Problems
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2013 Paper 2 Q8
The function \(\f\) satisfies \(\f(x)>0\) for \(x\ge0\) and is strictly decreasing (which means that \(\f(b)<\f(a)\) for \(b>a\)).
- For \(t\ge0\), let \(A_0(t)\) be the area of the largest rectangle with sides parallel to the coordinate axes that can fit in the region bounded by the curve \(y=\f(x)\), the \(y\)-axis and the line \(y=\f(t)\). Show that \(A_0(t)\) can be written in the form \[ A_0(t) =x_0\left( \f(x_0) -\f(t)\right), \] where \(x_0\) satisfies \(x_0 \f'(x_0) +\f(x_0) = \f(t)\,\).
- The function g is defined, for \(t> 0\), by \[ \g(t) =\frac 1t \int_0^t \f(x) \d x\,. \] Show that \(t \g'(t) = \f(t) -\g(t)\,\). Making use of a sketch show that, for \(t>0\), \[ \int_0^t \left( \f(x) - \f(t)\right) \d x > A_0(t) \] and deduce that \(-t^2 \g'(t)> A_0(t)\).
- In the case \(\f(x)= \dfrac 1 {1+x}\,\), use the above to establish the inequality \[ \ln \sqrt{1+t} > 1 - \frac 1 {\sqrt{1+t}} \,, \] for \(t>0\).
2013 Paper 2 Q9
The diagram shows three identical discs in equilibrium in a vertical plane. Two discs rest, not in contact with each other, on a horizontal surface and the third disc rests on the other two. The angle at the upper vertex of the triangle joining the centres of the discs is \(2\theta\).
- Show that the normal reaction between the horizontal surface and a disc in contact with the surface is \(\frac32 W\,\).
- Find the normal reaction between two discs in contact and show that the magnitude of the frictional force between two discs in contact is \(\dfrac{W\sin\theta}{2(1+\cos\theta)}\,\).
- Show that if \(\mu <2- \surd3\,\) there is no value of \(\theta\) for which equilibrium is possible.
2013 Paper 2 Q10
A particle is projected at an angle of elevation \(\alpha\) (where \(\alpha>0\)) from a point \(A\) on horizontal ground. At a general point in its trajectory the angle of elevation of the particle from \(A\) is \(\theta\) and its direction of motion is at an angle \(\phi\) above the horizontal (with \(\phi\ge0\) for the first half of the trajectory and \(\phi\le0\) for the second half). Let \(B\) denote the point on the trajectory at which \(\theta = \frac12 \alpha\) and let \(C\) denote the point on the trajectory at which \(\phi = -\frac12\alpha\).
- Show that, at a general point on the trajectory, \(2\tan\theta = \tan \alpha + \tan\phi\,\).
- Show that, if \(B\) and \(C\) are the same point, then \( \alpha = 60^\circ\,\).
- Given that \(\alpha < 60^\circ\,\), determine whether the particle reaches the point \(B\) first or the point \(C\) first.
2013 Paper 2 Q11
Three identical particles lie, not touching one another, in a straight line on a smooth horizontal surface. One particle is projected with speed \(u\) directly towards the other two which are at rest. The coefficient of restitution in all collisions is \(e\), where \(0 < e < 1\,\).
- Show that, after the second collision, the speeds of the particles are \(\frac12u(1-e)\), \(\frac14u (1-e^2)\) and \(\frac14u(1+e)^2\). Deduce that there will be a third collision whatever the value of \(e\).
- Show that there will be a fourth collision if and only if \(e\) is less than a particular value which you should determine.
Solution:
- First Collision:
By NEL, \(v_2 = v_1 + eu\), so \begin{align*} \text{COM}: && mu &= mv_1 + m(v_1 + eu) \\ \Rightarrow && 2mv_1 &= mu(1-e) \\ \Rightarrow && v_1 &= \frac12 u(1-e) \\ && v_2 &= \frac12 u(1-e) + eu \\ &&&= \frac12 u(1+e) \end{align*} The second collision is identical to the first except replacing \(u\) with \(\frac12u(1+e)\), therefore after that collision: \begin{align*} && \text{first particle} &= \frac12 u(1-e) \\ && \text{second particle} &= \frac12 \left (\frac12 u(1+e) \right)(1-e) \\ &&&= \frac14 u(1-e^2) \\ && \text{third particle} &= \frac12 \left (\frac12 u(1+e) \right)(1+e) \\ &&&= \frac14 u(1+e)^2 \end{align*} After all these collisions, all particles are moving in the same direction (since they all have positive velocity), but the first particle is now travelling faster than the second particle (as \(\frac12(1-e) < 1\)). Therefore they will collide again.
- The third collision:
The speed of approach will be \(\frac12u(1-e) - \frac14u(1-e^2) = \frac14u(1-e)(2 - (1+e)) = \frac14 u(1-e)^2\), therefore by NEL, \(w_2 = w_1 + \frac14ue(1-e)^2\) \begin{align*} \text{COM}: && m\frac12u(1-e) + m \frac14u(1-e^2) &= mw_1 + m\left (w_1 + \frac14ue(1-e)^2 \right) \\ \Rightarrow && \frac14u(1-e)(2+(1+e)) &= 2w_1 + \frac14ue(1-e)^2 \\ \Rightarrow && 2w_1 &= \frac14u(1-e)(3+e)-\frac14ue(1-e)^2 \\ &&&= \frac14u(1-e)(3+e-e(1-e)) \\ &&&= \frac14u(1-e)(3+e^2) \\ \Rightarrow && w_1 &= \frac18 u(1-e)(3+e^2) \\ && w_2 &= \frac18 u(1-e)(3+e^2) + \frac14ue(1-e)^2 \\ &&&= \frac18u(1-e)(3+e^2+2e(1-e)) \\ &&&= \frac18u(1-e)(3+2e-e^2) \\ &&&= \frac18u(1-e)(1+e)(3-e) \\ \end{align*} A fourth collision is possible, iff \begin{align*} && \frac18u(1-e)(1+e)(3-e)&> \frac14 u(1+e)^2 \\ \Leftrightarrow && (1-e)(3-e)&> 2 (1+e) \\ \Leftrightarrow &&3-4e-e^2&> 2+2e \\ \Leftrightarrow &&1-5e-e^2&>0 \\ \Leftrightarrow && e &< 3-\sqrt{2} \end{align*}
2013 Paper 2 Q12
The random variable \(U\) has a Poisson distribution with parameter \(\lambda\). The random variables \(X\) and \(Y\) are defined as follows. \begin{align*} X&= \begin{cases} U & \text{ if \(U\) is 1, 3, 5, 7, \(\ldots\,\)} \\ 0 & \text{ otherwise} \end{cases} \\ Y&= \begin{cases} U & \text{ if \(U\) is 2, 4, 6, 8, \(\ldots\,\) } \\ 0 & \text{ otherwise} \end{cases} \end{align*}
- Find \(\E(X)\) and \(\E(Y)\) in terms of \(\lambda\), \(\alpha\) and \(\beta\), where \[ \alpha = 1+\frac{\lambda^2}{2!}+\frac{\lambda^4}{4!} +\cdots\, \text{ \ \ and \ \ } \beta = \frac{\lambda}{1!} + \frac{\lambda^3}{3!} + \frac{\lambda^5}{5!} +\cdots\,. \]
- Show that \[ \var(X) = \frac{\lambda\alpha+\lambda^2\beta}{\alpha+\beta} - \frac{\lambda^2\alpha^2}{(\alpha+\beta)^2} \] and obtain the corresponding expression for \(\var(Y)\). Are there any non-zero values of \(\lambda\) for which \( \var(X) + \var(Y) = \var(X+Y)\,\)?
Solution:
- \begin{align*} \mathbb{E}(X) &= \sum_{r=1}^\infty r \mathbb{P}(X = r) \\ &= \sum_{j=1}^{\infty} (2j-1)\mathbb{P}(U=2j-1) \\ &= \sum_{j=1}^{\infty}(2j-1) \frac{e^{-\lambda} \lambda^{2j-1}}{(2j-1)!} \\ &= \sum_{j=1}^{\infty} e^{-\lambda} \frac{\lambda^{2j-1}}{(2j-2)!} \\ &= \lambda e^{-\lambda} \sum_{j=1}^{\infty} \frac{\lambda^{2j-2}}{(2j-2)!} \\ &= \lambda e^{-\lambda} \alpha \end{align*} Since \(\mathbb{E}(X+Y) = \lambda, \mathbb{E}(Y) = \lambda(1-e^{-\lambda}\alpha) = \lambda(e^{-\lambda}(\alpha+\beta) - e^{-\lambda}\alpha) = \lambda e^{-\lambda} \beta\). Alternatively, as \(\beta + \alpha = e^{\lambda}\), \(\mathbb{E}(X) = \frac{\lambda \alpha}{\alpha+\beta}, \mathbb{E}(Y) = \frac{\lambda \beta}{\alpha+\beta}\)
- \begin{align*} \textrm{Var}(X) &= \mathbb{E}(X^2) - [\mathbb{E}(X) ]^2 \\ &= \sum_{odd} r^2 \mathbb{P}(U = r) - \left [ \mathbb{E}(X) \right]^2 \\ &= \sum_{odd} (r(r-1)+r)\frac{e^{-\lambda}\lambda^r}{r!} - \frac{\lambda^2 \alpha^2}{(\alpha+\beta)^2} \\ &= \sum_{odd} \frac{e^{-\lambda}\lambda^r}{(r-2)!}+\sum_{odd} \frac{e^{-\lambda}\lambda^r}{(r-1)!} - \frac{\lambda^2 \alpha^2}{(\alpha+\beta)^2} \\ &= e^{-\lambda}\lambda^2 \beta + e^{-\lambda}\lambda \alpha - \frac{\lambda^2 \alpha^2}{(\alpha+\beta)^2} \\ &= \frac{\lambda \alpha + \lambda^2 \beta}{\alpha+\beta}- \frac{\lambda^2 \alpha^2}{(\alpha+\beta)^2} \end{align*} Similarly, \begin{align*} \textrm{Var}(Y) &= \mathbb{E}(Y^2) - [\mathbb{E}(Y) ]^2 \\ &= \sum_{even} r^2 \mathbb{P}(U = r) - \left [ \mathbb{E}(Y) \right]^2 \\ &= \sum_{even} (r(r-1)+r)\frac{e^{-\lambda}\lambda^r}{r!} - \frac{\lambda^2 \beta^2}{(\alpha+\beta)^2} \\ &= e^{-\lambda}\lambda^2\alpha + e^{-\lambda}\lambda \beta - \frac{\lambda^2 \beta^2}{(\alpha+\beta)^2} \\ &= \frac{\lambda \beta + \lambda^2 \alpha}{\alpha+\beta}- \frac{\lambda^2 \beta^2}{(\alpha+\beta)^2} \end{align*} Since \(\textrm{Var}(X+Y) = \textrm{Var}(U) = \lambda\), we are interested in solving: \begin{align*} \lambda &= \frac{\lambda \alpha + \lambda^2 \beta}{\alpha+\beta}- \frac{\lambda^2 \alpha^2}{(\alpha+\beta)^2} + \frac{\lambda \beta + \lambda^2 \alpha}{\alpha+\beta}- \frac{\lambda^2 \beta^2}{(\alpha+\beta)^2} \\ &= \frac{\lambda(\alpha+\beta) + \lambda^2(\alpha+\beta)}{\alpha+\beta} - \frac{\lambda^2(\alpha^2+\beta^2)}{(\alpha+\beta)^2} \\ &= \lambda + \lambda^2 \frac{(\alpha+\beta)^2 - (\alpha^2+\beta^2)}{(\alpha+\beta)^2} \\ &= \lambda + \lambda^2 \frac{2\alpha\beta}{(\alpha+\beta)^2} \end{align*} which is clearly not possible if \(\lambda \neq 0\)
2013 Paper 2 Q13
A biased coin has probability \(p\) of showing a head and probability \(q\) of showing a tail, where \(p\ne0\), \(q\ne0\) and \(p\ne q\). When the coin is tossed repeatedly, runs occur. A straight run of length \(n\) is a sequence of \(n\) consecutive heads or \(n\) consecutive tails. An alternating run of length \(n\) is a sequence of length \(n\) alternating between heads and tails. An alternating run can start with either a head or a tail. Let \(S\) be the length of the longest straight run beginning with the first toss and let \(A\) be the length of the longest alternating run beginning with the first toss.
- Explain why \(\P(A=1)=p^2+q^2\) and find \(\P(S=1)\). Show that \(\P(S=1)<\P(A=1)\).
- Show that \(\P(S=2)= \P(A=2)\) and determine the relationship between \(\P(S=3)\) and \( \P(A=3)\).
- Show that, for \(n>1\), \(\P(S=2n)>\P(A=2n)\) and determine the corresponding relationship between \(\P(S=2n+1)\) and \(\P(A=2n+1)\). [You are advised not to use \(p+q=1\) in this part.]
Solution:
- The only way \(A = 1\) is if we get \(HH\) or \(TT\) which has probability \(p^2+q^2\). The only way we get \(S=1\) is if we have \(HT\) to \(TH\), ie \(2pq\). Since \((p-q)^2 = p^2 + q^2 - 2pq >0\) we must have \(\mathbb{P}(A=1) > \mathbb{P}(S=1)\).
- \(\,\) \begin{align*} \mathbb{P}(S=2) &= p^2q + q^2p \\ \mathbb{P}(A=2) &= pq^2 + qp^2 = \mathbb{P}(S=2) \\ \\ \mathbb{P}(S=3) &= p^3q + q^3p = pq(p^2+q^2) \\ \mathbb{P}(A=3) &= pqp^2 + qpq^2 = pq(p^2+q^2) = \mathbb{P}(S=3) \end{align*}
- For \(n > 1\) we must have \begin{align*} && \mathbb{P}(S = 2n) &= p^{2n}q + q^{2n}p \\ && \mathbb{P}(A=2n) &= (pq)^{n}q + (qp)^{n}p \\ &&&= p^nq^{n+1} + q^np^{n+1} \\ && \mathbb{P}(S = 2n) &> \mathbb{P}(A = 2n) \\ \Leftrightarrow && p^{2n}q + q^{2n}p & > p^nq^{n+1} + q^np^{n+1}\\ \Leftrightarrow && 0 & < p^{2n}q+q^{2n}p - p^nq^{n+1} -q^np^{n+1}\\ &&&= (p^n-q^n)(qp^n - pq^n) \end{align*} which is clearly true. \begin{align*} && \mathbb{P}(S=2n+1) &= p^{2n+1}q + q^{2n+1}p \\ && \mathbb{P}(A=2n+1) &= (pq)^{n}p^2 + (qp)^{n}q^2 \\ &&&= p^{n+2}q^n + q^{n+2}p^n \end{align*} The same factoring logic shows that \(\mathbb{P}(S = 2n+1) > \mathbb{P}(A=2n+1)\)
2013 Paper 3 Q1
Given that \(t= \tan \frac12 x\), show that \(\dfrac {\d t}{\d x} = \frac12(1+t^2)\) and \( \sin x = \dfrac {2t}{1+t^2}\,\). Hence show that \[ \int_0^{\frac12\pi} \frac 1{1+a \sin x}\, \d x = \frac2 {\sqrt{1-a^2}} \arctan \frac{\sqrt{1-a}}{\sqrt{1+a}}\, \qquad \quad (0 < a < 1). \] Let \[ I_n = \int_0^{\frac12\pi} \frac{ \sin ^nx}{2+\sin x} \, \d x \qquad \quad (n\ge0). \] By considering \(I_{n+1}+2I_{n}\,\), or otherwise, evaluate \(I_3\).
Solution: Let \(t = \tan \frac12 x\), then \begin{align*} \frac{\d t}{\d x} &= \tfrac12 \sec^2 \tfrac12 t \\ &= \tfrac12 (1 + \tan^2 \tfrac12 ) \\ &= \tfrac12 (1 + t^2) \\ \\ \sin x &= 2 \sin \tfrac12 x \cos \tfrac12 \\ &= \frac{2 \frac{\sin \tfrac12 x}{ \cos \tfrac12x}}{\frac{1}{\cos^2 \tfrac12 x}} \\ &= \frac{2 \tan \tfrac12 x}{\sec^2 \tfrac12 } \\ &= \frac{2t }{1+t^2} \end{align*} Now consider \begin{align*} t = \tan \tfrac12 x: && \int_0^{\frac12\pi} \frac 1{1+a \sin x}\, \d x &= \int_{t=0}^{t = 1} \frac{1}{1 + a \frac{2t}{1+t^2}} \frac{2}{1+t^2} \d t \\ &&&= \int_0^1 \frac{2}{1+2at+t^2} \d t \\ &&&= \int_0^1 \frac{2}{(t+a)^2 + 1-a^2} \d t \\ (1-a^2) > 0: &&&= \left [ \frac{2}{\sqrt{1-a^2}} \arctan \frac{t+a}{\sqrt{1-a^2}} \right]_0^1 \\ &&&= \frac{2}{\sqrt{1-a^2}} \left ( \arctan \frac{1+a}{\sqrt{1-a^2}} - \arctan \frac{a}{\sqrt{1-a^2}} \right) \\ &&&= \frac{2}{\sqrt{1-a^2}} \arctan \left ( \frac{\frac{1+a}{\sqrt{1-a^2}}-\frac{a}{\sqrt{1-a^2}}}{1+\frac{1+a}{\sqrt{1-a^2}}\frac{a}{\sqrt{1-a^2}}} \right) \\ &&&= \frac{2}{\sqrt{1-a^2}} \arctan \left ( \frac{\frac{1}{\sqrt{1-a^2}}}{\frac{1+a}{1-a^2}} \right) \\ &&&= \frac{2}{\sqrt{1-a^2}} \arctan \left ( \frac{\sqrt{1-a}}{\sqrt{1+a}} \right) \end{align*} as required. Let \[ I_n = \int_0^{\frac12\pi} \frac{ \sin ^nx}{2+\sin x} \, \d x \qquad \quad (n\ge0). \] and consider \begin{align*} I_{n+1} + 2I_n &= \int_0^{\frac12\pi} \frac{ \sin ^{n+1}x+2\sin^{n} x}{2+\sin x} \, \d x \\ &= \int_0^{\frac12\pi} \frac{ \sin^n x (2 + \sin x)}{2+\sin x} \, \d x \\ &= \int_0^{\frac12\pi} \sin^n x \d x \end{align*} Therefore we can compute \begin{align*} I_0 &= \int_0^{\pi/2} \frac{1}{2 + \sin x} \d x \\ &= \frac12 \int_0^{\pi/2} \frac{1}{1 + \frac12 \sin x} \d x \\ &= \frac{1}{\sqrt{3/4}} \arctan \frac{\sqrt{1/2}}{\sqrt{3/2}} \\ &= \frac{2}{\sqrt{3}} \arctan \frac{1}{\sqrt{3}} \\ &= \frac{\pi}{3\sqrt{3}} \\ \\ I_1 &= \int_0^{\pi/2} 1 \d x - 2 I_0 \\ &= \frac{\pi}{2} - \frac{2\pi}{3\sqrt{3}} \\ I_2 &= \int_0^{\pi/2} \sin x \d x - 2I_1 \\ &= 1 - \pi + \frac{4\pi}{3\sqrt{3}} \\ I_3 &= \int_0^{\pi/2} \sin^2 x \d x - 2I_2 \\ &= \frac12 \int_0^{\pi/2} \sin^2 + \cos^2 x \d x - 2I_2 \\ &= \frac{\pi}{4} - 2 + 2\pi - \frac{8\pi}{3\sqrt{3}} \\ &= -2 + \frac{9\pi}{4} - \frac{8\pi}{3\sqrt{3}} \end{align*}
2013 Paper 3 Q2
In this question, you may ignore questions of convergence. Let \(y= \dfrac {\arcsin x}{\sqrt{1-x^2}}\,\). Show that \[ (1-x^2)\frac {\d y}{\d x} -xy -1 =0 \] and prove that, for any positive integer \(n\), \[ (1-x^2) \frac{\d^{n+2}y}{\d x^{n+2}} - (2n+3)x \frac{\d ^{n+1}y}{\d x ^{n+1}} -(n+1)^2 \frac{\d^ny}{\d x^n}=0\, . \] Hence obtain the Maclaurin series for \( \dfrac {\arcsin x}{\sqrt{1-x^2}}\,\), giving the general term for odd and for even powers of \(x\). Evaluate the infinite sum \[ 1 + \frac 1 {3!} + \frac{2^2}{5!} + \frac {2^2\times 3^2}{7!}+\cdots + \frac {2^2\times 3^2\times \cdots \times n^2}{(2n+1)!} + \cdots\,. \]
Solution: \begin{align*} && y &= \frac{\arcsin x}{\sqrt{1-x^2}} \\ && \frac{\d y}{\d x} &= \frac{(1-x^2)^{-1/2} \cdot (1-x^2)^{1/2}-\arcsin x \cdot (-x)(1-x^2)^{-1/2}}{1-x^2} \\ &&&= \frac{1+ xy}{1-x^2} \\ \Rightarrow && 0 &= (1-x^2) \frac{\d y}{\d x} -xy-1\\ \\ \frac{\d^n}{\d x^{n+1}}: && 0 &= \left ( (1-x^2) y' \right)^{(n+1)} - (xy)^{(n+1)} \\ \Rightarrow && 0 &= (1-x^2)y^{(n+2)} + \binom{n+1}{1}(1-x^2)^{(1)}y^{(n+1)}+\binom{n+1}{2} (1-x^2)^{(2)}y^{(n)} - (xy^{(n+1)} +\binom{n+1}{1} y^{(n)} ) \\ &&&= (1-x^2)y^{(n+2)}+\left ( (n+1)\cdot(-2x)-x \right)y^{(n+1)} + \left ( \frac{(n+1)n}{2} \cdot (-2)-(n+1) \right)y^{(n)} \\ &&&= (1-x^2)y^{(n+2)}-\left ( 2n+3 \right)xy^{(n+1)} - \left ( (n+1)n+(n+1)\right)y^{(n)} \\ &&&= (1-x^2)y^{(n+2)}-\left ( 2n+3 \right)xy^{(n+1)} - \left ( n+1\right)^2y^{(n)} \\ \end{align*} Since \(y(0) = 0, y'(0) = 1\) we can look at the recursion: \(y^{(n+2)} - (n+1)^2y^{(n)}\) for larger terms, ie \(y^{(2k)}(0) = 0\) \(y^{(1)}(0) = 1, y^{(3)}(0) = (1+1)^2 \cdot 1 = 2^2, y^{(5)}(0) = (3+1)^2 y^{(3)} = 4^2 \cdot 2^2\) and \(y^{(2k+1)}(0) = (2k)^2 \cdot (2k-2)^2 \cdots 2^2 \cdot 1^2 = 2^{2k} \cdot (k!)^2\). Therefore \begin{align*} && \frac{\arcsin x}{\sqrt{1-x^2}} &= \sum_{k=0}^{\infty} \frac{2^{2k} \cdot (k!)^2}{(2k+1)!} x^{2k+1} \\ \\ \Rightarrow && \frac{\arcsin \frac12}{\sqrt{1-\left (\frac12 \right)^2}} &= \sum_{k=0}^{\infty} \frac{2^{2k} \cdot (k!)^2}{(2k+1)!} 2^{-2k-1}\\ &&&= \frac12 \sum_{k=0}^{\infty} \frac{ (k!)^2}{(2k+1)!} \\ &&&= \frac12 \left ( 1 + \frac1{3!} + \frac{2^2}{5!} + \cdots+ \right) \\ \Rightarrow&& S &= 2 \frac{2\frac{\pi}{6}}{\sqrt{3}} = \frac{2\pi}{3\sqrt{6}} \end{align*}
2013 Paper 3 Q3
The four vertices \(P_i\) (\(i= 1, 2, 3, 4\)) of a regular tetrahedron lie on the surface of a sphere with centre at \(O\) and of radius 1. The position vector of \(P_i\) with respect to \(O\) is \({\bf p}_i\) (\(i= 1, 2, 3, 4\)). Use the fact that \({\bf p}_1+ {\bf p}_2+{\bf p}_3+{\bf p}_4={\bf 0}\,\) to show that \({\bf p}_i \,.\, {\bf p}_j =-\frac13\,\) for \(i\ne j\). Let \(X\) be any point on the surface of the sphere, and let \(XP_i\) denote the length of the line joining \(X\) and \(P_i\) (\(i= 1, 2, 3, 4\)).
- By writing \((XP_i) ^2\) as \(({\bf p}_i- {\bf x)}\,.\,({\bf p}_i- {\bf x})\), where \({\bf x}\) is the position vector of \(X\) with respect to \(O\), show that \[ \sum_{i=1}^4(XP_i) ^2 =8\,. \]
- Given that \(P_1\) has coordinates \((0,0,1)\) and that the coordinates of \(P_2\) are of the form \((a,0,b)\), where \(a>0\), show that \(a=2\sqrt2/3\) and \(b=-1/3\), and find the coordinates of \(P_3\) and \(P_4\).
- Show that \[ \sum_{i=1}^4 (XP_i)^4 = 4 \sum_{i=1}^4 (1- {\bf x}\,.\,{\bf p}_i)^2\,. \] By letting the coordinates of \(X\) be \( (x,y,z)\), show further that $ \sum\limits_{i=1}^4 (XP_i)^4\( is independent of the position of \)X$.
2013 Paper 3 Q4
Show that \((z-\e^{i\theta})(z-\e^{-i\theta})=z^2 -2z\cos\theta +1\,\). Write down the \((2n)\)th roots of \(-1\) in the form \(\e^{i\theta}\), where \(-\pi <\theta \le \pi\), and deduce that \[ z^{2n} +1 = \prod_{k=1}^n \left(z^2-2z \cos\left( \tfrac{(2k-1)\pi}{2n}\right) +1\right) \,. \] Here, \(n\) is a positive integer, and the \(\prod\) notation denotes the product.
- By substituting \(z=i\) show that, when \(n\) is even, \[ \cos \left(\tfrac \pi {2n}\right) \cos \left(\tfrac {3\pi} {2n}\right) \cos \left(\tfrac {5\pi} {2n}\right) \cdots \cos \left(\tfrac{(2n-1) \pi} {2n}\right) = {(-1\vphantom{\dot A})}^{\frac12 n} 2^{1-n} \,. \]
- Show that, when \(n\) is odd, \[ \cos^2 \left(\tfrac \pi {2n}\right) \cos ^2 \left(\tfrac {3\pi} {2n}\right) \cos ^2 \left(\tfrac {5\pi} {2n}\right) \cdots \cos ^2 \left(\tfrac{(n-2) \pi} {2n}\right) = n 2^{1-n} \,. \] You may use without proof the fact that \(1+z^{2n}= (1+z^2)(1-z^2+z^4 - \cdots + z^{2n-2})\,\) when \(n\) is odd.
Solution: \begin{align*} && (z-e^{i \theta})(z-e^{-i\theta}) &= z^2 - (e^{i\theta}+e^{-i\theta})z + 1 \\ &&&= z^2-2\cos \theta z + 1 \end{align*} The \(2n\)th roots of \(-1\) are \(e^{\frac{i (2k+1)\pi}{2n}}, k \in \{-n, \cdots, n-1 \}\) or \(e^{\frac{i k \pi}{2n}}, k \in \{-2n+1, -2n+3, \cdots, 2n-3, 2n-1 \}\) \begin{align*} && z^{2n}+1 &= (z-e^{-i(2n-1)/2n})\cdot (z-e^{-i(2n-3)/2n})\cdots (z-e^{i(2n-3)/2n})\cdot (z-e^{i(2n-1)/2n}) \\ &&&= \prod_{k=1}^n \left (z - e^{i \frac{2k-1}{2n}\pi} \right)\left (z - e^{-i \frac{2k-1}{2n}\pi} \right)\\ &&&= \prod_{k=1}^n \left (z^2 - 2z \cos \left ( \frac{(2k-1)\pi}{2n}\right) + 1 \right) \end{align*}
- \begin{align*} && i^{2n} + 1 &= \prod_{k=1}^n \left (i^2 - 2i \cos \left ( \frac{(2k-1)\pi}{2n}\right) + 1 \right) \\ \Rightarrow && (-1)^n + 1 &= (-1)^n2^ni^n\prod_{k=1}^n \cos \left ( \frac{(2k-1)\pi}{2n}\right) \\ \Rightarrow && \prod_{k=1}^n \cos \left ( \frac{(2k-1)\pi}{2n}\right) &= 2^{1-n}(-1)^{n/2} \tag{if \(n\equiv 0\pmod{2}\)} \end{align*}
- When \(n\) is odd, we notice that two of the roots are \(i\) and \(-i\), if we exclude those, (ie by factoring out \(z^2+1\), we see that \begin{align*} && 1-z^2+z^4-\cdots + z^{2n-2} &= \prod_{k=1, 2k-1\neq n}^n \left (z^2-2z \cos \left ( \frac{(2k-1)\pi}{2n} \right)+1 \right) \\ &&&= \prod_{k=1}^{(n-1)/2} \left (z^2-2z \cos \left ( \frac{(2k-1)\pi}{2n} \right)+1 \right) \prod_{k=(n+1)/2}^{n} \left (z^2-2z \cos \left ( \frac{(2k-1)\pi}{2n} \right)+1 \right)\\ &&&= \prod_{k=1}^{(n-1)/2} \left (z^2-2z \cos \left ( \frac{(2k-1)\pi}{2n} \right)+1 \right) \prod_{k=1}^{(n-1)/2} \left (z^2+2z \cos \left ( \frac{(2k-1)\pi}{2n} \right)+1 \right)\\ \Rightarrow && 1-i^2 + i^4 + \cdots + i^{2n-2} &= \prod_{k=1}^{(n-1)/2} \left (2 \cos \left ( \frac{(2k-1)\pi}{2n} \right) \right) \prod_{k=1}^{(n-1)/2} \left (2 \cos \left ( \frac{(2k-1)\pi}{2n} \right) \right)\\ \Rightarrow && n &= 2^{n-1} \prod_{k=1}^{(n-1)/2}\cos^2 \left ( \frac{(2k-1)\pi}{2n} \right) \\ \Rightarrow && \prod_{k=1}^{(n-1)/2}\cos^2 \left ( \frac{(2k-1)\pi}{2n} \right) &= n2^{1-n} \end{align*}