10273 problems found
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.
A circle of radius \(a\) rolls round the outside of a closed oval curve whose total perimeter is \(s\) and whose area is \(S\). Show that the locus of the centre of the circle is an oval curve of perimeter \(s+2\pi a\) enclosing an area \(S+as+\pi a^2\). \newline If the circle rolls on the inside of the oval curve and is sufficiently small to be always entirely within it, show that the locus of the centre is another oval of perimeter \(s-2\pi a\) enclosing an area \(S-as+\pi a^2\).
A particle of mass \(m\) moves under gravity in a medium that opposes the motion with a resisting force \(kmv\), where \(k\) is a constant and \(v\) is the speed. If it is projected up vertically with speed \(v_0\), show that after time \(t\) its height above the point of projection is \[ \left(v_0 + \frac{g}{k}\right) \frac{(1-e^{-kt})}{k} - \frac{gt}{k}. \] Hence find the greatest height reached by the particle. \newline Show that subsequently the speed cannot exceed a certain finite value however far the particle falls.