Problems

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Solution:

1. \begin{align*} && f(\beta) &= \beta - \frac1{\beta}-\frac1{\beta^2} \\ \Rightarrow && f'(\beta) &= 1 +\frac{1}{\beta^2}+\frac{2}{\beta^3} \\ \Rightarrow && 0 &= f'(\beta) \\ &&&= 1 + \frac1{\beta^2} + \frac{2}{\beta^3} \\ \Rightarrow && 0 &= \beta^3 + \beta + 2 \\ &&&= (\beta+1)(\beta^2-\beta+2) \end{align*} Therefore the only stationary point is at \(\beta = -1, f(-1) = -1\)
   

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   \begin{align*} && g(\beta) &= \beta + \frac3{\beta}-\frac1{\beta^2} \\ \Rightarrow && g'(\beta) &= 1 -\frac{3}{\beta^2}+\frac{2}{\beta^3} \\ \Rightarrow && 0 &= f'(\beta) \\ &&&= 1 - \frac3{\beta^2} + \frac{2}{\beta^3} \\ \Rightarrow && 0 &= \beta^3 - 3\beta + 2 \\ &&&= (\beta-1)^2(\beta+2) \end{align*} Therefore there are stationary points at \(\beta=1,f(1) = 3, \beta=-2, f(-2) = \frac14\)
   

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2. Let \(u,v\) be the roots of \(x^2 + \alpha x + \beta = 0\), then since \((x-u)(x-v) = 0\) we must have \(\alpha = -(u+v), \beta = uv\). Therefore: \begin{align*} && u+v +\frac{1}{uv} &= -\alpha + \frac{1}{\beta} \\ && \frac1u+\frac1v + uv &= \frac{u+v}{uv} + uv \\ &&&= -\frac{\alpha}{\beta} + \beta \end{align*} Given \(u+v + \frac 1 {uv} = -1\), ie \(-\alpha + \frac{1}{\beta} = -1\). Since the roots are real, we must also have that \(\alpha^2 - 4\beta \geq 0\), so \begin{align*} && -\alpha + \frac1\beta &= -1 \\ \Rightarrow && \alpha &= 1 +\frac1\beta \\ \Rightarrow && -\frac{\alpha}{\beta}+\beta &= -\frac{1}{\beta} \l1+\frac1{\beta}\r + \beta \\ &&&=\beta - \frac{1}{\beta}-\frac{1}{\beta^2} \end{align*} So we want to maximise \(f(\beta)\) subject to \(\alpha ^2 - 4\beta \geq 0\) \begin{align*} && 0 &\leq \alpha^2 -4 \beta \\ &&&= \l 1 + \frac1{\beta} \r^2 - 4\beta \\ &&&= 1+ \frac2{\beta} + \frac1{\beta^2} - 4\beta \\ \Leftrightarrow && 0 &\leq -4\beta^3+\beta^2 + 2\beta + 1 \\ &&&=-(\beta-1)(4\beta^2+3\beta+1)\\ \Leftrightarrow && \beta &\leq 1 \end{align*} But we know \(f(\beta) \leq -1\) on \((-\infty,1]\) so we're done.
3. Given that \(-\alpha + \frac{1}{\beta} = 3\) we have \begin{align*} && -\alpha + \frac1\beta &= 3 \\ \Rightarrow && \alpha &= -3 +\frac1\beta \\ \Rightarrow && -\frac{\alpha}{\beta}+\beta &= -\frac{1}{\beta} \l-3+\frac1{\beta}\r + \beta \\ &&&=\beta + \frac{3}{\beta}-\frac{1}{\beta^2} \end{align*} which we want to maximise, subject to: \begin{align*} && 0 &\leq \alpha^2 -4 \beta \\ &&&= \l -3 + \frac1{\beta} \r^2 - 4\beta \\ &&&= 9- \frac6{\beta} + \frac1{\beta^2} - 4\beta \\ \Leftrightarrow && 0 &\leq -4\beta^3+9\beta^2 - 6\beta + 1 \\ &&&=-(\beta-1)^2(4\beta-1)\\ \Leftrightarrow && \beta &\leq \frac14 \end{align*} Therefore the maximum will either be \(f(-2) = \frac14\) or \(f(\frac14) = -\frac{15}4\). Therefore the maximum is \(\frac14\)

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Solution:

1. \begin{align*} \frac{\d y_n}{\d x} &= \frac{\d}{\d x} \l (-1)^n e^{x^2} \frac{\d^n}{\d x^{n}} \l e^{-x^2}\r \r \\ &= (-1)^n 2xe^{x^2} \frac{\d^n}{\d x^{n}} \l e^{-x^2}\r + (-1)^n e^{x^2} \frac{\d^{n+1}}{\d x^{n+1}} \l e^{-x^2}\r \\ &= 2xy_n - (-1)^{n+1} e^{x^2} \frac{\d^{n+1}}{\d x^{n+1}} \l e^{-x^2}\r \\ &= 2xy_n - y_{n+1} \end{align*}
2. \(y_0 = 1\), \(y_1 = (-1) e^{x^2} \cdot (-2x) \cdot e^{-x^2} = 2x\), \(y_2 = e^{x^2} \frac{\d^2}{\d x^2} \l e^{-x^2}\r = e^{x^2} \frac{\d }{\d x}\l -2xe^{-x^2} \r = e^{x^2} \l -2e^{-x^2}+4x^2e^{-x^2}\r = 4x^2-2\). Therefore \(2xy_1 - 2y_0 = 2x \cdot 2x - 2\cdot1 = 4x^2-2 = y_2\) so our statement is true for \(n=1\). Assume the statement is true for \(n=k\), then \begin{align*} && y_{k+1} &= 2xy_k - 2ky_{k-1} \\ \frac{\d }{\d x}: && \frac{\d y_{k+1}}{\d x} &= 2\frac{\d}{\d x}\l xy_k \r - 2k\frac{\d y_{k-1}}{\d x} \\ \Rightarrow && 2xy_{k+1}-y_{k+2} &= 2y_k+2x \l 2xy_k-y_{k+1}\r - 2k \l 2xy_{k-1}-y_k \r \\ \Rightarrow && y_{k+2} &=2y_k+ 4x \cdot y_{k+1}-(4x^2+2k)y_k+2x \cdot 2k y_{k-1} \\ &&&= 4x \cdot y_{k+1}-(4x^2+2(k+1))y_k+2x \l2xy_k - y_{k+1} \r \\ &&&= 2x \cdot y_{k+1} -2(k+1)y_k \end{align*} Therefore since our statement is true for \(n=1\) and if it is true for \(n=k\) it is true for \(n=k=1\), therefore by the principle of mathematical induction it is true for all \(n \geq 1\). Since \(2x = \frac{y_{n+1}+2ny_{n-1}}{y_n}\) for all \(n\), we must have \begin{align*} && \frac{y_{n+1}+2ny_{n-1}}{y_n} &= \frac{y_{n+2}+2(n+1)y_{n}}{y_{n+1}} \\ \Leftrightarrow && y_{n+1}^2+2ny_{n-1}y_{n+1} &= y_ny_{n+2}+2ny_n^2+2y_n^2 \\ \Leftrightarrow && y_{n+1}^2-y_ny_{n+2} &= 2n(y_n^2-y_{n-1}y_{n+1})+2y_n^2 \end{align*}
3. Consider the functions \(f_n(x) = y_{n}^2-y_{n-1}y_{n+1}\) then clearly \(f_{n+1} = 2nf_{n} + 2y_n^2 \geq f_{n}\) so to prove \(f_n(x) > 0\) for \(n \geq 1\) it suffices to prove it for \(n = 1\). But \(f_1 = y_1^2 - y_0y_{2} = (2x)^2-(4x^2-2) = 2 > 0\) so we are done.

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Solution: Consider $x^a \big(x^b (x^cy)'\big)'$ then \begin{align*} x^a \big(x^b (x^cy)'\big)' &= x^a \big (bx^{b-1}(x^c y)'+x^b(x^cy)'' \big ) \\ &= x^a \big (bx^{b-1} (cx^{c-1}y + x^c y') + x^b(c(c-1)x^{c-2}y + 2cx^{c-1}y' + x^cy'') \\ &= x^{a+b+c}y'' + (2cx^{c-1+b+a}+bx^{c+b-1+a})y'+(c(b+c-1))x^{a+b+c-2} y \end{align*} So we need: \begin{align*} &&& \begin{cases} a+b+c &= 2 \\ 2c+b &= 1-2p \\ c(b+c-1) &= p^2-q^2 \end{cases} \\ \Rightarrow && c((1-2p)-2c+c-1) &=p^2-q^2 \\ \Rightarrow && c^2+2pc &= q^2-p^2 \end{align*}

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Solution: Note that \begin{align*} && \frac{\d a \sec \theta}{\d \theta} &= a \sec \theta \tan \theta \\ && \frac{\d b \tan \theta}{\d \theta} &= b \sec^2 \theta \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{b \sec^2 \theta}{a \sec \theta \tan \theta} \\ &&&= \frac{b}{a} \frac{1}{\sin \theta} \\ \Rightarrow && \frac{y-b \tan \theta}{x - a \sec \theta} &= \frac{b}{a} \frac{1}{\sin \theta} \\ \Rightarrow && a \sin \theta y - ab \tan \theta \sin \theta &= bx -ab \sec \theta \\ \Rightarrow && bx-ay\sin \theta &= ab \sec x (1 - \sin ^2 \theta) \\ &&&= ab \cos \theta \end{align*}

1. \begin{align*} S: &&& \begin{cases} bx-ay &= 0 \\ bx-ay \sin \theta &= ab \cos \theta \end{cases} \\ \Rightarrow && ay(1-\sin \theta) &= ab \cos \theta \\ \Rightarrow && y &= \frac{b \cos \theta}{1-\sin \theta} \\ &&x &=\frac{a\cos \theta}{1-\sin \theta} \\ T: &&& \begin{cases} bx+ay &= 0 \\ bx-ay \sin \theta &= ab \cos \theta \end{cases} \\ \Rightarrow && ay(1+\sin \theta) &= -ab \cos \theta \\ \Rightarrow && y &= \frac{-b \cos \theta}{1+\sin \theta} \\ &&x &=\frac{a\cos \theta}{1+\sin \theta} \\ M: && x &= \frac{a \cos \theta}{2} \frac{2}{1-\sin^2 \theta} \\ &&&= a \sec \theta \\ && y &= \frac{b \cos \theta}{2} \frac{2 \sin \theta}{1-\sin^2 \theta} \\ &&&= b \tan \theta \end{align*} The midpoint of \(ST\) is the same as \(P\).
2. The tangents are perpendicular, therefore \(\frac{b}{a} \cosec \theta = - \frac{a}{b} \sin \phi\), ie \(b^2 = -a^2 \sin \phi \sin \theta\) The will intersect at: \begin{align*} &&& \begin{cases} bx - ay \sin \theta &= ab \cos \theta \\ bx - ay \sin \phi &= ab \cos \phi \end{cases} \\ \Rightarrow && ay ( \sin \theta - \sin \phi) &= ab(\cos \phi - \cos \theta) \\ \Rightarrow && y &= \frac{b(\cos \phi - \cos \theta)}{(\sin \theta - \sin \phi)} \\ && y^2 &= \frac{-a^2 \sin \phi \sin \theta (\cos\phi - \cos \theta)^2}{(\sin \theta - \sin \phi)^2} \\ \Rightarrow && bx(\sin \phi - \sin \theta) &= ab(\cos \theta \sin \phi - \cos \phi \sin \theta) \\ \Rightarrow && x &= \frac{a(\cos \theta \sin \phi - \cos \phi \sin \theta)}{\sin \phi - \sin \theta} \\ &&&= \frac{a^2(\cos \theta \sin \phi - \cos \phi \sin \theta)^2}{(\sin \phi - \sin \theta)^2} \end{align*} Therefore \begin{align*} && x^2+y^2 &= \frac{a^2}{(\sin \phi - \sin \theta)^2} \l (\cos \theta \sin \phi- \cos \phi \sin \theta)^2 - \sin \phi \sin \theta (\cos\phi - \cos \theta)^2 \r \\ &&&= \frac{a^2}{(\sin \phi - \sin \theta)^2} \l (\sin \phi - \sin \theta)(\cos^2 \theta \sin \phi - \sin \theta \cos^2 \phi) \r \\ &&&=a^2-b^2 \end{align*}

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Solution:

1. \begin{align*} && (k+1) (A_{k+1} - G_{k+1}) & \geq k(A_k - G_k) \\ \Leftrightarrow && \sum_{i=1}^{k+1} a_i - (k+1)G_{k+1} &\geq \sum_{i=1}^k a_i - kG_k \\ \Leftrightarrow && a_{k+1} -(k+1)G_k^{k/(k+1)}a_{k+1}^{1/(k+1)} & \geq - k G_k \\ \Leftrightarrow && a_{k+1} -(k+1)G_k^{k/(k+1)}a_{k+1}^{1/(k+1)} + k G_k& \geq 0\\ \Leftrightarrow && \frac{a_{k+1}}{G_k} -(k+1)G_k^{k/(k+1)-1}a_{k+1}^{1/(k+1)} + k & \geq 0\\ \Leftrightarrow && \lambda_k^{k+1} -(k+1)\lambda_k+ k & \geq 0\\ \end{align*} as required.
2. \begin{align*} && f'(x) &= (k+1)x^k - (k+1) \\ &&&= (k+1)(x^k-1) \end{align*} Therefore \(f(x)\) is strictly decreasing on \((0,1)\) and strictly increasing on \((1,\infty)\) and so the minimum will be \(f(1) = 1 - (k+1) + k = 0\), so \(f(x) \geq 0\) with equality only at \(x = 1\).
3. 1. We can proceed by induction to show since the inequality holds for \(n=1\) and since if it holds for \(n=k\) it will hold for \(n=k+1\) as \(A_{k+1}-G_{k+1}\) must have the same sign as \(A_k - G_k\).
   2. The only way for equality to hold is if \(\lambda_k = 1\) for \(k = 1, \cdots n\), ie \(a_{k+1} = G_k\), but this means \(a_2 = a_1, a_3 = a_1\) etc. Therefore all values are equal.

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Solution:

1. \(A\), \(Q\), \(C\) lie on a straight line if \(q = \lambda a + (1-\lambda)c\) for some \(\lambda \in \mathbb{R}\), \begin{align*} && q &= \lambda a + (1-\lambda)c \\ \Leftrightarrow && q - a &= (\lambda - 1)a + (1-\lambda)c \\ \Leftrightarrow && q - a &= (\lambda - 1)(a-c) \\ \Leftrightarrow && \frac{q - a}{c-a} &= 1-\lambda \\ \end{align*} therefore \(\frac{q-a}{c-a} \in \mathbb{R}\) \begin{align*} && \frac{q-a}{c-a} & \in \mathbb{R} \\ \Leftrightarrow && \left (\frac{q-a}{c-a} \right)^* &= \frac{q-a}{c-a} \\ \Leftrightarrow && (q^*-a^*)(c-a) &= (q-a)(c^*-a^*) \\ \end{align*} Given \(aa^* = cc^* = 1\), \begin{align*} && (q^*-a^*)(c-a) &= (q-a)(c^*-a^*) \\ \Leftrightarrow && q^*(c-a) - \frac{c}{a}+1 &= q \frac{a-c}{ca} - \frac{a}{c}+1 \\ \Leftrightarrow && (c-a)\l q^* +\frac{q}{ca}\r &= \frac{c}{a} - \frac{a}{c} \\ &&&= \frac{c^2-a^2}{ac} \\ \Leftrightarrow && q^* +\frac{q}{ca} &= \frac{c+a}{ac} \\ \Leftrightarrow && q^*ac +q &= a+c \end{align*}
2. Since \(Q\) lies on \(AC\) and \(BD\) we must have \begin{align*} &&& \begin{cases} q^*ac +q &= a+c \\ q^*bd +q &= b+d \\ \end{cases} \\ \Rightarrow && q^*(ac-bd) &= (a+c)-(b+d) \\ \Rightarrow && q(ac-bd) &= (b+d)ac-(a+c)bd \\ \Rightarrow && (q+q^*)(ac-bd) &= (a+c)(1-bd)+(b+d)(ac-1) \\ &&&=a-abd+c-bcd+abc-b+acd-d \\ &&&= a(1+cd)-b(1+cd)+c(1+ab)-d(1+ab) \\ &&&= (a-b)(1+cd)+(c-d)(1+ab) \end{align*}
3. If \(AB\) and \(CD\) meet at \(p\) we must have \(p^*ab + p = a+b\), ie \(p(1+ab) = a+b\) amd \(p(1+cd) = c+d\), so \begin{align*} && (q+q^*)(ac-bd) &= (a-b) \frac{c+d}{p} + (c-d) \frac{a+b}{p} \\ \Leftrightarrow && p(q+q^*)(ac-bd) &= (a-b)(c+d)+(c-d)(a+b) \\ &&&= ac+ad-bc-bd+ac+bc-ad-bd \\ &&&= 2(ac-bd) \\ \Leftrightarrow && p(q+q^*) &= 2 \end{align*}

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Solution:

1. \begin{align*} \frac{\left(\cot \theta + i\right)^{2n+1} -\left(\cot \theta - i\right)^{2n+1}}{2i} &= \frac{1}{\sin^{2n+1} \theta}\frac{\left(\cos \theta + i \sin \theta \right)^{2n+1} -\left(\cos\theta - i\sin \theta\right)^{2n+1}}{2i} \\ &= \frac{1}{\sin^{2n+1} \theta} \frac{e^{i(2n+1) \theta} - e^{-i(2n+1) \theta} }{2i} \\ &=\frac{\sin (2n+1) \theta}{\sin^{2n+1} \theta} \end{align*} Notice that: \begin{align*} (\cot \theta + i)^{2n+1} - (\cot \theta - i)^{2n+1} &= \sum_{k=0}^{2n+1} \binom{2n+1}{k}(i)^k \cdot \cot^{2n+1-k} \theta - \sum_{k=0}^{2n+1} \binom{2n+1}{k}(-i)^k \cdot \cot^{2n+1-k} \theta \\ &= \sum_{k=0}^{2n+1} \binom{2n+1}{k} \l i^k - (-i)^k \r \cdot \cot^{2n+1-k} \theta \\ &= \sum_{l=0}^{n} \binom{2n+1}{2l+1} \l i^{2l+1} - (-i)^{2l+1} \r \cdot \cot^{2n+1-(2l+1)} \theta \\ &= \sum_{l=0}^{n} \binom{2n+1}{2l+1} 2i \cdot \cot^{2(n-l)} \theta \\ &= 2i\sum_{l=0}^{n} \binom{2n+1}{2l+1} \cot^{2(n-l)} \theta \\ \end{align*} Therefore if \(\theta\) satisfies \(\frac{\sin (2n+1) \theta}{\sin^{2n+1} \theta} = 0\) then \(z = \cot^2 \theta\) satisfies the equation. But \(\theta = \frac{m \pi}{2n+1}, m = 1, 2, \ldots, n\) are \(n\) distinct all the roots must be \(\cot^2 \frac{m \pi}{2n+1}\).
2. Notice that the sum of the roots will be \(\displaystyle \frac{\binom{2n+1}{3}}{\binom{2n+1}{1}} = \frac{(2n+1)\cdot 2n \cdot (2n-1)}{3! \cdot (2n+1)} = \frac{n \cdot (2n-1)}{3}\) and so \[ \sum_{m=1}^n \cot^{2}\left(\frac{m\pi}{2n+1}\right) =\frac{n\left(2n-1\right)}{3}. \]
3. For \(0 < \theta < \frac{1}{2}\pi\), \begin{align*} && 0 < \sin \theta < \theta < \tan \theta \\ \Leftrightarrow && 0 < \cot \theta < \frac{1}{\theta} < \frac{1}{\sin \theta} \\ \Leftrightarrow && 0 < \cot^2 \theta < \frac{1}{\theta^2} < \cosec^2 \theta = 1 + \cot^2 \theta\\ \end{align*} Therefore \begin{align*} && \sum_{n=1}^N \cot^2 \frac{n \pi}{2N+1} < \sum_{n=1}^N \frac{(2N+1)^2}{n^2 \pi^2} < N + \sum_{n=1}^N \cot^2 \frac{n \pi}{2N+1} \\ \Rightarrow && \frac{1}{(2N+1)^2} \frac{N(2N-1)}{3} < \sum_{n=1}^N \frac{1}{n^2 \pi^2} < \frac{1}{(2N+1)^2} \l \frac{N(2N-1)}{3} + 1 \r \\ \Rightarrow && \lim_{N \to \infty}\frac{1}{(2N+1)^2} \frac{N(2N-1)}{3} < \lim_{N \to \infty}\sum_{n=1}^N \frac{1}{n^2 \pi^2} < \lim_{N \to \infty}\frac{1}{(2N+1)^2} \l \frac{N(2N-1)}{3} + 1 \r \\ \Rightarrow && \frac{1}{6} \leq \lim_{N \to \infty}\sum_{n=1}^N \frac{1}{n^2 \pi^2} \leq \frac16 \\ \Rightarrow && \sum_{n=1}^N \frac{1}{n^2} = \frac{\pi^2}{6} \end{align*}

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Solution:

1. \begin{align*} && I &= \int_0^1 \frac{f(x^{-1})}{1+x} \d x \\ u = x^{-1}, \d u = -x^{-2} \d x: &&&= \int_{\infty}^1 \frac{f(u)}{1+\frac{1}{u}} \frac{-1}{u^2} \d u \\ &&&= \int_1^{\infty} \frac{f(u)}{u(1+u)} \d u \\ &&&= \sum_{n=1}^{\infty} \int_n^{n+1} \frac{f(u)}{u(u+1)} \d u \\ \\ \text{if} f(x) = f(x+1)\, \forall x && &=\sum_{n=1}^{\infty} \int_{0}^1 \frac{f(x+n)}{(x+n)(x+n+1)} \d x \\ &&&= \sum_{n=1}^\infty \int_0^1 \frac{f(x)}{(x+n)(x+n+1)} \d x \\ &&&= \int_0^1 f(x) \l \sum_{n=1}^{\infty} \frac{1}{(x+n)(x+n+1)}\r \d x \\ &&&= \int_0^1 f(x) \l \sum_{n=1}^{\infty} \l \frac{1}{x+n} - \frac{1}{x+n+1} \r\r \d x \\ &&&= \int_0^1 f(x) \l \frac{1}{x+1} \r \d x \\ &&&= \int_0^1\frac{f(x)}{x+1} \d x \\ \end{align*}
2. Since the fractional part is periodic with period \(1\), we can say \begin{align*} && \int_0^1 \frac{\{ x^{-1} \} }{1+x} \d x &= \int_0^1 \frac{\{ x\}}{x+1} \d x \\ &&&= \int_0^1 \frac{x}{x+1} \d x \\ &&&= \int_0^1 1-\frac{1}{x+1} \d x \\ &&&= [x - \ln (1+x) ]_0^1 \\ &&&= 1 - \ln 2 \end{align*} \begin{align*} && \int_0^1 \frac{\{ 2x^{-1}\}}{1+x} \d x &= \int_0^1 \frac{\{2x\}}{x+1} \d x \\ &&&= \int_0^{1/2} \frac{2x}{x+1} \d x +\int_{1/2}^{1} \frac{2x-1}{x+1} \d x \\ &&&= 2\int_0^1 \frac{x}{x+1} \d x + \int_{1/2}^1 \frac{-1}{x+1} \d x \\ &&&= 2 - 2\ln 2 - \l \ln 2 - \ln \tfrac32 \r \\ &&&= 2 - 4 \ln 2 + \ln 3 \\ &&&= 2 + \ln \tfrac {3}{16} \end{align*}
3. \begin{align*} && \int_0^1 \frac{x \{ x^{-1} \} }{1-x^2} \d x &= \frac12 \l \int_0^1 \frac{ \{ x^{-1} \}}{1-x} - \frac{\{x^{-1} \}}{1+x} \d x\r \end{align*} Consider for \(f\) periodic with period \(1\) \begin{align*} \int_0^1 \frac{f(x^{-1})}{1-x} \d x &= \int_1^{\infty} \frac{f(u)}{u(u-1)} \d u \\ &= \sum_{n=1}^{\infty}\int_{n}^{n+1} \frac{f(u)}{u(u-1)} \d u \\ &= \sum_{n=1}^{\infty}\int_{0}^{1} \frac{f(u)}{(u+n)(u+n-1)} \d u \\ &= \int_{0}^{1} \sum_{n=1}^{\infty}\frac{f(u)}{(u+n)(u+n-1)} \d u \\ &= \int_{0}^{1} f(u) \sum_{n=1}^{\infty}\l\frac{1}{u+n-1} - \frac{1}{u+n} \r\d u \\ &= \int_0^1 \frac{f(u)}{u} \d u \end{align*} So we have \begin{align*} && \int_0^1 \frac{x \{ x^{-1} \} }{1-x^2} \d x &= \frac12 \l \int_0^1 \frac{ \{ x^{-1} \}}{1-x} - \frac{\{x^{-1} \}}{1+x} \d x \r \\ &&&= \frac12 \int_0^1 \frac{\{ x \}}{x} \d x - \frac12 (1 - \ln 2) \\ &&&= \frac12 - \frac12 + \frac12 \ln 2 \\ &&&= \frac12 \ln 2 \end{align*}

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Solution:

1. Just before collision \(n-1\): Velocity of \(P\) is \(u_{n-2}\) Velocity of \(Q\) is \(-v_{n-2}\) \begin{align*} COM: && mu_{n-2}+km(-v_{n-2}) &= mu_{n-1}+kmv_{n-1} \\ \Rightarrow && u_{n-2}-kv_{n-2} &= u_{n-1}+kv_{n-1} \\ NEL: && v_{n-1}-u_{n-1} &= -e((-v_{n-2})-u_{n-2}) \\ \Rightarrow && v_{n-1}-u_{n-1} &= e(v_{n-2}+u_{n-2}) \end{align*} \begin{align*} &&kv_{n-1} &= u_{n-2} - kv_{n-2}-u_{n-1} \\ &&kv_{n-1}&= ku_{n-1}+kev_{n-2}+keu_{n-2} \\ \Rightarrow && kv_{n-2}(1+e) &= u_{n-2}(1-ke)-u_{n-1}(1+k) \\ \Rightarrow && kv_{n-1}(1+e) &= u_{n-1}(1-ke)-u_{n}(1+k) \\ && k(1+e)v_{n-1}-k(1+e)u_{n-1} &= k(1+e)e(v_{n-2}+u_{n-2}) \\ \Rightarrow && u_{n-1}(1-ke)-u_{n}(1+k)-k(1+e)u_{n-1} &= e(u_{n-2}(1-ke)-u_{n-1}(1+k))+k(1+e)eu_{n-2} \\ \Rightarrow && 0 &= (1+k)u_n + ((ke-1)+k(1+e)-e(1+k))u_{n-1} \\ &&& \quad \quad + (e(1-ke)+k(1+e)e)u_{n-2} \\ \Rightarrow && 0 &= (1+k)u_n- (1-k)(1+e)u_{n-1} +e(1+k)u_{n-2} \end{align*}
2. \(u_0 = A + B\) \begin{align*} &&& \begin{cases}u_0 - kv_0 &= kv_1 + u_1 \\ \frac12 (u_0+v_0) &= v_1 - u_1 \\ \end{cases} \\ \Rightarrow && (1+k)u_1 &= u_0 - kv_0 - \frac{k}{2}(u_0 + v_0) \\ \Rightarrow && u_1 &= \frac{1}{k+1} \l u_0 (1-\frac{k}{2}) - \frac32 k v_0 \r \\ &&&= \frac{67}{70} u_0 - \frac{3}{70} v_0 \end{align*} Therefore \(A+B = u_0, \frac{49A+50B}{70} = \frac{67}{70} u_0 - \frac{3}{70} v_0\) \begin{align*} && A+B &= u_0 \\ && 49A+50B &= 67u_0 - 3v_0 \\ \Rightarrow && 50u_0 - A &= 67u_0 - 3v_0 \\ \Rightarrow && A &= -17u_0 + 3v_0 \\ && B &= 18u_0 - 3v_0 \end{align*} If \(0 < 6u_0 < v_0\), then \(B < 0\) and as \(n \to \infty\) we will find that \(\l \frac57 \r^n\) dominates \(\l \frac7{10} \r^n\) and so our velocity will be negative and the particle will change direction

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Solution:

First, notice that the centre of mass will lie directly below \(A\) and will be \(\frac{m}{M+m}\) of the way between \(O\) and \(P\). Therefore \(\sin \beta = \frac{m}{M+m}\). The cosine rule states that: \begin{align*} && AP^2 &= a^2 + a^2 - 2a^2 \cos \angle AOP \\ \Rightarrow && \frac{AP^2}{a^2} &= 2 - 2 \sin \beta \\ &&&= \frac{2M+2m-2m}{M+m} \\ &&&= \frac{2M}{M+m} \\ \Rightarrow && \frac{AP}{a} &= \sqrt{\frac{2M}{M+m}} \end{align*} Considering conservation of energy, we have: Rotational kinetic energy for the disc: \(\frac12 I \dot{\theta}^2\) Kinetic energy for the particle: \(\frac12 m (\dot{\theta} \sqrt{2-2\sin \beta} a)^2 = (1- \sin \beta)ma^2 \dot{\theta}^2\) GPE: The important thing is the vertical location of \(G\). The triangle \(OAG\) will still have angle \(\beta\) at \(A\). The vertical height below is: \(\cos \theta \cdot AG = \cos \theta a \cos \beta\). The distance from when \(\theta = 0\) will be \(a \cos \beta (1- \cos \theta)\) and so the GPE will be \((M+m)ga \cos \beta ( 1- \cos \theta)\) we can therefore say by conservation of energy: \[ \frac12 I \dot{\theta}^2 + (1- \sin \beta)ma^2 \dot{\theta}^2+(M+m)ga \cos \beta ( 1- \cos \theta) \] is constant. Suppose \(m = \frac32 M\) and \(I = \frac32 Ma^2\) then differentiating the constant wrt to \(\theta\) gives \(\sin \beta = \frac{m}{M+m} = \frac{3}{5}, \cos \beta = \frac45\) \begin{align*} && 0 &= \frac12 \frac32 M a^2 \cdot 2 \dot{\theta}\ddot{\theta} + (1- \sin \beta)\frac32M a^2 2 \dot{\theta}\ddot{\theta} + (M+\frac32M) ga \cos \beta \sin \theta \cdot \dot{\theta} \\ \Rightarrow && 0 &= \frac32 \ddot{\theta} + 3(1-\sin \beta) \ddot{\theta} + \frac{5}{2}\frac{g}{a} \cos \beta \sin \theta \\ &&&= (\frac32 + \frac65) \ddot{\theta} + \frac{2g}{a} \sin \theta \\ &&&= \frac{27}{10} \ddot{\theta} + \frac{2g}{a} \sin \theta \end{align*} If \(\theta\) is small, we can approximate this by: \(\frac{27}{10} \ddot{\theta} + \frac{2g}{a} \theta = 0\) which will have period \(\displaystyle 2 \pi \sqrt{\frac{27a}{10\cdot2 g}} = 2 \pi \sqrt{\frac{3a}{5g}}\) as required.