Problems

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Solution:

1. \begin{align*} \lim_{n \to \infty} \frac{n}{n+1} &= \lim_{n \to \infty} \left (1 - \frac{1}{n+1} \right ) \\ &\underbrace{=}_{\text{sum of limits}} \lim_{n \to \infty} 1 - \lim_{n \to \infty} \frac{1}{n+1}\\ &= 1 \end{align*}
2. \begin{align*} \lim_{n \to \infty} \frac{5n+1}{n^2-3n+4} &= \lim_{n \to \infty} \frac{5/n + 1/n^2}{1-3/n+ 4/n^2} \\ &\underbrace{=}_{\text{ratio of limits}} \frac{\displaystyle \lim_{n \to \infty}(5/n + 1/n^2)}{\displaystyle \lim_{n \to \infty}(1-3/n+ 4/n^2)} \\ &= \frac{0}{1} = 0 \end{align*}
3. \begin{align*} && \lvert \frac{\sin n}{n} \rvert &\leq \frac{1}{n} \quad \quad (n \geq 1) \\ \Rightarrow && \lim_{n \to \infty} \lvert \frac{\sin n}{n} \rvert &\leq \lim_{n \to \infty}\frac{1}{n} \\ &&&= 0\\ \Rightarrow && \lim_{n \to \infty} \frac{\sin n}{n} &= 0 \end{align*}
4. First note that \(\displaystyle \lim_{x \to 0} \frac{\sin x}{x} \to 1\), then \(\frac1n\) is a sequence converging to zero, therefore \(\frac{\sin 1/n}{1/n}\) also must tend to \(1\).
5. Note that \(\lim_{x \to \infty} \tan^{-1} x = \frac{\pi}{2}\) and since \(n\) is a sequence tending to infinity we must have \(\lim_{n \to \infty} \tan^{-1} n = \frac{\pi}{2}\)
6. \begin{align*} \lim_{n \to \infty} \dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+2}-\sqrt{n}} &= \lim_{n \to \infty} \dfrac{\frac{1}{\sqrt{n+1}+\sqrt{n}}}{\frac{2}{\sqrt{n+2}+\sqrt{n}}} \\ &= \frac12 \lim_{n \to \infty} \dfrac{\sqrt{n+2}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}\\ &= \frac12 \lim_{n \to \infty} \dfrac{\sqrt{1+2/n}+\sqrt{1}}{\sqrt{1+1/n}+\sqrt{1}}\\ &= \frac12 \end{align*}

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Solution: \begin{align*} && y & =x\frac{\mathrm{d}y}{\mathrm{d}x}-\cosh\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right) \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{\d y}{\d x} + x\frac{\d ^2 y}{\d x^2} - \sinh \left ( \frac{\d y}{\d x} \right) \frac{\d^2 y}{\d x^2} \\ \Rightarrow && 0 &= \frac{\d^2 y}{\d x^2} \left ( x - \sinh \left ( \frac{\d y}{\d x}\right)\right) \end{align*} Therefore \(\frac{\d^2y}{\d x^2} = 0\) or \( x - \sinh \left ( \frac{\d y}{\d x}\right) = 0\) as required. \begin{align*} && \frac{\d ^2 y}{\d x^2} &= 0 \\ \Rightarrow && y &= ax + b \\ \\ && 0 &= x - \sinh \left ( \frac{\d y}{\d x}\right) \\ \Rightarrow && \frac{\d y}{\d x} &= \sinh^{-1} (x) \\ \Rightarrow && y &= x \sinh^{-1} x - \sqrt{x^2+1} + C \end{align*} Since it is necessary the solution satisfies one of those equations, we just need to check if either of these types of solutions work for our differential equation, ie \begin{align*} && ax + b &\stackrel{?}{=} ax - \cosh(a) \\ \Rightarrow && b &= -\cosh(a) \\ \Rightarrow && y &= ax -\cosh(a) \\ \\ && x \sinh^{-1} x - \sqrt{x^2+1} + C &\stackrel{?}{=} x\sinh^{-1} x - \cosh ( \sinh^{-1} x) \\ &&&= \sinh^{-1} x -\sqrt{x^2+1} \\ \Rightarrow && C &= 0 \end{align*} Therefore the general solutions are, \(y = ax - \cosh(a)\) and \(y = x \sinh^{-1} x - \sqrt{x^2+1}\)

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Solution: Claim: If \(f(x) = c\) then \(\lozenge f(x) = 0\) Proof: Consider \(g(x) = x\) then \begin{align*} (1) && \lozenge(f(x)g(x)) &= g(x) \lozenge f(x) + f(x) \lozenge g(x) \\ \Rightarrow && \lozenge(c x) &= x \lozenge f(x) + c \lozenge x \\ (4) && \lozenge(c x) &= c \lozenge x \\ \Rightarrow && 0 &= x \lozenge f(x) \\ \Rightarrow && \lozenge f(x) &= 0 \end{align*} \begin{align*} (1) && \lozenge(x^2) &= x \lozenge x + x \lozenge x \\ (3) &&&= 2 x \cdot 1 \\ &&&= 2x \\ \\ (1) && \lozenge (x^3) &= x^2 \lozenge x + x \lozenge (x^2) \\ &&&= x^2 \cdot \underbrace{1}_{(3)} + x \cdot\underbrace{ 2x}_{\text{previous part}} \\ &&&= 3x^2 \end{align*} Claim: \(\lozenge h(x) = \frac{\d }{\d x} ( h(x))\) for any polynomial \(h\). Proof: (By (strong) induction on the degree of \(h\)). Base case: True, we proved this in the first part of the question. Inductive step: Assume true for all polynomials of degree less than or equal to \(k\). Then consider \(n = k+1\). We can write \(h(x) = ax^{k+1} + h_k(x)\) where \(h_k(x)\) is a polynomial of degree less than or equal to \(k\). Then notice: \begin{align*} && \lozenge (h(x)) &= \lozenge (ax^{k+1} + h_k(x)) \\ (2) &&&= \lozenge (ax^{k+1})+ \lozenge (h_k(x)) \\ &&&=\underbrace{a\lozenge (x^{k+1})}_{(4)}+ \underbrace{\frac{\d}{\d x} (h_k(x))}_{\text{inductive hypothesis}}\\ &&&= a \underbrace{\left (x \lozenge x^k + x^k \lozenge x \right)}_{(1)} + \frac{\d}{\d x} (h_k(x)) \\ &&&= a \left ( x \cdot \underbrace{k x^{k-1}}_{\text{inductive hyp.}} + x^k \cdot \underbrace{1}_{(3)} \right) + \frac{\d}{\d x} (h_k(x)) \\ &&&= (k+1)a x^k + \frac{\d}{\d x} (h_k(x)) \\ &&&= \frac{\d }{\d x} \left ( ax^{k+1} + h_k(x) \right) \\ &&&= \frac{\d }{\d x} (h(x)) \end{align*} Therefore since our statement is true for \(n=0\) and if it is true for \(n=k\) it is true for \(n=k+1\) by the principle of mathematical induction it is true for all \(n \geq 0\)

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Solution: The order of an element \(g\) is the smallest positive number \(k\) such that \(g^k = e\). $\mathbf{A}^{-1} = \begin{pmatrix}z & -x\\ -y & w \end{pmatrix}$. Claim, \(S\) is a group. \begin{enumerate} \item (Closure) The product of two \(2\times2\) matrices is always a \(2\times 2\) matrix so we only need to check the determinant. Suppose \(\det(\mathbf{A}) = \det (\mathbf{B}) = 1\), then \(\det(AB) = \det(A)\det(B) = 1\), so our operation is closed \item (Associativity) Inherited from matrix multiplication \item (Identity) $\mathbf{I} =\begin{pmatrix}1 & 0\\ 1 & 1 \end{pmatrix}\( has determinant \)1$. \item (Inverses) The inverse is always fine since the matrix of cofactors always contains integers and the determinant is one, so we never end up with anything which isn't an integer. \end{itemize} If \(\mathbf{A}^-1 = \mathbf{A}\) then assuming $\mathbf{A} = \begin{pmatrix}a & b\\ c & d \end{pmatrix}\( then \)\mathbf{A}^{-1} = \begin{pmatrix}d & - b\\ -c & a \end{pmatrix}\( so we must have \)a=d, -b=b, -c=c\(, so \)b = c = 0\( and \)a = d\(. For the determinant to be \)1\( we must have \)ad = a^2 = 1\(, ie \)a = \pm 1\(. Therefore we must have \)\mathbf{A} = \begin{pmatrix}1 & 0\\ 0 & 1 \end{pmatrix}\( or \)\mathbf{A} = \begin{pmatrix}-1 & 0\\ 0 & -1 \end{pmatrix}$. For an element to have order \(2\) then \(\mathbf{A}^2 = \mathbf{I}\) ie, \(\mathbf{A} = \mathbf{A}^{-1}\) and \(\mathbf{A} \neq \mathbf{I}\) therefore the only element of order \(2\) is $\begin{pmatrix}-1 & 0\\ 0 & -1 \end{pmatrix}$. For an element to have order \(3\) we must have \(\mathbf{A}^2 = \mathbf{A}^{-1}\), ie $\begin{pmatrix}w^2 + xy & x(w+z)\\ y(w+z) & z^2 + xy \end{pmatrix} = \begin{pmatrix}z & -x\\ -y & w \end{pmatrix}$. Therefore \(w^2 + xy = z, x(w+z) = -x, y(w+z) = -y, z^2+xy = w\). The second and third equations are satisfied iff \(w+z+1 = 0\) or \(x = 0\) and \(y = 0\), but if \(x = 0\) and \(y = 0\) then we aren't order \(3\), so we just need to check this is sufficient for the first and last equations. Since \(\det(\mathbf{A}) = 1\) we have \(wz =xy +1\), so the first and last equations are equivalent to \(w^2 + wz - 1 = z\) and \(x^2 + wz-1 = w\) which are equivalent to \(w(w+z) = z+1\) or \(w + z+ 1 = 0\) as required

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Solution:

The red line is \(y = \ln (2 \sec x)\), blue is \(y = \sec x\). We can see that if the gradient is too small it never touches the red line. If it is large it will cross the red line twice in that interval. For some value it will be perfectly tangent. Since the line is tangent we must have \begin{align*} && y &= \ln (2 \sec x) \\ \Rightarrow && \frac{\d y}{ \d x} &= \frac{1}{2 \sec x} \cdot 2\sec x \tan x \\ &&&= \tan x \\ \Rightarrow && k_0 &=\tan x_0 \\ \Rightarrow && k_0 x_0 &= \ln(2 \sec x_0 ) \\ \Rightarrow && x_0 &= \cot x_0 \ln (2 \sec x_0) \end{align*} If \(f(x) =x- \cot x \ln (2 \sec x)\), then \(f'(x) =1 - 1+\ln(2\sec x) \cosec^2x = \ln(2 \sec x)\cosec^2x \) so we should look at \begin{align*} x_{n+1} &= x_n - \frac{f(x_n)}{f'(x_n)} \\ &= x_n - \frac{x_n- \cot x_n \ln (2 \sec x_n)}{\ln(2 \sec x_n)\cosec^2x_n } \\ &= x_n \left (1 - \frac{\sin^2 x_n}{\ln (2 \sec x_n)}\right) +\sin x_n \cos x_n \end{align*} \begin{array}{c|c} n & x_n \\ \hline 1 & \frac{\pi}{4} \\ 2 & 0.907701\ldots \\ 3 & 0.91439340\ldots \\ 4 & 0.914403867\ldots \\ 5 & 0.91440386\ldots \\ 6 & 0.91440386\ldots \\ \end{array} The sign change test shows that \(x_0 \approx 0.91\) and \(k_0 = \tan(x_0) \approx 1.30\)

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Solution: \(p = a+b+c, q = ab+bc+ca, r = abc\)

1. Suppose two roots are equal to each other, this means that one of the roots is also a root of the derivative. ie \begin{align*} && 0 &= x^3+qx - r \\ && 0 &= 3x^2+q \end{align*} have a common root, but this root must satisfy \(x^2 = -\frac{q}{3}\). Then \begin{align*} &&0 &= x^3 + qx - r \\ &&&= x^3 -3x^3 - r \\ &&&= -2x^3 -r \\ \Rightarrow && r^2 &= 4x^6 \\ &&&= 4 \left ( -\frac{q}{3}\right)^3 \\ \Rightarrow && 0 &= 27r^2+4q^3 \end{align*}
2. Consider \(x = z + \frac{p}{3}\), then the equation is: \begin{align*} x^{3}-px^{2}+qx-r &= (z + \frac{p}{3})^3 - p(z + \frac{p}{3})^2 + q(z + \frac{p}{3}) - r \\ &= z^3 + pz^2 + \frac{p^2}{3}z + \frac{p^3}{27} - \\ &\quad -pz^2-\frac{2p^2}{3}z-\frac{p^3}{9} + \\ &\quad\quad qz + \frac{pq}{3} - r \\ &= z^3+\left (\frac{p^2}{3}-\frac{2p^2}{3}+q \right)z + \left (\frac{p^3}{27}-\frac{p^3}{9}+\frac{pq}{3}-r \right) \\ &= z^3+\left (-\frac{p^2}{3}+q \right)z + \left (-\frac{2p^3}{27}+\frac{pq}{3}-r \right) \\ \end{align*} Since this equation must also have repeated roots we must have: \begin{align*} 4\left (-\frac{p^2}{3}+q \right)^3 + 27 \left (-\frac{2p^3}{27}+\frac{pq}{3}-r \right)^2 = 0 \end{align*} which is exactly our desired result

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Solution:

1. \begin{align*} \int\frac{x}{(x-1)(x^{2}-1)}\,\mathrm{d}x &= \int \frac{x}{(x-1)^2 (x+1)} \d x \\ &= \int \frac{1}{2(x-1)^2} + \frac{1}{4(x-1)} - \frac{1}{4(x+1)} \d x \\ &= -\frac12 (x-1)^{-1} + \frac14 \ln(x-1) - \frac14 \ln (x+1) + C \end{align*}
2. \begin{align*} \int \frac{1}{3 \cos x + 4 \sin x } \d x &= \int \frac{1}{5 \cos (x - \cos^{-1}(3/5))} \d x \\ &= \frac15 \int \sec (x - \cos^{-1}(3/5)) \d x\\ &= \frac15 \left (\ln | \sec (x - \cos^{-1}(3/5)) + \tan (x - \cos^{-1}(3/5)) | \right) + C \end{align*}
3. \begin{align*} \int \frac{1}{\sinh x} \d x &= \int \frac{2}{e^x - e^{-x}} \\ &= \int \frac{2e^x}{e^{2x}-1} \d x \\ &=\int \frac{e^x}{e^x-1} - \frac{e^x}{e^x+1} \d x \\ &= \ln (e^x - 1) + \ln (e^x+1) + C \end{align*}

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Solution: \(\mathbf{r}=(1-\lambda-\mu)\mathbf{a}+\lambda\mathbf{b}+\mu\mathbf{c} = \mathbf{a} + \lambda(\mathbf{b}-\mathbf{a})+\mu(\mathbf{c}-\mathbf{a})\) Therefore it is the plane through \(\mathbf{a}\) with direction vectors \(\mathbf{b}-\mathbf{a}\) and \(\mathbf{c}-\mathbf{a}\), ie it is the plane through \(\mathbf{a},\mathbf{b},\mathbf{c}\). The normal to this plane will be \((\mathbf{b}-\mathbf{a} ) \times (\mathbf{c}-\mathbf{a}) = \mathbf{b}\times \mathbf{c}-\mathbf{a} \times \mathbf{c}-\mathbf{b}\times \mathbf{a}\), so we must have: \begin{align*} && \mathbf{r} \cdot \left (\mathbf{b}\times \mathbf{c}-\mathbf{a} \times \mathbf{c}-\mathbf{b}\times \mathbf{a} \right) &= \mathbf{a} \cdot \left (\mathbf{b}\times \mathbf{c}-\mathbf{a} \times \mathbf{c}-\mathbf{b}\times \mathbf{a} \right) \\ &&&= \mathbf{a} \cdot (\mathbf{b}\times \mathbf{c}) \end{align*} Therefore, \begin{align*} && \mathbf{a} \cdot (\mathbf{b}\times \mathbf{c}) &= \mathbf{r} \cdot \left (\mathbf{b}\times \mathbf{c}-\mathbf{a} \times \mathbf{c}-\mathbf{b}\times \mathbf{a} \right) \\ &&&= \left ( (1-\lambda-\mu)\mathbf{a}+\lambda\mathbf{b}+\mu\mathbf{c} \right)\cdot \left (\mathbf{b}\times \mathbf{c}-\mathbf{a} \times \mathbf{c}-\mathbf{b}\times \mathbf{a} \right) \\ &&&= (1-\lambda- \mu) \mathbf{a}\cdot (\mathbf{b} \times \mathbf{c})-\lambda \mathbf{b}\cdot(\mathbf{a} \times \mathbf{c})-\mu \mathbf{c}\cdot(\mathbf{b} \times \mathbf{a}) \\ \Rightarrow && 0 &= (-\lambda- \mu) \mathbf{a}\cdot (\mathbf{b} \times \mathbf{c})-\lambda \mathbf{b}\cdot(\mathbf{a} \times \mathbf{c})-\mu \mathbf{c}\cdot(\mathbf{b} \times \mathbf{a}) \\ &&&= -(\lambda+ \mu) \mathbf{a}\cdot (\mathbf{b} \times \mathbf{c})+\lambda \mathbf{b}\cdot(\mathbf{c} \times \mathbf{a})+\mu \mathbf{c}\cdot(\mathbf{a} \times \mathbf{b}) \\ \end{align*} The result follows from setting \(\mu = 0, \lambda = 1\) and \(\mu = 1, \lambda = 0\). If they all lie in the same plane then the plane described is through the origin, and those values are all the same, but equal to \(0\).

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Solution:

\begin{align*} \arg w &= \arg \frac{(z_{1}-z_{2})(z_{3}-z_{4})}{(z_{4}-z_{1})(z_{2}-z_{3})} \\ &= \arg \frac{(z_{1}-z_{2})(z_{3}-z_{4})}{(z_{2}-z_{3})(z_{4}-z_{1})} \\ &= \arg \frac{(z_{1}-z_{2})}{(z_{3}-z_{2})}\frac{(z_{3}-z_{4})}{(z_{1}-z_{4})} \\ &= \arg \frac{(z_{1}-z_{2})}{(z_{3}-z_{2})} + \arg \frac{(z_{3}-z_{4})}{(z_{1}-z_{4})}\\ &= \beta + \pi - \beta = \pi \end{align*} Therefore \(w\) is real