Problems

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Solution:

1. \begin{align*} \mathcal{L}\{e^{-bt}f(t)\}(s) &= \int_0^{\infty}e^{-st}\{ e^{-bt}f(t) \} \d t \\ &= \int_0^{\infty} e^{-(s+b)t}f(t) \d t \\ &= F(s+b) \end{align*}
2. \begin{align*} \mathcal{L}\{f(at)\}(s) &= \int_0^{\infty} e^{-st}f(at) \d t \\ &= \int_{u=0}^{\infty}e^{-s \frac{u}{a}} f\left(a \tfrac{u}{a}\right)\frac{1}{a} \d u \\ &= \int_0^{\infty}e^{-su/a}f(u) a^{-1} \d u \\ &= a^{-1} \int_0^{\infty} e^{-(s/a)u}f(u) \d u \\ &= a^{-1} F\left (\frac{s}{a} \right) \end{align*}
3. \begin{align*} \mathcal{L}\{f'(t)\}(s) &= \int_0^{\infty} e^{-st}f'(t) \d t \\ &= \left [e^{-st} f(t) \right]_0^{\infty} - \int_0^{\infty} -s e^{-st} f(t) \d t\\ &= -f(0)+sF(s) \\ &= sF(s) - f(0) \end{align*}
4. Since \(f''(t) = -f(t)\) we must have: \begin{align*} && -\mathcal{L}(f)&= \mathcal{L}(f'') \\ &&&= s\mathcal{L}(f') -f'(0) \\ &&&= s(s\mathcal{L}(f)-f(0)) - f'(0) \\ &&&= s^2\mathcal{L}(f) - 1 \\ \Rightarrow && (1+s^2) \mathcal{L}(f) &= 1 \\ \Rightarrow && F(s) &= \frac{1}{1+s^2} \end{align*}

\begin{align*} \mathcal{L}\{e^{-pt}\cos qt\}(s) &= \mathcal{L}\{\cos qt\}(s+p) \\ &= q^{-1}\mathcal{L}\{\cos t\}\left (\frac{s+p}{q} \right) \\ &= q^{-1}\mathcal{L}\{\sin'\}\left (\frac{s+p}{q} \right) \\ &= q^{-1} \left (\frac{s+p}{q} \right) \mathcal{L}\{\sin\} \left (\frac{s+p}{q} \right) - q^{-1}\sin \left (0\right) \\ &= \frac{s+p}{q^2} \frac{1}{1+\left (\frac{s+p}{q} \right)^2 } \\ &= \frac{s+p}{q^2+(s+p)^2} \end{align*}

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Solution:

1. \(\,\) \begin{align*} && \lim_{x \to 0} x^m(\ln x)^n &= \lim_{t \to \infty} (e^{-t})^m (\ln e^{-t})^n \\ &&&= \lim_{t \to \infty} e^{-mt} (-t)^n \\ &&&= (-1)^n \lim_{t \to \infty} e^{-mt} t^n = 0 \\ \\ && \lim_{x \to 0} x^x &= \lim_{x \to 0} e^{x \ln x} \\ &&&= \exp \left (\lim_{x \to 0} x \ln x \right) \\ &&&= \exp \left (0 \right) = 1 \end{align*}
2. \(\,\) \begin{align*} && I_{n} &= \int_0^1 x^m (\ln x)^n \d x \\ && I_{n+1} &= \int_0^1 x^m (\ln x)^{n+1} \d x \\ &&&= \left [\frac{x^{m+1}}{m+1} (\ln x)^{n+1} \right]_0^1 - \frac{1}{m+1} \int_0^1 x^{m+1} (n+1) (\ln x)^n \cdot x^{-1} \d x \\ &&&= 0 - \frac{1}{m+1} \lim_{x \to 0} \left (x^{m+1} (\ln x)^{n+1} \right) - \frac{n+1}{m+1} \int_0^1 x^m (\ln x)^n \d x \\ &&&= - \frac{n+1}{m+1} I_n \\ \\ && I_0 &= \int_0^1 x^m \d x = \frac{1}{m+1} \\ && I_1 &= -\frac{1}{(m+1)^2} \\ && I_2 &= \frac{2}{(m+1)^3} \\ && I_n &= (-1)^n \frac{n!}{(m+1)^{n+1}} \end{align*}
3. \(\,\) \begin{align*} && \int_0^1 x^x \d x &= \int_0^1 e^{x \ln x} \d x \\ &&&= \int_0^1 \sum_{k=0}^{\infty} \frac{x^k(\ln x)^k}{k!} \d x \\ &&&= \sum_{k=0}^{\infty} \int_0^1 \frac{x^k(\ln x)^k}{k!} \d x\\ &&&= \sum_{k=0}^{\infty} (-1)^k \frac{k!}{k!(k+1)^{k+1}}\\ &&&= \sum_{k=0}^{\infty} \frac{(-1)^k}{(k+1)^{k+1}}\\ &&&= 1 - \frac{1}{2^2} + \frac{1}{3^3} - \frac{1}{4^4} + \cdots \end{align*}

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Solution:

1. 

   + - ↺
   \begin{align*} && f(x) &= \exp\left (\tfrac23\sin x \right) \\ && f'(x) &= \exp\left (\tfrac23\sin x \right) \cdot \tfrac23 \cos x \\ && f''(x) &= \left ( \exp\left (\tfrac23\sin x \right) \cdot \tfrac23\right) \left ( \tfrac23 \cos^2 x - \sin x \right) \end{align*} Therefore \(f(x)\) is convex when \(\frac23 \cos^2 x \geq \sin x\). Note that we can find the equality points when \begin{align*} && \sin x &= \frac23 \cos^2 x \\ &&&= \frac23 (1- \sin^2 x) \\ \Rightarrow && 0 &= 2\sin^2 x + 3 \sin x - 2 \\ &&&= (2 \sin x -1) (\sin x+2) \end{align*} ie \(\sin x = \frac12 \Rightarrow x = \frac{\pi}{6}, \frac{5\pi}{6}\). Therefore \(f\) is convex on \([0, \frac{\pi}{6}] \cup [\frac{5\pi}{6}, 2\pi]\)
2. Suppose \(g(x) = \exp \left ( -k \tan x \right)\) then \begin{align*} && g'(x) &= \exp \left ( -k \tan x \right) \cdot (-k \sec^2 x ) \\ && g''(x) &= \left ( -k \exp \left ( -k \tan x \right) \right) \left ( -k\sec^4 x + 2 \sec x \cdot \sec x \tan x\right) \\ &&&= -k \exp \left ( -k \tan x \right) \sec^4 x \left ( -k + 2\sin x \cos x \right) \\ &&&= -k \exp \left ( -k \tan x \right) \sec^4 x \left ( -k + \sin 2x \right) \\ \end{align*} If \(0 < \alpha < \frac{\pi}{4}\) then \(k > 0\) so \(g\) is convex if \(-k + \sin 2x < 0\), ie \(\sin 2x < \sin 2\alpha\), ie on \((0, \alpha)\) and \((\frac{\pi}{2} - \alpha, \frac{\pi}{2})\)

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Solution: Let \(X \sim Po(\lambda)\), then

1. \(\,\)
   

   + - ↺
   Suppose \(\mathbb{P}(X \geq 2) = 1-p\) then \(\mathbb{P}(X=0) + \mathbb{P}(X=1) = p\), ie \(e^{-\lambda} +\lambda e^{-\lambda} = p\) If \(f(x) = (1+x)e^{-x}\) we have see it is strictly decreasing on \(x \geq 0\) and takes all values from \(1\) to \(0\), therefore we can find a unique value such that \(f(\lambda) = p\) which is our desired \(\lambda\).
2. Note that \(\mathbb{P}(X = 1) = \lambda e^{-\lambda}\)
   

   + - ↺
   Sketching \(y = xe^{-x}\) and finding it's turning point we can see that there is a unique value of \(\lambda = 1\) when \(q = \frac{1}{e}\), otherwise there is either no solution (\(p > \frac1{e}\)) or two solutions (\(0 < q > \frac1{e}\)).
3. Suppose \(\mathbb{P}(X = 1 | X \leq 2) = r\), ie \begin{align*} && r &= \frac{\lambda e^{-\lambda}}{e^{-\lambda} + \lambda e^{-\lambda} + \frac12 \lambda^2 e^{-\lambda}} \\ &&&= \frac{2\lambda}{2+2\lambda+\lambda^2} \\ \Rightarrow && 0 &= r\lambda^2 + 2(r-1) \lambda + 2r\\ \Rightarrow && \Delta &= 4(r-1)^2 - 4\cdot r \cdot 2 r \\ &&&= 4((r-1)^2-2r^2) \\ &&&= 4(r-1-\sqrt{2}r)(r-1+\sqrt{2}r) \\ &&&= -4((\sqrt{2}-1)r + 1)((1+\sqrt{2})r-1) \end{align*} Therefore our quadratic in \(r\) has a unique solution if \(r = \frac{1}{1+\sqrt{2}}\). If it has a positive solution then note since \(2r > 0\) both solutions are positive, so \(\lambda\) is not unique by excluding other solutions.

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Solution:

1. \begin{align*} && I &= \int \frac{3}{(x-4)\sqrt{2x+1}}\, \d x \\ u^2 =2x+1, 2u \frac{\d u}{\d x}=2: && &= \int \frac{3}{\left(\frac{u^2-1}{2}-4\right)u} u \d u \\ &&&= \int \frac{6}{u^2-9} \d u \\ &&&= \int \frac{6}{(u-3)(u+3)} \d u\\ &&&= \int \left ( \frac{1}{u-3} - \frac{1}{u+3} \right )\d u \\ &&&= \ln (u-3) - \ln (u+3) + C \\ &&&= \ln \frac{u-3}{u+3} + C \\ &&&= \ln \left (\frac{\sqrt{2x+1}-3}{\sqrt{2x+1}+3} \right) + C \end{align*}
2. \begin{align*} && I &= \int_{\ln 3}^{\ln 8} \frac{2}{e^x\sqrt{e^x+1}} \d x \\ u = e^x, \frac{\d u}{\d x} = e^x: &&&= \int_{u=3}^{u=8} \frac{2}{u\sqrt{u+1}} \frac{1}{u} \d u \\ v^2=u+1, 2v \frac{\d v}{\d u} = 1: &&&= \int_{v=2}^{v=3} \frac{2}{v(v^2-1)^2} \d v \\ &&&= \int_2^3 \left ( \frac{2}{v} - \frac{}{v-1} - \frac{1}{2(v-1)^2} - \frac{1}{v+1} - \frac{1}{2(v+1)^2}\right) \d v \\ &&&= \left [2\ln v - \ln(v^2-1)+\frac12(v-1)^{-1}+\frac12(v+1)^{-1} \right]_2^3 \\ &&&= \left ( 2\ln3-\ln 8+\frac14+\frac18\right)-\left ( 2\ln2-\ln 3+\frac12+\frac16\right) \\ &&&= \end{align*}

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Solution: \begin{align*} &&2f(x) &\equiv f(x) + f(x) \\ &&&\equiv f(x+x) \\ &&&\equiv f(2x) \\ \\ \Rightarrow && 2f(0) &= f(0) \\ \Rightarrow && f(0) &= 0 \\ && f''(0) &= \lim_{h \to 0} \frac{f(2h)-2f(0)+f(-2h)}{h^2} \\ &&&= \lim_{h \to 0} \frac{f(2h)+f(-2h)}{h^2} \\ &&&= \lim_{h \to 0} \frac{f(0)}{h^2} \\ &&&= 0 \\ \Rightarrow && f''(0) &= 0 \end{align*} If \(f(x)\) satisfies the equation, then \(f'(x)\) satisfies the equation. In particular this means that \(f^{(n)}(0) = 0\) for all \(n \geq 2\). Therefore the only non-zero term in the Maclaurin series is \(x^1\). Therefore \(f(x) = cx\)

1. Suppose \(g(x)g(y) \equiv g(x+y)\), then if \(G(x) = \ln g(x)\) we must have \(G(x)+G(y) \equiv G(x+y)\), ie \(G(x) = cx \Rightarrow g(x) = e^{cx}\)
2. Suppose \(h(x)+h(y) \equiv h(xy)\), then if \(h(e^u) = H(u)\) we must have that \(H(u)+H(v) \equiv h(e^u) + h(e^v) \equiv h(e^{u+v}) \equiv H(u+v)\).Therefore \(H(u) = cu\), ie \(h(e^u) = cu\) or \(h(x) = h(e^{\ln x}) = c \ln x\).
3. Finally if \(t(x) + t(y) \equiv t(z)\), the considering \(T(w) = t(\tan w)\) then \(T(x) + T(y) \equiv t(\tan x) + t(\tan y) \equiv t( \frac{\tan x + \tan y}{1- \tan x \tan y}) \equiv t (\tan (x+y)) \equiv T(x+y)\). Therefore \(T(x) = cx\) Therefore \(t(\tan w) = c w \Rightarrow t(x) = c \tan^{-1} x\)

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Solution: \begin{align*} && y &= x^2e^{-x^2} \\ \Rightarrow && y' &= 2xe^{-x^2} +x^2 \cdot (-2x)e^{-x^2} \\ &&&= e^{-x^2}(2x-2x^3) \\ &&&= 2e^{-x^2}x(1-x^2) \end{align*} Therefore the derivative is zero iff \(x = 0, \pm 1\) \begin{align*} && y &= \P(x) e^{-x^2} \\ \Rightarrow && y' &= e^{-x^2} (\P'(x)-2x\P(x)) \end{align*} Therefore we want \(\P'(x) - 2x\P(x) = Kx(x^2-a^2)(x^2-b^2)\) Since this has degree \(5\), we should look at polynomials degree \(4\) for \(\P\). We can also immediately see that \(0\) is a root of \(\P'(x)\), so \(\P(x) = a_4x^4+a_3x^3+a_2x^2+a_0\). WLOG \(a_4 = 1\) and \(K = -2\), so \begin{align*} && -2(x^5-(a^2+b^2)x^3+a^2b^2x) &= 4x^3+3a_3x^2+2a_2x- 2x(x^4+a_3x^3+a_2x^2+a_0) \\ &&&= -2x^5-2a_3 x^4+(4-2a_2)x^3+(2a_2-2a_0)x \\ \Rightarrow && a_3 &= 0 \\ && a^2+b^2 &= 2-a_2 \\ \Rightarrow && a_2 &= 2-a^2-b^2 \\ && a^2b^2 &= a_0-a_2 \\ \Rightarrow && a_0 &= a^2b^2 + 2-a^2-b^2 \\ \Rightarrow && \P(x) &= x^4+(2-a^2-b^2)x^2+(a^2-1)(b^2-1)x \end{align*}

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Solution:

\begin{align*} && 1 &= \int_0^1 f(x) \d x \\ &&&= \int_0^1 kx e^{-ax^2} \d x \\ &&&= \left [-\frac{k}{2a}e^{-ax^2} \right]_0^1 \\ &&&= \frac{k(1-e^{-a})}{2a} \\ \Rightarrow && k &= \frac{2a}{1-e^{-a}} \end{align*} To find the mode, we want \(f'(x) = 0\), ie \begin{align*} && 0 &= f'(x) \\ &&&= -2kax^2e^{-ax^2} + k e^{-ax^2} \\ &&&= ke^{-ax^2} \left (1-2ax^2 \right)\\ \end{align*} So either \(m = \frac{1}{\sqrt{2a}}\) (if \(a > \frac12\)) or \(f(x)\) is increasing and the mode is \(m = 1\) (if \(a < \frac12\)). \begin{align*} && \frac12 &= \int_0^h f(x) \d x \\ &&&= \left [ -\frac{e^{-ax^2}}{1-e^{-a}} \right]_0^h \\ &&&= \frac{1-e^{-ah^2}}{1-e^{-a}} \\ \Rightarrow && e^{-ah^2}&= 1-\frac12(1-e^{-a}) \\ \Rightarrow && -a h^2 &= \ln \left ( \frac12(1+e^{-a}) \right) \\ \Rightarrow && h &= \sqrt{-\frac1a \ln (\tfrac12(1+e^{-a}))} \end{align*} \(h > m\) already means \(a > \frac12\) so \begin{align*} && h &> m \\ \Leftrightarrow &&\sqrt{-\frac1a \ln (\tfrac12(1+e^{-a}))} &> \frac{1}{\sqrt{2a}} \\ \Leftrightarrow && -\ln (\tfrac12(1+e^{-a})) &> \frac12 \\ \Leftrightarrow && e^{-1/2} & > \frac12(1+e^{-a}) \\ \Leftrightarrow && 2e^{-1/2}-1 &>e^{-a} \\ \Leftrightarrow && \ln(2e^{-1/2}-1) &>-a \\ \Leftrightarrow && a& > -\ln(2e^{-1/2}-1) \\ \end{align*} Noting that \begin{align*} && -\ln(2e^{-1/2} - 1) &= -\ln \left (\frac{2-\sqrt{e}}{e^{1/2}} \right) \\ &&&= \frac12 -\ln(\underbrace{2 - \sqrt{e}}_{<1}) \\ &&&> \frac12 \end{align*} If \(a > -\ln(2e^{-1/2}-1)\) then \begin{align*} && \mathbb{P}(X > m | X < h) &= \frac{\mathbb{P}(m < X < h)}{\mathbb{P}(X < h)} \\ &&&= \frac{e^{-am^2}-e^{-ah^2}}{1-e^{-ah^2}} \\ &&&= \frac{e^{-a\frac{1}{2a}}-e^{\ln \left ( \frac12(1+e^{-a}) \right)}}{1-e^{\ln \left ( \frac12(1+e^{-a}) \right)}} \\ &&&= \frac{e^{-1/2}-\frac12(1+e^{-a})}{1-\frac12(1+e^{-a})} \\ &&&= \frac{2e^{-1/2}-1-e^{-a}}{1-e^{-a}} \end{align*} as required.

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Solution: \begin{align*} && \frac{\d y}{\d t} &= - k \left (\frac{t^2-3t+2}{t+1} \right) y \\ \Rightarrow && \int \frac1y \d y &= -k\int \left (t-4 + \frac{6}{t+1}\right) \d t \\ \Rightarrow && \ln y &= -k \left ( \frac12 t^2 -4t + 6\ln (t+1) \right) + C \\ (t,y) = (0,A): && \ln A &=C \\ \Rightarrow && \ln y &= -k \left ( \frac12 t^2 -4t + 6\ln (t+1) \right) + \ln A \\ && \ln \left ( \frac{y}{A}(t+1)^{6k} \right) &= -k \l \frac12 t^2 - 4t \r \\ \Rightarrow && y &= A\frac{\exp \l -k(\frac12 t^2-4t)\r}{(t+1)^{6k}} \end{align*} \(y\) wil have stationary values when \(\frac{\d y}{\d t} = 0\), ie \begin{align*} k \left (\frac{t^2-3t+2}{t+1} \right) y &= 0 \\ k \left ( \frac{(t-2)(t-1)}{t+1} \right) y &= 0 \end{align*} ie when \(y = 0, t = 1, t =2\). Clearly \(y = 0\) is not a solution, so \(y\) has the values: \begin{align*} t = 1: && y &= A\frac{\exp \l -k(\frac12 -4)\r}{(2)^{6k}} \\ &&&= A \frac{e^{7/2 k}}{2^{6k}} \\ t = 2: && y &= A\frac{\exp \l -k(2 -8)\r}{(3)^{6k}} \\ &&&= A \frac{e^{6 k}}{3^{6k}} \\ \text{ratio}: && \frac{e^{7/2k}}{2^{6k}} \cdot \frac{3^{6k}}{e^{6k}} &= (3/2)^{6k} e^{-5k/2} \end{align*} If \(k > 0\) as \(t \to \infty\) \(y \to 0\) since the \(e^{-kt^2/2}\) term dominates everything. If \(k < 0\) as \(t \to \infty\) \(y \to \infty\) as since the \(e^{-kt^2}\) term also dominates but now it growing to infinity faster than everything else.

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Solution: \begin{align*} && S_2(x) &= e^{x^3} \frac{d^2}{\d x^2} \left [e^{-x^3} \right] \\ &&&= e^{x^3} \frac{d}{\d x} \left [e^{-x^3}(-3x^2) \right] \\ &&&= e^{x^3} \left [e^{-x^3}(9x^4-6x) \right] \\ &&&=9x^4-6x \\ \\ && S_3(x) &= e^{x^3} \frac{\d^3}{\d x^3} \left [ e^{-x^3} \right]\\ &&&= e^{x^3} \frac{\d}{\d x} \left [ e^{-x^3}(9x^4-6x) \right ] \\ &&&= e^{x^3} e^{-x^3}\left [ (-3x^2)(9x^4-6x)+(36x^3-6) \right ] \\ &&&= -27x^6 +54x^3-6 \end{align*} Claim: \(S_n\) is a polynomial of degree \(2n\) with leading coefficient \((-3)^n\). Proof: Clearly this is true for \(n = 1, 2, 3\) by demonstration. Suppose it is true for some \(n = k\), then \begin{align*} && S_k(x) &= e^{x^2} \frac{\d^k}{\d x^k} \left [ e^{x^3}\right] \\ && (-3)^kx^{2k} +\cdots &= e^{x^3} \frac{\d^k}{\d x^k} \left [ e^{x^3}\right] \\ \Rightarrow && \frac{\d^k}{\d x^k} \left [ e^{x^3}\right] &= e^{-x^3} \left ( (-3)^kx^{2k} +\cdots\right) \\ \Rightarrow && \frac{\d^k}{\d x^k}\left [ e^{x^3}\right] &= e^{-x^3} (-3x^2)\left ( (-3)^kx^{2k} +\cdots\right) + e^{-x^3} S_k'(x) \\ &&&= e^{-x^3} \left (\underbrace{ (-3)^{k+1}x^{2k+2} + \cdots + S_k'(x)}_{\deg =2k+2}\right) \\ \Rightarrow && S_{k+1}(x) &= (-3)^{k+1}x^{2k+2} + \cdots + S_k'(x) \end{align*} And therefore \(S_{k+1}\) is a polynomial degree \(2(k+1)\) with leading coefficient \((-3)^{k+1}\) so by induction it's true for all \(n\). If \(S'_n(a) = 0\) then \(S_{n+1}(a) = (-3a^2)S_n(a) + S_n'(a) \Rightarrow S_{n+1}(a)S_n(a) = -3 (aS_n(a))^2 \leq 0\)