Problems

Solution:

1. \(\,\)
   

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   Note that we have a right angled triangle, with the sides in a ratio of \(m\). So if our target length is \(x\) we have \(x^2 + (mx)^2 = c^2\) and so \(x = c(m^2+1)^{-\frac12}\)
2. The distance from the origin to \(L\) is \(a = c(m^2+1)^{-\frac12}\) so \begin{align*} && a^2(m^2+1) &= c^2 \\ && \frac{c-y(x)}{0-x} &= y' \\ \Rightarrow && c-y &= -xy' \\ \Rightarrow && a^2((y')^2+1) &= (y-xy')^2 \\ \\ && 2a^2y'y'' &= 2(y-xy')(y'-xy''-y') \\ &&&= 2(xy'-y)xy'' \\ \Rightarrow && y'' &= 0 \\ \text{ or } && 2a^2y' &= 2(xy'-y)x \end{align*} If \(y'' = 0\) then \(y = mx + c\) and the result follows immediately. \begin{align*} && 0 &= (a^2-x^2)y' + yx \\ \Rightarrow &&\frac1{y} y' &= -\frac{x}{a^2-x^2} \\ \Rightarrow && \ln y &= \frac12\ln (a^2-x^2) + K \\ \Rightarrow && y^2 &= M(a^2-x^2) \\ \Rightarrow && x^2 + y^2 &= a^2 \end{align*} Where in the last step we know the tangents from an ellipse are not all equidistant to the origin.
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Solution:

1. \begin{align*} && I &= \int_{1/b}^b \frac{x \ln x}{(a^2+x^2)(a^2x^2+1} \d x \\ u = 1/x, \d u = -1/x^2 \d x: &&&= \int_{u=b}^{u=1/b} \frac{1/u \ln(1/u)}{(a^2+u^{-2})(a^2u^{-2}+1)} (- \frac{1}{u^2}) \d u \\ &&&= \int_{1/b}^b \frac{-u\ln u}{(a^2u^2+1)(a^2+u^2)} \d u \\ &&&= -I \\ \Rightarrow && I &= 0 \end{align*}
2. \(\,\) \begin{align*} && I &= \int_{1/b}^b \frac{\arctan x}{x} \d x \\ u = 1/x, \d x = -1/u^2 \d u: &&&= \int_{u=b}^{u=1/b} \frac{\arctan \frac1u}{\frac1u} \frac{-1}{u^2} \d u \\ &&&= \int_{1/b}^b \frac{\arctan \frac1u}{u} \d u \\ \Rightarrow && 2I &= \int_{1/b}^b \frac{\arctan x + \arctan \frac1x}{x} \d x \\ &&&= \int_{1/b}^b \frac{\frac{\pi}2}{x} \d x \\ &&&= \pi \ln b \\ \Rightarrow && I &= \frac{\pi}{2} \ln b \end{align*}
3. \(\,\) \begin{align*} && I_a &= \int_0^{\infty} \frac{1}{(a^2+x^2)^2} \d x \\ u = a/x, \d x = -a/u^2 \d u:&&&= \int_{u=0}^{u=\infty} \frac{1}{\left (a^2+\frac{a^2}{u^2} \right)^2} \frac{a}{u^2} \d u \\ &&&= \frac1{a^3}\int_0^{\infty} \frac{1}{(u+1/u)^2} \d u \\ &&&= \frac{1}{a^3} \int_0^{\infty} \frac{u^2}{(u^2+1)^2} \d u \\ &&&= \frac{1}{a^3} \int_0^{\infty} \frac{u^2+1-1}{(u^2+1)^2} \d u \\ &&&= \frac{1}{a^3} \int_0^{\infty} \frac{1}{(u^2+1)} - \frac{1}{(u^2+1)^2} \d u \\ &&&= \frac1{a^3} \frac{\pi}{2} - \frac{1}{a^3} I_1 \\ \Rightarrow && 2I_1 &= \frac{\pi}{2} \\ \Rightarrow && I_1 &= \frac{\pi}{4} \\ \Rightarrow && I_a &= \frac{\pi}{4a^3} \end{align*}

Solution: \begin{align*} \frac{\d y}{\d x} &= \frac{\d }{\d x} \l y \r \\ &= \frac{\d }{\d x} \l xu \r \\ &\underbrace{=}_{\text{product rule}} \frac{\d}{\d x} \l x \r u + x \frac{\d}{\d x} \l u \r \\ &= u + x \frac{\d u}{\d x} \end{align*} \begin{questionparts} \item \begin{align*} && \frac{\d y}{\d x} &= \frac{2y + x}{y - 2x} \\ && u + x \frac{\d u}{\d x} &= \frac{2u + 1}{u - 2} \\ && x \frac{\d u}{\d x} &= \frac{2u-1-u^2+2u}{u-2} \\ \Rightarrow && \int \frac{2-u}{u^2-4u+1} \d u &= \int \frac{1}{x} \d x \\ && \int \frac{2-u}{(u-2)^2-5} \d u &= \int \frac1x \d x \\ && -\frac12\ln| (u-2)^2 - 5| &= \ln x + C \\ (x,y) = (1,1): && - \ln 2 &= C \\ \Rightarrow && \ln x^2 &= \ln 4 - \ln |5 - (u-2)^2| \\ \Rightarrow && x^2 &= \frac{4}{5- (u-2)^2} \\ \Rightarrow && 4 & = x^2(5 - (\frac{y}{x} - 2)^2) \\ &&&= 5x^2 - (y-2x)^2 \\ &&&= x^2+4xy-y^2 \end{align*} \item It would be convienient if \(x-2y -4 = X-2Y\) and \(2x+y-3 = 2X+Y\), ie \(a-2b = 4\) and \(2a+b = 3\), ie \(a = 2, b = -1\). Now our differential equation is: \begin{align*} && \frac{\d Y}{\d X} &= \frac{X - 2Y}{2X+Y} \\ && \frac{\d X}{\d Y} &= \frac{2X + Y}{X-2Y} \end{align*} This is the same differential equation we have already solved, just with the roles of \(x\) and \(y\) interchanged with \(Y\) and \(X\) and with the point \((0,3)\) being on the curve, ie: \(Y^2 + 4XY-X^2 = c\) and \(c = 9\), therefore our equation is: \[ (y-1)^2 + 4(y-1)(x+2)-(x+2)^2 = 9\]

Solution: \begin{align*} && \sin(r + \tfrac12)x - \sin(r - \tfrac12) x &= \sin rx \cos \tfrac12x + \cos r x\sin\tfrac12x - \sin r x \cos \tfrac12 x + \cos rx \sin \tfrac12 x \\ &&&= 2\cos r x \sin\tfrac12 x \\ \\ && S &= \cos x + \cos 2x + \cdots + \cos n x \\ && 2\sin \tfrac12 x S &= \sin(1 + \tfrac12)x - \sin \tfrac12 x + \\ &&&\quad+ \sin(2+\tfrac12)x - \sin(2- \tfrac12)x + \\ &&&\quad+ \sin(3+\tfrac12)x - \sin(3 - \tfrac12)x + \\ &&& \quad + \cdots + \\ &&&\quad + \sin(n+\tfrac12)x - \sin(n-\tfrac12)x \\ &&&=\sin(n+\tfrac12)x - \sin\tfrac12 x \\ \Rightarrow && S &= \frac{\sin(n+\tfrac12)x - \sin\tfrac12 x}{2 \sin \tfrac12 x} \end{align*}

1. \(\,\) \begin{align*} && S_2(x) &= \sin x + \tfrac12 \sin 2 x \\ && S'_2(x) &= \cos x + \cos 2x \\ &&&= \cos x + 2\cos^2 x - 1 \\ &&&= (2\cos x -1)(\cos x + 1) \\ \end{align*} Therefore the turning points are \(\cos x= \frac12 \Rightarrow x = \frac{\pi}{3}\) and \(\cos x = -1 \Rightarrow x = \pi\)
   

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2. Suppose \(S_n(x)\) has a stationary point at \(x_0\), then $$ therefore \begin{align*} &&0 &= S_n'(x_0) \\ &&&= \cos x_0 + \cos 2x_0 + \cdots + \cos n x_0 \\ &&&= \frac{\sin(n+\tfrac12)x_0 - \sin \tfrac12x_0}{2 \sin \tfrac12 x_0} \\ \Rightarrow &&\sin\tfrac12 x_0&= \sin nx_0 \cos \tfrac12 x_0 + \cos nx_0 \sin \tfrac12x_0 \\ \Rightarrow && \sin nx_0 &= (1-\cos nx_0)\tan \tfrac12 x_0 \end{align*} Therefore \(S_n(x_0) -S_{n-1}(x_0) = \tfrac1n \sin n x_0 = \tfrac1n \underbrace{(1-\cos nx_0)}_{\geq 0}\underbrace{\tan\tfrac12 x_0}_{\geq 0} \geq 0\). Therefore if \(S_{n-1}(x) > 0\) for all \(x\) on \(0 < x < \pi\) then since \(S_n(x) > S_{n-1}(x)\) at the turning points and since they agree at the end points, it must be larger at all points inbetween.
3. Notice that \(S_1(x) = \sin x \geq 0\) for all \(x \in [0,1]\) and by our previous argument we can show \(S_n > S_{n-1}\) inside the interval and equal on the boundary we must have \(S_n(x) \geq 0\) for \(x \in [0, \pi]\)

Solution:

1. \(\,\)
   

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   \((a,0)\), \((b,0)\), \((b,\f(b))\) and \((a, \f(a))\) forms a rectangle.
2. There are no solutions if \(a < c < b\):
   

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   There is one solution if \(a=c\) or \(a = b\)
   

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   And there are two solution if \(c \not \in [a,b]\)
   

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   There is exactly one solution unless....
   

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   ... there are infinitely many solutions when the gradients line up perfectly, ie when \(a+b=c+d\)
   

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Solution: \(c_r = \binom{n}{r}a^{n-r}b^r\) \begin{align*} && c_m &\geq c_{m+1} \\ \Rightarrow && \binom{n}{m} a^{n-m}b^m &\geq \binom{n}{m+1} a^{n-m-1}b^{m+1} \\ \Rightarrow && \frac{1}{(n-m)}a &\geq \frac{1}{m+1}b \\ \Rightarrow && (m+1)a &\geq (n-m)b \\ \Rightarrow && m(a+b) &\geq nb -a \\ \Rightarrow && m &\geq \frac{n(b+1)-a-b}{a+b}=\frac{n(b+1)}{a+b} - 1 \\ \\ && c_m &\geq c_{m-1} \\ \Rightarrow && \binom{n}{m} a^{n-m}b^m &\geq \binom{n}{m-1} a^{n-m+1}b^{m-1} \\ \Rightarrow && \frac{1}m b &\geq \frac{a}{(n-m+1)} \\ \Rightarrow && (n-m+1)b &\geq ma \\ \Rightarrow && (n+1)b &\geq m(a+b) \\ \Rightarrow && m &\leq \frac{(n+1)b}{a+b} \end{align*} Since \(m\) lies between two values \(1\) apart is is either equal to one of those two values or is the unique integer between them. Let \(\displaystyle G(n,a,b) = \left \lfloor \frac{b(n+1)}{a+b} \right \rfloor\), so

1. \(\,\) \begin{align*} && G(9,1,3) &= \left \lfloor \frac{3(9+1)}{1+3} \right \rfloor \\ &&&= \left \lfloor \frac{30}{4} \right \rfloor \\ &&&= 7 \\ \\ && G(9,2,3) &= \left \lfloor \frac{3(9+1)}{2+3} \right \rfloor \\ &&&= \left \lfloor \frac{30}{5} \right \rfloor \\ &&&= 6 \end{align*}
2. \(\,\) \begin{align*} && G(2k, a, a) &= \left \lfloor \frac{a(2k+1)}{a+a} \right \rfloor \\ && &= \left \lfloor \frac{2k+1}{2} \right \rfloor \\ &&&= k \\ \\ && G(2k-1, a, a) &= \left \lfloor \frac{a(2k-1+1)}{a+a} \right \rfloor \\ && &= \left \lfloor k\right \rfloor \\ &&&= k \\ \end{align*}
3. \(G(n,a,b)\) is decreasing in \(a\), therefore take \(a = 1\).
4. For fixed \(n\), we are looking at \(\frac{a(n+1)}{a+b}\) and we want this to be as large as possible. By considering \((n+1) - \frac{b(n+1)}{a+b}\) we can see this is increasing in \(b\) and the largest value possible is \(n\). This is achieved when \begin{align*} && \frac{b(n+1)}{a+b} & \geq n \\ \Leftrightarrow && bn + b &\geq na + bn \\ \Leftrightarrow && b& \geq na \end{align*}

Solution:

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1. \begin{align*} \text{N2}(\uparrow): && T \cos \theta + F -W &= 0 \\ && W &= T\cos \theta + \mu R \tag{1} \\ \text{N2}(\rightarrow): && R-T\sin \theta &= 0 \\ && R &= T \sin \theta \tag{2}\\ \\ (1)+(2): && W&=(\cos \theta + \mu \sin \theta)T \tag{3} \\ \overset{\curvearrowright}{A}: && 0 &= W(b\sin \phi + a \cos \phi) - Td\sin(\phi+\theta) \tag{4} \\ \\ (3)+(4): && 0 &= (\cos \theta + \mu \sin \theta)(b\sin \phi + a \cos \phi)-d\sin(\phi+\theta) \\ \Rightarrow && d\sin(\phi+\theta) &= (\cos \theta + \mu \sin \theta)(b\sin \phi + a \cos \phi) \end{align*} as required.
2. If \(F\) is operating downwards, it's equivalent to \(-\mu\), ie: \[d\sin(\phi+\theta) = (\cos \theta - \mu \sin \theta)(b\sin \phi + a \cos \phi)\]
3. For the frictional force to be acting upwards, we need \begin{align*} && d\sin(\phi+\theta) &\geq \cos \theta(b\sin \phi + a \cos \phi) \\ \Rightarrow && d &\geq \frac{\cos \theta(b\sin \phi + a \cos \phi)}{\sin(\phi + \theta)} \\ &&&= \frac{\cos \theta(b\sin \phi + a \cos \phi)}{\sin\phi \cos\theta+\cos\phi\sin \theta)}\\ &&&= \frac{(b\sin \phi + a \cos \phi)}{\sin\phi+\cos \phi \tan \theta)}\\ &&&= \frac{a+b\tan \phi}{\tan\theta+\tan\phi }\\ \end{align*} We know that \(d < 2b\), so \begin{align*} && 2b &>\frac{a+b\tan \phi}{\tan\theta+\tan\phi }\\ \Rightarrow && 2b \tan \theta + 2b \tan \phi &> a + b \tan \phi \\ \Rightarrow &&b(2 \tan \theta + \tan \phi) &> a\\ \end{align*} Therefore we will have problems if the inequality is reversed!

Solution:

1. \(\,\) \begin{align*} && \mathbb{P}(T > t + \delta t | T > t) &= \frac{\mathbb{P}(T < t + \delta t)}{\mathbb{P}(T > t )} \\ &&&= \frac{\int_t^{t+\delta t} f(s) \d s}{1-F(t)} \\ &&&\approx \frac{f(t)\delta t}{1-F(t)} \\ &&&= h(t) \delta t \end{align*}
2. If \(F(t) = t/a\) then \(f(t) = 1/a\) and \(h(t) = \frac{1/a}{1-t/a} = \frac{1}{a-t}\)
   

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3. \(\,\) \begin{align*} && \frac{F'}{1-F} &= \frac{1}{t} \\ \Rightarrow && -\ln (1-F) &= \ln t + C\\ \Rightarrow && 1-F &= \frac{A}{t} \\ && F &= 1 - \frac{A}{t} \\ F(a) = 0: && F &= 1 - \frac{a}{t} \\ && f(t) &= \frac{a}{t^2} \end{align*}
4. (\(\Rightarrow\)) \begin{align*} && \frac{F'}{1-F} &= k \\ \Rightarrow && -\ln(1-F) &= kt+C \\ \Rightarrow && 1-F &= Ae^{-kt} \\ F(b) = 0: && 1 &= Ae^{-kb} \\ \Rightarrow && 1-F &= e^{-k(t-b)}\\ \Rightarrow && f &= ke^{-k(t-b)} \\ \end{align*} (\(\Leftarrow\)) \(f(t) = ke^{-k(t-b)} \Rightarrow F(t) = 1-e^{-k(t-b)}\) and the result is clear.
5. \(\,\) \begin{align*} && \frac{F'}{1-F} &= \left ( \frac{\lambda}{\theta^{\lambda}} \right) t^{\lambda-1} \\ \Rightarrow && -\ln(1-F) &= \left ( \frac{t}{\theta} \right)^{\lambda} +C\\ \Rightarrow && F &= 1-A\exp \left (- \left ( \frac{t}{\theta} \right)^{\lambda} \right) \\ F(0) = 0: && 0 &= 1-A \\ \Rightarrow && F &= 1 - \exp \left (- \left ( \frac{t}{\theta} \right)^{\lambda} \right) \\ \Rightarrow && f &= \lambda t^{\lambda -1} \theta^{-\lambda} \exp \left (- \left ( \frac{t}{\theta} \right)^{\lambda} \right) \end{align*}

Solution: \begin{align*} && \mathbb{P}(X > 4) &= 1 \cdot \frac{n-1}{n} \cdot \frac{n-2}{n} \cdot \frac{n-3}{n} \\ && \mathbb{P}(X > 3) &= 1 \cdot \frac{n-1}{n} \cdot \frac{n-2}{n} \\ \Rightarrow && \mathbb{P}(X =4) &= \mathbb{P}(X > 3) - \mathbb{P}(X > 4) \\ &&&= \frac{(n-1)(n-2)}{n^2} \left (1 - \frac{n-3}{n} \right) \\ &&&= \frac{3(n-1)(n-2)}{n^3} \end{align*}

1. Notice that \begin{align*} && \mathbb{P}(X > r) &= \frac{n-1}{n} \cdots \frac{n-r+1}{n} \\ \Rightarrow && \mathbb{P}(X = r) &= \frac{n-1}{n} \cdots \frac{n-r+2}{n} \left (1 - \frac{n-r+1}{n} \right) \\ &&&= \frac{(n-1)\cdots(n-r+2)(r-1)}{n^{r-1}} \\ &&&= \left (1 - \frac{1}n \right)\left (1 - \frac{2}{n} \right) \cdots \left (1 - \frac{r-2}{n} \right) \frac{r-1}{n} \\ \Rightarrow && 1 &= \sum \mathbb{P}(X = r) \\ &&&= \sum_{r=2}^{n+1} \mathbb{P}(X = r) \\ &&&= \frac 1n + \left(1-\frac1n\right) \frac 2 n + \left(1-\frac1n\right)\left(1-\frac2n\right) \frac3 n + \cdots \end{align*}
2. \(\,\) \begin{align*} \mathbb{E}(X) &= \sum_{r=2}^{n+1} r\cdot\mathbb{P}(X = r) \\ &= \frac 2n + \left(1-\frac1n\right) \frac {2\cdot3} n + \left(1-\frac1n\right)\left(1-\frac2n\right) \frac{3\cdot4} n + \cdots \end{align*}
3. \(\displaystyle \mathbb{P}(X \geq k) = \frac{n-1}{n} \cdots \frac{n-r+2}{n}\)
4. \(\,\) \begin{align*} && \mathbb{E}(Y) &= \sum_{r=1}^N r \cdot \mathbb{P}(Y = r) \\ &&&= \sum_{r=1}^N \sum_{j=1}^r \mathbb{P}(Y = r) \\ &&&= \sum_{j=1}^N \sum_{r=j}^N \mathbb{P}(Y=r) \\ &&&= \sum_{j=1}^N \mathbb{P}(Y \geq j) \end{align*} Let \(P_k = \left(1-\frac1n\right)\left(1-\frac2n\right) \cdots \left(1-\frac1n\right)\left(1-\frac{k}n\right) \) \begin{align*} && \mathbb{E}(X) &= P_1 \frac{1 \cdot 2 }{n} + P_2 \cdot \frac{2 \cdot 3}{n} + \cdots + P_k \cdot \frac{k(k+1)}{n} + \cdots \\ && &= \sum_{k=1}^{n} \frac{k^2}{n}P_k + \sum_{k=1}^{n} \frac{k}{n}P_k \\ && \text{Using the identity } & \frac{k}{n}P_k = \frac{k}{n} \prod_{i=1}^{k-1} \left(1-\frac{i}{n}\right) = P_k - P_{k+1}: \\ && \sum_{k=1}^{n} \frac{k}{n}P_k &= (P_1 - P_2) + (P_2 - P_3) + \cdots + (P_n - P_{n+1}) \\ && &= P_1 - P_{n+1} = 1 - 0 = 1 \\ \\ \Rightarrow && \mathbb{E}(X) &= \sum_{k=1}^{n} \frac{k^2}{n}P_k + 1 \\ && &= \mathbb{P}(X \geq 1) + \mathbb{P}(X \geq 2) + \mathbb{P}(X \geq 3) + \cdots \\ && &= 1 + P_1 + P_2 + P_3 + \cdots \\ && &= 1 + \sum_{k=1}^{n} P_k \\ \\ \Rightarrow && 1 + \sum_{k=1}^{n} P_k &= \sum_{k=1}^{n} \frac{k^2}{n}P_k + 1 \\ \Rightarrow && \sum_{k=1}^{n} P_k &= \sum_{k=1}^{n} \frac{k^2}{n}P_k \\ \Rightarrow && 0 &= \sum_{k=1}^{n} P_k \left( 1 - \frac{k^2}{n} \right) \end{align*}

Solution: \begin{align*} (1+ax)(1+bx)(1+cx) &= (1+(a+b)x+abx^2)(1+cx) \\ &= 1+(a+b+c)x+(ab+bc+ca)x^2+abcx^3 \end{align*} Therefore by comparing coefficients, \(q = bc + ca + ab\) and \(r = abc\) as required.

1. \begin{align*} \ln (1+qx^2 + rx^3) &= \ln(1+ax) + \ln(1+bx)+\ln(1+cx) \\ &= -\sum_{n=1}^{\infty} \frac{(-ax)^n}{n}-\sum_{n=1}^{\infty} \frac{(-bx)^n}{n}-\sum_{n=1}^{\infty} \frac{(-cx)^n}{n} \\ &= \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(a^n+b^n+c^n)}{n} x^n \\ &= \sum_{n=1}^{\infty} (-1)^{n+1} S_n x^n \\ \end{align*}
2. \begin{align*} \ln (1 + qx^2 + rx^3) &= (qx^2+rx^3) -\frac{(qx^2+rx^3)^2}{2} + O(x^6) \\ &= qx^2 + rx^3 - \frac12 q^2 x^4 - qr x^5 + O(x^6) \\ \end{align*} Comparing coefficients we see that \(S_2 = -q\) and \(S_3 = r\), we also must have \(S_5 = -qr = S_2S_3\) as required.
3. \begin{align*} \ln (1 + qx^2 + rx^3) &= (qx^2+rx^3) -\frac{(qx^2+rx^3)^2}{2} +\frac{(qx^2+rx^3)^3}{3}+ O(x^8) \\ &= qx^2 + rx^3 - \frac12 q^2 x^4 - qr x^5 + \frac12 rx^6 + \frac13 q^3 x^6 + q^2r x^7 + O(x^8) \\ &= qx^2 + rx^3 - \frac12 q^2 x^4 - qr x^5 + \left ( \frac12 r+ \frac13 q^3 \right)x^6 + q^2r x^7 \end{align*} Comparing coefficients we see that \(S_2 = -q\) and \(S_5 =-qr\), we also must have \(S_7 = q^2r = S_2S_5\) as required.
4. Let \(a = b = 1, c = -2\), then \(S_2 = \frac{1^2+1^2 + (-2)^2}{2} = 3, S_7 = \frac{1^2+1^2+(-2)^7}{7} = -18, S_9 = \frac{1^1+1^2+(-2)^9}{9} = \frac{2 - 512}{9} \neq 3 \cdot (-18)\)

Solution:

1. \begin{align*} && \int \frac{\sinh x}{\cosh 2x} \d x &= \int \frac{\sinh x}{2\cosh^2 x -1} \d x \\ u = \cosh x, \d u = \sinh x \d x &&&= \int \frac{1}{2u^2 -1} \d u \\ &&&= \int\frac12 \left ( \frac{1}{\sqrt{2}u-1}-\frac{1}{\sqrt{2}u+1} \right) \d u \\ &&&= \frac1{2\sqrt{2}} \left (\ln (\sqrt{2}u-1) - \ln(\sqrt{2}u+1) \right) + C \\ &&&= \frac{1}{2\sqrt{2}} \ln \left ( \frac{\sqrt{2}\cosh x -1}{\sqrt{2}\cosh x +1} \right) + C \end{align*}
2. \begin{align*} && \int \frac{\cosh x}{\cosh 2x} \d x &= \int \frac{\cosh x}{1+2\sinh^2 x} \d x \\ u = \sinh x && &= \int \frac{1}{1+2u^2} \d u \\ &&&=\frac{1}{\sqrt{2}} \tan^{-1} (\sqrt{2}u) + C \\ &&&= \frac{1}{\sqrt{2}} \tan^{-1}(\sqrt{2}\sinh x) + C \end{align*}
3. \begin{align*} u = e^x : && \int_0^1 \frac{1}{1+u^4} \d u &= \int_{x=-\infty}^{x=0} \frac{1}{1+e^{4x}}e^{x} \d x \\ &&&= \int_{-\infty}^{0} \frac{e^{-x}}{e^{2x}+e^{-2x}} \d x \\ &&&= \int_{-\infty}^{0} \frac{\cosh x - \sinh x}{2\cosh 2x } \d x \\ &&&= \frac12 \int_{-\infty}^{0} \frac{\cosh x}{\cosh 2x} \d x - \frac12 \int_{-\infty}^{0} \frac{\sinh x}{\cosh 2x} \\ &&&= \frac12 \left [\frac{1}{\sqrt{2}} \tan^{-1}(\sqrt{2}\sinh x) \right]_{-\infty}^{0}-\frac12 \left [ \frac{1}{2\sqrt{2}}\ln \left ( \frac{\sqrt{2}\cosh x -1}{\sqrt{2}\cosh x +1} \right) \right]_{-\infty}^{0} \\ &&&= 0 - \frac1{2\sqrt{2}} \frac{-\pi}{2}-\left (\frac1{4\sqrt{2}} \ln \left (\frac{\sqrt{2}-1}{\sqrt{2}+1} \right) - 0 \right) \\ &&&= \frac{\pi - \ln((\sqrt{2}-1)^2)}{4\sqrt{2}} \\ &&&= \frac{\pi + 2 \ln(1+\sqrt{2})}{4\sqrt{2}} \end{align*}

Solution:

1. Suppose we have the shortest distance between the two curves, and the path between the points is not a normal to both curves. Then we could shift the endpoints to reduce the distance. (Assuming we're not at a point of intersection). Therefore, the normal to the curves must be the same (or in other words) the gradients of the curves must be the same. ie we are at a point where \(2y y' = 4a\) we must have \(y' = m\), so \(y = \frac{2a}{m}\) and \(x = \frac{a}{m^2}\) and the distance from this point to the line \(y=mx+c\) is \(\frac{|m \frac{a}{m^2} - \frac{2a}{m}+c|}{\sqrt{m^2+1}} = \frac{|mc-a|}{m\sqrt{m^2+1}} = \frac{mc-a}{m\sqrt{m^2+1}}\). If \(mc \leq a\) then we find \(\frac{a-mc}{m\sqrt{m^2+1}}\) However, we must check that the two curves do not intersect (otherwise the closest distace is \(0\)). ie we need to check if \((mx+c)^2 = 4ax\) has any solutions, this quadratic has discriminant \((2mc-4a)^2 - 4 \cdot m^2 \cdot c^2 = 16a^2-16amc = 16a(a-mc)\) which is clearly greater than \(0\) when \(a \geq mc\). Therefore the shortest distance in this case is \(0\).
2. The distance squared between the point \((p,0)\) and a point of the form \((at^2,2at)\) is \(D^2 = (at^2-p)^2+4a^2t^2 = a^2t^4+(4a^2-2ap)t^2+p^2\) \begin{align*} && \frac{D^2}{a^2} &= t^4 + 2\left(2-\frac{p}{a}\right)t^2 + \frac{p^2}{a^2} \\ &&&= \left (t^2 - \left (\frac{p}{a}-2 \right)\right)^2 + \frac{p^2}{a^2} - \left (2-\frac{p}{a} \right)^2 \\ &&&= \left (t^2 - \left (\frac{p}{a}-2 \right)\right)^2 +\frac{4p}{a} -4 \\ \end{align*} Therefore if \(2 \leq \frac{p}{a}\) then we can find a \(t\) such that we attain the minimum for \(D^2/a^2\) of \(\frac{4p}{a}-4\) and so \(D = \sqrt{4pa-4a^2} = 2\sqrt{a(p-a)}\) . If not the smallest value will be when \(t = 0\) and we will have \(|p|\) Now consider all the lines joining points on the parabola to the centre of the circle. The shortest distance from the parabola to the circle will be normal to the circle and therefore will also be a line through the center. Therefore we need only consider the shortest distance from \((p,0)\) to the parabola \(-b\). Case 1: If \(p \geq 2a\) we have \(2\sqrt{a(p-a)} - b\) or \(0\) if \(b \geq 2\sqrt{a(p-a)}\) Case 2: If \(p < 2a\) we have \(p-b\) or \(0\) if \(b \geq p\)

Solution:

1. \begin{align*} && I - I_1 &= \int_0^1 \left ( \left ( y' \right)^2 - y^2 \right) \d x - \int_0^1 \left ( y' + y \tan x \right)^2 \d x\\ &&&= \int_0^1 \left ( \left ( y' \right)^2 - y^2 \right) - \left ( y' + y \tan x \right)^2 \d x\\ &&&= \int_0^1 \left (-y^2-2yy' \tan x - y^2 \tan^2 x \right) \d x\\ &&&= \int_0^1 \left (-2yy' \tan x - y^2(1+ \tan^2 x )\right) \d x\\ &&&= \int_0^1 \left (-2yy' \tan x - y^2 \sec^2 x\right) \d x\\ &&&= \int_0^1 -\frac{\d}{\d x} \left (y^2 \tan x \right) \d x\\ &&&= \left [-y^2 \tan x \right]_0^1 \\ &&&= 0 \\ \\ \Rightarrow && I &= I_1 = \int_0^1 \left ( y' + y \tan x \right)^2 d x \geq 0 \end{align*} The only way \(I_0 = 0\) is is \(y' + y \tan x =0\), so \begin{align*} && \frac{\d y}{\d x} &= - y \tan x \\ \Rightarrow && \int \frac{1}{y} &= \int -\tan x \d x \\ \Rightarrow && \ln |y| &= \ln |\cos x| + C \\ \Rightarrow && y &= A \cos x \\ \Rightarrow && A &= 0 \Rightarrow y = 0 \end{align*}
2. Let \(J_1 = \int_0^1 (y'+ay\tan ax)^2 \d x\), then \begin{align*} && J-J_1 &= \int_0^1 \left ( \left ( y' \right)^2 - a^2y^2 \right) - \left ( y' + ya \tan ax \right)^2 \d x\\ &&&= \int_0^1 \left (-a^2y^2-2yy' a \tan a x-y^2a^2 \tan^2 ax \right) \d x \\ &&&= \int_0^1 \left (-2yy' a \tan ax - a^2y^2(1+\tan^2 ax) \right) \d x \\ &&&= \int_0^1 \left (-2yy' a \tan ax - a^2y^2\sec^2 ax \right) \d x \\ &&&= \left [ - a y^2 \tan a x \right]_0^1 = 0 \end{align*} This is true if \(a < \frac{\pi}{2}\), since otherwise we might care about the order of the zero for \(y\) at \(x = 1\). Consider \(y = \cos \frac{\pi}{2} x\), then \(y' = -\frac{\pi}{2} \sin^2\frac{\pi}{2} x\) and \begin{align*} && \int_0^1 \frac{\pi^2}{4} \left (\sin^2 \frac{\pi}{2}x - \cos^2 \frac{\pi}{2} x \right) \d x &= -\frac{\pi^2}{4} \int_0^1 \cos(\pi x) \d x \\ &&&= 0 \end{align*}