Show that the arithmetic mean \(A=(a_1+\dots+a_n)/n\) of \(n\) positive numbers \(a_1, \dots, a_n\) is never less than the geometric mean \(G=(a_1 a_2 \dots a_n)^{1/n}\). If, further, \(a_i \ge 1\) for \(1 \le i \le n\) show that \[ G \ge (nA-n+1)^{1/n}. \]
Let \(z_1, \dots, z_n\) be the zeros of \[ f(z) = z^n+c_1z^{n-1}+\dots+c_{n-1}z+c_n \] and let \(f'(z)\) be the derivative. Show that \[ (-1)^{\frac{1}{2}n(n-1)} \prod_{1\le i < j\le n} (z_i-z_j)^2 = \prod_{i=1}^n f'(z_i). \] Hence, or otherwise, show that if \(y_1, \dots, y_n\) are the zeros of \[ 1+\frac{x}{1!} + \frac{x^2}{2!} + \dots + \frac{x^n}{n!} \] then \[ \prod_{1\le i < j\le n} (y_i-y_j)^2 = (-1)^{\frac{1}{2}n(n-1)}(n!)^n. \]
Show that \[ (\cos\theta+i\sin\theta)(\cos\phi+i\sin\phi), \] where \(i^2=-1\), depends only on the sum \(\theta+\phi\). Hence, or otherwise, show that \[ (2\cos\theta)^n - \binom{n}{1}(2\cos\theta)^{n-1}\cos\theta + \dots + (-1)^r \binom{n}{r}(2\cos\theta)^{n-r}\cos r\theta + \dots + (-1)^n \cos n\theta = \cos n\theta, \] where \(\binom{n}{r}\) denotes the binomial coefficient \(\dfrac{n!}{r!(n-r)!}\).
Evaluate the determinant \[ \begin{vmatrix} \frac{1}{x_1+y_1} & \frac{1}{x_2+y_1} & \frac{1}{x_3+y_1} \\ \frac{1}{x_1+y_2} & \frac{1}{x_2+y_2} & \frac{1}{x_3+y_2} \\ \frac{1}{x_1+y_3} & \frac{1}{x_2+y_3} & \frac{1}{x_3+y_3} \end{vmatrix} \] where \(x_1, x_2, x_3, y_1, y_2, y_3\) are any numbers such that \(x_i+y_j\) is non-zero for all \(i,j\). Hence give the numerical value when \(x_1=y_1=1\), \(x_2=y_2=2\), \(x_3=y_3=3\).
Let \(u_0, u_1, \alpha, \beta\) be any real numbers and let \(u_2, u_3, u_4, \dots\) be given by the relation \[ u_n+\alpha u_{n-1} + \beta u_{n-2} = 0 \quad (n \ge 2). \] Put \(U_n = u_0+u_1+\dots+u_n\). Given that \(\alpha+\beta \neq -1\) show that there is a number \(\theta\) depending only on \(u_0, u_1, \alpha, \beta\) such that \(v_n = U_n-\theta\) satisfies the same relation \[ v_n+\alpha v_{n-1} + \beta v_{n-2} = 0 \quad (n \ge 2). \] Find \(U_n\) explicitly in terms of \(u_0\) and \(u_1\) when \(\alpha=-3, \beta=2\).
A square of side \(2x\) is drawn with its centre coincident with the centre of a circle of radius \(y\). The region of the plane comprising all points lying inside either the square or the circle but not inside both is denoted by \(R\). Obtain an expression for the area of \(R\) in terms of \(x\) and \(y\), distinguishing the cases (i) \(y \le x\), (ii) \(x \le y \le \sqrt{2}x\), (iii) \(y \ge \sqrt{2}x\). Show that, if the square is fixed, the concentric circle for which the area of \(R\) is minimum is that with exactly half its perimeter inside the square; and that, if the circle is fixed, the concentric square for which the area of \(R\) is minimum is that with exactly half its perimeter inside the circle.
(i) Evaluate \[ \int_0^{3\pi/2} \frac{dx}{2+\cos x}. \] (ii) If \[ I_n(X) = \int_0^X \frac{x^{2n}}{(1+x^2)^{n+1}} dx, \] where \(n > -\frac{1}{2}\), \(X>0\), show that \[ I_n(X) = \frac{2n-1}{2n} I_{n-1}(X) - \frac{X^{2n-1}}{2n(1+X^2)^n}, \] and hence evaluate \[ \int_0^\infty \frac{x^8}{(1+x^2)^5} dx. \]
The function \(f(x)\) is zero at the point \(\xi_0\) but is non-zero at \(\xi\). Show that \(\xi_0-\xi = -\{f(\xi_0)-f(\xi)\}/f'(\eta)\), where \(\eta\) is some point lying between \(\xi_0\) and \(\xi\). Deduce that, if \(f(x)\) has opposite signs at \(\xi_1\) and \(\xi_2\), and \(f'(x), f''(x)\) both have constant sign in the range \([\xi_1, \xi_2]\), then there is a zero of \(f(x)\) lying between \(\xi_1 - \{f(\xi_1)/f'(\xi_1)\}\) and \(\xi_2 - \{f(\xi_2)/f'(\xi_2)\}\). Show that these conditions are satisfied if \(f(x)=5\cos 2x + 6x - 5, \xi_1=0.70, \xi_2=0.72\). Deduce that there is a value of \(x\) lying between \(0.7128\) and \(0.7130\) for which \(f(x)=0\). [\(0.70 \text{ radians} = 40^\circ 6'\); \(0.72 \text{ radians} = 41^\circ 15'\).]
(i) Using the substitution \(y=xz\), or otherwise, obtain the general solution of the differential equation \[ x\frac{dy}{dx} = y + 2x\sqrt{(x^2+y^2)}. \] (ii) Find the solution of \[ \frac{d^2y}{dx^2}+\frac{dy}{dx}-2y=2 \] such that \(y=0\) and \(\dfrac{dy}{dx}=0\) when \(x=0\).
Find the area enclosed by the curve \[ (x/a)^{2/3} + (y/b)^{2/3} = 1 \] where \(a\) and \(b\) are both positive. Find also the length of the curve.