Problems

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Solution:

1. 

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   \begin{align*} \mathbb{E}(A \mid \text{exactly one point on diameter}) &= \int_{-1}^1\int_0^\pi \frac12 (x-0)\cdot 1 \cdot \sin(\pi - \theta) \frac{1}{\pi} \d \theta \frac{1}{2} \d x \\ &= \int_{-1}^1\frac1{2\pi} x \d x \cdot \left [ -\cos \theta \right]_0^\pi \\ &= \frac{1}{2\pi} \end{align*}
2. 

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   \begin{align*} \mathbb{E}(A \mid \text{no point on diameter}) &= \int_0^{\pi} \frac12 \cdot 1 \cdot 1 \cdot \sin x \cdot 2(\pi - x)/\pi^2 \d x \\ &= \frac1{\pi^2} \int_0^\pi \sin x (\pi - x) \d x \\ &= \frac1{\pi^2} \int_0^\pi x\sin x \d x \\ &= \frac1{\pi^2} \left [ \sin x - x \cos x \right]_0^{\pi} \\ &= \frac{1}{\pi} \end{align*}

If both points lie on the diameter the area of the triangle is \(0\). Therefore: \begin{align*} \mathbb{E}(A) &= \frac{1}{2\pi} \mathbb{P}(\text{exactly one point on diameter}) + \frac{1}{\pi}\mathbb{P}(\text{no points on diameter}) \\ &= \frac1{2\pi} \cdot \left (2 \cdot \frac{2}{2+\pi} \cdot \frac{\pi}{2+\pi} \right) + \frac{1}{\pi} \cdot \left ( \frac{\pi}{2+\pi} \cdot \frac{\pi}{2+\pi}\right) \\ &= \frac{1}{\pi} \frac{2\pi + \pi^2}{(2+\pi)^2} \\ &= \frac{1}{2+\pi} \end{align*}

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Solution: \begin{align*} \mathbb{P}(\text{exactly }k\text{ faults after test}) &= \mathbb{P}(k\text{ faults in non-tested, 0 in batch})+\mathbb{P}(k-1\text{ faults in non-tested, 1 in batch}) \\ &=\binom{N}{k}(1-p)^{N-k}p^k\binom{n}{0}(1-p)^n+\binom{N}{k-1}(1-p)^{N-k+1}p^{k-1}\binom{n}{1}(1-p)^{n-1}p \\ &= (1-p)^{N-k+n}p^k \cdot \left ( \binom{N}{k}+n\binom{N}{k-1} \right) \\ &= (1-p)^{N-k+n}p^k \cdot \left (\frac{N!}{k!(N-k)!}+\frac{N!n}{(k-1)!(N-k+1)!}\right) \\ &= (1-p)^{N-k+n}p^k \frac{N!}{k!(N-k+1)!} \cdot \left ((N-k+1)+nk \right) \\ &= \frac{\left[N+1+k\left(n-1\right)\right]N!}{\left(N-k+1\right)!k!}p^{k}\left(1-p\right)^{N+n-k} \end{align*} When \(k = N+1\) we get: \begin{align*} \frac{(N+1)n N!}{(N+1)!} p^{N+1}(1-p)^{N+n-k} &= np^{N+1}(1-p)^{N+n-k} \end{align*} and the probability is: \begin{align*} \mathbb{P}(\text{exactly }N+1\text{ faults after test}) &= \mathbb{P}(N\text{ faults in non-tested, 1 in batch}) \\ &= \binom{N}{N}p^N \cdot \binom{n}{1}p(1-p)^{N-1} \\ &= np^{N+1}(1-p)^{N+n-k} \end{align*} So the formula does work for \(k = N+1\). \begin{align*} \mathbb{E}(faults) &= \sum_{k=0}^{N+1} k \cdot \mathbb{P}(\text{exactly }k\text{ faults after test}) \\ &= \sum_{k=0}^{N+1} k \cdot \frac{\left[N+1+k\left(n-1\right)\right]N!}{\left(N-k+1\right)!k!}p^{k}\left(1-p\right)^{N+n-k} \\ &= \sum_{k=1}^{N+1} \frac{\left[N+1+k\left(n-1\right)\right]N!}{\left(N-k+1\right)!(k-1)!}p^{k}\left(1-p\right)^{N+n-k} \\ &= \sum_{k=1}^{N+1} \left[N+1+k\left(n-1\right)\right] p(1-p)^{n-1}\binom{N}{k-1}p^{k-1}\left(1-p\right)^{N-k+1} \\ &= p(1-p)^{n-1} \cdot \left ( (N+1+n-1)\sum_{k=1}^{N+1} \binom{N}{k-1}p^{k-1}\left(1-p\right)^{N-k+1}+ (n-1)\sum_{k=1}^{N+1} (k-1)\binom{N}{k-1}p^{k-1}\left(1-p\right)^{N-k+1} \right) \\ &= p(1-p)^{n-1} \left ((N+1+n-1) + (n-1)pN \right) \\ &= \left[N+n+pN(n-1)\right]p(1-p)^{n-1} \end{align*}

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Solution:

We can choose a coordinate frame where the parallel planes are parallel to the \(y-z\) axis. Then we can compute the surface area as an integral of the surface of revolution for \(x^2 + y^2 = r^2\). Using \(y = r \sin t, x = r \cos t\) we have: \begin{align*} S &= 2\pi\int_{\cos^{-1}a}^{\cos^{-1}b}y \sqrt{\left ( \frac{\d x}{\d t} \right)^2+\left ( \frac{\d y}{\d t} \right)^2} \d t \\ &=2\pi\int_{\cos^{-1}a}^{\cos^{-1}b} r^2 \sin t \d t \\ &= 2\pi \cdot r^2 \cdot (a - b) \\ &= 2 \pi \cdot r \cdot (ra-rb) \\ &= 2\pi\times(\mbox{radius of sphere})\times(\mbox{perpendicular distance between the planes}). \end{align*} We can view this surface as a sphere missing a cap of height \(XQ\) and adding a cone of slant height \(OP\) and radius \(PX\) The centre of the circle is at \((c,0)\) and \(OP^2 + a^2 = c^2 \Rightarrow OP = \sqrt{c^2-a^2}\) Since \(OPC \sim OXP\) we must have that \(\frac{OX}{OP} = \frac{OP}{OC} \Rightarrow OX = \frac{c^2-a^2}{c}\) and \(\frac{PX}{OP} = \frac{CP}{OC} \Rightarrow PX = \frac{a}{c}\sqrt{c^2-a^2}\) \(QX = OX - OQ = \frac{c^2-a^2}{c}-(c-a) = \frac{ac-a^2}{c}\) Therefore the surface area is: \begin{align*} S &= 4 \pi a^2 - 2\pi \cdot a \cdot QX+ \pi PX \cdot OP \\ &= 4 \pi a^2 - 2\pi a \cdot \frac{ac-a^2}{c}+\pi \frac{a}{c}\sqrt{c^2-a^2}\cdot \sqrt{c^2-a^2} \\ &= 4\pi a^2 -2\pi \frac{a^2c-a^3}{c}+\pi \frac{ac^2-a^3}{c} \\ &= \pi a \frac{(a+c)^2}{c} \end{align*}

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Solution: Set up a coordinate system so that \(x\) is E-W, \(y\) is N-S and \(z\) is the vertical direction. Also assume \(B\) is the origin, then, \(A = \begin{pmatrix} 0 \\ -100 \\ 5\end{pmatrix}, B = \begin{pmatrix} 0 \\ 0 \\ 0\end{pmatrix}, C= \begin{pmatrix} 60 \\ 0\\ -5\end{pmatrix},\). The coal seam has points: \(\begin{pmatrix} 0 \\ -100 \\ -90\end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ -45\end{pmatrix}, \begin{pmatrix} 60 \\ 0\\ -81\end{pmatrix},\) Therefore we can find the normal to the coal seam: \begin{align*} \mathbf{n} &= \left (\begin{pmatrix} 0 \\ -100 \\ -90\end{pmatrix} - \begin{pmatrix} 0 \\ 0 \\ -45\end{pmatrix}\right ) \times \left ( \begin{pmatrix} 60 \\ 0\\ -81\end{pmatrix} - \begin{pmatrix} 0 \\ 0 \\ -45\end{pmatrix}\right ) \\ &= \begin{pmatrix} 0 \\ - 100 \\ -45\end{pmatrix} \times \begin{pmatrix} 60 \\ 0 \\ -36\end{pmatrix} \\ &= \begin{pmatrix} 3600 \\ -60 \cdot 45 \\ 60 \cdot 100 \end{pmatrix} \\ &= 300\begin{pmatrix} 12 \\ -9 \\ 20\end{pmatrix} \end{align*} To measure the incline \(\theta\) to the horizontal we can take a dot with \(\hat{\mathbf{k}}\), to see: \begin{align*} \cos \theta &= \frac{20}{\sqrt{12^2+(-9)^2+20^2} \sqrt{1^2+0^2+0^2}} \\ &= \frac{20}{25} \\ &= \frac{4}{5} \end{align*} Therefore the angle is \(\cos^{-1} \tfrac 45\) The equation of the seam is \(12x - 9y + 20z = -900\). The equation of the surface is \(5x + 3y + 60z = 0\) We can compute the direction of the overlap again with a cross product: \begin{align*} \mathbf{d} &= \begin{pmatrix} 12 \\ -9 \\ 20\end{pmatrix} \times \begin{pmatrix} 5 \\ 3 \\ 60\end{pmatrix} \\ &= \begin{pmatrix} -600 \\ -620 \\ 81 \end{pmatrix} \end{align*} To get the bearing of this vector we just need to look at the \(x\) and \(y\) components, so it will be \(\tan^{-1} \frac{600}{620} = \tan^{-1} \frac{30}{31}\)

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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.

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Solution:

The gradient at the origin is \(\frac{5}{2}\) which we can observe by looking at the taylor series

1. \begin{align*} && 4x^{2}(x^{2}-5)&=(5x-2)(x^{2}-4) \\ && \frac{2x(x^2-5)}{x^2-4} &= \frac{5x-2}{2x} = \frac52 -\frac1x \end{align*} Therefore it will have two roots as the hyperbola will intersect our graph in two places. (In the upper left and lower right quadrants).
2. \begin{align*} && 4x^{2}(x^{2}-5)^{2}&=(x^{2}-4)^{2}(x^{2}+1) \\ && \frac{2x(x^2-5)}{x^2-4} &= \frac{(x^2-4)(x^2+1)}{2x} \end{align*}
   

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   No solutions \begin{align*} && 4z^{2}(z-5)^{2}&=(z-4)^{2}(z+1) \\ && \frac{2z(z^2-5)}{z^2-4} &= \frac{(z+5)(z-4)(z+1)}{(z-5)(z+4)} \end{align*}
   

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   5 solutions

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Solution: Recall, \(\cosh^{-1} x = \ln (x + \sqrt{x^2-1})\) \begin{align*} \cosh(n \cosh^{-1} x) &= \frac12 \left ( \exp(n \cosh^{-1} x) + \exp(-n\cosh^{-1}x) \right) \\ &= \frac12 \left ((x + \sqrt{x^2-1})^n + (x + \sqrt{x^2-1})^{-n} \right) \\ &= \frac12 \left ((x + \sqrt{x^2-1})^n + (x - \sqrt{x^2-1})^{n} \right) \\ \end{align*} When \(n = 2k+1\) \begin{align*} \cosh(n \cosh^{-1} x)&= \frac12 \left ((x + \sqrt{x^2-1})^n + (x - \sqrt{x^2-1})^{n} \right) \\ &= \frac12 \left (\sum_{i=0}^{2k+1}\binom{2k+1}{i}x^{2k+1-i}\left ( (\sqrt{x^2-1}^{i} + (-\sqrt{x^2-1})^{i} \right) \right) \\ &=\sum_{i=0}^{k} \binom{2k+1}{2i}x^{2k+1-2i}(x^2-1)^i \\ &=\sum_{i=0}^{k} \binom{2k+1}{2i}x^{2(k-i)+1}(x^2-1)^i \\ \end{align*} Which is clearly a polynomial with only odd degree terms. \begin{align*} a_0 &= \frac{\d y}{\d x} \vert_{x=0} \\ &= \sum_{i=0}^k\binom{2k+1}{2i} \left ( (2(k-i)+1)x^{2(k-i)}(x^2-1)^i + 2i\cdot x^{2(k-i)+2}(x^2-1) \right) \\ &= \binom{2k+1}{2k} (-1)^{k} \\ &= (-1)^k(2k+1) \end{align*}

1. \begin{align*} a_1 &= \binom{2k+1}{2k}\binom{k}{1}(-1)^{k-1}+\binom{2k+1}{2(k-1)}(-1)^{k-1} \\ &=(-1)^{k-1}\cdot ( (2k+1)k + \frac{(2k+1)\cdot 2k \cdot (2k-1)}{3!}) \\ &= (-1)^{k-1}(2k+1)k\frac{3 + 2k-1}{3} \\ &= (-1)^{k-1}2(2k+1)k (k+1) \end{align*}
2. \begin{align*} a_2 &= \binom{2k+1}{2k} \binom{k}{2}(-1)^{k-2} + \binom{2k+1}{2(k-1)} \binom{k-1}{1} (-1)^{k-2}+\binom{2k+1}{2(k-2)} (-1)^{k-2} \\ &= \binom{2k+1}{1} \binom{k}{2}(-1)^{k-2} + \binom{2k+1}{3} \binom{k-1}{1} (-1)^{k-2}+\binom{2k+1}{5} (-1)^{k-2} \\ &= (-1)^{k} \left (\binom{2k+1}{1} \frac{k(k-1)}{2} + \binom{2k+1}{3}(k-1)+\binom{2k+1}{5} \right) \\ &= (-1)^{k} \left ( \frac{(2k+1)k(k-1)}{2} + \frac{(2k+1)k(2k-1)}{3} + \frac{(2k+1)k(2k-1)(k-1)(2k-3)}{5\cdot2\cdot3} \right) \\ &= (-1)^k (2k+1)k\frac{1}{30} \left ( 15(k-1) + 10(2k-1)+(2k-1)(k-1)(2k-3) \right) \end{align*}

\begin{align*} \sum_{r=0}^k a_k &= \frac12 \left ((1 + \sqrt{1^2-1})^n + (1 - \sqrt{1^2-1})^{n} \right) \\ &= 1 \end{align*}

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Solution: \begin{align*} && \frac1{r} &= A \cos \theta + B \cos (\theta - \gamma) \\ &&&= A \cos \theta + B \cos \theta \cos \gamma + B \sin \theta \sin \gamma \\ &&&= (A+B \cos \gamma) \cos \theta + B \sin \gamma \sin \theta \\ \Longleftrightarrow && 1 &= (A+B \cos \gamma) x + B \sin \gamma y \end{align*} So if we choose \(B = \frac{m}{\sin \gamma}\) and \(A = l-m \cot \gamma\) we have the desired result. \begin{align*} && \frac{1 + e \cos (\gamma -\delta)}a &= A \cos (\gamma - \delta) + B \cos (\gamma - \delta - \gamma) \\ &&&= A \cos(\gamma-\delta) +B \cos \delta\\ && \frac{1 + e \cos (\gamma +\delta)}{a} &= A \cos (\gamma + \delta) + B \cos (\gamma + \delta - \gamma) \\ &&&= A \cos(\gamma + \delta) + B \cos \delta\\ \Rightarrow && \frac1{r} &= \frac{e}{a} \cos \theta + \frac{1}{a \cos \delta} \cos (\theta - \gamma) \\ \lim{\delta \to 0} &&\frac1{r} &= \frac{e}{a} \cos \theta+ \frac{1}{a} \cos (\theta - \gamma) \end{align*} Suppose we have have points \(L\) and \(M\) with \(\theta = \gamma_L, \gamma_M\) then our tangents are: \begin{align*} && \frac{a}{r} &= \cos \theta + \cos (\theta - \gamma_L) \\ && \frac{a}{r} &= \cos \theta + \cos (\theta - \gamma_M) \\ \Rightarrow && 0 &= \cos (\theta - \gamma_L) -\cos(\theta - \gamma_M) \\ &&&= - 2 \sin \frac{(\theta - \gamma_L)+(\theta - \gamma_M)}{2} \sin \frac{(\theta - \gamma_L)-(\theta - \gamma_M)}{2} \\ &&&= -2 \sin \left ( \theta - \frac{\gamma_L+\gamma_M}2 \right) \sin \left ( \frac{\gamma_M - \gamma_L}{2}\right) \\ \Rightarrow && \theta &= \frac{\gamma_L+\gamma_M}{2} \end{align*} Therefore clearly \(ST\) bisects \(LSM\). The line \(LM\) can be seen as the chord from the points \(\frac{\gamma_L+\gamma_M}{2} \pm \frac{\gamma_L-\gamma_M}{2}\), so the line is: \begin{align*} && \frac{a}{r} &= e \cos \theta + \frac{1}{\cos \left ( \frac{\gamma_L-\gamma_M}{2}\right)} \cos \left (\theta - \frac{\gamma_L+\gamma_M}{2} \right) \end{align*} and we want the point on the line where \(\theta =\frac{\gamma_L+\gamma_M}{2}\) so \begin{align*} && \frac{a}{r} &= e \cos \left ( \frac{\gamma_L+\gamma_M}{2} \right) + \frac{1}{\cos \left ( \frac{\gamma_L-\gamma_M}{2}\right)} \\ \Rightarrow && r &= \frac{a}{e \cos \left ( \frac{\gamma_L+\gamma_M}{2} \right) + \frac{1}{\cos \left ( \frac{\gamma_L-\gamma_M}{2}\right)}} \end{align*}

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Solution:

We can see that \(\mathbf{M}\) sends \(\begin{pmatrix} 1 \\ \tan \alpha \end{pmatrix}\) to itself, and \(\begin{pmatrix} -\tan \alpha \\ 1 \end{pmatrix}\) to \(\begin{pmatrix} -\tan \alpha \\ 1 \end{pmatrix} + k \begin{pmatrix} 1 \\ \tan \alpha \end{pmatrix}\) Therefore, we have: \begin{align*} && \mathbf{M} \begin{pmatrix} 1 & -\tan \alpha \\ \tan \alpha & 1 \end{pmatrix} &= \begin{pmatrix} 1 & k - \tan \alpha \\ \tan \alpha & 1 + k\tan \alpha \end{pmatrix} \\ && \sec \alpha \mathbf{M} \begin{pmatrix} \cos \alpha & -\sin\alpha \\ \sin \alpha & \cos \alpha \end{pmatrix} &= \begin{pmatrix} 1 & k - \tan \alpha \\ \tan \alpha & 1 + k\tan \alpha \end{pmatrix} \\ \Rightarrow && \mathbf{M} &= \cos \alpha\begin{pmatrix} 1 & k - \tan \alpha \\ \tan \alpha & 1 + k\tan \alpha \end{pmatrix}\begin{pmatrix} \cos \alpha & \sin\alpha \\ -\sin \alpha & \cos \alpha \end{pmatrix} \\ &&&= \cos\alpha \begin{pmatrix}\cos \alpha -k\sin\alpha + \frac{\sin^2 \alpha}{\cos \alpha} & \sin \alpha + k \cos \alpha - \sin \alpha \\ \sin \alpha - \sin \alpha - k\frac{\sin^2 \alpha}{\cos \alpha} & \frac{\sin^2 \alpha}{\cos \alpha} + \cos\alpha + k \sin \alpha \end{pmatrix} \\ &&&= \begin{pmatrix}1-k\sin\alpha\cos\alpha & k\cos^{2}\alpha\\ -k\sin^{2}\alpha & 1+k\sin\alpha\cos\alpha \end{pmatrix} \end{align*} Suppose $\begin{pmatrix}p & q\\ r & t \end{pmatrix} \mathbf{M} = \mathbf{M} \begin{pmatrix}p & q\\ r & t \end{pmatrix}$ then, \begin{align*} && \begin{pmatrix}p & q\\ r & t \end{pmatrix} \mathbf{M} &= \mathbf{M} \begin{pmatrix}p & q\\ r & t \end{pmatrix} \\ \Leftrightarrow && \small \begin{pmatrix} p(1-k\sin\alpha\cos\alpha) + q(-k\sin^{2}\alpha) & pk\cos^{2}\alpha + q(1+k\sin\alpha\cos\alpha)\\ r(1-k\sin\alpha\cos\alpha) + t(-k\sin^{2}\alpha) & rk\cos^{2}\alpha + t(1+k\sin\alpha\cos\alpha)\end{pmatrix} &= \\ && \qquad \small \begin{pmatrix} p(1-k\sin\alpha\cos\alpha) + rk\cos^{2}\alpha & q(1-k\sin\alpha\cos\alpha) + tk\cos^{2}\alpha \\ -pk\sin^{2}\alpha + r(1+k\sin\alpha\cos\alpha) & -qk\sin^{2}\alpha+t (1+k\sin\alpha\cos\alpha) \end{pmatrix} \\ \Leftrightarrow && \begin{cases} p(1-k\sin\alpha\cos\alpha) + q(-k\sin^{2}\alpha) &= p(1-k\sin\alpha\cos\alpha) + rk\cos^{2}\alpha \\ pk\cos^{2}\alpha + q(1+k\sin\alpha\cos\alpha) &=q(1-k\sin\alpha\cos\alpha) + tk\cos^{2}\alpha \\ r(1-k\sin\alpha\cos\alpha) + t(-k\sin^{2}\alpha) &=-pk\sin^{2}\alpha + r(1+k\sin\alpha\cos\alpha) \\ rk\cos^{2}\alpha + t(1+k\sin\alpha\cos\alpha) &= -qk\sin^{2}\alpha+t (1+k\sin\alpha\cos\alpha) \end{cases} \\ \Leftrightarrow && \begin{cases} -q\tan^{2}\alpha &= r \\ p\cos^{2}\alpha + q\sin\alpha\cos\alpha &=-q\sin\alpha\cos\alpha + t\cos^{2}\alpha \\ -r\sin\alpha\cos\alpha + -t\sin^{2}\alpha &=-p\sin^{2}\alpha + r\sin\alpha\cos\alpha \\ r &= -q\tan^{2}\alpha \end{cases} \\ \Leftrightarrow && \begin{cases} -q\tan^{2}\alpha &= r \\ 2q\sin\alpha\cos\alpha &=(t-p)\cos^{2}\alpha \\ (p-t)\sin^{2}\alpha &=2r\sin\alpha\cos\alpha \end{cases} \\ \Leftrightarrow && \begin{cases} -q\tan^{2}\alpha &= r \\ 2q\tan \alpha &=(t-p) \end{cases} \\ \end{align*} as required

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Solution: \begin{align*} && \frac{\d x}{\d t} &= 4(x-y) \\ && \frac{\d y}{\d t} &= x - 12(e^{2t}+e^{-2t}) \\ \Rightarrow && \frac{\d^2 x}{\d t^2} &= 4 \frac{\d x}{\d t}-4\frac{\d y}{\d t} \\ &&&= 4 \frac{\d x}{\d t}-4 \left ( x - 12(e^{2t}+e^{-2t}) \right) \\ \Rightarrow && \frac{\d^2 x}{\d t^2} - 4 \frac{\d x}{\d t}+4x &= 48 (e^{2t}+e^{-2t}) \end{align*} This differential equation has characteristic polynomial \(\lambda^2 - 4\lambda + 4 = (\lambda-2)^2\). Therefore we should expect a general solution of \((At+B)e^{2t}\). For particular integrals we should try \(ke^{-2t}\) and \(Ct^2 e^{2t}\). For the former, we have: \begin{align*} && 48 &= 4k+8k+k \\ \Rightarrow && k &= \frac{48}{13} \end{align*} For the latter we have: \begin{align*} &&4Ct^2e^{2t} -4C(2te^{2t}+2t^2e^{2t})+2C((1+2t)e^{2t}+2t^2e^{2t}) &= 48e^{2t} \\ \Rightarrow && 2C &= 48 \\ \Rightarrow && C &= 24 \end{align*} Therefore the solution should be: \begin{align*} x = (At+B)e^{2t} + \frac{48}{13}e^{-2t} + 24t^2 e^{2t} \\ x(0) = B + \frac{48}{13} \\ x'(0) = 2B+A-\frac{96}{13} \\ x =\frac{48}{13}((4t-1)e^{2t}+e^{-2t})+24t^2e^{2t} \\ y = x - \frac{1}{4} \frac{\d x}{\d t} \end{align*}

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Solution:

1. \begin{align*} && \frac1{1-x} &= \sum_{r=0}^{\infty} x^r \\ \underbrace{\Rightarrow}_{\frac{\d}{\d x}} && \frac{1}{(1-x)^2} &= \sum_{r=0}^\infty rx^{r-1} \\ \underbrace{\Rightarrow}_{x = \frac12} && 4 &= \sum_{r=0}^{\infty} \frac{r}{2^{r-1}} \end{align*}
2. \begin{align*} && \frac1{1-x} &= \sum_{r=1}^{\infty} x^{r-1} \\ \underbrace{\Rightarrow}_{\int} && -\ln (1-x) &= \sum_{r=1}^{\infty} \frac1{r} x^r \\ \underbrace{\Rightarrow}_{x = \frac12} && \ln 2 &= \sum_{r=1}^{\infty} \frac1{r } \times \frac{1}{ 2^{r}} \\ \Rightarrow && 2 \ln 2 &= \sum_{r=1}^{\infty} \frac1{r } \times \frac{1}{ 2^{r-1}} \\ \end{align*}
3. \begin{align*} && (1-x)^{-1/2} &= 1 + \frac{(-\tfrac12)}{1!} (-x) +\frac{(-\tfrac12)(-\tfrac32)}{2!}(-x)^2 + \cdots \\ &&&= \sum_{r=0}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots \cdot (2r-1)}{2^rr!} x^r \\ \underbrace{\Rightarrow}_{x = \frac23} && \sqrt{3} &= \sum_{r=0}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots \cdot (2r-1)}{r!} \frac1{3^r} \\ &&&= 1 + \frac{1}{1!} \frac23 + \frac13 \sum_{r=2}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots \cdot (2r-1)}{r!} \frac1{3^{r-1}} \\ \Rightarrow && 3\sqrt{3}-5 &= \sum_{r=2}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots \cdot (2r-1)}{r!} \frac1{3^{r-1}} \\ \end{align*}

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Solution:

1. Claim: \[ \sum_{r=1}^{n}r(r+1)(r+2)(r+3)(r+4)=\tfrac{1}{6}n(n+1)(n+2)(n+3)(n+4)(n+5) \] Proof: (By Induction) Base case: (n=1) \begin{align*} LHS &= 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 5! \\ RHS &= \frac16 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 = 5! \end{align*} Therefore the base case is true. Inductive step: Suppose our statement is true for some \(n=k\), then consider \(n = k+1\) \begin{align*} \sum_{r=1}^{k+1} r(r+1)(r+2)(r+3)(r+4) &= \sum_{r=1}^{k} r(r+1)(r+2)(r+3)(r+4) + (k+1)(k+2)(k+3)(k+4)(k+5) \\ &\underbrace{=}_{\text{assumption}} \frac16 k(k+1)(k+2)(k+3)(k+4)(k+5) + (k+1)(k+2)(k+3)(k+4)(k+5) \\ &= (k+1)(k+2)(k+3)(k+4)(k+5) \l \frac{k}{6} +1\r \\ &= \frac16 (k+1)(k+2)(k+3)(k+4)(k+5)(k+6) \end{align*} Therefore our statement is true for \(n = k+1\). Therefore since our statement is true for \(n=1\) and if it is true for \(n=k\) then it is true for \(n = k+1\) by the principle of mathematical induction it is true for all \(n \geq 1\) Since \begin{align*} \sum_{r=1}^{n}r^5 &< \sum_{r=1}^{n}r(r+1)(r+2)(r+3)(r+4) \\ &= \frac16 n(n+1)(n+2)(n+3)(n+4)(n+5) \end{align*}
2. \begin{align*}\sum_{r=0}^{n-1} r^5 &> \sum_{r=0}^{n-1} (r-4)(r-3)(r-2)(r-1)r \\ &= \sum_{r=0}^{n-5} r(r+1)(r+2)(r+3)(r+4) \\ &= \frac16 (n-5)(n-4)(n-3)(n-2)(n-1)n \end{align*}
3. Let \(f(x) = x^5\) \begin{align*} S_{1,n} &= \sum_{r=0}^{n-1}\frac{a}{n}f\left(\frac{ra}{n}\right) \\ &= \sum_{r=0}^{n-1}\frac{a}{n}\left(\frac{ra}{n}\right)^5 \\ &=\frac{a^6}{n^6} \sum_{r=0}^{n-1}r^5\\ \end{align*} Therefore \(\frac{a^6}6 \frac{(n-5)(n-4)(n-3)(n-2)(n-1)n}{n^6} < S_{1,n} < \frac{a^6}6 \frac{(n-1)n(n+1)(n+2)(n+3)(n+4)}{n^6}\) and so \(\lim_{n\to\infty} S_{1,n} = \frac{a^6}{6}\). Similarly, \begin{align*} S_{2,n} &= \sum_{r=1}^{n}\frac{a}{n}f\left(\frac{ra}{n}\right) \\ &= \sum_{r=1}^{n}\frac{a}{n}\left(\frac{ra}{n}\right)^5 \\ &= \frac{a^6}{n^6} \sum_{r=1}^{n} r^5 \end{align*} Therefore \(\frac{a^6}6 \frac{(n-4)(n-3)(n-2)(n-1)n(n+1)}{n^6} < S_{2,n} < \frac{a^6}6 \frac{n(n+1)(n+2)(n+3)(n+4)(n+5)}{n^6}\) and so \(\lim_{n\to\infty} S_{2,n} = \frac{a^6}{6}\). Since both limits exist are are equal, we have \[ \int_{0}^{a}x^{5}\,\mathrm{d}x=\tfrac{1}{6}a^6. \]

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Solution: Set up coordinate axes so that at time \(t\) \(OA\) is the \(x\)-axis, and all rotations are counter-clockwise. Then if \(OA = \mathbf{a}\), \(AP = \mathbf{x}\) and \(OP = \mathbf{p}\) we have: \begin{align*} \mathbf{a} &= \binom{a}{0} \\ \dot{\mathbf{a}} &= \binom{0}{-a \Omega} \\ \ddot{\mathbf{a}} &= \binom{-a \Omega^2}{0} \\ \\ \mathbf{x} &= \binom{b \cos \phi }{b \sin \phi } \\ \dot{\mathbf{x}} &= b \dot{\phi} \binom{-\sin \phi}{\cos \phi} \\ \ddot{\mathbf{x}} &= \binom{-b \ddot{\phi} \sin \phi-b \dot{\phi}^2 \cos \phi }{b \ddot{\phi} \cos \phi-b \dot{\phi}^2 \sin \phi} \\ \\ \ddot{\mathbf{p}} &= \binom{-a \Omega^2 +-b \ddot{\phi} \sin \phi-b \dot{\phi}^2 \cos \phi }{b \ddot{\phi} \cos \phi-b \dot{\phi}^2 \sin \phi} \end{align*} We can take a dot product with \(\mathbf{n} = \binom{-\sin \phi}{\cos \phi}\) to obtain the component perpendicular to \(AP\), which is: \begin{align*} && \binom{-\sin \phi}{\cos \phi} \cdot \ddot{x} &= a \Omega^2 \sin \phi + b \ddot{\phi} \end{align*} Noticing that this component must be \(0\) (since the only force acting on \(P\) is the rod), this must be equal to zero. \begin{align*} && 0 &= a \Omega^2 \sin \phi + b \ddot{\phi} \\ \Rightarrow && 0 &= a \Omega^2 \dot{\phi} \sin \phi + b\dot{\phi} \ddot{\phi} \\ \Rightarrow && C &= -a \Omega^2 \cos \phi + \tfrac12 b \dot{\phi}^2 \end{align*} Noticing that the initial conditions are \(\phi = 0\) and \(\dot{\phi} = 2\sqrt{\frac{a}{b}} \Omega\), so \begin{align*} && C &= -a \Omega^2+ \tfrac12 b \left ( 2\sqrt{\frac{a}{b}} \Omega \right)^2 \\ &&&= -a \Omega^2 + 2a \Omega^2 \\ &&&= a \Omega^2\\ \Rightarrow && \dot{\phi} &=\sqrt{\frac{2}{b} \left ( a \Omega^2 + a \Omega^2 \cos \phi \right)} \\ &&&= \Omega \sqrt{\frac{2a}{b}} \sqrt{1+ \cos \phi} \\ &&& = \Omega \sqrt{\frac{2a}{b}}\sqrt{2} \cos \tfrac{\phi}{2} \\ \Rightarrow && \int \sec \tfrac{\phi}{2} \d \phi &= 2 \Omega \sqrt{\frac{a}{b}}t \\ \Rightarrow && \tfrac12 \ln | \sec \tfrac{\phi}{2}+\tan \tfrac{\phi}{2} | &= 2 \Omega \sqrt{\frac{a}{b}}t + C \\ t = 0, \phi = 0: && C = 0 \\ \Rightarrow && \sec \tfrac{\phi}{2}+\tan \tfrac{\phi}{2} &= e^{4 \Omega \sqrt{\frac{a}{b}}t} \end{align*} Since when \(t > 0\) this is positive and larger than \(1\) we cannot have \(\phi = 0\) and since it remains below infinite \(\phi\) cannot reach \(\pi\). Therefore it cannot cross \(OA\)

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Solution:

The extension of \(OAP\) is \(|AP|\) and so the tension \(T_a = \frac{\lambda}{a} |AP|\). To simplify calculations, let \(A = a, B = a \omega, C = a \omega^2, \cdots\) where \(\omega = e^{2 \pi i/5}\) and let \(P = z\). then we can calculate the force as: \begin{align*} &&\sum_{p}T_p \mathbf{n}_{z \to p} &= \sum_{p} \frac{\lambda}{a} |z-p| \frac{p-z}{|p-z|} \\ &&&= \frac{\lambda}{a} \sum_{p} ( p - z) \\ &&&= -\frac{5\lambda}{a}z \end{align*} Therefore the force has magnitude \(\frac{5 \lambda}{a} |OP|\) directly towards the origin. Therefore if we set up our coordinate axis such that \(OP\) is the \(x\) axis, the particle will remain on the \(x\) axis and will move under the equation: \[ m \ddot{x} + \frac{5 \lambda}{a} x = 0 \] But then we can say that \(P\) moves under SHM with period \(\displaystyle 2 \pi \sqrt{\frac{am}{5 \lambda}}\) as required. Now suppose that \(PAO\) has been replaced with the string of modulus \(k \lambda\) but that \(P\) is along \(OA\). \begin{align*} F &= \frac{\lambda}{a}\left ( (a \omega - z) + (a \omega^2 - z)+ (a \omega^3 -z)+ (a \omega^4 - z) + k(a -z) \right) \\ &= \frac{\lambda}{a}(-a - 4z+ka -kz) \\ &= \frac{\lambda}{a}((k-1)a-(k+4)z) \end{align*} Notice that if \(z\) is real, this expression is also real, so all forces are acting along \(OA\). Therefore the particle will remain on the line \(OA\). We can also notice that the particle will move under the differential equation \[ m \ddot{x} + \frac{(k+4) \lambda}{a}x = \lambda(k-1) \] Therefore it will move with SHM about a point slightly displaced from the origin. The period will be: \(\displaystyle 2 \pi \sqrt{\frac{ma}{(k+4)\lambda}}\) which is equal to \(T/2\) if \((k+4) = 20 \Rightarrow k = 16\)