Problems

Solution: After \(5\) steps we can get to: \begin{array}{c|c} x & \text{ways} \\ \hline 5 & 1 \text { - go positive every time}\\ 4 & 0 \\ 3 & \binom{5}{1} \text { - go positive every time but 1} \\ 2 &0 \\ \hline & 6 \end{array} Therefore there are \(\frac{6}{2^5} = \frac{3}{16}\) ways to get to \(\{2,3,4,5\}\) After \(6\) steps we can get to: \begin{array}{c|c} x & \text{ways} \\ \hline 5 & 0 \\ 4 & \binom{6}{1} \text { - go positive every time but 1}\\ 3 & 0 \\ 2 & \binom{6}{2} - \text{ - go positive every time but 2} \\ \hline & 21 \end{array} Therefore there are \(\frac{21}{2^6} = \frac{21}{64}\) ways to get to \(\{2,3,4,5\}\) After \(2k\) steps we can reach \(2\) or \(4\). To get to \(2\) we must take \(k+1\) positive steps and \(k-1\) negative steps, ie \(\binom{2k}{k-1}\). To get to \(4\) we must take \(k+2\) positive steps and \(k-2\) negative steps, ie \(\binom{2k}{k-2}\) Therefore there are \(\binom{2k+1}{k-1}\) routes, ie a probability of \(\frac{1}{2^{2k}} \binom{2k+1}{k-1}\) After \(2k+1\) steps we can reach \(3\) or \(5\). To get to \(3\) we must take \(k+2\) positive steps and \(k-1\) negative steps, ie \(\binom{2k+1}{k-1}\). To get to \(5\) we must take \(k+3\) positive steps and \(k-2\) negative steps, ie \(\binom{2k+1}{k-2}\) Therefore there are \(\binom{2k+2}{k-1}\) routes, ie a probability of \(\frac{1}{2^{2k+1}} \binom{2k+2}{k-1}\) To find the maximum of \(P(s)\) notice that \begin{align*} && \frac{P(2k+1)}{P(2k)} &= \frac12 \frac{\binom{2k+2}{k-1}}{\binom{2k+1}{k-1}} \\ &&&= \frac12 \frac{(2k+2)!(k-1)!(k+2)!}{(2k+1)!(k-1)!(k+3)!} \\ &&&= \frac12 \frac{2k+2}{k+3} = \frac{k+1}{k+3} < 1 \end{align*} So we should only look at the even terms. \begin{align*} && \frac{P(2k+2)}{P(2k)} &= \frac14 \frac{\binom{2k+3}{k}}{\binom{2k+1}{k-1}} \\ &&&= \frac14 \frac{(2k+3)!(k-1)!(k+2)!}{(2k+1)!k!(k+3)!} \\ &&&= \frac14 \frac{(2k+3)(2k+2)}{k(k+3)} \\ &&&= \frac{(2k+3)(k+1)}{2k(k+3)} \geq 1 \\ \Leftrightarrow && (2k+3)(k+1) &\geq 2k(k+3) \\ \Leftrightarrow && 2k^2+5k+3 &\geq 2k^2+6k \\ \Leftrightarrow && 3 &\geq k \\ \end{align*} Therefore the maximum is when \(s = 2\cdot 3\) or \(s = 2\cdot 4\) which we computed earlier to be \(\frac{21}{64}\)

Solution: \begin{align*} \mathbb{P}(\text{run out reading for toffee on } N\text{th occassion}) &= \binom{N-1}{n}p^nq^{N-1-n}p \end{align*} Since out of the first \(N-1\) times, we need to choose toffee \(n\) times, and then choose it again for the \(N\)th time. Therefore: \begin{align*} \mathbb{P}(\text{reaching for toffee} | \text{run out on }N\text{th occassion}) &= \frac{\mathbb{P}(\text{reaching for toffee and run out on }N\text{th occassion})}{\mathbb{P}(\text{reaching for toffee and run out on }N\text{th occassion}) + \mathbb{P}(\text{reaching for mint and run out on }N\text{th occassion})} \\ &= \frac{ \binom{N-1}{n}p^nq^{N-1-n}p}{ \binom{N-1}{n}p^nq^{N-1-n}p + \binom{N-1}{m}q^mp^{N-1-m}q} \\ &= \frac{ \binom{N-1}{n}}{ \binom{N-1}{n} + \binom{N-1}{m} \l \frac{q}{p} \r^{m+ n+ 2-N}} \end{align*} Some conclusions we can draw from this are: As \(p \to 1, q \to 0\), the probability they were reaching for a Toffee tends to \(1\). (And vice versa). If \(p = q\), then the probability is: \begin{align*} \frac{ \binom{N-1}{n}}{ \binom{N-1}{n} + \binom{N-1}{m} } \end{align*} Since \(n+1 \leq N \leq n+m+1\) where \(n \geq m\) we can notice that: \begin{align*} \text{if } N = m + n + 1 && \binom{m+n+1 - 1}{n} &= \binom{m+n+1 - 1}{m} & \text{ so } \mathbb{P} = \frac12 \\ \text{if } N = n+k && \binom{n+k-1}{n} &< \binom{n+k-1}{m} & \text{ so } \mathbb{P} < \frac12 \\ \end{align*}

Solution:

1. \begin{align*} && f(x) &= \ln (1+e^x) \\ && f'(x) &= \frac{1}{1+e^x} \cdot e^x \\ &&&= \frac{e^x}{1+e^x} \\ \Rightarrow && \ln [f'(x)] &= x - \ln (1+e^x) \\ &&&= x - f(x) \\ \\ \Rightarrow && \frac{f''(x)}{f'(x)} &= 1 - f'(x) \\ \Rightarrow && f''(x) &= f'(x) - [f'(x)]^2 \\ && f'''(x) &= f''(x) - 2f'(x) f''(x) \\ && f^{(4)}(x) &= f'''(x) - 2[f''(x)]^2-2f'(x)f'''(x) \end{align*} \begin{align*} f(0) &= \ln 2 \\ f'(0) &= \tfrac12 \\ f''(0) &= \tfrac12 -\tfrac14 \\ &= \tfrac 14 \\ f'''(0) &= \tfrac14 - 2 \tfrac12 \tfrac 14 \\ &= 0 \\ f^{(4)}(0) &= -2 \cdot \tfrac1{16} \\ &= -\frac18 \end{align*} Therefore \(f(x) = \ln 2 + \tfrac12 x + \tfrac18 x^2 - \frac1{8 \cdot 4!} x^4 + O(x^5)\)
2. As \(x \to 0\), \(g(x) \to \infty\) therefore there can be no power series about \(0\). But as \(x \to 0, x g(x) \not \to \infty\) as \(\frac{x}{\sinh x}\) is well behaved. We can also notice that \(x g(x)\) is an even function, since \(\cosh x\) is even and \(\frac{x}{\sinh x}\) is even, therefore the power series will consist of even powers of \(x\) \begin{align*} \lim_{x \to 0} \frac{x}{\sinh x \cosh 2 x} &= \lim_{x \to 0} \frac{x}{\sinh x} \cdot \lim_{x \to 0} \frac{1}{\cosh2 x} \\ &= 1 \end{align*} Notice that \begin{align*} \frac{x}{\sinh x \cosh 2 x} &= \frac{4x}{(e^x - e^{-x})(e^{2x}+e^{-2x})} \\ &= \frac{4x}{(2x + \frac{x^3}{3} + \cdots)(2 + 4x^2 + \frac43 x^4 + \cdots )} \\ &= \frac{1}{1+\frac{x^2}{6}+\frac{x^4}{5!} + \cdots } \frac{1}{1 + 2x^2 + \frac23 x^4 + \cdots } \\ &= \left (1-(\frac{x^2}{6} + \frac{x^4}{5!})+ (\frac{x^2}{6} )^2 + O(x^6)\right) \left (1-(2x^2+\frac23 x^4)+ (2x^2)^2 + O(x^6)\right) \\ &= \left (1 - \frac16 x^2 + \frac{7}{360} x^4 + O(x^6) \right) \left (1 - 2x^2+ \frac{10}3x^4 + O(x^6) \right) \\ &= 1 - \frac{13}{6} x^2 + \frac{1327}{360}x^4 + O(x^6) \end{align*}

Solution: If $\mathbf{J}=\begin{pmatrix}1 & 1\\ 1 & 1 \end{pmatrix}\(, them \)\mathbf{J}^2=\begin{pmatrix}2 & 2\\ 2 & 2 \end{pmatrix} = 2\mathbf{J}\(. Therefore \)\mathbf{J}^n = 2\mathbf{J}^{n-1} = 2^{n-1}\mathbf{J}$ Let \(\mathbf{A}=\mathbf{I}+a\mathbf{J}\) then \begin{align*} \mathbf{A}^2 &=\l \mathbf{I}+a\mathbf{J}\r^2 \\ &= \mathbf{I}+2a\mathbf{J} + a^2\mathbf{J}^2 \\ &= \mathbf{I}+2a\mathbf{J} + 2a^2\mathbf{J} \\ &= \mathbf{I}+(2a+ 2a^2)\mathbf{J} \\ &= \mathbf{I}+\frac12(1+4a+ 4a^2-1)\mathbf{J} \\ &= \mathbf{I}+\frac12((1+2a)^2-1)\mathbf{J} \\ \end{align*} \begin{align*} \mathbf{A}^3 &=\l \mathbf{I}+a\mathbf{J}\r^3 \\ &= \mathbf{I}+3a\mathbf{J} + a^2\mathbf{J} + a^3\mathbf{J}^3 \\ &= \mathbf{I}+3a\mathbf{J} + 6a^2\mathbf{J} + 4a^3\mathbf{J} \\ &= \mathbf{I}+(3a+ 6a^3+4a^3)\mathbf{J} \\ &= \mathbf{I}+\frac12(1+3\cdot2a+3\dot4a^2+ 8a^3-1)\mathbf{J} \\ &= \mathbf{I}+\frac12((1+2a)^3-1)\mathbf{J} \\ \end{align*} \begin{align*} \mathbf{A}^4 &=\l \mathbf{I}+a\mathbf{J}\r^4 \\ &= \mathbf{I}+4a\mathbf{J} + 6a^2\mathbf{J}^2 + 4a^3\mathbf{J}^3+a^4\mathbf{J}^4 \\ &= \mathbf{I}+4a\mathbf{J} + 12a^2\mathbf{J} + 16a^3\mathbf{J}+8a^4\mathbf{J}\\ &= \mathbf{I}+(4a+ 12a^3+16a^3+8a^4)\mathbf{J} \\ &= \mathbf{I}+\frac12(1+4\cdot2a+6\cdot4a^2+ 4\cdot8a^3+16a^4-1)\mathbf{J} \\ &= \mathbf{I}+\frac12((1+2a)^4-1)\mathbf{J} \\ \end{align*} Claim: \(\mathbf{A}^k = \mathbf{I} + \frac12 ((1+2a)^{k}-1)\mathbf{J}\) Proof: Firstly, note that \(\mathbf{I}\) commutes with everything, so we can just apply the binomial theorem as if we were using real numbers: \begin{align*} \mathbf{A}^k &=\l \mathbf{I}+a\mathbf{J}\r^k \\ &= \sum_{i=0}^k \binom{k}{i}a^i\mathbf{J}^i \\ &= \mathbf{I} + \sum_{i=1}^k \binom{k}{i}a^i2^{i-1}\mathbf{J} \\ &= \mathbf{I} + \frac12\l\sum_{i=1}^k \binom{k}{i}a^i2^{i}\r\mathbf{J} \\ &= \mathbf{I} + \frac12\l\sum_{i=0}^k \binom{k}{i}a^i2^{i} - 1\r\mathbf{J} \\ &= \mathbf{I} + \frac12\l(1+2a)^k - 1\r\mathbf{J} \end{align*} as required

Solution:

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Solution: Since \(B\) and \(C\) are reflections of each other in the plane \(y+z=2\) (since that's what it means to be symmetrical), we must have that \(N\) also lies on the plane \(y+z=2\). Since \(\overrightarrow{AY}=3\overrightarrow{YN}.\) we must have \(\overrightarrow{AN}=\overrightarrow{AY}+\overrightarrow{YN} = \frac43\overrightarrow{AY} = \frac43\begin{pmatrix} 3\\-6\\6\end{pmatrix} = \begin{pmatrix} 4\\-8\\8\end{pmatrix}\) and \(N\) is the point \((4,-4,6)\) (which fortunately is on our plane as expected). \(Y\) is the point \((3,-2,4)\) \(|\overrightarrow{AY}| = \sqrt{3^2+(-6)^2+6^2} = 3\sqrt{1+4+4} = 9\)

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