A rod \(AB\) of length 0.81 m and mass 5 kg is in equilibrium with the end \(A\) on a rough floor and the end \(B\) against a very rough vertical wall. The rod is in a vertical plane perpendicular to the wall and is inclined at \(45^{\circ}\) to the horizontal. The centre of gravity of the rod is at \(G\), where \(AG = 0.21\) m. The coefficient of friction between the rod and the floor is 0.2, and the coefficient of friction between the rod and the wall is 1.0. Show that the friction cannot be limiting at both \(A\) and \(B\). A mass of 5 kg is attached to the rod at the point \(P\) such that now the friction is limiting at both \(A\) and \(B\). Determine the length of \(AP\).
I have \(k\) different keys on my key ring. When I come home at night I try one key after another until I find the key that fits my front door. What is the probability that I find the correct key in exactly \(n\) attempts in each of the following three cases?
Every person carries two genes which can each be either of type \(A\) or of type \(B\). It is known that \(81\%\) of the population are \(AA\) (i.e. both genes are of type \(A\)), \(18\%\) are \(AB\) (i.e. there is one gene of type \(A\) and one of type \(B\)) and \(1\%\) are \(BB\). A child inherits one gene from each of its parents. If one parent is \(AA\), the child inherits a gene of type \(A\) from that parent; if the parent is \(BB\), the child inherits a gene of type \(B\) from that parent; if the parent is \(AB\), the inherited gene is equally likely to be \(A\) or \(B\).
The random variable \(X\) is uniformly distributed on the interval \([-1,1]\). Find \(\E(X^2)\) and \(\var (X^2)\). A second random variable \(Y\), independent of \(X\), is also uniformly distributed on \([-1,1]\), and \(Z=Y-X\). Find \(\E(Z^2)\) and show that \(\var (Z^2) = 7 \var (X^2)\).
Solution: \(X \sim U(-1,1)\) \begin{align*} \E[X^2] &= \int_{-1}^1 \frac12 x^2 \, dx \\ &= \frac{1}{6} \left [ x^3 \right]_{-1}^1 \\ &= \frac{1}{3} \end{align*} \begin{align*} \E[X^4] &= \int_{-1}^1 \frac12 x^4 \, dx \\ &= \frac{1}{10} \left [ x^5 \right]_{-1}^1 \\ &= \frac{1}{5} \end{align*} \begin{align*} \var[X^2] &=\E[X^4] - \E[X^2]^2 \\ &= \frac{1}{5} - \frac{1}{9} \\ &= \frac{4}{45} \end{align*} \begin{align*} \E(Z^2) &= \E(Y^2 - 2XY+Z^2) \\ &= \E(Y^2) - 2\E(X)\E(Y)+\E(Z^2) \\ &= \frac{1}{3} - 0 + \frac{1}{3} \\ &= \frac{2}{3} \end{align*} \begin{align*} \E[Z^4] &= \E[Y^4 -4Y^3X+6Y^2X^2-4YX^3+X^4] \\ &= \E[Y^4]-4\E[Y^3]\E[X]+6\E[Y^2]\E[X^2]-4\E[Y]\E[X^3]+\E[X^4] \\ &= \frac{1}{5}+6 \frac{1}{3} \frac13 + \frac{1}{5} \\ &= \frac{2}{5} + \frac{2}{3} \\ &= \frac{16}{15} \end{align*} \begin{align*} \var(Z^2) &= \E(Z^4) - \E(Z^2) \\ &= \frac{16}{15} - \frac{4}{9} \\ &= \frac{28}{45} \\ &= 7 \var(X^2) \end{align*}
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 \[ \int_0^{\frac{1}{2}\pi} {1 \over {1 + \sin\alpha \cos \theta}} \,\d\theta \,. \] %$$