A number of the form \(1/N\), where \(N\) is an integer greater than 1, is called a unit fraction. Noting that \[ \frac1 2 =\frac13 + \frac16\\\ \mbox{ and } \frac13 = \frac14 + \frac1{12}, \] guess a general result of the form $$ \frac1N =\frac1a +\frac1b \tag{*} $$ and hence prove that any unit fraction can be expressed as the sum of two distinct unit fractions. By writing \((*)\) in the form \[ (a-N)(b-N)=N^2 \] and by considering the factors of \(N^2\), show that if \(N\) is prime, then there is only one way of expressing \(1/N\) as the sum of two distinct unit fractions. Prove similarly that any fraction of the form \(2/N\), where \(N\) is prime number greater than 2, can be expressed uniquely as the sum of two distinct unit fractions.
Solution: Notice that \(\frac{1}{N} = \frac{1}{N+1} + \frac{1}{N(N+1)}\), so any unit fraction can be expressed as the sum of two distinct unit fractions. \begin{align*} && \frac{1}N &= \frac1a + \frac1b \\ \Leftrightarrow && ab&= Nb+Na \\ \Leftrightarrow && 0 &= (a-N)(b-N)-N^2 \\ \Leftrightarrow && N^2 &= (a-N)(b-N) \end{align*} If \(N\) is prime then the only factors of \(N^2\) are \(1,N\) and \(N^2\). if \(a-N = b-N = N\) then \(a=b\) and we don't have distinct fractions. Therefore \(a-N = 1\) and \(b-N = N^2\) and we obtain the decomposition earlier (and it must be the only solution). \begin{align*} && \frac2N &= \frac1a+\frac1b \\ \Leftrightarrow && 2ab &= Nb+Na \\ \Leftrightarrow && 4ab &= 2Na+2Nb \\ \Leftrightarrow && N^2 &= (2a-N)(2b-N) \end{align*} Therefore for \(a,b\) to be distinct we must have \(2a = N+1\) and \(2b = N+N^2\) as the only possible factorisation. Both of the right hand sides are even so we can write \[ \frac{1}{N} = \frac{1}{\frac{N+1}{2}} + \frac{1}{\frac{N(N+1)}{2}} \] and this is unique
Prove that if \({(x-a)^{2}}\) is a factor of the polynomial \(\p(x)\), then \(\p'(a)=0\). Prove a corresponding result if \((x-a)^4\) is a factor of \(\p(x).\) Given that the polynomial $$ x^6+4x^5-5x^4-40x^3-40x^2+32x+k $$ has a factor of the form \({(x-a)}^4\), find \(k\).
Solution: First notice that \(p(x) = (x-a)^2q(x)\) so \(p'(x) = 2(x-a)q(x) + (x-a)^2q'(x) = (x-a)(2q(x)+(x-a)q'(x))\), in particular \(p'(a) = 0\) so \(x-a\) is a root of \(p'(x)\). If \((x-a)^4\) is a root of \(p(x)\) then \(p^{(3)}(a)= 0\). The proof is similar. Differentiating \(3\) times we obtain: \(6 \cdot 5 \cdot 4 x^3 + 4 \cdot 5 \cdot 4 \cdot 3 x^2 - 5\cdot4 \cdot 3 \cdot 2 x-40 \cdot 3 \cdot 2 \cdot 1 = 5!(x^3+2x^2-x-2) = 5!(x+2)(x^2-1)\). So our possible (repeated) roots are \(x=-2,-1,1\). We can check \(p'(x) = 6x^5+20x^4-20x^3-120x^2-80x+32\), and see \(p'(1) = 36 - 200 \neq 0\), \(p'(-1) = -6+20+20-120+80+32 \neq 0\), therefore \(a = -2\)
The lengths of the sides \(BC\), \(CA\), \(AB\) of the triangle \(ABC\) are denoted by \(a\), \(b\), \(c\), respectively. Given that $$ b = 8+{\epsilon}_1, \, c=3+{\epsilon}_2,\, A=\tfrac{1}{3}\pi + {\epsilon}_3, $$ where \({\epsilon}_1\), \({\epsilon}_2\), and \( {\epsilon}_3\) are small, show that \(a \approx 7 + {\eta}\), where ${\eta}= {\left(13 \, {{\epsilon}_1}-2\,{\epsilon}_2 + 24{\sqrt 3} \;{{\epsilon}_3}\right)}/14$. Given now that $$ {\vert {\epsilon}_1} \vert \le 2 \times 10^{-3}, \ \ \ {\vert {\epsilon}_2} \vert \le 4\cdot 9\times 10^{-2}, \ \ \ {\vert {\epsilon}_3} \vert \le \sqrt3 \times 10^{-3}, $$ find the range of possible values of \({\eta}\).
Solution: The cosine rule states that: \(a^2 = b^2 + c^2 - 2bc \cos (A)\) Therefore \begin{align*} a^2 &= (8 + \epsilon_1)^2 + (3 + \epsilon_2)^2 - 2(8 + \epsilon_1) (3 + \epsilon_2)\cos \l \frac{\pi}{3} + \epsilon_3 \r \\ &\approx 64 + 16\epsilon_1 + 9 + 6\epsilon_2- 2(24 + 3\epsilon_1+8\epsilon_2) \cos \l \frac{\pi}{3} + \epsilon_3 \r \\ &= 73 + 16\epsilon_1+ 6\epsilon_2 - 2(24 + 3\epsilon_1+8\epsilon_2) \l \cos \l \frac{\pi}{3} \r \cos \epsilon_3 - \sin \l \frac{\pi}{3} \r \sin \epsilon_3 \r \\ &\approx 73 + 16\epsilon_1+ 6\epsilon_2 - (24 + 3 \epsilon_1+8\epsilon_2) + 24\sqrt{3}\epsilon_3 \\ &= 49 + 13 \epsilon_1 - 2\epsilon_2+24\sqrt{3}\epsilon_3 \\ &= 7^2 + 2 \cdot 7 \cdot \frac{13 \epsilon_1 - 2\epsilon_2+24\sqrt{3}\epsilon_3}{14} \\ &\approx \l 7 + \frac{13 \epsilon_1 - 2\epsilon_2+24\sqrt{3}\epsilon_3}{14} \r^2 \end{align*} In this approximation, we are ignoring all terms of order \(2\), and using the approximations \(\cos \varepsilon \approx 1, \sin \varepsilon \approx \varepsilon\) Therefore \(a \approx 7 + \frac{ 13 \epsilon_1 - 2\epsilon_2+24\sqrt{3}\epsilon_3}{14}\). \(\eta\) is maximised if \(\epsilon_1, \epsilon_3\) are and \(\epsilon_2\) is minimized, ie: \begin{align*} \eta &\leq \frac{13 \cdot 2 \cdot 10^{-3} - 2 \cdot 4.9 \cdot 10^{-2} + 24 \sqrt{3} \cdot \sqrt{3} \cdot 10^{-3}}{14} \\ &= 10^{-3} \cdot \frac{26 - 98 + 74}{14} \\ &= 10^{-3} \cdot \frac{1}{7}\end{align*} Similarly, it is maximised when signs are reversed, ie: \(| \eta | \leq 10^{-3} \cdot \frac{1}{7}\)
Prove that \[ (\cos\theta +\mathrm{i}\sin\theta) (\cos\phi +\mathrm{i}\sin\phi) = \cos(\theta+\phi) +\mathrm{i}\sin(\theta+\phi) \] and that, for every positive integer \(n\), $$ {(\cos {\theta} + \mathrm{i}\sin {\theta})}^n = \cos{n{\theta}} + \mathrm{i}\sin{n{\theta}}. $$ By considering \((5-\mathrm{i})^2(1+\mathrm{i})\), or otherwise, prove that \[ \arctan\left(\frac{7}{17}\right)+2\arctan\left(\frac{1}{5}\right)=\frac{\pi}{4}\,. \] Prove also that \[ 3\arctan\left(\frac{1}{4}\right)+\arctan\left(\frac{1}{20}\right)+\arctan\left(\frac{1}{1985}\right)=\frac{\pi}{4}\,. \] [Note that \(\arctan\theta\) is another notation for \(\tan^{-1}\theta\).]
Solution: \begin{align*} && LHS &= (\cos\theta +\mathrm{i}\sin\theta) (\cos\phi +\mathrm{i}\sin\phi) \\ &&&= \cos \theta \cos \phi - \sin \theta \sin \phi + \mathrm{i}(\sin \theta \cos \phi + \cos \theta \sin \phi) \\ &&&= \cos (\theta + \phi) + \mathrm{i} \sin (\theta + \phi) \\ &&&= RHS \end{align*} Therefore we can see \((\cos \theta + i \sin \theta)^n = \cos n\theta + i \sin n \theta\). \begin{align*} && (5-i)^2(1+i) &= (24-10i)(1+i) \\ &&&= (24+10) + i(24-10) \\ &&&= 34+14i \\ \Rightarrow && 2\arg(5-i) +\arg(1+i) &= \arg(34+14i) \\ \Rightarrow && 2\arctan\left (-\frac{1}{5} \right) + \frac{\pi}{4} &= \arctan \left ( \frac{7}{17} \right) \\ \Rightarrow && 2\arctan\left (\frac{1}{5} \right) +\arctan \left ( \frac{7}{17} \right) &= \frac{\pi}{4} \\ \end{align*} Consider \((1+i)(4-i)^3(20-i)\) \begin{align*} && (1+i)(4-i)^3(20-i) &= (21+19i)(52-47i) \\ &&&= 1985+i \\ \Rightarrow && \frac{\pi}{4} - 3 \arctan \left ( \frac{1}{4} \right) -\arctan \left ( \frac{1}{20} \right) &= \arctan \left ( \frac{1}{1985} \right) \end{align*}
It is required to approximate a given function \(\f(x)\), over the interval \(0 \le x \le 1\), by the linear function \(\lambda x\), where \(\lambda\) is chosen to minimise \[ \int_0^1 \big(\f(x)-\lambda x \big)^{\!2} \,\d x . \] Show that \[ \lambda = 3 \int_0^1 x\f(x)\,\d x. \] The residual error, \(R\), of this approximation process is such that \[ R^2 = \int_0^1 \big(\f(x)-\lambda x \big)^{\!2}\,\d x. \] Show that \[ R^2 = \int_0^1 \big(\f(x)\big)^{\!2}\,\d x -\tfrac{1}{3} \lambda ^2. \] Given now that \(\f(x)= \sin (\pi x/n)\), show that (i) for large \(n\), \(\lambda \approx \pi/n\) and (ii) \(\lim_{n \to \infty}R = 0.\) Explain why, prior to any calculation, these results are to be expected. [You may assume that, when \(\theta\) is small, $\sin \theta \approx \theta-\frac{1}{6}\theta^3$ and \(\cos \theta \approx 1 - \frac{1}{2}\theta^2.\)]
Solution: \begin{align*} && g(\lambda) &= \int_0^1 \big(\f(x)-\lambda x \big)^{\!2} \,\d x \\ &&&= \int_0^1 \left ( f(x)^2 -2\lambda xf(x) + \lambda^2 x^2\right) \d x \\ &&&= \frac13\lambda^2 - 2\lambda \int_0^1 x f(x) \d x + \int_0^1 f(x)^2 \d x \\ \end{align*} Differentiating (or completing the square) it is clear the minimum occurs when \(\displaystyle \lambda = 3 \int_0^1 xf(x) \d x\) \begin{align*} && R^2 &= \int_0^1 (f(x) - \lambda x )^2 \d x \\ &&&= \frac13\lambda^2 - 2\lambda \int_0^1 x f(x) \d x + \int_0^1 f(x)^2 \d x \\ &&&= \frac13 \left (\lambda -3\int_0^1 xf(x) \d x \right)^2 -\frac13 \left ( 3\int_0^1 xf(x) \d x \right)^2+\int_0^1 f(x)^2 \d x \\ \end{align*} When \(\lambda = 3\int_0^1 xf(x) \d x \) clearly this is the desired result. \begin{align*} && \lambda &= 3\int_0^1 xf(x) \d x \\ &&&= 3\int_0^1 x \sin(\pi x /n) \d x \\ &&&= 3 \left [-x \frac{n}{\pi} \cos (\pi x /n) \right]_0^1 + \frac{3n}{\pi} \int_0^1 \cos(\pi x /n) \d x \\ &&&= -\frac{3n}{\pi}\cos(\pi/n) + \frac{3n}{\pi} \left [ \frac{n}{\pi} \sin(\pi x /n)\right]_0^1 \\ &&&= -\frac{3n}{\pi} \cos(\pi/n) + \frac{3n^2}{\pi^2} \sin(\pi /n) \\ \text{for large }n: &&&\approx -\frac{3n}{\pi}\left ( 1 - \frac12\frac{\pi^2}{n^2} + o(1/n^4)\right) + \frac{3n^2}{\pi^2} \left (\frac{\pi}{n} - \frac16 \frac{\pi^3}{n^3} +o(1/n^5) \right) \\ &&&= \left (\frac32 -\frac12\right)\frac{\pi}{n} + o(1/n^3) \\ &&&= \frac{\pi}{n} + o(1/n^2) \end{align*} Therefore for large \(n\), \(\lambda \approx \frac{\pi}n\) \begin{align*} && \int_0^1 \sin^2(\pi x/n) \d x &= \frac12\int_0^1(1- \cos(2\pi x/n)) \d x\\ &&&= \frac12\left ( 1 - \frac{n}{2\pi}\left[\sin(2\pi x/n) \right]_0^1 \right) \\ &&&= \frac12 -\frac{n}{4\pi}\sin(2\pi /n) \\ \\ && R^2 &= \frac12 -\frac{n}{4\pi}\sin(2\pi /n) - \frac13 \left ( \frac{\pi}{n}+o(1/n^2)\right)^2 \\ &&&= \frac12 - \left ( \frac{1}{2} -\frac16\frac{\pi}{n}+o(1/n^3) \right) - o(1/n^2) \\ &&& = \frac16 \frac{\pi}{n} + o(1/n^2) \\ &&&\to 0 \text{ as } n \to \infty \end{align*} We should expect these results as for \(n\) very large \(\sin(\pi x/n) \approx \frac{\pi }{n}x\) so the best linear approximation is likely to be \(\lambda = \frac{\pi}{n}\) and we should expect it to improve to the point that we cannot tell the difference, ie \(R^2 \to 0\)
Show that \[ \sin\theta = \frac {2t}{1+t^2}, \ \ \ \cos\theta = \frac{1-t^2}{1+t^2}, \ \ \ \frac{1+\cos\theta}{\sin\theta} = \tan (\tfrac{1}{2}\pi-\tfrac{1}{2}\theta), \] where \(t =\tan\frac{1}{2}\theta\). Use the substitution \(t =\tan\frac{1}{2}\theta\) to show that, for \(0<\alpha<\frac{1}{2}\pi\), \[ \int_0^{\frac{1}{2}\pi} {1 \over {1 + \cos\alpha \sin \theta}} \,\d\theta =\frac{\alpha}{\sin\alpha}\,, \] and deduce a similar result for \[ \int_0^{\frac{1}{2}\pi} {1 \over {1 + \sin\alpha \cos \theta}} \,\d\theta \,. \] %$$
The line \(l\) has vector equation \({\bf r} = \lambda {\bf s}\), where \[ {\bf s} = (\cos\theta+\sqrt3\,) \; {\bf i} +(\surd2\;\sin\theta)\;{\bf j} +(\cos\theta-\sqrt3\,)\;{\bf k} \] and \(\lambda\) is a scalar parameter. Find an expression for the angle between \(l\) and the line \mbox{\({\bf r} = \mu(a\, {\bf i} + b\,{\bf j} +c\, {\bf k})\)}. Show that there is a line \(m\) through the origin such that, whatever the value of \(\theta\), the acute angle between \(l\) and \(m\) is \(\pi/6\). A plane has equation \(x-z=4\sqrt3\). The line \(l\) meets this plane at \(P\). Show that, as \(\theta\) varies, \(P\) describes a circle, with its centre on \(m\). Find the radius of this circle.
In an aerobatics display, Jane and Karen jump from a great height and go through a period of free fall before opening their parachutes. While in free fall at speed \(v\), Jane experiences air resistance \(kv\) per unit mass but Karen, who spread-eagles, experiences air resistance \mbox{\(kv + (2k^2/g)v^2\)} per unit mass. Show that Jane's speed can never reach \(g/k\). Obtain the corresponding result for Karen. Jane opens her parachute when her speed is \(g/(3{k})\). Show that she has then been in free fall for time \(k^{-1}\ln (3/2)\). Karen also opens her parachute when her speed is \(g/(3{k})\). Find the time she has then been in free fall.
Solution: Looking at the forces on Jane, \(kv < g \Rightarrow v < \frac{g}{k}\). For Karen we have \begin{align*} kv + (2k^2/g)v^2 &< g\\ -g^2 + gkv + (2k^2)v^2 &< 0 \\ (2kv-g)(kv+g) &< 0\\ \Rightarrow v &< \frac{g}{2k} \end{align*} \begin{align*} && \dot{v} &= g - kv \\ \Rightarrow && \frac{\dot{v}}{g - kv} &= 1 \\ \Rightarrow && T &= \int_0^{g/(3k)} \frac{1}{g - kv} dv \\ && &= \int_0^{g/(3k)} \frac{1}{g - kv} dv\\ && &= \int_0^{g/(3k)} \frac{1}{g - kv} dv \\ && &= \left [-\frac{1}{k} \ln \l g - kv \r \right ]_0^{g/(3k)} \\ && &= \frac{1}{k} \ln \l g \r - \frac{1}{k} \ln \l \frac{2}{3}g \r\\ &&&= \frac{1}{k} \ln \l \frac{3}{2} \r \end{align*} \begin{align*} && \dot{v} &= g - kv - (2k^2/g)v^2 \\ \Rightarrow && \frac{\dot{v}}{g - kv - (2k^2/g)v^2} &= 1 \\ \Rightarrow && T &= \int_0^{g/(3k)} \frac{1}{g - kv - (2k^2/g)v^2} dv \\ && &= \int_0^{g/(3k)} \frac{g}{(g-2kv)(kv+g)} dv\\ && &= \int_0^{g/(3k)} \l \frac{2}{3(g-2kv)} + \frac{1}{3(kv+g)} \r dv\\ && &= \left [ \l -\frac{1}{3k} \ln (g-2kv) + \frac{1}{3k}\ln(kv+g) \r \right ]_0^{g/(3k)} \\ && &= \left [ \l -\frac{1}{3k}\ln \l \frac{g}{3} \r + \frac{1}{3k}\ln \l \frac{4g}{3} \r \r \right ] - \left [- \frac1{3k} \ln(g) + \frac{1}{3k} \ln (g) \right ] \\ && &= \frac{1}{3k} \ln \l 4 \r \end{align*} NB: \(\sqrt[3]{4} \approx 1.58 > \frac{3}{2}\) so Karen has been in free-fall for longer, but not \emph{much} longer than Jane.
A long light inextensible string passes over a fixed smooth light pulley. A particle of mass 4~kg is attached to one end \(A\) of this string and the other end is attached to a second smooth light pulley. A long light inextensible string \(BC\) passes over the second pulley and has a particle of mass 2 kg attached at \(B\) and a particle of mass of 1 kg attached at \(C\). The system is held in equilibrium in a vertical plane. The string \(BC\) is then released from rest. Find the accelerations of the two moving particles. After \(T\) seconds, the end \(A\) is released so that all three particles are now moving in a vertical plane. Find the accelerations of \(A\), \(B\) and \(C\) in this second phase of the motion. Find also, in terms of \(g\) and \(T\), the speed of \(A\) when \(B\) has moved through a total distance of \(0.6gT^{2}\)~metres.