Problems

Solution:

1. \begin{align*} && \int \frac {x^3-2}{(x+1)^2}\, \e ^x \d x &= \frac{\P(x)}{Q(x)}\,\e^x + \text{constant} \\ \underbrace{\Rightarrow}_{\frac{\d}{\d x}} && \frac{x^3-2}{(x+1)^2}e^x &= \frac{P'(x)Q(x)-Q'(x)P(x)}{Q(x)^2}e^x + \frac{P(x)}{Q(x)}e^x \\ \Rightarrow && \frac{x^3-2}{(x+1)^2} &= \frac{(P(x)+P'(x))Q(x)-Q'(x)P(x)}{Q(x)^2} \\ \Rightarrow && Q(x)^2(x^3-2) &= ((P(x)+P'(x))Q(x)-Q'(x)P(x))(x+1)^2 \\ \Rightarrow && Q(-1) &= 0 \\ \Rightarrow && x+1 &\mid Q(x) \end{align*} We have \(\frac{x^3-2}{(x+1)^2}\) has degree \(1\) (plus some remainder term). Therefore \begin{align*} 1 &= \deg \l (P(x)+P'(x))Q(x)-Q'(x)P(x)\r - 2 \deg Q(x) \\ &= \deg P(x) + \deg Q(x) - 2 \deg Q(x) \\ &= \deg P(x) - \deg Q(x) \end{align*} as required. Suppose \(Q(x) = x+1, P(x) = ax^2+bx+c\) then \begin{align*} && \frac{x^3-2}{(x+1)^2} &= \frac{(P(x)+P'(x))(x+1)-P(x)}{(x+1)^2} \\ \Rightarrow && x^3-2 &= (P(x)+P'(x))(x+1) - P(x) \\ \Rightarrow && x^3-2 &= (ax^2+bx+c+2ax+b)(x+1) - (ax^2+bx+c) \\ &&&= a x^3+ x^2 (2 a + b) + x (2 a + b + c)+b \\ \Rightarrow && a &= 1 \\ && b &= -2 \\ && c &= 0 \end{align*} So \(P(x) = x^2-2x\)
2. \begin{align*} && \int \frac1{x+1}e^x \d x &= \frac{P(x)}{Q(x)}e^x + c \\ \Rightarrow && \frac{1}{x+1} e^x &= \frac{P'(x)Q(x)-Q'(x)P(x)}{Q(x)^2}e^x + \frac{P(x)}{Q(x)}e^x \\ \Rightarrow && \frac{1}{x+1} &= \frac{(P(x)+P'(x))Q(x)-Q'(x)P(x)}{Q(x)^2} \end{align*} Therefore \(Q(-1) = 0\) and so \(x +1 \mid Q(x)\). Considering degrees, we must have that \(P(x)\) has degree \(1\) less than \(Q(x)\). Consider also the number of factors of \(x+1\) in the numerator and denominator. Since \(P(x)\) and \(Q(x)\) have no common factors, the \(Q(x)\) could have \(q\) factors and \(P(x)\) must have none. The denominator therefore has \(2q\) factors and the numerator must have \(q-1\) factors (coming from \(Q'(x)\)), we must have \(2q = (q-1) + 1\), but that implies \(q = 0\). Contradiction! \end{align*}

Solution: Notice that the coefficient of \(x^r\) is \((-1)^r\frac{(-3) \cdot (-3-1) \cdots (-3-r+1)}{r!} = (-1)^r \frac{(-1)(-2)(-3)(-4) \cdots (-(r+2))}{(-1)(-2)r!} = (-1)^r(-1)^{r+2}\frac{(r+2)!}{2r!} = \frac{(r+2)(r+1)}2\).

1. The coefficient of \(x^r\) is \begin{align*} && c_r &=\frac{(r+1)(r+2)}{2} - \frac{(r-1+1)(r-1+2)}{2} + 2 \frac{((r-2+1)(r-2+2)}{2} \\ &&&= \frac{r^2+3r+2}{r} - \frac{r^2+r}{2} + \frac{2r^2-2r}{2}\\ &&&= \frac{2r^2+2}{2} = r^2+1 \end{align*} \begin{align*} && S & = 1+\frac22+\frac54+\frac{10}8+\frac{17}{16}+\frac{26}{32}+\frac{37}{64} +\frac{50}{128}+ \cdots \\ &&&= \sum_{r=0}^{\infty} \frac{r^2+1}{2^r} \\ &&&= \frac{1-\tfrac12+2 \cdot \tfrac14}{(1-\tfrac12)^3} \\ &&&= 8 \end{align*}
2. \(\,\) \begin{align*} && S &= 1+2+\frac94+2+\frac{25}{16}+\frac{9}{8}+\frac{49}{64} + \cdots \\ &&&= \sum_{r=0}^{\infty} \frac{(r+1)^2}{2^r} \\ &&&= 2 \sum_{r=0}^{\infty} \frac{(r+1)^2}{2^{r+1}} \\ &&&= 2 \sum_{r=1}^{\infty} \frac{r^2}{2^{r}} \\ &&&= 2 \left (\sum_{r=0}^{\infty} \frac{r^2+1}{2^{r}} - \sum_{r=0}^{\infty} \frac{1}{2^{r}} \right) \\ &&&= 2 (8 - 1) = 14 \end{align*}

Solution: \begin{align*} && \int_0^1 \frac{x^4}{1+x^2} \d x &= \int_0^1 \frac{(x^2-1)(1+x^2)+1}{x^2+1} \d x\\ &&&= \int_0^1 \frac{1}{1+x^2} \d x -\int_0^1 (1-x^2) \d x \\ &&&= \left [\tan^{-1}x \right]_0^1 - \left [x - \tfrac13x^3 \right]_0^1 \\ &&&= \frac{\pi}{4} - \frac23 \end{align*}

1. \(\,\) \begin{align*} && I &= \int_0^1 x^3 \; \tan ^{-1} \left(\frac {1-x} {1+x} \right) \,\d x \\ &&&= \left [ \frac{x^4}{4}\tan ^{-1} \left(\frac {1-x} {1+x} \right) \right]_0^1 -\int_0^1 \frac{x^4}{4} \frac{1}{1 +\left(\frac {1-x} {1+x} \right) ^2 } \cdot \frac{-2}{(1+x)^2} \d x \\ &&&= \frac{1}{2} \int_0^1 \frac{x^4}{(1+x)^2+(1-x)^2} \d x \\ &&&= \frac{1}{4} \int_0^1 \frac{x^4}{1+x^2} \d x \\ &&&= \frac{\pi}{16} - \frac{1}{6} \end{align*}
2. \(\,\) \begin{align*} && J &= \int_0^1 \frac {(1-y)^3} {(1+y)^5} \; {{\tan}^{-1} y}\, \d y \\ &&&= \left [ \frac {(y(1+y^2)} {(1+y)^4} \tan^{-1}y \right]_0^1 - \int_0^1 \frac {(y(1+y^2)} {(1+y)^4} \frac{1}{1+y^2} \d y \\ &&&= \frac{\pi}{32} - \int_0^1 \frac{y}{(1+y)^4} \d y \\ &&&= \frac{\pi}{32} - \left[ - \frac{3y+1}{6(1+y)^3} \right]_0^1 \\ &&&= \frac{\pi}{32} +\frac{4}{6 \cdot 8} - \frac{1}{6} \\ &&&= \frac{\pi}{32} - \frac{1}{12} \end{align*}

Solution: \begin{align*} && x+ c &= f(x) \\ \Rightarrow && (x+c)(x^2-1) &= x(x-2)(x-a) \\ \Rightarrow && x^3 + cx^2-x-c &= x^3-(2+a)x^2+2ax \\ \Rightarrow && 0 &= (c+2+a)x^2-(1+2a)x-c \\ && 0 &\leq \Delta = (1+2a)^2 + 4(2+c+a)c \\ &&&= 4c^2+(4a+8)c + (1+2a)^2 \\ && \Delta_c &= 16(a+2)^2-16(1+2a)^2 \\ &&&= 16(1-a)(3a+3) \\ &&&= 48(1-a^2) \end{align*} Therefore if \(|a| \geq 1\) we must have \(\Delta_c \leq 0\) which means \(\Delta \geq 0\) and so there are always solutions. If \(|a| < 1\) there are values for \(c\) where \(\Delta < 0\) and there would be no solutions. \begin{align*} && y &= \frac{x^3-(2+a)x^2+2ax}{x^2-1} \\ &&&= \frac{(x^2-1)(x-(2+a))+(2a+1)x-(2+a)}{x^2-1} \\ &&&= x - (2+a) + \frac{(2a+1)x-(2+a)}{x^2-1} \end{align*} therefore the oblique asymptote has equation \(y = x - (2+a)\)

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2. 

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Solution: Notice by polynomial division, we can write \(p(x) = (x-a)q(x) + r(x)\) where degree \(r(x) < 1\), ie \(r(x)\) is a constant. Evaluating at \(x = a\), we have \(p(a) = (a-a)q(a) + r(a) = r(a)\). Therefore \(r(a) = p(a)\) and since \(r(x)\) is a constant, it is always \(p(a)\).

1. \(\,\) \begin{align*} && p(x) &= (x-1)(x-2)(x-3)q(x) + r(x) \\ && p(1) &= r(1) = 3 \\ && p(2) &= r(2) = 1 \\ && p(3) &= r(3) = 5 \end{align*} Therefore \(r(x)\) is a polynomial of degree \(2\) or less through \((1,3),(2,1), (3, 5)\) we can write this as \begin{align*} && r(x) &= 5\frac{(x-1)(x-2)}{(3-1)(3-2)} + 1\frac{(x-1)(x-3)}{(2-1)(2-3)} + 3\frac{(x-2)(x-3)}{(1-2)(1-3)} \\ &&&= \frac52(x^2-3x+2)-(x^2-4x+3) + \frac32(x^2-5x+6) \\ &&&= 3x^2-11x+11 \end{align*}
2. Let \(P_n(x) = x(x-1)\cdots(x-n) + x\), then \(P_n(a) = a\)