298 problems found
Let \[ \mathrm{C}_{n}(\theta)=\sum_{k=0}^{n}\cos k\theta \] and let \[ \mathrm{S}_{n}(\theta)=\sum_{k=0}^{n}\sin k\theta, \] where \(n\) is a positive integer and \(0<\theta<2\pi.\) Show that \[ \mathrm{C}_{n}(\theta)=\frac{\cos(\tfrac{1}{2}n\theta)\sin\left(\frac{1}{2}(n+1)\theta\right)}{\sin(\frac{1}{2}\theta)}, \] and obtain the corresponding expression for \(\mathrm{S}_{n}(\theta)\). Hence, or otherwise, show that for \(0<\theta<2\pi,\) \[ \left|\mathrm{C}_{n}(\theta)-\frac{1}{2}\right|\leqslant\frac{1}{2\sin(\frac{1}{2}\theta)}. \]
Solution: \begin{align*} && C_n(\theta) &= \sum_{k=0}^n \cos k \theta \\ &&&= \textrm{Re} \left ( \sum_{k=0}^n \exp (ik \theta)\right)\\ &&&= \textrm{Re} \left ( \frac{e^{i(n+1)\theta}-1}{e^{i\theta}-1}\right)\\ &&&= \textrm{Re} \left ( \frac{e^{i(n+1)\theta/2}}{e^{i\theta/2}}\frac{e^{i(n+1)\theta/2}-e^{-i(n+1)\theta/2}}{e^{i\theta/2}-e^{-i\theta/2}}\right)\\ &&&= \textrm{Re} \left ( e^{in\theta/2}\frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\right)\\ &&&= \frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\textrm{Re} \left ( e^{in\theta/2}\right)\\ &&&= \frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\cos \left ( \frac12n\theta\right)\\ \\ && S_n(\theta) &= \sum_{k=0}^n \sin k \theta \\ &&&= \textrm{Im} \left ( \sum_{k=0}^n \exp (ik \theta)\right)\\ &&&= \frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\textrm{Im} \left ( e^{in\theta/2}\right)\\ &&&= \frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\sin\left ( \frac12n\theta\right)\\ \\ && C_n(\theta) - \frac12 &= \frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\cos \left ( \frac12n\theta\right) - \frac12 \\ &&&= \frac{2\sin \left ( (n+1)\theta/2 \right)\cos\left ( n\theta/2 \right)-\sin (\theta/2)}{2 \sin (\theta/2)} \\ &&&= \frac{\sin\left ( (n+1)\theta/2+n\theta/2\right)+\sin\left ( (n+1)\theta/2-n\theta/2\right)-\sin (\theta/2)}{2 \sin (\theta/2)} \\ &&&= \frac{\sin\left ( (n+1)\theta/2+n\theta/2\right)+\sin\left ( \theta/2\right)-\sin (\theta/2)}{2 \sin (\theta/2)} \\ &&&= \frac{\sin\left ( (2n+1)\theta/2\right)}{2 \sin (\theta/2)} \leqslant\frac{1}{2 \sin (\theta/2)} \\ \end{align*}
A cannon-ball is fired from a cannon at an initial speed \(u\). After time \(t\) it has reached height \(h\) and is at a distance \(\sqrt{x^{2}+h^{2}}\) from the cannon. Ignoring air resistance, show that \[ \tfrac{1}{4}g^{2}t^{4}-(u^{2}-gh)t^{2}+h^{2}+x^{2}=0. \] Hence show that if \(u^{2}>2gh\) then the horizontal range for a given height \(h\) and initial speed \(u\) is less than or equal to \[ \frac{u\sqrt{u^{2}-2gh}}{g}. \] Show that there is always an angle of firing for which this value is attained.
Solution: Suppose it is fired with angle to the horizontal \(\alpha\), then \begin{align*} \rightarrow: && x &= u\cos \alpha \cdot t \\ \uparrow: && h &= u \sin \alpha \cdot t - \frac12 g t^2 \\ \Rightarrow && u\cos \alpha &= \frac{x}{t} \\ && u \sin \alpha &= \frac{h + \frac12 gt^2}{t} \\ \Rightarrow && u^2 &= \frac{x^2}{t^2} + \frac{(h + \frac12 gt^2)^2}{t^2} \\ \Rightarrow && 0 &= x^2+h^2-u^2t^2+ght^2+\tfrac14 g^2 t^4 \\ &&&= \tfrac14 g^2 t^4 - (u^2 - gh)t^2 + h^2 + x^2 \end{align*} For a distance \(x\) to be achievable there must be a root to this quadratic in \(t^2\), ie \begin{align*} && 0 &\leq \Delta = (u^2-gh)^2 - 4 \cdot \tfrac14 g^2 (h^2 + x^2) \\ \Rightarrow && x^2 &\leq \frac{(u^2-gh)^2}{g^2} - h^2 \\ &&&= \frac{u^4+g^2h^2 - 2ghu^2-g^2h^2}{g^2} \\ &&&= \frac{u^2(u^2-2gh)}{g^2} \\ \Rightarrow && x &\leq \frac{u\sqrt{u^2-2gh}}{g} \end{align*} This is achieved when \begin{align*} && t^2 &= \frac{u^2-gh}{\tfrac12g^2}\\ &&&= \frac{2(u^2-gh)}{g^2} \\ \Rightarrow && \cos \alpha &= \frac{u\sqrt{u^2-2gh}}{g} \cdot \frac{g}{\sqrt{2(u^2-gh)}} \frac{1}{u} \\ &&&= \frac{1}{\sqrt{2}} \end{align*} ie when \(\alpha = \frac{\pi}{4}\)
One end \(A\) of a light elastic string of natural length \(l\) and modulus of elasticity \(\lambda\) is fixed and a particle of mass \(m\) is attached to the other end \(B\). The particle moves in a horizontal circle with centre on the vertical through \(A\) with angular velocity \(\omega.\) If \(\theta\) is the angle \(AB\) makes with the downward vertical, find an expression for \(\cos\theta\) in terms of \(m,g,l,\lambda\) and \(\omega.\) Show that the motion described is possible only if \[ \frac{g\lambda}{l(\lambda+mg)}<\omega^{2}<\frac{\lambda}{ml}. \]
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`24 Hour Spares' stocks a small, widely used and cheap component. Every \(T\) hours \(X\) units arrive by lorry from the wholesaler, for which the owner pays a total \(\pounds (a+qX)\). It costs the owner \(\pounds b\) per hour to store one unit. If she has the units in stock she expects to sell \(r\) units per hour at \(\pounds(p+q)\) per unit. The other running costs of her business remain at \(\pounds c\) pounds an hour irrespective of whether she has stock or not. (All of the quantities \(T,X,a,b,r,q,p\) and \(c\) are greater than 0.) Explain why she should take \(X\leqslant rT\). Given that the process may be assumed continuous (the items are very small and she sells many each hour), sketch \(S(t)\) the amount of stock remaining as a function of \(t\) the time from the last delivery. Compute the total profit over each period of \(T\) hours. Show that, if \(T\) is fixed with \(T\geqslant p/b\), the business can be made profitable if \[ p^{2}>2\frac{(a+cT)b}{r}. \]
The makers of Cruncho (`The Cereal Which Cares') are giving away a series of cards depicting \(n\) great mathematicians. Each packet of Cruncho contains one picture chosen at random. Show that when I have collected \(r\) different cards the expected number of packets I must open to find a new card is \(n/(n-r)\) where \(0\leqslant r\leqslant n-1.\) Show by means of a diagram, or otherwise, that \[ \frac{1}{r+1}\leqslant\int_{r}^{r+1}\frac{1}{x}\,\mathrm{d}x\leqslant\frac{1}{r} \] and deduce that \[ \sum_{r=2}^{n}\frac{1}{r}\leqslant\ln n\leqslant\sum_{r=1}^{n-1}\frac{1}{r} \] for all \(n\geqslant2.\) My children will give me no peace until we have the complete set of cards, but I am the only person in our household prepared to eat Cruncho and my spouse will only buy the stuff if I eat it. If \(n\) is large, roughly how many packets must I expect to consume before we have the set?
The function \(\mathrm{f}\) is given by \(\mathrm{f}(x)=\sin^{-1}x\) for \(-1 < x < 1.\) Prove that \[ (1-x^{2})\mathrm{f}''(x)-x\mathrm{f}'(x)=0. \] Prove also that \[ (1-x^{2})\mathrm{f}^{(n+2)}(x)-(2n+1)x\mathrm{f}^{(n+1)}(x)-n^{2}\mathrm{f}^{(n)}(x)=0, \] for all \(n>0\), where \(\mathrm{f}^{(n)}\) denotes the \(n\)th derivative of \(\mathrm{f}\). Hence express \(\mathrm{f}(x)\) as a Maclaurin series. The function \(\mathrm{g}\) is given by \[ \mathrm{g}(x)=\ln\sqrt{\frac{1+x}{1-x}}, \] for \(-1 < x < 1.\) Write down a power series expression for \(\mathrm{g}(x),\) and show that the coefficient of \(x^{2n+1}\) is greater than that in the expansion of \(\mathrm{f},\) for each \(n > 0\).
Three points, \(P,Q\) and \(R\), are independently randomly chosen on the perimeter of a circle. Prove that the probability that at least one of the angles of the triangle \(PQR\) will exceed \(k\pi\) is \(3(1-k)^{2}\) if \(\frac{1}{2}\leqslant k\leqslant1.\) Find the probability if \(\frac{1}{3}\leqslant k\leqslant\frac{1}{2}.\)
Solution:
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 \(\left|x\right|>\frac{1}{2}.\)
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 a clay pigeon shoot the target is launched vertically from ground level with speed \(v\). At a time \(T\) later the competitor fires a rifle inclined at angle \(\alpha\) to the horizontal. The competitor is also at ground level and is a distance \(l\) from the launcher. The speed of the bullet leaving the rifle is \(u\). Show that, if the competitor scores a hit, then \[ l\sin\alpha-\left(vT-\tfrac{1}{2}gT^{2}\right)\cos\alpha=\frac{v-gT}{u}l. \] Suppose now that \(T=0\). Show that if the competitor can hit the target before it hits the ground then \(v < u\) and \[ \frac{2v\sqrt{u^{2}-v^{2}}}{g}>l. \]
In the game of ``Colonel Blotto'' there are two players, Adam and Betty. First Adam chooses three non-negative integers \(a_{1},a_{2}\) and \(a_{3},\) such that \(a_{1}+a_{2}+a_{3}=9,\) and then Betty chooses non-negative integers \(b_{1},b_{2}\) and \(b_{3}\), such that \(b_{1}+b_{2}+b_{3}=9.\) If \(a_{1} > b_{1}\) then Adam scores one point; if \(a_{1} < b_{1}\) then Betty scores one point; and if \(a_{1}=b_{1}\) no points are scored. Similarly for \(a_{2},b_{2}\) and \(a_{3},b_{3}.\) The winner is the player who scores the greater number of points: if the socres are equal then the game is drawn. Show that, if Betty knows the numbers \(a_{1},a_{2}\) and \(a_{3},\) she can always choose her numbers so that she wins. Show that Adam can choose \(a_{1},a_{2}\) and \(a_{3}\) in such a way that he will never win no matter what Betty does. Now suppose that Adam is allowed to write down two triples of numbers and that Adam wins unless Betty can find one triple that beats both of Adam's choices (knowing what they are). Confirm that Adam wins by writing down \((5,3,1)\) and \((3,1,5).\)
Suppose that \(a_{i}>0\) for all \(i>0\). Show that \[ a_{1}a_{2}\leqslant\left(\frac{a_{1}+a_{2}}{2}\right)^{2}. \] Prove by induction that for all positive integers \(m\) \[ a_{1}\cdots a_{2^{m}}\leqslant\left(\frac{a_{1}+\cdots+a_{2^{m}}}{2^{m}}\right)^{2^{m}}.\tag{*} \] If \(n<2^{m}\), put \(b_{1}=a_{2},\) \(b_{2}=a_{2},\cdots,b_{n}=a_{n}\) and \(b_{n+1}=\cdots=b_{2^{m}}=A\), where \[ A=\frac{a_{1}+\cdots+a_{n}}{n}. \] By applying \((*)\) to the \(b_{i},\) show that \[ a_{1}\cdots a_{n}A^{(2^{m}-n)}\leqslant A^{2^{m}} \] (notice that \(b_{1}+\cdots+b_{n}=nA).\) Deduce the (arithmetic mean)/(geometric mean) inequality \[ \left(a_{1}\cdots a_{n}\right)^{1/n}\leqslant\frac{a_{1}+\cdots+a_{n}}{n}. \]
Solution: \begin{align*} && 0 &\leqslant (a_1 - a_2)^2 \\ &&&= a_1^2 -2a_1a_2 + a_2^2 \\ &&&= (a_1+a_2)^2 -4a_1a_2 \\ \Leftrightarrow && a_1a_2 &\leqslant \left ( \frac{a_1+a_2}2 \right)^2 \end{align*} Claim: \((*)\) is true Proof: (By induction) We have already proven the base case. Suppose it is true for some \(m\), then consider \(m+1\) \begin{align*} && a_1 \cdots a_{2^m} &\leqslant \left ( \frac{a_1 + \cdots + a_{2^m}}{2^m} \right)^{2^m} \tag{by (*)} \\ && a_{2^m+1} \cdots a_{2^{m+1}} &\leqslant \left ( \frac{a_{2^m+1} + \cdots + a_{2^{m+1}}}{2^m} \right)^{2^m} \tag{by (*)} \\ \Rightarrow && (a_1 \cdots a_{2^m})^{1/2^m} &\leqslant \left ( \frac{a_1 + \cdots + a_{2^m}}{2^m} \right) \\ && (a_{2^m+1} \cdots a_{2^{m+1}})^{1/2^m} &\leqslant \left ( \frac{a_{2^m+1} + \cdots + a_{2^{m+1}}}{2^m} \right) \\ \Rightarrow && (a_1 \cdots a_{2^m})^{1/2^m} \cdot (a_{2^m+1} \cdots a_{2^{m+1}})^{1/2^m} &\leqslant \left ( \frac{ (a_1 \cdots a_{2^m})^{1/2^m} +(a_{2^m+1} \cdots a_{2^{m+1}})^{1/2^m} }{2} \right )^2 \\ &&&\leqslant \left ( \frac{ \frac{a_1 + \cdots + a_{2^m}}{2^m}+\frac{a_{2^m+1} + \cdots + a_{2^{m+1}}}{2^m} }{2} \right )^2 \\ &&&\leqslant \left ( \frac{ a_1 + \cdots + a_{2^m}+a_{2^m+1} + \cdots + a_{2^{m+1}} }{2^{m+1}} \right )^2 \\ \Rightarrow && a_1 \cdots a_{2^{m+1}} &\leqslant \left ( \frac{a_1 + \cdots + a_{2^{m+1}}}{2^{m+1}} \right)^{2^{m+1}} \end{align*} Which is precisely \((*)\) for \(m+1\). Therefore our statement is true by induction. Suppose \(n < 2^m\) and \(b_1 = a_1, b_2 = a_2, \cdots b_n = a_n\) and \(b_{n+1} = \cdots = b_{2^m} = A\) where \(A = \frac{a_1 + \cdots + a_n}{n}\) then \begin{align*} && b_1 \cdots b_n \cdot b_{n+1} \cdots b_{2^m} &\leq \left ( \frac{b_1 + \cdots + b_n + b_{n+1} + \cdots + b_{2^m}}{2^{m}} \right)^{2^m} \\ \Leftrightarrow && a_1 \cdots a_n \cdot A^{2^m-n} &\leq \left ( \frac{a_1 + \cdots + a_n + (2^m-n)A}{2^m} \right)^{2^m} \\ &&&= \left ( \frac{nA + (2^m - n)A}{2^m} \right)^{2^m} \\ &&&= A^{2^m} \\ \Rightarrow && a_1 \cdots a_n &\leq A^n \\ \Rightarrow && (a_1 \cdots a_n)^{1/n} &\leq A = \frac{a_1 + \cdots + a_n}{n} \end{align*}
\textit{In this question, the argument of a complex number is chosen to satisfy \(0\leqslant\arg z<2\pi.\)} Let \(z\) be a complex number whose imaginary part is positive. What can you say about \(\arg z\)? The complex numbers \(z_{1},z_{2}\) and \(z_{3}\) all have positive imaginary part and \(\arg z_{1}<\arg z_{2}<\arg z_{3}.\) Draw a diagram that shows why \[ \arg z_{1}<\arg(z_{1}+z_{2}+z_{3})<\arg z_{3}. \] Prove that \(\arg(z_{1}z_{2}z_{3})\) is never equal to \(\arg(z_{1}+z_{2}+z_{3}).\)