1531 problems found
If \[ \mathrm{I}_{n}=\int_{0}^{a}x^{n+\frac{1}{2}}(a-x)^{\frac{1}{2}}\,\mathrm{d}x, \] show that \(\mathrm{I}_{0}=\pi a^{2}/8.\) Show that \((2n+4)\mathrm{I}_{n}=(2n+1)a\mathrm{I}_{n-1}\) and hence evaluate \(\mathrm{I}_{n}\).
Solution: \begin{align*} && I_n &= \int_{0}^{a}x^{n+\frac{1}{2}}(a-x)^{\frac{1}{2}}\,\mathrm{d}x\\ && I_0 &= \int_0^a x^{\frac12}(a-x)^{\frac12} \d x \\ x = a \sin^2 \theta, \d x = 2a \sin \theta \cos \theta \d \theta &&&= \int_{\theta =0}^{\theta = \pi/2} \sqrt{a}\sin \theta\sqrt{a} \cos \theta 2a \sin \theta \cos \theta \d \theta \\ &&&= \frac{a^2}{2} \int_0^{\pi/2} \sin^2 2 \theta \d \theta \\ &&&= \frac{a^2}{4} \int_0^{\pi/2}(1- \underbrace{\cos 4\theta}_{\text{runs round the whole unit circle}}) \d \theta \\ &&&= \frac{\pi a^2}{8} \\ \\ && I_n &= \int_0^a x^{n+\frac12}(a-x)^{\frac12} \d x \\ &&&=\underbrace{\left [-\frac23x^{n+\frac12}(a-x)^\frac32 \right]_0^a}_{=0} + \frac23 \left(n+\frac12\right) \int_0^ax^{n-1+\frac12}(a-x)^\frac32 \d x \\ &&&= \frac23 \left(n+\frac12\right) \int_0^ax^{n-1+\frac12}(a-x)(a-x)^\frac12 \d x \\ &&&= \frac23 \left(n+\frac12\right)aI_{n-1}-\frac23 \left(n+\frac12\right)I_{n} \\ \Rightarrow && \left(n+\frac12+\frac32\right)I_{n} &= \left(n+\frac12\right)aI_{n-1}\\ \Rightarrow && (2n+4)I_n &= (2n+1)aI_{n-1} \\ \\ \Rightarrow && I_n &= \frac{2n+1}{2n+4}a I_{n-1} \\ &&&=\frac{2n+1}{2n+4}\frac{2n-1}{2n+2}a^2 I_{n-2} \\ &&&= \frac{(2n+1)!!}{(2n+4)!!} \pi a^{n+2} \end{align*}
What is the general solution of the differential equation \[ \frac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}}+2k\frac{\mathrm{d}x}{\mathrm{d}t}+x=0 \] for each of the cases: (i) \(k>1;\) (ii) \(k=1\); (iii) \(0 < x < 1\)? In case (iii) the equation represents damped simple harmonic motion with damping factor \(k\). Let \(x(0)=0\) and let \(x_{1},x_{2},\ldots,x_{n},\ldots\) be the sequence of successive maxima and minima, so that if \(x_{n}\) is a maximum then \(x_{n+1}\) is the next minimum. Show that \(\left|x_{n+1}/x_{n}\right|\) takes a value \(\alpha\) which is independent of \(n\), and that \[ k^{2}=\frac{(\ln\alpha)^{2}}{\pi^{2}+(\ln\alpha)^{2}}. \]
Solution: The auxiliary equation is \(\lambda^2 + 2k\lambda + 1 = (\lambda + k)^2+1-k^2 = 0\) (i) If \(k > 1\) then the solution is \(A\exp \left ({(-k + \sqrt{k^2-1})t} \right)+B\exp\left((-k-\sqrt{k^2-1})t \right)\). (ii) If \(k = 1\) then the solution is \(x = (A+Bt)e^{-kt}\) (iii) If \(k < 1\) then the solution is \(x = Ae^{-kt} \sin \left ( \sqrt{1-k^2} t \right)+Be^{-kt} \cos \left ( \sqrt{1-k^2} t \right)\) If \(x(0) = 0\) then \begin{align*} && x &= Ae^{-kt} \sin(\sqrt{1-k^2}t)\\ && \dot{x} &= Ae^{-kt} \left (-k \sin(\sqrt{1-k^2}t)+\sqrt{1-k^2} \cos(\sqrt{1-k^2}t) \right) \\ (\dot{x} =0): && \tan (\sqrt{1-k^2}t) &= \frac{\sqrt{1-k^2}}{k}\\ \end{align*} Therefore maxima and minima occur every \(\frac{\pi}{\sqrt{1-k^2}}\), so \begin{align*} && \frac{x_{n+1}}{x_n} &= \exp\left ( -\frac{k\pi}{\sqrt{1-k^2}} \right) \frac{\sin\left (\sqrt{1-k^2}\left(t+\frac{\pi}{\sqrt{1-k^2}}\right)\right)}{\sin(\sqrt{1-k^2}t)} \\ &&&= \exp\left ( -\frac{k\pi}{\sqrt{1-k^2}} \right) \left (-1+0 \right)\\ &&&= -\exp\left ( -\frac{k\pi}{\sqrt{1-k^2}} \right) \\ \Rightarrow && \ln \alpha &= - \frac{k\pi}{\sqrt{1-k^2}} \\ \Rightarrow && (\ln \alpha)^2 &= \frac{k^2\pi^2}{1-k^2} \\ \Rightarrow && (1-k^2)(\ln \alpha)^2 &= k^2 \pi^2 \\ \Rightarrow && k^2(\pi^2+(\ln \alpha)^2) &= (\ln \alpha)^2 \\ \Rightarrow && k^2 &= \frac{(\ln \alpha)^2}{\pi^2 + (\ln \alpha)^2} \end{align*}
Let \[ \mathrm{C}_{n}(\theta)=\sum_{k=0}^{n}\cos k\theta \] and let \[ \mathrm{S}_{n}(\theta)=\sum_{k=0}^{n}\sin k\theta, \] where \(n\) is a positive integer and \(0<\theta<2\pi.\) Show that \[ \mathrm{C}_{n}(\theta)=\frac{\cos(\tfrac{1}{2}n\theta)\sin\left(\frac{1}{2}(n+1)\theta\right)}{\sin(\frac{1}{2}\theta)}, \] and obtain the corresponding expression for \(\mathrm{S}_{n}(\theta)\). Hence, or otherwise, show that for \(0<\theta<2\pi,\) \[ \left|\mathrm{C}_{n}(\theta)-\frac{1}{2}\right|\leqslant\frac{1}{2\sin(\frac{1}{2}\theta)}. \]
Solution: \begin{align*} && C_n(\theta) &= \sum_{k=0}^n \cos k \theta \\ &&&= \textrm{Re} \left ( \sum_{k=0}^n \exp (ik \theta)\right)\\ &&&= \textrm{Re} \left ( \frac{e^{i(n+1)\theta}-1}{e^{i\theta}-1}\right)\\ &&&= \textrm{Re} \left ( \frac{e^{i(n+1)\theta/2}}{e^{i\theta/2}}\frac{e^{i(n+1)\theta/2}-e^{-i(n+1)\theta/2}}{e^{i\theta/2}-e^{-i\theta/2}}\right)\\ &&&= \textrm{Re} \left ( e^{in\theta/2}\frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\right)\\ &&&= \frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\textrm{Re} \left ( e^{in\theta/2}\right)\\ &&&= \frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\cos \left ( \frac12n\theta\right)\\ \\ && S_n(\theta) &= \sum_{k=0}^n \sin k \theta \\ &&&= \textrm{Im} \left ( \sum_{k=0}^n \exp (ik \theta)\right)\\ &&&= \frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\textrm{Im} \left ( e^{in\theta/2}\right)\\ &&&= \frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\sin\left ( \frac12n\theta\right)\\ \\ && C_n(\theta) - \frac12 &= \frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\cos \left ( \frac12n\theta\right) - \frac12 \\ &&&= \frac{2\sin \left ( (n+1)\theta/2 \right)\cos\left ( n\theta/2 \right)-\sin (\theta/2)}{2 \sin (\theta/2)} \\ &&&= \frac{\sin\left ( (n+1)\theta/2+n\theta/2\right)+\sin\left ( (n+1)\theta/2-n\theta/2\right)-\sin (\theta/2)}{2 \sin (\theta/2)} \\ &&&= \frac{\sin\left ( (n+1)\theta/2+n\theta/2\right)+\sin\left ( \theta/2\right)-\sin (\theta/2)}{2 \sin (\theta/2)} \\ &&&= \frac{\sin\left ( (2n+1)\theta/2\right)}{2 \sin (\theta/2)} \leqslant\frac{1}{2 \sin (\theta/2)} \\ \end{align*}
Show that \(y=\sin^{2}(m\sin^{-1}x)\) satisfies the differential equation \[ (1-x^{2})y^{(2)}=xy^{(1)}+2m^{2}(1-2y), \] and deduce that, for all \(n\geqslant1,\) \[ (1-x^{2})y^{(n+2)}=(2n+1)xy^{(n+1)}+(n^{2}-4m^{2})y^{(n)}, \] where \(y^{(n)}\) denotes the \(n\)th derivative of \(y\). Derive the Maclaurin series for \(y\), making it clear what the general term is.
Solution: \begin{align*} && y &= \sin^2 (m \sin^{-1} x) \\ \Rightarrow && y' &= 2 \sin (m \sin^{-1} x) \cdot \cos (m \sin^{-1} x) \cdot m \cdot \frac1{\sqrt{1-x^2}} \\ \Rightarrow && y'' &= 2 \cos^2(m \sin^{-1} x) \cdot m^2 \cdot \frac{1}{1-x^2} + \\ &&&\quad\quad-2\sin^2(m \sin^{-1} x) m^2 \frac{1}{1-x^2} + \\ &&&\quad\quad\quad-\sin(m \sin^{-1} x) \cdot \cos(m \sin^{-1} x) \cdot m \cdot (1-x^2)^{-\frac32} \cdot (-2x) \\ \Rightarrow && (1-x^2)y^{(2)} &= 2m^2-4m^2y+xy' \\ &&&= xy^{(1)} + 2m^2(1-2y) \\ \\ \Rightarrow && (1-x^2)y^{(n+2)}-2nxy^{(n+1)}-2\binom{n}{2}y^{(n)} &= xy^{(n+1)}+ny^{(n)} -4m^2y^{(n)} \\ \Rightarrow && (1-x^2)y^{(n+2)} &= (2n+1)xy^{(n+1)}+(n(n-1)+n-4m^2)y^{(n)} \\ &&&= (2n+1)xy^{(n+1)}+(n^2-4m^2)y^{(n)} \\ \end{align*} \begin{align*} && y(0) &= \sin^2(m \sin^{-1} 0) \\ &&&= \sin^2 0 = 0 \\ \\ && y'(0) &= 0 \\ && (1-0^2)y^{(2)}(0) &= 2m^2(1-2y(0)) \\ \Rightarrow && y^{(2)}(0) &= 2m^2 \\ \\ && y^{(n+2)} (0) &= (2n+1) \cdot 0 \cdot y^{(n+1)} +(n^2-4m^2)y^{(n)}(0) \\ &&&= (n^2-4m^2)y^{(n)}(0) \\ \\ && y^{(2)}(0) &= 2m^2 \\ && y^{(4)}(0) &= (4-4m^2) \cdot 2m^2 \\ &&&= -8m(m+1)m(m-1) \\ && y^{(6)}(0) &= 32m(m+2)(m+1)m(m-1)(m-2) \\ && y^{(2k)}(0) &= (-1)^{k+1}2^{2k-1}m (m+k)\cdots(m-k) \text{ if }k < m \\ \\ && y &= m^2x^2 -2m\binom{m+1}{3} x^4 + \frac{16}{3}m\binom{m+2}{5}x^6 - \cdots \\ &&&+ (-1)^{k}\frac{2^{2k}}{k+1} m \binom{m+k}{2k+1}x^{2k+2}+\cdots \\ &&&= mx^2\sum_{k=0}^{m-1} \frac{(-1)^k2^{2k}}{k+1}\binom{m+k}{2k+1}x^{2k} \end{align*}
The variable non-zero complex number \(z\) is such that \[ \left|z-\mathrm{i}\right|=1. \] Find the modulus of \(z\) when its argument is \(\theta.\) Find also the modulus and argument of \(1/z\) in terms of \(\theta\) and show in an Argand diagram the loci of points which represent \(z\) and \(1/z\). Find the locus \(C\) in the Argand diagram such that \(w\in C\) if, and only if, the real part of \((1/w)\) is \(-1\).
Solution:
Consider the following sets with the usual definition of multiplication appropriate to each. In each case you may assume that the multiplication is associative. In each case state, giving adequate reasons, whether or not the set is a group.
Solution:
| 1 | 3 | 5 | 7 | |
| 1 | 1 | 3 | 5 | 7 |
| 3 | 3 | 1 | 7 | 5 |
| 5 | 5 | 7 | 1 | 3 |
| 7 | 7 | 5 | 3 | 1 |
| 1 | 2 | 3 | 4 | |
| 1 | 1 | 2 | 3 | 4 |
| 2 | 2 | 4 | 1 | 3 |
| 3 | 3 | 1 | 4 | 2 |
| 4 | 4 | 3 | 2 | 1 |
| (i) | (iii) | (iv) | (vi) | |
| (i) | \(\checkmark\) | \(\checkmark\) consider \(z \mapsto \begin{pmatrix} \cos \arg (z) | - \sin \arg(z) | |
| \sin \arg(z) | \cos \arg(z) \end{pmatrix}\) | not finite | not finite | |
| (iii) | \(\checkmark\) | not finite | not finite | |
| (iv) | \(\checkmark\) | no element order \(4\) | ||
| (vi) | \(\checkmark\) |
A plane \(\pi\) in 3-dimensional space is given by the vector equation \(\mathbf{r}\cdot\mathbf{n}=p,\) where \(\mathbf{n}\) is a unit vector and \(p\) is a non-negative real number. If \(\mathbf{x}\) is the position vector of a general point \(X\), find the equation of the normal to \(\pi\) through \(X\) and the perpendicular distance of \(X\) from \(\pi\). The unit circles \(C_{i},\) \(i=1,2,\) with centres \(\mathbf{r}_{i},\) lie in the planes \(\pi_{i}\) given by \(\mathbf{r}\cdot\mathbf{n}_{i}=p_{i},\) where the \(\mathbf{n}_{i}\) are unit vectors, and \(p_{i}\) are non-negative real numbers. Prove that there is a sphere whose surface contains both circles only if there is a real number \(\lambda\) such that \[ \mathbf{r}_{1}+\lambda\mathbf{n}_{1}=\mathbf{r}_{2}\pm\lambda\mathbf{n}_{2}. \] Hence, or otherwise, deduce the necessary conditions that \[ (\mathbf{r}_{1}-\mathbf{r}_{2})\cdot(\mathbf{n}_{1}\times\mathbf{n}_{2})=0 \] and that \[ (p_{1}-\mathbf{n}_{1}\cdot\mathbf{r}_{2})^{2}=(p_{2}-\mathbf{n}_{2}\cdot\mathbf{r}_{1})^{2}. \] Interpret each of these two conditions geometrically.
Solution: The equation of the normal to \(\pi\) through \(X\) is \(\mathbf{x} + \lambda \mathbf{n}\). The distance is \(|\mathbf{x}\cdot \mathbf{n}-p|\) We know that the centre of the sphere must lie above the centre of the circle following the normal, ie \(\mathbf{c} = \mathbf{r}_1+\lambda_1 \mathbf{n}_1 = \mathbf{r}_2+\lambda_2 \mathbf{n}_2\)
A thin circular disc of mass \(m\), radius \(r\) and with its centre of mass at its centre \(C\) can rotate freely in a vertical plane about a fixed horizontal axis through a point \(O\) of its circumference. A particle \(P\), also of mass \(m,\) is attached to the circumference of the disc so that the angle \(OCP\) is \(2\alpha,\) where \(\alpha\leqslant\pi/2\).
A cannon is situated at the bottom of a plane inclined at angle \(\beta\) to the horizontal. A (small) cannon ball is fired from the cannon at an initial speed \(u.\) Ignoring air resistance, find the angle of firing which will maximise the distance up the plane travelled by the cannon ball and show that in this case the ball will land at a distance \[ \frac{u^{2}}{g(1+\sin\beta)} \] from the cannon.
A ship is sailing due west at \(V\) knots while a plane, with an airspeed of \(kV\) knots, where \(k>\sqrt{2},\) patrols so that it is always to the north west of the ship. If the wind in the area is blowing from north to south at \(V\) knots and the pilot is instructed to return to the ship every thirty minutes, how long will her outward flight last? Assume that the maximum distance of the plane from the ship during the above patrol was \(d_{w}\) miles. If the air now becomes dead calm, and the pilot's orders are maintained, show that the ratio \(d_{w}/d_{c}\) of \(d_{w}\) to the new maximum distance, \(d_{c}\) miles, of the plane from the ship is \[ \frac{k^{2}-2}{2k(k^{2}-1)}\sqrt{4k^{2}-2}. \]
The random variables \(X\) and \(Y\) are independently normally distributed with means 0 and variances 1. Show that the joint probability density function for \((X,Y)\) is \[ \mathrm{f}(x,y)=\frac{1}{2\pi}\mathrm{e}^{-\frac{1}{2}(x^{2}+y^{2})}\qquad-\infty < x < \infty,-\infty < y < \infty. \] If \((x,y)\) are the coordinates, referred to rectangular axes, of a point in the plane, explain what is meant by saying that this density is radially symmetrical. The random variables \(U\) and \(V\) have a joint probability density function which is radially symmetrical (in the above sense). By considering the straight line with equation \(U=kV,\) or otherwise, show that \[ \mathrm{P}\left(\frac{U}{V} < k\right)=2\mathrm{P}(U < kV,V > 0). \] Hence, or otherwise, show that the probability density function of \(U/V\) is \[ \mathrm{g}(k)=\frac{1}{\pi(1+k^{2})}\qquad-\infty < k < \infty. \]
A message of \(10^{k}\) binary digits is sent along a fibre optic cable with high probabilities \(p_{0}\) and \(p_{1}\) that the digits 0 and 1, respectively, are received correctly. If the probability of a digit in the original message being a 1 is \(\alpha,\) find the probability that the entire message is received correctly. Find the probability \(\beta\) that a randomly chosen digit in the message is received as a 1 and show that \(\beta=\alpha\) if, and only if \[ \alpha=\frac{q_{0}}{q_{1}+q_{0}}, \] where \(q_{0}=1-p_{0}\) and \(q_{1}=1-p_{1}.\) If this condition is satisfied and the received message consists entirely of zeros, what is the probability that it is correct? If now \(q_{0}=q_{1}=q\) and \(\alpha=\frac{1}{2},\) find the approximate value of \(q\) which will ensure that a message of one million binary digits has a fifty-fifty chance of being received entirely correctly. The probability of error \(q\) is proportional to the square of the length of the cable. Initially the length is such that the probability of a message of one million binary bits, among which 0 and 1 are equally likely, being received correctly is \(\frac{1}{2}.\) What would this probability become if a booster station were installed at its mid-point, assuming that the booster station re-transmits the received version of the message, and assuming that terms of order \(q^{2}\) may be ignored?
A candidate finishes examination questions in time \(T\), where \(T\) has probability density function \[ \mathrm{f}(t)=t\mathrm{e}^{-t}\qquad t\geqslant0, \] the probabilities for the various questions being independent. Find the moment generating function of \(T\) and hence find the moment generating function for the total time \(U\) taken to finish two such questions. Show that the probability density function for \(U\) is \[ \mathrm{g}(u)=\frac{1}{6}u^{3}\mathrm{e}^{-u}\qquad u\geqslant0. \] Find the probability density function for the total time taken to answer \(n\) such questions.
Solution: \begin{align*} && M_T(x) &= \mathbb{E}[e^{xT}] \\ &&&= \int_0^{\infty} e^{xt}te^{-t} \d t \\ &&&= \int_0^{\infty}te^{(x-1)t} \d t \\ &&&= \left [ \frac{t}{x-1} e^{(x-1)t} \right]_0^{\infty} - \int_0^\infty \frac{e^{(x-1)t}}{x-1} \d t \\ &&&= \left [ \frac{e^{(x-1)t}}{(x-1)^2} \right]_0^{\infty} \\ &&&= \frac{1}{(x-1)^2} \\ \\ && M_U(x) &= M_{T_1+T_2}(x) \\ &&&= \frac1{(x-1)^4} \\ \\ && I_n &= \int_0^{\infty} t^ne^{(x-1)t} \d t \\ &&&= \left[ \frac{1}{(x-1)}t^ne^{(x-1)t} \right]_0^{\infty} - \frac{n}{(x-1)} \int_0^{\infty}t^{n-1}e^{(x-1)t} \d t \\ &&&= -\frac{n}{(x-1)}I_{n-1} \\ \Rightarrow && I_n &= \frac{n!}{(1-x)^{n+1}} \\ \\ \Rightarrow && \int_0^{\infty} e^{xt} \frac16u^3e^{-u} \d u &= \int_0^{\infty} \frac16u^3e^{(x-1)u} \d u \\ &&&= \frac{1}{(1-x)^4} \\ \Rightarrow && f_U(u) &= \frac16u^3e^{-u} \\ \\ && M_{X_1+\cdots+X_n}(x) &= \frac{1}{(x-1)^{2n}} \\ \Rightarrow && f_{X_1+\cdots+X_n}(t) &= \frac1{(2n-1)!} t^{2n-1}e^{-t} \end{align*} (NB: This is the gamma distribution)
My house has an attic consisting of a horizontal rectangular base of length \(2q\) and breadth \(2p\) (where \(p < q\)) and four plane roof sections each at angle \(\theta\) to the horizontal. Show that the length of the roof ridge is independent of \(\theta\) and find the volume of the attic and the surface area of the roof.
Solution:
Given that \(a\) is constant, differentiate the following expressions with respect to \(x\):
Solution: \begin{align*} && y &= x^a \\ && \frac{\d y}{\d x} &= \begin{cases} ax^{a-1} & a \neq 0 \\ 0 & a = 0 \end{cases} \\ \\ && y &= a^x \\ &&&= e^{(\ln a) \cdot x} \\ && \frac{\d y}{\d x} &= \ln a e^{(\ln a) x} \\ &&&= \ln a \cdot a^ x \\ \\ && y &= x^x \\ &&&= e^{x \ln x}\\ && \frac{\d y}{\d x} &= e^{x \ln x} \cdot \left ( \ln x + x \cdot \frac1x \right) \\ &&&= x^x \left (1 + \ln x \right) \\ \\ && y&= x^{(x^x)} \\ &&&= e^{x^ x \cdot \ln x} \\ && \frac{\d y}{\d x} &= e^{x^x \cdot \ln x} \left ( x^x \left (1 + \ln x \right) \cdot \ln x + x^x \cdot \frac1x\right) \\ &&&= x^{x^x} \left (x^x (1+ \ln x) \ln x +x^{x-1} \right) \\ &&&= x^{x^x+x-1} \left (1 + x \ln x + x (\ln x)^2 \right) \\ \\ && y &= (x^x)^x \\ &&&= x^{2x} \\ &&&= e^{2x \ln x} \\ && \frac{\d y}{\d x} &= e^{2 x \ln x} \left (2 \ln x + 2 \right) \\ &&&= 2(x^x)^x(1 + \ln x) \end{align*}