87 problems found
Show that setting \(z - z^{-1}=w\) in the quartic equation \[ z^4 +5z^3 +4z^2 -5z +1=0 \] results in the quadratic equation \(w^2+5w+6=0\). Hence solve the above quartic equation. Solve similarly the equation \[ 2z^8 -3z^7-12z^6 +12z^5 +22z^4-12z^3 -12 z^2 +3z +2=0 \;. \]
Solution: \begin{align*} && 0 &= z^4 +5z^3 +4z^2 -5z +1 \\ &&0 &= z^2 + z^{-2} + 5(z-z^{-1}) + 4 \\ &&&= (z-z^{-1})^2+2+5(z-z^{-1})+4 \\ &&&= w^2 + 5w + 6 \\ &&&= (w+3)(w+2) \\ \Rightarrow && 0 &= z-z^{-1}+3 \\ \Rightarrow && 0 &= z^2+3z-1 \\ \Rightarrow && z &= \frac{-3 \pm \sqrt{3^2+4}}{2} = \frac{-3 \pm \sqrt{13}}{2} \\ \Rightarrow && 0 &= z-z^{-1}+2 \\ \Rightarrow && 0 &= z^2+2z-1 \\ \Rightarrow && z &= \frac{-2 \pm \sqrt{2^2+4}}{2} = - 1 \pm \sqrt{2} \\ \end{align*} \begin{align*} &&0 &= 2z^8 -3z^7-12z^6 +12z^5 +22z^4-12z^3 -12 z^2 +3z +2 \\ && 0 &= 2(z^4+z^{-4}) - 3(z^3-z^{-3})-12(z^2+z^{-2})+12(z-z^{-1})+22 \\ &&&= 2\left ((z-z^{-1})^4+4(z^2+z^{-2})-6\right)-3 \left ((z-z^{-1})^3+3(z-z^{-1}) \right)-12 \left ((z-z^{-1})^2+2 \right)+12(z-z^{-1})+22 \\ &&&= 2(w^4+4(w^2+2)-6)-3w^3-9w-12w^2-24+12w+22 \\ &&&= 2 w^4-3w^3-4w^2+3w+2 \\ \Rightarrow && 0 &= 2(w^2+w^{-2})-3(w-w^{-1})-4 \\ &&&= 2((w-w^{-1})^2+2)-3(w-w^{-1})-4 \\ &&&= 2x^2-3x \\ &&&= x(2x-3) \\ \Rightarrow && 0 &= w -w^{-1} \\ \Rightarrow && w &= \pm 1 \\ \Rightarrow && \pm 1 &= z-z^{-1} \\ \Rightarrow && 0 &= z^2 \mp z-1 \\ \Rightarrow && z &= \frac{\pm 1 \pm \sqrt{5}}{2} \\ \Rightarrow && \frac32 &= w-w^{-1} \\ \Rightarrow && 0 &= 2w^2-3w -2 \\ &&&= (2w+1)(w-2) \\ \Rightarrow && 2 &= z-z^{-1} \\ \Rightarrow && 0 &= z^2-2z-1 \\ \Rightarrow && z &= 1 \pm \sqrt{2} \\ \Rightarrow && -\frac12 &= z-z^{-1} \\ \Rightarrow && 0 &= 2z^2+z-2 \\ \Rightarrow && z &= \frac{-1 \pm \sqrt{17}}{4} \\ \Rightarrow && z &\in \left \{ \frac{\pm 1 \pm \sqrt{5}}{2}, 1 \pm \sqrt{2}, \frac{-1 \pm \sqrt{17}}{4} \right \} \end{align*}
Show that if \(x\) and \(y\) are positive and \(x^3 + x^2 = y^3 - y^2\) then \(x < y\,\). Show further that if \(0 < x \le y - 1\), then \(x^3 + x^2 < y^3 - y^2\). Prove that there does not exist a pair of {\sl positive} integers such that the difference of their cubes is equal to the sum of their squares. Find all the pairs of integers such that the difference of their cubes is equal to the sum of their squares.
Give a condition that must be satisfied by \(p\), \(q\) and \(r\) for it to be possible to write the quadratic polynomial \(px^2 + qx + r\) in the form \(p \l x + h \r^2\), for some \(h\). Obtain an equation, which you need not simplify, that must be satisfied by \(t\) if it is possible to write \[ \l x^2 + \textstyle{{1 \over 2}} bx + t \r^2 - \l x^4 + bx^3 + cx^2 +dx +e \r \] in the form \(k \l x + h \r^2\), for some \(k\) and \(h\). Hence, or otherwise, write \(x^4 + 6x^3 + 9x^2 -2x -7\) as a product of two quadratic factors.
Show that the coefficient of \(x^{-12}\) in the expansion of \[ \left(x^{4}-\frac{1}{x^{2}}\right)^{5} \left(x-\frac{1}{x}\right)^{6} \] is \(-15\), and calculate the coefficient of \(x^2\). Hence, or otherwise, calculate the coefficients of \(x^4\) and \(x^{38}\) in the expansion of \[ (x^2-1)^{11}(x^4+x^2+1)^5. \]
Solution: The powers of \(x\) in the first bracket will be \(x^{20}, x^{14}, \cdots, x^{-10}\). The powers of \(x\) in the second bracket will be \(x^6, x^4, \cdots, x^{-6}\). Therefore we can achieve \(x^{-12}\) in only one way: \begin{array}{c|c|c|c|c} 1\text{st bracket} & 2\text{nd bracket} & 1\text{st coef} & 2\text{nd coef} & \text{prod} \\ \hline x^{-10} & x^{-2} & \binom{5}{5}(-1)^5 = -1 & \binom{6}{4}(-1)^4 = 15& -15 \\ \end{array} We can achieve \(x^2\) as follows: \begin{array}{c|c|c|c|c} 1\text{st bracket} & 2\text{nd bracket} & 1\text{st coef} & 2\text{nd coef} & \text{prod} \\ \hline x^{-4} & x^{6} & \binom{5}{4}(-1)^4 = 5 & \binom{6}{0}(-1)^0 = 1& 5 \\ x^{2} & x^{0} & \binom{5}{3}(-1)^3 = -10 & \binom{6}{3}(-1)^3 = -20 & 200 \\ x^{8} & x^{-6} & \binom{5}{2}(-1)^2 = 10 & \binom{6}{6}(-1)^6 = 1 & 10 \end{array} Therefore the coefficient is \(215\) \((x^2-1)(x^4+x^2+1) = x^6-1\), therefore \begin{align*} (x^2-1)^{11}(x^4+x^2+1)^5 &= (x^2-1)^6(x^6-1)^5 \\ &= x^6\left(x-\frac1x\right)^6(x^6-1)^6 \\ &= x^6\left(x-\frac1x\right)^6\left(x^2\left(x^4-\frac{1}{x^2}\right)\right)^5 \\ &= x^6\left(x-\frac1x\right)^6x^{10}\left(x^4-\frac{1}{x^2}\right)^5 \\ &= x^{16}\left(x-\frac1x\right)^6\left(x^4-\frac{1}{x^2}\right)^6 \\ \end{align*} Therefore the coefficient of \(x^4\) is the coefficient of \(x^{4-16} = x^{-12}\) in our original expression, ie \(-15\). Similarly, the coefficient of \(x^{38}\) is the coefficient of \(x^{38-16} = x^{22}\), which can only be achieved in one way: \begin{array}{c|c|c|c|c} 1\text{st bracket} & 2\text{nd bracket} & 1\text{st coef} & 2\text{nd coef} & \text{prod} \\ \hline x^{20} & x^{2} & \binom{5}{0}(-1)^0 = 1 & \binom{6}{2}(-1)^2 = 15& 15 \\ \end{array} Therefore the coefficient is \(15\)
A number of the form \(1/N\), where \(N\) is an integer greater than 1, is called a unit fraction. Noting that \[ \frac1 2 =\frac13 + \frac16\\\ \mbox{ and } \frac13 = \frac14 + \frac1{12}, \] guess a general result of the form $$ \frac1N =\frac1a +\frac1b \tag{*} $$ and hence prove that any unit fraction can be expressed as the sum of two distinct unit fractions. By writing \((*)\) in the form \[ (a-N)(b-N)=N^2 \] and by considering the factors of \(N^2\), show that if \(N\) is prime, then there is only one way of expressing \(1/N\) as the sum of two distinct unit fractions. Prove similarly that any fraction of the form \(2/N\), where \(N\) is prime number greater than 2, can be expressed uniquely as the sum of two distinct unit fractions.
Solution: Notice that \(\frac{1}{N} = \frac{1}{N+1} + \frac{1}{N(N+1)}\), so any unit fraction can be expressed as the sum of two distinct unit fractions. \begin{align*} && \frac{1}N &= \frac1a + \frac1b \\ \Leftrightarrow && ab&= Nb+Na \\ \Leftrightarrow && 0 &= (a-N)(b-N)-N^2 \\ \Leftrightarrow && N^2 &= (a-N)(b-N) \end{align*} If \(N\) is prime then the only factors of \(N^2\) are \(1,N\) and \(N^2\). if \(a-N = b-N = N\) then \(a=b\) and we don't have distinct fractions. Therefore \(a-N = 1\) and \(b-N = N^2\) and we obtain the decomposition earlier (and it must be the only solution). \begin{align*} && \frac2N &= \frac1a+\frac1b \\ \Leftrightarrow && 2ab &= Nb+Na \\ \Leftrightarrow && 4ab &= 2Na+2Nb \\ \Leftrightarrow && N^2 &= (2a-N)(2b-N) \end{align*} Therefore for \(a,b\) to be distinct we must have \(2a = N+1\) and \(2b = N+N^2\) as the only possible factorisation. Both of the right hand sides are even so we can write \[ \frac{1}{N} = \frac{1}{\frac{N+1}{2}} + \frac{1}{\frac{N(N+1)}{2}} \] and this is unique
Given that \[ x^4 + p x^2 + q x + r = ( x^2 - a x + b ) ( x^2 + a x + c ) , \] express \(p\), \(q\) and \(r\) in terms of \(a\), \(b\) and \(c\). Show also that \( a^2\) is a root of the cubic equation $$ u^3 + 2 p u^2 + ( p^2 - 4 r ) u - q^2 = 0 . $$ Explain why this equation always has a non-negative root, and verify that \(u = 9\) is a root in the case \(p = -1\), \(q = -6\), \(r = 15\) . Hence, or otherwise, express $$y^4 - 8 y^3 + 23 y^2 - 34 y + 39$$ as a product of two quadratic factors.
Solution: \begin{align*} && ( x^2 - a x + b ) ( x^2 + a x + c ) &= x^4 + (b+c-a^2)x^2 + a(b-c)x + bc \\ \Rightarrow && x^4 + p x^2 + q x + r &= x^4 + (b+c-a^2)x^2 + a(b-c)x + bc \\ \Rightarrow && p &= b+c-a^2 \tag{1}\\ && q &= a(b-c) \tag{2}\\ && r &= bc \tag{3} \end{align*} \begin{align*} (1): && p+a^2 &= b+ c \\ (2): && \frac{q}{a} &= b - c \\ \Rightarrow && b &= \frac12 (p+a^2 + \frac{q}{a}) \\ && c &= \frac12 (p+a^2 - \frac{q}{a}) \\ (3): && r &= \frac12 (p+a^2 + \frac{q}{a}) \frac12 (p+a^2 - \frac{q}{a}) \\ \Rightarrow && 4ra^2 &= (pa + a^3 + q)(pa+a^3-q) \\ &&&= (pa+a^3)^2 - q^2 \\ &&&= a^2(p+a^2)^2 -q^2 \\ &&&= a^2(p^2 + 2pa^2 + a^4) - q^2 \\ &&&= pa^2 + 2pa^4 + a^6 - q^2 \\ \end{align*} Therefore \(a^2\) is a root of \(u^3 + 2pu^2 + pu - q^2 = 4ru\), ie the given equation. When \(u = 0\), this equation is \(-q^2\), therefore the cubic is negative. But as \(u \to \infty\) the cubic tends to \(\infty\), therefore it must cross the \(x\)-axis and have a positive root. If \(p=-1, q = -6, r = 15\) then the cubic is: \(u^3 - 2u^2 + (1-60)u -36\) and so when \(u = 9\) we have \begin{align*} 9^3 - 2\cdot 9^2 -59 \cdot 9 -36 &= 9(9^2-2\cdot 9 - 29 -4) \\ &= 9(81 -18-59-4) \\ &= 0 \end{align*} so \(u = 9\) is a root Let \(y=z + 2\) \begin{align*} &&y^4 - 8 y^3 + 23 y^2 - 34 y + 39 &= (z+2)^4-8(z+2)^3 + 23(z+2)^2 - 34(z+2) + 39 \\ &&&= z^4+8z^3+24z^2+32z+16 - \\ &&&\quad -8z^3-48z^2-96z-64 \\ &&&\quad\quad +23z^2+92z+92 \\ &&&\quad\quad -34z-68 + 39 \\ &&&= z^4-z^2-6z+15 \end{align*} So conveniently this is \(p = -1, q = -6, r = 15\), so we know that \(a = 3\) is a sensible thing to true. \(b = \frac12(-1 + 9 + \frac{-6}{3}) = 3\) \(c = \frac12(-1+9-\frac{-6}{3}) = 5\) so \begin{align*} && z^4-z^2-6z+15 &= (z^2-3z+3)(z^2+3z+5) \\ &&y^4 - 8 y^3 + 23 y^2 - 34 y + 39 &= ((y-2)^2-3(y-2)+3)((y-2)^2+3(y-2)+5) \\ &&&= (y^2-4y+4-3y+6+3)(y^2-4y+4+3y-6+5) \\ &&&= (y^2-7y+13)(y^2-y+3) \end{align*}
A tortoise and a hare have a race to the vegetable patch, a distance \(X\) kilometres from the starting post, and back. The tortoise sets off immediately, at a steady \(v\) kilometers per hour. The hare goes to sleep for half an hour and then sets off at a steady speed \(V\) kilometres per hour. The hare overtakes the tortoise half a kilometre from the starting post, and continues on to the vegetable patch, where she has another half an hour's sleep before setting off for the return journey at her previous pace. One and quarter kilometres from the vegetable patch, she passes the tortoise, still plodding gallantly and steadily towards the vegetable patch. Show that \[ V= \frac{10}{4X-9} \] and find \(v\) in terms of \(X\). Find \(X\) if the hare arrives back at the starting post one and a half hours after the start of the race.
Solution: If \(T_1\) is the time they meet after \(\frac12\)km from the starting points and \(T_2\) is the time they meet a second time, then \begin{align*} && \frac12 &= vT_1 \\ &&&= V(T_1-\tfrac12) \\ && X - \frac54 &= vT_2 \\ && X + \frac54 &= V(T_2 - 1) \\ && \frac{T_2}{T_1} &= \frac{4X-5}{2} \\ && X + \frac54 + V &= VT_2 \\ && \frac12 + \frac12 V &= VT_1 \\ \Rightarrow && \frac{T_2}{T_1} &= \frac{4X + 5 + 4V}{2(1+V)} \\ \Rightarrow && \frac{4X-5}{2}&=\frac{4X + 5 + 4V}{2(1+V)} \\ \Rightarrow && V(4X-9) &= 10 \\ \Rightarrow && V &= \frac{10}{4X-9} \\ \\ && T_1 &= \frac{1}{2V} + \frac12 \\ &&&= \frac{4X+1}{20} \\ && v &= \frac{1}{2T_1} \\ &&&= \frac{10}{4X+1} \end{align*} \begin{align*} && 2X &= \frac12 V \\ \Rightarrow && 2X(4X-9) &= 5 \\ \Rightarrow && 0 &= 8X^2-18X-5 \\ &&&= (4X+1)(2X-5) \\ \Rightarrow && X &= -\frac14, \frac52 \end{align*} Since \(X\) is positive, we must have \(X = \frac52\)km
Show, by means of a suitable change of variable, or otherwise, that \[ \int_{0}^{\infty}\mathrm{f}((x^{2}+1)^{1/2}+x)\,{\mathrm d}x =\frac{1}{2} \int_{1}^{\infty}(1+t^{-2})\mathrm{f}(t)\,{\mathrm d}t. \] Hence, or otherwise, show that \[ \int_{0}^{\infty}((x^{2}+1)^{1/2}+x)^{-3}\,{\mathrm d}x =\frac{3}{8}. \]
Solution: \begin{align*} && t &= (x^2+1)^{1/2}+x \\ && 1&=t^2-2tx \\ && x &= \frac{t^2-1}{2t} = \frac12 \left (t - \frac1t\right) \\ && \frac{\d x}{\d t} &= \frac12 \left ( 1+ \frac{1}{t^2} \right) \\ \Rightarrow && \int_0^{\infty} f((x^2+1)^{1/2}+x) \d x &= \int_{t=1}^{t = \infty}f(t) \frac12(1 + t^{-2}) \d t\\ &&&= \frac12 \int_1^{\infty}(1+t^{-2})f(t) \d t \end{align*} \begin{align*} \int_{0}^{\infty}((x^{2}+1)^{1/2}+x)^{-3}\,{\mathrm d}x &= \frac12 \int_1^{\infty}(1+t^{-2})t^{-3} \d t \\ &= \frac12 \left [\frac{-1}{2}t^{-2}-\frac{1}{4}t^{-4} \right]_{1}^{\infty} \\ &= \frac12 \left ( \frac12 + \frac14\right) = \frac38 \end{align*}
My bank pays \(\rho\%\) interest at the end of each year. I start with nothing in my account. Then for \(m\) years I deposit \(\pounds a\) in my account at the beginning of each year. After the end of the \(m\)th year, I neither deposit nor withdraw for \(l\) years. Show that the total amount in my account at the end of this period is \[\pounds a\frac{r^{l+1}(r^{m}-1)}{r-1}\] where \(r=1+{\displaystyle \frac{\rho}{100}}\). At the beginning of each of the \(n\) years following this period I withdraw \(\pounds b\) and this leaves my account empty after the \(n\)th withdrawal. Find an expression for \(a/b\) in terms of \(r\), \(l\), \(m\) and \(n\).
Solution: Rather than putting the deposits in the same account, imagine they are all put in separate accounts. Then for example, the first \(\pounds a\) will go on to become \(\pounds r^m \cdot r^l a\) from \(m\) years of compound interest as more money is deposited, followed by \(l\) years where no money is deposited. Therefore the total amount at the end is: \begin{align*} && S &= r^{m}r^l a + r^{m-1}r^l a + \cdots + r r^l a \\ &&&= r^l a(r^m + \cdots + r) \\ &&&= ar^l\frac{r^{m+1}-r}{r-1} \\ &&&= a \frac{r^{l+1}(r^m-1)}{r-1} \end{align*} Rather than withdrawing \(b\) each time, imagine that in the \(n\)th year we withdraw each \(b\) with the appropriate additional interest, ie \begin{align*} && \underbrace{a \frac{r^{l+1}(r^m-1)}{r-1}}_{\text{amount before \(n\) years}} \underbrace{r^{n-1}}_{\text{accounting for interest}} &= b r^{n-1} + br^{n-2} + \cdots + b \\ &&&= b \frac{r^n-1}{r-1} \\ \Rightarrow && \frac{a}{b} &= \frac{r^n-1}{r^{l+n}(r^m-1)} \end{align*}
Suppose that $$3=\frac{2}{ x_1}=x_1+\frac{2}{ x_2} =x_2+\frac{2}{ x_3}=x_3+\frac{2}{ x_4}=\cdots.$$ Guess an expression, in terms of \(n\), for \(x_n\). Then, by induction or otherwise, prove the correctness of your guess.
Solution: \begin{align*} x_1 &= \frac{2}{3} \\ x_n &= \frac{2}{3-x_{n-1}} \\ x_2 &= \frac{2}{3 - \frac23} \\ &= \frac{6}7 \\ x_3 &= \frac{2}{3-\frac67} \\ &= \frac{14}{15} \\ x_4 &= \frac{2}{3 - \frac{14}{15}} \\ &= \frac{30}{31} \end{align*} Guess: \(x_n = \frac{2^{n+1}-2}{2^{n+1}-1}\). Proof: (By induction) (Base case): We have checked several initial cases. (Inductive step): Suppose our formula is true for \(n = k\), then consider: \begin{align*} x_{k+1} &= \frac{2}{3 - x_{k}} \\ &= \frac{2}{3 - \frac{2^{k+1}-2}{2^{k+1}-1}}\tag{assumption} \\ &= \frac{2\cdot(2^{k+1}-1)}{3 \cdot(2^{k+1}-1) - (2^{k+1}-2) } \\ &= \frac{2^{k+2}-2}{2\cdot 2^{k+1} - 3 + 2 } \\ &= \frac{2^{k+2}-2}{ 2^{k+2} - 1 } \\ \end{align*} Therefore, if our formula is true for \(n = k\) it is true for \(n = k+1\). Therefore by the principle of mathematical induction it is true for \(n \geq 1, n \in \mathbb{Z}\)
A particle is projected under the influence of gravity from a point \(O\) on a level plane in such a way that, when its horizontal distance from \(O\) is \(c\), its height is \(h\). It then lands on the plane at a distance \(c+d\) from \(O\). Show that the angle of projection \(\alpha\) satisfies \[ \tan\alpha=\frac{h(c+d)}{cd} \] and that the speed of projection \(v\) satisfies \[ v^{2}=\frac{g}{2}\left(\frac{cd}{h}+\frac{(c+d)^{2}h}{cd}\right)\,. \]
Show that \(\cos 4u=8\cos^{4}u-8\cos^{2}u+1\). If \[ I=\int_{-1}^{1} \frac{1}{\vphantom{{\big(}^2}\; \surd(1+x)+\surd(1-x)+2\; }\;{\rm d}x ,\] show, by using the change of variable \(x=\cos t\), that \[ I= \int_0^\pi \frac{\sin t}{4\cos^{2}\left(\frac{t}{4}-\frac{\pi}{8}\right)}\,{\rm d}t.\] By using the further change of variable \(u=\frac{t}{4}-\frac{\pi}{8}\), or otherwise, show that \[I=4\surd{2}-\pi-2.\] \noindent[You may assume that \(\tan\frac{\pi}{8}=\surd{2}-1\).]
Solution: \begin{align*} && \cos 4u &= 2\cos^2 2u - 1 \\ &&&= 2 (2\cos^2 u - 1)^2 - 1 \\ &&&= 2(4\cos^4u - 4\cos^2 u + 1) - 1\\ &&&= 8\cos^4u - 8\cos^2 u + 1 \end{align*} \begin{align*} && I &= \int_{-1}^1 \frac{1}{\sqrt{1+x}+\sqrt{1-x}+2} \d x \\ x = \cos t, \d x = - \sin t \d t: &&&= \int_{t = \pi}^{t=0} \frac{1}{\sqrt{1+\cos t} + \sqrt{1-\cos t} + 2} (- \sin t ) \d t \\ &&&= \int_0^\pi \frac{\sin t}{\sqrt{2 \cos^2 \frac{t}{2}}+\sqrt{2 \sin^2 \frac{t}{2}}+2} \d t \\ &&&= \int_0^\pi \frac{\sin t}{\sqrt{2}(\cos \frac{t}{2} + \sin \frac{t}{2}) + 2} \d t \\ &&&= \int_0^\pi \frac{\sin t}{\sqrt{2}(\sqrt{2} \cos (\frac{t}{2}-\frac{\pi}{4})) + 2} \d t \\ &&&= \int_0^\pi \frac{\sin t}{2(1+\cos (\frac{t}{2}-\frac{\pi}{4}))} \d t \\ &&&= \int_0^\pi \frac{\sin t}{4\cos^2(\frac{t}{4}-\frac{\pi}{8})} \d t \\ \\ u = \tfrac{t}{4} -\tfrac{\pi}{8}, \d u = \tfrac14 \d t:&&&=\int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \frac{\sin (4u+\frac{\pi}{2})}{4 \cos^2 u} 4 \d u \\ &&&= \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \frac{\cos4u}{\cos^2 u} \d u \\ &&&= \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} 4 (2 \cos^2 u-1)-4 + \sec^2 u \d u \\ &&&= \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} 4 \cos 2u-4 + \sec^2 u \d u \\ &&&= \left [2\sin 2u - 4u + \tan u \right]_{-\pi/8}^{\pi/8} \\ &&&= 4 \sin \frac{\pi}{4} - \pi+ 2\tan \frac{\pi}{8} \\ &&&= \frac{4}{\sqrt{2}} - \pi + 2\sqrt{2}-2 \\ &&&= 4\sqrt{2} - \pi - 2 \end{align*}
Define \(\cosh x\) and \(\sinh x\) in terms of exponentials and prove, from your definitions, that \[ \cosh^{4}x-\sinh^{4}x=\cosh2x \] and \[ \cosh^{4}x+\sinh^{4}x=\tfrac{1}{4}\cosh4x+\tfrac{3}{4}. \] Find \(a_{0},a_{1},\ldots,a_{n}\) in terms of \(n\) such that \[ \cosh^{n}x=a_{0}+a_{1}\cosh x+a_{2}\cosh2x+\cdots+a_{n}\cosh nx. \] Hence, or otherwise, find expressions for \(\cosh^{2m}x-\sinh^{2m}x\) and \(\cosh^{2m}x+\sinh^{2m}x,\) in terms of \(\cosh kx,\) where \(k=0,\ldots,2m.\)
Solution: \begin{align*} \cosh x &= \frac12 (e^x + e^{-x}) \\ \sinh x &= \frac12 (e^x - e^{-x}) \\ \end{align*} \begin{align*} \cosh^4x -\sinh^4 x &= (\cosh^2x -\sinh^2 x)(\cosh^2x +\sinh^2 x) \\ &= \left ( \frac14 \left (e^{2x}+2+e^{-2x} \right)- \frac14 \left (e^{2x}-2+e^{-2x} \right) \right)(\cosh^2x +\sinh^2 x) \\ &= (\cosh^2x +\sinh^2 x) \\ &= \left ( \frac14 \left (e^{2x}+2+e^{-2x} \right)+ \frac14 \left (e^{2x}-2+e^{-2x} \right) \right) \\ &= \frac{1}{4} \left (2e^{2x}+2e^{-2x} \right) \\ &= \frac12 \left ( e^{2x}+e^{-2x} \right) \\ &= \cosh 2x \\ \\ \cosh^4x +\sinh^4 x &= \frac1{2^4}\left (e^{4x}+4e^{2x}+6+4e^{-2x}+e^{-4x} \right)+\frac1{2^4}\left (e^{4x}-4e^{2x}+6-4e^{-2x}+e^{-4x} \right) \\ &= \frac18 (e^{4x}+e^{-4x}) + \frac{3}{4} \\ &= \frac14 \cosh 4x + \frac34 \end{align*} \begin{align*} \cosh^n x &=\frac{1}{2^n} \left ( e^{x}+e^{-x} \right)^n \\ &= \frac{1}{2^n} \sum_{k=0}^n \binom{n}{k} e^{kx}e^{-(n-k)x} \\ &= \frac{1}{2^n} \sum_{k=0}^n \binom{n}{k} e^{2kx-nx} \\ &= \frac{1}{2^n} \left ( \binom{n}{n} \left(e^{nx}+e^{-nx} \right) + \binom{n}{n-1}\left(e^{(n-2)x}+e^{-(n-2)x} \right) + \cdots + \binom{n}{n-k} \left( e^{(n-2k)x}+e^{-(n-2k)x} \right) + \cdots \right) \\ &= \frac{1}{2^{n-1}} \cosh nx + \frac{1}{2^{n-1}} \binom{n}{n-1} \cosh (n-2)x + \cdots + \frac{1}{2^{n-1}} \binom{n}{n-k} \cosh (n-2k)x + \cdots \end{align*} ie \begin{align*} \cosh^{2m} x &= \frac{1}{2^{2m-1}} \cosh 2m x + \frac{2m}{2^{2m-1}} \cosh(2(m-1)x) + \cdots + \frac{1}{2^{2m-1}}\binom{2m}{k} \cosh (2(m-k)x) +\cdots+ \frac{1}{2^{2m-1}} \binom{2m}{m} \\ \sinh^{2m} x &= \frac{1}{2^{2m-1}} \cosh 2m x - \frac{2m}{2^{2m-1}} \cosh(2(m-1)x) + \cdots + (-1)^{k}\frac{1}{2^{2m-1}}\binom{2m}{k} \cosh (2(m-k)x) +\cdots+ (-1)^m\frac{1}{2^{2m-1}} \binom{2m}{m} \\ \cosh^{2m} x -\sinh^{2m} x &= \frac{m}{2^{2m-3}} \cosh (2(m-1)x) + \cdots + \frac{1}{2^{2m-2}} \binom{2m}{2k+1}\cosh(2(m-2k-1)x) + \cdots\\ \cosh^{2m} x +\sinh^{2m} x &= \frac{1}{2^{2m-2}} \cosh (2mx) + \cdots + \frac{1}{2^{2m-2}} \binom{2m}{2k}\cosh(2(m-2k)x) + \cdots \end{align*}
Solution: