Problems

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Solution:

1. Notice that \(P\) lies on \(C_1C_2\), and that the triangles formed from \(C_iPT_i\) where \(T_i\) are the tangent points are similar, with ratios \(\frac{r_1}{r_2}\). Therefore \(\frac{C_1P}{r_1} = \frac{C_2P}{r_2}\), and hence \(\frac{C_1P}{C_1C_2} = \frac{C_1P}{C_1P-C_2P} = \frac{1}{1-\frac{r_2}{r_1}} = \frac{r_1}{r_1-r_2}\) So we have \(\mathbf{p} = \mathbf{x_1} + (\mathbf{x}_2 - \mathbf{x}_1)\cdot\frac{r_1}{r_1-r_2} = \frac{r_1\mathbf{x}_2 - r_2\mathbf{x}_1}{r_1-r_2}\)
2. Suppose \(\mathbf{x}_3 = \binom{\alpha}{\beta}\) in the basis of \(\{ \mathbf{x}_1, \mathbf{x}_2 \}\), then we can see that \begin{align*} && \mathbf{p} &= \frac{1}{r_1-r_2}\binom{-r_2}{r_1} \\ && \mathbf{q} &= \frac{r_1(\alpha \mathbf{x}_1 +\beta \mathbf{x}_2) - r_3\mathbf{x}_1}{r_1-r_3} \\ &&&= \frac{1}{r_1-r_3} \binom{r_1\alpha -r_3}{r_1\beta} \\ && \mathbf{r} &=\frac{1}{r_2-r_3} \binom{r_2\alpha}{r_2\beta - r_3} \\ && \mathbf{p}-\mathbf{q} &= \frac{1}{r_1-r_2}\binom{-r_2}{r_1} - \frac{1}{r_1-r_3} \binom{r_1\alpha -r_3}{r_1\beta} \\ &&&= \frac{1}{(r_1-r_2)(r_1-r_3)} \binom{(r_1-r_3)(-r_2)-(r_1-r_2)(r_1\alpha-r_3)}{(r_1-r_3)r_1 - (r_1-r_2)r_1\beta} \\ &&&= \frac{r_1}{(r_1-r_2)(r_1-r_3)} \binom{(r_3-r_2)-\alpha(r_1-r_2)}{(r_1-r_3)-\beta(r_1-r_2)} \\ && \mathbf{q} - \mathbf{r} &= \frac{1}{r_1-r_3} \binom{r_1\alpha -r_3}{r_1\beta} - \frac{1}{r_2-r_3} \binom{r_2\alpha}{r_2\beta - r_3} \\ &&&= \frac{1}{(r_1-r_3)(r_2-r_3)}\binom{(r_2-r_3)(r_1\alpha-r_3) - (r_1-r_3)r_2\alpha)}{(r_2-r_3)r_1\beta - (r_1-r_3)(r_2\beta - r_3)} \\ &&&= \frac{1}{(r_1-r_3)(r_2-r_3)}\binom{(-r_2r_3+r_3^2) - \alpha(r_1r_3-r_3r_2)}{r_3(r_1-r_3)-\beta(r_1-r_2)} \\ &&&= \frac{r_3}{(r_1-r_3)(r_2-r_3)}\binom{(r_3-r_2)-\alpha(r_1-r_2)}{(r_1-r_3)-\beta(r_1-r_2)} \end{align*} Therefore they are clearly parallel, and hence lie on a line.
3. \(Q\) is halfway between \(P\) and \(R\) if \begin{align*} && \frac{r_1}{(r_1-r_2)(r_1-r_3)} &= \frac{r_3}{(r_1-r_3)(r_2-r_3)} \\ \Leftrightarrow && r_1(r_2-r_3) &= r_3(r_1-r_2) \\ \Leftrightarrow && r_1r_2 - r_1r_3 &= r_1r_3 - r_2r_3 \\ \Leftrightarrow && r_2 &= \frac{2r_1r_3}{r_1+r_3} \end{align*}

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Solution:

1. The sine rule tells us: \begin{align*} && \frac{\frac58 a}{\sin(30^\circ + \theta)} &= \frac{a}{\sin(90^{\circ}-\theta)} \\ \Rightarrow &&\frac58 \cos \theta &= \frac12 \cos \theta+ \frac{\sqrt{3}}2 \sin \theta \\ \Rightarrow && \frac{1}{4\sqrt{3}} &= \tan \theta \\ \Rightarrow && \sin \theta &= \sqrt{\frac{1}{48+1}} = \frac17 \end{align*}
2. \(\,\) \begin{align*} && \text{initial energy} &= \frac12(5m)v^2 + \frac12 (3m)v^2 - 3m \cdot g \cdot a \cos(30^{\circ}+\theta) -5m \cdot g \cdot a\cos(30^\circ - \theta) \\ &&&= 4m v^2 - amg(4\sqrt{3} \cos \theta + \sin \theta) \\ &&&= 4mv^2 - 7amg \\ && \text{energy at top} &= \frac12 m v_{top}^2 + 7amg \end{align*} We need this equation to be positive for all values of \(v_{top} \geq 0\), so \(4mv^2 \geq 14amg \Rightarrow v_0 = \sqrt{\frac{7ag}2}\)

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Solution:

The length of the string is \(h/\cos \theta\), and it is decreasing at a rate \(V\). The distance along the ground is decreasing at a rate of \(V/\sin \theta\). Note that \(-V = \frac{\d}{\d t} \left ( \frac{h}{\cos \theta} \right) = \frac{h} {\cos^2 \theta} \sin \theta \cdot \dot{\theta} \Rightarrow \dot{\theta} = -\frac{V\cos^2\theta}{h \sin \theta}\). Note that \(a = \frac{\d}{\d t} \left ( \frac{V}{\sin \theta} \right) = -\frac{V}{\sin^2 \theta} \cos \theta \cdot \dot{\theta} = \frac{V^2 \cos^3 \theta}{h\sin^3 \theta}\). Resolving horizontally we must have \(T \sin \theta = ma \Rightarrow T = \frac{V^2m \cos^3 \theta}{h \sin^4 \theta}\). Resolving vertically at the point where we are about to leave the ground, we must have \(T\cos \theta = mg \Rightarrow \frac{V^2m \cos^4 \theta}{h \sin^4 \theta} = mg \Rightarrow \tan^4 \theta = \frac{V^2}{gh}\)

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Solution:

1. \(A\) has coordinates \((x-a\cos \theta, a\sin \theta)\). Differentiating with respect to \(t\) the velocity of \(A\) is \((\dot{x}+a\sin \theta \cdot \dot{\theta}, a \cos \theta \cdot \dot{\theta})\)
2. By considervation of momentum \(\rightarrow\) we must have \(mu = m(\dot{x}+a\dot{\theta}\sin \theta) + m\dot{x} + m(\dot{x}+a\dot{\theta}\sin \theta) = m(3\dot{x} + 2a \dot{\theta} \sin \theta)\) and the first equation follows. By conservation of energy, we must have \begin{align*} && \frac12 m u^2 &= \frac12 m \dot{x}^2 + \frac12m((\dot{x}+a\dot{\theta}\sin \theta)^2 + a^2 \dot{\theta}^2 \cos^2\theta ) + \frac12m((\dot{x}+a\dot{\theta}\sin \theta)^2 + a^2 \dot{\theta}^2 \cos^2\theta ) \\ &&&= \frac32m\dot{x}^2 + 2m a\dot{x}\dot{\theta}\sin \theta + ma^2\dot{\theta}^2(\sin^2\theta+\cos^2\theta) \\ \Rightarrow && u^2 &= \dot{x}(3\dot{x} + 4a \dot{\theta} \sin \theta) + 2a^2\dot{\theta}^2 \\ &&&= \left ( \frac{u-2a\dot{\theta}\sin \theta}{3}\right)\left ( 3\left ( \frac{u-2a\dot{\theta}\sin \theta}{3}\right)+ 4a \dot{x}\dot{\theta} \sin \theta \right) + 2a^2\dot{\theta}^2 \\ \Rightarrow && 3u^2 &= (u - 2a\dot{\theta} \sin \theta)^2 + 4a(u - 2 a \dot{\theta} \sin \theta) \dot{\theta}\sin \theta + 6a^2 \dot{\theta}^2 \\ &&&= u^2 + 4a^2\dot{\theta}^2 \sin^2 \theta - 8a^2\dot{\theta}^2\sin^2\theta + 6a^2 \dot{\theta}^2 \\ \Rightarrow && \dot{\theta}^2 &= \frac{u^2}{a^2(3-2\sin^2\theta)} \end{align*}
3. Since \(\dot{\theta}^2 > 0\) \(\theta\) is strictly increasing or decreasing, therefore the first collision will be when \(\theta = 0\), the second when \(\theta = \pi\)
4. If \(v = 0\), from our first equation we have \(2a \dot{\theta} \sin \theta = u \Rightarrow \dot{\theta}^2 = \frac{u^2}{4a^2 \sin^2 \theta} = \frac{u^2}{a^2(3-2\sin^2\theta)}\) so \(4\sin^2 \theta = 3 - 2\sin^2 \theta \Rightarrow \sin^2 \theta = \frac{1}{2}\) therefore the angles are all the multiples of \(\frac{\pi}{4}\).

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Solution:

1. The only way \(A\) can win is if the sequence starts HH, if it does not start like this, then the only way HHT can appear is after a sequence of THH...H, but then THH has already appeared and \(B\) has won. Therefore the probability is \(\frac14\)
2. If HH appears before TT then either \(A\) or \(B\) will win. If HH appears first, then \(A\) has a \(\frac14\) probability of winning. So \(A\): \(\frac18\), \(B:\), \(\frac38\), \(C:\), \(\frac18\), \(D: \frac38\)
3. If the first two tosses are TT then \(C\) will win. If the first two tosses are HT, then either the next toss is T and \(C\) wins, or the next toss is H, and it's as if we started TH. ie \(p = \frac12 + \frac12 q\). If the first two tosses are TH, then either the next toss is H and \(C\) losses or the next toss is T and it's like starting HT. So \(q = \frac12 p\). Therefore \(p = \frac12 + \frac14p \Rightarrow p = \frac13\) If the first two tosses are HH, then eventually a T appears, and it's the same as starting HT. Therefore the probability \(C\) wins is: \(\frac14 + \frac14 \cdot \frac13 + \frac14 \cdot \frac16 + \frac14 \cdot \frac13 = \frac{11}{24}\)

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Solution:

1. \(\,\) \begin{align*} && \mathbb{E}(C) &= \int_0^\infty \text{cost}(x) f(x) \d x \\ &&&= ky + \int_y^{\infty} a(x-y) \lambda e^{-\lambda x} \d x\\ &&&= ky + \int_0^{\infty} a u \lambda e^{-\lambda u -\lambda y} \d x \\ &&&= ky + ae^{-\lambda y} \left( \left [ -ue^{-\lambda u} \right]_0^\infty -\int_0^\infty e^{-\lambda u} \d u\right) \\ &&&= ky + \frac{a}{\lambda}e^{-\lambda y} \\ \\ && \frac{\d \mathbb{E}(C)}{\d y} &= k - ae^{-\lambda y} \\ \Rightarrow && y &= \frac{1}{\lambda}\ln \left ( \frac{a}{k} \right) \end{align*} Since \(\mathbb{E}(C)\) is clearly increasing when \(y\) is very large, the optimal value will be \(\frac{1}{\lambda}\ln \left ( \frac{a}{k} \right)\), if \(\frac{a}{k} > 1\), otherwise you should spend nothing on flood defenses.
2. \begin{align*} && \mathbb{E}(C^2) &= \int_0^{\infty} \text{cost}(x)^2 f(x) \d x \\ &&&= \int_0^{\infty}(ky + a(x-y)\mathbb{1}_{x > y})^2 f(x) \d x \\ &&&= k^2y^2 + \int_y^{\infty}2kya(x-y)f(x)\d x + \int_y^{\infty}a^2 (x-y)^2 f(x) \d x \\ &&&= k^2y^2 + \frac{2kya}{\lambda}e^{- \lambda y}+a^2e^{-\lambda y}\int_{u=0}^\infty u^2 \lambda e^{-\lambda u} \d u \\ &&&= k^2y^2 + \frac{2kya}{\lambda}e^{-\lambda y}+a^2e^{-\lambda y}(\textrm{Var}(Exp(\lambda)) + \mathbb{E}(Exp(\lambda))^2\\ &&&= k^2y^2 + \frac{2kya}{\lambda}e^{-\lambda y} + a^2e^{-\lambda y} \frac{2}{\lambda^2} \\ && \textrm{Var}(C) &= k^2y^2 + \frac{2kya}{\lambda}e^{-\lambda y} + a^2e^{-\lambda y} \frac{2}{\lambda^2} - \left ( ky + \frac{a}{\lambda} e^{-\lambda y}\right)^2 \\ &&&= a^2e^{-\lambda y} \frac{2}{\lambda^2} - a^2 e^{-2\lambda y}\frac{1}{\lambda^2} \\ &&&= \frac{a^2}{\lambda^2} e^{-\lambda y}\left (2 - e^{-\lambda y} \right) \\ \\ && \frac{\d \textrm{Var}(C)}{\d y} &= \frac{a^2}{\lambda^2} \left (-2\lambda e^{-\lambda y} +2\lambda e^{-2\lambda y} \right) \\ &&&= \frac{2a^2}{\lambda} e^{-\lambda y}\left (e^{-\lambda y}-1 \right) \leq 0 \end{align*} so \(\textrm{Var}(C)\) is decreasing in \(y\).

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Solution: \begin{align*} I_n - I_{n+1} &= \int_0^\infty \frac 1 {(1+u^2)^n}\, \d u - \int_0^\infty \frac 1 {(1+u^2)^{n+1}}\, \d u \\ &= \int_0^\infty \l \frac 1 {(1+u^2)^n}- \frac 1 {(1+u^2)^{n+1}} \r\, \d u \\ &= \int_0^\infty \frac {u^2} {(1+u^2)^{n+1}} \, \d u \\ &= \left [ u \frac{u}{(1+u^2)^{n+1}} \right]_0^{\infty} - \frac{-1}{2n}\int_0^{\infty} \frac{1}{(1+u^2)^n} \d u \tag{\(IBP: u = u, v' = \frac{u}{(1+u^2)^{n+1}}\)}\\ &= \frac{1}{2n} I_n \end{align*} \(\displaystyle I_1 = \int_0^{\infty} \frac{1}{1+u^2} \d u = \left [ \tan^{-1} u \right]_0^\infty = \frac{\pi}{2}\) as expected. We also have, \(I_{n+1} = \frac{2n(2n-1)}{2n \cdot 2n} I_n \), by rearranging the recurrence relation. Therefore, when we multiply out the top we will have \(2n!\) and the bottom we will have two factors of \(n!\) and two factors of \(2^n\) combined with the original \(\frac{\pi}{2}\) we get \[ I_{n+1} = \frac{(2n)! \pi}{2^{2n+1} (n!)^2} \] \begin{align*} J = \int_0^\infty f\big( (x- x^{-1})^2\big ) \, \d x &= \int_{u = \infty}^{u = 0} f((u^{-1}-u)^2)(-u^{-2} )\d u \tag{\(u = x^{-1}, \d u = -x^{-2} \d x\)} \\ &= \int^{u = \infty}_{u = 0} f((u^{-1}-u)^2)u^{-2} \d u \\ &= \int^{\infty}_{0} u^{-2}f((u-u^{-1})^2) \d u \\ \end{align*} Therefore adding the two forms for \(J\) we have \begin{align*} 2 J &= \int_0^\infty f\big( (x- x^{-1})^2\big ) \, \d x + \int_0^\infty x^{-2} f\big( (x- x^{-1})^2\big ) \, \d x \\ &= \int_0^\infty (1+x^{-2}) f\big( (x- x^{-1})^2\big ) \, \d x \end{align*} And letting \(u = x - x^{-1}\), we have \(\d u = (1 + x^{-2}) \d x\), and \(u\) runs from \(-\infty\) to \(\infty\) so we have: \begin{align*} \int_0^\infty (1+x^{-2}) f\big( (x- x^{-1})^2\big ) \, \d x &= \int_{-\infty}^\infty f(u^2) \, \d u \\ &=2 \int_{0}^\infty f(u^2) \, \d u \end{align*} Since both of these are \(2J\) we have the result we are after. Finally, \begin{align*} \int_0^\infty \frac {x^{2n-2}}{(x^4-x^2+1)^n} \, \d x &= \int_0^{\infty} \frac{x^{2n-2}}{x^{2n}(x^2-1+x^{-2})^n} \d x \\ &= \int_0^{\infty} \frac{x^{-2}}{((x-x^{-1})^2+1)^n} \d x \\ &= \int_0^{\infty} \frac{1}{(x^2+1)^n} \d x \tag{Where \(f(x) = (1+x^2)^{-n}\) in \(J\) integral} \\ &= I_n = \frac{(2n-2)! \pi}{2^{2n-1} ((n-1)!)^2} \end{align*}

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Solution:

1. If \(m = 1000\), then \(n \geq m \Rightarrow n^2 \geq 1000n \Rightarrow (1000n) \leq (n^2)\)
2. This is false. Let \(s_i = 1,2,1,2,\cdots\) and \(t_i = 2,1,2,1,\cdots\).
3. Suppose that for \(n \geq m_1, s_n \le t_n\) and for \(n \geq m_2, s_t \le u_n\), then for \(n \geq m = \max(m_1, m_2), s_n \leq t_n \leq u_n \Rightarrow s_n \leq u_n \Rightarrow (s_n) \leq (u_n)\)
4. Let \(m = 6\), then if \(n \geq m, 2^n \geq 1 + n + \frac{n(n-1)}{2} + \frac{n(n-1)}{2} + n + 1 = n^2 + n + 2 \geq n^2\), so \((2^n) \geq (n^2)\)

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Solution:

1. \(r = a \sec \theta \pm b\). The points on \(r = a \sec \theta \Leftrightarrow r \cos \theta = a \Leftrightarrow x = a\) are points on the line \(x = a\). Therefore points on the curve \(r = a \sec \theta \pm b\) are points which are a distance \(b\) from the line \(x = a\) measured towards \(O\). So \(A\) is the origin and \(d = b\).
   

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2. 

   + - ↺
   The loop starts and ends when \(r = a \sec \theta - b = 0 \Rightarrow \cos \theta = \frac{a}{b}\), so when \(a = 1, b = 2\), this is \(-\frac{\pi}{3}\) to \(\frac{\pi}{3}\) \begin{align*} && A &= \frac12 \int r^2 \d \theta \\ &&&= \frac12 \int_{-\pi/3}^{\pi/3} \left ( \sec \theta - 2 \right)^2 \d \theta \\ &&&= \frac12 \int_{-\pi/3}^{\pi/3} \left (\sec^2 \theta - 4 \sec \theta + 4\right)\d \theta \\ &&&= \frac12 \left [ \tan \theta -4 \ln | \sec \theta + \tan \theta| + 4 \theta \right]_{-\pi/3}^{\pi/3} \\ &&&= \frac12 \left (\left (\tan \frac{\pi}3 - 4 \ln | \sec \frac{\pi}3 + \tan \frac{\pi}3 | + 4\left ( \frac{\pi}3 \right)\right) - \left (\tan \left (-\frac{\pi}3 \right) - 4 \ln | \sec \left (-\frac{\pi}3 \right)+ \tan\left ( -\frac{\pi}3 \right) | + 4\left ( -\frac{\pi}3 \right)\right) \right) \\ &&&= \frac12 \left ( 2\sqrt{3} - 4 \ln |2 + \sqrt{3}| + 4 \ln |2-\sqrt{3}| + \frac{8\pi}3 \right) \\ &&&= \sqrt{3} + 2\ln \frac{2-\sqrt{3}}{2+\sqrt{3}} + \frac{4\pi}3 \\ &&&= \sqrt{3} + 4 \ln (2 - \sqrt{3})+ \frac{4\pi}3 \end{align*}

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Solution:

1. Let \(z \in \mathbb{R}\) and let \(z \to \pm \infty\) then \(z^3 + az^2 + bz + c\) changes sign, therefore somewhere it must have a real root.
2. \begin{align*} &&z^3 + az^2 + bz + c &= (z-z_1)(z-z_2)(z-z_3) \\ && &= z^3 - (z_1+z_2+z_3)z^2 + (z_1z_2 + z_2z_3+z_3z_1)z - (z_1z_2z_3) \\ \\ \Rightarrow && S_1 &= z_1+z_2+z_3 \\ &&&= -a \\ \\ \Rightarrow && S_2 &= z_1^2+z_2^2+z_3^2 \\ &&&= (z_1+z_2+z_3)^2 - 2(z_1z_2 + z_2z_3+z_3z_1) \\ &&&= a^2 - 2b \\ \Rightarrow && a &= -S_1 \\ && b &= \frac12 \l S_1^2 - S_2\r \\ \\ && 0 &= z_i^3 + az_i^2+bz_i+c \\ \Rightarrow && 0 &= S_3 + aS_2+bS_1+3c \\ &&&= S_3 -S_1S_2 + \frac12 \l S_1^2 - S_2\r S_1 + 3c \\ \Rightarrow && 0 &= 2S_3 - 3S_1S_2 + S_1^3 + 6c \end{align*}
3. Let \(z_k= r_ke^{i \theta_k}\), then we have \(\textrm{Im}(S_k) = 0\) and so the polynomial with roots \(z_k\) has real coefficients, and therefore at least one root is real. This root will have \(\theta_k = 0\). Moreover, since if \(w\) is a root of a real polynomial \(\overbar{w}\) is also a root, and therefore if \(\theta_1 = 0\), we must have that \(z_2\) and \(z_3\) are complex conjugate, ie \(\theta_2 = - \theta_3\)

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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 \(r = Ae^{-\theta}\)
   

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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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