Problems

Solution:

1. \begin{align*} && \mathbb{P}(X+Y = n) &= \sum_{i = 0}^n \mathbb{P}(X = i, Y = n-i) \\ &&&= \sum_{i = 0}^n e^{-1} \frac{n \lambda^n}{i! (n-i)!} \\ &&&=e^{-1} n \lambda^n \sum_{i = 0}^n\frac{1}{i! (n-i)!} \\ &&&=\frac{e^{-1} n}{n!} \lambda^n \sum_{i = 0}^n\frac{n!}{i! (n-i)!} \\ &&&= \frac{n\lambda^n}{e n!} 2^n \\ &&&= \frac{n (2 \lambda)^n}{e \cdot n!} \end{align*}
2. \begin{align*} && 1 &= \sum_{n = 0}^{\infty} \mathbb{P}(X+Y =n ) \\ &&&= \sum_{n = 0}^{\infty}\frac{n (2 \lambda)^n}{e \cdot n!} \\ &&&= \sum_{n = 1}^\infty \frac{ (2 \lambda)^n}{e \cdot (n-1)!} \\ &&&= \frac{2 \lambda}{e}\sum_{n = 0}^\infty \frac{ (2 \lambda)^n}{n!} \\ &&&= \frac{2 \lambda}{e} e^{2\lambda} \\ &&&= 2 \lambda e^{2\lambda - 1} \end{align*} \\
3. Consider \(f(x) = 2xe^{2x-1}\), then \begin{align*} && f'(x) &= 2e^{2x-1} + 2xe^{2x-1} \cdot 2 \\ &&&= e^{2x-1} (2 + 4x) > 0 \end{align*} Therefore \(f(x)\) is an increasing function of \(x\), which means \(f(x) = 1\) has at most one solution for \(\lambda\). Therefore \(\lambda = \frac12\)
4. \begin{align*} \mathbb{E}(2^{X+Y}) &= \sum_{n = 0}^\infty \mathbb{P}(X+Y = n) 2^n \\ &= \sum_{n = 1}^\infty \frac{1}{e(n-1)!} 2^{n} \\ &= \frac{2}{e} \sum_{n=0}^\infty \frac{2^n}{n!} \\ &= \frac{2}{e} e^2 \\ &= 2e \end{align*}

Solution: First consider \(5\) digit numbers containing at most \(2\) non-zero digits. Then there are \(\binom{9}{2}\) ways to choose the two digits, and \(2^{5}-2\) different ways to arrange them, removing the ones which are all the same. Considering all the pairs including zero, there are \(9\) ways to choose the non-zero (first) digit. There are \(2^4-1\) remaining digits where not all the numbers are the same. Finally we must not forget \(100\,000\). Therefore there are \(\binom{9}{2}(2^5-2) +9\cdot(2^4-1) + 1 = 1216\)

Solution: \begin{align*} && t &= (x^2+1)^{1/2}+x \\ && 1&=t^2-2tx \\ && x &= \frac{t^2-1}{2t} = \frac12 \left (t - \frac1t\right) \\ && \frac{\d x}{\d t} &= \frac12 \left ( 1+ \frac{1}{t^2} \right) \\ \Rightarrow && \int_0^{\infty} f((x^2+1)^{1/2}+x) \d x &= \int_{t=1}^{t = \infty}f(t) \frac12(1 + t^{-2}) \d t\\ &&&= \frac12 \int_1^{\infty}(1+t^{-2})f(t) \d t \end{align*} \begin{align*} \int_{0}^{\infty}((x^{2}+1)^{1/2}+x)^{-3}\,{\mathrm d}x &= \frac12 \int_1^{\infty}(1+t^{-2})t^{-3} \d t \\ &= \frac12 \left [\frac{-1}{2}t^{-2}-\frac{1}{4}t^{-4} \right]_{1}^{\infty} \\ &= \frac12 \left ( \frac12 + \frac14\right) = \frac38 \end{align*}

Solution:

1. True. \begin{align*} && \ln a \cdot \ln b &= \ln b \cdot \ln a \\ \Leftrightarrow && \exp ( \ln a \cdot \ln b) &= \exp ( \ln b \cdot \ln a) \\ \Leftrightarrow && \exp ( \ln a )^{\ln b} &= \exp ( \ln b )^{\ln a} \\ \Leftrightarrow && a^{\ln b} &= b^{\ln a} \\ \end{align*}
2. False. Consider \(\theta = 0\). We'd need \(\cos 0 = 1 = \sin 1\), but \(0 < 1 < \frac{\pi}{2}\) so \(\sin 1 \neq 1\)
3. False. If the polynomial has positive degree, then as \(n \to \infty\), \(\P(x) \to \pm \infty\), in particular it must be well outside the interval \([-1,1]\). Therefore it can't be within \(10^{-6}\) of \(\cos \theta\) which is confined to that interval. The only polynomial which is restricted to that range are constants, but then \(|\cos 0 - c| \leq 10^{-6}\) and \(|\cos \pi - c| \leq 10^{-6}\) \(2 = |1-(-1)| \leq |1-c| + |-1-c| \leq 2\cdot 10^{-6}\) contradiction.
4. True. \begin{align*} && (x^2-x^{-2})^2 &\geq 0 \\ \Leftrightarrow && x^4-2+x^{-4} &\geq0 \\ \Leftrightarrow && x^4+3+x^{-4} &\geq 5 \\ \end{align*}

Solution: We can always rotate the circle so that sides are parallel to the \(x\) and \(y\) axes. Therefore if one corner is \((a,b)\) the other coordinates are \((-a,b), (a,-b), (-a,-b)\) and the perimeter will be \(4(a+b)\). Therefore we wish to maximise \(4(a+b)\) subject to \(a^2+b^2 = \text{some constant}\). Notice that \(\frac{a+b}{2} \leq \sqrt{\frac{a^2+b^2}{2}}\) with equality when \(a = b\), therefore the maximum is a square. If \(n = 2\) then we are looking at \(2((2a)^2+(2b)^2) = 8(a^2+b^2)\) which is constant for all rectangles. If \(n=3\) we are maximising \(16(a^3+b^3) = 16(a^3+(c^2-a^2)^{3/2})\) which is maximised when \(a = 0, c\)

Solution:

1. 

   + - ↺
   We are looking for \((10+2i) - (4+6i) = 6 - 4i\) rotated by \(\frac{\pi}{3}\) and then added to \(4+6i\), which is \begin{align*} (6-4i)(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}) &= (6-4i)\left(\tfrac12 +\tfrac{\sqrt{3}}2i\right) \\ &= 3+2\sqrt{3} + (3\sqrt{3}-2)i \end{align*}
2. \begin{align*} &&& \frac{a}{1+i}+\frac{b}{1+2i}+\frac{c}{1+3i} &\in \mathbb{R} \\ \Longleftrightarrow && \frac{a}{1+i}+\frac{b}{1+2i}+\frac{c}{1+3i} &= \frac{a}{1-i}+\frac{b}{1-2i}+\frac{c}{1-3i} \\ && 0 &= a\left ( \frac{1}{1+i} - \frac{1}{1-i} \right)+ b\left ( \frac{1}{1+2i} - \frac{1}{1-2i} \right)+ c\left ( \frac{1}{1+3i} - \frac{1}{1-3i} \right) \\ &&&= a\left ( \frac{(1-i)-(1+i)}{1^2+1^2} \right) + b\left ( \frac{(1-2i)-(1+2i)}{1^2+2^2} \right) + c\left ( \frac{(1-3i)-(1+3i)}{1^2+3^2} \right) \\ &&&= -\frac{2i}{2}a-\frac{4i}{5}b-\frac{-6i}{10}c \\ \Longleftrightarrow && 0 &= a+\tfrac45b+\tfrac35c \end{align*}