Problems

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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 \((k+4) = 20 \Rightarrow k = 16\)

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Solution:

1. 

   + - ↺
   \begin{align*} m &= \int_0^a \underbrace{(\rho r)}_{\text{mass per area}} \underbrace{\pi r^2}_{\text{area}} \d r \\ &= \rho \pi \frac{a^3}{3} \\ \\ I &= \sum m r^2 \\ &= \sum (\rho r) \pi r^2 \cdot r^2 \\ &\to \int_0^a \rho \pi r^4 \\ &= \frac15 \rho \pi a^5 \\ &= \frac35 m a^2 \end{align*}
2. 

   + - ↺
   \begin{align*} \text{N2}(\downarrow, D): && Mg -T_C &= Mf \\ \overset{\curvearrowright}{C} && (T_C - T_B)a &= I \frac{f}{a} \\ &&&= \frac35 m a f \\ \text{N2}(\uparrow, B): && 2T_B-mg &= \frac12 m f \\ \\ \Rightarrow && Mg-T_B &= \left (M + \frac35 m \right)f \\ \Rightarrow && Mg - \frac12 mg &= \left (M + \frac35 m + \frac14 m \right)f \\ \Rightarrow && f &= \frac{(M-\frac12 m)g}{M + \frac{17}{20} m} \\ &&&= \frac{(2M-m)g}{2M +\frac{17}{10}m} \end{align*} Therefore the speed after a time \(t\) is \(\displaystyle \frac{(2M-m)g}{2M +\frac{17}{10}m} t\) and the impulse will be the change in momentum, ie \(\displaystyle \frac{(2M-m)g}{2M +\frac{17}{10}m} Mt\)

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Solution:

1. \begin{align*} \mathbb{P}(n\text{ independent values of }X > k) &= \prod_{i=1}^n \mathbb{P}(X > k) \\ &= \left ( \frac{c-k}{2c}\right)^n \end{align*}
2. \begin{align*} \mathbb{P}(n\text{ independent values of }X < k) &= \prod_{i=1}^n \mathbb{P}(X < k) \\ &= \left ( \frac{k+c}{2c}\right)^n \end{align*}

\begin{align*} &&\mathbb{P}(\text{median} < z+\delta \text{ and median} > z - \delta) &= \mathbb{P}(n\text{ values } < z - \delta \text{ and } n \text{ values} > z + \delta) \\ &&&= \binom{2n+1}{n,n,1} \left ( \frac{c-(z+\delta)}{2c}\right)^n\left ( \frac{(z-\delta)+c}{2c}\right)^n \frac{2 \delta}{2 c} \\ &&&= \frac{(2n+1)!}{n! n!} \frac{((c-(z+\delta))(c+(z-\delta)))^n 2\delta}{2^n c^n} \\ &&&= \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}}((c-(z+\delta))(c+(z-\delta)))^n 2\delta \\ \Rightarrow && \lim_{\delta \to 0} \frac{\mathbb{P}(\text{median} < z+\delta \text{ and median} > z - \delta)}{2 \delta} &= \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}}((c-z)(c+z))^n \\ &&&= \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}}(c^2-z^2) \\ \end{align*} \begin{align*} && 1 &= \int_{-c}^c \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}}(c^2-z^2)^n \d z \\ \Rightarrow && \frac{(n!)^2 (2c)^{2n+1}}{(2n+1)!} &= \int_{-c}^c (c^2-z^2)^n \d z \end{align*} \begin{align*} \mathbb{E}(Z) &= \int_{-c}^c z \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}}(c^2-z^2)^n \d z \\ &=\frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}} \int_{-c}^c z (c^2-z^2)^n \d z \\ &= 0 \end{align*} \begin{align*} \mathrm{Var}(Z) &= \mathbb{E}(Z^2) - \mathbb{E}(Z)^2 \\ &= \mathbb{E}(Z^2) \\ &= \int_{-c}^c z^2 \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}}(c^2-z^2)^n \d z \\ &=\frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}} \int_{-c}^c z^2 (c^2-z^2)^n \d z \\ &=\frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}} \left ( \left [ -\frac{1}{2(n+1)}z(c^2-z^2)^{n+1} \right]_{-c}^c + \frac{1}{2(n+1)}\int_{-c}^c (c^2-z^2)^{n+1} \d z \right) \\ &= \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}} \frac{1}{2(n+1)} \frac{((n+1)!)^2 (2c)^{2n+3}}{(2n+3)!} \\ &= \frac{(n+1)^2(2c)^2}{(n+1)(2n+2)(2n+3)} \\ &= \frac{2c^2}{2n+3} \end{align*}

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Solution:

Conditions tell us: \begin{align*} && a+b+d+e &= 0.25 \\ && b+c+e+f &= 0.2 \\ && e &= 0 \\ && b+c &= e + f \\ && b &= c \\ && c+h &= 0.5 \\ && a &= g \\ \end{align*} So \(4b = 0.2 \Rightarrow b = 0.05\) And \begin{align*} && 0.25 &= a + d + 0.05 \\ && 1 &= 2a + d + 0.65 \\ \Rightarrow && a &= 0.15 \\ && d &= 0.05 \end{align*} So the proportion who are narrow-headed is \(30\%\). It's obviously relatively unlikely for your six Ruritanian friends to all be broad-headed if it's a random sample, but friendship groups are are likely to be biased so it's not too surprising. Assuming there is a sufficiently large number of Ruritanians, we might model the number of narrow-headed Ruritanians from a sample of \(200\) as \(X \sim B(200, 0.3)\). Computing \(\mathbb{P}(X \leq 50)\) by hand is tricky, so let's use a binomial approximation to obtain: \(X \approx N(60, 42)\) and \begin{align*} \mathbb{P}(X \leq 50) &\approx \mathbb{P} \left (Z \leq \frac{50 - 60+0.5}{\sqrt{42}} \right) \\ &\approx \mathbb{P} \left (Z \leq -\frac{9.5}{6.5} \right) \\ &\approx \mathbb{P} \left (Z \leq -\frac{3}{2} \right) \\ &\approx 5\% \end{align*} (actually this approximation gives \(7.1\%\) and the binomial value gives \(7.0\%\)). This also seems somewhat surprising