Show that \[\cos 3\theta = 4\cos^3\theta - 3\cos\theta\] (thus expressing \(\cos 3\theta\) as a cubic in \(\cos\theta\)). Show that if we can express \(\cos m\theta\) and \(\sin\theta\sin(m-1)\theta\) as polynomials of degree at most \(m\) in \(\cos\theta\) for all \(m\) with \(1 \leq m \leq n\), then we can express \(\cos(n+1)\theta\) and \(\sin\theta\sin n\theta\) as polynomials of degree at most \(n+1\) in \(\cos\theta\). Deduce that \[\cos n\theta = \sum_{r=0}^{n} a_{nr}(\cos\theta)^r\] for suitable real numbers \(a_{n0}, a_{n1}, \ldots, a_{nn}\). If we write \[T_n(x) = \sum_{r=0}^{n} a_{nr}x^r,\] show, using the fact that \(T_n(x) = \cos(n\cos^{-1}x)\) for \(|x| \leq 1\), or otherwise, that \begin{align} \text{(i)} \quad & |T_n(x)| \leq 1 \text{ for } |x| \leq 1,\\ \text{yet (ii)} \quad & |T_n'(1)| = n^2. \end{align} [Hint for (ii): If \(f\) is continuous then, automatically, \(f(1) = \lim_{x \to 1}f(x)\).]
Find a solution to the differential equation \[\frac{dy}{dt} = 2(2y^{\frac{1}{2}} - y^2)^{\frac{1}{2}}\] for which \(y = 1\) when \(t = 0\).
Prove that the curves \(y = \frac{3x}{2}\) and \(y = \sin^{-1}x\) intersect precisely once in the range \(0 < x \leq 1\); \(\sin^{-1}x\) is to be interpreted as the value of \(\theta\) between 0 and \(\frac{1}{2}\pi\) for which \(\sin\theta = x\). Sketch, on the same axes, these two functions for this range of \(x\). Use this sketch to illustrate graphically the sequence of numbers \(q_n\) governed by \[q_{n+1} = \sin\left(\frac{3q_n}{2}\right), \quad q_0 = \frac{1}{2},\] and deduce from the picture that the sequence converges as \(n \to \infty\) to a number less than 1.
Evaluate the indefinite integral \[\int \frac{d\theta}{a + \cos\theta},\] where \(a > 1\), using the substitution \(t = \tan\frac{1}{2}\theta\) or otherwise. What is the value of \[\int_0^{2\pi}\frac{d\theta}{a+\cos\theta}?\] What happens to the latter integral as \(a \to 1\) from above?
Let \((a,b)\) be a fixed point, and \((x,y)\) a variable point, on the curve \(y = f(x)\) (where \(z > a\), \(f'(x) \geq 0\)). The curve divides the rectangle with vertices \((a,b)\), \((a,y)\), \((x,y)\) and \((x,b)\) into two portions, the lower of which has always half the area of the upper. Show that the curve is a parabola with its vertex at \((a,b)\).
By using diagrams or otherwise, explain why \[\sum_{r=n}^{\infty} r^{-2} > \int_{n}^{\infty} x^{-2}dx > \sum_{r=n+1}^{\infty} r^{-2}.\] If we write \(A = \sum_{r=1}^{\infty} r^{-2}\), show that \[n^{-1} > A - \sum_{r=1}^{n} r^{-2} > (n+1)^{-1}.\] How large must we take \(n\) to ensure that \(\sum_{r=1}^{n} r^{-2}\) approximates \(A\) with an error of less than \(10^{-4}\)? Show that, for the same \(n\), \[(n+1)^{-1} + \sum_{r=1}^{n} r^{-2}\] approximates \(A\) with an error of less than about \(10^{-8}\).
The barrel of a gun may be considered as a tube of length \(L\), closed at one end, and of uniform circular cross section of area \(A\). The rear surface of the bullet is at a distance \(x\) from the closed end, and \(x = x_0\) when the gun is fired. The pressure in the gun is \(P\), and \(P = P_0\) immediately after firing. Subsequently \(P\) obeys the equation \[PV^{\gamma} = \text{constant},\] where \(V = Ax\) is the volume of propellant gas, and \(\gamma\) is a constant \(\geq 1\). The equation of motion of the bullet is \[m\frac{d^2x}{dt^2} = AP.\] Find the velocity of the bullet when it leaves the barrel, for all values of \(\gamma \geq 1\).
Evaluate \[\int_0^1 \frac{u^{\frac{1}{2}}}{(1+u)^{\frac{1}{2}}}\,du.\]
Prove that the rectangle of greatest perimeter which can be inscribed in a given circle is a square. This result changes if, instead of maximizing the sum of lengths of sides of the rectangle, we seek to maximize the sum of \(n\)th powers of the lengths of those sides, for an integer \(n > 1\). What happens? Justify your answer.
By considering the derivative of \(x - \sin x\) show that \(x \geq \sin x\) for all \(x \geq 0\). By considering the repeated derivatives of \(\sin x - x + x^3/3!\) show that \(\sin x \geq x - x^3/3!\) for all \(x \geq 0\). More generally, show that \[\sum_{r=0}^{2m} (-1)^r \frac{x^{2r+1}}{(2r+1)!} \geq \sin x \geq \sum_{r=0}^{2m-1} (-1)^r \frac{x^{2r+1}}{(2r+1)!}\] for all \(x \geq 0\) and \(m \geq 1\). Deduce that \[\left|\sum_{r=0}^{n-1} (-1)^r \frac{x^{2r+1}}{(2r+1)!} - \sin x\right| \leq \frac{x^{2n+1}}{(2n+1)!}\] for all \(x \geq 0\). [The power series expansion of \(\sin x\) must not be used.]