Problems

Solution: This is just the result that all of the area beneath \(g(t)\) is also below \(f(t)\) If \(p > q > 0, x \geq 1 \Rightarrow x^p \geq x^q\), therefore applying the result we have \begin{align*} && \int_1^x x^p\, \d t & \geq \int_1^x x^q\, \d t \\ \Rightarrow && \frac{x^p-1}{p} & \geq \frac{x^q-1}{q} \end{align*} When \(p > q > 0, 0 \leq x \leq 1\) we have \(x^p \leq x^q\), ie \begin{align*} && \int_x^1 x^q\, \d t & \geq \int_{x}^1 x^p\, \d t \\ \Rightarrow && \frac{1-x^q}{q} & \geq \frac{1-x^p}{p} \\ \Rightarrow && \frac{x^p-1}{p} &\geq \frac{x^q-1}{q} \end{align*} Now looking at the functions \(f(x) = \frac{x^p-1}{p}, g(x) = \frac{x^q-1}{q}\) and \(x \geq 1\) we have \begin{align*} && \int_0^x \frac{t^p-1}{p} \d t & \geq \int_0^x \frac{t^q-1}{q} \d t\\ \Rightarrow &&\frac1p \left[\frac{t^{p+1}}{p+1} - t \right]_0^x &\geq\frac1q \left[\frac{t^{q+1}}{q+1} - t \right]_0^x \\ \Rightarrow &&\frac1p \left(\frac{x^{p+1}}{p+1} -x\right) &\geq\frac1q \left(\frac{x^{q+1}}{q+1} - x \right)\\ \Rightarrow &&\frac1p \left(\frac{x^{p}}{p+1} -1\right) &\geq\frac1q \left(\frac{x^{q}}{q+1} - 1 \right)\\ \end{align*}

Solution: We have \begin{align*} \rightarrow: && a &= v\cos \alpha t_{net} \\ \Rightarrow && t_{net} &= \frac{a}{v \cos \alpha} \\ \downarrow: && H-h &= v\sin \alpha t_{net} + \frac12 g t_{net}^2 \\ &&&= a \tan \alpha + \frac12 g \frac{a^2}{v^2} \sec^2 \alpha \\ &&&= a \tan \alpha + \frac{a^2g}{2v^2}(1 + \tan^2 \alpha) \tag{*}\\ \\ \rightarrow: && a+b &= v \cos \alpha t_{ground} \\ && t_{ground} &= \frac{a+b}{v \cos \alpha}\\ \downarrow: && H &= v\sin \alpha t_{ground} + \frac12 g t_{ground}^2 \\ &&&= (a+b)\tan \alpha + \frac{(a+b)^2g}{2v^2}(1+\tan^2\alpha) \tag{**} \\ \\ (**): && v^2 &= \frac{g(a+b)^2(1+\tan^2\alpha)}{2[H-(a+b)\tan \alpha]} \\ (a+b)^2(*) - a^2(**): && (a+b)^2(H-h) -a^2H &= [(a+b)^2a - a^2(a+b)]\tan \alpha \\ \Rightarrow && (2ab+b^2)H - (a+b)^2h &= ab(a+b) \tan \alpha \\ \Rightarrow && \tan \alpha &= \frac{2a+b}{a(a+b)}H - \frac{a+b}{ab} h \end{align*} Noting that \(v^2 \geq 0\) and the numerator is positive, we must have \begin{align*} && H &> (a+b)\tan \alpha \\ &&&= \frac{2a+b}{a}H - \frac{(a+b)^2}{ab} h \\ \Rightarrow && \frac{a+b}{a}H &< \frac{(a+b)^2}{ab} h \\ \Rightarrow && H &< \frac{a+b}{b} h \end{align*} Noting that \(\tan \alpha > 0\) we must have \begin{align*} && \frac{2a+b}{a(a+b)} H & > \frac{a+b}{ab} h \\ \Rightarrow && H &> \frac{(a+b)^2}{b(2a+b)}h \end{align*}

Solution:

1. Suppose \(pqr = qrs\), since the roots are positive, we can divide by \(qr\) to obtain \(p=s\) (a contradiction. Therefore all those terms are distinct.
2. \(4c^3 = pqr+qrs+rsp+spq\), \(d^4 = pqrs\). Applying AM-GM, we obtain: \begin{align*} && c^3 = \frac{ pqr+qrs+rsp+spq}{4} & > \sqrt[4]{p^3q^3r^3s^3} = d^{3} \\ \Rightarrow && c &> d \end{align*}
3. There must be a turning point between each root (since there are no repeated roots).
4. \(f'(x) = 4x^3-12ax^2+12b^2-4c^3 = 4(x^3-3ax^2+3b^2-c^3)\). Letting the roots of this polynomial be \(\alpha, \beta, \gamma\) and again applying AM-GM, we must have: \begin{align*} && b^2 = \frac{\alpha\beta + \beta \gamma+\gamma \alpha}{3} &> \sqrt[3]{\alpha^2\beta^2\gamma^2} = c^2 \\ \Rightarrow && b &> c \end{align*}
5. Again, since there are turning points between the roots of \(f'(x)\) we must have distinct roots for \(f''(x)\), ie: \(f''(x) = 3x^2-6ax+6b^2 = 3(x^2-2ax+b^2)\) has distinct real roots. But for this to occur we must have that \((2a)^2-4b^2 = 4(a^2-b^2) > 0\), ie \(a>b\)

Solution: \begin{align*} && \E[Y^n] &= \frac14 \cdot \frac1{4^n} + \frac14 \cdot \frac{3^n}{4^n} + \frac12 \int_{1/4}^{3/4}2 y^n \d y \\ &&&= \frac{3^n+1}{4^{n+1}} + \left [ \frac{y^{n+1}}{n+1} \right]_{1/4}^{3/4} \\ &&&= \frac{3^n+1}{4^{n+1}} + \frac{3^{n+1}-1}{(n+1)4^{n+1}} \end{align*} \begin{align*} && \E[Y] &= \frac{3+1}{16} + \frac{9-1}{2 \cdot 16} \\ &&&= \frac{1}{4} + \frac{1}{4} = \frac12 = \E[X] \end{align*} \begin{align*} && \E[X^2] &= \int_0^1 x^2 \d x = \frac13 \\ && \E[Y^2] &= \frac{9+1}{64} + \frac{27-1}{3 \cdot 64} = \frac{56}{3 \cdot 64} = \frac{7}{24} \\ \Rightarrow && \E[X^2] - \E[Y^2] &= \frac13 - \frac{7}{24} = \frac{1}{24} \end{align*} \begin{align*} && \E[X^n] &= \frac{1}{n+1} \\ && \E[Y^n] &= \frac{1}{n+1} \frac{1}{4^{n+1}}\left ( (n+1)(3^n+1)+3^{n+1}-1 \right) \\ &&&= \frac{1}{n+1} \frac{1}{4^{n+1}}\left ( 3^{n+1} + (n+1)3^n +n \right) \\ \\ && (3+1)^{n+1} &= 3^{n+1} + (n+1)3^n + \cdots + (n+1) \cdot 3 + 1 \\ &&&> 3^{n+1} + (n+1)3^n + n + 1 \end{align*} if \(n \geq 2\) Notice that by the central limit theorem: \begin{align*} &&\frac{1}{240\,000} \sum_{i=1}^{240\,000} Y_i &\sim N \left ( \frac12, \frac{1}{24 \cdot 240\,000}\right) \\ \Rightarrow && \mathbb{P}\left (\frac{\frac{1}{240\,000} \sum_{i=1}^{240\,000} Y_i - \frac12}{\frac1{24} \frac{1}{100}} \leq \frac23 \right) &\approx 0.75 \\ \Rightarrow && \mathbb{P} \left ( \sum_i Y_i \leq 240\,000 \cdot \left ( \frac2{3} \frac1{2400}+\frac12 \right) \right ) & \approx 0.75 \\ \Rightarrow && K &= 120\,000 + 66 \\ &&&\approx 120\,066 \end{align*}

Solution: \begin{align*} && k &\leq 2(k-2) \\ \Rightarrow && 4 &\leq k \end{align*} Lemma: Suppose we construct \(N \neq \) (optimally) as a sum out of \(a_1 + \cdots +a_k\), then \(a_i \in \{2, 3\}\). Proof: Suppose not, suppose some \(a_i > 3\). Then from our earlier inequality, the sum \(a_1 + \cdots +a_{i-1} + 2 + (a_i - 2) + \cdots \) has the same sum, but a larger product. Therefore \(a_i \leq 3\). Suppose also some \(a_i = 1\), then we could replace \(a_1\) with \(a_1+1\) and remove \(a_i\), leaving us again with the same sum but larger product. (Assuming \(N \neq 1\)) \(5 = 2+3\) is the only way to write \(5\) as a sum of \(2\)s and \(3\)s, therefore \(P(5) = 2\times 3\) \(6 = 2 + 2 + 2 = 3 + 3\) and we can immediately see that \(2^3 = 8 < 3^2 = 9\), so \(P(6) = 3^2\) and whenever we have three \(2\)s we should replace them with two \(3\)s. So \(7 = 2 + 2 + 3 \Rightarrow P(7) = 2^2 \times 3\) \(8 = 3 + 3 + 2 \Rightarrow P(8) = 2\times 3^2\) \(9 = 3 + 3 + 3 \Rightarrow P(9) = 3^3\) Suppose \(1000 = 2n + 3m\), considered \(\pmod{3}\) we can see that \(n \equiv 2 \pmod{3}\) therefore we should have \(1000 = 2 + 2 + \underbrace{3 + \cdots + 3}_{332\text{ }3\text{s}}\) and so \(P(1000) = 2^2 \times 3^{332}\)

Solution:

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1. \begin{align*} \overset{\curvearrowright}{O_2}: && 0 &= F_2 - F \\ \Rightarrow && F_2 &= F \\ \overset{\curvearrowright}{O_1}: && 0 &= F_1- F \\ \Rightarrow && F_1 &= F \\ \text{N2}(\swarrow, 2): && 0 &= R+W_2\sin\alpha -F \tag{1}\\ \text{N2}(\swarrow, 1): && 0 &= W_1\sin\alpha -F-R\tag{2}\\ \Rightarrow && W_1 \sin \alpha-R &= W_2 \sin \alpha+R \\ \Rightarrow && W_1 &\geq W_2 \end{align*}
2. \begin{align*} (1)+(2)\Rightarrow && F &= \frac12 \sin \alpha (W_1 + W_2) \\ (1)-(2) \Rightarrow && R &= \frac12 \sin \alpha (W_1-W_2) \\ \Rightarrow && \frac{F}{R} &= \frac{W_1+W_2}{W_1-W_2} \\ \underbrace{\Rightarrow}_{F \leq \mu R} && \mu &\geq \frac{W_1+W_2}{W_1-W_2}\\ \end{align*}
3. \begin{align*} \text{N2}(\nwarrow, 1): && 0 &= F+R_1-W_1\cos \alpha \\ \Rightarrow && R_1 &= W_1\cos \alpha - F \\ &&&= W_1 \cos \alpha - \frac12 \sin \alpha (W_1 + W_2) \\ \Rightarrow && \frac{R_1}{F_1} &= \frac{R_1}{F} \\ &&&= \frac{W_1 \cos \alpha - \frac12 \sin \alpha (W_1 + W_2)}{\frac12 \sin \alpha (W_1 + W_2) } \\ &&&= \frac{2W_1 \cot \alpha}{W_1+W_2} - 1 \\ \Rightarrow && \mu_1 & \geq \left ( \frac{2W_1 \cot \alpha}{W_1+W_2} - 1 \right)^{-1} \end{align*}

Solution: Definition: \(\displaystyle \mathbb{E}(X) = \sum_{i=1}^n x_i \mathbb{P}(X = x_i)\) Definition: \(\displaystyle \mathrm{Var}(X) = \sum_{i=1}^n (x_i-\mathbb{E}(X))^2 \mathbb{P}(X = x_i)\) Claim: \(\mathbb{E}(aX+b) = a\mathbb{E}(X)+b\) Proof: \begin{align*} \mathbb{E}(aX+b) &= \sum_{i=1}^n (ax_i+b) \mathbb{P}(X = x_i) \\ &= a\sum_{i=1}^n x_i \mathbb{P}(X = x_i) + b\sum_{i=1}^n \mathbb{P}(X = x_i)\\ &= a \mathbb{E}(X) + b \end{align*} Claim: \(\mathrm{Var}(aX+b) = a^2 \mathrm{Var}(X)\) Claim: \(\mathrm{P}\left(\left|X-\mathrm{E}(X)\right|\geqslant\lambda\right)\leqslant\frac{\mathrm{var}(X)}{\lambda^{2}}\) Proof: \begin{align*} \mathrm{Var}(X) &= \sum_{i=1}^n (x_i-\mathbb{E}(X))^2 \mathbb{P}(X = x_i) \\ &\geq \sum_{|x_i - \mathbb{E}(X)| \geq \lambda} (x_i-\mathbb{E}(X))^2 \mathbb{P}(X = x_i) \\ &\geq \sum_{|x_i - \mathbb{E}(X)| \geq \lambda} \lambda^2 \mathbb{P}(X = x_i) \\ &= \lambda^2 \sum_{|x_i - \mathbb{E}(X)| \geq \lambda} \mathbb{P}(X = x_i) \\ &= \lambda^2 \mathrm{P}\left(\left|X-\mathrm{E}(X)\right|\geqslant\lambda\right) \end{align*} Claim: \[ \mathrm{P}\left(\left|X-\mathrm{E}(X)\right|\geqslant\lambda\right)\geqslant\frac{\mathrm{var}(X)-\lambda^{2}}{k^{2}-\lambda^{2}}\,. \] Proof: \begin{align*} && \mathrm{Var}(X) &= \sum_{i=1}^n (x_i-\mathbb{E}(X))^2 \mathbb{P}(X = x_i) \\ &&&= \sum_{|x_i - \mathbb{E}(X)| \geq \lambda} (x_i-\mathbb{E}(X))^2 \mathbb{P}(X = x_i) + \sum_{|x_i - \mathbb{E}(X)| < \lambda} (x_i-\mathbb{E}(X))^2 \mathbb{P}(X = x_i) \\ &&& \leq \sum_{|x_i - \mathbb{E}(X)| \geq \lambda} k^2 \mathbb{P}(X = x_i) + \sum_{|x_i - \mathbb{E}(X)| < \lambda} \lambda^2 \mathbb{P}(X = x_i) \\ &&&= k^2 \mathbb{P}\left(\left|X-\mathrm{E}(X)\right|\geqslant\lambda\right) + \lambda^2 \mathbb{P}\left(\left|X-\mathrm{E}(X)\right| < \lambda\right) \\ &&&= k^2 \mathbb{P}\left(\left|X-\mathrm{E}(X)\right|\geqslant\lambda\right) + \lambda^2(1- \mathbb{P}\left(\left|X-\mathrm{E}(X)\right| \leq \lambda\right) \\ &&&= (k^2 - \lambda^2) \mathbb{P}\left(\left|X-\mathrm{E}(X)\right|\geqslant\lambda\right) + \lambda^2 \\ \Rightarrow&& \frac{\mathrm{Var}(X)-\lambda^2}{k^2 - \lambda^2} &\leq \mathbb{P}\left(\left|X-\mathrm{E}(X)\right|\geqslant\lambda\right) \end{align*} [Note: This result is known as Chebyshev's inequality, and is an important starting point to understanding the behaviour of tails of random variables]

Solution: The position of the particle is projected at angle \(\theta\) is \((x,y) = (v \cos \theta t, v \sin \theta t - \frac12 g t^2)\), ie \(t = \frac{x}{v \cos \theta}\), \begin{align*} && y &= x\tan \theta -\frac12 g \frac{x^2}{v^2} \sec^2 \theta \\ && y &= x \tan \theta -\frac{1}{4h} (1+\tan^2 \theta) x^2 \\ && 0 &= \frac{1}{4h} x^2\tan^2 \theta - x \tan \theta + \frac{x^2}{4h} +y \\ \Delta \geq 0: && 0 &\leq \Delta = x^2-4\frac{x^2}{4h}\left (\frac{x^2}{4h}+y \right) \\ &&&=1-\frac{1}{4h^2}(x^2+4hy) \\ \Rightarrow && x^2+4hy &\leq 4h^2 \\ \Rightarrow && x^2 &\leq 4h(h-y) \end{align*} We are assuming that there are no forces acting other than gravity (eg air resistance)

Solution:

1. \(\,\) \begin{align*} && \sqrt{p(1-p)} &= \sqrt{p-p^2} \\ &&&= \sqrt{\tfrac14-(\tfrac12-p)^2} \\ &&&\leq \sqrt{\tfrac14} = \tfrac12 \end{align*} Therefore the maximum is \(\tfrac12\) when \(p=\frac12\)
2. Notice that our estimate \(\hat{p}\) will (for large \(n\)) be follow a normal distribution \(N(p, pq/n)\) by either the normal approximation to the binomial or central limit theorem. We would like \(0.01 > \mathbb{P}\left ( |\hat{p}-p| < 0.01 \right)\) or in other words \begin{align*} && 0.01 &> \mathbb{P}\left ( |\hat{p}-p| > 0.01 \right) \\ &&&=\mathbb{P}\left ( |\sqrt{\frac{pq}{n}}Z+p-p| > 0.01 \right) \\ &&&= \mathbb{P} \left (|Z|>\frac{0.01\sqrt{n}}{\sqrt{pq}}\right) \end{align*} therefore we need \(\frac{0.01\sqrt{n}}{\sqrt{pq}}> 2.58 \Rightarrow n > 258^2 pq \approx 2^{14} \approx 16\,000\), where we are using \(pq = \frac14\) as the worst case possibility and \(258 \approx 256 = 2^8\)
3. If we were looking at when we are looking at left handed people (maybe ~\(10\%\), we would be looking at \(pq = \frac{9}{100}\) so we need a smaller sample). If we are looking at millionaires (an even smaller again percentage), we would need an even smaller sample. This is surprising since you would expect you would need a larger sample to accurately gauge smaller proportions. However, this surprise can be resolved by considering that this is an absolute error. For smaller values the relative error is larger, but the absolute error is smaller.

Solution: \begin{align*} && u_n &= \int_0^{\tfrac12 \pi} \sin^{n} t \, \d t \\ && &= \int_0^{\tfrac12 \pi} \sin t \sin^{n-1} t \, \d t \\ && &= \left [ -\cos t \sin^{n-1} t \right]_0^{\tfrac12 \pi} + \int_0^{\tfrac12 \pi} \cos t (n-1) \sin^{n-2} t \cos t \d t \\ && &= 0 + (n-1)\int_0^{\tfrac12 \pi} \cos^2 t \sin^{n-2} t \d t \\ && &= (n-1) \int_0^{\tfrac12 \pi}(1-\sin^2 t) \sin^{n-2} t \d t \\ && &= (n-1)u_{n-2} - (n-1)u_n \\ \Rightarrow && n u_n &= (n-1)u_{n-2} \\ \end{align*} Mutplying both sides by \(u_{n-1}\) we obtain \(nu_{n}u_{n-1}=\left(n-1\right)u_{n-1}u_{n-2}\). Therefore \(nu_nu_{n-1}\) is constant, ie is equal to \(\displaystyle u_1u_0 = \int_0^{\tfrac12 \pi} \sin^{1} t \, \d t \int_0^{\tfrac12 \pi} \sin^{0} t \, \d t = 1 \cdot \frac{\pi}{2} = \frac{\pi}{2}\)

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