Problems

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 \(S_{n+1}(a) = (-3a^2)S_n(a) + S_n'(a) \Rightarrow S_{n+1}(a)S_n(a) = -3 (aS_n(a))^2 \leq 0\)

Solution: \begin{align*} && y &= \frac x {\sqrt{x^2-2x+a}} \\ && y' &= \frac{\sqrt{x^2-2x+a} - \frac{x(x-1)}{\sqrt{x^2-2x+a}}}{x^2-2x+a} \\ &&&= \frac{-x+a}{(x^2-2x+a)^{3/2}} \end{align*} Since the denominator is always positive, the only stationary point is when \(x = a\)

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Solution: \begin{align*} && \sum_{k=0}^n \sin k\theta &= \textrm{Im} \left ( \sum_{k=0}^n e^{i k \theta}\right)\\ &&&= \textrm{Im} \left ( \frac{e^{i(n+1)\theta}-1}{e^{i \theta}-1} \right)\\ &&&= \textrm{Im} \left ( \frac{e^{i(n+\tfrac12)\theta}-e^{-i\theta/2}}{e^{i \theta/2}-e^{-i \theta/2}} \right)\\ &&&= \textrm{Im} \left ( \frac{(e^{i(n+\tfrac12)\theta}-e^{-i\theta/2})/2i}{(e^{i \theta/2}-e^{-i \theta/2})/2i} \right)\\ &&&= \textrm{Im} \left ( \frac{(e^{i(n+\tfrac12)\theta}-e^{-i\theta/2})/i}{2\sin \tfrac12 \theta} \right)\\ &&&= \frac{\cos \tfrac12 \theta - \cos(n+ \tfrac12)\theta}{2\sin \tfrac12 \theta} \end{align*}

1. When \(n\) is large we have \begin{align*} &&\sum_{k=0}^n \sin \left(\frac{k\pi}{n}\right) &= \frac{\cos \frac{\pi}{2n} - \cos \frac{(2n + 1)\pi}{2n}}{2 \sin \frac{\pi}{2n}} \\ &&&= \frac{\cos \frac{\pi}{2n} +\cos \frac{\pi}{2n}}{2 \sin \frac{\pi}{2n}} \\ &&&\approx \frac{1- \frac{\pi^2}{4n^2}}{\frac{\pi}{2n}} \\ &&&= \frac{2n}{\pi} - \frac{\pi}{2n} \\ &&&\approx \frac{2n}{\pi} \end{align*}
2. \(\,\) \begin{align*} && \sum_{k=0}^n \sin k\theta &= \frac { \cos \tfrac12\theta - \cos (n+ \tfrac12) \theta} {2\sin \tfrac12\theta} \\ \frac{\d }{\d \theta}: && \sum_{k=0}^n k\cos k\theta &= \frac { (-\tfrac12\sin\tfrac12\theta +(n+\tfrac12) \sin (n+ \tfrac12) \theta)2 \sin \tfrac12 \theta - (\cos \tfrac12\theta - \cos (n+ \tfrac12) \theta) \cos \tfrac12 \theta} {4\sin^2 \tfrac12\theta} \\ &&&= \frac{-\sin^2 \tfrac12 \theta+(2n+1)\sin(n+\tfrac12)\theta \sin \tfrac12 \theta-\cos^2\tfrac12 \theta+\cos(n+\frac12)\theta \cos\tfrac12 \theta}{4 \sin^2 \tfrac12 \theta} \\ &&&= \frac{(2n+1)\sin(n+\tfrac12)\theta \sin \tfrac12 \theta+\cos(n+\frac12)\theta \cos\tfrac12 \theta-1}{4 \sin^2 \tfrac12 \theta} \\ &&&= \frac{2n\sin(n+\tfrac12)\theta \sin \tfrac12 \theta+\cos n \theta-1}{4 \sin^2 \tfrac12 \theta} \\ \\ \Rightarrow && \sum_{k=0}^n k\left ( 1-2\sin^2 \left ( \frac{k\theta}{2} \right) \right) &= \frac{2n\sin(n+\tfrac12)\theta \sin \tfrac12 \theta+\cos n \theta-1}{4 \sin^2 \tfrac12 \theta} \\ \Rightarrow && \sum_{k=0}^n k\sin^2 \left ( \frac{k\theta}{2} \right) &= \frac{n(n+1)}{4} - \frac{2n\sin(n+\tfrac12)\theta \sin \tfrac12 \theta+\cos n \theta-1}{8 \sin^2 \tfrac12 \theta} \\ \theta = \frac{\pi}{n}: && \sum_{k=0}^n k\sin^2 \left ( \frac{k\pi}{2n} \right) &= \frac{n(n+1)}{4} - \frac{2n\sin(n+\tfrac12)\frac{\pi}{n}\sin \tfrac12 \frac{\pi}{n}+\cos n \frac{\pi}{n}-1}{8 \sin^2 \tfrac12 \frac{\pi}{n}} \\ &&&= \frac{n(n+1)}{4} - \frac{-2n \sin^2 \frac{\pi}{2n}-2}{8 \sin^2 \tfrac12 \frac{\pi}{n}} \\ &&&= \frac{n^2+n}{4} + \frac{n}{4} + \frac{1}{4\sin^2\frac{\pi}{2n}} \\ &&&\approx \frac{n^2}4 + \frac{n}{2}+ \frac{n^2}{\pi^2} \\ &&&= \left ( \frac14 + \frac{1}{\pi^2} \right)n^2 + \frac{n}{4} \\ &&&\approx \left ( \frac14 + \frac{1}{\pi^2} \right)n^2 \end{align*}

Solution:

1. \(\,\) \begin{align*} && f(x) &= (1+x^2)e^x \\ && f'(x) &= 2xe^x + (1+x^2)e^x \\ &&&= (1+x)^2e^x \geq 0 \end{align*}
   

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   As \(x \to -\infty\), \(f(x) \to 0\) and as \(x \to \infty\), \(f(x) \to +\infty\) and since \(f\) is strictly increasing we have exactly one solution to \(f(x) = k\) on \((0,\infty)\). Since \(f(x) > 0\) there are no solutions if \(k \leq 0\).
2. Considering the function \(g(x) = (e^x-1)-k\tan^{-1} x \) then \(g'(x) = e^x - \frac{k}{1+x^2}\) therefore \(g'(x)\) has exactly one turning point when \(k > 0\) and \(0\) otherwise at the root of \(f(x) = k\) Notice also that \(g(0) = 0\) so we already have one solution, and \(g'(0) = 1 - k > 0\). Notice from our sketch that if \(0 < k < 1\) the root for \(f(x) = k\) has \(x \leq 0\), so our turning point is to the left of the origin and we are interested in the behaviour of \(g(x)\) as \(x \to -\infty\). (ie do we cross the axis again). \(\lim_{x \to -\infty} \left [ e^x - 1 - k \tan^{-1} x \right] = 0 - 1 +k \frac{\pi}{2} = k \frac{\pi}{2} - 1\). if this is positive, ie if \(k > \frac{2}{\pi}\) there are two solutions, otherwise there is only one real root.

Solution: \begin{align*} && y &= \ln x - x \ln a \\ \Rightarrow && y' &= \frac1x - \ln a \\ && y'' &= -\frac{1}{x^2} \end{align*} Therefore the maximum is when \(x = \frac{1}{\ln a}\) and \(y_{max} = -\ln \ln a - 1\). If \(y_{max} < 0\) then \(y \neq 0\). But that's equivalent to \(a > e^{1/e}\). \begin{align*} && 0 &> -\ln \ln a - 1 \\ \Leftrightarrow && 1 &> - \ln \ln a \\ \Leftrightarrow && \ln \ln a &>-1 \\ \Leftrightarrow && \ln a &> e^{-1} \\ \Leftrightarrow && a & > e^{1/e} \end{align*} If \(0 < a < 1\) then, when \(x\) is small, \(\ln x - x \ln a\) is large and negative. When \(x\) is large and positive \(\ln x\) is positive and \(-x \ln a\) is positive. We also notice there is no turning point. Hence exactly one solution

Solution: Suppose we have height \(h\) and radius \(r\), then: \(V = \pi r^2 h\) and \(S = 2\pi r^2 + 2\pi r h\). \(h = \frac{V}{\pi r^2}\), so \begin{align*} S &= 2 \pi r^2 + 2 \pi r\frac{V}{\pi r^2} \\ &= 2\pi r^2 +V \frac1{r}+V \frac1{r} \\ &\underbrace{ \geq }_{\text{AM-GM}} 3 \sqrt[3]{2\pi r^2 \frac{V^2}{r^2} } = 3 (2 \pi V^2)^{1/3} \end{align*} Equality holds when \(r = \sqrt[3]{\frac{V}{2 \pi}}, h = \frac{V}{\pi (V/2\pi)^{2/3}} = \sqrt[3]{\frac{4V}{\pi}}\) Since \(h > r\) the sphere has a maximum radius of \(r\) and so it's largest volume is \(\frac43 \pi r^3 = \frac43 \pi \frac{V}{2 \pi} = \frac23 V\).

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Solution: \begin{align*} f(x) & =nx-\binom{n}{2}\frac{x^{2}}{2}+\binom{n}{3}\frac{x^{3}}{3}-\cdots+(-1)^{r+1}\binom{n}{r}\frac{x^{r}}{r}+\cdots+(-1)^{n+1}\frac{x^{n}}{n} \\ f'(x) &= n - \binom{n}{2} x + \binom{n}{3}x^2 - \cdots (-1)^{r+1} \binom{n}{r} + \cdots + (-1)^{n+1} x^{n-1} \\ &= \frac{1-(1-x)^n}{x} \end{align*} Therefore, since \(\displaystyle f(x) = \int_0^xf'(t)\,dt\) \begin{align*} f(x) &= \int_0^x \frac{1 - (1-t)^n}{t} \, dt \\ &= \int_{1}^{1-x} \frac{1-y^n}{1-y} (-1)\, dy \tag{Let \(y = 1-t, \frac{dy}{dt} = -1\)} \\ &= \boxed{\int_{1-x}^1 \frac{1-y^n}{1-y} dy} \\ &= \int_{1-x}^1 \l 1 + y + y^2 + \cdots + y^{n-1} \r \, dy \\ &= \left [ y + \frac{y^2}{2} + \frac{y^3}{3} + \cdots + \frac{y^n}{n} \right]_{1-x}^1 \\ \end{align*} So when \(x = 1, 1-x = 0\) so we exactly have the sum required.

Solution:

1. \begin{align*} && (1+x+x^{2}+\cdots+x^{n})(1-x) &= 1-x+x-x^2+\cdots -x^n+x^n-x^{n+1} \\ &&&= 1-x^{n+1} \\ \Rightarrow && 1+x+x^2+\cdots+x^n &= \frac{1-x^{n+1}}{1-x} \tag{dividing by \(1-x\)} \end{align*}
2. \begin{align*} \frac{\d}{\d x}: && 0+1+2x+\cdots+nx^{n-1} &= \frac{(n+1)(1-x)x^n+(1-x^{n+1})}{(1-x)^2} \\ \Rightarrow && 1-2x+\cdots+(-1)^n n &= \frac{-(n+1)2(-1)^n+(1-(-1)^{n+1})}{4} \\ &&&= \begin{cases} \frac{-(n+1)\cdot2\cdot1+(1-(-1)}{4} & \text{if }n\text{ is even} \\ \frac{-(n+1)\cdot 2 \cdot(-1)+(1-1)}{4} & \text{if }n\text{ is odd}\end{cases} \\ &&&= \begin{cases} \frac{-n}{2} & \text{if }n\text{ is even}\\ \frac{n+1}{2} & \text{if }n\text{ is odd}\end{cases} \\ \end{align*}
3. \begin{align*} x: && x+2x^2+\cdots+nx^{n} &= \frac{(n+1)(1-x)x^{n+1}+x(1-x^{n+1})}{(1-x)^2} \\ &&&= \frac{x+(n+1)x^{n+1}-nx^{n+2}}{(1-x)^2}\\ \frac{\d}{\d x}: && 1^2+2^2x + \cdots + n^2x^{n-1} &= \frac{(1-x)^2(1+(n+1)^2x^{n}-n(n+2)x^{n+1}) +2(1-x)(x+(n+1)x^{n+1}-nx^{n+2})}{(1-x)^4} \\ &&&= \frac{1 + x - (1 + n)^2 x^n + (2 n^2+2n-1) x^{n+1} - n^2 x^{n+2}}{(1-x)^3} \\ \Rightarrow && 1^2-2^2 + \cdots + (-1)^{n-1}n^2 &= \frac{(-1)^n \l - (1 + n)^2- (2 n^2+2n-1) - n^2 \r}{8} \\ &&&= \frac{(-1)^n(-4n^2-4n}{8} \\ &&&= \frac{(-1)^{n-1}(n^2+n)}{2} \end{align*}

Solution: \begin{align*} && y &= x^a \\ && \frac{\d y}{\d x} &= \begin{cases} ax^{a-1} & a \neq 0 \\ 0 & a = 0 \end{cases} \\ \\ && y &= a^x \\ &&&= e^{(\ln a) \cdot x} \\ && \frac{\d y}{\d x} &= \ln a e^{(\ln a) x} \\ &&&= \ln a \cdot a^ x \\ \\ && y &= x^x \\ &&&= e^{x \ln x}\\ && \frac{\d y}{\d x} &= e^{x \ln x} \cdot \left ( \ln x + x \cdot \frac1x \right) \\ &&&= x^x \left (1 + \ln x \right) \\ \\ && y&= x^{(x^x)} \\ &&&= e^{x^ x \cdot \ln x} \\ && \frac{\d y}{\d x} &= e^{x^x \cdot \ln x} \left ( x^x \left (1 + \ln x \right) \cdot \ln x + x^x \cdot \frac1x\right) \\ &&&= x^{x^x} \left (x^x (1+ \ln x) \ln x +x^{x-1} \right) \\ &&&= x^{x^x+x-1} \left (1 + x \ln x + x (\ln x)^2 \right) \\ \\ && y &= (x^x)^x \\ &&&= x^{2x} \\ &&&= e^{2x \ln x} \\ && \frac{\d y}{\d x} &= e^{2 x \ln x} \left (2 \ln x + 2 \right) \\ &&&= 2(x^x)^x(1 + \ln x) \end{align*}

Solution: \begin{align*} && p_0(x) &= (1-x)(1-x^2)(1-x^4) \\ &&&= (1-x)(1-x)(1+x)(1-x^2)(1+x^2) \\ &&&= (1-x)^2 (1+x)(1-x)(1+x)(1+x^2) \\ &&&= (1-x)^3 (1+x)^2 (1+x^2) \end{align*} \begin{align*} && p_0'(x) &= 3(1-x)^2(1+x)^2(1+x^2) + (1-x)^3 q(x) \\ \Rightarrow && p_0'(1) &= 3 \cdot 0 \cdots + 0 \cdots \\ &&&= 0 \\ && p_1'(x) &= p_0(x) + xp'_0(x) \\ \Rightarrow && p_1'(1) &= p_0(1) + 1\cdot p_0'(1) \\ &&&= 0 + 1 \cdot 0 \\ &&&= 0 \end{align*} Notice that \(p_0(x) = (1-x-x^2+x^3)(1-x^4) = 1-x-x^2+x^3-x^4+x^5+x^6-x^7\), so: \(p'_0(x) = -1-2x+3x^2-4x^3+5x^4+6x^5-7x^6 \Rightarrow p'_0(1) = 0 = -1 -2 -4 -7 + 3 + 5+6\). \((xp'_1(1))' = 0 = -1^2-2^2-4^2-7^2 + 3^2 + 5^2 + 6^2\). Consider \(q_0(x) = (1-x)(1-x^2)(1-x^4)(1-x^8)\), then \((1-x)^4\) is a factor, so in particular we know \(q_0(1), (xq_0(x))'|_{x=1} = 0,(x(xq_0(x))')'|_{x=1} = 0\), and so: \(q_0(x) = 1-x-x^2+x^3-x^4+x^5+x^6-x^7 - x^8+x^9+x^{10}-x^{11}+x^{12}-x^{13}-x^{14}+x^{15}\), and so: \(1^r+2^r+4^r+7^r+8^r+11^r+13^r+14^r = 3^r+5^r+6^r+9^r+10^r+12^r+15^r\) for \(r = 1,2,3\)