Problems

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Solution:

1. If \(a,b,c\) were all real then \(a^2+b^2+c^2 = 0 \Rightarrow a,b,c = 0\) but they are non-zero. Therefore they cannot all be real. Since \((a+b+c)^2 = 0\) we must have \(ab+bc+ca = 0\). Therefore \(a,b,c\) must satisfy \(x^3 -abc = 0 \Rightarrow\) they all have the same modulus, since they are all cube roots of the same number.
2. Notice that \(a^3+b^3+c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - ab-bc-ca) \Rightarrow abc = 0\) but therefore they cannot all be non-zero.
3. Suppose \(a+b+c+d = 0\) then note that \(\displaystyle a^2+b^2+c^2+d^2 = (a+b+c+d)^2 - 2\sum_{sym} ab\) and \(\displaystyle a^3+b^3+c^3+d^3 = (a+b+c+d)^3 - 3(a+b+c+d)(ab+ac+ad+bc+bd+cd) + 3(abc+abd+acd+bcd) \Rightarrow abc+abd+acd+bcd = 0\). Therefore \(a,b,c,d\) are roots of a polynomial of the form \(x^4 -kx^2 + l = 0\), but this means they must come in pairs with the same modulus.
4. Suppose \(c = 1, d = -2, e = 3\) so \(c+d+e = 2\) and \(c^3 + d^3 + e^3 = 1 - 8 + 27 = 20\), so we need to find \(a,b\) satisfying \(a+b = -2, a^2+b^2 = -20\), ie \(4 = (a+b)^2 = -20 + 2ab \Rightarrow ab = 12\), so we need the roots of \(x^2 +2x + 12= 0\) which clearly have different modulus.

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Solution:

1. \(\,\) \begin{align*} && I &= \int_{-a}^a \frac{1}{1+e^x} \d x \\ &&&= \int_{-a}^a \frac{e^{-x}}{e^{-x}+1} \d x \\ &&&= \left [ -\ln(1 + e^{-x} ) \right ]_{-a}^a \\ &&&= \ln(1 + e^a) - \ln(1 + e^{-a}) \\ &&&= \ln \left ( \frac{1+e^a}{1+e^{-a}} \right) \\ &&&= \ln \left ( e^a \frac{1+e^a}{e^a+1} \right) \\ &&& = a \end{align*}
2. Suppose \(g\) is continuous and \(\int_0^a g(x) \d x = 0\) for all \(a \geq 0\) then \(g(x) = 0\) for all \(x\). Proof: Differentiate with respect to \(a\) to obtain \(g(a) = 0\) for all \(a\) as required. \begin{align*} && a &= \int_{-a}^a \frac{1}{1+ f(x)} \d x \\ \Leftrightarrow && 1 &= \frac{1}{1 + f(a)} + \frac{1}{1 + f(-a)} \tag{FTC} \\ \Leftrightarrow && (1+f(x))(1+f(-x)) &= 2+f(-x) + f(x) \\ \Leftrightarrow && f(x) f(-x) & = 1 \end{align*}
3. \(\,\) \begin{align*} && J &= \int_{-a}^a \frac{h(x)}{1 + f(x)} \d x \\ y = - x: &&&=\int_{-a}^a \frac{h(-y)}{1 + f(-y)} \d y \\ &&&= \int_{-a}^a \frac{h(y)}{1 + f(-y)} \d y \\ \Rightarrow && 2J &= \int_{-a}^a h(x) \left ( \frac{1}{1+f(x)} + \frac{1}{1+f(-x)} \right) \d x \\ &&&= \int_{-a}^a h(x) \d x \\ &&&= \int_{-a}^0 h(x) \d x + \int_0^a h(x) \d x\\ &&&= \int_0^a h(-x) \d x + \int_0^a h(x) \d x \\ &&&= 2 \int_0^a h(x) \d x \\ \Rightarrow && J &= \int_0^a h(x) \d x \end{align*}
4. First notice that \(h(x) = \cos x = h(-x)\). Also notice that if \(f(x) = e^{2x}\) then \(f(x)f(-x) = 1\) so \begin{align*} && K &= \int_{-\frac12 \pi}^{\frac12\pi} \frac{e^{-x}\cos x}{\cosh x} \d x \\ &&&= \int_{-\frac12 \pi}^{\frac12\pi} \frac{2h(x)}{1+f(x)} \d x \\ &&&= 2 \int_0^{\frac12 \pi} h(x) \d x \\ &&&= 2 \int_0^{\frac12 \pi} \cos x \d x \\ &&&= 2 \end{align*}

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Solution:

1. Suppose \(n = 1\), then all polynomials are reflexive (since \(x - a_1\) has the root \(a_1\). Suppose \(n = 2\), then we want \begin{align*} && x^2-a_1x+a_2 &= (x-a_1)(x-a_2) \\ &&&= x^2-(a_1+a_2)x+a_1a_2 \\ \Rightarrow && a_2 &= 0 \\ \end{align*} So all polynomials of the form \(x^2-a_1x\) work and no others. Suppose \(n = 3\) then we want \begin{align*} && x^3-a_1x^2+a_2x-a_3 &= (x-a_1)(x-a_2)(x-a_3) \\ &&&= x^3-(a_1+a_2+a_3)x+(a_1a_2+a_1a_3+a_2a_3)x-a_1a_2a_3 \\ \Rightarrow && a_2+a_3 &= 0 \\ && a_2a_3 &= a_2 \\ \Rightarrow && -a_2^2 &= a_2 \\ \Rightarrow && a_2 &= 0, -1 \\ && -a_1a_2^2 &= -a_2 \\ \Rightarrow && a_2 &= 0, a_2 = 1/a_1 \end{align*} So we need either \(x^3-a_1x\) or \((x+1)^2(x-1) = x^3+x^2-x-1\)
2. Suppose \(n > 3\) then \begin{align*} && \sum a_i^2 &= \left (\sum a_i \right)^2 - 2 \sum_{i < j} a_i a_j \\ && &= a_1^2 - 2a_2 \\ \Rightarrow && 2a_2 &= a_1^2 - \sum a_i^2 \\ &&&= -a_2^2 - a_3^2 - \cdots - a_n^2 \end{align*} So \((a_2+1)^2 = 1-a_3^2 -\cdots -a_n^2\) so if \(a_n > 0\) (or any other \(a_i, i > 2\) for that matter) then we must have \(a_n = \pm 1, a_{3}, \ldots a_{n-1} = 0\), but if \(a_n = \pm 1\) \(x = 0\) is not a root. Therefore we must have \(a_0\) and \(a_i = 0\) for all \(i > 3\)
3. The only reflexive polynomials therefore must be \(x^n - kx^{n-1}\) and \(x^{n+3}+x^{n+2}-x^{n+1}-x^n\)

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Solution: Let \(f(x) = x^4 + ax^3 + bx^2 + ax + 1\), and suppose \(f(k) = 0\). Since \(f(0) = 1\), \(k \neq 0\), therefore we can talk about \(k^{-1}\). \begin{align*} && f(k^{-1}) &= k^{-4} + ak^{-3} + bk^{-2} + ak^{-1} + 1 \\ &&&= k^{-4}(1 + ak + bk^2 + ak^3 + k^4) \\ &&&= k^{-4}(k^4+ak^3+bk^2+ak+1) \\ &&&= k^{-4}f(k) = 0 \end{align*} Therefore \(k^{-1}\) is also a root of \(f\)

1. If \(f\) has only on distinct root, we must have \(f(x) = (x+k)^4\) therefore \(k = k^{-1} \Rightarrow k^2 = 1 \Rightarrow k = \pm1\), or \(a = 4, b = 6\) or \(a = -4, b = 6\)
2. If \(f\) has exactly three distinct roots then one of the roots must be a repeated \(1\) or \(-1\), ie \(0 = f(1) = 1 + a + b + a + 1 = 2 + b +2a \Rightarrow b = -2a-2\) or \(0 = f(-1) = 1 -a + b -a + 1 \Rightarrow b = 2a - 2\)
3. If \(b = 2a-2\), we have \begin{align*} && f(x) &= 1 + ax + (2a-2)x^2 + ax^3 + x^4 \\ &&&= (x^2+2x+1)(1+(a-2)x+x^2) \\ \Rightarrow && x &= \frac{2-a \pm \sqrt{(a-2)^2 - 4}}{2} \\ &&&= \frac{2-a \pm \sqrt{a^2-4a}}{2} \end{align*}

\(f\) has exactly three distinct real roots iff \(b = \pm 2a - 2\) and \(b \neq 6\)

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Solution:

1. \begin{align*} && f(\beta) &= \beta - \frac1{\beta}-\frac1{\beta^2} \\ \Rightarrow && f'(\beta) &= 1 +\frac{1}{\beta^2}+\frac{2}{\beta^3} \\ \Rightarrow && 0 &= f'(\beta) \\ &&&= 1 + \frac1{\beta^2} + \frac{2}{\beta^3} \\ \Rightarrow && 0 &= \beta^3 + \beta + 2 \\ &&&= (\beta+1)(\beta^2-\beta+2) \end{align*} Therefore the only stationary point is at \(\beta = -1, f(-1) = -1\)
   

   + - ↺
   \begin{align*} && g(\beta) &= \beta + \frac3{\beta}-\frac1{\beta^2} \\ \Rightarrow && g'(\beta) &= 1 -\frac{3}{\beta^2}+\frac{2}{\beta^3} \\ \Rightarrow && 0 &= f'(\beta) \\ &&&= 1 - \frac3{\beta^2} + \frac{2}{\beta^3} \\ \Rightarrow && 0 &= \beta^3 - 3\beta + 2 \\ &&&= (\beta-1)^2(\beta+2) \end{align*} Therefore there are stationary points at \(\beta=1,f(1) = 3, \beta=-2, f(-2) = \frac14\)
   

   + - ↺
2. Let \(u,v\) be the roots of \(x^2 + \alpha x + \beta = 0\), then since \((x-u)(x-v) = 0\) we must have \(\alpha = -(u+v), \beta = uv\). Therefore: \begin{align*} && u+v +\frac{1}{uv} &= -\alpha + \frac{1}{\beta} \\ && \frac1u+\frac1v + uv &= \frac{u+v}{uv} + uv \\ &&&= -\frac{\alpha}{\beta} + \beta \end{align*} Given \(u+v + \frac 1 {uv} = -1\), ie \(-\alpha + \frac{1}{\beta} = -1\). Since the roots are real, we must also have that \(\alpha^2 - 4\beta \geq 0\), so \begin{align*} && -\alpha + \frac1\beta &= -1 \\ \Rightarrow && \alpha &= 1 +\frac1\beta \\ \Rightarrow && -\frac{\alpha}{\beta}+\beta &= -\frac{1}{\beta} \l1+\frac1{\beta}\r + \beta \\ &&&=\beta - \frac{1}{\beta}-\frac{1}{\beta^2} \end{align*} So we want to maximise \(f(\beta)\) subject to \(\alpha ^2 - 4\beta \geq 0\) \begin{align*} && 0 &\leq \alpha^2 -4 \beta \\ &&&= \l 1 + \frac1{\beta} \r^2 - 4\beta \\ &&&= 1+ \frac2{\beta} + \frac1{\beta^2} - 4\beta \\ \Leftrightarrow && 0 &\leq -4\beta^3+\beta^2 + 2\beta + 1 \\ &&&=-(\beta-1)(4\beta^2+3\beta+1)\\ \Leftrightarrow && \beta &\leq 1 \end{align*} But we know \(f(\beta) \leq -1\) on \((-\infty,1]\) so we're done.
3. Given that \(-\alpha + \frac{1}{\beta} = 3\) we have \begin{align*} && -\alpha + \frac1\beta &= 3 \\ \Rightarrow && \alpha &= -3 +\frac1\beta \\ \Rightarrow && -\frac{\alpha}{\beta}+\beta &= -\frac{1}{\beta} \l-3+\frac1{\beta}\r + \beta \\ &&&=\beta + \frac{3}{\beta}-\frac{1}{\beta^2} \end{align*} which we want to maximise, subject to: \begin{align*} && 0 &\leq \alpha^2 -4 \beta \\ &&&= \l -3 + \frac1{\beta} \r^2 - 4\beta \\ &&&= 9- \frac6{\beta} + \frac1{\beta^2} - 4\beta \\ \Leftrightarrow && 0 &\leq -4\beta^3+9\beta^2 - 6\beta + 1 \\ &&&=-(\beta-1)^2(4\beta-1)\\ \Leftrightarrow && \beta &\leq \frac14 \end{align*} Therefore the maximum will either be \(f(-2) = \frac14\) or \(f(\frac14) = -\frac{15}4\). Therefore the maximum is \(\frac14\)

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Solution: \begin{align*} A &= -(\alpha \beta + \gamma\delta + \alpha\gamma+\beta\delta+\alpha \delta + \beta\gamma) \\ &= -q \end{align*}

1. The corresponding cubic equation is: \begin{align*} && 0 &= y^3 - 3y^2-40y+(120-36) \\ &&&= y^3 -3y^2 - 40y + 84 \\ &&&= (y-7)(y-2)(y+6) \end{align*} Therefore \(\alpha\beta + \gamma \delta = 7\)
2. \begin{align*}(\alpha+\beta)(\gamma+\delta) &= \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta \\ &= 3 -(\alpha\beta + \gamma\delta) \\ &=3-7 = -4 \end{align*} Let \(\alpha\beta\) and \(\gamma\delta\) be the roots of a quadratic; then the quadratic will be \(t^2-7t+10 = 0 \Rightarrow t = 2,5\) so \(\alpha\beta = 5\)
3. \(\alpha\beta = 5, \gamma\delta = 2\) Consider the quadratic with roots \(\alpha+\beta\) and \(\gamma+\delta\), then \(t^2-4 = 0 \Rightarrow t = \pm 2\). Suppose \(\alpha+\beta = 2, \gamma+\delta=-2\) then \(\alpha, \beta = 1 \pm 2i, \gamma,\delta = -1 \pm i\) \(\alpha \beta \gamma + \beta\gamma\delta + \gamma\delta\alpha + \delta\alpha\beta = 5\gamma + 2\beta + 2\alpha + 5\delta = -6 \neq 6\) Suppose \(\alpha+\beta = -2, \gamma+\delta=2\) then \(\alpha, \beta = -1 \pm 2i, \gamma,\delta = 1 \pm i\) \(\alpha \beta \gamma + \beta\gamma\delta + \gamma\delta\alpha + \delta\alpha\beta = 5\gamma + 2\beta + 2\alpha + 5\delta = 6\), therefore these are there roots. (In some order): \(1 \pm i, -1 \pm 2i\)

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Solution: Suppose \(c = a+b\) then \begin{align*} (a+b+c)^3 &-6(a+b+c)(a^2+b^2+c^2) +8(a^3+b^3+c^3) \\ &= (2(a+b))^3-6(2(a+b))(a^2+b^2+(a+b)^2) + 8(a^3+b^3+(a+b)^3) \\ &=16(a+b)^3 - 24(a+b)(a^2+b^2+ab)+8(a^3+b^3) \\ &= 8(a+b)(2(a+b)^2-3(a^2+b^2+ab)+(a^2-ab+b^2)) \\ &= 0 \end{align*} Therefore \(a+b-c\) is a factor. By symmetry \(a-b+c\) and \(-a+b+c\) are also factors. Since our polynomial is degree \(3\) it must be \(K(a+b-c)(b+c-a)(c+a-b)\) for some \(K\). Since the coefficient of \(a^3\) is \(3\), \(K = 3\). so we have: \(3(a+b-c)(b+c-a)(c+a-b)\)

1. We want \(x + a + b = x+1\), \(x^2 + a^2 + b^2 = x^3+\frac52, x^3 + a^3 + b^3 = x^3+ \frac{13}{4}\). \(a+b = 1, a^2 + b^2 = 5/2\) so \(a = \frac32, b = -\frac12\) \begin{align*} 0 &= (x+1)^3 - 3(x+1)(2x^2+5)+2(4x^3+13) \\ &= 3(x +\frac{3}{2}+\frac{1}{2})(x - \frac{3}{2} - \frac{1}{2})(-x + \frac{3}{2} - \frac{1}{2}) \\ &= 3(x+2)(x-2)(1-x) \end{align*} and so the roots are \(x = 1, 2, -2\)
2. Letting \(c = d+e\) we have \begin{align*} (a+b+d+e)^3 &-6(a+b+d+e)(a^2+b^2+d^2+e^2) +8(a^3+b^3+d^3+e^3) \\ &= (a+b+c)^3 -6(a+b+c)(a^2+b^2+c^2-2de) +8(a^3+b^3+c^3 - 3cde) \\ &= (a+b+c)^3 -6(a+b+c)(a^2+b^2+c^2)+8(a^3+b^3+c^3)+12(a+b+c)de - 24cde \\ &= \underbrace{(a+b+c)^3 -6(a+b+c)(a^2+b^2+c^2)+8(a^3+b^3+c^3)}_{\text{has a factor of }a+b-c} + 12(a+b-c)de \end{align*} Therefore there is a factor of \(a+b-c\) or \(a+b-d-e\). By symmetry we must have the factors: \((a+b-d-e)(a-b-d+e)(a-b+d-e)\) and so the final expression must be: \(K(a+b-d-e)(a-b-d+e)(a-b+d-e)\) The coefficient of \(a^3\) is \(3\), therefore \(K = 3\) We want \(x+a+b+c = x + 6\), \(x^2+a^2+b^2+c^2 = 14\) and \(x^3 + a^3+b^3+c^3 = 36\), ie \(a = 1,b=2,c=3\) would work, so \begin{align*} 0 &= (x+6)^3 - 6(x+6)(x^2+14) +8(x^3+36) \\ &= 3(x+1-2-3)(x-1+2-3)(x-1-2+3) \\ &= 3x(x-4)(x-2) \end{align*} ie the roots are \(x = 0, 2, 4\)

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Solution:

1. \(\,\) \begin{align*} && S & = 1 +\quad 2\quad \;\;+ \quad 3 \quad+ \cdots + \quad n \\ && S &= n + (n-1) + (n-2) + \cdots + 1 \\ && 2S &= (n+1) + (n+1) + \cdots + (n+1) \\ \Rightarrow && S &= \frac12n(n+1) \end{align*}
2. \(\,\) \begin{align*} && (N-m)^{k} + m^k&= \sum_{i=0}^k \binom{k}{i} N^{k-i} (-m)^{i} + m^k \\ &&&= N\sum_{i=0}^{k-1} \binom{k}{i}N^{k-i-1}(-m)^i -m^k+m^k \\ &&&= N\sum_{i=0}^{k-1} \binom{k}{i}N^{k-i-1}(-m)^i \end{align*} which is clearly divisible by \(N\).

\begin{align*} 2S &= 2\sum_{i=1}^n i^k \\ &= \sum_{i=0}^n (\underbrace{(n-i)^k + i^k}_{\text{divisible by }n}) \\ \end{align*} Therefore \(2S\) is divisible by \(n\) and so if \(n\) is odd, \(n\) divides \(S\) and if \(n\) is even, \(\frac{n}{2}\) divides \(S\). Also notice \begin{align*} 2S &= 2\sum_{i=1}^n i^k \\ &= \sum_{i=1}^{n} (\underbrace{(n+1-i)^k + i^k}_{\text{divisible by }n+1}) \\ \end{align*} Therefore if \(n+1\) is odd, \(n+1 \mid S\) otherwise \(\frac{n+1}{2} \mid S\), and in either case \(\frac{n(n+1)}{2} \mid S\) (since they are both coprime) but this is the same as \(1 + 2 + \cdots + n \mid S\)

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Solution:

1. Let \(z \in \mathbb{R}\) and let \(z \to \pm \infty\) then \(z^3 + az^2 + bz + c\) changes sign, therefore somewhere it must have a real root.
2. \begin{align*} &&z^3 + az^2 + bz + c &= (z-z_1)(z-z_2)(z-z_3) \\ && &= z^3 - (z_1+z_2+z_3)z^2 + (z_1z_2 + z_2z_3+z_3z_1)z - (z_1z_2z_3) \\ \\ \Rightarrow && S_1 &= z_1+z_2+z_3 \\ &&&= -a \\ \\ \Rightarrow && S_2 &= z_1^2+z_2^2+z_3^2 \\ &&&= (z_1+z_2+z_3)^2 - 2(z_1z_2 + z_2z_3+z_3z_1) \\ &&&= a^2 - 2b \\ \Rightarrow && a &= -S_1 \\ && b &= \frac12 \l S_1^2 - S_2\r \\ \\ && 0 &= z_i^3 + az_i^2+bz_i+c \\ \Rightarrow && 0 &= S_3 + aS_2+bS_1+3c \\ &&&= S_3 -S_1S_2 + \frac12 \l S_1^2 - S_2\r S_1 + 3c \\ \Rightarrow && 0 &= 2S_3 - 3S_1S_2 + S_1^3 + 6c \end{align*}
3. Let \(z_k= r_ke^{i \theta_k}\), then we have \(\textrm{Im}(S_k) = 0\) and so the polynomial with roots \(z_k\) has real coefficients, and therefore at least one root is real. This root will have \(\theta_k = 0\). Moreover, since if \(w\) is a root of a real polynomial \(\overbar{w}\) is also a root, and therefore if \(\theta_1 = 0\), we must have that \(z_2\) and \(z_3\) are complex conjugate, ie \(\theta_2 = - \theta_3\)

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Solution:

1. Notice that \(u^2 = v+2wz\), so \(w,z\) are roots of the quadratic \(t^2 -ut+\frac{u^2-v}{2}\). Therefore they are both real if \(u^2 \geq 2(u^2-v) \Rightarrow 2v \geq u^2\).
2. \begin{align*} && w+z &= u \\ && w^2+z^2 &= u^2 - \tfrac23 \\ && w^3+z^3 &= \lambda(u-1) \\ \\ && wz &= \frac{u^2 - (u^2-\tfrac23)}{2} = \tfrac13\\ \\ && (w+z)(w^2+z^2) &= w^3+z^3+wz(w+z) \\ &&u(u^2-\tfrac23)&= \lambda(u-1)+\frac13u \\ \Rightarrow && u^3-u&= \lambda (u-1) \\ \Rightarrow && 0 &= (u-1)(u(u+1) - \lambda) \\ \Rightarrow && 0 &= (u-1)(u^2+u - \lambda) \end{align*} Therefore there will be at most 3 values for \(u\), unless \(1\) is a root of \(u^2+u-\lambda\), ie \(\lambda = 2\). Suppose \(u = 1\), then we have: \(w+z = 1, wz = 1/3 \Rightarrow w,z = \frac{-1 \pm \sqrt{-1/3}}{2}\) which are clearly complex.

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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 \(S_2 = \frac{1^2+1^2 + (-2)^2}{2} = 3, S_7 = \frac{1^2+1^2+(-2)^7}{7} = -18, S_9 = \frac{1^1+1^2+(-2)^9}{9} = \frac{2 - 512}{9} \neq 3 \cdot (-18)\)

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Solution: \(p(x) = q(x)(x-1)^5 + 1\) where \(q(x)\) has degree \(4\).

1. \(p(1) = q(1)(1-1)^5 + 1 = 1\).
2. \(p'(x) = q'(x)(x-1)^5 + 5(x-1)^4q(x) + 0 = (x-1)^4((x-1)q'(x) + 5q(x))\) so \(p'(x)\) is divisible by \((x-1)^4\)
3. \(p(x)+1\) divisible by \((x+1)^5\) means that \(p(-1) = -1\) and \(p'(x)\) is divisible by \((x+1)^4\). Since \(p'(x)\) is degree \(8\) it must be \(c(x+1)^4(x-1)^4 = c(x^2 - 1)^4\). Expanding and integrating, we get \(p(x) = c(\frac{1}{9}x^9 -\frac{4}{7}x^7 + \frac{6}{5}x^5 - \frac{4}{3}x^3 + x) + d\). When \(x = 1\) we get \(c \frac{128}{315} + d = 1\) and when \(x = -1\) we get \(-c \frac{128}{315} + d = -1\) so \(2d = 0 \Rightarrow d = 0, c = \frac{315}{128}\) and \[ p(x) =\frac{315}{128} \l \frac{1}{9}x^9 -\frac{4}{7}x^7 + \frac{6}{5}x^5 - \frac{4}{3}x^3 + x\r \]