1](1+k)^m-d[(1+k)^m-1]}{k} \\ &= \frac{(1+k)^m [c ((1+k)^T-1)-d]+d}{k} \end{align*} Therefore they will eventually pay back their debts if \( [c ((1+k)^T-1)-d]\) is negative. ie \(d > c((1+k)^T-1) \Rightarrow d/c > (1+k)^T-1\)
Show that the sum of the infinite series \[ \log_{2}\mathrm{e}-\log_{4}\mathrm{e}+\log_{16}\mathrm{e}-\ldots+(-1)^{n}\log_{2^{2^{n}}}\mathrm{e}+\ldots \] is \[ \frac{1}{\ln(2\sqrt{2})}. \] {[}\(\log_{a}b=c\) is equivalent to \(a^{c}=b\).{]}
Solution: Let \(S = \log_{2}\mathrm{e}-\log_{4}\mathrm{e}+\log_{16}\mathrm{e}-\ldots+(-1)^{n}\log_{2^{2^{n}}}\mathrm{e}+\ldots\) then \begin{align*} S &= \sum_{n=0}^{\infty} (-1)^n \log_{2^{2^n}} e \\ &= \sum_{n=0}^{\infty} (-1)^n \frac{\log e}{\log {2^{2^n}}} \\ &= \sum_{n=0}^{\infty} (-1)^n \frac{\log e}{2^n\log {2}} \\ &= \frac{\log e}{\log 2} \sum_{n=0}^{\infty} \frac{(-1)^n}{2^n} \\ &= \frac{1}{\log_e 2} \frac{1}{1+\frac12} \\ &= \frac{1}{\ln (2^{3/2})} \\ &= \frac{1}{\ln (2 \sqrt{2})} \end{align*}
Sum each of the series \[ \sin\left(\frac{2\pi}{23}\right)+\sin\left(\frac{6\pi}{23}\right)+\sin\left(\frac{10\pi}{23}\right)+\cdots+\sin\left(\frac{38\pi}{23}\right)+\sin\left(\frac{42\pi}{23}\right) \] and \[ \sin\left(\frac{2\pi}{23}\right)-\sin\left(\frac{6\pi}{23}\right)+\sin\left(\frac{10\pi}{23}\right)-\cdots-\sin\left(\frac{38\pi}{23}\right)+\sin\left(\frac{42\pi}{23}\right), \] giving each answer in terms of the tangent of a single angle. {[}No credit will be given for a numerical answer obtained purely by use of a calculator.{]}
Solution: \(\sin x = \frac{e^{ix} - e^{-ix}}{2i}\). Also let \(z = e^{ \frac{2\pi i}{23}}\) \begin{align*} \sum_{k=0}^{10} \sin \l \frac{(4k +2)\pi}{23} \r &= \sum_{k=0}^{10} \textrm{Im} \l \exp\l \frac{(4k +2)\pi i}{23} \r \r \\ &= \textrm{Im} \l \sum_{k=0}^{10} \exp\l \frac{(4k +2)\pi i}{23} \r \r \\ &= \textrm{Im} \l e^{ \frac{2\pi i}{23}} \sum_{k=0}^{10} z^{2k} \r \\ &= \textrm{Im} \l z \l \frac{z^{22}-1}{z^2-1} \r \r \\ &= \textrm{Im} \l z \l \frac{z^{11}(z^{11}-z^{-11})}{z(z-z^{-1})} \r \r \\ &= \textrm{Im} \l \frac{z^{11}2i \sin \frac{22 \pi}{23} }{2i \sin \frac{2 \pi}{23}} \r \r \\ &= \frac{\sin \frac{22 \pi}{23}}{\sin \frac{2 \pi}{23}} \textrm{Im} ( z^{11}) \\ &= \frac{\sin^2 \frac{22 \pi}{23}}{\sin \frac{2 \pi}{23}} \\ &= \frac{\sin^2 \frac{\pi}{23}}{2\sin \frac{\pi}{23}\cos \frac{\pi}{23}} \\ &= \frac12 \tan \frac{\pi}{23} \end{align*} Similarly, \begin{align*} \sum_{k=0}^{10} (-1)^k\sin \l \frac{(4k +2)\pi}{23} \r &= \sum_{k=0}^{10} \textrm{Im} \l (-1)^k\exp\l \frac{(4k +2)\pi i}{23} \r \r \\ &= \textrm{Im} \l \sum_{k=0}^{10} (-1)^k\exp\l \frac{(4k +2)\pi i}{23} \r \r \\ &= \textrm{Im} \l e^{ \frac{2\pi i}{23}} \sum_{k=0}^{10} (-1)^kz^{2k} \r \\ &= \textrm{Im} \l z \l \frac{z^{22}+1}{z^2+1} \r \r \\ &= \textrm{Im} \l z \l \frac{z^{11}(z^{11}+z^{-11})}{z(z+z^{-1})} \r \r \\ &= \textrm{Im} \l \frac{z^{11}2 \cos \frac{22 \pi}{23} }{2 \cos\frac{2 \pi}{23}} \r \r \\ &= \frac{\cos\frac{22 \pi}{23}}{\cos \frac{2 \pi}{23}} \textrm{Im} ( z^{11}) \\ &= \frac{\cos \frac{22 \pi}{23}\sin \frac{22 \pi}{23}}{\cos\frac{2 \pi}{23}} \\ &= \frac12 \frac{\sin \frac{44 \pi}{23}}{\cos\frac{2 \pi}{23}} \\ &= \frac12 \frac{-\sin \frac{2\pi}{23}}{\cos\frac{2 \pi}{23}} \\ &= -\frac12 \tan \frac{2\pi}{23} \end{align*}
A patient arrives with blue thumbs at the doctor's surgery. With probability \(p\) the patient is suffering from Fenland fever and requires treatment costing \(\pounds 100.\) With probability \(1-p\) he is suffering from Steppe syndrome and will get better anyway. A test exists which infallibly gives positive results if the patient is suffering from Fenland fever but also has probability \(q\) of giving positive results if the patient is not. The test cost \(\pounds 10.\) The doctor decides to proceed as follows. She will give the test repeatedly until either the last test is negative, in which case she dismisses the patient with kind words, or she has given the test \(n\) times with positive results each time, in which case she gives the treatment. In the case \(n=0,\) she treats the patient at once. She wishes to minimise the expected cost \(\pounds E_{n}\) to the National Health Service.
Solution: