Problems

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Solution: \begin{align*} && \f(x) &= \frac1{(x-1)(x-2)} \\ &&&=\frac{1}{x-2} -\frac{1}{x-1} \\ \end{align*} Therefore, for \(|x| < 1\) \begin{align*} && \f(x) &=\frac{1}{x-2} -\frac{1}{x-1} \\ &&&= -\frac12 \frac{1}{1-\frac{x}{2}} + \frac{1}{1-x} \\ &&&= \sum_{n=0}^{\infty} x^n - \frac12 \sum_{n=0}^\infty \l \frac{x}2 \r^2 \end{align*} where both geometric series converge since \(|x| < 1\) and \(|\frac{x}{2}| < 1\) When \(1 < |x|< 2 \Rightarrow |\frac{1}{x}| < 1\), we must have: \begin{align*} && \f(x) &=\frac{1}{x-2} -\frac{1}{x-1} \\ &&&= -\frac12 \frac{1}{1-\frac{x}{2}} + \frac1{x}\frac{1}{1-\frac{1}{x}} \\ &&&= - \frac12 \sum_{n=0}^\infty \l \frac{x}2 \r^2 - \frac{1}{x} \sum_{n=0}^{\infty} x^{-n} \\ &&&= - \frac12 \sum_{n=0}^\infty \l \frac{x}2 \r^2 - \sum_{n=1}^{\infty} x^{-n} \\ \end{align*} Finally, when \(|x| > 2\), ie \(|\frac{2}{x}| < 1\) we have \begin{align*} && \f(x) &=\frac{1}{x-2} -\frac{1}{x-1} \\ &&& =\frac1{x} \frac{1}{1-\frac{2}{x}} - \frac{1}{x}\frac{1}{1-\frac{1}{x}} \\ &&&= \frac1{x} \sum_{n=0}^{\infty} \l \frac{2}{x} \r^n - \sum_{n=1}^{\infty}x^{-n} \\ &&&= \sum_{n=1}^{\infty} 2^{n-1} x^{-n} - \sum_{n=1}^{\infty}x^{-n} \\ \end{align*}

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Solution: \begin{align*} && \begin{cases} 2a-3y=\frac{\left(z-x\right)^{2}}{y} \\ 2a-3z=\frac{\left(x-y\right)^{2}}{z} \end{cases} \\ \Rightarrow && \begin{cases} 2ay-3y^2=\left(z-x\right)^{2} \\ 2az-3z^2=\left(x-y\right)^{2} \end{cases} \\ \Rightarrow && 2a(y-z)-3(y+z)(y-z) &= (z-x+x-y)(z-x-x+y) \\ \Rightarrow && (y-z)(2a-3y-3z) &= (z-y)(z-2x+y) \\ \Rightarrow && 2a-3y-3z &= 2x-y-z \tag{\(y \neq z\)} \\ \Rightarrow && a &= x+y+z \\ \end{align*} This is is our first result. \begin{align*} && 2a-3y-3z &= 2x-y-z \\ \Rightarrow && 2a-3y-3x &= 3z-y-x \\ \Rightarrow && (y-x)2a-3(y-x)(y+x) &= (y-x)(2z-x-y) \\ \Rightarrow && 2a(y-x)-3(y^2-x^2) &= (z-y)^2-(x-z)^2 \\ \Rightarrow && 2ax - 3x^2 &= (y-z)^2 \\ \Rightarrow && 2a - 3x &= \frac{(y-z)^2}{x} \end{align*} Suppose \(x = \frac23 a, y = z = \frac16 a\) then all equations are satisfied, but \(y = z\).

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Solution: Suppose \(k\) is a root of our quadratic. There are two possibilities, if \(k^{-1} = k\) then we must have \(k^2 = 1\) so either \(\pm 1\) is a root or we must have \((x-k)(x-k^{-1}) = x^2+bx+c\). In the first case we can have: \begin{align*} x^2+bx +c = (x-1)^2 &\Rightarrow b = -2, c = 1 \\ x^2+bx +c = (x+1)^2 &\Rightarrow b = 2, c = 1 \\ x^2+bx +c = (x-1)(x+1) &\Rightarrow b = 0, c = 1 \\ \end{align*} In the other cases, \(c = 1\) and \(b = k^{-1}+k\). Therefore we must have \(c = 1\) and \(b\) can take any values. The statement "if \(k\) is a root then \(1-k\) is a root" implies these two roots are different, so we must have \(1-k = k^{-1} \Rightarrow k-k^2 = 1 \Rightarrow k^2-k+1 = 0\) so \(b = -1, c = 1\). Suppose \(x^3+px^2+qx+r = 0\) has the first property, then for any root \(k\) we must have: \(k^3 + pk^2 + qk + r = 0\) and \(1 + pk^{-1} + qk^{-2} + rk^{-3} = 0\) therefore \(x^3+px^2+qx+r\) and \(rx^3+qx^2+px+1 = 0\) must have identical roots (since \(x = \pm\) either wont work here since they imply having the roots \(1, 0\) or \(-1, 2, \frac12\) which is exactly our equation. Therefore \(r = 1, p = q\). Suppose \(x^3 + px^2 + px+1 = 0\) has the property that if \(k\) is a root \(1-k\) is a root, therefore: \begin{align*} 0 &= (1-k)^3+p(1-k)^2 + p(1-k) + 1 \\ &= 1 -3k+3k^2-k^3+p-2kp+pk^2+p-pk+1 \\ &= -k^3+(3+p)k^2+(-3-3p)k+(2+2p) \end{align*} Since these roots must be the same as the original roots, we must have \(3+p = -p, -3-3p = -p, 2+2p = -1 \Rightarrow p = -\frac32\)

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Solution:

1. Suppose \(w = u+ vi\) then \(w^2 - 2w = u^2-v^2-2u+(2uv-2v)i\) so to be purely real we must have \(2uv-2v = 2v(u-1) = 0\) ie either \(v = 0\) or \(u = 1\). Therefore the locus is the real axis and the line \(1 + ti\):
   

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2. If \(w^2 = x+yi\) then we must have \(x = u^2-v^2\) and \(y = 2uv\), so either \(v = 0, y = 0, x = u^2-2u \geq -1\) or \(u = 1, x = 1-v^2, y = 2v\) which is a parabola:
   

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If \(t = u+iv\) then \(t^2-2t = u^2-v^2-2u + (2uv-2v)i\). For this to be purely imaginary, we need \(u^2-v^2 - 2u = 0 \Rightarrow (u-1)^2-v^2 = 1\), ie points on a hyperbola. Then \(p = u^2-v^2\) and \(q = 2uv\). We can parameterise our hyperbola as \(u = 1 \pm \cosh s, v = \sinh s\) and so \(p = 1 + 2 \cosh s\) and \(q = \sinh 2s\) or \(q = \pm (p-1) \sqrt{(\frac{p-1}{2})^2-1}\) where \(p \geq 3\)

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Solution: Suppose \(z^7 = 1\), then we can write \(z = \cos \theta + i \sin \theta\) and we must have that: \begin{align*} 0 &= \textrm{Im}((\cos \theta + i \sin \theta)^7) \\ &= \binom{7}{6}\cos^6 \theta \sin \theta - \binom{7}{4} \cos^4 \theta \sin^3 \theta + \binom{7}{2} \cos^2 \theta \sin^5 \theta - \sin^7 \theta \\ &= 7 \cos^6 \sin \theta - 35 \cos^4 \theta \sin ^3 \theta + 21 \cos^2 \theta \sin^5 \theta - \sin^7 \theta \\ &= -\cos^7 \theta \l \tan^7 \theta - 21 \tan^5 \theta + 35 \tan^3 \theta - 7 \tan \theta\r \\ &= \cos^7 \theta \cdot t (t^7-21t^4+35t^2-7) \end{align*} Where \(t = \tan \theta\). So if \(z\) is a root of \(z^7 = 1\) and \(\cos \theta \neq 0, \tan \theta \neq 0\) then \(t\) is a root of the equation. Thererefore the roots are: \(\tan \frac{2\pi k}{7}\) where \(k = 1, 2, \ldots 6\). Noting that \(\tan \frac{\pi}7 = -\tan \frac{6\pi}{7}, \tan \frac{3\pi}{7} = -\tan \frac{4 \pi}{7}, \tan \frac{5\pi}{7} = -\tan \frac{2 \pi}{7}\) we can conclude that: \begin{align*} && 7 &= \prod_{k=1}^k \tan \frac{k \pi}{6} \\ &&&= \l \tan\frac{2\pi}{7}\tan\frac{4\pi}{7}\tan\frac{6\pi}{7} \r^2 \\ \Rightarrow&& \pm \sqrt{7} &= \tan\frac{2\pi}{7}\tan\frac{4\pi}{7}\tan\frac{6\pi}{7} \end{align*} However, we know that \(\tan \frac{2\pi}{7}\) is positive, \(\tan \frac{4\pi}{7},\tan \frac{6\pi}{7}\) are negative, therefore the result must be positive, ie \(+\sqrt{7}\) Using a similar method, we notice that: \begin{align*} 0 &= \textrm{Im} \l (\cos \theta + i \sin \theta)^n \r \\ &= \cos^n \theta \cdot t (t^{n-1} + \cdots - \binom{n}{n-1}) \end{align*} Therefore \(\prod_{k=0}^{n-1} \tan \frac{k \pi}{n} = n\) and since \(\tan \frac{(2k+1) \pi}{n} = \tan \frac{(n-2k-1)\pi}{n}\) is a map of all the odd numbers to the even numbers (and vice versa) when \(n\) is odd. We also know that the terms less where \(\tan \theta\) has \(\theta < \frac{\pi}{2}\) are positive, and the others even, we can determine the signs: \begin{align*} \tan \frac{2 \pi}{9} \tan \frac{4 \pi}{9} \tan \frac{6 \pi}{9} \tan \frac{8 \pi}{9} & = 3 \\ \tan \frac{2 \pi}{11} \tan \frac{4 \pi}{11} \tan \frac{6 \pi}{11} \tan \frac{8 \pi}{11} \tan \frac{10 \pi}{11} &= -\sqrt{11} \end{align*}

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Solution:

1. Notice that \(f(x) = x - \tanh x\) has \(f'(x) = 1-\textrm{sech}^2 x = \tanh^2 x > 0\) so \(f(x)\) is strictly increasing on \((0, \infty)\) and \(f(0) = 0\) therefore \(f(x)\) is positive for all \(x\) positive
2. Let \(f(x) = x\sinh x-2\cosh x+2\) then \(f'(x) = \sinh x +x \cosh x - 2 \sinh x = x \cosh x -\sinh x = \cosh x ( x - \tanh x) > 0\) by the first part. \(f(0) = 0\) so \(f(x)\) is positive for all \(x\) positive.
3. Let \(f(x) = 2x\cosh2x-3\sinh2x+4x\) then \begin{align*} f'(x) &= 2\cosh 2x +4x\sinh 2x - 6 \cosh 2x + 4 \\ &= 4( x\sinh 2x-\cosh 2x +1) \\ &= 4(x2\cosh x \sinh x -2\cosh^2x ) \\ &= 8 \cosh^2 x (x - \tanh x) \end{align*} Which is always positive when \(x\) > 0, \(f(0) = 0\) so \(f(x) > 0\) for all positive \(x\).

Let \(f(x) = \frac{x(\cosh x)^{\frac{1}{3}}}{\sinh x}\) then \begin{align*} f'(x) &= \frac{(\cosh x)^{\frac13}\sinh x+\frac13 x \cosh^{-\frac23} x \sinh^2 x - x(\cosh x)^{\frac13} \cosh x}{\sinh^2 x} \\ &= \frac{\cosh x \sinh x + \frac13 x \sinh^2 x - x \cosh^2 x}{\cosh x^{\frac23} x \sinh^2 x} \\ &= \frac{3\cosh x \sinh x + x( \sinh^2 x - 3 \cosh^2 x)}{3\cosh x^{\frac23} x \sinh^2 x} \\ &= \frac{\frac32 \sinh 2x + x( -2\cosh 2x - 2)}{3\cosh x^{\frac23} x \sinh^2 x} \\ &= \frac{3 \sinh 2x -4x\cosh 2x - 4x}{6\cosh x^{\frac23} x \sinh^2 x} \\ \end{align*} which from the earlier part is always negative.

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Solution: \begin{align*} I &=\int_{1}^{2}\frac{(2-2x+x^{2})^{k}}{x^{k+1}}\,\mathrm{d}x \\ u = x-1 &, \quad \d u = \d x \\ &= \int_{u = 0}^{u=1} \frac{(u^2+1)^k}{(u+1)^{k+1}} \d u \\ &= \boxed{\int_0^1 \frac{(1+x^2)^k}{(1+x)^{k+1}} \d x} \\ x = \tan \theta &, \quad \d x = \sec^2 \theta \d \theta \\ &= \int_{\theta = 0}^{\theta = \pi/4} \frac{\sec^{2k+2} \theta }{(1 + \tan \theta)^{k+1}} \d \theta \\ &= \int_0^{\pi/4} \frac{\d \theta}{\cos^{2k+2} \theta (\frac{\sin \theta + \cos \theta}{\cos \theta})^{k+1}} \\ &= \int_0^{\pi/4} \frac{\d \theta}{\cos^{k+1} \theta ({\sin \theta + \cos \theta})^{k+1}} \\ &= \int_0^{\pi/4} \frac{\d \theta}{\cos^{k+1} \theta (\sqrt{2} \cos (\frac{\pi}{4} - \theta))^{k+1}} \\ I &= \boxed{ \int_0^{\pi/4} \frac{\d \theta}{(\sqrt{2}\cos \theta \cos (\frac{\pi}{4} - \theta))^{k+1}}} \\ \end{align*} Since \(f(\theta) = \cos \theta \cos (\frac{\pi}{4} - \theta)\) is symmetric about \(\frac{\pi}{8}\) this integral is twice the integral to \(\frac{\pi}{8}\). \(\tan 2 \theta = \frac{2\tan \theta}{1 - \tan^2 \theta} \Rightarrow 1 = \frac{2 \tan \frac{\pi}{8}}{1 - \tan^2 \frac{\pi}{8}} \Rightarrow \tan \frac{\pi}{8} = \sqrt{2}-1\). Therefore, using the same substitution we must have: \[ I=2\int_{0}^{\sqrt{2}-1}\frac{(1+x^{2})^{k}}{(1+x)^{k+1}}\,\mathrm{d}x \] Let \(u = x^2\), then \(\d u = 2 x\d x\) \begin{align*} \int_{1}^{\sqrt{2}}\left(\frac{2-2x^{2}+x^{4}}{x^{2}}\right)^{k}\frac{1}{x}\,\mathrm{d}x &= \int_{u = 1}^{u = 2} \l \frac{2-2u+u^2}{u}\r^k \frac{1}{2u} \d u \\ &= \frac12 I \\ &= \int_{0}^{\sqrt{2}-1}\frac{(1+x^{2})^{k}}{(1+x)^{k+1}}\,\mathrm{d}x \\ u = 1+x & \quad \d u = \d x \\ &= \int_1^{\sqrt{2}} \frac{(1+(u-1)^2)^k}{u^{k+1}} \d u \\ &= \int_{1}^{\sqrt{2}}\left(\frac{2-2u+u^{2}}{u}\right)^{k}\frac{1}{u}\,\mathrm{d}x \\ &= \int_{1}^{\sqrt{2}}\left(\frac{2-2x+x^{2}}{x}\right)^{k}\frac{1}{x}\,\mathrm{d}x \end{align*}

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Solution: If the population in a given year is \(\mathbf{v} = \begin{pmatrix}A \\ B \end{pmatrix}\) then the population the next year is \(\mathbf{Mv}\) where \(\mathbf{M} = \begin{pmatrix} \frac14 & \frac34 \\ -\frac12 &\frac32 \end{pmatrix}\) The ratio is the same if \(\mathbf{Mv} = \lambda \mathbf{v}\) ie if \(\mathbf{v}\) is an eigenvector of \(\mathbf{M}\). The eigenvalues will be \(1\) and \(\frac38\) (by inspection) so we should be able to solve for the eigenvectors: \(\lambda = 1\) we have \(\frac14A + \frac34B = A \Rightarrow A = B\) a ratio of \(1\). \(\lambda = \frac38\) we have \(\frac14A + \frac34B = \frac38A \Rightarrow \frac34B = \frac18A \Rightarrow A = 6B\) a ratio of \(6\). If we write \(\begin{pmatrix} a \\ b \end{pmatrix}\) as \(x_1 \begin{pmatrix} 1 \\ 1 \end{pmatrix} + x_6 \begin{pmatrix} 6 \\ 1 \end{pmatrix}\) we find that after \(n\) years, we have: \(x_1 \begin{pmatrix} 1 \\ 1 \end{pmatrix} + \l \frac38 \r^n x_6 \begin{pmatrix} 6 \\ 1 \end{pmatrix}\) for the populations. Therefore if \(x_1\) is \(< 0\) then in finite time we will end up with one population being 0. If \(x_1 > 0\) are positive we tend to a finite population and if \(x_1 = 0\) then over time the population will tend to \(0\) at infinity. In our diagram these areas correspond to (red) - die out in finite time, (green) population stable and the thick black line where the population goes extinct as \(t \to \infty\)

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Solution: Suppose \(|G_1|, |G_2| < |G|\) for sake of contraction. Then by Lagrange's theorem \(|G_1| \mid |G|\) and \(|G_2| \mid |G|\), so \(|G_1|, |G_2| \leq \frac{|G|}{2}\). But \(|G_1 \cup G_2| = |G_1| + |G_2| - |G_1 \cap G_2|\). \(|G_1 \cup G_2| = |G|\) by assumption, and \(e \in G_1 \cap G_2\), so \(|G_1 \cap G_2| \geq 1\). Therefore \(|G| = |G_1| + |G_2| - |G_1 \cap G_2| \leq \frac{|G|}{2} + \frac{|G|}{2} - 1 = |G| - 1 < |G|\), contradiction! Let \(H = K_4 = \{e, a, b, c\}\) with \(a^2 = b^2 = c^2 = e\). then \(H = \{e, a\} \cup \{e, b\} \cup \{e, c\}\) with all subgroups distinct

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Solution: The plane is the same as the plane \((\mathbf{r} - \mathbf{a}) \cdot \mathbf{a} = 0\), ie the plane through \(\mathbf{a}\) whose normal is parallel to \(\mathbf{a}\) The distance from \(\mathbf{r}\) to the plane therefore is \(\lambda\) where \(\mathbf{r}+\lambda \frac{1}{a}\mathbf{a}\) must be on the plane, ie \((\mathbf{r}+\frac{\lambda}{a} \mathbf{a} - \mathbf{a})\cdot \mathbf{a} = 0 \Rightarrow \lambda = \frac{a^2-\mathbf{a} \cdot \mathbf{r}}{a}\) But if \(\mathbf{a} \cdot \mathbf{r} = a^2 - ar\) then \(\lambda = r\), ie the distance to the plane is the same as the distance to the origin. \(\mathbf{q} = k \mathbf{p}\) and so \(\mathbf{a} \cdot k \mathbf{p} + a |k|p = a^2\) if \(k > 0\) we will find \(k = 1\) the position vector we already know about, therefore suppose \(k < 0\) so: \begin{align*} && \mathbf{a} \cdot k \mathbf{p} - ka p &= a^2 \\ \Rightarrow && k(a^2-ap)-kap &= a^2 \\ \Rightarrow && k(a^2-2ap) &= a^2 \\ \Rightarrow && k &= \frac{a^2}{a^2-2ap} \end{align*} Therefore \(\mathbf{q} = \frac{a^2}{a^2-2ap} \mathbf{p}\) The line through \(O\) orthogonal to \(\mathbf{p}\) and \(\mathbf{a}\) will be parallel to \(\mathbf{a} \times \mathbf{p}\). Therefore we should consider points of the from \(s \mathbf{a} \times \mathbf{p}\) on the surface \(S\). \begin{align*} && s\mathbf{a} \cdot ( \mathbf{a} \times \mathbf{p}) + sa^2p |\sin \theta| &= a^2 \end{align*} The angle between \(\cos \theta = \frac{\mathbf{a} \cdot \mathbf{p}}{ap} = \frac{a^2-ap}{ap} \Rightarrow |\sin \theta| = \sqrt{1-\frac{(a-p)^2}{p^2}} = \frac{1}{p} \sqrt{2ap-a^2}\) Therefore \(sa^2 \sqrt{2ap-a^2} = a^2 \Rightarrow s = \frac{1}{\sqrt{2ap-a^2}}\) and so the points are as required. Noting that \(|\mathbf{p} \times \mathbf{t}| = |\frac{1}{p \sin \theta}\mathbf{p} \times (\mathbf{p} \times \mathbf{a}) | = |\frac{1}{p \sin \theta}p^2a \sin \theta | = pa\) The area of triangle \(PQT\) is : \begin{align*} \frac12 | (\mathbf{p} - \mathbf{t}) \times (\mathbf{q} - \mathbf{t}) | &= \frac12 |\mathbf{p} \times \mathbf{q} - \mathbf{t} \times \mathbf{q} - \mathbf{p} \times \mathbf{t} - \mathbf{t} \times \mathbf{t}| \\ &= \frac12 |\mathbf{t} \times (\mathbf{p} - \mathbf{q})| \\ &= \frac12 \cdot (1 - \frac{a^2}{a^2-2ap})| \mathbf{t} \times \mathbf{p}| \\ &= \frac12 \frac{2ap}{a^2-2ap} \cdot ap \\ &= \frac{ap^2}{a^2-ap} \end{align*}