Problems

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Solution: Claim: \(n \not \mid (n-1)!\) if and only if \(n\) is prime or \(4\) Proof: \((\Leftarrow)\)

1. \(4 \not \mid 3! = 6\).
2. If \(p\) is prime, then \(p \not \mid k\) for \(k < n\), therefore \(p \not \mid (n-1)!\)

\((\Rightarrow)\) If \(n = 1\) then \(1 \mid 0! = 1\) so \(1\) is not in our set. The numbers less than \(6\) are all accounted for (either primes, \(4\) or \(1\)), so let \(n\) be a composite number larger than \(6\), ie \(n = ab\). Suppose first \(a \neq b\) then \((n-1) = 1 \cdots a \cdots b \cdots (n-1)\) so \(n \mid (n-1)!\). Suppose instead that \(a = b\), then \(n = a^2\). Since we know \(a \geq 3\) we must have \(1 \cdots a \cdots (2a) \cdots (a^2-1)\) so \(a^2 \mid (n-1)!\) and we're done.

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Solution: Let \({\displaystyle I_{m,n}=\int\cos^{m}x\sin nx\,\mathrm{d}x,}\) Then \begin{align*} && I_{m,n} &= \int\cos^{m}x\sin nx\,\mathrm{d}x \\ &&&= \left [ -\frac1n \cos^m x \cos n x \right] - \frac{m}{n} \int \sin^{m-1} x \cos x \cos n x \d x \\ &&&= \left [ -\frac1n \cos^m x \cos n x \right] + \frac{m}{n} \int \sin^{m-1} x (\cos (n-1)x -\sin x \sin nx) \d x\\ &&&= \left [ -\frac1n \cos^m x \cos n x \right] + \frac{m}{n} \int \sin^{m-1} x \cos (n-1)x \d x-\frac{m}{n} I_{m,n} \\ &&&= \left [ -\frac1n \cos^m x \cos n x \right] + \frac{m}{n} I_{m-1,n-1} -\frac{m}{n} I_{m,n} \\ \Rightarrow && nI_{m,n} &= -\cos^m x \cos n x + mI_{m-1,n-1} -mI_{m,n}\\ \Rightarrow && (m+n)I_{m,n} &= -\cos^m x \cos n x + mI_{m-1,n-1} \end{align*}

1. Note that \(I_{2m,0} = 0\) (the integrand is 0) and \(I_{0, 2m} = 0\) (symmetry for our limits). \(\displaystyle \left [-\cos^m x \cos n x \right]_0^\pi = \l - (-1)^m (-1)^n \r - \l -1 \r = 1 - (-1)^{m+n} = 0\) since \(m+n\) is even. Therefore all reductions are \(I_{m,n} = \frac{I_{m-1,n-1}}{m+n}\) terminating at \(0\), so all values are zero
2. \begin{align*} \int_{0}^{\frac{\pi}{2}}\sin^{2}x\sin3x\,\mathrm{d}x &= \int_{0}^{\frac{\pi}{2}}(1-\cos^2x)\sin3x\,\mathrm{d}x \\ &= I_{0,3} - I_{2,3} \\ &= \frac13 - \frac15 \l \left [-\cos^2 x \cos 3 x \right]_0^{\pi/2} + 2 \cdot I_{1,2} \r \\ &= \frac13 - \frac15 \l 1 + \frac23 \l \left [-\cos x \cos 2 x \right]_0^{\pi/2} + 1\cdot I_{0,1} \r \r \\ &= \frac13 - \frac15 -\frac2{15} - \frac2{15} \\ &= \frac{5}{15} - \frac{3}{15} - \frac{4}{15} \\ &= -\frac2{15} \end{align*}

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Solution:

1. If \(z=x+iy\) then \(|z|^2 = x^2 + y^2 \leq x^2 + y^2 + 2|x||y| \leq (|x|+|y|)^2\). The LHS is Cauchy-Schwarz with the vectors \(\begin{pmatrix} |x| \\ |y| \end{pmatrix}\) and \(\begin{pmatrix} \cos \alpha \\ \sin \alpha \end{pmatrix}\), although that's not in the spirit of the question. Consider \(e^{i \alpha}z = (\cos \alpha x - \sin \alpha y) + i(\sin \alpha x + \cos \alpha y)\) then \(\left | \textrm{Re}(e^{i \alpha} z) \right | \leq |z|\) for all values of \(\alpha\) and in particular we can choose \(\alpha\) to match the signs of the \(x\) and \(y\) to prove the result in question.
2. Consider \((5-\mathrm{i})^{4}(1+\mathrm{i})\), then \begin{align*} \arg \l (5-\mathrm{i})^{4}(1+\mathrm{i}) \r &= \arg (5-i)^4 + \arg (1+i) \\ &= 4 \arg (5-i) + \arg (1+i) \\ &= -4 \tan^{-1} \frac{1}{5} + \tan^{-1} 1 \\ \\ &= \arg ( (24 - 10i)^2 (1+i)) \\ &= \arg (4 (12 - 5i)^2(1+i)) \\ &= \arg ((119 - 120i)(1+i)) \\ &= \arg (239 - i) \\ &= -\tan^{-1} \frac{1}{239} \end{align*} Therefore \(\displaystyle \frac{\pi}{4} =4 \tan^{-1} \frac{1}{5}- \tan^{-1} \frac{1}{239}\) Consider \((4-i)^3(1+i)(20-i)\) then \begin{align*} \arg \l(4-i)^3(1+i)(20-i) \r &= -3 \tan^{-1} \frac14 + \tan^{-1} 1 -\tan^{-1} \frac1{20} \\ \\ &= \arg \l(15-8i)(4-i)(1+i)(20-i) \r \\ &= \arg \l (52 - 47i)(1+i)(20-i) \r \\ &= \arg \l (99 + 5i)(20-i) \r \\ &= \arg (1985+i) \\ &= \tan^{-1} \frac1{1985} \end{align*} Therefore \(\displaystyle \frac{\pi}{4}=3\tan^{-1}\frac{1}{4}+\tan^{-1}\frac{1}{20}+\tan^{-1}\frac{1}{1985}\)

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Solution: The initial volume of water in \(A\) is: \begin{align*} \pi \int_0^4 x^2 \, \d y &= \pi \int_0^4 y \d y \\ &= \pi [ \frac{y^2}{2}]_0^4 \\ &= 8\pi \end{align*} We assume that no water is in the tube as it is `thin'. Therefore we must have: \begin{align*} && 8\pi &= \pi \int_0^{h-1} x^2 \d y +\pi \int_0^{h} x^2 \d y \\ &&&= \pi \int_0^{h-1} y \d y +\pi \int_0^{h} \l \sinh \frac{x}{2}\r^2 \d y \\ &&&= \pi \left [\frac{y^2}{2} \right]_0^{h-1} + \pi \int_0^h \frac{-1+\cosh y}{2}\d y \\ &&&= \pi \frac{(h-1)^2}{2} + \pi \left [ -\frac{y}{2} +\frac{\sinh y}{2}\right]_0^h \\ &&&= \pi \frac{(h-1)^2}{2} -\pi \frac{h}{2} + \pi \frac{\sinh h}{2} \\ \Rightarrow && 0 &= h^2-2h+1-h+\sinh h -16 \\ &&&= h^2 -3h+\sinh h - 15 \\ \Rightarrow && 15 &= h^2 -3h+\sinh h \end{align*}

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Solution: Considering the odd numbers, we have \begin{array}{l|rrrrrr} b_{2j+1} & 35 & 2 & 15 & 8 & 40 & 24 \\ a_{2j+1}+n_1 & 35 & 1 & 13 & 5 & 36 & 19 \end{array} Considering the even numbers, we have \begin{array}{l|rrrrrr} b_{2j} & 27& 36 & 35 & 40 & 37 & 48 \\ a_{2j}+n_0 & 27 & 35 & 33 & 37 & 33 & 43 \end{array} There are three numbers in the original sequence which are repeated (\(43\)). By the pigeonhole principle, one of the odds or evens must have at least two of them. We can see that the even numbers have some number repeated twice (\(33\)). Therefore these must be the \(43\)s. Therefore \(n_0 = -10\) \begin{array}{l|rrrrrr} b_{2j} & 27& 36 & 35 & 40 & 37 & 48 \\ a_{2j}+n_0 & 27 & 35 & 33 & 37 & 33 & 43 \\ a_{2j} & 37 & 45 & 43 & 47 & 43 & 3 \end{array} This leaves the remaining numbers to be decoded from the original sequence as \(1,7,23,24,39,43\). Two of these numbers are consecutive (\(23\) and \(24\)), and two numbers in our sequence are \(35\) and \(36\). Therefore \(n_1\) must be \(12\). \begin{array}{l|rrrrrr} b_{2j+1} & 35 & 2 & 15 & 8 & 40 & 24 \\ a_{2j+1}+n_1 & 35 & 1 & 13 & 5 & 36 & 19 \\ a_{2j+1} & 23 & 39 & 1 &43 & 24& 7 \end{array} Therefore the original sequence was: \(23, 37, 39, 45, 1, 43, 43, 47, 24, 43, 7, 3\)

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Solution: Letting \(\mathbf{x} =\begin{pmatrix} x(t) \\ y(t) \end{pmatrix}\) and \(\mathbf{A} = \begin{pmatrix} -2 & 5 \\ -a & 2 \end{pmatrix}\) then our differential equation is \(\mathbf{x}' = \mathbf{Ax} + \begin{pmatrix} 0 \\2 \cos t \end{pmatrix}\). Looking at the eigenvalues of \(\mathbf{A}\), we find: \begin{align*} && \det \begin{pmatrix} -2-\lambda & 5 \\ -a & 2 -\lambda \end{pmatrix} &= (\lambda^2-4)+5a\\ &&&= \lambda^2 +5a-4 \end{align*} Therefore if \(a = 1\), \(\lambda = \pm i\). In which case we should expect the complementary solutions to be of the form \(\mathbf{x} = \begin{pmatrix} A \sin t + B \cos t \\ C \sin t + D \cos t \end{pmatrix}\). The first equation tells us that \((A-5D+B)\cos t + (-B+5C)\sin t=0\) so the complementary solution is:\(\mathbf{x} = \begin{pmatrix} 5(D-C) \sin t + 5C \cos t \\ C \sin t + D \cos t \end{pmatrix}\). Looking for a particular integral, we should expect to try something like \(\mathbf{x} = \begin{pmatrix} Et\cos t+Ft\sin t\\ Gt\cos t+Ht \sin t\end{pmatrix}\) and we find

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Solution: We start by noticing that \(\displaystyle \tan^{-1} t = \int_0^t \frac{1}{1+x^2} \d x\). Consider the geometric series \(1-x^2+(-x^2)^2+ \cdots + (-x^2)^n = \frac{1-(-x^2)^{n+1}}{1+x^2}\). Therefore, \((1+x^2)(1-x^2+(-x^2)^2+ \cdots + (-x^2)^n) = 1-(-x^2)^{n+1}\) or \(1 = (1+x^2)(1-x^2+x^4-\cdots+(-1)^nx^{2n}) +(-1)^{n+1}x^{2n+2}\) \begin{align*} \tan^{-1} t &= \int_0^t \frac{1}{1+x^2} \d x \\ &= \int_0^t \frac{(1+x^2)(1-x^2+x^4-\cdots+(-1)^nx^{2n}) +(-1)^{n+1}x^{2n+2}}{x^2+1} \d x \\ &= \int_0^t (1-x^2+x^4-\cdots+(-1)^nx^{2n})\d x + \int_0^t \frac{(-1)^{n+1}x^{2n+2}}{x^2+1} \d x \\ &= t - \frac{t^3}{3}+\frac{t^5}{5}-\cdots + (-1)^n \frac{t^{2n+1}}{2n+1}+\int_0^t \frac{(-1)^{n+1}x^{2n+2}}{x^2+1} \d x \\ &= \sum_{r=0}^n \frac{(-1)^r t^{2r+1}}{2r+1} + \int_0^t \frac{(-1)^{n+1}x^{2n+2}}{x^2+1} \d x \\ \end{align*} Therefore we can say (for \(0 \leq t \leq 1\)) \begin{align*} \left | \tan^{-1} t - \sum_{r=0}^n \frac{(-1)^r t^{2r+1}}{2r+1} \right | &= \left | \int_0^t \frac{(-1)^{n+1}x^{2n+2}}{x^2+1} \d x \right | \\ &\leq \left | \int_0^t x^{2n+2} \d x \right | \\ &= \frac{t^{2n+3}}{2n+3} \\ \\ \left | \tan^{-1} t - \sum_{r=0}^n \frac{(-1)^r t^{2r+1}}{2r+1} \right | &= \left | \int_0^t \frac{(-1)^{n+1}x^{2n+2}}{x^2+1} \d x \right | \\ &\geq \left | \int_0^t \frac{(-1)^{n+1}x^{2n+2}}{1+1} \d x \right | \\ &= \frac{t^{2n+3}}{2(2n+3)} \\ \end{align*} Since \(\tan^{-1} 1 = \frac{\pi}{4}\) we must have that: \begin{align*} \lim_{n \to \infty} \left | \frac{\pi}{4} - \sum_{r=0}^{n}\frac{(-1)^{r}}{(2r+1)} \right | \to 0 \Rightarrow \lim_{n \to \infty} 4\sum_{r=0}^{n}\frac{(-1)^{r}}{(2r+1)} \to \pi \end{align*} However, \begin{align*} && \left | 4\sum_{r=0}^{n}\frac{(-1)^{r}}{(2r+1)} - \pi \right | &\geq 4 \frac{1}{2(2n+3)} \\ && &= \frac{2}{2n+3} \\ \\ && \frac{2}{2n+3} \geq 10^{-2} \\ \Leftrightarrow && 200 \geq 2n+3 \\ \Leftrightarrow && 197 \geq 2n \\ \Leftrightarrow && 98.5 \geq n \\ \end{align*} Therefore we need more than \(98\) terms to get two decimal places of accuracy. Not great!

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Solution: In a parallelogram the diagonals meet at their mid points. Fixing one diagonal, we can look at the two triangles formed by the other diagonal. Suppose the angle between them is \(\theta\). Then the area of the triangles will be \(\frac12 \frac{l_1}{2} \frac{l_2}2 \sin \theta+\frac12 \frac{l_1}{2} \frac{l_2}2 \sin (\pi -\theta) = \frac{l_1l_2}{4} \sin \theta\). This will be true on both sides. Therefore we can maximise this area by setting \(\theta = \frac{\pi}{2}\).

Consider the (darker) shaded area. This is our set \(A\). The area of the set is indifferent to a parallel shift in the lines, so without loss of generality, we can consider \(c_1 = 0, c_2 = 0\), so our lines meet at the origin. Now also consider the linear transformation \(\begin{pmatrix} a_1 & b_1 \\ a_2 & b_2 \end{pmatrix}^{-1}\) which takes the coordinate axes to these lines. This will take the square \([-\delta, \delta] \times [-\delta, \delta]\) which has area \(4\delta^2\) to our square, which will have area \(\frac{4 \delta^2}{ |a_1b_2 - a_2b_1|}\). If we consider the length of the two diagonals of this area, \(l_1, l_2\) we know that \(\frac{l_1l_2}2 \sin \theta = \frac{4 \delta^2}{|a_1b_2 - a_2b_1|}\), if we consider the larger of \(l_1\) and \(l_2\) (wlog \(l_1\)) we must have that \(\frac{l_1^2}{2} \geq \frac{4 \delta^2}{|a_1b_2 - a_2b_1|}\) and so points on opposite ends of the diagonal will satisfy the inequality in the question.

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Solution: Claim: \(G \times H\) is a group. (Called the product group). Proof: Checking the group axioms:

1. (Closure) is inherited from \(G\) and \(H\), since \(g_1 * g_2 \in G\) and \(h_1 \circ h_2 \in H\)
2. (Associativity) \begin{align*} (g_1, h_1)\l (g_2, h_2)(g_3,h_3)\r &= (g_1, h_1)(g_2 *g_3, h_2 \circ h_3) \\ &= (g_1*(g_2 *g_3), h_1 \circ (h_2 \circ h_3)) \\ &= ((g_1*g_2) *g_3), (h_1 \circ h_2) \circ h_3) \\ &= (g_1*g_2, h_1 \circ h_2)(g_3, h_3) \\ &= \l(g_1, h_1)(g_2, h_2) \r(g_3,h_3) \end{align*}
3. (Identity) Consider \((e_G, e_H)\), then \((e_G, e_H)(g,h) = (g,h) = (g,h)(e_G, e_H)\)
4. (Inverses) If \((g,h) \in G \times H\) then consider \((g^{-1}, h^{-1})\) and we have \((g^{-1}, h^{-1})(g,h) = (e_G,e_H) = (g,h)(g^{-1}, h^{-1})\)

• Claim: \(G \times H\) is abelian iff \(G\) and \(H\) are. Proof: \(\Rightarrow\) Suppose \(g_1, g_2 \in G\) and \(h_1, h_2 \in H\) then \((g_1, g_2)(h_1,h_2) = (g_1 * g_2, h_1 \circ h_2) = (h_1,h_2)(g_1, g_2) = (g_2 * g_1, h_2 \circ h_1)\) so \(g_1*g_2 = g_2*g_1\) and \(h_1 \circ h_2 = h_2 \circ h_1\), therefore \(G\) and \(H\) are commutative. \(\Leftarrow\) If \(H\) and \(G\) are commutative then: \((g_1, g_2)(h_1,h_2) = (g_1 * g_2, h_1 \circ h_2) = (g_2 * g_1, h_2 \circ h_1) = (h_1,h_2)(g_1, g_2)\) so \(G \times H\) is commutative.
• Claim: \(G\times H\) contains a subgroup isomorphic to \(G\). Consider the subset \(S = \{(g,e_H) : g \in G \}\). Then this is a subgroup isomorphic to \(G\) with isomorphism given by \(\phi : S \to G\) by \(\phi((g,e_H)) = g\)
• If \(x \in \mathbb{Z}_2 \times \mathbb{Z}_2\) then \(x^2 = e\), but \(1\) does not have order 2 in \(\mathbb{Z}_4\)
• \(S_2 \times S_3\) has order \(2 \times 6 = 12\). \(S_6\) has order \(6! \neq 12\)

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Solution: \(\displaystyle \int_x^{x+1} nP_{n-1}(x) \, dx = P_n(x+1) - P_n(x)\) Claim: \(P_{n}(x+1)-P_{n}(x)=nx^{n-1},\) for \(n \geq 1\) Proof: (By induction). (Base case, \(n=1\)). \(P_1(x) = x - \frac12\), \(P_1(x+1) - P_1(x) = 1 x^{0}\) as required. Assume the equation is true for \(n = k\). So \(P_k(x+1) - P_k(x) = kx^{k-1}\) now consider \begin{align*} P_{k+1}(x+1) - P_{k+1}(x) &= \int_0^{x+1} (k+1) P_k(t) \d t + P_{k+1}(0)- \int_0^{x} (k+1) P_k(t) \d t - P_{k+1}(0) \\ &= \int_0^x (k+1)(P_k(t+1)-P_k(t)) \d t + \int_0^1 (k+1)P_k(t) \d t \\ &= (k+1)x^{k} + 0 \end{align*} So by induction we are done. \begin{align*} n\sum_{m=0}^{k}m^{n-1} &= \sum_{m=0}^{k}n \cdot m^{n-1} \\ &= \sum_{m=0}^{k}\l P_n(m+1)-P_n(m) \r \\ &= P_n(k+1) - P_n(0) \end{align*} We need to find \(P_4\) \begin{align*} P_0(x) &= 1 \\ P_1(x) &= x - \frac12 \\ P_2(x) &= x^2 -x - \int_0^1 \l x^2 - x \r \d x \\ &= x^2 - x + \frac16 \\ P_3(x) &= x^3 -\frac{3}{2}x^2 + \frac12x - \int_0^1 \l x^3 -\frac{3}{2}x^2 + \frac12x \r \d x \\ &= x^3 -\frac{3}{2}x^2 + \frac12x \\ P_4(x) &= x^4 - 2x^3 + x^2 + c \end{align*} Therefore the sum we are interested in is \(\frac14 \l P_4(1001) - P_4(0) \r = \frac14 (1001)^2 (1001-1)^2 = (1001 \cdot 500)^2 = (500500)^2\)