14 problems found
Show that, if \(n\) is an integer such that $$(n-3)^3+n^3=(n+3)^3,\quad \quad {(*)}$$ then \(n\) is even and \(n^2\) is a factor of \(54\). Deduce that there is no integer \(n\) which satisfies the equation \((*)\). Show that, if \(n\) is an integer such that $$(n-6)^3+n^3=(n+6)^3, \quad \quad{(**)}$$ then \(n\) is even. Deduce that there is no integer \(n\) which satisfies the equation \((**)\).
Solution: \begin{align*} && n^3 &= (n+3)^3 - (n-3)^3 \\ &&&= n^3 + 9n^2+27n + 27 - (n^3 - 9n^2+27n-27) \\ &&&= 18n^2+54 \end{align*} Therefore since \(2 \mid 2(9n^2 + 27)\), \(2 \mid n^3 \Rightarrow 2 \mid n\), so \(n\) is even. Since \(n^2 \mid n^3\), \(n^2 \mid 54 = 2 \cdot 3^3\), therefore \(n = 1\) or \(n = 3\). \((1-3)^3 + 1^3 < 0 < (1+3)^3\). So \(n = 1\) doesn't work. \((3 - 3)^3 + 3^3 < (3+3)^3\) so \(n = 3\) doesn't work. Therefore there are no solutions. \begin{align*} && n^3 &= (n+6)^3 - (n-6)^3 \\ &&&= n^3 + 18n^2 + 180n + 6^3 - (n^3 - 18n^2 + 180n - 6^3 ) \\ &&&= 36n^2+2 \cdot 6^3 \end{align*} Therefore \(n^2 \mid 2 \cdot 6^3 = 2^4 \cdot 3^3\), therefore \(n = 1, 2, 3, 4, 6, 12\). \(n = 1\), \(1^3 <36+2\cdot 6^3\) \(n = 2\), \(2^3 <36 \cdot 4 + 2 \cdot 6^3\) \(n = 3\), \(3^3 <36 \cdot 9 + 2 \cdot 6^3\) \(n = 4\), \(4^3 < 36 \cdot 16 + 2 \cdot 6^3\) \(n = 6\), \(6^3 < 36\cdot 6^2+ 2 \cdot 6^3\) \(n = 12\), \(12^3 < 36 \cdot 12^2 + 2 \cdot 6^3\) Therefore there are no solutions \(n\) to the equation. These are both special cases of Fermat's Last Theorem, when \(n = 3\)
Use the first four terms of the binomial expansion of \((1-1/50)^{1/2}\), writing \(1/50 = 2/100\) to simplify the calculation, to derive the approximation \(\sqrt 2 \approx 1.414214\). Calculate similarly an approximation to the cube root of 2 to six decimal places by considering \((1+N/125)^a\), where \(a\) and \(N\) are suitable numbers. [You need not justify the accuracy of your approximations.]
Solution: \begin{align*} && (1-1/50)^{1/2} &= 1 + \frac12 \cdot \left ( -\frac1{50} \right) + \frac1{2!} \frac12 \cdot \left ( -\frac12 \right)\cdot \left ( -\frac1{50} \right)^2 + \frac1{3!} \frac12 \cdot \left ( -\frac12 \right) \cdot \left ( -\frac32 \right)\cdot \left ( -\frac1{50} \right)^3 + \cdots \\ &&&=1-\frac{1}{100} - \frac12 \frac1{10000} -\frac12 \frac1{1000000} +\cdots \\ &&&= 0.9899495 + \cdots \\ \Rightarrow && \frac{7\sqrt{2}}{10} &\approx 0.9899495 \\ \Rightarrow && \sqrt{2} &\approx \frac{9.899495}{7} \\ &&&\approx 1.414214 \end{align*} \begin{align*} && (1 + 3/125)^{1/3} &= \frac{\sqrt[3]{125+3}}{5} \\ &&& = \frac{8\sqrt[3]{2}}{10} \\ && (1 + 3/125)^{1/3} &= 1 + \frac13 \left ( \frac{3}{125} \right) + \frac1{2!} \cdot \frac{1}{3} \cdot \left ( -\frac23\right) \left ( \frac{3}{125}\right)^2 +\cdots \\ &&&= 1+ \frac{8}{1000} - \frac{64}{1000000} \\ &&&= 1.007936 \\ \Rightarrow && \sqrt[3]{2} &= \frac{10.07936}{8} \\ &&&= 1.259920 \end{align*}
Show that the sum \(S_N\) of the first \(N\) terms of the series $$\frac{1}{1\cdot2\cdot3}+\frac{3}{\cdot3\cdot4}+\frac{5}{3\cdot4\cdot5}+\cdots +\frac{2n-1}{n(n+1)(n+2)}+\cdots$$ is $${1\over2}\left({3\over2}+{1\over N+1}-{5\over N+2}\right).$$ What is the limit of \(S_N\) as \(N\to\infty\)? The numbers \(a_n\) are such that $$\frac{a_n}{a_{n-1}}=\frac{(n-1)(2n-1)}{(n+2)(2n-3)}.$$ Find an expression for \(a_n/a_1\) and hence, or otherwise, evaluate \(\sum\limits_{n=1}^\infty a_n\) when \(\displaystyle a_1=\frac{2}{9}\;\).
Solution: First notice by partial fractions: \begin{align*} \frac{2n-1}{n(n+1)(n+2)} &= \frac{-1/2}{n} + \frac{3}{n+1} + \frac{-5/2}{n+2} \\ &= \frac{-1}{2n} + \frac{3}{n+1} - \frac{5}{2(n+2)} \end{align*} And therefore: \begin{align*} \sum_{n = 1}^N \frac{2n-1}{n(n+1)(n+2)} &= -\frac12 \sum_{n=1}^N \frac1n +3\sum_{n=1}^N \frac1{n+1} -\frac52 \sum_{n=1}^N \frac1{n+2} \\ &= -\frac12-\frac14 + \frac{3}{2}+ \sum_{n=3}^N (3-\frac12 -\frac52)\frac1n + \frac{3}{N+1} - \frac{5}{2(N+1)} - \frac{5}{2(N+2)} \\ &= \frac12 \l \frac32+\frac1{N+1}-\frac{5}{N+2} \r \end{align*} As \(N \to \infty, S_N \to \frac{3}{4}\). \begin{align*} && \frac{a_n}{a_{n-1}}&=\frac{(n-1)(2n-1)}{(n+2)(2n-3)} \\ \Rightarrow && \frac{a_n}{a_1} &= \frac{a_n}{a_{n-1}} \cdot \frac{a_{n-1}}{a_{n-2}} \cdots \frac{a_2}{a_1} \\ &&&= \frac{(n-1)(2n-1)}{(n+2)(2n-3)} \cdot \frac{(n-2)(2n-3)}{(n+1)(2n-5)} \cdots \frac{(1)(3)}{(4)(1)} \\ &&&= \frac{(2n-1)3\cdot 2\cdot 1}{(n+2)(n+1)n} \\ &&& = \frac{6(2n-1)}{n(n+1)(n+2)} \end{align*} Therefore \(a_n = \frac{4}{3} \frac{2n-1}{n(n+1)(n+2)}\) and so our sequence is \(\frac43\) the earlier sum, ie \(1\)
The integral \(I_n\) is defined by $$I_n=\int_0^\pi(\pi/2-x)\sin(nx+x/2)\,{\rm cosec}\,(x/2)\,\d x,$$ where \(n\) is a positive integer. Evaluate \(I_n-I_{n-1}\), and hence evaluate \(I_n\) leaving your answer in the form of a sum.
Solution: \begin{align*} && I_n - I_{n-1} &= \int_0^\pi \left ( \frac{\pi}{2} - x \right) \left ( \sin\left(nx + \frac{x}{2}\right) - \sin \left ((n-1)x + \frac{x}{2} \right)\right) \cosec \frac{x}{2} \d x \\ &&&= \int_0^\pi \left ( \frac{\pi}{2} - x \right) \left (2 \sin \left ( \frac{nx + \frac{x}{2} - (n-1)x - \frac{x}{2} }{2}\right)\cos nx \right) \cosec \frac{x}{2} \d x \\ &&&= \int_0^\pi \left ( \frac{\pi}{2} - x \right) \left (2 \sin \left ( \frac{x}{2}\right)\cos nx \right) \cosec \frac{x}{2} \d x \\ &&&=2 \int_0^\pi \left ( \frac{\pi}{2} - x \right) \cos nx \d x \\ &&&=\pi \left [ \frac{\sin nx}{n}\right]_0^{\pi} - 2\int_0^\pi x \cos n x \d x \\ &&&= 0 - 2\left[ \frac{x \sin nx}{n} \right]_0^{\pi} + 2\int_0^\pi \frac{\sin nx}{n} \d x \\ &&&= 2\left[ -\frac{\cos nx}{n^2} \right]_0^{\pi} \\ &&&=2 \frac{1-(-1)^{n}}{n^2} \\ \\ && I_0 &= \int_0^\pi (\pi/2 - x) \d x =0 \\ \Rightarrow && I_{2k+2} = I_{2k+1} &= 4 \left (\frac{1}{1^2} + \frac{1}{3^2} + \cdots + \frac{1}{(2k+1)^2} \right) \end{align*}
Define the modulus of a complex number \(z\) and give the geometric interpretation of \(\vert\,z_1-z_2\,\vert\) for two complex numbers \(z_1\) and \(z_2\). On the basis of this interpretation establish the inequality $$\vert\,z_1+z_2\,\vert\le \vert\,z_1\,\vert+\vert\,z_2\,\vert.$$ Use this result to prove, by induction, the corresponding inequality for \(\vert\,z_1+\cdots+z_n\,\vert\). The complex numbers \(a_1,\,a_2,\,\ldots,\,a_n\) satisfy \(|a_i|\le 3\) (\(i=1, 2, \ldots , n\)). Prove that the equation $$a_1z+a_2z^2\cdots +a_nz^n=1$$ has no solution \(z\) with \(\vert\,z\,\vert\le 1/4\).
Solution: Suppose \(z = a+ib\), where \(a,b \in \mathbb{R}\) then the modulus of \(z\), \(|z| = \sqrt{a^2+b^2}\). Noting the similarity to the Pythagorean theorem, we can say that \(|z_1 - z_2|\) is the distance between \(z_1\) and \(z_2\) in the Argand diagram. \begin{align*} |z_1 + z_2| &= |(z_1 - 0) + (0 -z_2)| \\ &\underbrace{\leq}_{\text{the direct distance is shorter than going via }0} |z_1 - 0| + |0 - z_2| \\ &= |z_1| + |-z_2| \\ &= |z_1| + |z_2| \end{align*} Claim: \(\displaystyle \vert\,z_1+\cdots+z_n\,\vert \leq \sum_{i=1}^n |z_i|\) Proof: (By Induction) Base Case: \(n = 1, 2\) have been proven. Inductive step, suppose it is true for \(n = k\), then consider \(n = k+1\), ie \begin{align*} \vert\,z_1+\cdots+z_k+z_{k+1}\,\vert &\leq \vert\,z_1+\cdots+z_k\vert + \vert z_{k+1}\,\vert \\ &\underbrace{\leq}_{\text{inductive hypothesis}} \sum_{i=1}^k |z_i| + |z_{k+1}| \\ &= \sum_{i=1}^{k+1} |z_i| \end{align*} Therefore if our hypothesis is true for \(n = k\) it is true for \(n = k+1\), and so since it is true for \(n = 1\) it is true by the principle of mathematical induction for all integers \(n \geq 1\). Suppose \(|z| \leq 1/4\), then consider: \begin{align*} \vert a_1z+a_2z^2+\cdots +a_nz^n \vert &\leq \vert a_1 z\vert + \vert a_2z^2\vert + \cdots + \vert a_n z_n\ \vert \\ &= \vert a_1\vert\vert z\vert + \vert a_2\vert\vert z^2\vert + \cdots + \vert a_n\vert\vert z^n\ \vert \\ &\leq 3\left ( |z| + |z|^2 + \cdots + |z|^n \right) \\ &\leq 3 \left ( \frac{1}{4} + \frac1{4^2} + \cdots + \frac{1}{4^n} \right) \\ &< 3 \frac{1/4}{1-1/4} \\ &= 1 \end{align*} Therefore we cannot have equality and there are no solutions.
Two curves are given parametrically by \[ x_{1}=(\theta+\sin\theta),\qquad y_{1}=(1+\cos\theta),\tag{1} \]and \[ x_{2}=(\theta-\sin\theta),\qquad y_{1}=-(1+\cos\theta),\tag{2} \] Find the gradients of the tangents to the curves at the points where \(\theta= \pi/2\) and \(\theta=3\pi/2\). Sketch, using the same axes, the curves for \(0\le\theta \le 2\pi\). Find the equation of the normal to the curve (1) at the point with parameter \(\theta\). Show that this normal is a tangent to the curve (2).
\begin{eqnarray*} {\rm f}(x)&=& \tan x-x,\\ {\rm g}(x)&=& 2-2\cos x-x\sin x,\\ {\rm h}(x)&=& 2x+x\cos 2x-\tfrac{3}{2}\sin 2x,\\ {\rm F}(x)&=& {x(\cos x)^{1/3}\over\sin x}. \end{eqnarray*} \vspace{1mm}
Points \(\mathbf{A},\mathbf{B},\mathbf{C}\) in three dimensions have coordinate vectors \(\mathbf{a},\mathbf{b},\mathbf{c}\), respectively. Show that the lines joining the vertices of the triangle \(ABC\) to the mid-points of the opposite sides meet at a point \(R\). \(P\) is a point which is {\bf not} in the plane \(ABC\). Lines are drawn through the mid-points of \(BC\), \(CA\) and \(AB\) parallel to \(PA\), \(PB\) and \(PC\) respectively. Write down the vector equations of the lines and show by inspection that these lines meet at a common point \(Q\). Prove further that the line \(PQ\) meets the plane \(ABC\) at \(R\).
A light smoothly jointed planar framework in the form of a regular hexagon \(ABCDEF\) is suspended smoothly from \(A\) and a weight 1kg is suspended from \(C\). The framework is kept rigid by three light rods \(BD\), \(BE\) and \(BF\). What is the direction and magnitude of the supporting force which must be exerted on the framework at \(A\)? Indicate on a labelled diagram which rods are in thrust (compression) and which are in tension. Find the magnitude of the force in \(BE\).
A wedge of mass \(M\) rests on a smooth horizontal surface. The face of the wedge is a smooth plane inclined at an angle \(\alpha\) to the horizontal. A particle of mass \(m\) slides down the face of the wedge, starting from rest. At a later time \(t\), the speed \(V\) of the wedge, the speed \(v\) of the particle and the angle \(\beta\) of the velocity of the particle below the horizontal are as shown in the diagram.
A fielder, who is perfectly placed to catch a ball struck by the batsman in a game of cricket, watches the ball in flight. Assuming that the ball is struck at the fielder's eye level and is caught just in front of her eye, show that \(\frac{ {\rm d}}{{\rm d t}} (\tan\theta ) \) is constant, where \(\theta\) is the angle between the horizontal and the fielder's line of sight. In order to catch the next ball, which is also struck towards her but at a different velocity, the fielder runs at constant speed \(v\) towards the batsman. Assuming that the ground is horizontal, show that the fielder should choose \(v\) so that \(\frac{ {\rm d}}{{\rm d t}} (\tan\theta ) \) remains constant.
Solution: Set up a coordinate frame such that the position of the catch is the origin and the time of the catch is \(t = 0\) . We must have then that the trajectory of the ball is \(\mathbf{s} =\mathbf{u} t + \frac12 \mathbf{g} t^2 = \binom{u_x t}{u_y t - \frac12 gt^2}\). We must then have: \begin{align*} && \tan \theta &= \frac{u_y t - \frac12 gt^2}{u_x t} \\ &&&= \frac{u_y}{u_x} - \frac{g}{2u_x} t \\ \Rightarrow && \frac{\d}{\d \theta} \left ( \tan \theta \right) &= 0 - \frac{g}{2 u_x} \end{align*} which is clearly constant. Set coordinates so \(y\)-axis starts from eye-level and \(t = 0\) the first time the ball reaches that level. (Or move the trajectory backwards if that's not the case). Then the ball has trajectory \(\binom{u_xt}{u_yt - \frac12 gt^2}\). The ball reaches eye level a second time when \(t = \frac{2u_y}{g}\), ie at a point \(\frac{2u_xu_y}{g}\). The fielder therefore needs to have position \(f + (u_x-\frac{g}{2u_y}f)t\) at all times. Therefore \begin{align*} && \tan \theta &= \frac{u_y t - \frac12 gt^2}{f + (u_x-\frac{g}{2u_y}f)t - u_x t} \\ &&&= \frac{u_y t - \frac12 gt^2}{f(1-\frac{g}{2u_y}t)} \\ &&&= \frac{u_yt ( 1- \frac{g}{2u_y}t)}{f( 1- \frac{g}{2u_y}t)} \\ &&&= u_y t \\ \Rightarrow && \frac{\d}{\d \theta} \left ( \tan \theta \right) &= u_y \end{align*} Ie \( \frac{\d}{\d \theta} \left ( \tan \theta \right) \) is constant as required.
The diagnostic test AL has a probability 0.9 of giving a positive result when applied to a person suffering from the rare disease mathematitis. It also has a probability 1/11 of giving a false positive result when applied to a non-sufferer. It is known that only \(1\%\) of the population suffer from the disease. Given that the test AL is positive when applied to Frankie, who is chosen at random from the population, what is the probability that Frankie is a sufferer? In an attempt to identify sufferers more accurately, a second diagnostic test STEP is given to those for whom the test AL gave a positive result. The probablility of STEP giving a positive result on a sufferer is 0.9, and the probability that it gives a false positive result on a non-sufferer is \(p\). Half of those for whom AL was positive and on whom STEP then also gives a positive result are sufferers. Find \(p\).
Solution: \begin{align*} \mathbb{P}(M | P_{AL}) &= \frac{\mathbb{P}(M \cap P_{AL})}{\mathbb{P}(P_{AL})} \\ &= \frac{\frac{1}{100} \frac{9}{10}}{\frac{1}{100} \frac{9}{10} + \frac{99}{100} \frac{1}{11}} \\ &= \frac{\frac{9}{10}}{\frac{9}{10} + \frac{9}{1}} \\ &= \frac{9}{99} = \frac{1}{11} \\ \end{align*} \begin{align*} && \frac12 &= \mathbb{P}(M | P_{STEP}, P_{AL}) \\ &&&= \frac{\frac{1}{100} \frac{9}{10} \frac{9}{10}}{\frac{1}{100} \frac{9}{10} \frac{9}{10} + \frac{99}{100} \frac{1}{11}p} \\ &&&= \frac{81}{81+900p} \\ \Rightarrow && p &= \frac{81}{900} = \frac{9}{100} \end{align*} Therefore \(p = 9\%\)
A random variable \(X\) has the probability density function \[ \mathrm{f}(x)=\begin{cases} \lambda\mathrm{e}^{-\lambda x} & x\geqslant0,\\ 0 & x<0. \end{cases} \] Show that $${\rm P}(X>s+t\,\vert X>t) = {\rm P}(X>s).$$ The time it takes an assistant to serve a customer in a certain shop is a random variable with the above distribution and the times for different customers are independent. If, when I enter the shop, the only two assistants are serving one customer each, what is the probability that these customers are both still being served at time \(t\) after I arrive? One of the assistants finishes serving his customer and immediately starts serving me. What is the probability that I am still being served when the other customer has finished being served?
Solution: \begin{align*} && \mathbb{P}(X > t) &= \int_t^{\infty} \lambda e^{-\lambda x} \d x\\ &&&= \left[ -e^{-\lambda x} \right]_t^\infty \\ &&&= e^{-\lambda t}\\ \\ && \mathbb{P}(X > s + t | X > t) &= \frac{\mathbb{P}(X > s + t)}{\mathbb{P}(X > t)} \\ &&&= \frac{e^{-(s+t)\lambda}}{e^{-t\lambda}} \\ &&&= e^{-s\lambda} = \mathbb{P}(X > s) \end{align*} The probability both are still being served (independently) is \(\mathbb{P}(X > t)^2 = e^{-2\lambda t}\). The probability is exactly \(\frac12\). The property we proved in the first part of the questions shows the distribution is memoryless, ie we are both experiencing samples from the same distribution. Therefore we are equally likely to finish first.
The staff of Catastrophe College are paid a salary of \(A\) pounds per year. With a Teaching Assessment Exercise impending it is decided to try to lower the student failure rate by offering each lecturer an alternative salary of \(B/(1+X)\) pounds, where \(X\) is the number of his or her students who fail the end of year examination. Dr Doom has \(N\) students, each with independent probability \(p\) of failure. Show that she should accept the new salary scheme if $$A(N+1)p < B(1-(1-p)^{N+1}).$$ Under what circumstances could \(X\), for Dr Doom, be modelled by a Poisson random variable? What would Dr Doom's expected salary be under this model?
Solution: \begin{align*} && \E[\text{salary}] &= B\sum_{k=0}^N \frac{1}{1+k}\binom{N}{k}p^k(1-p)^{N-k} \\ \\ && (q+x)^N &= \sum_{k=0}^N \binom{N}{k}x^kq^{N-k} \\ \Rightarrow && \int_0^p(q+x)^N \d x &= \sum_{k=0}^N \binom{N}{k} \frac{p^{k+1}}{k+1}q^{N-k} \\ && \frac{(q+p)^{N+1}-q^{N+1}}{N+1} &= \frac{p}{B} \E[\text{salary}] \\ \Rightarrow && \E[\text{salary}] &= B\frac{1-q^{N+1}}{p(N+1)} \end{align*} Therefore if \(Ap(N+1) < B(1-(1-p)^{N+1})\) the expected value of the new salary is higher. (Whether or not the new salary is worth it in a risk adjusted sense is for the birds). We could model \(X\) by a Poisson random variable if \(N\) is large and \(Np = \lambda \) is small. Suppose \(X \approx Po(\lambda)\) then \begin{align*} \E \left [\frac{B}{1+X} \right] &= B\sum_{k=0}^\infty \frac{1}{1+k}\frac{e^{-\lambda}\lambda^k}{k!} \\ &= \frac{B}{\lambda} \sum_{k=0}^\infty e^{-\lambda} \frac{\lambda^{k+1}}{(k+1)!} \\ &= \frac{B}{\lambda}e^{-\lambda}(e^{\lambda}-1) \\ &= \frac{B(1-e^{-\lambda})}{\lambda} = B \frac{1-e^{-Np}}{Np} \end{align*}