Problems

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Solution:

1. For step 3, since we have assumed \(\sqrt{2}\) is rational we can write it in the form \(p/q\) with \(p, q\) coprime with \(q \geq 1\). Then \(q \in S\) since \(q\sqrt{2} = p\) which is an integer. For step 5, notice that \((\sqrt{2}-1)k\) is an integer (since \(\sqrt{2}k\) is an integer and so is \(-k\). It is also positive since \(\sqrt{2} > 1\). We must check that \((\sqrt{2}-1)k \cdot \sqrt{2} = 2k - \sqrt{2}k\) is also an integer, but clearly it is as both \(2k\) and \(-\sqrt{2}k\) are integers. Therefore \((\sqrt{2}-1)k \in S\). For step 6, notice that \((\sqrt{2}-1) < 1\) and therefore \((\sqrt{2}-1)k < k\), contradicting that \(k\) is the smallest element in our set. (And all non-empty sets of positive integers have a smallest element)
2. Claim: \(2^{\frac13}\) is irrational \(\Leftrightarrow 2^{\frac23}\) is irrational. Proof: Since \(2^{\frac13} \cdot 2^{\frac23} = 2\) if one of them is rational, then the other one must also be rational. Which is the same as them both being irrational at the same time.
   1. Assume that \(\sqrt[3]{2}\) is rational, ie \(\sqrt[3]{2} = p/q\) for some integers.
   2. \(S := \{ n \in \mathbb{Z}_{>0} : n \sqrt[3]{2} \text{ and } n \sqrt[3]{4}\in \mathbb{Z}\}\)
   3. Suppose \(k\) is the smallest element in \(S\) (which must exist, consider \(q^2\)
   4. Consider \((\sqrt[3]{2}-1)k\) then clearly this is an integer, and \((\sqrt[3]{2}-1)\sqrt[3]{2}k = \sqrt[3]{4}k - \sqrt[3]{2}k \in \mathbb{Z}\) and \((\sqrt[3]{2}-1)\sqrt[3]{4}k = 2 k -\sqrt[3]{4}k \in \mathbb{Z}\).
   5. But this is a smaller element of \(S\), contradicting that \(k\) is the smallest element. Therefore, we have a contradiction.

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Solution:

1. Notice that \(u^2 = v+2wz\), so \(w,z\) are roots of the quadratic \(t^2 -ut+\frac{u^2-v}{2}\). Therefore they are both real if \(u^2 \geq 2(u^2-v) \Rightarrow 2v \geq u^2\).
2. \begin{align*} && w+z &= u \\ && w^2+z^2 &= u^2 - \tfrac23 \\ && w^3+z^3 &= \lambda(u-1) \\ \\ && wz &= \frac{u^2 - (u^2-\tfrac23)}{2} = \tfrac13\\ \\ && (w+z)(w^2+z^2) &= w^3+z^3+wz(w+z) \\ &&u(u^2-\tfrac23)&= \lambda(u-1)+\frac13u \\ \Rightarrow && u^3-u&= \lambda (u-1) \\ \Rightarrow && 0 &= (u-1)(u(u+1) - \lambda) \\ \Rightarrow && 0 &= (u-1)(u^2+u - \lambda) \end{align*} Therefore there will be at most 3 values for \(u\), unless \(1\) is a root of \(u^2+u-\lambda\), ie \(\lambda = 2\). Suppose \(u = 1\), then we have: \(w+z = 1, wz = 1/3 \Rightarrow w,z = \frac{-1 \pm \sqrt{-1/3}}{2}\) which are clearly complex.

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Solution: \begin{align*} {\mathrm D}^2 x^a &= x\frac{\d\ }{\d x}\left( x\frac{\d}{\d x} \left ( x^a \right) \right) \\ &= x\frac{\d\ }{\d x}\left( ax^a \right) \\ &= a^2 x^a \end{align*}

1. Claim: \({\mathrm D^n}(x^a) =a^n x^a\) Proof: Induct on \(n\). Base cases we have already seen, so consider \(D^{k+1}(x^a) = D(a^k x^a) = a^{k+1}x^a\) as required. Claim: \({\mathrm D}\) is linear, ie \({\mathrm D}(f(x) + g(x)) = {\mathrm D}(f(x)) + {\mathrm D}(g(x))\) Proof: \begin{align*} {\mathrm D}(f(x) + g(x)) &= x\frac{\d\ }{\d x}\left(f(x) + g(x) \right) \\ &= x\frac{\d\ }{\d x}f(x) + x\frac{\d\ }{\d x}g(x) \\ &= {\mathrm D}(f(x)) + {\mathrm D}(g(x)) \end{align*} Claim: If \(p(x)\) is a polynomial degree \(r\) then \({\mathrm D}^n p(x)\) is a polynomial degree \(n\). Proof: Since \({\mathrm D}\) is linear, it suffices to prove this for a monomial of degree \(n\), but this was already proven in the first question.
2. Claim: If \(f(x)\) is some polynomial, \({\mathrm D}((1-x)^m f(x))\) is divisible by \((1-x)^{m-1}\) Proof: \({\mathrm D}((1-x)^mf(x)) = -xm(1-x)^{m-1}f(x) + (1-x)^mxf'(x) = x(1-x)^{m-1}((1-x)f'(x)-xf(x))\) as required. Therefore repeated application of \({\mathrm D}\) will reduce the factor of \(1-x\) by at most \(1\) each time as required.
3. \begin{align*} {\mathrm D}^n(1-x)^m &= {\mathrm D}^n \left ( \sum_{r=0}^m \binom{m}{r}(-1)^r x^r\right) \\ &= \sum_{r=0}^m {\mathrm D}^n \left ( \binom{m}{r}(-1)^r x^r \right ) \\ &= \sum_{r=0}^m\binom{m}{r}(-1)^r r^n x^r \end{align*} Since the left-hand side is divisible by \(1-x\), if we substitute \(x = 1\), the sum must be \(0\), i.e., we get the desired result.
4. On each application of \({\mathrm D}\) to \((1-x)^m f(x)\) we end up with a term in the form \(x(1-x)^{m-1}(x)\) and a term of the form \((1-x)^m\). After the latter term will be annihilated once we evaluate at \(x = 1\) because there will be insufficient applications to remove the factors of \(1-x\). Therefore we only need to focus on the term which does not get annihilated. This term is will be \((-x)^n n \cdot (n-1) \cdots 1\), so \(f_n(1) = (-1)^n n!\) as required. Alternatively: \begin{align*} {\mathrm D}^n((1-x)^n) &= D^{n-1}(-nx(1-x)^{n-1}) \\ &= -n{\mathrm D}^{n-1}(x(1-x)^{n-1}) \\ &= -n{\mathrm D}^{n-1}((x-1+1)(1-x)^{n-1}) \\ &= -n{\mathrm D}^{n-1}(-(1-x)^{n}+(1-x)^{n-1}) \\ &= -n{\mathrm D}^{n-1}(-(1-x)^{n})-n{\mathrm D}^{n-1}((1-x)^{n-1}) \\ \end{align*} Therefore, when this is evaluated at \(x = 1\), recursively, we will have \(f_n(1) = -nf_{n-1}(1)\), in particular, \(f_n(1) = (-1)^n n!\)

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Solution:

1. \begin{align*} && (y+x)\frac{\d y}{\d x} &= y-x \\ \Rightarrow && (r \sin \theta + r \cos\theta) \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} &= (r \sin\theta - r \cos\theta) \\ \Rightarrow && ( \sin \theta + \cos\theta) \frac{dy}{d\theta} &= (\sin\theta - \cos\theta){\frac{dx}{d\theta}} \\ \Rightarrow && ( \sin \theta + \cos\theta) \left ( \frac{dr}{d\theta} \cos \theta - r \sin \theta\right ) &= (\sin\theta - \cos\theta)\left ( \frac{dr}{d\theta} \sin\theta + r \cos \theta\right) \\ \Rightarrow && \frac{dr}{d\theta} \left (\sin \theta \cos \theta + \cos^2 \theta - \sin^2 \theta + \sin \theta \cos \theta \right)&= r \left (\sin \theta \cos \theta - \cos^2 \theta + \sin^2 \theta + \sin\theta \cos \theta\right) \\ \Rightarrow && \frac{dr}{d\theta}&= -r \\ \end{align*} Therefore \(r = Ae^{-\theta}\)
   

   + - ↺
2. \begin{align*} && \left( y+x -x(x^2+y^2) \right) \, \frac{\d y }{\d x} &= y-x - y(x^2+y^2) \\ \Rightarrow && \left( r \sin \theta+r\cos \theta -r^3\cos \theta \right) \, \frac{\d y }{\d \theta} &= \left ( r \sin \theta- r \cos \theta- r^3\sin \theta \right)\frac{\d x }{\d \theta} \\ \Rightarrow && \left( r \sin \theta+r\cos \theta -r^3\cos \theta \right) \, \left (\frac{\d r}{\d \theta} \sin \theta + r \cos \theta \right) &= \\ && \left ( r \sin \theta- r \cos \theta- r^3\sin \theta \right)&\left (\frac{\d r}{\d \theta} \cos \theta - r \sin \theta \right) \\ \Rightarrow && \frac{\d r}{\d \theta} \left (\sin \theta ( \sin \theta + \cos \theta - r^2 \cos \theta) - \cos \theta (\sin \theta - \cos \theta - r^2 \sin \theta) \right) &= \\ && r ( -\sin \theta (\sin \theta - \cos \theta - r^2 \sin \theta) - \cos \theta ( \sin \theta + \cos \theta &- r^2 \cos \theta)) \\ \Rightarrow && \frac{\d r}{\d \theta} &= r ( -1 +r^2) \\ \Rightarrow && \int \frac{1}{r(r-1)(r+1)} \d r &= \int \d \theta \\ \Rightarrow && \int \l \frac{-1}{r} + \frac{1}{2(r-1)} + \frac{1}{2(r+1)} \r \d r &= \int \d \theta \\ \Rightarrow && \l -\log r+ \frac12 \log (1+r)+ \frac12 \log (1-r)\r + C &= \theta \\ \Rightarrow && \frac12 \log \left (\frac{1-r^2}{r^2} \right) + C &= \theta \\ \Rightarrow && \log \left (\frac{1}{r^2}-1 \right) + C &= 2\theta \\ \Rightarrow && r &= \frac{1}{1 + Ae^{2\theta}} \\ \end{align*}
   

   + - ↺

   

   + - ↺

   

   + - ↺

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Solution:

1. Since the particle is not moving (relative to the hinge) there is no moment about the hinge and in particular the only forces must be directed towards the hinge, ie parallel to the rod.
   

   + - ↺
   \begin{align*} \text{N2}(\uparrow): && R \cos \alpha &= mg \\ \\ \text{N2}(\leftarrow, \text{radially}): && R \sin \alpha &= m (r-d\sin \alpha) \omega^2 \\ \Rightarrow && \cot \alpha &= \frac{g}{(r-d\sin \alpha) \omega^2} \\ \Rightarrow && r\cot \alpha-d \cos \alpha &= a \\ \Rightarrow && r \cot \alpha &= a + d \cos \alpha \end{align*} The force of the hinge is acting in the same direction and magnitude as the rod on the particle (the force \(R\) in the diagram). It has magnitude \(mg \sec \alpha\)
2. \(\,\)
   

   + - ↺
   \begin{align*} \overset{\curvearrowleft}{\text{hinge}}: && gm_1d_1 \sin \beta+gm_2d_2 \sin \beta &= m_1 (r-d_1 \sin \beta) \omega^2 d_1 \cos \beta + m_2 (r-d_2 \sin \beta) \omega^2 d_2 \cos \beta \\ \Rightarrow && a(m_1d_1+m_2d_2) \tan \beta &= r(m_1d_1+m_2d_2) - (m_1d_1^2+m_2d_2^2) \sin \beta \\ \Rightarrow && r\cot \beta &= a + \frac{m_1d_1^2+m_2d_2^2}{m_1d_1+m_2d_2} \cos \beta \end{align*}

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Solution:

1. \(G(x) = \frac{1}{6} (1 + x + x^2 + x^3 + x^4 + x^5)\) The pgf for \(R_2\) is: \begin{align*} \frac1{36}x^2 + \frac{2}{36}x^3 + \frac{3}{36}x^4 + \frac{4}{36}x^5 + \frac{5}{36} +\\ \quad \quad + \frac{6}{36}x^1 + \frac{5}{36}x^2 + \frac4{36}x^3 + \frac3{36}x^4 + \frac{2}{36}x^5 + \frac{1}{36} \\ = \frac{1}{6}(1 + x + x^2 + x^3 + x^4 + x^5) = G(x) \end{align*} Since rolling the dice twice is the same as rolling the dice once, rolling the dice \(n\) times will be the same as rolling it once, ie the pgf for \(R_n\) will be \(G(x)\) and the probability \(S_n\) is divisible by \(6\) is \(\frac16\)
2. \(G_1(x) = \frac{1}{6} + \frac{1}{3}x^1 + \frac{1}{6}x^2 + \frac16x^3+ \frac16x^4 = \frac16(1 + 2x+x^2+x^3+x^4)\). If \(G_n\) is the probability generating function for \(T_n\) then we can obtain \(G_n\) by multiplying \(G_{n-1}\) by \(G(x)\) and replacing any terms of order higher than \(5\) with their remainder on division by \(5\). (Or equivalently, working over \(\mathbb{R}[x]/(x^5-1)\). If \(y = 1 + x + x^2 + x^3 + x^4\) then: \begin{align*} xy &= x + x^2 + x^3 + x^4 +x^5 \\ &= x + x^2 + x^3 + x^4 + 1 \\ &= y \\ \\ y^2 &= (1 + x+x^2 + x^3+x^4)^2 \\ &= 1 + 2x + 3x^2 + 4x^3+5x^4+4x^5+3x^6 + 2x^7 + x^8 \\ &= (1+4) + (2+3)x+(3+2)x^2 + (4+1)x^3 + 5x^4 \\ &= 5y \end{align*} \begin{align*} \frac{1}{36}(y+x)(y+x) &= \frac1{36}(y^2 + 2xy + x^2) \\ &= \frac1{36}(5y + 2y + x^2 ) \\ &= \frac1{36}(7y + x^2) \end{align*} Similarly, \begin{align*} G_n(x) &= \l\frac{1}{6}(x+y) \r^n \\ &= \frac1{6^n} \l \sum_{i=0}^n \binom{n}{i} y^ix^{n-i} \r \\ &= \frac1{6^n} \l \sum_{i=1}^n \binom{n}{i} y^i + x^n \r \\ &= \frac1{6^n} \l \sum_{i=1}^n \binom{n}{i} 5^{i-1}y + x^n \r \\ &= \frac1{6^n} \l \frac{1}{5}y((5+1)^n-1) + x^n \r \\ &= \frac1{6^n} \l \frac{1}{5}y(6^n-1) + x^n \r \\ \end{align*} Therefore if \(n \not \equiv 0 \pmod{5}\), we can find the probability of \(T_n = 0\) by looking at the constant coefficient, ie plugging in \(x = 0\), which is: \[\frac1{6^n} \l \frac{1}{5}(6^n-1) \r = \frac{1}{5} \l 1- \frac{1}{6^n} \r \] When \(n \equiv 0 \pmod{5}\) we can also find the constant coefficient by plugging in \(x = 0\), which is: \[\frac1{6^n} \l \frac{1}{5}(6^n-1) + 1 \r = \frac{1}{5} \l 1+ \frac{4}{6^n} \r \]

Note: this whole question can be considered a "roots-of-unity" filter in disguise. Our computations in \(\mathbb{R}[x]/(x^5 - 1)\) are the same as computations using \(\omega\), in fact \(\mathbb{R}[x]/(x^5 - 1) \cong \mathbb{R}[\omega]\) where \(\omega\) is a primitive \(5\)th root of unity

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Solution:

1. \(\mathbb{P}(X + Y \leq c) \) is the area between the \(x\)-axis, \(y\)-axis and the line \(x + y = c\). There are two cases for this: \[\mathbb{P}(X + Y \leq c) = \begin{cases} 0 & \text{ if } c \leq 0 \\ \frac{c^2}{2} & \text{ if } c \leq 1 \\ 1- \frac{(2-c)^2}{2} & \text{ if } 1 \leq c \leq 2 \\ 1 & \text{ otherwise} \end{cases}\]
2. \begin{align*} && \mathbb{P}((X + Y)^{-1} \leq t) &= 1- \mathbb{P}(X + Y \leq \frac1{t}) \\ \Rightarrow && f_{(X+Y)^{-1}}(t) &= 0 -\begin{cases} 0 & \text{ if } \frac1{t} \leq 0 \\ \frac{\d}{\d t}\frac{1}{2t^2} & \text{ if } \frac{1}{t} \leq 1 \\ \frac{\d}{\d t} \l 1- \frac{(2-\frac1t)^2}{2} \r & \text{ if } 1 \leq \frac{1}{t} \leq 2 \\ 0 & \text{ otherwise}\end{cases} \\ && &= \begin{cases} t^{-3} & \text{ if } t \geq 1 \\ (2-\frac1t)t^{-2} & \text{ if } \frac12 \leq t \leq 1\\ 0 & \text{ otherwise}\end{cases} \\ && &= \begin{cases} t^{-3} & \text{ if } t \geq 1 \\ 2t^{-2}-t^{-3} & \text{ if } \frac12 \leq t \leq 1\\ 0 & \text{ otherwise}\end{cases} \end{align*} Therefore, \begin{align*} \E \Big(\dfrac1{X+Y}\Big) &= \int_{\frac12}^{\infty} t f_{(X+Y)^{-1}}(t) \, \d t \\ &= \int_{\frac12}^{1} t f_{(X+Y)^{-1}}(t) \, \d t + \int_{1}^{\infty} t f_{(X+Y)^{-1}}(t) \d t\\ &= \int_{\frac12}^{1} \l 2t^{-1} - t^{-2} \r \, \d t + \int_{1}^{\infty} t^{-2} \d t\\ &= \left [ 2 \ln (t) + t^{-1} \right]_{\frac12}^{1} + \left [ -t^{-1} \right ]_{1}^{\infty} \\ &= 1 + 2 \ln 2 -2 + 1 \\ &= 2 \ln 2 \end{align*}
3. \begin{align*} &&\mathbb{P} \l \frac{Y}{X} \leq c \r &= \mathbb{P}( Y \leq c X) \\ &&&= \begin{cases} 0 & \text{if } c \leq 0 \\ \frac{c}{2} & \text{if } 0 \leq c \leq 1 \\ 1-\frac{1}{2c} & \text{if } 1 \leq c \end{cases} \\ \\ \Rightarrow && \mathbb{P} \l \frac{X}{X+Y} \leq t\r &= \mathbb{P} \l \frac{1}{1+\frac{Y}{X}} \leq t\r \\ &&&= \mathbb{P} \l \frac{1}{t} \leq 1+\frac{Y}{X}\r \\ &&&= \mathbb{P} \l \frac{1}{t} - 1\leq \frac{Y}{X}\r \\ &&&= 1- \mathbb{P} \l \frac{Y}{X} \leq \frac{1}{t} - 1\r \\ &&&= 1 - \begin{cases} 0 & \text{if } \frac1{t} \leq 0 \\ \frac{1}{2t} - \frac{1}{2} & \text{if } 0 \leq \frac1{t} \leq 1 \\ 1-\frac{t}{2-2t} & \text{if } 1 \leq \frac1{t} \end{cases} \\ && f_{\frac{X}{X+Y}}(t) &= \begin{cases} 0 & \text{if } \frac1{t} \leq 0 \\ \frac{1}{2t^2} & \text{if } t \geq 1 \\ \frac{1}{2(1-t)^2} & \text{if } 0 \leq t \leq 1 \end{cases} \\ \Rightarrow && \mathbb{E} \l \frac{X}{X+Y} \r &= \int_0^\infty t f(t) \d t \\ &&&= \int_0^1 \frac{1}{2(1-t)^2} \d t + \int_1^\infty \frac{1}{t^2} \d t \\ &&& = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \\ \\ && \mathbb{E} \l \frac{X}{X+Y} \r &= \int_0^1 \int_0^1 \frac{x}{x+y} \d y\d x \\ &&&= \int_0^1 \l x \ln (x+1) - x \ln x \r \d x \\ &&&= \left [\frac{x^2}2 \ln(x+1) - \frac{x^2}{2} \ln(x) \right]_0^1 -\int_0^1 \l \frac{x^2}{2(x+1)} - \frac{x}{2} \r \d x \\ &&&= \frac{\ln 2}{2} + \frac{1}{4} - \int_0^1 \frac{x^2-1+1}{2(x+1)}\d x \\ &&&= \frac{\ln 2}{2} + \frac{1}{4} - \int_0^1 \frac{x -1}{2} + \frac{1}{2(x+1)}\d x \\ &&&= \frac{\ln 2}{2} + \frac{1}{4} - \frac{1}{4} + \frac{1}{2} - \frac{\ln 2}{2} \\ &&&= \frac{1}{2} \end{align*} We can also notice that \(1 = \mathbb{E} \l \frac{X+Y}{X+Y} \r = \mathbb{E} \l \frac{X}{X+Y} \r + \mathbb{E} \l \frac{Y}{X+Y} \r = 2 \mathbb{E} \l \frac{X}{X+Y} \r\) so it's clearly true as long as we can show that the integral converges.