Problems

Give a rough sketch of the function \(\tan^{k}\theta\) for \(0\leqslant\theta\leqslant\frac{1}{4}\pi\) in the two cases \(k=1\) and \(k\gg1\) (i.e. \(k\) is much greater than 1). Show that for any positive integer \(n\) \[ \int_{0}^{\frac{1}{4}\pi}\tan^{2n+1}\theta\,\mathrm{d}\theta=(-1)^{n}\left(\tfrac{1}{2}\ln2+\sum_{m=1}^{n}\frac{(-1)^{m}}{2m}\right), \] and deduce that \[ \sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{2m}=\tfrac{1}{2}\ln2. \] Show similarly that \[ \sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{2m-1}=\frac{\pi}{4}. \]

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Solution:

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Let \(\displaystyle I_n = \int_0^{\pi/4} \tan^{n} \theta \, \d \theta\), then \begin{align*} I_0 &= \int_0^{\pi/4} \tan \theta \d \theta \\ &= \left [ -\ln \cos \theta \right]_0^{\pi/4} \\ &= -\ln \frac{1}{\sqrt{2}} - 0 \\ &= \frac12 \ln 2 \\ \\ \\ I_{2n+1} &= \int_0^{\pi/4} \tan^{2n+1} \theta \, \d \theta \\&= \int_0^{\pi/4} \tan^{2n-1} \theta \tan ^2 \theta \, \d \theta \\ &= \int_0^{\pi/4} \tan^{2n-1} \theta (\sec^2 \theta - 1) \, \d \theta \\ &= \int_0^{\pi/4} \tan^{2n-1} \theta \sec^2 \theta - \tan^{2n-1} \theta \, \d \theta \\ &= \left[ \frac{1}{2n} \tan^{2n} \theta \right]_0^{\pi/4} - I_{2n-1} \\ &= \frac1{2n} - I_{2n-1} \end{align*} Therefore we can see that \(\displaystyle I_{2n+1} = (-1)^{n}\left(\tfrac{1}{2}\ln2+\sum_{m=1}^{n}\frac{(-1)^{m}}{2m}\right)\). As we can see as \(n \to \infty\), \(I_n \to 0\) Therefore \begin{align*} && 0 &= \tfrac{1}{2}\ln2+\lim_{n \to \infty} \sum_{m=1}^{n}\frac{(-1)^{m}}{2m} \\ \Rightarrow && \tfrac{1}{2}\ln2 &= \sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{2m} \end{align*} \begin{align*} && I_{-1} &= \int_0^{\pi/4} 1 \d \theta \\ &&&= \frac{\pi}{4} \end{align*} Therefore \(\displaystyle I_{2n} = (-1)^n \left ( \frac{\pi}{4} + \sum_{m=1}^n \frac{(-1)^m}{2m-1} \right)\) and since \(I_{2m} \to 0\) the same result follows.

Show by means of a sketch, or otherwise, that if \(0\leqslant\mathrm{f}(y)\leqslant\mathrm{g}(y)\) for \(0\leqslant y\leqslant x\) then \[ 0\leqslant\int_{0}^{x}\mathrm{f}(y)\,\mathrm{d}y\leqslant\int_{0}^{x}\mathrm{g}(y)\,\mathrm{d}y. \] Starting from the inequality \(0\leqslant\cos y\leqslant1,\) or otherwise, prove that if \(0\leqslant x\leqslant\frac{1}{2}\pi\) then \(0\leqslant\sin x\leqslant x\) and \(\cos x\geqslant1-\frac{1}{2}x^{2}.\) Deduce that \[ \frac{1}{1800}\leqslant\int_{0}^{\frac{1}{10}}\frac{x}{(2+\cos x)^{2}}\,\mathrm{d}x\leqslant\frac{1}{1797}. \] Show further that if \(0\leqslant x\leqslant\frac{1}{2}\pi\) then \(\sin x\geqslant x-\frac{1}{6}x^{3}.\) Hence prove that \[ \frac{1}{3000}\leqslant\int_{0}^{\frac{1}{10}}\frac{x^{2}}{(1-x+\sin x)^{2}}\,\mathrm{d}x\leqslant\frac{2}{5999}. \]

The function \(\mathrm{g}\) satisfies, for all positive \(x\) and \(y\), \[ \mathrm{g}(x)+\mathrm{g}(y)=\mathrm{g}(z),\tag{\ensuremath{*}} \] where \(z=xy/(x+y+1).\) By treating \(y\) as a constant, show that \[ \mathrm{g}'(x)=\frac{y^{2}+y}{(x+y+1)^{2}}\mathrm{g}'(z)=\frac{z(z+1)}{x(x+1)}\mathrm{g}'(z), \] and deduce that \(2\mathrm{g}'(1)=(u^{2}+u)\mathrm{g}'(u)\) for all \(u\) satisfying \(0 < u < 1.\) Now by treating \(u\) as a variable, show that \[ \mathrm{g}(u)=A\ln\left(\frac{u}{u+1}\right)+B, \] where \(A\) and \(B\) are constants. Verify that \(\mathrm{g}\) satisfies \((*)\) for a suitable value of \(B\). Can \(A\) be determined from \((*)\)? The function \(\mathrm{f}\) satisfies, for all positive \(x\) and \(y\), \[ \mathrm{f}(x)+\mathrm{f}(y)=\mathrm{f}(z) \] where \(z=xy.\) Show that \(\mathrm{f}(x)=C\ln x\) where \(C\) is a constant.

Solve the quadratic equation \(u^{2}+2u\sinh x-1=0\), giving \(u\) in terms of \(x\). Find the solution of the differential equation \[ \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{2}+2\frac{\mathrm{d}y}{\mathrm{d}x}\sinh x-1=0 \] which satisfies \(y=0\) and \(y'>0\) at \(x=0\). Find the solution of the differential equation \[ \sinh x\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{2}+2\frac{\mathrm{d}y}{\mathrm{d}x}-\sinh x=0 \] which satisfies \(y=0\) at \(x=0\).

Let \(G\) be the set of all matrices of the form \[ \begin{pmatrix}a & b\\ 0 & c \end{pmatrix}, \] where \(a,b\) and \(c\) are integers modulo 5, and \(a\neq0\neq c\). Show that \(G\) forms a group under matrix multiplication (which may be assumed to be associative). What is the order of \(G\)? Determine whether or not \(G\) is commutative. Determine whether or not the set consisting of all elements in \(G\) of order \(1\) or \(2\) is a subgroup of \(G\).

A straight stick of length \(h\) stands vertically. On a sunny day, the stick casts a shadow on flat horizontal ground. In cartesian axes based on the centre of the Earth, the position of the Sun may be taken to be \(R(\cos\theta,\sin\theta,0)\) where \(\theta\) varies but \(R\) is constant. The positions of the base and tip of the stick are \(a(0,\cos\phi,\sin\phi)\) and \(b(0,\cos\phi,\sin\phi)\), respectively, where \(b-a=h\). Show that the displacement vector from the base of the stick to the tip of the shadow is \[ Rh(R\cos\phi\sin\theta-b)^{-1}\begin{pmatrix}-\cos\theta\\ -\sin^{2}\phi\sin\theta\\ \cos\phi\sin\phi\sin\theta \end{pmatrix}. \] {[}`Stands vertically' means that the centre of the Earth, the base of the stick and the tip of the stick are collinear, `horizontal' means perpendicular to the stick.

The Ruritanian army is supplied with shells which may explode at any time in flight but not before the shell reaches its maximum height. The effect of the explosion on any observer depends only on the distance between the exploding shell and the observer (and decreases with distance). Ruritanian guns fire the shells with fixed muzzle speed, and it is the policy of the gunners to fire the shell at an angle of elevation which minimises the possible damages to themselves (assuming the ground is level) - i.e. they aim so that the point on the descending trajectory that is nearest to them is as far away as possible. With that intention, they choose the angle of elevation that minimises the damage to themselves if the shell explodes at its maximum height. What angle do they choose? Does the shell then get any nearer to the gunners during its descent?

A particle is attached to one end \(B\) of a light elastic string of unstretched length \(a\). Initially the other end \(A\) is at rest and the particle hangs at rest at a distance \(a+c\) vertically below \(A\). At time \(t=0\), the end \(A\) is forced to oscillate vertically, its downwards displacement at time \(t\) being \(b\sin pt\). Let \(x(t)\) be the downwards displacement of the particle at time \(t\) from its initial equilibrium position. Show that, while the string remains taut, \(x(t)\) satisfies \[ \frac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}}=-n^{2}(x-b\sin pt), \] where \(n^{2}=g/c\), and that if \(0 < p < n\), \(x(t)\) is given by \[ x(t)=\frac{bn}{n^{2}-p^{2}}(n\sin pt-p\sin nt). \] Write down a necessary and sufficient condition that the string remains taut throughout the subsequent motion, and show that it is satisfied if \(pb < (n-p)c.\)

A non-uniform rod \(AB\) of mass \(m\) is pivoted at one end \(A\) so that it can swing freely in a vertical plane. Its centre of mass is a distance \(d\) from \(A\) and its moment of inertia about any axis perpendicular to the rod through \(A\) is \(mk^{2}.\) A small ring of mass \(\alpha m\) is free to slide along the rod and the coefficient of friction between the ring and rod is \(\mu.\) The rod is initially held in a horizontal position with the ring a distance \(x\) from \(A\). If \(k^{2} > xd\), show that when the rod is released, the ring will start to slide when the rod makes an angle \(\theta\) with the downward vertical, where \[ \mu\tan\theta=\frac{3\alpha x^{2}+k^{2}+2xd}{k^{2}-xd}. \] Explain what will happen if (i) \(k^{2}=xd\) and (ii) \(k^{2} < xd\).

The current in a straight river of constant width \(h\) flows at uniform speed \(\alpha v\) parallel to the river banks, where \(0<\alpha<1\). A boat has to cross from a point \(A\) on one bank to a point \(B\) on the other bank directly opposite to \(A\). The boat moves at constant speed \(v\) relative to the water. When the position of the boat is \((x,y)\), where \(x\) is the perpendicular distance from the opposite bank and \(y\) is the distance downstream from \(AB\), the boat is pointing in a direction which makes an angle \(\theta\) with \(AB\). Determine the velocity vector of the boat in terms of \(v,\theta\) and \(\alpha.\) The pilot of the boat steers in such a way that the boat always points exactly towards \(B\). Show that the velocity vector of the boat is \[ \begin{pmatrix}\dfrac{\mathrm{d}x}{\mathrm{d}t}\\ \tan\theta\dfrac{\mathrm{d}x}{\mathrm{d}t}+x\sec^{2}\theta\dfrac{\mathrm{d}\theta}{\mathrm{d}t} \end{pmatrix}. \] By comparing this with your previous expression deduce that \[ \alpha\frac{\mathrm{d}x}{\mathrm{d}\theta}=-x\sec\theta \] and hence show that \[ (x/h)^{\alpha}=(\sec\theta+\tan\theta)^{-1}. \] Let \(s(t)\) be a new variable defined by \(\tan\theta=\sinh(\alpha s).\) Show that \(x=h\mathrm{e}^{-s},\) and that \[ h\mathrm{e}^{-s}\cosh(\alpha s)\frac{\mathrm{d}s}{\mathrm{d}t}=v. \] Hence show that the time of crossing is \(hv^{-1}(1-\alpha^{2})^{-1}.\)