Problems

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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 \(n \geq 1, n \in \mathbb{Z}\)

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Solution: First notice that \((x^2+2+2x)(x^2+2-2x) = (x^2+2)^2-4x^2 = x^2+4\) and so \begin{align*} && \frac{1}{x^4+4} &= \frac{ax+b}{ x^2+2x+2}+\frac{cx+d}{ x^2-2x+2} \\ \Rightarrow && 1 &= (ax+b)(x^2-2x+2) + (cx+d)(x^2+2x+2) \\ &&&= (a+c)x^3+(b-2a+d+2c)x^2+(2a-2b+2c+2d)x+2b+2d \\ \Rightarrow && 0 &= a+c \\ && 2a &= b+d+2c \\ && b &= a+c+d \\ && \frac12 &= b+d \\ \Rightarrow && c &= -a \\ && 2a &= \frac12+2(-a) \\ \Rightarrow && a &= \frac18, c = -\frac18 \\ && b &= d \\ \Rightarrow && b &= \frac14, d = \frac14 \end{align*} Therefore \begin{align*} \int_0^1\frac {\d x}{ x^4+4} &= \int_0^1 \frac18\left ( \frac{x+2}{x^2+2x+2} -\frac{x-2}{x^2-2x+2}\right) \d x \\ &= \frac1{16}\int_0^1 \left ( \frac{2x+2+2}{x^2+2x+2} -\frac{2x-2-2}{x^2-2x+2}\right) \d x \\ &= \frac1{16} \left [ \ln(x^2+2x+2) -\ln(x^2-2x+2)\right]_0^1 + \frac1{16} \int_0^1 \left ( \frac{2}{(x+1)^2+1} + \frac{2}{(x-1)^2+1} \right) \d x \\ &= \frac1{16} \left (\ln 5 - \ln 1 -(\ln 2-\ln 2) \right) + \frac18 \left [ \tan^{-1} (x+1)+\tan^{-1}(x-1) \right]_0^1 \\ &= \frac1{16} \ln 5 + \frac18 \left (\tan^{-1} 2+\tan^{-1} 0 - \tan^{-1}1-\tan^{-1} (-1) \right) \\ &= \frac1{16} \ln 5+ \frac18 \tan^{-1} 2 \end{align*}

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Solution: Notice by polynomial division, we can write \(p(x) = (x-a)q(x) + r(x)\) where degree \(r(x) < 1\), ie \(r(x)\) is a constant. Evaluating at \(x = a\), we have \(p(a) = (a-a)q(a) + r(a) = r(a)\). Therefore \(r(a) = p(a)\) and since \(r(x)\) is a constant, it is always \(p(a)\).

1. \(\,\) \begin{align*} && p(x) &= (x-1)(x-2)(x-3)q(x) + r(x) \\ && p(1) &= r(1) = 3 \\ && p(2) &= r(2) = 1 \\ && p(3) &= r(3) = 5 \end{align*} Therefore \(r(x)\) is a polynomial of degree \(2\) or less through \((1,3),(2,1), (3, 5)\) we can write this as \begin{align*} && r(x) &= 5\frac{(x-1)(x-2)}{(3-1)(3-2)} + 1\frac{(x-1)(x-3)}{(2-1)(2-3)} + 3\frac{(x-2)(x-3)}{(1-2)(1-3)} \\ &&&= \frac52(x^2-3x+2)-(x^2-4x+3) + \frac32(x^2-5x+6) \\ &&&= 3x^2-11x+11 \end{align*}
2. Let \(P_n(x) = x(x-1)\cdots(x-n) + x\), then \(P_n(a) = a\)

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Solution: \begin{align*} && y^2 &= x^2a^2-x^2 \\ &&&= \frac{a^4}{4} -\left ( x^2 -\frac{a^2}{2} \right)^2 \end{align*} Therefore the maximum and minimum values of \(y\) are \(\pm \frac{a^2}2\)

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Solution: This is just the result that all of the area beneath \(g(t)\) is also below \(f(t)\) If \(p > q > 0, x \geq 1 \Rightarrow x^p \geq x^q\), therefore applying the result we have \begin{align*} && \int_1^x x^p\, \d t & \geq \int_1^x x^q\, \d t \\ \Rightarrow && \frac{x^p-1}{p} & \geq \frac{x^q-1}{q} \end{align*} When \(p > q > 0, 0 \leq x \leq 1\) we have \(x^p \leq x^q\), ie \begin{align*} && \int_x^1 x^q\, \d t & \geq \int_{x}^1 x^p\, \d t \\ \Rightarrow && \frac{1-x^q}{q} & \geq \frac{1-x^p}{p} \\ \Rightarrow && \frac{x^p-1}{p} &\geq \frac{x^q-1}{q} \end{align*} Now looking at the functions \(f(x) = \frac{x^p-1}{p}, g(x) = \frac{x^q-1}{q}\) and \(x \geq 1\) we have \begin{align*} && \int_0^x \frac{t^p-1}{p} \d t & \geq \int_0^x \frac{t^q-1}{q} \d t\\ \Rightarrow &&\frac1p \left[\frac{t^{p+1}}{p+1} - t \right]_0^x &\geq\frac1q \left[\frac{t^{q+1}}{q+1} - t \right]_0^x \\ \Rightarrow &&\frac1p \left(\frac{x^{p+1}}{p+1} -x\right) &\geq\frac1q \left(\frac{x^{q+1}}{q+1} - x \right)\\ \Rightarrow &&\frac1p \left(\frac{x^{p}}{p+1} -1\right) &\geq\frac1q \left(\frac{x^{q}}{q+1} - 1 \right)\\ \end{align*}

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Solution:

The ball is on the ground when \(\cos \theta = \frac12 \Rightarrow \theta = 60^\circ\) and ball will make it over the step when \(\theta = 0^\circ\). It is also worth emphasising we are moving \emph{very slowly}, so we can treat the system as static at any given point. \begin{align*} \overset{\curvearrowleft}{X}: && mg \frac{d}{2}\sin \theta - F \frac{d}{2} \l 1 + \cos \theta \r &= 0\\ \Rightarrow && \frac{mg \sin \theta}{1 + \cos \theta} &= F& \\ \Rightarrow && mg \tan \frac{\theta}{2} &= F& \\ \end{align*} Therefore \(F\) is maximised when \(\theta = 60^\circ\), ie \(F_{max} = \frac{mg}{\sqrt{3}}\) \begin{align*} \text{N2}(\parallel OX): && mg \cos \theta - R + F \sin \theta &= 0 \\ \Rightarrow && mg \cos \theta - R + \frac{mg\sin \theta}{1 + \cos \theta} \sin \theta &= 0 \\ \Rightarrow && mg &= R \\ \\ \text{N2}(\perp OX): && F_X - mg \sin \theta + F \cos \theta &= 0 \\ \Rightarrow && mg \sin \theta - \frac{mg\sin \theta}{1 + \cos \theta} \cos \theta &= F_X \\ \Rightarrow && \frac{mg\sin \theta}{1 + \cos \theta} &= F_X \tag{We could also see this taking moments about \(O\)}\\ % \text{N2}(\rightarrow): && F + \mu R \cos \theta - R \sin \theta &\geq 0 \\ % \text{N2}(\uparrow): && -mg +\mu R \sin \theta + R \cos \theta &\geq 0 \\ % \Rightarrow && R \l \sin \theta - \mu \cos \theta\r &\leq F \\ % \Rightarrow && R \l \mu \sin \theta + \cos \theta\r &\geq mg \\ % \Rightarrow && \l \frac{\sin \theta - \mu \cos \theta}{\mu \sin \theta + \cos \theta} \r mg & \leq F \\ % \Rightarrow && \l \frac{\tan \theta - \mu }{1+\mu \tan \theta} \r mg & \leq F \\ % \Rightarrow && \tan \l \theta - \alpha \r mg & \leq F \tag{where \(\tan \alpha = \mu\)} \end{align*} Therefore since \(F_X \leq \mu R\), \(\displaystyle \frac{mg\sin \theta}{1 + \cos \theta} \leq \mu mg \Rightarrow \mu \geq \tan \frac{\theta}{2}\) which is maximised at \(\theta = 60^\circ\) and implies \(\mu \geq \frac{1}{\sqrt{3}}\)

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Solution: We have \begin{align*} \rightarrow: && a &= v\cos \alpha t_{net} \\ \Rightarrow && t_{net} &= \frac{a}{v \cos \alpha} \\ \downarrow: && H-h &= v\sin \alpha t_{net} + \frac12 g t_{net}^2 \\ &&&= a \tan \alpha + \frac12 g \frac{a^2}{v^2} \sec^2 \alpha \\ &&&= a \tan \alpha + \frac{a^2g}{2v^2}(1 + \tan^2 \alpha) \tag{*}\\ \\ \rightarrow: && a+b &= v \cos \alpha t_{ground} \\ && t_{ground} &= \frac{a+b}{v \cos \alpha}\\ \downarrow: && H &= v\sin \alpha t_{ground} + \frac12 g t_{ground}^2 \\ &&&= (a+b)\tan \alpha + \frac{(a+b)^2g}{2v^2}(1+\tan^2\alpha) \tag{**} \\ \\ (**): && v^2 &= \frac{g(a+b)^2(1+\tan^2\alpha)}{2[H-(a+b)\tan \alpha]} \\ (a+b)^2(*) - a^2(**): && (a+b)^2(H-h) -a^2H &= [(a+b)^2a - a^2(a+b)]\tan \alpha \\ \Rightarrow && (2ab+b^2)H - (a+b)^2h &= ab(a+b) \tan \alpha \\ \Rightarrow && \tan \alpha &= \frac{2a+b}{a(a+b)}H - \frac{a+b}{ab} h \end{align*} Noting that \(v^2 \geq 0\) and the numerator is positive, we must have \begin{align*} && H &> (a+b)\tan \alpha \\ &&&= \frac{2a+b}{a}H - \frac{(a+b)^2}{ab} h \\ \Rightarrow && \frac{a+b}{a}H &< \frac{(a+b)^2}{ab} h \\ \Rightarrow && H &< \frac{a+b}{b} h \end{align*} Noting that \(\tan \alpha > 0\) we must have \begin{align*} && \frac{2a+b}{a(a+b)} H & > \frac{a+b}{ab} h \\ \Rightarrow && H &> \frac{(a+b)^2}{b(2a+b)}h \end{align*}

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Solution:

1. Suppose the needle's center is \(x\) cm from the nearest line and makes an angle of \(\theta\). Then if \(\sin \theta > x\) it will cross the line, otherwise it will not. Given that \(\theta \sim U(0, \frac{\pi}{2})\), we can see that \begin{align*} && \mathbb{P}(\text{needle crosses}) &= \mathbb{P}(\sin \theta > x) \\ &&&= \mathbb{P}(\theta > \sin^{-1} x) \\ &&&= 1-\frac{2\sin^{-1} x}{\pi} \end{align*}
2. The length of the short segment is \(L = 1 - \frac{x}{\sin \theta}\) and \(\theta \sim U(\sin^{-1} x, \frac{\pi}{2})\). So \begin{align*} && F_L(l) &= \mathbb{P}(L < l) \\ &&&= \mathbb{P}\left (1 - \frac{x} {\sin \theta} < l\right) \\ &&&= \mathbb{P}\left ( \sin \theta < \frac{x}{1-l}\right) \\ &&&= \mathbb{P}\left (\theta < \sin^{-1} \frac{x}{1-l}\right) \\ &&&= \frac{ \sin^{-1} \frac{x}{1-l} - \sin^{-1} x }{\frac{\pi}{2} - \sin^{-1}x} \end{align*}
3. The needle (with probability \(1\)) cannot hit \(2\) lines, so let's only consider the line it's nearest too. The distance to this line is uniform on \([0,1]\), and the so we want to calculate. \begin{align*} && \mathbb{P}(\text{needle crosses}) &= \int_0^1 \left (1 - \frac{2\sin^{-1}x}{\pi} \right) \d x \\ &&&= 1 - \frac{2}{\pi} \int_0^1 \sin^{-1} x \d x\\ &&&= 1 - \frac{2}{\pi} \left ( \frac{\pi}{2} - 1 \right) \\ &&&= \frac{2}{\pi} \end{align*} Therefore the probability it misses is \(1 - \frac{\pi}{2}\)

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Solution:

1. \begin{align*} (x^2+5x+4)e^x &= \sum_{k=0}^\infty \frac{1}{k!} x^{k+2}+\sum_{k=0}^\infty \frac{5}{k!} x^{k+1}+\sum_{k=0}^\infty \frac{4}{k!} x^{k} \\ &= \sum_{k=0}^{\infty} \l \frac{1}{k!}+\frac{5}{(k+1)!}+\frac{4}{(k+2)!} \r x^{k+2} + 5x+4+4x \\ &= 4 + 9x + \sum_{k=0}^{\infty} \l \frac{(k+2)(k+1)}{(k+2)!}+\frac{5(k+2)}{(k+2)!}+\frac{4}{(k+2)!} \r x^{k+2}\\ &= 4 + 9x + \sum_{k=0}^{\infty} \l \frac{k^2+3k+2+5k+10+4}{(k+2)!} \r x^{k+2}\\ &= 4 + 9x + \sum_{k=0}^{\infty} \frac{(k+4)^2}{(k+2)!} x^{k+2}\\ &= 4 + 9x + \sum_{k=2}^{\infty} \frac{(k+2)^2}{k!} x^{k}\\ \end{align*} So when \(x = 1\) we have \[10e = 4 + \frac{3^2}{1!} + \frac{4^2}{2!} + \frac{5^2}{3!} + \cdots \]
2. \begin{align*} (x^2+3x+1)e^x &= \sum_{k=0}^\infty \frac{1}{k!}x^{k+2}+\sum_{k=0}^\infty 3\frac{1}{k!}x^{k+1} + \sum_{k=0}^{\infty} \frac{1}{k!} x^k \\ &= 1+3x+\sum_{k=1}^{\infty} \l \frac1{(k-1)!}+\frac{3}{k!} + \frac{1}{(k+1)!} \r x^{k+1} \\ &= 1+3x+\sum_{k=1}^{\infty} \frac{(k+1)k + 3(k+1)+1}{(k+1)!}x^{k+1} \\ &= 1+3x+\sum_{k=1}^{\infty} \frac{k^2+4k+4}{(k+1)!}x^{k+1} \\ &= 1+3x+\sum_{k=0}^{\infty} \frac{(k+2)^2}{(k+1)!}x^{k+1} \\ &=1+3x+ \sum_{k=1}^{\infty} \frac{(k+1)^2}{k!}x^k \end{align*} Plugging in \(x=1\) we get the desired result.
3. \begin{align*} && xe^x &= \sum_{k=0}^{\infty} \frac{x^{k+1}}{k!} \\ x\frac{\d}{\d x} : && x(1+x)e^x &= \sum_{k=0}^{\infty} \frac{(k+1)x^{k+1}}{k!} \\ x\frac{\d}{\d x} : && x(x(1+x)+1+2x)e^x &= \sum_{k=0}^{\infty} \frac{(k+1)^2x^{k+1}}{k!} \\ &&(x^3+3x^2+x)e^x &= \sum_{k=0}^{\infty} \frac{(k+1)^2x^{k+1}}{k!} \\ \frac{\d}{\d x} : && e^x(x^3+3x^2+x+3x^2+6x+1) &=\sum_{k=0}^{\infty} \frac{(k+1)^3x^{k}}{k!} \\ \Rightarrow && 15e &= 1 + \frac{2^3}{1!} + \frac{3^3}{2!} + \cdots \end{align*}