Problems

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Solution:

1. \(\,\) \begin{align*} && f(x) &= f(1-x) \\ \Rightarrow && f'(x) &= -f'(1-x) \\ \Rightarrow && f'(\tfrac12) &= -f'(\tfrac12) \\ \Rightarrow && f'(\tfrac12) &= 0 \\ \\ && f(x) &= f(\tfrac1x) \\ \Rightarrow && f'(x) &= f'(\tfrac1x) \cdot \frac{-1}{x^2} \\ \Rightarrow && f'(-1) &= -f'(-1) \\ \Rightarrow && f'(-1) &= 0 \\ \\ && f'(2) &= -\frac{1}{4}f'(\tfrac12) \\ &&&= 0 \end{align*}
2. Suppose \begin{align*} && f(x) &= \frac{(x^2-x+1)^3}{(x^2-x)^2} \\ && f(1/x) &= \frac{(x^{-2}-x^{-1}+1)^3}{(x^{-2}-x^{-1})^2} \\ &&&= \frac{(1-x+x^2)^3/x^6}{((x-x^2)^2/x^6} \\ &&&= f(x) \\ \\ && f(1-x) &= \frac{((1-x)^2-(1-x)+1)^3}{((1-x)^2-(1-x))^2} \\ &&&= \frac{(1-x+x^2)^3}{(x^2-x)^2} = f(x) \end{align*}
   

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3. Clearly \(x = -1\) is a root of \(f(x) = \frac{27}{4}\), so we must also have \(x=2\) and \(x = \frac12\), therefore \(f(x) > \frac{27}{4}\) if \(x \in \mathbb{R} \setminus \{-1, 2, \tfrac12, 0, 1 \}\). Clearly \(x = 3\) and \(x = -2\) are solutions so we also have: \(\frac13, -\frac12, \frac32, \frac23\) and these must be all solutions so we must have: \(f(x) > \frac{343}{36} \Leftrightarrow x \in (-\infty, -2) \cup (-\frac12, 0) \cup (0, \frac13) \cup (\frac23, 1) \cup (1, \frac32) \cup (3, \infty)\)

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Solution:

1. \(\,\) \begin{align*} && u_{n+2} - u_n &= 1 + \frac{1}{u_{n+1}} - \left (1 + \frac{1}{u_{n-1}} \right) \\ &&&= \frac{1}{1 + \frac1{u_n}} - \frac{1}{1 + \frac{1}{u_{n-2}}} \\ &&&= \frac{u_n}{u_n+1} - \frac{u_{n-2}}{1+u_{n-2}} \\ &&&= \frac{u_n(1+u_{n-2}) - u_{n-2}(1+u_n)}{(1+u_n)(1+u_{n-2})} \\ &&&= \frac{u_n - u_{n-2}}{(1+u_n)(1+u_{n-2})} \end{align*}
2. Claim: \(u_n \in [1,2]\) Proof: (By induction). Note that \(u_1 = 1, u_2 = 2\) so our claim is true for the first few terms. Note that if \(u_n \in [1,2]\), \(\frac{1}{u_n} \in [\tfrac12, 1]\) and \(1+\frac{1}{u_{n}} \in [\tfrac32,2] \subset [1,2]\). Therefore \(u_{n+1} \in [1,2]\). Therefore since \(u_1 \in [1,2]\) and \(u_n \in [1,2] \Rightarrow u_{n+1} \in [1,2]\) \(u_n \in [1,2]\) for all \(n \ge 1\)
3. First notice that \(u_3 = \frac32 > u_1\) and therefore by the recursion we found in the first part, \(u_{2n+1}-u_{2n-1} > 0\) so \(u_{2k+1}\) is increasing and bounded, and so by our theorem converges to a limit. Suppose this limit is \(L\), then we must have \(L = 1 + \frac1{L} \Rightarrow L^2 - L - 1 = 0 \Rightarrow L = \frac{1+\sqrt5}{2}\) since it must be in \([1,2]\). Similarly, not that \(u_4 = \frac{5}{3} < u_2\) and so \(u_{2k+2} - u_{2k} < 0\) and \(-u_{2k}\) is increasing and bounded above. Therefore it tends to a limit (and so does \(u_{2k}\)). By the same reasoning as before, it's the same limit, \(\frac{1+\sqrt5}{2}\) and therefore the sequence converges. If \(u_1 = 3, u_2 = \frac43 \in [1,2]\) so we have our sequence being bounded and all the same logic will follow through.

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Solution:

1. \((x,y) = (1,0)\) we have Suppose \(p^2-2q^2 = 1\), then \begin{align*} && (3p+4q)^2-2\cdot(2p+3q)^2 &= 9p^2+24pq + 16q^2 - 2\cdot(4p^2+12pq+9q^2) \\ &&&= p^2(9-8) + pq(24-24) + q^2(16-18) \\ &&&= p^2 - 2q^2 = 1 \end{align*} So we have: \begin{array}{c|c} x & y \\ \hline 1 & 0 \\ 3 & 2 \\ 17 & 12 \\ \end{array}
2. Suppose \(x = 2m+1\) and \(y = 2n\) then \begin{align*} && 1 & = x^2 - 2y^2 \\ &&&= (2m+1)^2 - 2(2n)^2 \\ &&&= 4m^2 + 4m + 1 - 8n^2 \\ \Leftrightarrow && n^2 &= \frac{m(m+1)}{2} \end{align*}
3. Suppose \(b^3 = c^4 - a^2 =(c^2-a)(c^2+a)\), since \(b\) is prime and \(c^2 + a > c^2-a\) we must have: \begin{align*} && p = c^2-a && p^2 =c^2 +a \\ \Rightarrow && c^2 = \frac{p+p^2}{2} && a = \frac{p^2-p}2\\ && 1 = c^2-a && p^3 = c^2+a \\ \Rightarrow && c^2 = \frac{p^3+1}{2} && a = \frac{p^3-1}{2} \end{align*} Note that \(c^2 = \frac{p(p+1)}{2}\) is reminicent of our first equation, so suppose \(n = c = 6\) and \(p = m = 8\) then \(6^4 = 8^3 + 28^2\)

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Solution:

1. First, not that the point must be ony the curve:
   

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   Since otherwise it's pretty clear we could make the area of the rectangle larger by moving the point onto the curve. Therefore \(A = x(f(x)-f(t))\). To maximise this we need \(xf'(x) + f(x)-f(t) = 0\), ie \(x_0f'(x_0) + f(x_0) = f(t)\)
2. Suppose \(\displaystyle \g(t) =\frac 1t \int_0^t \f(x) \d x\) then \begin{align*} && \g(t) &=\frac 1t \int_0^t \f(x) \d x\\ \Rightarrow && tg(t) &= \int_0^t \f(x) \d x \\ \Rightarrow && tg'(t) +g(t) &= f(t) \\ \Rightarrow && tg'(t) &= f(t) - g(t) \end{align*}
   

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   Clearly the blue area + green area is larger than the green area. So \(\displaystyle \int_0^t (f(x) - f(t))\d x > A_0(t)\). Notice that \(f(t) = \frac1{t} \int_0^t f(t) \d x \) so \(-t^2g'(t) = \int_0^t f(x) \d x > A_0(t)\)
3. Not that if \(f(x) = \dfrac{1}{1+x}\), the \(f'(x) = -\frac{1}{(1+x)^2}\) and so \begin{align*} && -\frac{x_0}{(1+x_0)^2} + \frac{1}{1+x_0} &= \frac{1}{1+t} \\ && \frac{1}{(1+x_0)^2} &= \frac{1}{1+t} \\ \Rightarrow && x_0 &= \sqrt{1+t} - 1 \\ && A_0(t) &= (\sqrt{1+t} - 1) \left ( \frac{1}{\sqrt{1+t}} - \frac{1}{t+1} \right) \\ &&&= 1 - \frac{1}{\sqrt{1+t}} - \frac{1}{\sqrt{1+t}} + \frac{1}{1+t} \\ &&&= \frac{2+t}{1+t} - \frac{2}{\sqrt{1+t}} \\ && g(t) &= \frac{1}{t} \int_0^t \frac{1}{1+x} \d x \\ &&&= \frac{\ln(1+t)}{t} \\ \Rightarrow && g'(t) &= \frac{\frac{t}{1+t} - \ln(1+t)}{t^2} \\ \Rightarrow && -t^2g(t) &= \ln(1+t) - \frac{t}{1+t} \\ \Rightarrow && \ln(1+t) - \frac{t}{1+t} &> \frac{2+t}{1+t} - \frac{2}{\sqrt{1+t}} \\ \Rightarrow && \ln \sqrt{1+t} & > 1 - \frac{1}{\sqrt{1+t}} \end{align*}

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Solution:

1. First Collision:
   

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   By NEL, \(v_2 = v_1 + eu\), so \begin{align*} \text{COM}: && mu &= mv_1 + m(v_1 + eu) \\ \Rightarrow && 2mv_1 &= mu(1-e) \\ \Rightarrow && v_1 &= \frac12 u(1-e) \\ && v_2 &= \frac12 u(1-e) + eu \\ &&&= \frac12 u(1+e) \end{align*} The second collision is identical to the first except replacing \(u\) with \(\frac12u(1+e)\), therefore after that collision: \begin{align*} && \text{first particle} &= \frac12 u(1-e) \\ && \text{second particle} &= \frac12 \left (\frac12 u(1+e) \right)(1-e) \\ &&&= \frac14 u(1-e^2) \\ && \text{third particle} &= \frac12 \left (\frac12 u(1+e) \right)(1+e) \\ &&&= \frac14 u(1+e)^2 \end{align*} After all these collisions, all particles are moving in the same direction (since they all have positive velocity), but the first particle is now travelling faster than the second particle (as \(\frac12(1-e) < 1\)). Therefore they will collide again.
2. The third collision:
   

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   The speed of approach will be \(\frac12u(1-e) - \frac14u(1-e^2) = \frac14u(1-e)(2 - (1+e)) = \frac14 u(1-e)^2\), therefore by NEL, \(w_2 = w_1 + \frac14ue(1-e)^2\) \begin{align*} \text{COM}: && m\frac12u(1-e) + m \frac14u(1-e^2) &= mw_1 + m\left (w_1 + \frac14ue(1-e)^2 \right) \\ \Rightarrow && \frac14u(1-e)(2+(1+e)) &= 2w_1 + \frac14ue(1-e)^2 \\ \Rightarrow && 2w_1 &= \frac14u(1-e)(3+e)-\frac14ue(1-e)^2 \\ &&&= \frac14u(1-e)(3+e-e(1-e)) \\ &&&= \frac14u(1-e)(3+e^2) \\ \Rightarrow && w_1 &= \frac18 u(1-e)(3+e^2) \\ && w_2 &= \frac18 u(1-e)(3+e^2) + \frac14ue(1-e)^2 \\ &&&= \frac18u(1-e)(3+e^2+2e(1-e)) \\ &&&= \frac18u(1-e)(3+2e-e^2) \\ &&&= \frac18u(1-e)(1+e)(3-e) \\ \end{align*} A fourth collision is possible, iff \begin{align*} && \frac18u(1-e)(1+e)(3-e)&> \frac14 u(1+e)^2 \\ \Leftrightarrow && (1-e)(3-e)&> 2 (1+e) \\ \Leftrightarrow &&3-4e-e^2&> 2+2e \\ \Leftrightarrow &&1-5e-e^2&>0 \\ \Leftrightarrow && e &< 3-\sqrt{2} \end{align*}

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Solution:

1. \begin{align*} \mathbb{E}(X) &= \sum_{r=1}^\infty r \mathbb{P}(X = r) \\ &= \sum_{j=1}^{\infty} (2j-1)\mathbb{P}(U=2j-1) \\ &= \sum_{j=1}^{\infty}(2j-1) \frac{e^{-\lambda} \lambda^{2j-1}}{(2j-1)!} \\ &= \sum_{j=1}^{\infty} e^{-\lambda} \frac{\lambda^{2j-1}}{(2j-2)!} \\ &= \lambda e^{-\lambda} \sum_{j=1}^{\infty} \frac{\lambda^{2j-2}}{(2j-2)!} \\ &= \lambda e^{-\lambda} \alpha \end{align*} Since \(\mathbb{E}(X+Y) = \lambda, \mathbb{E}(Y) = \lambda(1-e^{-\lambda}\alpha) = \lambda(e^{-\lambda}(\alpha+\beta) - e^{-\lambda}\alpha) = \lambda e^{-\lambda} \beta\). Alternatively, as \(\beta + \alpha = e^{\lambda}\), \(\mathbb{E}(X) = \frac{\lambda \alpha}{\alpha+\beta}, \mathbb{E}(Y) = \frac{\lambda \beta}{\alpha+\beta}\)
2. \begin{align*} \textrm{Var}(X) &= \mathbb{E}(X^2) - [\mathbb{E}(X) ]^2 \\ &= \sum_{odd} r^2 \mathbb{P}(U = r) - \left [ \mathbb{E}(X) \right]^2 \\ &= \sum_{odd} (r(r-1)+r)\frac{e^{-\lambda}\lambda^r}{r!} - \frac{\lambda^2 \alpha^2}{(\alpha+\beta)^2} \\ &= \sum_{odd} \frac{e^{-\lambda}\lambda^r}{(r-2)!}+\sum_{odd} \frac{e^{-\lambda}\lambda^r}{(r-1)!} - \frac{\lambda^2 \alpha^2}{(\alpha+\beta)^2} \\ &= e^{-\lambda}\lambda^2 \beta + e^{-\lambda}\lambda \alpha - \frac{\lambda^2 \alpha^2}{(\alpha+\beta)^2} \\ &= \frac{\lambda \alpha + \lambda^2 \beta}{\alpha+\beta}- \frac{\lambda^2 \alpha^2}{(\alpha+\beta)^2} \end{align*} Similarly, \begin{align*} \textrm{Var}(Y) &= \mathbb{E}(Y^2) - [\mathbb{E}(Y) ]^2 \\ &= \sum_{even} r^2 \mathbb{P}(U = r) - \left [ \mathbb{E}(Y) \right]^2 \\ &= \sum_{even} (r(r-1)+r)\frac{e^{-\lambda}\lambda^r}{r!} - \frac{\lambda^2 \beta^2}{(\alpha+\beta)^2} \\ &= e^{-\lambda}\lambda^2\alpha + e^{-\lambda}\lambda \beta - \frac{\lambda^2 \beta^2}{(\alpha+\beta)^2} \\ &= \frac{\lambda \beta + \lambda^2 \alpha}{\alpha+\beta}- \frac{\lambda^2 \beta^2}{(\alpha+\beta)^2} \end{align*} Since \(\textrm{Var}(X+Y) = \textrm{Var}(U) = \lambda\), we are interested in solving: \begin{align*} \lambda &= \frac{\lambda \alpha + \lambda^2 \beta}{\alpha+\beta}- \frac{\lambda^2 \alpha^2}{(\alpha+\beta)^2} + \frac{\lambda \beta + \lambda^2 \alpha}{\alpha+\beta}- \frac{\lambda^2 \beta^2}{(\alpha+\beta)^2} \\ &= \frac{\lambda(\alpha+\beta) + \lambda^2(\alpha+\beta)}{\alpha+\beta} - \frac{\lambda^2(\alpha^2+\beta^2)}{(\alpha+\beta)^2} \\ &= \lambda + \lambda^2 \frac{(\alpha+\beta)^2 - (\alpha^2+\beta^2)}{(\alpha+\beta)^2} \\ &= \lambda + \lambda^2 \frac{2\alpha\beta}{(\alpha+\beta)^2} \end{align*} which is clearly not possible if \(\lambda \neq 0\)

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Solution:

1. The only way \(A = 1\) is if we get \(HH\) or \(TT\) which has probability \(p^2+q^2\). The only way we get \(S=1\) is if we have \(HT\) to \(TH\), ie \(2pq\). Since \((p-q)^2 = p^2 + q^2 - 2pq >0\) we must have \(\mathbb{P}(A=1) > \mathbb{P}(S=1)\).
2. \(\,\) \begin{align*} \mathbb{P}(S=2) &= p^2q + q^2p \\ \mathbb{P}(A=2) &= pq^2 + qp^2 = \mathbb{P}(S=2) \\ \\ \mathbb{P}(S=3) &= p^3q + q^3p = pq(p^2+q^2) \\ \mathbb{P}(A=3) &= pqp^2 + qpq^2 = pq(p^2+q^2) = \mathbb{P}(S=3) \end{align*}
3. For \(n > 1\) we must have \begin{align*} && \mathbb{P}(S = 2n) &= p^{2n}q + q^{2n}p \\ && \mathbb{P}(A=2n) &= (pq)^{n}q + (qp)^{n}p \\ &&&= p^nq^{n+1} + q^np^{n+1} \\ && \mathbb{P}(S = 2n) &> \mathbb{P}(A = 2n) \\ \Leftrightarrow && p^{2n}q + q^{2n}p & > p^nq^{n+1} + q^np^{n+1}\\ \Leftrightarrow && 0 & < p^{2n}q+q^{2n}p - p^nq^{n+1} -q^np^{n+1}\\ &&&= (p^n-q^n)(qp^n - pq^n) \end{align*} which is clearly true. \begin{align*} && \mathbb{P}(S=2n+1) &= p^{2n+1}q + q^{2n+1}p \\ && \mathbb{P}(A=2n+1) &= (pq)^{n}p^2 + (qp)^{n}q^2 \\ &&&= p^{n+2}q^n + q^{n+2}p^n \end{align*} The same factoring logic shows that \(\mathbb{P}(S = 2n+1) > \mathbb{P}(A=2n+1)\)

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Solution: Let \(t = \tan \frac12 x\), then \begin{align*} \frac{\d t}{\d x} &= \tfrac12 \sec^2 \tfrac12 t \\ &= \tfrac12 (1 + \tan^2 \tfrac12 ) \\ &= \tfrac12 (1 + t^2) \\ \\ \sin x &= 2 \sin \tfrac12 x \cos \tfrac12 \\ &= \frac{2 \frac{\sin \tfrac12 x}{ \cos \tfrac12x}}{\frac{1}{\cos^2 \tfrac12 x}} \\ &= \frac{2 \tan \tfrac12 x}{\sec^2 \tfrac12 } \\ &= \frac{2t }{1+t^2} \end{align*} Now consider \begin{align*} t = \tan \tfrac12 x: && \int_0^{\frac12\pi} \frac 1{1+a \sin x}\, \d x &= \int_{t=0}^{t = 1} \frac{1}{1 + a \frac{2t}{1+t^2}} \frac{2}{1+t^2} \d t \\ &&&= \int_0^1 \frac{2}{1+2at+t^2} \d t \\ &&&= \int_0^1 \frac{2}{(t+a)^2 + 1-a^2} \d t \\ (1-a^2) > 0: &&&= \left [ \frac{2}{\sqrt{1-a^2}} \arctan \frac{t+a}{\sqrt{1-a^2}} \right]_0^1 \\ &&&= \frac{2}{\sqrt{1-a^2}} \left ( \arctan \frac{1+a}{\sqrt{1-a^2}} - \arctan \frac{a}{\sqrt{1-a^2}} \right) \\ &&&= \frac{2}{\sqrt{1-a^2}} \arctan \left ( \frac{\frac{1+a}{\sqrt{1-a^2}}-\frac{a}{\sqrt{1-a^2}}}{1+\frac{1+a}{\sqrt{1-a^2}}\frac{a}{\sqrt{1-a^2}}} \right) \\ &&&= \frac{2}{\sqrt{1-a^2}} \arctan \left ( \frac{\frac{1}{\sqrt{1-a^2}}}{\frac{1+a}{1-a^2}} \right) \\ &&&= \frac{2}{\sqrt{1-a^2}} \arctan \left ( \frac{\sqrt{1-a}}{\sqrt{1+a}} \right) \end{align*} as required. Let \[ I_n = \int_0^{\frac12\pi} \frac{ \sin ^nx}{2+\sin x} \, \d x \qquad \quad (n\ge0). \] and consider \begin{align*} I_{n+1} + 2I_n &= \int_0^{\frac12\pi} \frac{ \sin ^{n+1}x+2\sin^{n} x}{2+\sin x} \, \d x \\ &= \int_0^{\frac12\pi} \frac{ \sin^n x (2 + \sin x)}{2+\sin x} \, \d x \\ &= \int_0^{\frac12\pi} \sin^n x \d x \end{align*} Therefore we can compute \begin{align*} I_0 &= \int_0^{\pi/2} \frac{1}{2 + \sin x} \d x \\ &= \frac12 \int_0^{\pi/2} \frac{1}{1 + \frac12 \sin x} \d x \\ &= \frac{1}{\sqrt{3/4}} \arctan \frac{\sqrt{1/2}}{\sqrt{3/2}} \\ &= \frac{2}{\sqrt{3}} \arctan \frac{1}{\sqrt{3}} \\ &= \frac{\pi}{3\sqrt{3}} \\ \\ I_1 &= \int_0^{\pi/2} 1 \d x - 2 I_0 \\ &= \frac{\pi}{2} - \frac{2\pi}{3\sqrt{3}} \\ I_2 &= \int_0^{\pi/2} \sin x \d x - 2I_1 \\ &= 1 - \pi + \frac{4\pi}{3\sqrt{3}} \\ I_3 &= \int_0^{\pi/2} \sin^2 x \d x - 2I_2 \\ &= \frac12 \int_0^{\pi/2} \sin^2 + \cos^2 x \d x - 2I_2 \\ &= \frac{\pi}{4} - 2 + 2\pi - \frac{8\pi}{3\sqrt{3}} \\ &= -2 + \frac{9\pi}{4} - \frac{8\pi}{3\sqrt{3}} \end{align*}

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Solution: \begin{align*} && y &= \frac{\arcsin x}{\sqrt{1-x^2}} \\ && \frac{\d y}{\d x} &= \frac{(1-x^2)^{-1/2} \cdot (1-x^2)^{1/2}-\arcsin x \cdot (-x)(1-x^2)^{-1/2}}{1-x^2} \\ &&&= \frac{1+ xy}{1-x^2} \\ \Rightarrow && 0 &= (1-x^2) \frac{\d y}{\d x} -xy-1\\ \\ \frac{\d^n}{\d x^{n+1}}: && 0 &= \left ( (1-x^2) y' \right)^{(n+1)} - (xy)^{(n+1)} \\ \Rightarrow && 0 &= (1-x^2)y^{(n+2)} + \binom{n+1}{1}(1-x^2)^{(1)}y^{(n+1)}+\binom{n+1}{2} (1-x^2)^{(2)}y^{(n)} - (xy^{(n+1)} +\binom{n+1}{1} y^{(n)} ) \\ &&&= (1-x^2)y^{(n+2)}+\left ( (n+1)\cdot(-2x)-x \right)y^{(n+1)} + \left ( \frac{(n+1)n}{2} \cdot (-2)-(n+1) \right)y^{(n)} \\ &&&= (1-x^2)y^{(n+2)}-\left ( 2n+3 \right)xy^{(n+1)} - \left ( (n+1)n+(n+1)\right)y^{(n)} \\ &&&= (1-x^2)y^{(n+2)}-\left ( 2n+3 \right)xy^{(n+1)} - \left ( n+1\right)^2y^{(n)} \\ \end{align*} Since \(y(0) = 0, y'(0) = 1\) we can look at the recursion: \(y^{(n+2)} - (n+1)^2y^{(n)}\) for larger terms, ie \(y^{(2k)}(0) = 0\) \(y^{(1)}(0) = 1, y^{(3)}(0) = (1+1)^2 \cdot 1 = 2^2, y^{(5)}(0) = (3+1)^2 y^{(3)} = 4^2 \cdot 2^2\) and \(y^{(2k+1)}(0) = (2k)^2 \cdot (2k-2)^2 \cdots 2^2 \cdot 1^2 = 2^{2k} \cdot (k!)^2\). Therefore \begin{align*} && \frac{\arcsin x}{\sqrt{1-x^2}} &= \sum_{k=0}^{\infty} \frac{2^{2k} \cdot (k!)^2}{(2k+1)!} x^{2k+1} \\ \\ \Rightarrow && \frac{\arcsin \frac12}{\sqrt{1-\left (\frac12 \right)^2}} &= \sum_{k=0}^{\infty} \frac{2^{2k} \cdot (k!)^2}{(2k+1)!} 2^{-2k-1}\\ &&&= \frac12 \sum_{k=0}^{\infty} \frac{ (k!)^2}{(2k+1)!} \\ &&&= \frac12 \left ( 1 + \frac1{3!} + \frac{2^2}{5!} + \cdots+ \right) \\ \Rightarrow&& S &= 2 \frac{2\frac{\pi}{6}}{\sqrt{3}} = \frac{2\pi}{3\sqrt{6}} \end{align*}

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Solution: Note that \({\bf p}_i \cdot {\bf p}_i = 1\) and \({\bf p}_i \cdot {\bf p}_j\) are all equal when \(i \neq j\) by symmetry and commutativity. \begin{align*} && 0 &= {\bf p}_i \cdot \left ( {\bf p}_1+ {\bf p}_2+{\bf p}_3+{\bf p}_4 \right) \\ &&&= 1 + \sum_{j \neq i} {\bf p}_i \cdot {\bf p}_j \\ &&&= 1 + 3 {\bf p}_i \cdot {\bf p}_j \\ \Rightarrow && {\bf p}_i \cdot {\bf p}_j &= -\frac13 \end{align*}

1. \(\,\) \begin{align*} && (XP_i)^2 &= ({\bf p}_i- {\bf x)}\,.\,({\bf p}_i- {\bf x}) \\ &&&= {\bf p}_i \cdot {\bf p}_i - 2 {\bf p}_i \cdot {\bf x} + {\bf x} \cdot {\bf x} \\ &&&= 2 - 2 {\bf p}_i \cdot {\bf x} \\ \Rightarrow && \sum_i (XP_i)^2 &= \sum_i \left (2 - 2 {\bf p}_i \cdot {\bf x} \right) \\ &&&= 8 - 2 \sum_i {\bf p}_i \cdot {\bf x} \\ &&&= 8 - 2 \left ( {\bf p}_1+ {\bf p}_2+{\bf p}_3+{\bf p}_4 \right) \cdot {\bf x} \\ &&&= 8 \end{align*}
2. Notice we have \(1 = \left \|\begin{pmatrix} a \\0 \\b \end{pmatrix} \right \|= a^2 + b^2\) and \(-\frac13 = \begin{pmatrix} a \\0 \\b \end{pmatrix} \cdot \begin{pmatrix} 0 \\0 \\ 1 \end{pmatrix} = b\). So \(b = -1/3\) and \(a = \sqrt{1-b^2} = 2\sqrt{2}/3\). Suppose another of the vertices has coordinates \((u,v,w)\) we must have \begin{align*} && 1 &= u^2+v^2+w^2 \\ && -\frac13&=w \\ && -\frac13 &= \frac{2\sqrt{2}}3 u +\frac19 \\ \Rightarrow && u &= -\frac{\sqrt2}3 \\ \Rightarrow && 1 &= \frac19 + \frac29 + v^2 \\ \Rightarrow && v &= \pm \sqrt{\frac{2}{3}} \end{align*} So \(P_3, P_4 = (-\frac{\sqrt2}3, \pm \frac{\sqrt{6}}3, -\frac13)\)
3. \(\,\) \begin{align*} && \sum_{i=1}^4 (XP_i)^4 &= \sum_i \left (2 - 2 {\bf p}_i \cdot {\bf x} \right)^2 \\ &&&= 4 \sum_i \left (1 - {\bf p}_i \cdot {\bf x} \right)^2 \\ &&&= 4 \sum_i (1 - 2{\bf p}_i \cdot {\bf x} + ({\bf p}_i \cdot {\bf x})^2) \\ &&&= 16 + 4\sum_i ({\bf p}_i \cdot {\bf x})^2 \\ &&&=16+ 4\left ( z^2+\left (\frac{2\sqrt{2}}3x-\frac13z \right)^2 +\left (-\frac{\sqrt{2}}3x-\frac{\sqrt{6}}3y-\frac13z \right)^2 +\left (-\frac{\sqrt{2}}3x+\frac{\sqrt{6}}3y-\frac13z \right)^2 \right) \\ &&&= 16+4 \left ( \frac43z^2 + \left (\frac89 + \frac29+\frac29 \right)x^2+\left (\frac69 + \frac69 \right)y^2 + 0xz + 0yz + 0zx \right) \\ &&&= 16+ 4\cdot\frac43(x^2+y^2+z^2) \\ &&&=16+\frac{16}{3}=\frac{64}{3} \end{align*}

Note: It may be better to view the last part of this question in terms of linear transformations. There are two possible approaches. One is to show \(T:{\bf x} \mapsto \sum_i ({\bf p}_i \cdot x) {\bf p}_i\) is \(\frac43I\) (easy since it has three eigenvectors with the same eigenvalue which span \(\mathbb{R}^3\) and we are interested in the value \({\bf x} \cdot T\mathbf{x} = \frac43 \lVert {\bf x} \rVert^2\). The second is to consider \(\sum_I ({\bf p}_i \cdot {\bf x})^2 = {\bf x}^TM{\bf x}\) where \(M = \sum_i {\bf p}_i{\bf p}_i^T\) and note that this matrix is invariant under rotations.

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Solution: \begin{align*} && (z-e^{i \theta})(z-e^{-i\theta}) &= z^2 - (e^{i\theta}+e^{-i\theta})z + 1 \\ &&&= z^2-2\cos \theta z + 1 \end{align*} The \(2n\)th roots of \(-1\) are \(e^{\frac{i (2k+1)\pi}{2n}}, k \in \{-n, \cdots, n-1 \}\) or \(e^{\frac{i k \pi}{2n}}, k \in \{-2n+1, -2n+3, \cdots, 2n-3, 2n-1 \}\) \begin{align*} && z^{2n}+1 &= (z-e^{-i(2n-1)/2n})\cdot (z-e^{-i(2n-3)/2n})\cdots (z-e^{i(2n-3)/2n})\cdot (z-e^{i(2n-1)/2n}) \\ &&&= \prod_{k=1}^n \left (z - e^{i \frac{2k-1}{2n}\pi} \right)\left (z - e^{-i \frac{2k-1}{2n}\pi} \right)\\ &&&= \prod_{k=1}^n \left (z^2 - 2z \cos \left ( \frac{(2k-1)\pi}{2n}\right) + 1 \right) \end{align*}

1. \begin{align*} && i^{2n} + 1 &= \prod_{k=1}^n \left (i^2 - 2i \cos \left ( \frac{(2k-1)\pi}{2n}\right) + 1 \right) \\ \Rightarrow && (-1)^n + 1 &= (-1)^n2^ni^n\prod_{k=1}^n \cos \left ( \frac{(2k-1)\pi}{2n}\right) \\ \Rightarrow && \prod_{k=1}^n \cos \left ( \frac{(2k-1)\pi}{2n}\right) &= 2^{1-n}(-1)^{n/2} \tag{if \(n\equiv 0\pmod{2}\)} \end{align*}
2. When \(n\) is odd, we notice that two of the roots are \(i\) and \(-i\), if we exclude those, (ie by factoring out \(z^2+1\), we see that \begin{align*} && 1-z^2+z^4-\cdots + z^{2n-2} &= \prod_{k=1, 2k-1\neq n}^n \left (z^2-2z \cos \left ( \frac{(2k-1)\pi}{2n} \right)+1 \right) \\ &&&= \prod_{k=1}^{(n-1)/2} \left (z^2-2z \cos \left ( \frac{(2k-1)\pi}{2n} \right)+1 \right) \prod_{k=(n+1)/2}^{n} \left (z^2-2z \cos \left ( \frac{(2k-1)\pi}{2n} \right)+1 \right)\\ &&&= \prod_{k=1}^{(n-1)/2} \left (z^2-2z \cos \left ( \frac{(2k-1)\pi}{2n} \right)+1 \right) \prod_{k=1}^{(n-1)/2} \left (z^2+2z \cos \left ( \frac{(2k-1)\pi}{2n} \right)+1 \right)\\ \Rightarrow && 1-i^2 + i^4 + \cdots + i^{2n-2} &= \prod_{k=1}^{(n-1)/2} \left (2 \cos \left ( \frac{(2k-1)\pi}{2n} \right) \right) \prod_{k=1}^{(n-1)/2} \left (2 \cos \left ( \frac{(2k-1)\pi}{2n} \right) \right)\\ \Rightarrow && n &= 2^{n-1} \prod_{k=1}^{(n-1)/2}\cos^2 \left ( \frac{(2k-1)\pi}{2n} \right) \\ \Rightarrow && \prod_{k=1}^{(n-1)/2}\cos^2 \left ( \frac{(2k-1)\pi}{2n} \right) &= n2^{1-n} \end{align*}

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Solution:

1. Suppose \(q^nN = p^n\) then since \((p,q) =1\) we must have \(p \mid N\), and then by dividing both \(p^n\) and \(N\) by \(p\) we can repeat this process \(n\) times to find that \(N = kp^n\) and in particular \(q = 1\). Suppose \(\sqrt[n]{N} = \frac{p}q\) for \(p,q\) coprime positive integers (ie it is not irrational), then \(q^nN = p^n\) and so \(q = 1\) and in fact \(\sqrt[n]{N}\) is an integer so \(N\).
2. Suppose \((a,b) = 1, (c,d) = 1\) and \(a^ad^b = b^ac^b\), then since \((a,b) = 1\) we must have \((b^a, a) = 1\) so \(b^a \mid d^b\). Similarly since \((c,d) = 1\) we must have \((d^b, c) = 1\) so \(d^b \mid b^a\). Therefore \(d^b = b^a\). Suppose \(p \mid d\) then \(p \mid d^b = b^a \Rightarrow p \mid b\). Suppose \(\nu_p(d) = m, \nu_p(b) = n\) we must have \(bm = \nu_p(d^b) = \nu_p(b^a) = an\), ie \(b = \frac{an}{m}\). Note that \(p^n \mid b \Rightarrow p^n \mid n \frac{a}{m} \Rightarrow p^n \mid n \Rightarrow p^n \leq n\). Since \((p,a) = 1\).. But since \(p^n > n\) if \(p \geq 2\) we must have that \(b = 1\). Therefore suppose \(r = \frac{a}{b}\) with \((a,b) = 1\) an \(r^r = \frac{c}{d}\) we must ahve \(a^ac^b = b^ad^b\) and so \(b = 1\) implying \(r\) is an integer.

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Solution:

1. \(\,\) \begin{align*} && |z-w|^2 &= (z-w)(z^*-w^*) \\ &&&= zz^* - wz^*-zw^* + ww^* \\ &&&= |z|^2+|w|^2 - E + 2|zw| \\ &&&= (|z|+|w|)^2 - E \\ \Rightarrow && E &= (|z|+|w|)^2 - |z-w|^2 &\in \mathbb{R} \end{align*} and by the triangle inequality \(|z|+|w| \geq |z-w|\), so \(E \geq 0\)
2. \(\,\) \begin{align*} && |1-zw^*|^2 &= (1-zw^*)(1-z^*w) \\ &&&= 1 - zw^*-z^*w + |zw|^2 \\ &&&= 1 - E + 2|zw| + |zw|^2 \\ &&&= (1+|zw|)^2 - E \end{align*} \begin{align*} && \frac{|z-w|^2}{|1-zw^*|^2} &= \frac{(|z|+|w|)^2-E}{(1+|zw|)^2-E} \\ \Leftrightarrow && (1+|zw|^2)|z-w|^2 -E|z-w|^2 &= (|z|+|w|)^2|1-zw^*|^2-E|1-zw^*|^2\\ \Leftrightarrow && (1+|zw|^2)|z-w|^2-(|z|+|w|)^2|1-zw^*|^2 &= E(|z-w|^2-|1-zw^*|^2)\\ &&&= E(|z|^2-zw^*-z^*w+|w|^2-1+zw^*+z^*w-|z|^2|w|^2) \\ &&&= E(|z|^2+|w|^2-1-|z|^2|w|^2) \\ &&&= -E(1-|z|^2)(1-|w|^2) \\ &&&\leq 0 \\ \Leftrightarrow&& (1+|zw|^2)|z-w|^2& \leq (|z|+|w|)^2|1-zw^*|^2\\ \Leftrightarrow&& \frac{|z-w|^2}{|1-zw^*|^2} &\leq \frac{(|z|+|w|)^2}{(1+|zw|)^2}\\ \Leftrightarrow && \frac{|z-w|}{|1-zw^*|} &\leq \frac{(|z|+|w|)}{(1+|zw|)}\\ \end{align*} Yes, this inequality holds if \(|z|, |w|\) are the same side of \(1\) and is reversed otherwise.