Problems

Solution:

1. \begin{align*} && \frac{6}{r(r+1)(r+3)} &= \frac{2}{r} - \frac{3}{r+1} + \frac{1}{r+3} \\ \Rightarrow && \sum_{r=1}^n \frac{6}{r(r+1)(r+3)} &= \sum_{r=1}^n \l \frac{2}{r} - \frac{3}{r+1} + \frac{1}{r+3} \r \\ &&&= \sum_{r=1}^n \frac{2}{r} - \sum_{r=1}^n \frac{3}{r+1} + \sum_{r=1}^n \frac{1}{r+3} \\ &&&= \sum_{r=1}^n \frac{2}{r} - \sum_{r=2}^{n+1} \frac{3}{r} + \sum_{r=3}^{n+2} \frac{1}{r} \\ &&& = \frac{2}{1} + \frac{2}{2} - \frac{3}{2} - \frac{3}{n+1} + \frac{1}{n+1} + \frac{1}{n+2} \\ &&& = \frac{3}{2} - \frac{2}{n+1} + \frac{1}{n+2} \end{align*}
2. \begin{align*} && \ln (1 + x+ x^2 + x^3) &= \ln \l \frac{1-x^4}{1-x} \r \\ &&&= \ln (1-x^4) - \ln(1-x) \\ &&&= \sum_{k=1}^{\infty} -\frac{x^{4k}}{k} - \sum_{k=1}^{\infty} - \frac{x^k}{k} \\ &&&= x + \frac12x^2+\frac13x^3-\frac34x^4+\frac15x^5 + \cdots \\ &&&= \sum_{k=1}^{\infty}a_k x^k \end{align*} Where \(a_k = \frac{1}{k}\) if \(k \neq 0 \pmod{4}\) otherwise \(a_k = -\frac{3}{k}\) if \(k \equiv 0 \pmod{4}\)
3. \begin{align*} \exp(x \ln (1+x) ) &= \exp\l x \l x-\frac12x^2+\frac13x^3-\cdots \r \r \\ &= \exp\l x^2-\frac12x^3+\frac13x^4 \r \\ &= 1 + \l x^2-\frac12x^3+\frac13x^4 \r + \frac12 \l x^2-\frac12x^3+\frac13x^4 \r^2 + \cdots \\ &= 1 + x^2-\frac12x^3+\frac13x^4 + \frac12x^4 + \cdots \\ &= 1 + x^2 -\frac12x^3+\frac56x^4+\cdots \end{align*}

Solution:

1. \begin{align*} \int_{1}^{3}\frac{1}{6x^{2}+19x+15}\,\mathrm{d}x &= \int_1^3 \frac1{(2x+3)(3x+5)} \d x \\ &= \int_1^3 \l \frac{2}{2x+3} - \frac{3}{3x+5} \r \d x \\ &= \left [\ln(2x+3) - \ln(3x+5) \right ]_1^3 \\ &= \l \ln9 - \ln14 \r - \l \ln 5 - \ln 8 \r \\ &= \ln \frac{72}{70} \\ &= \ln \frac{36}{35} \end{align*}
2. 

   + - ↺
   When \(q = 0\) the roots are \(-1, 0, 1\) There can be \(0, 2, 3, 4\) roots. There will be no roots if \(q > -\min (x^{1760} - x^{220})\) since the whole graph will be above the axis. There will be \(2\) roots if \(q = -\min (x^{1760} - x^{220})\) or \(q > 0\) There will be \(4\) roots if \(0 > q > -\min (x^{1760} - x^{220})\). There will be \(3\) roots if \(q =0\)

Solution: \begin{align*} \frac{1}{(1-ax)(1-bx)} &=\frac{1}{b-a} \l \frac{b}{1-bx}-\frac{a}{1-ax}\r \\ &= \frac{1}{b-a} \l \sum_{k=0}^{\infty} b(bx)^k-\sum_{k=0}^{\infty} a(ax)^k \r \\ &= \frac{1}{b-a} \sum_{k=0}^{\infty} \l b^{k+1} - a^{k+1} \r x^k \end{align*} Therefore \(\displaystyle c_m = \frac{b^{k+1}-a^{k+1}}{b-a}\). \begin{align*} c_m^2 &= \frac{(b^{m+1}-a^{m+1})^2}{(b-a)^2} \\ &= \frac{a^{2m+2} - 2(ab)^{m+1} + b^{2m+2}}{(b-a)^2} \end{align*} \begin{align*} \sum_{m=0}^{\infty} c_m x^m &= \sum_{m=0}^{\infty} \l \frac{a^{2m+2} - 2(ab)^{m+1} + b^{2m+2}}{(b-a)^2} \r x^m \\ &= \frac{1}{(b-a)^2} \l \sum_{m=0}^{\infty} a^{2m+2} x^m-2\sum_{m=0}^{\infty} (ab)^{m+1} x^m+\sum_{m=0}^{\infty} b^{2m+2} x^m \r \\ &= \frac{1}{(b-a)^2} \l a^2\sum_{m=0}^{\infty} a^{2m} x^m-2ab\sum_{m=0}^{\infty} (ab)^{m} x^m+b^2\sum_{m=0}^{\infty} b^{2m} x^m \r \\ &= \frac{1}{(b-a)^2} \l \frac{a^2}{1-a^2x^2} - \frac{2ab}{1-abx} + \frac{b^2}{1-b^2x^2}\r \\ &= \frac{1+ab}{(1-a^2x)(1-abx)(1-b^2x)} \end{align*} Where geometric series will converge if \(|a^2x| < 1, |b^2x| < 1, |abx| < 1\), ie \(|x| < \min (\frac{1}{a^2}, \frac{1}{b^2} )\)

Solution:

1. \begin{align*} \int_{1}^{\mathrm{e}}\frac{\ln x}{x^{2}}\,\mathrm{d}x &= \left [-\frac{\ln x}{x} \right]_1^e + \int_1^e \frac{1}{x^2} \, \d x \\ &= -\frac{1}{e} + \left [ -\frac{1}{x} \right]_1^e \\ &= 1 - \frac{2}{e} \end{align*}
2. \begin{align*} \int\frac{\cos x}{\sin x\sqrt{1+\sin x}}\,\mathrm{d}x &= \int \frac{2u}{(u^2-1)u} \d u \tag{\(u^2 = 1+\sin x\)} \\ &= \int \frac{1}{u-1} - \frac{1}{u+1} \d u \\ &= \ln(u-1) - \ln (u+1) + C \\ &= \ln \l \frac{u-1}{u+1} \r + C \\ &= \ln \l \frac{\sqrt{\sin x + 1} + 1}{\sqrt{\sin x + 1} -1} \r + C \end{align*}

Solution: \begin{align*} && \f(x) &= \frac1{(x-1)(x-2)} \\ &&&=\frac{1}{x-2} -\frac{1}{x-1} \\ \end{align*} Therefore, for \(|x| < 1\) \begin{align*} && \f(x) &=\frac{1}{x-2} -\frac{1}{x-1} \\ &&&= -\frac12 \frac{1}{1-\frac{x}{2}} + \frac{1}{1-x} \\ &&&= \sum_{n=0}^{\infty} x^n - \frac12 \sum_{n=0}^\infty \l \frac{x}2 \r^2 \end{align*} where both geometric series converge since \(|x| < 1\) and \(|\frac{x}{2}| < 1\) When \(1 < |x|< 2 \Rightarrow |\frac{1}{x}| < 1\), we must have: \begin{align*} && \f(x) &=\frac{1}{x-2} -\frac{1}{x-1} \\ &&&= -\frac12 \frac{1}{1-\frac{x}{2}} + \frac1{x}\frac{1}{1-\frac{1}{x}} \\ &&&= - \frac12 \sum_{n=0}^\infty \l \frac{x}2 \r^2 - \frac{1}{x} \sum_{n=0}^{\infty} x^{-n} \\ &&&= - \frac12 \sum_{n=0}^\infty \l \frac{x}2 \r^2 - \sum_{n=1}^{\infty} x^{-n} \\ \end{align*} Finally, when \(|x| > 2\), ie \(|\frac{2}{x}| < 1\) we have \begin{align*} && \f(x) &=\frac{1}{x-2} -\frac{1}{x-1} \\ &&& =\frac1{x} \frac{1}{1-\frac{2}{x}} - \frac{1}{x}\frac{1}{1-\frac{1}{x}} \\ &&&= \frac1{x} \sum_{n=0}^{\infty} \l \frac{2}{x} \r^n - \sum_{n=1}^{\infty}x^{-n} \\ &&&= \sum_{n=1}^{\infty} 2^{n-1} x^{-n} - \sum_{n=1}^{\infty}x^{-n} \\ \end{align*}

Solution: \begin{align*} \int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi}\frac{\cos^{2}\theta}{1-\sin\theta\sin2\alpha}\,\mathrm{d}\theta &= \int_{u = \frac12 \pi}^{u = -\frac12 \pi} \frac{\cos^2 (-u)}{1-\sin(-u) \sin 2 \alpha} -\d u \tag{\(u = -\theta\)} \\ &= \int_{\frac12 \pi}^{-\frac12 \pi} \frac{\cos^2 u}{1+\sin u \sin 2 \alpha} -\d u \\ &= \int_{-\frac12 \pi}^{\frac12 \pi} \frac{\cos^2 u}{1+\sin u \sin 2 \alpha} \d u \\ &= \int_{-\frac12 \pi}^{\frac12 \pi} \frac{\cos^2 \theta}{1+\sin \theta \sin 2 \alpha} \d \theta \\ \end{align*} Since \(\displaystyle \frac{1}{(1-a^2u^2)} = \frac12 \l \frac{1}{1+au} + \frac1{1-au} \r\) \begin{align*} \int_{-\frac12 \pi}^{\frac12 \pi} \frac{\cos^2 \theta}{1+\sin \theta \sin 2 \alpha} \d \theta &= \int_{-\frac12 \pi}^{\frac12 \pi} \frac{\cos^2 \theta}{1-\sin ^2\theta \sin^2 2 \alpha} \d \theta \\ &= \int_{-\frac12 \pi}^{\frac12 \pi} \frac{1-\sin^2 \theta}{1-\sin ^2\theta \sin^2 2 \alpha} \d \theta \\ &= \int_{-\frac12 \pi}^{\frac12 \pi} \frac{(1-\sin ^2\theta \sin^2 2 \alpha) \frac{1}{\sin^2 2\alpha} + 1 - \cosec^2 2\alpha}{1-\sin ^2\theta \sin^2 2 \alpha} \d \theta \\ &= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha \int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{1}{1 - \sin^2 \theta \sin^2 2 \alpha} \d \theta \\ &= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha \int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{1}{1 - \sin^2 \theta (1-\cos^2 2 \alpha)} \d \theta \\ &= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha \int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{1}{\cos^2 \theta +\sin^2 \theta \cos^2 2 \alpha} \d \theta \\ &= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha \int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{\sec^2 \theta}{1 +\tan^2 \theta \cos^2 2 \alpha} \d \theta \\ \end{align*} Finally, using the substitution \(u =|\cos 2 \alpha | \tan \theta, \d u = |\cos 2 \alpha |\sec^2 \theta \d \theta\) \begin{align*} \int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{\sec^2 \theta}{1 +\tan^2 \theta \cos^2 2 \alpha} \d \theta &= |\sec 2\alpha|\int_{u = -\infty}^{u = \infty} \frac{1}{1 + u^2} \d u \\ &= |\sec 2 \alpha|\pi \end{align*} and so \begin{align*} I &= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha |\sec 2 \alpha|\pi \\ &= \frac{\pi}{\sin^2 2\alpha} \l 1-\cos 2\alpha \r \\ &= \frac{\pi}{4\sin^2 \alpha\cos^2 \alpha} \l 2 \sin^2 \alpha \r \\ &= \frac{\pi}{2 \cos^2 \alpha} = \frac{\pi}{2} \sec^2 \alpha \end{align*} When \(\alpha\) small enough that the modulus doesn't flip the sign. When if \(\frac{1}{4}\pi<\alpha<\frac{1}{2}\pi\) we have: \begin{align*} I &= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha |\sec 2 \alpha|\pi \\ &= \frac{\pi}{\sin^2 2\alpha} \l 1+\cos 2\alpha \r \\ &= \frac{\pi}{4\sin^2 \alpha\cos^2 \alpha} \l 2 \cos^2 \alpha \r \\ &= \frac{\pi}{2 \sin^2 \alpha} = \frac{\pi}{2} \cosec^2 \alpha \end{align*}