Problems

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Solution: \begin{align*} && a^{2}x+ay+z & =a^{2} \tag{1}\\ && ax+y+bz & =1 \tag{2}\\ && a^{2}bx+y+bz & =b \tag{3} \\ \\ (1) - a(2): && (1-ba)z &= a^2-a \\ \Rightarrow && z &= \frac{a^2-a}{1-ab} \tag{if \(ab \neq 1\)} \\ \\ (2) - (3): && (a-a^2b)x &= b - 1 \\ \Rightarrow && x &= \frac{b-1}{a(1-ab)} \tag{if \(a \neq 0, ab \neq 1\)} \\ \\ b(1) - (3): && (ab-1)y &= a^2 - b^2 \\ \Rightarrow && y &= \frac{a^2-b^2}{ab-1} \end{align*} Let's consider the cases where \(a = 0\), then \begin{align*} && z &= 0 \\ && y + bz &= 1 \\ && y+bz &= b \\ \Rightarrow && y &= 1 = b \end{align*} So if \(a = 0\) then \(b = 1\) and \(x \in \mathbb{R}, y = 1, z = 0\). If \(a \neq 0, ab = 1\), then \begin{align*} && a^2 x + ay + z &= a^2 \\ && ax + y + \frac1{a}z &= a \\ && ax + y + \frac{1}{a}z &= b \\ \end{align*} The last two equations imply \(a = b = \pm 1\). \(a = 1 \Rightarrow x+y+z = 1\), so we have a lot of solutions. \(a = -1 \Rightarrow x -y +z = 1\) so again, lots of solutions. Conclusion: If \(ab \neq 1, a \neq 0\), we have: \[ (x,y,z) = \left (\frac{b-1}{a(1-ab)}, \frac{a^2-b^2}{ab-1}, \frac{a^2-a}{1-ab} \right)\] If \(a = 0\) then \(b = 1\) and we have: \((x,y,z) = (t, 1, 0)\). If \(ab = 1\) then \(a = 1\) or \(a = -1\). If \(a = 1\) then \((x,y,z) = (t, s, 1-t-s)\) If \(a = -1\) then \((x,y,z) = (t,s,1-t+s)\)

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Solution: \begin{align*} && I_n &= \int_{0}^{a}x^{n+\frac{1}{2}}(a-x)^{\frac{1}{2}}\,\mathrm{d}x\\ && I_0 &= \int_0^a x^{\frac12}(a-x)^{\frac12} \d x \\ x = a \sin^2 \theta, \d x = 2a \sin \theta \cos \theta \d \theta &&&= \int_{\theta =0}^{\theta = \pi/2} \sqrt{a}\sin \theta\sqrt{a} \cos \theta 2a \sin \theta \cos \theta \d \theta \\ &&&= \frac{a^2}{2} \int_0^{\pi/2} \sin^2 2 \theta \d \theta \\ &&&= \frac{a^2}{4} \int_0^{\pi/2}(1- \underbrace{\cos 4\theta}_{\text{runs round the whole unit circle}}) \d \theta \\ &&&= \frac{\pi a^2}{8} \\ \\ && I_n &= \int_0^a x^{n+\frac12}(a-x)^{\frac12} \d x \\ &&&=\underbrace{\left [-\frac23x^{n+\frac12}(a-x)^\frac32 \right]_0^a}_{=0} + \frac23 \left(n+\frac12\right) \int_0^ax^{n-1+\frac12}(a-x)^\frac32 \d x \\ &&&= \frac23 \left(n+\frac12\right) \int_0^ax^{n-1+\frac12}(a-x)(a-x)^\frac12 \d x \\ &&&= \frac23 \left(n+\frac12\right)aI_{n-1}-\frac23 \left(n+\frac12\right)I_{n} \\ \Rightarrow && \left(n+\frac12+\frac32\right)I_{n} &= \left(n+\frac12\right)aI_{n-1}\\ \Rightarrow && (2n+4)I_n &= (2n+1)aI_{n-1} \\ \\ \Rightarrow && I_n &= \frac{2n+1}{2n+4}a I_{n-1} \\ &&&=\frac{2n+1}{2n+4}\frac{2n-1}{2n+2}a^2 I_{n-2} \\ &&&= \frac{(2n+1)!!}{(2n+4)!!} \pi a^{n+2} \end{align*}

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Solution: The auxiliary equation is \(\lambda^2 + 2k\lambda + 1 = (\lambda + k)^2+1-k^2 = 0\) (i) If \(k > 1\) then the solution is \(A\exp \left ({(-k + \sqrt{k^2-1})t} \right)+B\exp\left((-k-\sqrt{k^2-1})t \right)\). (ii) If \(k = 1\) then the solution is \(x = (A+Bt)e^{-kt}\) (iii) If \(k < 1\) then the solution is \(x = Ae^{-kt} \sin \left ( \sqrt{1-k^2} t \right)+Be^{-kt} \cos \left ( \sqrt{1-k^2} t \right)\) If \(x(0) = 0\) then \begin{align*} && x &= Ae^{-kt} \sin(\sqrt{1-k^2}t)\\ && \dot{x} &= Ae^{-kt} \left (-k \sin(\sqrt{1-k^2}t)+\sqrt{1-k^2} \cos(\sqrt{1-k^2}t) \right) \\ (\dot{x} =0): && \tan (\sqrt{1-k^2}t) &= \frac{\sqrt{1-k^2}}{k}\\ \end{align*} Therefore maxima and minima occur every \(\frac{\pi}{\sqrt{1-k^2}}\), so \begin{align*} && \frac{x_{n+1}}{x_n} &= \exp\left ( -\frac{k\pi}{\sqrt{1-k^2}} \right) \frac{\sin\left (\sqrt{1-k^2}\left(t+\frac{\pi}{\sqrt{1-k^2}}\right)\right)}{\sin(\sqrt{1-k^2}t)} \\ &&&= \exp\left ( -\frac{k\pi}{\sqrt{1-k^2}} \right) \left (-1+0 \right)\\ &&&= -\exp\left ( -\frac{k\pi}{\sqrt{1-k^2}} \right) \\ \Rightarrow && \ln \alpha &= - \frac{k\pi}{\sqrt{1-k^2}} \\ \Rightarrow && (\ln \alpha)^2 &= \frac{k^2\pi^2}{1-k^2} \\ \Rightarrow && (1-k^2)(\ln \alpha)^2 &= k^2 \pi^2 \\ \Rightarrow && k^2(\pi^2+(\ln \alpha)^2) &= (\ln \alpha)^2 \\ \Rightarrow && k^2 &= \frac{(\ln \alpha)^2}{\pi^2 + (\ln \alpha)^2} \end{align*}

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Solution: \begin{align*} && C_n(\theta) &= \sum_{k=0}^n \cos k \theta \\ &&&= \textrm{Re} \left ( \sum_{k=0}^n \exp (ik \theta)\right)\\ &&&= \textrm{Re} \left ( \frac{e^{i(n+1)\theta}-1}{e^{i\theta}-1}\right)\\ &&&= \textrm{Re} \left ( \frac{e^{i(n+1)\theta/2}}{e^{i\theta/2}}\frac{e^{i(n+1)\theta/2}-e^{-i(n+1)\theta/2}}{e^{i\theta/2}-e^{-i\theta/2}}\right)\\ &&&= \textrm{Re} \left ( e^{in\theta/2}\frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\right)\\ &&&= \frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\textrm{Re} \left ( e^{in\theta/2}\right)\\ &&&= \frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\cos \left ( \frac12n\theta\right)\\ \\ && S_n(\theta) &= \sum_{k=0}^n \sin k \theta \\ &&&= \textrm{Im} \left ( \sum_{k=0}^n \exp (ik \theta)\right)\\ &&&= \frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\textrm{Im} \left ( e^{in\theta/2}\right)\\ &&&= \frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\sin\left ( \frac12n\theta\right)\\ \\ && C_n(\theta) - \frac12 &= \frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\cos \left ( \frac12n\theta\right) - \frac12 \\ &&&= \frac{2\sin \left ( (n+1)\theta/2 \right)\cos\left ( n\theta/2 \right)-\sin (\theta/2)}{2 \sin (\theta/2)} \\ &&&= \frac{\sin\left ( (n+1)\theta/2+n\theta/2\right)+\sin\left ( (n+1)\theta/2-n\theta/2\right)-\sin (\theta/2)}{2 \sin (\theta/2)} \\ &&&= \frac{\sin\left ( (n+1)\theta/2+n\theta/2\right)+\sin\left ( \theta/2\right)-\sin (\theta/2)}{2 \sin (\theta/2)} \\ &&&= \frac{\sin\left ( (2n+1)\theta/2\right)}{2 \sin (\theta/2)} \leqslant\frac{1}{2 \sin (\theta/2)} \\ \end{align*}

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Solution: \begin{align*} && y &= \sin^2 (m \sin^{-1} x) \\ \Rightarrow && y' &= 2 \sin (m \sin^{-1} x) \cdot \cos (m \sin^{-1} x) \cdot m \cdot \frac1{\sqrt{1-x^2}} \\ \Rightarrow && y'' &= 2 \cos^2(m \sin^{-1} x) \cdot m^2 \cdot \frac{1}{1-x^2} + \\ &&&\quad\quad-2\sin^2(m \sin^{-1} x) m^2 \frac{1}{1-x^2} + \\ &&&\quad\quad\quad-\sin(m \sin^{-1} x) \cdot \cos(m \sin^{-1} x) \cdot m \cdot (1-x^2)^{-\frac32} \cdot (-2x) \\ \Rightarrow && (1-x^2)y^{(2)} &= 2m^2-4m^2y+xy' \\ &&&= xy^{(1)} + 2m^2(1-2y) \\ \\ \Rightarrow && (1-x^2)y^{(n+2)}-2nxy^{(n+1)}-2\binom{n}{2}y^{(n)} &= xy^{(n+1)}+ny^{(n)} -4m^2y^{(n)} \\ \Rightarrow && (1-x^2)y^{(n+2)} &= (2n+1)xy^{(n+1)}+(n(n-1)+n-4m^2)y^{(n)} \\ &&&= (2n+1)xy^{(n+1)}+(n^2-4m^2)y^{(n)} \\ \end{align*} \begin{align*} && y(0) &= \sin^2(m \sin^{-1} 0) \\ &&&= \sin^2 0 = 0 \\ \\ && y'(0) &= 0 \\ && (1-0^2)y^{(2)}(0) &= 2m^2(1-2y(0)) \\ \Rightarrow && y^{(2)}(0) &= 2m^2 \\ \\ && y^{(n+2)} (0) &= (2n+1) \cdot 0 \cdot y^{(n+1)} +(n^2-4m^2)y^{(n)}(0) \\ &&&= (n^2-4m^2)y^{(n)}(0) \\ \\ && y^{(2)}(0) &= 2m^2 \\ && y^{(4)}(0) &= (4-4m^2) \cdot 2m^2 \\ &&&= -8m(m+1)m(m-1) \\ && y^{(6)}(0) &= 32m(m+2)(m+1)m(m-1)(m-2) \\ && y^{(2k)}(0) &= (-1)^{k+1}2^{2k-1}m (m+k)\cdots(m-k) \text{ if }k < m \\ \\ && y &= m^2x^2 -2m\binom{m+1}{3} x^4 + \frac{16}{3}m\binom{m+2}{5}x^6 - \cdots \\ &&&+ (-1)^{k}\frac{2^{2k}}{k+1} m \binom{m+k}{2k+1}x^{2k+2}+\cdots \\ &&&= mx^2\sum_{k=0}^{m-1} \frac{(-1)^k2^{2k}}{k+1}\binom{m+k}{2k+1}x^{2k} \end{align*}

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Solution:

\(z\) is a point on the circle shown: Therefore using the cosine rule \(|z|^2 = 1^2 + 1^2 - 2\cdot 1 \cdot 1 \cdot \cos (2 \theta) = 2 -2\cos 2\theta = 2\sin^2 \theta \Rightarrow |z| = \sqrt{2}|\sin \theta|\) \(\frac{1}{z}\) has modulus \(\frac{1}{\sqrt{2}|\sin \theta|}\) and argument \(-\theta\). \(|\frac{1}{z} - i| = 1 \Rightarrow |1-iz| = |z| \Rightarrow |-i-z| = |z|\) ie we're looking for the points on the perpendicular bisector of \(0\) and \(-i\). \(\textrm{Re}\left (\frac{1}{w}\right) = -1 \Rightarrow -1 = \textrm{Re} \left (\frac{1}{a+ib} \right) = \frac{a-ib}{a^2+b^2} = \frac{a}{a^2+b^2} \Rightarrow a^2+b^2 = -a \Rightarrow (a+\tfrac12)^2+b^2 = \tfrac14\) so we are looking at a circle radius \(\tfrac12\) centre \(-\frac12\)

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Solution:

1. \(\{ z \in \mathbb{C} : |z| = 1\}\) is a group.
   1. (Closure) \(|z_1z_2| = |z_1||z_2| = 1\). Set is closed under multiplication
   2. (Associativity) Multiplication of complex numbers is associative
   3. (Identity) \(|1| = 1\)
   4. (Inverses) \(| \frac{1}{z} | = \frac{1}{|z|} = \frac{1}{1} = 1\), the set contains inverses
2. the integers \(\pmod{4}\) are not a group under multiplication, \(2\) has no inverse, since \(0 \times k \equiv 0 \pmod{4}\)
3. The set of rotation matrices is a group:
   1. (Closure) \begin{align*} \begin{pmatrix}\cos\theta_1 & -\sin\theta_1\\ \sin\theta_1 & \cos\theta_1 \end{pmatrix} \begin{pmatrix}\cos\theta_2 & -\sin\theta_2\\ \sin\theta_2 & \cos\theta_2 \end{pmatrix} &= {\scriptscriptstyle\begin{pmatrix}\cos\theta_1 \cos \theta_2 - \sin \theta_1\sin \theta_2 & -\sin\theta_1\ \cos \theta_1 - \sin \theta_2\cos\theta_1\\ \sin\theta_1\ \cos \theta_1 + \sin \theta_2\cos\theta_1 & \cos\theta_1 \cos \theta_2 - \sin \theta_1\sin \theta_2 \end{pmatrix}} \\ &= \begin{pmatrix}\cos(\theta_1+\theta_2) & -\sin(\theta_1+\theta_2)\\ \sin(\theta_1+\theta_2) & \cos(\theta_1+\theta_2) \end{pmatrix} \end{align*} Since \(\cos, \sin\) are periodic with period \(2\pi\), we can find \(\theta_3 = \theta_1 + \theta_2 + 2k\pi\) such that \(0 \leq \theta_3 < 2 \pi\), so our set is closed
   2. (Associativity) Matrix multiplication is associative
   3. (Identity) Consider \(\theta = 0\)
   4. (Inverses) Consider \(2\pi - \theta\)
4. \(\{1, 3, 5, 7\} \pmod{8}\) is a group: \begin{tabular}{c|cccc} & 1 & 3 & 5 & 7 \\ \hline 1 & 1 & 3 & 5 & 7 \\ 3 & 3 & 1 & 7 & 5 \\ 5 & 5 & 7 & 1 & 3 \\ 7 & 7 & 5 & 3 & 1 \\ \end{tabular}
   1. (Closure) See Cayley table
   2. (Associativity) Integer multiplication is associative
   3. (Identity) \(1\)
   4. (Inverses) \(x \mapsto x\) (See Cayley table)
5. \(2\times2\) matrices are not a group, consider $0 = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\(, then \)\mathbf{0}\mathbf{M} = \mathbf{0}$ for all other matrices.
6. \begin{tabular}{c|cccc} & 1 & 2 & 3 & 4 \\ \hline 1 & 1 & 2 & 3 & 4 \\ 2 & 2 & 4 & 1 & 3 \\ 3 & 3 & 1 & 4 & 2 \\ 4 & 4 & 3 & 2 & 1 \\ \end{tabular}
   1. (Closure) See Cayley table
   2. (Associativity) Integer multiplication is associative
   3. (Identity) \(1\)
   4. (Inverses) \(1 \mapsto 1, 2 \mapsto 3, 3 \mapsto 2, 4 \mapsto 4\) (See Cayley table)

\begin{tabular}{c|c|c|c|c} & (i) & (iii) & (iv) & (vi) \\ \hline (i) & \(\checkmark\) & \(\checkmark\) consider \(z \mapsto \begin{pmatrix} \cos \arg (z) & - \sin \arg(z) \\ \sin \arg(z) & \cos \arg(z) \end{pmatrix}\) & not finite & not finite \\ (iii) & & \(\checkmark\) & not finite & not finite \\ (iv) & & & \(\checkmark\) & no element order \(4\) \\ (vi) & & & & \(\checkmark\)\\ \end{tabular}

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Solution: The equation of the normal to \(\pi\) through \(X\) is \(\mathbf{x} + \lambda \mathbf{n}\). The distance is \(|\mathbf{x}\cdot \mathbf{n}-p|\) We know that the centre of the sphere must lie above the centre of the circle following the normal, ie \(\mathbf{c} = \mathbf{r}_1+\lambda_1 \mathbf{n}_1 = \mathbf{r}_2+\lambda_2 \mathbf{n}_2\)

We can also see that \(R^2 = 1 + \lambda_1^2 = 1 + \lambda_2^2 \Rightarrow \lambda_1 = \pm \lambda_2 \), from which we obtain the desired result. Therefore the condition is \begin{align*} && \mathbf{r}_1+\lambda \mathbf{n}_1 &= \mathbf{r}_2 \pm \lambda \mathbf{n}_2 \tag{1}\\ && \mathbf{r}_1 - \mathbf{r}_2 &= \lambda(\pm \mathbf{n}_1 - \mathbf{n}_2) \\ \Rightarrow && (\mathbf{r}_{1}-\mathbf{r}_{2})\cdot(\mathbf{n}_{1}\times\mathbf{n}_{2}) &= (\lambda(\pm \mathbf{n}_1 - \mathbf{n}_2))\cdot(\mathbf{n}_{1}\times\mathbf{n}_{2}) \\ &&&= \lambda \left (\pm \mathbf{n}_1 \cdot ( \mathbf{n}_{1}\times\mathbf{n}_{2}) - \mathbf{n}_2 \cdot (\mathbf{n}_{1}\times\mathbf{n}_{2})\right) \\ &&&= 0 \\ \\ \mathbf{n}_1 \cdot (1)&& \mathbf{r}_1 \cdot \mathbf{n}_1+\lambda \mathbf{n}_1 \cdot \mathbf{n}_1 &= \mathbf{r}_2 \cdot \mathbf{n}_1 \pm \lambda \mathbf{n}_2 \cdot \mathbf{n}_1 \\ && p_1 + \lambda &= \mathbf{r}_2 \cdot \mathbf{n}_1 \pm \lambda \mathbf{n}_2 \cdot \mathbf{n}_1 \\ \\ \mathbf{n}_2 \cdot (1)&& \mathbf{r}_1 \cdot \mathbf{n}_2+\lambda \mathbf{n}_1 \cdot \mathbf{n}_2 &= \mathbf{r}_2 \cdot \mathbf{n}_2 \pm \lambda \mathbf{n}_2 \cdot \mathbf{n}_2 \\ && \mathbf{r}_1 \cdot \mathbf{n}_2+\lambda \mathbf{n}_1 \cdot \mathbf{n}_2 &= p_2 \pm \lambda \\ && \pm \lambda -\lambda \mathbf{n}_1\cdot\mathbf{n}_2 &= \mathbf{r}_1 \cdot \mathbf{n}_2 - p_2 \\ &&&= \pm (\mathbf{r}_2\cdot \mathbf{n}_1 - p_1) \\ \Rightarrow && (p_{1}-\mathbf{n}_{1}\cdot\mathbf{r}_{2})^{2}&=(p_{2}-\mathbf{n}_{2}\cdot\mathbf{r}_{1})^{2} \end{align*} The first condition means the line between the centres lies in the plane spanned by the normal of the two planes \(\pi_1\) and \(\pi_2\). The second condition means that the distance of the center to the other plane is the same for both centres/planes.