Problems

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Solution: \begin{align*} \sum_{m=1}^n a_m(b_{m+1} -b_m) +\sum_{m=1}^n b_{m+1}(a_{m+1} -a_m) &= \sum_{m=1}^n \left (a_{m+1}b_{m+1}-a_mb_m \right) \\ &= a_{n+1}b_{n+1} - a_1b_1 \end{align*} And the result follows.

1. Let \(b_m = \sin m x \), \(a_m = \cosec \frac{x}{2}\), so \begin{align*} && \sum_{m=1}^n \cosec \frac{x}{2} \left (\sin (m+1)x - \sin mx \right) &= \sum_{m=1}^n \cosec \frac{x}{2} 2 \cos \left ( \frac{2m+1}{2}x \right) \sin \left ( \frac{(m+1)-m}{2}x \right) \\ &&&=2 \sum_{m=1}^n\cos \left ( (m + \tfrac12)x \right)\\ \\ \Rightarrow && \sum_{m=1}^n\cos \left ( (m + \tfrac12)x \right) &= \tfrac12 \cosec \tfrac{x}{2}\left ( \sin(n+1)x - \sin x \right) \end{align*}
2. \(\,\) \begin{align*} && b_{m+1}-b_m &= \sin m x \sin \tfrac12 x \\ &&&= \frac12 \left ( \cos (m-\tfrac12)x - \cos (m+\tfrac12)x \right)\\ \Rightarrow && b_m &= -\tfrac12 \cos (m - \tfrac12)x\\ && a_m &= m \\ \Rightarrow && \sum_{m=1}^n m \sin m x \sin \tfrac12 x &= (n+1) b_{n+1} - 1 \cdot b_1 - \sum_{m=1}^n b_{m+1} \cdot 1 \\ &&&= -(n+1) \tfrac12\cos(n+1-\tfrac12)x+\tfrac12\cos(\tfrac12x) + \tfrac12\sum_{m=1}^n \cos(m+\tfrac12)x \\ &&&= -(n+1) \tfrac12\cos(n+1-\tfrac12)x+\tfrac12\cos(\tfrac12x) + \tfrac14 \cosec \tfrac{x}{2}\left ( \sin(n+1)x - \sin x \right) \\ &&&= -(n+1) \tfrac12\cos(n+1-\tfrac12)x+ \tfrac14 \cosec \tfrac{x}{2}\sin(n+1)x \\ &&&= \tfrac12\cosec\tfrac{x}2 \left (\tfrac12 \sin (n+1)x-(n+1)\cos(n+\tfrac12)x\sin\tfrac12x \right) \\ &&&= \tfrac12\cosec\tfrac{x}2 \left (\tfrac12 \sin (n+1)x-(n+1)\tfrac12 \left ( \sin (n+1)x - \sin nx \right) \right) \\ &&&= \tfrac14 \cosec \tfrac{x}{2} \left ( -n \sin (n+1)x +(n+1) \sin n x \right) \end{align*} Therefore \(p = -\frac{n}4, q = \frac{n+1}{4}\)

Notice the connection here to integration by parts.

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Solution:

By energy considerations, the initial energy is \(0\). Therefore: \begin{align*} && \l 3a^2 + \ell^2\r \dot{\theta}^2 &= \l 3a + 2\ell \r g \sin \theta \\ \Rightarrow && \l 3a^2 + \ell^2\r 2\dot{\theta} \ddot{\theta} &= \l 3a + 2\ell \r g \cos\theta \dot{\theta} \tag{\(\frac{\d}{\d t}\)} \\ \Rightarrow && 2\l 3a^2 + \ell^2\r \ddot{\theta} &= \l 3a + 2\ell \r g \cos\theta \\ \Rightarrow && \ddot{\theta} &= \boxed{\frac{3a + 2\ell }{2(3a^2 + \ell^2)}g \cos\theta} \\ \end{align*} \begin{align*} \text{N}2(\perp PQ): && mg \cos \theta - R &= m \ell \ddot{\theta} \\ && R &= mg \cos \theta - m \ell \l \frac{3a + 2\ell }{2(3a^2 + \ell^2)}g \cos\theta \r \\ && &= mg\cos \theta \l 1 - \ell \frac{3a + 2\ell }{2(3a^2 + \ell^2)} \r \\ && &= mg \cos \theta \l \frac{6a^2 + 2\ell^2 - 3a\ell - 2\ell^2}{2(3a^2 + \ell^2)} \r \\ && &= mg \cos \theta \l \frac{3a(2a - \ell)}{2(3a^2 + \ell^2)} \r > 0 \tag{since \(2a > \ell\)} \end{align*} At limiting equilibrium, \(F = \mu R\). \begin{align*} \text{N}2(\parallel PQ): && \mu R - mg \sin \theta &= m \ell \dot{\theta}^2 \\ \Rightarrow && \mu mg \cos \theta \l \frac{3a(2a - \ell)}{2(3a^2 + \ell^2)} \r - mg \sin \theta &= m \ell \frac{(3a+2\ell)}{(3a^2+\ell^2)} g \sin \theta \\ \Rightarrow && \mu \l 3a(2a - \ell) \r - \l 2(3a^2 + \ell^2) \r \tan \theta &= 2\ell (3a+2\ell) \tan \theta \\ \Rightarrow && \mu \l 3a(2a - \ell) \r &= \l 6a\ell + 6a^2 + 6\ell^2 \r \tan \theta \\ \Rightarrow && \tan \theta &= \boxed{\frac{\mu a(2a-\ell)}{2(a^2 + a\ell + \ell^2)}} \end{align*} If \(\ell > 2a\), then the initial reaction force will be \(0\), ie the particle will have no contact with the rod. In other words, the rod will rotate faster than the particle will free-fall and the particle immediately loses contact with the rod.

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Solution:

1. \(\,\) \begin{align*} && \mathbb{P}(X = x) &= \sum_{y=1}^n \mathbb{P}(X=x,Y=y) \\ &&&= \sum_{y=1}^n k(x+y) \\ &&&= nkx + k\frac{n(n+1)}2 \\ \\ && 1 &= \sum_{x=1}^n \mathbb{P}(X=x) \\ &&&= nk\frac{n(n+1)}{2} + kn\frac{n(n+1)}2 \\ &&&= kn^2(n+1) \\ \Rightarrow && k &= \frac{1}{n^2(n+1)} \\ \Rightarrow && \mathbb{P}(X = x) &= \frac{nx}{n^2(n+1)} + \frac{n(n+1)}{2n^2(n+1)} \\ &&&= \frac{n+1+2x}{2n(n+1)} \\ \\ && \mathbb{P}(X=x)\mathbb{P}(Y=y) &= \frac{(n+1)^2+2(n+1)(x+y)+4xy}{4n^2(n+1)^2} \\ &&&\neq \frac{x+y}{n^2(n+1)} \end{align*} Therefore \(X\) and \(Y\) are not independent.
2. \(\,\) \begin{align*} && \E[X] &= \sum_{x=1}^n x \mathbb{P}(X=x) \\ &&&= \sum_{x=1}^n x \mathbb{P}(X=x)\\ &&&= \sum_{x=1}^n x \frac{n+1+2x}{2n(n+1)} \\ &&&= \frac{1}{2n(n+1)} \left ( (n+1) \sum x + 2\sum x^2\right)\\ &&&= \frac{1}{2n(n+1)} \left ( \frac{n(n+1)^2}{2} + \frac{n(n+1)(2n+1)}{3} \right) \\ &&&= \frac{1}{2} \left ( \frac{n+1}{2} + \frac{2n+1}{3} \right)\\ &&&= \frac{1}{2} \left ( \frac{7n+5}{6} \right)\\ &&&= \frac{7n+5}{12} \\ \\ && \textrm{Cov}(X,Y) &= \mathbb{E}\left[XY\right] - \E[X] \E[Y] \\ &&&= \sum_{x=1}^n \sum_{y=1}^n xy \frac{x+y}{n^2(n+1)} - \E[X]^2 \\ &&&= \frac{1}{n^2(n+1)} \sum \sum (x^2 y+xy^2) - \E[X]^2 \\ &&&= \frac{1}{n^2(n+1)} \left (\sum y \right )\left (\sum x^2\right ) - \E[X]^2 \\ &&&=\frac{(n+1)(2n+1)}{12} - \left ( \frac{7n+5}{12}\right)^2 \\ &&&= \frac1{144} \left (12(2n^2+3n+1) - (49n^2+70n+25) \right)\\ &&&= \frac{1}{144} \left (-25n^2-34n-13 \right) \\ &&& < 0 \end{align*} since \(\Delta = 34^2 - 4 \cdot 25 \cdot 13 = 4(17^2-25 \times 13) = -4 \cdot 36 < 0\)

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Solution: \begin{align*} {\rm V}(x) &= \E \big( (X-x)^2\big) \\ &= \E \l X^2 - 2xX + x^2\r \\ &= \E [ X^2 ]- 2x\E[X] + x^2 \\ &= \sigma^2+\mu^2 - 2x\mu + x^2 \\ &= \sigma^2 + (\mu - x)^2 \end{align*} \begin{align*} \E[Y] &= \E[\sigma^2 + (\mu - X)^2] \\ &= \sigma^2 + \E[(\mu - X)^2]\\ &= \sigma^2 + \sigma^2 \\ &= 2\sigma^2 \end{align*} If \(X \sim U(0,1)\) then \(V(x) = \frac{1}{12} + (\frac12 - x)^2\). \begin{align*} \P(Y \leq y) &= \P(\frac1{12} + (\frac12 - X)^2 \leq y) \\ &= \P((\frac12 -X)^2 \leq y - \frac1{12}) \\ &= \P(|\frac12 -X| \leq \sqrt{y - \frac1{12}}) \\ &= \begin{cases} 1 & \text{if } y - \frac1{12} > \frac14 \\ 2 \sqrt{y - \frac1{12}} & \text{if } \frac14 > y - \frac1{12} > 0 \\ \end{cases} \\ &= \begin{cases} 1 & \text{if } y> \frac13 \\ \sqrt{4y - \frac1{3}} & \text{if } \frac13 > y > \frac1{12} \\ \end{cases} \end{align*} Therefore $f_Y(y) = \begin{cases} \frac{2}{\sqrt{4y-\frac{1}{3}}} & \text{if } \frac1{12} < y < \frac13 \\ 0 & \text{otherwise} \end{cases}$ \begin{align*} \E[Y] &= \int_{1/12}^{1/3} \frac{2x}{\sqrt{4x-\frac13}} \, dx \\ &= 2\int_{u = 0}^{u=1} \frac{\frac{1}{4}u +\frac1{12}}{\sqrt{u}} \,\frac{1}{4} du \tag{\(u = 4x - \frac13, \frac{du}{dx} = 4\)}\\ &= \frac{1}{2 \cdot 12}\int_{u = 0}^{u=1} 3\sqrt{u} +\frac{1}{\sqrt{u}} \, du \\ &= \frac{1}{2 \cdot 12} \left [2 u^{3/2} + 2u^{1/2} \right ]_0^1 \\ &= \frac{1}{2 \cdot 12} \cdot 4 \\ &= \frac{2}{12} \end{align*} as required

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Solution:

1. \(n=1\): \begin{align*} && p_1(x) &= (x+1)^2 - 3x(x^2+x+1)^0 \\ &&&= x^2+2x+1-3x \\ &&&= x^2-x+1\\ && q_1(x) &= \frac{x^3+1}{x+1} \\ &&&= x^2-x+1 = p_1(x) \\ \\ && p_2(x) &= (x+1)^4-5x(x^2+x+1)^1 \\ &&&= x^4+4x^3+6x^2+4x+1 - 5x^3-5x^2-5x \\ &&&= x^4-x^3+x^2-x+1 \\ &&q_2(x) &= \frac{x^5+1}{x+1} \\ &&&= x^4-x^3+x^2-x+1 = p_2(x) \\ \\ && p_3(x) &= (x+1)^6-7x(x^2+x+1)^2 \\ &&&= x^6+6x^5+15x^4+20x^3+15x^2+6x+1 - 7x(x^4+2x^3+3x^2+2x+1) \\ &&&= x^6-x^5+x^4-x^3+x^2-x+1 \\ && q_3(x) &= \frac{x^7+1}{x+1} \\ &&&= x^6-x^5+x^4-x^3+x^2-x+1 = p_3(x) \\ \\ && p_4(1) &= 2^8 - 9 \cdot 1 \cdot 3^3 \\ &&&= 256 - 243 = 13 \\ && q_4(1) &= \frac{2}{2} = 1 \neq 13 \end{align*}
2. • [\bf (a)] \(\,\) \begin{align*} && \frac{300^3+1}{300+1} &= (300+1)^2 - 3 \cdot 300 \\ &&&= 301^2 - 30^2 \\ &&&= 271 \cdot 331 \end{align*}
   • [\bf (b)] \(\,\) \begin{align*} && \dfrac {7^{49}+1}{7^7+1} &= (7^7+1)^6 - 7 \cdot 7^7 \cdot (7^2+7+1)^2 \\ &&&= (7^7+1)^6 - 7^8 \cdot (7^2+7+1)^2 \\ &&&= ((7^7+1)^3 - 7^4(7^2+7+1)) \cdot ((7^7+1)^3 + 7^4(7^2+7+1)) \end{align*}

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Solution: \begin{align*} && y &= (ax^2+bx+c)\,\ln \big( x+\sqrt{1+x^2}\big) +\big(dx+e\big)\sqrt{1+x^2} \\ && y' &= (2ax+b)\,\ln \big( x+\sqrt{1+x^2}\big) + (ax^2+bx+c) \frac{1}{x + \sqrt{1+x^2}} \cdot \left(1 + \frac{x}{\sqrt{1+x^2}} \right) + d\sqrt{1+x^2} + \frac{x(dx+e)}{\sqrt{1+x^2}} \\ &&&= (2ax+b)\,\ln \big( x+\sqrt{1+x^2}\big) + \frac{1}{\sqrt{1+x^2}} \left ( (ax^2+bx+c) + d(1+x^2) + x(dx+e) \right) \\ &&&= (2ax+b)\,\ln \big( x+\sqrt{1+x^2}\big) + \frac{1}{\sqrt{1+x^2}} \left ( (a+2d)x^2+(b+e)x+(d+c) \right) \\ \end{align*}

1. We want \(a = 0, b = 1, d = 0, e = -1, c =0\), so \begin{align*} I &= \int \ln \big( x+\sqrt{1+x^2}\,\big) \,\d x \\ &= x\ln(x+\sqrt{1+x^2})-\sqrt{1+x^2}+C \end{align*}
2. We want \(a = b =0, e = 0, d = \frac12, c = \frac12\), so \begin{align*} I &= \int \sqrt{1+x^2}\, \d x \\ &= \frac12\ln(x+\sqrt{1+x^2}) + \frac12x\sqrt{1+x^2}+C \end{align*}
3. We want \(a = \frac12, b = 0, d = -\frac14, e = 0, c = \frac14\), so \begin{align*} I &= \int x \ln (x+\sqrt{1+x^2}) \, \d x \\ &= \left (\frac12 x^2+\frac14 \right)\ln(x+\sqrt{1+x^2}) -\frac14x\sqrt{1+x^2}+C \end{align*}

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Solution:

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Solution:

1. Let \(\displaystyle y = \frac z {(1+z^2)^{\frac12}}\) then \(\frac{d y}{d x} = \frac{(1+z^2)^{\frac12} - z^2(1+z^2)^{-\frac12}}{1+z^2} = \frac{(1+z^2)-z^2}{(1+z^2)^\frac32} = \frac{1}{(1+z^2)^\frac32}\)
2. \(\kappa = \frac {f''(x)}{\big({1+ (f'(x))^2\big)^{\frac32}}}\) then \begin{align*} && \int \kappa \, dx &= \int \frac{f''(x)}{( 1 + (f'(x))^2)^{\frac32}} \, dx \\ && \kappa x &= \frac{f'(x)}{(1 + (f'(x))^2)^\frac12} + C \\ \Rightarrow && (\kappa x-C)^2 &= \frac{f'(x)^2}{1 + (f'(x))^2} \\ \Rightarrow && f'(x)^2((\kappa x - C)^2 - 1) &= -(\kappa x-C)^2 \\ \Rightarrow && f'(x) &= \frac{\kappa x - C}{\sqrt{1-(\kappa x - C)^2 }} \\ \Rightarrow && f(x) &= \frac{1}{\kappa} \sqrt{1 - (\kappa x - C)^2} \\ \Rightarrow && (\kappa y)^2 + (\kappa x - C)^2 &= 1 \\ \Rightarrow && y^2 + (x - C')^2 &= \frac{1}{\kappa^2} \end{align*} Therefore all the curves are circles and \(\kappa\) is the reciprocal of the radius.