1531 problems found
Sum the following infinite series.
Solution:
The set \(S\) consists of ordered pairs of complex numbers \((z_1,z_2)\) and a binary operation \(\circ\) on \(S\) is defined by $$ (z_1,z_2)\circ(w_1,w_2)= (z_1w_1-z_2w^*_2, \; z_1w_2+z_2w^*_1). $$ Show that the operation \(\circ\) is associative and determine whether it is commutative. Evaluate \((z,0)\circ(w,0)\), \((z,0)\circ(0,w)\), \((0,z)\circ(w,0)\) and \((0,z)\circ(0,w)\). The set \(S_1\) is the subset of \(S\) consisting of \(A\), \(B\), \(\ldots\,\), \(H\), where \(A=(1,0)\), \(B=(0,1)\), \(C=(i,0)\), \(D=(0,i)\), \(E=(-1,0)\), \(F=(0,-1)\), \(G=(-i,0)\) and \(H=(0,-i)\). Show that \(S_1\) is closed under \(\circ\) and that it has an identity element. Determine the inverse and order of each element of \(S_1\). Show that \(S_1\) is a group under \(\circ\). \hfil\break [You are not required to compute the multiplication table in full.] Show that \(\{A,B,E,F\}\) is a subgroup of \(S_1\) and determine whether it is isomorphic to the group generated by the \(2\times2\) matrix $\begin{pmatrix}0 & 1\\ -1 & 0 \end{pmatrix}$ under matrix multiplication.
The point in the Argand diagram representing the complex number \(z\) lies on the circle with centre \(K\) and radius \(r\), where \(K\) represents the complex number \(k\). Show that $$ zz^* -kz^* -k^*z +kk^* -r^2 =0. $$ The points \(P\), \(Q_1\) and \(Q_2\) represent the complex numbers \(z\), \(w_1\) and \(w_2\) respectively. The point \(P\) lies on the circle with \(OA\) as diameter, where \(O\) and \(A\) represent \(0\) and \(2i\) respectively. Given that \(w_1=z/(z-1)\), find the equation of the locus \(L\) of \(Q_1\) in terms of \(w_1\) and describe the geometrical form of \(L\). Given that \(w_2=z^*\), show that the locus of \(Q_2\) is also \(L\). Determine the positions of \(P\) for which \(Q_1\) coincides with \(Q_2\).
The real numbers \(x\) and \(y\) satisfy the simultaneous equations $$ \sinh (2x) = \cosh y \qquad\hbox{and}\qquad \sinh(2y) = 2 \cosh x. $$ Show that \(\sinh^2 y\) is a root of the equation $$ 4t^3 + 4t^2 -4t -1=0 $$ and demonstrate that this gives at most one valid solution for \(y\). Show that the relevant value of \(t\) lies between \(0.7\) and \(0.8\), and use an iterative process to find \(t\) to 6 decimal places. Find \(y\) and hence find \(x\), checking your answers and stating the final answers to four decimal places.
Solution: Let \(t = \sinh^2 y\), then \begin{align*} && \sinh(2x) &= \cosh y \tag{1}\\ && \sinh(2y) &= 2 \cosh x \tag{2} \\ \\ && \cosh(2x) &= 2 \cosh^2 x -1 \\ (2): &&&= \frac12 \sinh^2(2y) -1 \\ && 1 &= \left (\frac12 \sinh^2(2y) -1 \right)^2 - \cosh^2 y \\ &&&= \frac14 \sinh^4(2y)-\sinh^2(2y)+1-\cosh^2 y \\ \Rightarrow && 0 &= \frac14 (\cosh^2 (2y)-1)^2- (\cosh^2 (2y)-1) - \cosh^2 y \\ &&&= \frac14 \left ( \left (1+2\sinh^2 y \right)^2-1 \right)^2 -\left ( \left (1+2\sinh^2 y \right)^2 -1\right) - (1 + \sinh^2 y ) \\ &&&= \frac14 \left ( 1 + 4t+4t^2 -1\right)^2 - \left ( 1+4t+4t^2-1\right) - (1 + t) \\ &&&= \frac14 (4t + 4t^2)^2 - (4t+4t^2)-1-t \\ &&&= 4(t+t^2)^2 - 4t^2-5t-1 \\ &&&= 4t^4+8t^3+4t^2-4t^2-5t-1 \\ &&& = 4t^4+8t^3-5t-1 \\ &&&= (t+1)(4t^3+4t^2-4t-1) \end{align*} Since \(\sinh^2 y\) is positive, we must be a root of the second cubic. Let \(f(t) = 4t^3+4t^2-4t-1\), then \(f(0) = -1\) and \(f'(t) = 12t^2+8t-4 = 4(t+1)(3t-1)\), so we have turning points at \(-1\) and \(\frac13\). Since \(f(-1) = 3 > 0\) and \(f(0) < 0\) we must have exactly one root larger than zero. Therefore there is a unique root. \(f(0.7) = -0.468 < 0\) \(f(0.8) = 0.408 > 0\) since \(f\) is continuous and changes sign, the root must fall in the interval \((0.7, 0.8)\). Let \(t_{n+1} = t_n - \frac{f(t_n)}{f'(t_n)}\), and \(t_0 = 0.75\), then \begin{align*} t_0 &= 0.75 \\ t_1 &= 0.7571428571 \\ t_2 &= 0.7570684728 \\ t_3 &= 0.7570684647 \end{align*} So \(t \approx 0.757068\), \(\sinh y \approx 0.870097\), \(y \approx 0.786474\), \(x \approx 0.546965\)
A square pyramid has its base vertices at the points \(A\) \((a,0,0)\), \(B\) \((0,a,0)\), \(C\) \((-a,0,0)\) and \(D\) \((0,-a,0)\), and its vertex at \(E\) \((0,0,a)\). The point \(P\) lies on \(AE\) with \(x\)-coordinate \(\lambda a\), where \(0<\lambda<1\), and the point \(Q\) lies on \(CE\) with \(x\)-coordinate \(-\mu a\), where \(0<\mu<1\). The plane \(BPQ\) cuts \(DE\) at \(R\) and the \(y\)-coordinate of \(R\) is \(-\gamma a\). Prove that $$ \gamma = {\lambda \mu \over \lambda + \mu - \lambda \mu}. $$ Show that the quadrilateral \(BPRQ\) cannot be a parallelogram.
For the real numbers \(a_1\), \(a_2\), \(a_3\), \(\ldots\),
The transformation \(T\) of the point \(P\) in the \(x\),\(y\) plane to the point \(P'\) is constructed as follows: \hfil\break Lines are drawn through \(P\) parallel to the lines \(y=mx\) and \(y=-mx\) to cut the line \(y=kx\) at \(Q\) and \(R\) respectively, \(m\) and \(k\) being given constants. \(P'\) is the fourth vertex of the parallelogram \(PQP'R\). Show that if \(P\) is \((x_1,y_1)\) then \(Q\) is $$ \left( {mx_1-y_1 \over m-k}, {k(mx_1-y_1)\over m-k}\right). $$ Obtain the coordinates of \(P'\) in terms of \(x_1\), \(y_1\), \(m\) and \(k\), and express \(T\) as a matrix transformation. Show that areas are transformed under \(T\) into areas of the same magnitude.
In this question, all gravitational forces are to be neglected. A rigid frame is constructed from 12 equal uniform rods, each of length \(a\) and mass \(m,\) forming the edges of a cube. Three of the edges are \(OA,OB\) and \(OC,\) and the vertices opposite \(O,A,B\) and \(C\) are \(O',A',B'\) and \(C'\) respectively. Forces act along the lines as follows, in the directions indicated by the order of the letters: \begin{alignat*}{3} 2mg\mbox{ along }OA, & \qquad & mg\mbox{ along }AC', & \qquad & \sqrt{2}mg\mbox{ along }O'A,\\ \sqrt{2}mg\mbox{ along }OA', & & 2mg\mbox{ along }C'B, & & mg\mbox{ along }A'C. \end{alignat*}
\(ABCD\) is a horizontal line with \(AB=CD=a\) and \(BC=6a\). There are fixed smooth pegs at \(B\) and \(C\). A uniform string of natural length \(2a\) and modulus of elasticity \(kmg\) is stretched from \(A\) to \(D\), passing over the pegs at \(B\) and \(C\). A particle of mass \(m\) is attached to the midpoint \(P\) of the string. When the system is in equilibrium, \(P\) is a distance \(a/4\) below \(BC\). Evaluate \(k\). The particle is pulled down to a point \(Q\), which is at a distance \(pa\) below the mid-point of \(BC\), and is released from rest. \(P\) rises to a point \(R\), which is at a distance \(3a\) above \(BC\). Show that \(2p^2-p-17=0\). Show also that the tension in the strings is less when the particle is at \(R\) than when the particle is at \(Q\).
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A particle rests at a point \(A\) on a horizontal table and is joined to a point \(O\) on the table by a taut inextensible string of length \(c\). The particle is projected vertically upwards at a speed \(64\surd(6gc)\). It next strikes the table at a point \(B\) and rebounds. The coefficient of restitution for any impact between the particle and the table is \({1\over 2}\). After rebounding at \(B\), the particle will rebound alternately at \(A\) and \(B\) until the string becomes slack. Show that when the string becomes slack the particle is at height \(c/2\) above the table. Determine whether the first rebound between \(A\) and \(B\) is nearer to \(A\) or to \(B\).
Solution: \begin{align*} \text{N2}(radially): && T + mg \cos \theta &= m\frac{v^2}{r} \\ \Rightarrow && v^2-gc \cos \theta &\geq 0 \\ \text{COE}: && \frac12 m u^2 &= \frac12mv^2 + mgc\cos \theta \\ \Rightarrow && u^2 &= v^2 + 2gc\cos \theta \\ && u^2 &\geq gc \cos \theta+2gc\cos \theta \\ \Rightarrow && u^2 &\geq 3gc\cos \theta \end{align*} Therefore it will complete bounces with the string taught if it leaves the table with \(u^2 \geq 3gc\). After \(6\) bounces it will leave the table with speed \(\sqrt{6gc} > \sqrt{3gc}\) and after \(7\) bounces it will leave the table with speed \(\sqrt{\frac{3}{2} gc} < \sqrt{3 gc}\). When it leaves the table with speed \(\sqrt{\tfrac32 gc}\), the string will go slack when \begin{align*} && \tfrac32 gc &= 3gc \cos \theta \\ \Rightarrow && \cos \theta &= \frac{1}{2} \end{align*} ie at a height \(c\cos \theta = \frac12c\) above the table. Once the string goes slack, the particle travels under circular motion, \begin{align*} && u^2 &= \frac12 gc \\ \Rightarrow && u_\rightarrow &= \sqrt{\tfrac12 gc} \cos \theta \\ && u_{\uparrow} &= \sqrt{\tfrac12 gc} \sin \theta \\ \Rightarrow && s &= ut - \tfrac12 gt^2 \\ \Rightarrow && -\frac{c}{2} &= \sqrt{\tfrac12 gc} \frac{\sqrt{3}}{2} t - \tfrac12 g t^2 \\ \Rightarrow && t &= \sqrt{\frac{c}{g}} \left ( \frac{\sqrt{6}+\sqrt{22}}{4} \right) \\ \Rightarrow && s_{\rightarrow} &= \tfrac12 \sqrt{\tfrac12 gc} \sqrt{\frac{c}{g}} \left ( \frac{\sqrt{6}+\sqrt{22}}{4} \right) - \frac{\sqrt{3}}{2}c \\ &&&= \left ( \frac{\sqrt{11}-3\sqrt{3}}{8} \right)c \\ \end{align*} We need to establish whether this position is positive or negative (ie if we cross the centre line. Clearly \(\sqrt{11} < 3\sqrt{3}\) so we haven't crossed the centre line and we land closer to where we took off. Since it's the 7th take off, this is closer to \(B\).
The probability of throwing a head with a certain coin is \(p\) and the probability of throwing a tail is \(q=1-p\). The coin is thrown until at least two heads and at least two tails have been thrown; this happens when the coin has been thrown \(N\) times. Write down an expression for the probability that \(N=n\). Show that the expectation of \(N\) is $$ 2\bigg({1\over pq} -1-pq\bigg). $$
Solution: This can either occur via \(N-2\) heads and \(1\) tail in the first \(N-1\) flips, followed by a tail, or \(N-2\) tails and \(1\) head in the first \(N-1\) flips, followed by another head, ie \begin{align*} \mathbb{P}(N = n) &= \underbrace{\binom{n-1}{1}}_{\text{ways to choose when the first tail occurs}}p^{n-2}q^2 + \underbrace{\binom{n-1}{1}}_{\text{ways to choose when the first head occurs}}q^{n-2}p^2 \\ &= (n-1)p^2q^2(p^{n-4}+q^{n-4}) \\ \\ \mathbb{E}(N) &= \sum_{n=4}^{\infty} n \cdot \mathbb{P}(N = n) \\ &= \sum_{n=4}^{\infty} n \cdot (n-1)p^2q^2(p^{n-4}+q^{n-4}) \\ &= \sum_{n=4}^{\infty} n \cdot (n-1)(p^{n-2}q^2+q^{n-2}p^2) \\ &= q^2\sum_{n=4}^{\infty} n(n-1)p^{n-2}+p^2\sum_{n=4}^{\infty} n(n-1)q^{n-2} \\ &= q^2\left ( \sum_{n=2}^{\infty} n(n-1)p^{n-2} -2 \cdot 1 - 3 \cdot 2 \cdot p\right)+p^2\left ( \sum_{n=2}^{\infty} n(n-1)q^{n-2} - 2-6q\right) \\ &= q^2\left ( 2(1-p)^{-3} -2 - 6 p\right)+p^2\left ( 2(1-q)^{-3} - 2-6q\right) \\ &= q^2\left ( 2q^{-3} -2 - 6 p\right)+p^2\left ( 2p^{-3} - 2-6q\right) \\ &= \frac{2}{q} - 2q^2 - 6pq^2+\frac{2}{p} -2p^2-6p^2q \\ &= \frac{2}{q}+\frac2p - 2(p^2+q^2) - 6pq \\ &= \frac{2}{pq} - 2((p+q)^2-2pq) - 6pq \\ &= \frac{2}{pq} - 2 -2pq \\ &= 2 \left (\frac1{pq} - 1 - pq \right) \end{align*}
The time taken for me to set an acceptable examination question it \(T\) hours. The distribution of \(T\) is a truncated normal distribution with probability density \(\f\) where \[ \mathrm{f}(t)=\begin{cases} \dfrac{1}{k\sigma\sqrt{2\pi}}\exp\left(-\dfrac{1}{2}\left(\dfrac{t-\sigma}{\sigma}\right)^{2}\right) & \mbox{ for }t\geqslant0\\ 0 & \mbox{ for }t<0. \end{cases} \] Sketch the graph of \(\f(t)\). Show that \(k\) is approximately \(0.841\) and obtain the mean of \(T\) as a multiple of \(\sigma\). Over a period of years, I find that the mean setting time is 3 hours.
Today's date is June 26th 1992 and the day of the week is Friday. Find which day of the week was April 3rd 1905, explaining your method carefully. {[}30 days hath September, April, June and November. All the rest have 31, excepting February alone which has 28 days clear and 29 in each leap year.{]}
Solution: There are \(87\) years between 1905 and 1992. Of those years, every 4th is a leap years, starting with 1908, and ending with 1992, so there are 22 leap years. There are 30 days between Apr 3 and May 3, 31 days between May 3 and Jun 3 and a further 23 days to the 26th. Therefore \(87 \times 365 + 22 \cdot 1 + 61 + 23\) total days. \(\pmod{7}\) this is \(3 \times 1 + 1 + 5 + 2 = 3\), therefore it is \(4\) week days before Friday, ie Monday.
A \(3\times3\) magic square is a \(3\times3\) array \[ \begin{array}{ccc} a & b & c\\ d & e & f\\ g & h & k \end{array} \] whose entries are the nine distinct integers \(1,2,3,4,5,6,7,8,9\) and which has the property that all its rows, columns and main diagonals add up to the same number \(n\). (Thus \(a+b+c=d+e+f=g+h+k=a+d+g=b+e+h=c+f+k=a+e+k=c+e+g=n.)\)
Solution: