The numbers \(x\), \(y\) and \(z\) satisfy \begin{align*} x+y+z&= 1\\ x^2+y^2+z^2&=2\\ x^3+y^3+z^3&=3\,. \end{align*} Show that \[ yz+zx+xy=-\frac12 \,.\] Show also that \(x^2y+x^2z+y^2z+y^2x+z^2x+z^2y=-1\,\), and hence that \[ xyz=\frac16 \,.\] Let \(S_n=x^n+y^n+z^n\,\). Use the above results to find numbers \(a\), \(b\) and \(c\) such that the relation \[ S_{n+1}=aS_{n}+bS_{n-1}+cS_{n-2}\,, \] holds for all \(n\).
Solution: \begin{align*} && (x+y+z)^2 &= x^2 + y^2 + z^2 + 2(xy+yz+zx) \\ \Rightarrow && 1^2 &= 2 + 2(xy+yz+zx) \\ \Rightarrow && xy+yz+zx &= -\frac12 \end{align*} \begin{align*} && 1 \cdot 2 &= (x+y+z)(x^2+y^2+z^2) \\ &&&= x^3 + y^3 + z^3 + x^2y+x^2z+y^2z+y^2x+z^2x+z^2y \\ &&&= 3 + x^2y+x^2z+y^2z+y^2x+z^2x+z^2y\\ \Rightarrow && -1 &= x^2y+x^2z+y^2z+y^2x+z^2x+z^2y \end{align*} \begin{align*} && (x+y+z)^3 &= x^3 + y^3 + z^3 + \\ &&&\quad \quad 3xy^2 + 3xz^2 + \cdots + 3zx^2 + 3zy^2 + \\ &&&\quad \quad \quad 6xyz \\ \Rightarrow && 1 &= 3 + 3(-1) + 6xyz \\ \Rightarrow && xyz &= \frac16 \end{align*} Since we have \(f(t) = (t-x)(t-y)(t-z) = t^3-t^2-\frac12 t - \frac16\) is zero for \(x,y,z\) we can notice that: \(t^{n+1} = t^n +\frac12 t^{n-1} + \frac16 t^{n-2}\) is also true for \(x,y,z\) (by multiplying by \(t^{n-2}\). Therefore: \(S_{n+1} = S_n + \frac12 S_{n-1} + \frac16 S_{n-2}\)
Show that $\big\vert \e^{\i\beta} -\e^{\i\alpha}\big\vert = 2\sin\frac12 (\beta-\alpha)\,\( for \)0<\alpha<\beta<2\pi\,$. Hence show that \[ \big\vert \e^{\i\alpha} -\e^{\i\beta}\big\vert \; \big\vert \e^{\i\gamma} -\e^{\i\delta}\big\vert + \big\vert \e^{\i\beta} -\e^{\i\gamma}\big\vert \; \big\vert \e^{\i\alpha} -\e^{\i\delta}\big\vert = \big\vert \e^{\i\alpha} -\e^{\i\gamma}\big\vert \; \big\vert \e^{\i\beta} -\e^{\i\delta}\big\vert \,, \] where \(0<\alpha<\beta<\gamma<\delta<2\pi\). Interpret this result as a theorem about cyclic quadrilaterals.
Let \(m\) be a positive integer and let \(n\) be a non-negative integer.
Solution:
A particle is projected under gravity from a point \(P\) and passes through a point \(Q\). The angles of the trajectory with the positive horizontal direction at \(P\) and at \(Q\) are \(\theta\) and \(\phi\), respectively. The angle of elevation of \(Q\) from \(P\) is \(\alpha\).
A light spring is fixed at its lower end and its axis is vertical. When a certain particle \(P\) rests on the top of the spring, the compression is \(d\). When, instead, \(P\) is dropped onto the top of the spring from a height \(h\) above it, the compression at time \(t\) after \(P\) hits the top of the spring is \(x\). Obtain a second-order differential equation relating \(x\) and \(t\) for \(0\le t \le T\), where \(T\) is the time at which \(P\) first loses contact with the spring. Find the solution of this equation in the form \[ x= A + B\cos (\omega t) + C\sin(\omega t)\,, \] where the constants \(A\), \(B\), \(C\) and \(\omega\) are to be given in terms of \(d\), \(g\) and \(h\) as appropriate. Show that \[ T = \sqrt{d/g\;} \left (2 \pi - 2 \arctan \sqrt{2h/d\;}\;\right)\,. \]
A comet in deep space picks up mass as it travels through a large stationary dust cloud. It is subject to a gravitational force of magnitude \(M\!f\) acting in the direction of its motion. When it entered the cloud, the comet had mass \(M\) and speed \(V\). After a time \(t\), it has travelled a distance \(x\) through the cloud, its mass is \(M(1+bx)\), where~\(b\) is a positive constant, and its speed is \(v\).