A uniform rod of mass \(M\) and length \(2a\) lies on a smooth horizontal table, and is free to rotate about a vertical axis through its mid-point. When the rod is at rest, a small frog of mass \(m\) leaps from one end of the rod with velocity \(\sqrt{(\lambda ag)}\), in such a direction that the angle between the plane of his trajectory and the initial position of the rod is \(\theta\), where \(0 < \theta < \pi/2\). Show that if he is able, by suitable choice of the angle of elevation of his jump, to alight on the other end of the rod as it comes round for the first time, then \[ M<3m, \quad \lambda \ge 2 \cos\theta, \] and \(\theta\) is the angle defined by \[ \theta = \frac{3m}{2M}\sin 2\theta. \]
(i) Given that the product of two of the roots is 2, solve the equation \[ x^4+2x^3-14x^2-11x-2=0. \] (ii) Show that if \(a\) is a root of the equation \[ x^4+3x^3-6x^2-3x+1=0 \] then so also is \(\frac{a-1}{a+1}\). Express the remaining roots in terms of \(a\), and hence, or otherwise, solve the equation completely.
Prove that if \[ 1+c_1x+c_2x^2+c_3x^3+\dots = (ax^2+2bx+1)^{-1}, \] then \[ 1+c_1^2x+c_2^2x^2+c_3^2x^3+\dots = \frac{1+ax}{1-ax}\{a^2x^2+2(a-2b^2)x+1\}^{-1}. \]
Solution: Suppose the roots of \(ax^2 + bx + 1\) are \(r_\pm = \frac{-b \pm \sqrt{b^2-a}}{a}\) and note that \(r_+r_- = \frac{1}{a}\) \begin{align*} \frac{1}{ax^2+2bx+1} &= \frac{1}{a(x-r_+)(x-r_-)} \\ &= \frac{1}{a(r_+-r_-)} \left ( \frac{1}{x-r_+} - \frac{1}{x-r_-} \right) \\ &= \frac{1}{a(r_+-r_-)} \left ( \frac{1/r_-}{1-\frac{x}{r_-}} - \frac{1/r_+}{1-\frac{x}{r_+}} \right) \\ &= \frac{1}{a(r_+-r_-)} \sum_{n=0}^{\infty} \left ( \frac{x^n}{r_-^{n+1}}- \frac{x^n}{r_+^{n+1}} \right) \\ &= \frac{1}{a(r_+-r_-)} \sum_{n=0}^{\infty} \left ( \frac{r_+^{n+1}-r_-^{n+1}}{(r_-r_+)^{n+1}}\right)x^n \\ &= \frac{1}{(r_+-r_-)} \sum_{n=0}^{\infty} \left ( r_+^{n+1}-r_-^{n+1}\right)a^nx^n \\ \end{align*} Therefore, \begin{align*} && c_n &= \left ( \frac{r_+^{n+1}-r_-^{n+1}}{r_+-r_-}\right)a^n \\ \Rightarrow && c_n^2 &=\left ( \frac{r_+^{n+1}-r_-^{n+1}}{r_+-r_-}\right)^2a^{2n} \\ &&&= \left ( \frac{r_+^{2n+2}+r_-^{2n+2}-2(r_+r_-)^{n+1})}{(r_+-r_-)^2}\right)a^{2n} \\ &&&= \left ( \frac{r_+^{2n+2}+r_-^{2n+2}-2a^{-(n+1)})}{\frac{4(b^2-a)}{a^2}}\right)a^{2n} \\ &&&= \frac{a^{2n+2}}{4(b^2-a)} \left (r_+^{2n+2}+r_-^{2n+2} \right) - \frac{a^{n+1}}{2(b^2-a)} \\ \Rightarrow && \sum_{n=0}^{\infty} c_n^2 x^n &= \frac{a^{2}}{4(b^2-a)} \left ( \frac{r_+^2}{1-(a^2r_+^2x)} + \frac{r_-^2}{1-(a^2r_-^2x)} \right) - \frac{a}{2(b^2-a)} \frac{1}{1-ax} \\ \end{align*}
If \(a, b\) and \(h(>0)\) are real constants, prove that the roots \(x_1, x_2 (x_1>x_2)\) of the equation \[ (a-x)(b-x)=h^2 \] lie outside the range between \(a\) and \(b\). \newline If \(\phi(x)\) denotes the polynomial \[ \begin{vmatrix} a-x & h & g \\ h & b-x & f \\ g & f & c-x \end{vmatrix} \] show that \begin{align*} \phi(x_1) &= (g\sqrt{x_1-b}+f\sqrt{x_1-a})^2, \\ \phi(x_2) &= -(g\sqrt{b-x_2}-f\sqrt{a-x_2})^2. \end{align*} Deduce that if \(g, f\) and \(c\) are real the equation \(\phi(x)=0\) has three real unequal roots.
A number, \(n\), of different objects are divided into two groups containing \(r\) and \(n-r\) members. If the objects of the two separate groups so formed are then permuted amongst themselves and give \(N(n,r)\) different permutations of the \(n\) objects, prove that \[ \sum_{r=0}^n \{N(n,r)\}^{-1} = \frac{2^n}{n!}. \]
Explain what is meant by the statement that ``\(f(n)\) tends to the limit \(l\) as \(n\) tends to infinity'' where \(n\) is a positive integer. \newline A positive quantity \(a_n\) satisfies the relationship \[ a_n = \frac{1}{2}\left(a_{n-1} + \frac{a^2}{a_{n-1}}\right), \] where \(n\) is a positive integer greater than unity and \(a\) is positive. Prove that, provided \(a_1>0\),
Trace the curve \(16a^3y^2=b^2x^2(a-2x)\), where \(a\) and \(b\) are positive, and find the area enclosed by the loop. \newline If \(16a^2=3b^2\), show that the perimeter of the loop is \(\frac{1}{2}b\).
Establish Newton's formula for the radius of curvature of a curve, namely that if rectangular axes are taken with the origin at the point of the curve, and the tangent and normal as the \(x\) and \(y\) axes respectively, then the radius of curvature is \(\lim_{x\to 0} \frac{x^2}{2y}\). \newline Find the radius of curvature at the origin, of the curve \[ y = 2x+3x^2-2xy+y^2+2y^3, \] and show that the circle of curvature at the origin has equation \[ 3(x^2+y^2)=5(y-2x). \]
If \(I_n = \int_0^\infty x^n e^{-ax}\cos bx \, dx\), \(J_n = \int_0^\infty x^n e^{-ax}\sin bx \, dx\), where \(n\) is a positive integer and \(a\) and \(b\) are positive, prove that: \begin{align*} I_n(a^2+b^2) &= n(aI_{n-1}-bJ_{n-1}), \\ J_n(a^2+b^2) &= n(bI_{n-1}+aJ_{n-1}). \end{align*} Show that \begin{align*} (a^2+b^2)^{\frac{n+1}{2}} I_n &= n! \cos(n+1)\alpha, \\ (a^2+b^2)^{\frac{n+1}{2}} J_n &= n! \sin(n+1)\alpha, \end{align*} where \(\tan\alpha = \frac{b}{a}\) and \(0 < \alpha < \frac{\pi}{2}\).
Prove that the mean value with respect to area over the surface of a sphere centre \(O\) and radius \(a\) of the reciprocal of the distance from a fixed point \(C\) is equal to the reciprocal of \(OC\) if \(C\) is outside the sphere, but equal to the reciprocal of the radius \(a\) if \(C\) is inside the sphere.