Problems

Solution:

1. \begin{align*} && b(p,q) &= \int_0^1 x^{p-1}(1-x)^{q-1}\, \d x \\ u = 1-x, \d u = -\d x && &= \int_{u=1}^{u = 0} (1-u)^{p-1}u^{q-1} (-1) \, \d u \\ &&&= \int_0^1 (1-u)^{p-1}u^{q-1} \d u \\ &&&= \int_0^1 u^{q-1}(1-u)^{p-1} \d u \\ &&&= b(q,p) \end{align*}
2. \begin{align*} b(p+1,q) + b(p,q+1) &= \int_0^1 x^p(1-x)^{q-1} \d x + \int_0^1 x^{p-1}(1-x)^{q} \d x \\ &= \int_0^1 \left (x^p(1-x)^{q-1} + x^{p-1}(1-x)^{q}\right) \d x \\ &= \int_0^1 x^{p-1}(1-x)^{q-1} \left (x + (1-x) \right) \d x \\ &= \int_0^1 x^{p-1}(1-x)^{q-1} \d x \\ &= b(p,q) \end{align*} Therefore \(b(p+1,q) = b(p,q) - b(p,q+1)\), in particular \(2b(p+1,p) = b(p+1,p)+b(p,p+1) = b(p,p) \Rightarrow b(p+1,p) = \frac12 b(p,p)\) as required.
3. \begin{align*} && b(p,q) &= \int_0^1 x^{p-1} (1-x)^{q-1} \d x \\ x = \sin^2 \theta, \d x = 2 \sin \theta \cos \theta \d \theta && &= \int_{u=0}^{u = \pi/2} \sin^{2p-2} \theta (1-\sin^2 \theta)^{q-1} \cdot 2 \sin \theta \cos \theta \d \theta \\ &&&= 2 \int_0^{\pi/2} \sin^{2p-1} \theta \cos^{2q-2} \cos \theta \d \theta \\ &&&= 2 \int_0^{\pi/2} \sin^{2p-1} \theta \cos^{2q-1} \theta \d \theta \end{align*} \begin{align*} b(p,p) &= 2\int_0^{\pi/2} (\sin \theta)^{2p-1}(\cos \theta)^{2p-1} \d \theta \\ &= 2 \int_0^{\pi/2} \left (\frac12 \sin 2\theta \right)^{2p-1} \d \theta \\ &= \frac1{2^{2p-1}} 2 \int_0^{\pi/2} (\sin 2 \theta)^{2p-1} \d \theta \\ &= \frac1{2^{2p-1}} 2 \int_{x=0}^{x=\pi} (\sin x)^{2p-1} 2 \d x\\ &= \frac1{2^{2p-1}} 2 \int_{x=0}^{x=\pi/2} (\sin x)^{2p-1} \d x\\ &= \frac1{2^{2p-1}} 2 \int_{0}^{\pi/2} (\sin x)^{2p-1} (\cos x)^{0} \d x\\ &= \frac1{2^{2p-1}} b(p,\tfrac12) \end{align*}
4. \begin{align*} &&b(p,q) &= \int_0^1 x^{p-1}(1-x)^{q-1} \d x \\ t = \frac{x}{1-x}, \d t = (1-x)^{-2} \d x &&&= \int_{t=0}^{t = \infty} \left ( \frac{t}{1+t} \right)^{p-1} \left ( 1-\frac{t}{1+t} \right)^{q+1} \d t\\ x = \frac{t}{1+t} && &=\int_0^\infty t^{p-1} (1+t)^{-(p-1)-(q+1)} \d t \\ &&&= \int_0^{\infty} \frac{t^{p-1}}{(1+t)^{p+q}} \d t \end{align*}
5. \begin{align*} I &= \int_0^\infty \frac{t^{3/2}}{(1+t)^6} \, dt \\ &= b( \tfrac52, \tfrac72) \\ &= b( \tfrac52, \tfrac52+1) \\ &= \tfrac12 b( \tfrac52, \tfrac52) \\ &= \frac12 \cdot \frac1{2^{4}} b(\tfrac52, \tfrac12) \\ &= \frac{1}{2^5} \cdot 2 \int_0^{\pi/2} (\sin \theta)^{4} \d \theta \\ &= \frac1{2^4} \int_0^{\pi/2}\left (\frac{1-\cos 2 \theta}{2} \right)^2 \d \theta \\ &= \frac1{2^6} \int_0^{\pi/2}\left (1 - 2 \cos 2 \theta + \cos^{2} 2 \theta \right) \d \theta \\ &= \frac1{2^6} \int_0^{\pi/2}\left (1 - 2 \cos 2 \theta + \frac{\cos 4 \theta + 1}{2} \right) \d \theta \\ &= \frac1{2^6} \left [\frac32 \theta - \sin 2 \theta + \frac18 \sin 4 \theta \right]_0^{\pi/2} \\ &= \frac1{2^6} \frac{3 \pi}{4} \\ &= \frac{3 \pi}{2^8} \end{align*}

Solution:

1. 1. 

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   2. If \(a = 1\) then \(u_1 = f(a) = 7-2 = 5\), \(u_2 = f(5) = -3\), \(u_3 = f(-3) = 7-6 = 1\). Therefore it must be the case that \(f(f(f(x))) = x\) for \(x = 1, 5, -3\). Similarly, if \(a = -\tfrac79\) then \(u_1 = f(-\tfrac79) = \tfrac{49}{9}\), \(u_2 = f(\tfrac{49}{9}) = -\tfrac{35}{9}\) and \(u_3 = f(-\tfrac{35}{9}) = -\tfrac79\). Therefore we must also have roots \(x = -\tfrac79, \tfrac{49}{9}, -\tfrac{35}9\). We also have the roots \(x = -7, \tfrac73\) from the first part so we have found all \(8\) roots.
   3. We need \(f(f(f(x))) = 1\) but \(f(f(x)) \neq -3, f(x) \neq 5, x \neq 1\). Suppose \(f(y) = 1 \Rightarrow 7-2|y| = 1 \Rightarrow y = \pm 3\). So \(y = 3\), ie \(f(f(x)) = 3\). Suppose \(f(z) = 3 \Rightarrow 7-2|z| = 3 \Rightarrow z = \pm 2\). Finally we need \(f(x) = \pm 2\), so say \(7-2|x| = 2 \Rightarrow x = \tfrac52\), so we have the sequence \(\tfrac52, 2, 3, 1, 5, -3, 1, \cdots\)as required.

Solution:

1. The line \(AB\) has equation: \begin{align*} && \frac{y+f(b)}{x-b} &= \frac{f(a)+f(b)}{a-b} \\ \Rightarrow && \frac{f(b)}{m-b} &= \frac{f(a)+f(b)}{a-b} \\ \Rightarrow && m &= \frac{a-b}{f(a)+f(b)}f(b) + b \\ &&&= \frac{af(b)+bf(a)}{f(a)+f(b)} \end{align*}
2. Suppose \(f(x) = \sqrt{x}\) then \begin{align*} m &= \frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{a}+\sqrt{b}} \\ &= \frac{\sqrt{ab}(\sqrt{a}+\sqrt{b})}{\sqrt{a}+\sqrt{b}} \\ &= \sqrt{ab} \end{align*} Suppose \(f(x) = x^{-n}\) then \begin{align*} m &= \frac{a b^{-n}+ba^{-n}}{a^{-n}+b^{-n}} \\ &= \frac{a^{n+1}+b^{n+1}}{b^n + a^n} \\ \end{align*}
3. Without loss of generality, we can scale \(g_1(x)\) and \(g_2(x)\) so that \(g_1(a) = g_2(a)\) and \(m\) won't change for either of them. Then since \(\frac{g_1(b)}{g_2(b)} < 1\) (this function is decreasing) our line connecting \((a,g_i(a))\) and \((b,-g_i(b))\) must interect the axis first for \(g_2\), in particular \(M_1 > M_2\). Suppose \(g_1(x) =1, g_2(x) = \sqrt{x}, g_3(x) = x^{-1}\), the notice that \(\frac{g_1(x)}{g_2(x)} =\frac{g_2(x)}{g_3(x)}= x^{-1/2}\) are decreasing, therefore: \begin{align*} \frac{a+b}{1+1} &> \sqrt{ab} > \frac{1+1}{a^{-1}+b^{-1}} \\ \frac{a+b}{2} &> \sqrt{ab} > \frac{2ab}{a+b} \\ \end{align*}
4. We must have: \begin{align*} && p(c-a)^3 &= f(a) \\ && p(c-b)^3 &= -f(b) \\ \Rightarrow &&\left ( \frac{c-a}{c-b} \right)^3 &= -\frac{f(a)}{f(b)} \\ \Rightarrow && \frac{c-a}{b-c} &= \left (\frac{f(a)}{f(b)} \right)^{\tfrac13} \\ \Rightarrow && c-a &= \left (\frac{f(a)}{f(b)} \right)^{\tfrac13}(b-c)\\ \Rightarrow && c \left (1 + \left (\frac{f(a)}{f(b)} \right)^{\tfrac13} \right) &= \left (\frac{f(a)}{f(b)} \right)^{\tfrac13}b + a \\ \Rightarrow && c &= \frac{\left (\frac{f(a)}{f(b)} \right)^{\tfrac13}b + a}{1 + \left (\frac{f(a)}{f(b)} \right)^{\tfrac13}} \\ &&&= \frac{b[f(a)]^\tfrac13+a[f(b)]^\tfrac13}{[f(a)]^\tfrac13+[f(b)]^\tfrac13} \end{align*} We have that \(\frac{c-a}{b-c} = \left (\frac{f(a)}{f(b)} \right)^{\tfrac13} \) and \(\frac{m-a}{b-c} = \frac{f(a)}{f(b)}\). Since \(f\) is decreasing, \(\frac{f(a)}{f(b)} > 1\) and so \(\left (\frac{f(a)}{f(b)} \right)^{\tfrac13} < \frac{f(a)}{f(b)}\), therefore \(m > c\).

Solution:

1. Suppose \begin{align*} && \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} &= \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} x_1 \\ y_1 \end{pmatrix} \\ \Rightarrow && \binom{x_1}{y_1} &= \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \binom{x_2}{y_2} \end{align*} Therefore if \(\frac{x_1^2}9+\frac{y_1^2}{4} = 1\) we must have \begin{align*} \frac{(\frac{x_2}{\sqrt{2}}+\frac{y_2}{\sqrt{2}})^2 }{9} + \frac{(-\frac{x_2}{\sqrt{2}}+\frac{y_2}{\sqrt{2}})^2}{4} = 1 \end{align*} but this is precisely the statement that \((x_1, y_1)\) is on \(G_1\) is equivalent to \((x_2,y_2)\) being on the \(G_2\). Since the point \((x_2,y_2)\) is a \(45^{\circ}\) rotation of \((x_1,y_1)\) anticlockwise about the origin, this means \(G_2\) is a \(45^{\circ}\) anticlockwise rotation of \(G_1\).
2. 1. \begin{align*} && \begin{pmatrix} -0.6 & 0.8 \\ 0.8 & 0.6 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} x \\ y \end{pmatrix} \\ \Rightarrow && \begin{pmatrix} -0.6 x + 0.8y \\ 0.8x + 0.6y \end{pmatrix} &= \begin{pmatrix} x \\ y \end{pmatrix} \\ \Rightarrow && \begin{pmatrix} -1.6 x + 0.8y \\ 0.8x -0.4y \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\ \Rightarrow && y &=2 x \end{align*}
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3. Consider the transformation \(\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\) which is a shear, leaving the \(x\)-axis invariant. Then we must have:
   

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   Since the shear leaves lines of the form \(y = k\) invariant, the points where \(\frac{\d y}{\d x} = 0\) must also map to points where this is true, ie \((\tfrac{\pi}{2}, 1), (\tfrac{3\pi}{2}, -1)\) map to points \((\tfrac{\pi}{2}+2,1), (\tfrac{3\pi}{2} -2,-1)\) where the tangent is horizontal. The line \(x = c\) map back to lines \(\begin{pmatrix} 1 & -2 \\ 0 & 1\end{pmatrix} \begin{pmatrix} c \\ t\end{pmatrix} = \begin{pmatrix}c - 2t \\ t \end{pmatrix}\), ie \(y = -\frac12 x- \frac{c}{2}\). Therefore we are interested in points on the original curve where the gradient is \(-\frac12\), ie \((\frac{2\pi}{3}, \frac{\sqrt{3}}{2}), (\frac{4\pi}{3}, -\frac{\sqrt{3}}{2})\), these map to \((\frac{2\pi}{3}+\sqrt{3},\frac{\sqrt{3}}{2}), (\frac{4\pi}{3}-\sqrt{3}, -\frac{\sqrt{3}}{2})\)

Solution:

1. \(\overrightarrow{OL} = \lambda \mathbf{a}\) and \(\overrightarrow{OX} = \mu \mathbf{b} + (1-\mu) \mathbf{c}\). Since \(LX\) passes through \(P\), and \(A,B,C\) not colinear we must have that \(s \lambda = \alpha, (1-s)\mu = \beta, (1-s)(1-\mu) = \gamma\)

Solution:

1. If \(a,b,c\) were all real then \(a^2+b^2+c^2 = 0 \Rightarrow a,b,c = 0\) but they are non-zero. Therefore they cannot all be real. Since \((a+b+c)^2 = 0\) we must have \(ab+bc+ca = 0\). Therefore \(a,b,c\) must satisfy \(x^3 -abc = 0 \Rightarrow\) they all have the same modulus, since they are all cube roots of the same number.
2. Notice that \(a^3+b^3+c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - ab-bc-ca) \Rightarrow abc = 0\) but therefore they cannot all be non-zero.
3. Suppose \(a+b+c+d = 0\) then note that \(\displaystyle a^2+b^2+c^2+d^2 = (a+b+c+d)^2 - 2\sum_{sym} ab\) and \(\displaystyle a^3+b^3+c^3+d^3 = (a+b+c+d)^3 - 3(a+b+c+d)(ab+ac+ad+bc+bd+cd) + 3(abc+abd+acd+bcd) \Rightarrow abc+abd+acd+bcd = 0\). Therefore \(a,b,c,d\) are roots of a polynomial of the form \(x^4 -kx^2 + l = 0\), but this means they must come in pairs with the same modulus.
4. Suppose \(c = 1, d = -2, e = 3\) so \(c+d+e = 2\) and \(c^3 + d^3 + e^3 = 1 - 8 + 27 = 20\), so we need to find \(a,b\) satisfying \(a+b = -2, a^2+b^2 = -20\), ie \(4 = (a+b)^2 = -20 + 2ab \Rightarrow ab = 12\), so we need the roots of \(x^2 +2x + 12= 0\) which clearly have different modulus.

Solution:

1. \begin{align*} \sqrt{x^2+1} &= |x|\sqrt{1+\frac{1}{x^2}} \\ &=|x| \left (1 + \frac12 \frac{1}{x^2} + \cdots \right) & \text{if } \left (\frac{1}{x^2} < 1 \right) \\ &= |x| + \frac12 \frac{1}{|x|} + \cdots \\ &\approx |x| + \frac{1}{2|x|} \end{align*}
   

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2. 1. \begin{align*} && \tan( g(x) + g(-x)) &= \tan \left ( \tan^{-1}(\sqrt{x^2+1}-x) + \tan^{-1}(\sqrt{x^2+1}+x) \right) \\ &&&= \frac{\sqrt{x^2+1}-x+\sqrt{x^2+1}+x}{1-1} \\ \Rightarrow && g(x) + g(-x) &\in \left \{\cdots, -\frac{\pi}{2}, \frac{\pi}{2}, \cdots \right\} \end{align*} But \(g(x), g(-x) > 0\) and \(g(x), g(-x) \in (-\frac{\pi}{2}, \frac{\pi}{2})\), therefore it must be \(\frac{\pi}{2}\).
   2. \begin{align*} && \tan(2(k(x) + k(-x))) &= \tan(\tan^{-1}x + \tan^{-1}(-x)) \\ &&&= 0 \\ \Rightarrow && k(x)+k(-x) &\in \left \{\cdots, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \cdots \right\} \\ \end{align*} But \(k(x) \in (-\frac{\pi}{4}, \frac{\pi}{4})\), therefore \(k(x) + k(-x) = 0\).
   3. Let \(t = \tan k(x)\). \begin{align*} && \tan \left ( \tan^{-1} \frac{1}{x} \right) &= \frac{2 \tan\left ( \frac12 \tan^{-1} \frac1x \right)}{ 1-\tan^2\left ( \frac12 \tan^{-1} \frac1x \right)} \\ \Rightarrow && \frac1x &= \frac{2t}{1-t^2} \\ \Rightarrow && 1-t^2 &= 2tx \\ \Rightarrow && 0 &= t^2+2tx - 1 \\ \Rightarrow && 0 &= (t+x)^2 - 1-x^2 \\ \Rightarrow && t &= -x \pm \sqrt{1+x^2} \end{align*} Since \(t > 0\), \(t = \sqrt{1+x^2}-x = f(x) = \tan g(x)\)
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   5. \begin{align*} \int_0^1 k(x) \d x &= \int_0^1 \frac12 \tan^{-1} \left ( \frac1x \right) \d x \\ &= \left [ \frac{x}{2} \tan^{-1}\left ( \frac1x \right) \right]_0^1 - \int_0^1 \frac{x}{2} \frac{-1/x^2}{1+1/x^2} \d x \\ &= \left [ \frac{x}{2} \tan^{-1}\left ( \frac1x \right) \right]_0^1 + \frac14 \int_0^1 \frac{2x}{1+x^2} \d x \\ &= \frac12 \frac{\pi}{4} + \frac14 \ln(2) \\ &= \frac{\pi + \ln 4}{8}\end{align*} Therefore \(\displaystyle \int_{-1}^0 g(x) \d x = -\frac{\pi + \ln 4}{8}\)

Solution:

1. \begin{align*} RHS &= \left(z - \frac{1}{z}\right)\left(z^m + \frac{1}{z^m}\right) + \left(z^{m-1} - \frac{1}{z^{m-1}}\right) \\ &= z^{m+1} + \frac{1}{z^{m-1}} - z^{m-1} - \frac{1}{z^{m+1}} + z^{m-1} - \frac{1}{z^{m-1}} \\ &= z^{m+1} - \frac{1}{z^{m+1}} \\ &= LHS \end{align*}. Claim: For \(n \geq 1\), $$z^{2n} - \frac{1}{z^{2n}} = \left(z - \frac{1}{z}\right)\sum_{r=1}^n \left(z^{2r-1} + \frac{1}{z^{2r-1}}\right)$$ Proof: (By Induction) Base Case: (\(n=1\)). \begin{align*} LHS &= z^2 - \frac{1}{z^2} \\ &= (z-\frac1z)(z + \frac{1}{z}) \\ &= (z - \frac1z) \sum_{r=1}^1 \left ( z + \frac{1}{z} \right) \\ &= (z - \frac1z) \sum_{r=1}^1 \left ( z^{2r-1} + \frac{1}{z^{2r-1}} \right) \\ &= RHS \end{align*} as required. Inductive step: Suppose our result is true for some \(n=k\), then consider \(n = k+1\). \begin{align*} RHS &= \left(z - \frac{1}{z}\right)\sum_{r=1}^{k+1} \left(z^{2r-1} + \frac{1}{z^{2r-1}}\right) \\ &= \left(z - \frac{1}{z}\right)\sum_{r=1}^{k} \left(z^{2r-1} + \frac{1}{z^{2r-1}}\right) + \left(z - \frac{1}{z}\right)\left(z^{2k+1} + \frac{1}{z^{2k+1}}\right) \\ &= z^{2k} - \frac{1}{z^{2k}} + \left(z - \frac{1}{z}\right)\left(z^{2k+1} + \frac{1}{z^{2k+1}}\right) \\ &= z^{2k+2} - \frac{1}{z^{2k+2}} \\ &= LHS \end{align*}. Therefore if our result is true for \(n=k\) is true, it is true for \(n=k+1\). Since it is also true for \(n=1\) it is true for all \(n \geq 1\) but the principle of mathematical induction. Since \(\displaystyle z^{m+1} - \frac{1}{z^{m+1}} = \left(z + \frac{1}{z}\right)\left(z^m - \frac{1}{z^m}\right) + \left(z^{m-1} - \frac{1}{z^{m-1}}\right)\), we must have \(\displaystyle z^{2n}-\frac{1}{z^{2n}} = \left ( z + \frac{1}{z} \right) \sum_{r=1}^n \left (z^{2r-1}-\frac{1}{z^{2r-1}} \right)\)
2. 1. Since $$z^{2n} - \frac{1}{z^{2n}} = \left(z - \frac{1}{z}\right)\sum_{r=1}^n \left(z^{2r-1} + \frac{1}{z^{2r-1}}\right)$$ we have \begin{align*} && e^{2n\theta i} - e^{-2n\theta i} &= \left(e^{\theta i} - e^{-\theta i}\right)\sum_{r=1}^n \left(e^{(2r-1)\theta i} + e^{-(2r-1)\theta i}\right) \\ \Rightarrow && 2i \sin 2n \theta &= 2i \sin \theta \sum_{r=1}^n 2 \cos (2r-1) \theta \\ \Rightarrow && \sin 2n \theta &= 2\sin \theta \sum_{r=1}^n \cos (2r-1) \theta \end{align*}
   2. When \(n = 2, \theta = \frac{\pi}{5}\) we have: \begin{align*} &&\sin \frac{4\pi}{5} &= 2 \sin \frac{\pi}{5} (\cos \frac{\pi}{5} + \cos \frac{3\pi}{5}) \\ &&\sin \frac{\pi}{5} &= 2 \sin \frac{\pi}{5} (\cos \frac{\pi}{5} - \cos \frac{2\pi}{5}) \\ &&\frac12 &= \cos \frac{\pi}{5} - \cos \frac{2 \pi}{5} \\ \Rightarrow && \cos \frac{2\pi}{5} &= \cos \frac{\pi}{5} - \frac12 \end{align*}
   3. When \(n = 7, \theta = \frac{\pi}{15}\) we have: \begin{align*} && \sin \frac{14 \pi}{15} &= 2 \sin \frac{\pi}{15} \sum_{r=1}^7 \cos (2r-1) \frac{\pi}{15} \\ \Rightarrow && \frac12 &= \cos \frac{\pi}{15} + \cos \frac{3 \pi}{15} + \cos \frac{5 \pi}{15}+ \cos \frac{7 \pi}{15}+ \cos \frac{9 \pi}{15}+ \cos \frac{11 \pi}{15}+ \cos \frac{13 \pi}{15} \\ &&&= -\cos \frac{16\pi}{15} + \cos \frac{3 \pi}{15} + \cos \frac{5 \pi}{15}- \cos \frac{8 \pi}{15}+ \cos \frac{9 \pi}{15}- \cos \frac{4 \pi}{15}- \cos \frac{2\pi}{15} \\ &&&= - \left ( \cos \frac{2\pi}{15}+\cos \frac{4\pi}{15}+\cos \frac{8\pi}{15}+\cos \frac{16\pi}{15}\right) + \cos \frac{\pi}{5} + \cos \frac{\pi}{3} + \cos \frac{3 \pi}{5} \\ &&&= - \left ( \cos \frac{2\pi}{15}+\cos \frac{4\pi}{15}+\cos \frac{8\pi}{15}+\cos \frac{16\pi}{15}\right) + \frac12 + \frac12 \\ \Rightarrow && \frac12 &= cos \frac{2\pi}{15}+\cos \frac{4\pi}{15}+\cos \frac{8\pi}{15}+\cos \frac{16\pi}{15} \end{align*}
3. By using \(z = e^{i \theta}\) we have that: \begin{align*} && z^{2n}-\frac{1}{z^{2n}} &= \left ( z + \frac{1}{z} \right) \sum_{r=1}^n \left (z^{2r-1}-\frac{1}{z^{2r-1}} \right ) \\ \Rightarrow && e^{2n \theta i} - e^{-2n \theta i} &= (e^{\theta i} + e^{-\theta i}) \sum_{r=1}^n (e^{(2r-1)\theta i} - e^{(2r-1) \theta i}) \\ \Rightarrow && 2i \sin 2n \theta &= 2 \cos \theta \sum_{r=1}^n 2i \sin(2r-1) \theta \\ \Rightarrow && \sin 2n \theta &= 2 \cos \theta \sum_{r=1}^n \sin(2r-1) \theta \end{align*} When \(n = 3, \theta = \frac{\pi}{14}\) we must have: \begin{align*} &&\sin \frac{3 \pi}{7} &= 2 \cos \frac{\pi}{14}( \sin \frac{\pi}{14}+\sin \frac{3\pi}{14}+\sin \frac{5\pi}{14}) \\ &&&= 2 \sin \left (\frac{\pi}{2} - \frac{\pi}{14} \right)( \sin \frac{\pi}{14}+\sin \frac{3\pi}{14}+\sin \frac{5\pi}{14}) \\ &&&= 2 \sin \frac{3\pi}{7} ( \sin \frac{\pi}{14}+\sin \frac{3\pi}{14}+\sin \frac{5\pi}{14}) \\ \Rightarrow && \frac12 &= \sin \frac{\pi}{14}+\sin \frac{3\pi}{14}+\sin \frac{5\pi}{14} \end{align*} as required.

Solution:

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   By symmetry, the centre of mass will lie on the \(y\) axis. Notice that a single slice (when revolved around the \(y\)-axis) has volume \(y \cdot \pi \cdot ((x+ \delta x)^2 - x^2) = 2 \pi x y \delta x\), and COM at height \(\frac12 y\) so we can conclude: \[ \overline{y} \sum_{\delta x} 2 \pi x y \delta x = \sum_{\delta x} \pi xy^2 \delta x\] \begin{align*} && \overline{y} \int_0^r 2xy \d x &= \int_0^r y^2 x \d x \\ \Rightarrow && \overline{y} 2\int_0^r \left (r - \frac{x^n}{r^{n-1}} \right)x \d x &= \int_0^r \left (r - \frac{x^n}{r^{n-1}} \right)^2 x \d x \\ \Rightarrow && \overline{y} \left [r \frac{x^2}{2} - \frac{1}{r^{n-1}} \frac{x^{n+2}}{n+2} \right]_0^r &= \left [r^2 \frac{x^2}{2} - \frac{2}{r^{n-2}} \frac{x^{n+2}}{n+2} + \frac{1}{r^{2n-2}} \frac{x^{2n+2}}{2n+2} \right]_0^r \\ \Rightarrow && 2\overline{y} \left (\frac{r^3}{2} - \frac{r^3}{n+2} \right) &= \left (\frac12 r^4 - \frac{2}{n+2}r^4 + \frac{1}{2n+2}r^4 \right) \\ \Rightarrow && \overline{y}r^3 \frac{n}{(n+2)} &= r^4\frac{(n+1)(n+2)-2\cdot2\cdot(n+1)+(n+2)}{2(n+1)(n+2)} \\ \Rightarrow && \overline{y} \frac{n}{(n+2)} &= r \left ( \frac{n^2}{2(n+1)(n+2)} \right) \\ \Rightarrow && \overline{y} &= \frac{nr}{2(n+1)} \\ &&&= r \left (1 -p^n \right) \end{align*} as required.
2. \begin{align*} && r^{n-1}y &= r^n - x^n \\ \frac{\d}{\d x}: && r^{n-1} \frac{\d y}{\d x} &= -n x^{n-1} \\ && \frac{\d y}{\d x} &= -np^{n-1} \end{align*} Therefor the normal has the equation: \begin{align*} && \frac{y-r(1-p^n)}{x-rp} &= \frac{1}{np^{n-1}} \\ \Rightarrow && Y &= \frac{-rp}{np^{n-1}} + r(1-p^n) \\ &&&= r \left (1 - p^n - \frac{1}{np^{n-2}} \right) \end{align*} If \(n = 4\) then \begin{align*} && Y &= r\left (1 - p^4 - \frac{1}{4p^{2}} \right) \\ \Rightarrow && \frac{\d Y}{\d p} &= r \left (-4p^3 + \frac{1}{2p^3} \right) \end{align*} Therefore there is a stationary point if \(p^6 = \frac18 \Rightarrow p =2^{-1/2}\). Clearly this will be a maximum (sketch or second derivative) therefore, \(Y = r(1 - \frac14 - \frac{2}{4}) = \frac14 r\)
3. The centre of mass of this shape can be found using this table: \begin{array}{|c|c|c|} \hline \text{} & \overline{y} & \text{mass} \\ \hline r^3y = r^4 - x^4 & \frac{2r}{5} & \frac{4\pi r^3}{6} = \frac23 \pi r^3\\ ry = -(r^2 - x^2) & -\frac{r}{3}& \frac{2 \pi r^3}{4}=\frac12\pi r^3 \\ \text{combined} & \frac{(\frac25 \cdot \frac23-\frac13 \cdot \frac12)r^4}{\frac76 r^3} = \frac3{35}r & \frac76 \pi r^3\\ \hline \end{array} Normals to the surface through points on the upper surface will meet the \(y\)-axis between \((-\infty, \frac14 r)\), and since \(p = 0 \to -\infty\) and \(p = 1 \to -\frac14 r\), so normals will pass through \((0, \frac3{35}r)\) from two different points. Normals to the surface through points on the lower surface will go through \(-r(1 - p^2 - \frac12) =- r(\frac12 -p^2)\) which ranges monotonically from \(\frac12 r \to -\frac12 r\) so there will be one point where the normal goes through \(\frac3{35}r\). Therefore there are three angles where the vector \(\overrightarrow{AB}\) is not vertical but the normal to the surfaces runs through the centre of mass (ie the the solid can rest in equilibrium)

Solution:

1. Since we have \(h'' + \omega^2 h = 0\) we must have that \(h(t) = A \cos \omega t + B \sin \omega t\). The initial conditions tell us that \(A = 0\) and \(B = L\), so \(h(t) = L \sin \omega t\).
   

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   Therefore we can see the angle at \(B\) is \(\omega t\) and so \(P\) has \(y\)-coordinate \((1-s)L \sin \omega t\) and \(x\)-coordinate \(sL \cos \omega t\)
2. If the position is \(\binom{sL \cos \omega t}{(1-s) L \sin \omega t}\) then the acceleration is \(-\omega^2 \binom{sL \cos \omega t}{(1-s) L \sin \omega t}\)
3. 

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   \begin{align*} \text{N2}(\rightarrow): && - F\cos \omega t + N \sin \omega t &= -m\omega^2 sL \cos \omega t\\ \text{N2}(\uparrow): && -mg + F\sin \omega t + N \cos \omega t &= -m\omega^2 (1-s) L \sin \omega t \\ \Rightarrow && \begin{pmatrix} \cos \omega t & -\sin \omega t \\ \sin \omega t & \cos \omega t \end{pmatrix} \begin{pmatrix} F \\ N \end{pmatrix} &= \begin{pmatrix} m\omega^2 s L \cos \omega t \\ mg - m\omega^2(1-s)L \sin \omega t \end{pmatrix} \\ \Rightarrow && \begin{pmatrix}F \\ N \end{pmatrix} &= \begin{pmatrix} \cos \omega t & \sin \omega t \\ -\sin \omega t & \cos \omega t \end{pmatrix} \begin{pmatrix} m\omega^2 s L \cos \omega t \\ mg - m\omega^2(1-s)L \sin \omega t \end{pmatrix} \\ \Rightarrow && N &= m \omega^2 s L (-\sin \omega t \cos \omega t) + mg \cos \omega t - m \omega^2 (1-s)L \sin \omega t \cos \omega t \\ &&&=mg \cos \omega t - m \omega^2 L \sin \omega t \cos \omega t \\ &&&= mg \cos \omega t \left (1 - \frac{L \omega^2}{g} \sin \omega t \right) \\ &&&= mg (1 - k \sin \omega t) \cos \omega t \\ \Rightarrow && F &= m \omega^2 s L \cos^2 \omega t + mg \sin \omega t - m \omega^2 (1-s) L \sin ^2 \omega t \\ &&&= m \omega^2 s L + mg \sin \omega t - m \omega^2 L \sin^2 \omega t \\ &&&= mg \frac{\omega^2 L}{g} s + mg(1-\frac{\omega^2 L}{g} \sin \omega t)\sin \omega t \\ &&&= mg sk + mg(1-k \sin \omega t) \cos \omega t \tan \omega t \\ &&&= mgsk + N \tan \omega t \end{align*}
4. The particle will not slip if \(F < \tan \alpha N\). When \(t = 0\), \(N = mg, F = mgsk\), but clearly \(sk < \tan \alpha \Rightarrow mgsk = F < \tan \alpha mg = \tan \alpha N\). The particle will slip when: \(F > \tan \alpha N\), but we have \(F = mgsk + N \tan \omega t\). Clearly when \(\omega t = \alpha\) we have reached a point where \(F > \tan \alpha N\). Therefore we must slip before we reach this point, ie at a point where the plank makes an angle of less than \(\alpha\) to the horizontal. Notice also that \(N\) changes sign when \(1-k \sin \omega t = 0\), however, to do this \(N\) must become very small, smaller than \(mgsk\), therefore we must slip before this point too. Since we slip before either condition occurs, we must be in a position when \(N\) is positive AND the plank still makes a shallow angle.