77 problems found
Show that \(y=\sin^{2}(m\sin^{-1}x)\) satisfies the differential equation \[ (1-x^{2})y^{(2)}=xy^{(1)}+2m^{2}(1-2y), \] and deduce that, for all \(n\geqslant1,\) \[ (1-x^{2})y^{(n+2)}=(2n+1)xy^{(n+1)}+(n^{2}-4m^{2})y^{(n)}, \] where \(y^{(n)}\) denotes the \(n\)th derivative of \(y\). Derive the Maclaurin series for \(y\), making it clear what the general term is.
Solution: \begin{align*} && y &= \sin^2 (m \sin^{-1} x) \\ \Rightarrow && y' &= 2 \sin (m \sin^{-1} x) \cdot \cos (m \sin^{-1} x) \cdot m \cdot \frac1{\sqrt{1-x^2}} \\ \Rightarrow && y'' &= 2 \cos^2(m \sin^{-1} x) \cdot m^2 \cdot \frac{1}{1-x^2} + \\ &&&\quad\quad-2\sin^2(m \sin^{-1} x) m^2 \frac{1}{1-x^2} + \\ &&&\quad\quad\quad-\sin(m \sin^{-1} x) \cdot \cos(m \sin^{-1} x) \cdot m \cdot (1-x^2)^{-\frac32} \cdot (-2x) \\ \Rightarrow && (1-x^2)y^{(2)} &= 2m^2-4m^2y+xy' \\ &&&= xy^{(1)} + 2m^2(1-2y) \\ \\ \Rightarrow && (1-x^2)y^{(n+2)}-2nxy^{(n+1)}-2\binom{n}{2}y^{(n)} &= xy^{(n+1)}+ny^{(n)} -4m^2y^{(n)} \\ \Rightarrow && (1-x^2)y^{(n+2)} &= (2n+1)xy^{(n+1)}+(n(n-1)+n-4m^2)y^{(n)} \\ &&&= (2n+1)xy^{(n+1)}+(n^2-4m^2)y^{(n)} \\ \end{align*} \begin{align*} && y(0) &= \sin^2(m \sin^{-1} 0) \\ &&&= \sin^2 0 = 0 \\ \\ && y'(0) &= 0 \\ && (1-0^2)y^{(2)}(0) &= 2m^2(1-2y(0)) \\ \Rightarrow && y^{(2)}(0) &= 2m^2 \\ \\ && y^{(n+2)} (0) &= (2n+1) \cdot 0 \cdot y^{(n+1)} +(n^2-4m^2)y^{(n)}(0) \\ &&&= (n^2-4m^2)y^{(n)}(0) \\ \\ && y^{(2)}(0) &= 2m^2 \\ && y^{(4)}(0) &= (4-4m^2) \cdot 2m^2 \\ &&&= -8m(m+1)m(m-1) \\ && y^{(6)}(0) &= 32m(m+2)(m+1)m(m-1)(m-2) \\ && y^{(2k)}(0) &= (-1)^{k+1}2^{2k-1}m (m+k)\cdots(m-k) \text{ if }k < m \\ \\ && y &= m^2x^2 -2m\binom{m+1}{3} x^4 + \frac{16}{3}m\binom{m+2}{5}x^6 - \cdots \\ &&&+ (-1)^{k}\frac{2^{2k}}{k+1} m \binom{m+k}{2k+1}x^{2k+2}+\cdots \\ &&&= mx^2\sum_{k=0}^{m-1} \frac{(-1)^k2^{2k}}{k+1}\binom{m+k}{2k+1}x^{2k} \end{align*}
From the facts \begin{alignat*}{2} 1 & \quad=\quad & & 0\\ 2+3+4 & \quad=\quad & & 1+8\\ 5+6+7+8+9 & \quad=\quad & & 8+27\\ 10+11+12+13+14+15+16 & \quad=\quad & & 27+64 \end{alignat*} guess a general law. Prove it. Hence, or otherwise, prove that \[ 1^{3}+2^{3}+3^{3}+\cdots+N^{3}=\tfrac{1}{4}N^{2}(N+1)^{2} \] for every positive integer \(N\). [Hint. You may assume that \(1+2+3+\cdots+n=\frac{1}{2}n(n+1)\).]
Solution: \begin{align*} && (n^2+1) + (n^2+2) + \cdots + (n+1)^2 &= n^3+(n+1)^3 \\ \Leftrightarrow && \sum_{i=n^2+1}^{(n+1)^2} i &= n^3 + (n+1)^3 \\ && \sum_{i=n^2+1}^{(n+1)^2} i &= \sum_{i=1}^{(n+1)^2} i- \sum_{i=1}^{n^2} i \\ &&&= \frac{(n+1)^2((n+1)^2+1)}{2} - \frac{n^2(n^2+1)}{2} \\ &&&= \frac{(n+1)^2(n^2+2n+2) - n^2(n^2+1)}{2} \\ &&&= \frac{2(n+1)^3+n^2(n^2+2n+1) - n^2(n^2+1)}{2}\\ &&&= \frac{2(n+1)^3+2n^3 + n^2(n^2+1) - n^2(n^2+1)}{2}\\ &&&= (n+1)^3+n^3 \end{align*} \begin{align*} && \sum_{i=1}^{N^2} i &=(0^3+1^3)+ (1^3+2^3)+(2^3+3^3) + \cdots + ((N-1)^3+N^3) \\ &&&= 2 \left (1^3+2^3 + 3^3 + \cdots + (N-1)^3 \right) + N^3 \\ \Rightarrow && \sum_{i=1}^N i^3 &= \frac12 \left ( N^3+ \sum_{i=1}^{N^2} i \right) \\ &&&= \frac12 \left ( N^3 + \frac{N^2(N^2+1)}{2} \right) \\ &&&= \frac{N^2(N^2+1)+2N^3}{4} \\ &&&= \frac{N^2(N^2+2N+1)}{4} \\ &&&= \frac{N^2(N+1)^2}{4} \\ \end{align*}
Prove by induction, or otherwise, that, if \(0<\theta<\pi\), \[ \frac{1}{2}\tan\frac{\theta}{2}+\frac{1}{2^{2}}\tan\frac{\theta}{2^{2}}+\cdots+\frac{1}{2^{n}}\tan\frac{\theta}{2^{n}}=\frac{1}{2^{n}}\cot\frac{\theta}{2^{n}}-\cot\theta. \] Deduce that \[ \sum_{r=1}^{\infty}\frac{1}{2^{r}}\tan\frac{\theta}{2^{r}}=\frac{1}{\theta}-\cot\theta. \]
Solution: Claim: \(\displaystyle \sum_{r=1}^n \frac1{2^r} \tan \tfrac{\theta}{2^r} = \frac1{2^n}\cot \tfrac{\theta}{2^n} - \cot \theta\) Proof: (By Induction) Base case: \(n = 1\) \begin{align*} && LHS &= \sum_{r=1}^1 \frac1{2^r} \tan \frac{\theta}{2^r} \\ &&&= \frac1{2} \tan \frac{\theta}{2}\\ \\ && RHS &= \frac12 \cot \frac{\theta}{2} - \cot \theta \\ &&&= \frac12 \frac{1}{\tan \frac{\theta}{2}} - \frac{1-\tan^2 \frac{\theta}{2}}{2 \tan \frac{\theta}{2}} \\ &&&= \frac{1}{2} \tan \frac{\theta}{2} = LHS \end{align*} Therefore our base case is true. Assume our statement is true for some \(n=k\), then consider \(n = k+1\), ie \begin{align*} \sum_{r=1}^{k+1} \frac1{2^r} \tan \tfrac{\theta}{2^r} &= \sum_{r=1}^{k} \frac1{2^r} \tan \tfrac{\theta}{2^r} + \frac1{2^{k+1}} \tan \frac{\theta}{2^{k+1}} \\ &= \frac{1}{2^k} \cot \frac{\theta}{2^k} - \cot \theta + \frac{1}{2^{k+1}}\tan \frac{\theta}{2^{k+1}} \\ &= \frac{1}{2^{k+1}} \left (2 \cot \frac{\theta}{2^k} +\tan \frac{\theta}{2^{k+1}} \right) - \cot \theta \\ &= \frac{1}{2^{k+1}} \left (2\frac{1-\tan^2 \frac{\theta}{2^{k+1}}}{2 \tan \frac{\theta}{2^{k+1}}} + \tan \frac{\theta}{2^{k+1}} \right) - \cot \theta \\ &= \frac{1}{2^{k+1}} \cot \frac{\theta}{2^{k+1}} - \cot \theta \\ \end{align*} Therefore, since as \(x \to 0, x\cot x \to 1\) or \(x \cot \theta x \to \frac{1}{\theta}\) \begin{align*} \sum_{r=1}^{\infty}\frac{1}{2^{r}}\tan\frac{\theta}{2^{r}} &= \lim_{k\to \infty} \sum_{r=1}^{k}\frac{1}{2^{r}}\tan\frac{\theta}{2^{r}} \\ &= \lim_{k\to \infty} \left ( \frac{1}{2^{k+1}} \cot \frac{\theta}{2^{k+1}} - \cot \theta \right) \\ &= \lim_{k\to \infty}\frac{1}{2^{k+1}} \cot \frac{\theta}{2^{k+1}} - \cot \theta \\ &= \lim_{k\to \infty}\frac{1}{2^{k+1}} \cot \frac{\theta}{2^{k+1}} - \cot \theta \\ &= \lim_{k\to \infty}\frac{1}{2^{k+1}} \cot \frac{\theta}{2^{k+1}} - \cot \theta \\ &= \frac{1}{\theta} - \cot \theta \end{align*}
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 \(n > 0\).
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*}
Let \(\mathrm{g}(x)=ax+b.\) Show that, if \(\mathrm{g}(0)\) and \(\mathrm{g}(1)\) are integers, then \(\mathrm{g}(n)\) is an integer for all integers \(n\). Let \(\mathrm{f}(x)=Ax^{2}+Bx+C.\) Show that, if \(\mathrm{f}(-1),\mathrm{f}(0)\) and \(\mathrm{f}(1)\) are integers, then \(\mathrm{f}(n)\) is an integer for all integers \(n\). Show also that, if \(\alpha\) is any real number and \(\mathrm{f}(\alpha-1),\) \(\mathrm{f}(\alpha)\) and \(\mathrm{f}(\alpha+1)\) are integers, then \(\mathrm{f}(\alpha+n)\) is an integer for all integers \(n\).
Solution: If \(g(0) \in \mathbb{Z} \Rightarrow b \in \mathbb{Z}\). If \(g(1) \in \mathbb{Z} \Rightarrow a+b \in \mathbb{Z} \Rightarrow a \in \mathbb{Z}\), therefore \(a \cdot n + b \in \mathbb{Z}\), in particular \(g(n) \in \mathbb{Z}\) for all integers \(n\). \(f(0) \in \mathbb{Z} \Rightarrow C \in \mathbb{Z}\), \(f(1) \in \mathbb{Z} = A+ B + C \in \mathbb{Z} \Rightarrow A+ B \in \mathbb{Z}\) \(f(-1) \in \mathbb{Z} = A- B + C \in \mathbb{Z} \Rightarrow A- B \in \mathbb{Z}\) \(\Rightarrow 2A, 2B \in \mathbb{Z}\) \begin{align*} f(n) &= An^2 + Bn + C \\ &= An^2-An + An+Bn + C \\ &= 2A \frac{n(n-1)}2 + (A+B)n + C \\ &\in \mathbb{Z} \end{align*} Consider \(g(x) = f(x + \alpha)\), therefore \(g(0), g(1), g(-1) \in \mathbb{Z} \Rightarrow g(n) \in \mathbb{Z} \Rightarrow f(n+\alpha) \in \mathbb{Z}\)
\(\lozenge\) is an operation which take polynomials in \(x\) to polynomials in \(x\); that is, given a polynomial \(\mathrm{h}(x)\) there is another polynomial called \(\lozenge\mathrm{h}(x)\). It is given that, if \(\mathrm{f}(x)\) and \(\mathrm{g}(x)\) are any two polynomials in \(x\), the following are always true:
Solution: Claim: If \(f(x) = c\) then \(\lozenge f(x) = 0\) Proof: Consider \(g(x) = x\) then \begin{align*} (1) && \lozenge(f(x)g(x)) &= g(x) \lozenge f(x) + f(x) \lozenge g(x) \\ \Rightarrow && \lozenge(c x) &= x \lozenge f(x) + c \lozenge x \\ (4) && \lozenge(c x) &= c \lozenge x \\ \Rightarrow && 0 &= x \lozenge f(x) \\ \Rightarrow && \lozenge f(x) &= 0 \end{align*} \begin{align*} (1) && \lozenge(x^2) &= x \lozenge x + x \lozenge x \\ (3) &&&= 2 x \cdot 1 \\ &&&= 2x \\ \\ (1) && \lozenge (x^3) &= x^2 \lozenge x + x \lozenge (x^2) \\ &&&= x^2 \cdot \underbrace{1}_{(3)} + x \cdot\underbrace{ 2x}_{\text{previous part}} \\ &&&= 3x^2 \end{align*} Claim: \(\lozenge h(x) = \frac{\d }{\d x} ( h(x))\) for any polynomial \(h\). Proof: (By (strong) induction on the degree of \(h\)). Base case: True, we proved this in the first part of the question. Inductive step: Assume true for all polynomials of degree less than or equal to \(k\). Then consider \(n = k+1\). We can write \(h(x) = ax^{k+1} + h_k(x)\) where \(h_k(x)\) is a polynomial of degree less than or equal to \(k\). Then notice: \begin{align*} && \lozenge (h(x)) &= \lozenge (ax^{k+1} + h_k(x)) \\ (2) &&&= \lozenge (ax^{k+1})+ \lozenge (h_k(x)) \\ &&&=\underbrace{a\lozenge (x^{k+1})}_{(4)}+ \underbrace{\frac{\d}{\d x} (h_k(x))}_{\text{inductive hypothesis}}\\ &&&= a \underbrace{\left (x \lozenge x^k + x^k \lozenge x \right)}_{(1)} + \frac{\d}{\d x} (h_k(x)) \\ &&&= a \left ( x \cdot \underbrace{k x^{k-1}}_{\text{inductive hyp.}} + x^k \cdot \underbrace{1}_{(3)} \right) + \frac{\d}{\d x} (h_k(x)) \\ &&&= (k+1)a x^k + \frac{\d}{\d x} (h_k(x)) \\ &&&= \frac{\d }{\d x} \left ( ax^{k+1} + h_k(x) \right) \\ &&&= \frac{\d }{\d x} (h(x)) \end{align*} Therefore since our statement is true for \(n=0\) and if it is true for \(n=k\) it is true for \(n=k+1\) by the principle of mathematical induction it is true for all \(n \geq 0\)
The matrices \(\mathbf{I}\) and \(\mathbf{J}\) are \[ \mathbf{I}=\begin{pmatrix}1 & 0\\ 0 & 1 \end{pmatrix}\quad\mbox{ and }\quad\mathbf{J}=\begin{pmatrix}1 & 1\\ 1 & 1 \end{pmatrix} \] respectively and \(\mathbf{A}=\mathbf{I}+a\mathbf{J},\) where \(a\) is a non-zero real constant. Prove that \[ \mathbf{A}^{2}=\mathbf{I}+\tfrac{1}{2}[(1+2a)^{2}-1]\mathbf{J}\quad\mbox{ and }\quad\mathbf{A}^{3}=\mathbf{I}+\tfrac{1}{2}[(1+2a)^{3}-1]\mathbf{J} \] and obtain a similar form for \(\mathbf{A}^{4}.\) If \(\mathbf{A}^{k}=\mathbf{I}+p_{k}\mathbf{J},\) suggest a suitable form for \(p_{k}\) and prove that it is correct by induction, or otherwise.
Solution: If $\mathbf{J}=\begin{pmatrix}1 & 1\\ 1 & 1 \end{pmatrix}\(, them \)\mathbf{J}^2=\begin{pmatrix}2 & 2\\ 2 & 2 \end{pmatrix} = 2\mathbf{J}\(. Therefore \)\mathbf{J}^n = 2\mathbf{J}^{n-1} = 2^{n-1}\mathbf{J}$ Let \(\mathbf{A}=\mathbf{I}+a\mathbf{J}\) then \begin{align*} \mathbf{A}^2 &=\l \mathbf{I}+a\mathbf{J}\r^2 \\ &= \mathbf{I}+2a\mathbf{J} + a^2\mathbf{J}^2 \\ &= \mathbf{I}+2a\mathbf{J} + 2a^2\mathbf{J} \\ &= \mathbf{I}+(2a+ 2a^2)\mathbf{J} \\ &= \mathbf{I}+\frac12(1+4a+ 4a^2-1)\mathbf{J} \\ &= \mathbf{I}+\frac12((1+2a)^2-1)\mathbf{J} \\ \end{align*} \begin{align*} \mathbf{A}^3 &=\l \mathbf{I}+a\mathbf{J}\r^3 \\ &= \mathbf{I}+3a\mathbf{J} + a^2\mathbf{J} + a^3\mathbf{J}^3 \\ &= \mathbf{I}+3a\mathbf{J} + 6a^2\mathbf{J} + 4a^3\mathbf{J} \\ &= \mathbf{I}+(3a+ 6a^3+4a^3)\mathbf{J} \\ &= \mathbf{I}+\frac12(1+3\cdot2a+3\dot4a^2+ 8a^3-1)\mathbf{J} \\ &= \mathbf{I}+\frac12((1+2a)^3-1)\mathbf{J} \\ \end{align*} \begin{align*} \mathbf{A}^4 &=\l \mathbf{I}+a\mathbf{J}\r^4 \\ &= \mathbf{I}+4a\mathbf{J} + 6a^2\mathbf{J}^2 + 4a^3\mathbf{J}^3+a^4\mathbf{J}^4 \\ &= \mathbf{I}+4a\mathbf{J} + 12a^2\mathbf{J} + 16a^3\mathbf{J}+8a^4\mathbf{J}\\ &= \mathbf{I}+(4a+ 12a^3+16a^3+8a^4)\mathbf{J} \\ &= \mathbf{I}+\frac12(1+4\cdot2a+6\cdot4a^2+ 4\cdot8a^3+16a^4-1)\mathbf{J} \\ &= \mathbf{I}+\frac12((1+2a)^4-1)\mathbf{J} \\ \end{align*} Claim: \(\mathbf{A}^k = \mathbf{I} + \frac12 ((1+2a)^{k}-1)\mathbf{J}\) Proof: Firstly, note that \(\mathbf{I}\) commutes with everything, so we can just apply the binomial theorem as if we were using real numbers: \begin{align*} \mathbf{A}^k &=\l \mathbf{I}+a\mathbf{J}\r^k \\ &= \sum_{i=0}^k \binom{k}{i}a^i\mathbf{J}^i \\ &= \mathbf{I} + \sum_{i=1}^k \binom{k}{i}a^i2^{i-1}\mathbf{J} \\ &= \mathbf{I} + \frac12\l\sum_{i=1}^k \binom{k}{i}a^i2^{i}\r\mathbf{J} \\ &= \mathbf{I} + \frac12\l\sum_{i=0}^k \binom{k}{i}a^i2^{i} - 1\r\mathbf{J} \\ &= \mathbf{I} + \frac12\l(1+2a)^k - 1\r\mathbf{J} \end{align*} as required
Given that \({\displaystyle I_{n}=\int_{0}^{\pi}\frac{x\sin^{2}(nx)}{\sin^{2}x}\,\mathrm{d}x,}\) where \(n\) is a positive integer, show that \(I_{n}-I_{n-1}=J_{n},\) where \[ J_{n}=\int_{0}^{\pi}\frac{x\sin(2n-1)x}{\sin x}\,\mathrm{d}x. \] Obtain also a reduction formula for \(J_{n}.\) The curve \(C\) is given by the cartesian equation \[ y=\dfrac{x\sin^{2}(nx)}{\sin^{2}x}, \] where \(n\) is a positive integer and \(0\leqslant x\leqslant\pi.\) Show that the area under the curve \(C\) is \(\frac{1}{2}n\pi^{2}.\)
Solution: \begin{align*} I_n - I_{n-1} &= \int_0^{\pi} \frac{x \sin^2(nx)}{\sin ^2 x} \d x-\int_0^{\pi} \frac{x \sin^2((n-1)x)}{\sin ^2 x} \d x \\ &= \int_0^{\pi} \frac{x}{\sin^2 x} \left ( \sin^2 (nx) - \sin^2((n-1)x) \right) \d x \\ &= \int_0^{\pi} \frac{x}{\sin^2 x}\frac12 \left ( \cos (2(n-1)x) - \cos(2nx) \right) \d x \\ &= \int_0^{\pi} \frac{x}{\sin^2 x}\frac12 2 \sin ((2n-1)x )\sin x \d x \\ &= \int_0^{\pi} \frac{x\sin ((2n-1)x )}{\sin x}d x \\ &= J_n \\ \\ J_{n+1} - J_{n} &= \int_0^{\pi} \frac{x \left (\sin ((2n+1)x )-\sin ((2n-1)x )\right)}{\sin x} \d x \\ &= \int_0^{\pi} \frac{x \left ( 2 \cos (\frac{4n x}{2}) \sin \frac{2x}{2} \right)}{\sin x} \d x \\ &= \int_0^{\pi}2x \cos (2n x) \d x \\ &= \left [ \frac{x}{2n} \sin (2n x) \right]_0^{\pi} - \int_0^{\pi} \frac{1}{2n} \sin (2n x) \d x \\ &= \left [ \frac{1}{4n^2} \cos (2n x)\right]_0^{\pi} \\ &= 0 \\ \\ J_1 &= \int_0^\pi x \d x \\ &= \frac{\pi^2}{2} \\ \Rightarrow J_n &= \frac{\pi^2}{2} \\ \end{align*} And so \(I_n = I_1 + (n-1) \frac{\pi^2}{2}\) and \(I_1 = \frac{\pi^2}{2}\) so \(I_n = \frac12 n \pi^2\). But \(I_n\) is exactly the area under the curve described.
The equation \[ x^{n}-qx^{n-1}+r=0, \] where \(n\geqslant5\) and \(q\) and \(r\) are real constants, has roots \(\alpha_{1},\alpha_{2},\ldots,\alpha_{n}.\) The sum of the products of \(m\) distinct roots is denoted by \(\Sigma_{m}\) (so that, for example, \(\Sigma_{3}=\sum\alpha_{i}\alpha_{j}\alpha_{k}\) where the sum runs over the values of \(i,j\) and \(k\) with \(n\geqslant i>j>k\geqslant1\)). The sum of \(m\)th powers of the roots is denoted by \(S_{m}\) (so that, for example, \(S_{3}=\sum\limits_{i=1}^{n}\alpha_{i}^{3}\)). Prove that \(S_{p}=q^{p}\) for \(1\leqslant p\leqslant n-1.\) You may assume that for any \(n\)th degree equation and \(1\leqslant p\leqslant n\) \[ S_{p}-S_{p-1}\Sigma_{1}+S_{p-2}\Sigma_{2}-\cdots+(-1)^{p-1}S_{1}\Sigma_{p-1}+(-1)^{p}p\Sigma_{p}=0.] \] Find expressions for \(S_{n},\) \(S_{n+1}\) and \(S_{n+2}\) in terms of \(q,r\) and \(n\). Suggest an expression for \(S_{n+m},\) where \(m < n\), and prove its validity by induction.
Solution: Claim: \(S_p = q^p\) for \(1 \leq p \leq n-1\) Proof: When \(p = 1\), \(S_p = \Sigma_1 = q\) as expected. Note that \(\Sigma_i = 0\) for \(i = 2, \cdots, n-1\). Using \(S_p = S_{p-1}\Sigma_{1}-S_{p-2}\Sigma_{2}+\cdots+(-1)^{p-1+1}S_{1}\Sigma_{p-1}+(-1)^{p+1}p\Sigma_{p}\), we can see that \(S_p = qS_{p-q}\) when \(1 \leq p \leq n-1\), ie \(S_p = q^p\). Note that \begin{align*} S_n &= \sum \alpha_i^n \\ &= q\sum \alpha_i^{n-1} - \sum r \\ &= qS_{n-1} - nr \\ &= q^n - nr \\ \\ S_{n+1} &= \sum \alpha_i^{n+1} \\ &= q \sum \alpha_i^{n} - r \sum \alpha_i \\ &= q^{n+1} - rq \\ \\ S_{n+2} &= \sum \alpha_i^{n+2} \\ &= q \sum \alpha_i^{n+1} - r \sum \alpha_i^2 \\ &= q^{n+2} - rq^2 \\ \end{align*} Claim: \(S_{n+m} = q^{n+m} - rq^{m}\) Proof: The obvious
Show that \[ \cos\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{4}\right)=\frac{\sin\alpha}{4\sin\left(\dfrac{\alpha}{4}\right)}\,, \] where \(\alpha\neq k\pi\), \(k\) is an integer. Prove that, for such \(\alpha\), \[ \cos\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{4}\right)\cdots\cos\left(\frac{\alpha}{2^{n}}\right)=\frac{\sin\alpha}{2^{n}\sin\left(\dfrac{\alpha}{2^{n}}\right)}\,, \] where \(n\) is a positive integer. Deduce that \[ \alpha=\frac{\sin\alpha}{\cos\left(\dfrac{\alpha}{2}\right)\cos\left(\dfrac{\alpha}{4}\right)\cos\left(\dfrac{\alpha}{8}\right)\cdots}\,, \] and hence that \[ \frac{\pi}{2}=\frac{1}{\sqrt{\frac{1}{2}}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}}}\cdots}\,. \]
Solution: \begin{align*} &&\sin \alpha &= 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \\ &&&= 4 \sin \frac{\alpha}{4} \cos \frac{\alpha}{4} \cos \frac{\alpha}{2} \\ \Rightarrow && \frac{\sin \alpha}{4 \sin \frac{\alpha}{4}} &= \cos \frac{\alpha}{2} \cos \frac{\alpha}{4} \end{align*} We proceed by induction on \(n\). Clearly this is true for \(n = 1\) (as we just established). Assume it is true for \(n=k\). Then: \begin{align*} && \frac{\sin \alpha}{2^n \sin \frac{\alpha}{2^n}} &= \cos\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{4}\right)\cdots\cos\left(\frac{\alpha}{2^{n}}\right) \\ \Rightarrow && \frac{\sin \alpha}{2^{n+1} \sin \frac{\alpha}{2^{n+1}} \cos \frac{\alpha}{2^{n+1}}} &= \cos\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{4}\right)\cdots\cos\left(\frac{\alpha}{2^{n}}\right) \\ \Rightarrow && \frac{\sin \alpha}{2^{n+1} \sin \frac{\alpha}{2^{n+1}} } &= \cos\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{4}\right)\cdots\cos\left(\frac{\alpha}{2^{n}}\right)\cos \left ( \frac{\alpha}{2^{n+1}} \right) \\ \end{align*} Therefore it is true for \(n=k+1\) Therefore since it is true for \(n=1\) and if it is true for \(n=k\) it is also true for \(n=k+1\) by the principle of mathematical induction it is true for all \(n \geq 1\) \begin{align*} \lim_{n \to \infty} \frac{\sin \alpha}{\cos\left(\frac{\alpha}{2}\right)\cdots\cos\left(\frac{\alpha}{2^{n}}\right)} &= \lim_{n \to \infty} 2^n \sin \frac{\alpha}{2^n} \\ &= \lim_{n \to \infty} \alpha \frac{\sin \frac{\alpha}{2^n}}{\frac{\alpha}{2^n}} \\ &= \alpha \lim_{t \to 0} \frac{\sin t}{t} \\ &= \alpha \end{align*} When \(\alpha = \frac{\pi}{2}\) notice that \(\sin \alpha =1\), \(\cos \frac{\alpha}{2} = \sqrt{\frac12}\) and \(2\cos^2 \frac{\alpha}{2^{n+1}}-1 = \cos \frac{\alpha}{2} \Rightarrow \cos \frac{\alpha}{2^{n+1}} = \sqrt{\frac12 + \cos \frac{\alpha}{2^n}}\) exactly the series we see.
A plane contains \(n\) distinct given lines, no two of which are parallel, and no three of which intersect at a point. By first considering the cases \(n=1,2,3\) and \(4\), provide and justify, by induction or otherwise, a formula for the number of line segments (including the infinite segments). Prove also that the plane is divided into \(\frac{1}{2}(n^{2}+n+2)\) regions (including those extending to infinity).
Solution: With \(n=1\) line, the plane is divided in half. With \(n=2\) lines the plane is divided into four pieces. (Each of the previous pieces are split in half) With \(n=3\) lines the plane is divided into up to \(7\) pieces. (The new line crosses two lines in two places dividing \(3\) regions into \(2\), thus increasing the number of regions by \(3\)). With \(n=4\) lines the plane is divided into \(11\) pieces. (The new line crosses three lines in three places doubling the number of regions of \(4\) places). Claim: With \(n\) lines the plane is divided into \(\frac12(n^2+n+2)\) regions. Proof: (By induction) (Base case) When \(n=1\) clearly the line is divided into \(2\) regions, and \(\frac12 (1^2 + 1^2 + 2) = 2\) so the base case is true. (Inductive step) Suppose our formula is true for \(n=k\), so we have placed \(k\) lines in general position and divided the plane into \(\frac12(k^2+k+2)\) regions. When we place a new line it will meet those \(k\) lines in \(k\) places (since no lines are parallel) and there will be k+1 regions the line will run through (since no three lines meet at a point). Each of those \(k+1\) regios is now split in half, so there are \(k+1\) "new regions". Therefore there are now \(\frac12(k^2+k+2)+(k+1) = \frac12(k^2+k+1+2k+2) = \frac12 ((k+1)^2+(k+1)+1)\) regions, ie our hypothesis is true for \(n=k+1\). (Conclusion) Therefore since our statement is true for \(n=1\) and since if it is true for some \(n=k\) it is true for \(n=k+1\) by the principle of mathematical induction it is true for all \(n \geq 1\) Proof: (Alternative). There are \(\binom{n}{2}\) places where the lines meet. Each intersection is the most extreme point (say lowest) for one region (if two are tied, rotate by a very small amount) so this is a unique mapping. There will be \(n+1\) regions which are infinite and don't have a most extreme point, hence \(\binom{n}{2} + n+1 = \frac12(n^2-n)+n+1 = \frac12(n^2+n+2)\)
Prove that, for any integers \(n\) and \(r\), with \(1\leqslant r\leqslant n,\) \[ \binom{n}{r}+\binom{n}{r-1}=\binom{n+1}{r}. \] Hence or otherwise, prove that \[ (uv)^{(n)}=u^{(n)}v+\binom{n}{1}u^{(n-1)}v^{(1)}+\binom{n}{2}u^{(n-2)}v^{(2)}+\cdots+uv^{(n)}, \] where \(u\) and \(v\) are functions of \(x\) and \(z^{(r)}\) means \(\dfrac{\mathrm{d}^{r}z}{\mathrm{d}x^{r}}\). Prove that, if \(y=\sin^{-1}x,\) then \((1-x^{2})y^{(n+2)}-(2n+1)xy^{(n+1)}-n^{2}y^{(n)}=0.\)
Solution: \begin{align*} \binom{n}{r} + \binom{n}{r-1} &= \frac{n!}{r!(n-r)!} + \frac{n!}{(r-1)!(n-r+1)!} \\ &= \frac{n!}{(r-1)!(n-r)!} \left ( \frac{1}{r} + \frac{1}{n-r+1} \right) \\ &= \frac{n!}{(r-1)!(n-r)!} \frac{(n-r+1)+r}{r(n-r+1)} \\ &= \frac{n! (n+1)}{r! (n-r+1)!} \\ &= \frac{(n+1)!}{r!(n+1-r)!} \\ &= \binom{n+1}{r} \end{align*} Claim: \(\displaystyle (uv)^{(n)} = \sum_{r=0}^n \binom{n}{r} u^{(n-r)} v^{(r)}\) Proof: (By induction on \(n\)). Base case: \(n = 0\) is clear. Inductive step: Suppose it is true for \(n = k\), then consider \begin{align*} (uv)^{(k+1)} &= \left ( (uv)^{(k)} \right)' \\ &= \left ( \sum_{r=0}^k \binom{k}{r} u^{(k-r)} v^{(r)} \right)' \tag{by assumption} \\ &=\sum_{r=0}^k \binom{k}{r} \left ( u^{(k-r)} v^{(r)}\right)' \tag{linearity} \\ &=\sum_{r=0}^k \binom{k}{r} \left ( u^{(k-r+1)} v^{(r)} + u^{(k-r)}v^{(r+1)}\right) \\ &= \sum_{r=0}^{k} \binom{k}{r} u^{(k-r+1)} v^{(r)} + \sum_{r=0}^{k} \binom{k}{r} u^{(k-r)}v^{(r+1)} \\ &= \sum_{r=0}^{k} \binom{k}{r} u^{(k-r+1)} v^{(r)} + \sum_{r=1}^{k+1} \binom{k}{r-1} u^{(k-r+1)}v^{(r)} \\ &= u^{(k+1)}v + \sum_{r=1}^k \left (\binom{k}{r} + \binom{k}{r-1} \right)u^{(k-r+1)}v^{(r)} + u v^{(k+1)}\\ &= u^{(k+1)}v + \sum_{r=1}^k \binom{k+1}{r} u^{(k-r+1)}v^{(r)} + u v^{(k+1)}\\ &= \sum_{r=0}^{k+1} \binom{k+1}{r} u^{(k-r+1)}v^{(r)}\\ \end{align*} Therefore if our statement is true for \(n = k\) it is true for \(n = k+1\). Since it is true for \(n = 0\) by the principle of mathematical induction it is true for all integer \(n \geq 0\) Suppose \( y = \sin^{-1} x\), then \(y' = \frac{1}{\sqrt{1-x^2}}\), \(y'' = \frac{x}{(1-x^2)^{3/2}}\). Not that this means that \((1-x^2)y'' - xy' = 0\) (which is our formula when \(n = 0\)). Now apply Leibniz's formula to this. \begin{align*} 0 &= \left ( (1-x^2)y'' - xy' \right)^{(n)} \\ &= \left ( (1-x^2)y'' \right)^{(n)} -\left ( xy' \right)^{(n)} \\ &= \left ( (1-x^2)y^{(n+2)} - n\cdot 2x \cdot y^{(n+1)}-\binom{n}{2} \cdot 2 \cdot y^{(n)} \right )- \left (xy^{(n+1)}+ny^{(n)} \right) \\ &= (1-x^2)y^{(n+2)} - (2n+1)y^{(n+1)} - \left ( n(n-1)+n \right)y^{(n)} \\ &= (1-x^2)y^{(n+2)} - (2n+1)y^{(n+1)} - n^2y^{(n)} \\ \end{align*} as required
The matrix \(\mathbf{M}\) is given by \[ \mathbf{M}=\begin{pmatrix}\cos(2\pi/m) & -\sin(2\pi/m)\\ \sin(2\pi/m) & \cos(2\pi/m) \end{pmatrix}, \] where \(m\) is an integer greater than \(1.\) Prove that \[ \mathbf{M}^{m-1}+\mathbf{M}^{m-2}+\cdots+\mathbf{M}^{2}+\mathbf{M}+\mathbf{I}=\mathbf{O}, \] where $\mathbf{I}=\begin{pmatrix}1 & 0\\ 0 & 1 \end{pmatrix}\( and \)\mathbf{O}=\begin{pmatrix}0 & 0\\ 0 & 0 \end{pmatrix}.$ The sequence \(\mathbf{X}_{0},\mathbf{X}_{1},\mathbf{X}_{2},\ldots\) is defined by \[ \mathbf{X}_{k+1}=\mathbf{PX}_{k}+\mathbf{Q}, \] where \(\mathbf{P,Q}\) and \(\mathbf{X}_{0}\) are given \(2\times2\) matrices. Suggest a suitable expression for \(\mathbf{X}_{k}\) in terms of \(\mathbf{P},\) \(\mathbf{Q}\) and \(\mathbf{X}_{0},\) and justify it by induction. The binary operation \(*\) is defined as follows: \[ \mathbf{X}_{i}*\mathbf{X}_{j}\mbox{ is the result of substituting \ensuremath{\mathbf{X}_{j}}for \ensuremath{\mathbf{X}_{0}}in the expression for \ensuremath{\mathbf{X}_{i}}. } \] Show that if \(\mathbf{P=M},\) the set \(\{\mathbf{X}_{1},\mathbf{X}_{2},\mathbf{X}_{3},\ldots\}\) forms a finite group under the operation \(*\).
Solution: \(\mathbf{M}^m = \mathbf{I}\), we also have \(\mathrm{det}(\mathbf{M - I}) = \cos^2(2\pi/m) - 2\cos(2\pi/m) + 1 + \sin^2(2\pi/m) = 2(1-\cos(2\pi/m))\) therefore \(\mathbf{M-I}\) is invertible. Therefore since \(\mathbf{(M-I)(M^{m-1} + M^{m-1} + \cdots + M^2 + M + I)= M^m-I = 0}\) we can cancel the \(\mathbf{M-I}\) to obtain the desired result. \(\mathbf{X_0 = X_0}\) \(\mathbf{X_1 = PX_0+Q}\) \(\mathbf{X_2 = P(PX_0+Q)+Q = P^2X_0 + PQ + Q}\) Claim: \(\mathbf{X_k = P^k X_0 + (P^{k-1} + P^{k-2} + \cdots + I)Q}\) Proof: (By induction on \(k\)). Base case \(k = 0\) is true. Assume it's true for some \(k = l\), then consider \(k = l+1\) \(\mathbf{X_{l+1} = PX_l + Q = P( P^l X_0 + (P^{l-1} + P^{l-2} + \cdots + I)Q) + Q = P^{l+1}X_0 + (P^l + P^{l-1} + \cdots + P)Q + Q = P^{l+1}X_0 + (P^l + P^{l-1} + \cdots + P + I)Q}\) Suppose \(\mathbf{P} = \mathbf{M}\), then consider the set \(\{\mathbf{X_1, X_2}, \ldots\}\) with the operation \(*\) as defined. \(\mathbf{X_i * X_j} = M^{i}(X_j) + (M^{i-1} + M^{i-2} + \cdots + M + I)Q = M^{i}(M^jX_0 + (M^{j-1} + M^{j-2} + \cdots + M + I)Q) + (M^{i-1} + M^{i-2} + \cdots + M + I)Q = M^{i+j}X_0 + (M^{i+j-1}+\cdots + M + I)Q = X_{i+j}\) Since \(X_m = X_0\) we can check all the requirements of the group, but this is going to be isomorphic to the cyclic group with \(m\) elements.
Solution: