I have two dice whose faces are all painted different colours. I number the faces of one of them \(1,2,2,3,3,6\) and the other \(1,3,3,4,5,6.\) I can now throw a total of 3 in two different ways using the two number \(2\)'s on the first die once each. Show that there are seven different ways of throwing a total of 6. I now renumber the dice (again only using integers in the range 1 to 6) with the results shown in the following table \noindent
If \(\left|r\right|\neq1,\) show that \[ 1+r^{2}+r^{4}+\cdots+r^{2n}=\frac{1-r^{2n+2}}{1-r^{2}}\,. \] If \(r\neq1,\) find an expression for \(\mathrm{S}_{n}(r),\) where \[ \mathrm{S}_{n}(r)=r+r^{2}+r^{4}+r^{5}+r^{7}+r^{8}+r^{10}+\cdots+r^{3n-1}. \] Show that, if \(\left|r\right|<1,\) then, as \(n\rightarrow\infty,\) \[ \mathrm{S}_{n}(r)\rightarrow\frac{1}{1-r}-\frac{1}{1-r^{3}}\,. \] If \(\left|r\right|\neq1,\) find an expression for \(\mathrm{T}_{n}(r),\) where \[ \mathrm{T}_{n}(r)=1+r^{2}+r^{3}+r^{4}+r^{6}+r^{8}+r^{9}+r^{10}+r^{12}+r^{14}+r^{15}+r^{16}+\cdots+r^{6n}. \] If \(\left|r\right|<1,\) find the limit of \(\mathrm{T}_{n}(r)\) as \(n\rightarrow\infty.\) What happens to \(\mathrm{T}_{n}(r)\) as \(n\rightarrow\infty\) in the three cases \(r>1,r=1\) and \(r=-1\)? In each case give reasons for your answer.
Solution: \begin{align*} && S &= 1 + r^2 + r^4 + \cdots + r^{2n} \\ && r^2S &= \quad \,\,\,\, r^2 + r^4 + \cdots+r^{2n}+r^{2n+2} \\ \Rightarrow && (1-r^2)S &= 1 - r^{2n+2} \\ \Rightarrow && S &= \frac{1-r^{2n+2}}{1-r^2} \end{align*} \begin{align*} && S_n(r) &= r + r^2 + r^4 + r^5 + r^7 + \cdots + r^{3n-1} \\ &&&= 1 + r + r^2 + \cdots + r^{3n} - (1 + r^3 + r^6 + r^{3n}) \\ &&&= \frac{1-r^{3n+1}}{1-r} - \frac{1-r^{3n+3}}{1-r^3} \\ \\ \Rightarrow && \lim_{n \to \infty} S_n(r) &= \frac{1-0}{1-r} - \frac{1-0}{1-r^3} = \frac{1}{1-r} - \frac{1}{1-r^3} \end{align*} \begin{align*} && T_n(r) &= 1 + r^2 + r^3 + r^4 + r^6 + \cdots + r^{6n} \\ &&&= \frac{1-r^{6n+6}}{1-r^6} + \frac{r^2-r^{6n+2}}{1-r^6} + \frac{r^3-r^{6n+3}}{1-r^6} + \frac{r^4-r^{6n+4}}{1-r^6} \\ &&&= \frac{1+r^2+r^3+r^4-r^{6n}(r^2+r^3+r^4+r^6))}{1-r^6} \\ \\ &&\lim_{n \to \infty} T_n(r) &= \frac{1+r^2+r^3+r^4}{1-r^6} \end{align*} If \(r > 1\) clear it diverges. if \(r = 1\) same story. if \(r = -1\) the sums in blocks of \(4\) are all \(1+1-1+1 = 2 > 0\) and so it also diverges.
Solution:
By making the change of variable \(t=\pi-x\) in the integral \[ \int_{0}^{\pi}x\mathrm{f}(\sin x)\,\mathrm{d}x, \] or otherwise, show that, for any function \(\mathrm{f},\) \[ \int_{0}^{\pi}x\mathrm{f}(\sin x)\,\mathrm{d}x=\frac{\pi}{2}\int_{0}^{\pi}\mathrm{f}(\sin x)\,\mathrm{d}x\,. \] Evaluate \[ \int_{0}^{\pi}\frac{x\sin x}{1+\cos^{2}x}\,\mathrm{d}x\quad\mbox{ and }\quad\int_{0}^{2\pi}\frac{x\sin x}{1+\cos^{2}x}\,\mathrm{d}x\,. \]
Solution: \begin{align*} && I &= \int_0^{\pi} x f(\sin x) \d x \\ t = \pi - x, \d t = -\d t : &&&= \int_{t = \pi}^{t = 0} (\pi - t) f(\sin (\pi - t)) -\d t \\ &&&= \int_0^{\pi} (\pi - t) f(\sin t) \d t \\ \Rightarrow && 2 I &= \pi \int_0^\pi f(\sin t) \d t \\ \Rightarrow && I &= \frac{\pi}{2} \int_0^{\pi} f(\sin x) \d x \end{align*} \begin{align*} && I &= \int_{0}^{\pi}\frac{x\sin x}{1+\cos^{2}x}\,\mathrm{d}x \\ &&&= \frac{\pi}{2}\int_0^\pi \frac{\sin x}{1 + \cos^2 x} \d x \\ &&&= \frac{\pi}{2}\left [ -\tan^{-1} \cos x\right]_0^{\pi} \\ &&&= \tan 1 - \tan (-1) = \frac{\pi^2}{4} \\ \\ && I &= \int_{0}^{2\pi}\frac{x\sin x}{1+\cos^{2}x}\,\mathrm{d}x \\ &&&= \int_{0}^{\pi}\frac{x\sin x}{1+\cos^{2}x}\,\mathrm{d}x + \int_{\pi}^{2\pi} \frac{x\sin x}{1+\cos^{2}x}\,\mathrm{d}x \\ u = x - \pi, \d u = \d x: &&&= \frac{\pi^2}{4} + \int_{0}^{\pi} \frac{(u+\pi)(-\sin u)}{1 + \cos^2 u}\d u \\ &&&= \frac{\pi^2}{4} -\frac{3\pi}{2} \int_0^{\pi} \frac{\sin u}{1+\cos^2 u} \d u \\ &&&= - \frac{\pi^2}2 \end{align*}
If \(z=x+\mathrm{i}y\) where \(x\) and \(y\) are real, define \(\left|z\right|\) in terms of \(x\) and \(y\). Show, using your definition, that if \(z_{1},z_{2}\in\mathbb{C}\) then \(\left|z_{1}z_{2}\right|=\left|z_{1}\right|\left|z_{2}\right|.\) Explain, by means of a diagram, or otherwise, why \(\left|z_{1}+z_{2}\right|\leqslant\left|z_{1}\right|+\left|z_{2}\right|.\) Suppose that \(a_{j}\in\mathbb{C}\) and \(\left|a_{j}\right|\leqslant1\) for \(j=1,2,\ldots,n.\) Show that, if \(\left|z\right|\leqslant\frac{1}{2},\) then \[ \left|a_{n}z^{n}+a_{n-1}z^{n-1}+\cdots+a_{1}z\right|<1, \] and deduce that any root \(w\) of the equation \[ a_{n}z^{n}+a_{n-1}z^{n-1}+\cdots+a_{1}z+1=0 \] must satisfy \(\left|x\right|>\frac{1}{2}.\)
Let \(N=10^{100}.\) The graph of \[ \mathrm{f}(x)=\frac{x^{N}}{1+x^{N}}+2 \] for \(-3\leqslant x\leqslant3\) is sketched in the following diagram. \noindent
Sketch the curve \[ \mathrm{f}(x)=x^{3}+Ax^{2}+B \] first in the case \(A>0\) and \(B>0\), and then in the case \(A<0\) and \(B>0.\) Show that the equation \[ x^{3}+ax^{2}+b=0, \] where \(a\) and \(b\) are real, will have three distinct real roots if \[ 27b^{2}+3a^{3}b<0, \] but will have fewer than three if \[ 27b^{2}+4a^{3}b<0. \]
In the manufacture of Grandma's Home Made Ice-cream, chemicals \(A\) and \(B\) pour at constant rates \(a\) and \(b-a\) litres per second (\(0 < a < b\)) into a mixing vat which mixes the chemicals rapidly and empties at a rate \(b\) litres per second into a second mixing vat. At time \(t=0\) the first vat contains \(K\) litres of chemical \(B\) only. Show that the volume \(V(t)\) (in litres) of the chemical \(A\) in the first vat is governed by the differential equation \[ \dot{V}(t)=-\frac{bV(t)}{K}+a, \] and that \[ V(t)=\frac{aK}{b}(1-\mathrm{e}^{-bt/K}) \] for \(t\geqslant0.\) The second vat also mixes chemicals rapidly and empties at the rate of \(b\) litres per second. If at time \(t=0\) it contains \(L\) litres of chemical \(C\) only (where \(L\neq K\)), how many litres of chemical \(A\) will it contain at a later time \(t\)?
Solution: The total volume in the first vat at time \(t\) is always \(K\), since \(b\) litres per second are coming in and \(b\) litres per second are going out. \begin{align*} &&\frac{\d V}{\d t} &= \underbrace{a}_{\text{incoming chemical }A} - \underbrace{b}_{\text{outgoing volume}} \cdot \underbrace{\frac{V(t)}{K}}_{\text{fraction of outgoing which is }A} \\ &&&= a - b \frac{V}{K} \\ \Rightarrow && \int \frac{1}{a-b\frac{V}{K}}\d V &= \int \d t \\ && - \frac{K}{b} \ln |a - b \frac{V}{K}| &= t +C\\ (t,V) = (0,0): && -\frac{K}{b} \ln a &= C \\ \Rightarrow && 1-\frac{b}{a} \frac{V}{K} &= e^{-bt/K} \\ \Rightarrow && V &= \frac{aK}{b} (1 - e^{-bt/K}) \end{align*} \begin{align*} &&\frac{\d W}{\d t} &= \underbrace{b}_{\text{incoming volume}} \cdot \underbrace{\frac{a}{b} (1 - e^{-bt/K})}_{\text{incoming fraction }A} - \underbrace{b}_{\text{outgoing volume}} \cdot \underbrace{ \frac{W(t)}{L}}_{\text{fraction of outgoing which is }A} \\ &&&= a (1 - e^{-bt/K}) - b \frac{W}{L} \\ \Rightarrow && \frac{\d W}{\d t} + \frac{b}{L} W &= a (1-e^{-bt/K}) \\ && \frac{\d}{\d t} \left ( e^{b/L t} W\right) &= ae^{b/L t}(1-e^{-bt/K}) \\ \Rightarrow && W &= e^{-bt/L} \left ( \frac{aL}{b}e^{b/Lt} - \frac{a}{\frac{b}{L} - \frac{b}{K}}e^{b/L t - b/K} \right) + Ce^{-bt/L} \\ &&&= \frac{aL}{b} \left (1 - \frac{K}{K-L}e^{-b/Kt} \right)+ Ce^{-bt/L} \\ (t,W) = (0,0): && 0 &= \frac{aL}{b} \frac{-L}{K-L} + C \\ \Rightarrow && C &= \frac{aL^2}{b(K-L)} \\ \Rightarrow && W &= \frac{aL}{b} \left (1 - \frac{K}{K-L} e^{-bt/K} + \frac{L}{K-L} e^{-bt/L} \right) \end{align*}
A small lamp of mass \(m\) is at the end \(A\) of a light rod \(AB\) of length \(2a\) attached at \(B\) to a vertical wall in such a way that the rod can rotate freely about \(B\) in a vertical plane perpendicular to the wall. A spring \(CD\) of natural length \(a\) and modulus of elasticity \(\lambda\) is joined to the rod at its mid-point \(C\) and to the wall at a point \(D\) a distance \(a\) vertically above \(B\). The arrangement is sketched below. \noindent