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2022 Paper 2 Q2
D: 1500.0 B: 1500.0
sequences and series recurrence relation polynomial sequences difference equations degree of polynomial

A sequence \(u_n\), where \(n = 1, 2, \ldots\), is said to have \emph{degree} \(d\) if \(u_n\), as a function of \(n\), is a polynomial of degree \(d\).

  1. Show that, in any sequence \(u_n\) \((n = 1, 2, \ldots)\) that satisfies \(u_{n+1} = \frac{1}{2}(u_{n+2} + u_n)\) for all \(n \geqslant 1\), there is a constant difference between successive terms. Deduce that any sequence \(u_n\) for which \(u_{n+1} = \frac{1}{2}(u_{n+2} + u_n)\), for all \(n \geqslant 1\), has degree at most 1.
  2. The sequence \(v_n\) \((n = 1, 2, \ldots)\) satisfies \(v_{n+1} = \frac{1}{2}(v_{n+2} + v_n) - p\) for all \(n \geqslant 1\), where \(p\) is a non-zero constant. By writing \(v_n = t_n + pn^2\), show that the sequence \(v_n\) has degree 2. Given that \(v_1 = v_2 = 0\), find \(v_n\) in terms of \(n\) and \(p\).
  3. The sequence \(w_n\) \((n = 1, 2, \ldots)\) satisfies \(w_{n+1} = \frac{1}{2}(w_{n+2} + w_n) - an - b\) for all \(n \geqslant 1\), where \(a\) and \(b\) are constants with \(a \neq 0\). Show that the sequence \(w_n\) has degree 3. Given that \(w_1 = w_2 = 0\), find \(w_n\) in terms of \(n\), \(a\) and \(b\).

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2012 Paper 2 Q2
D: 1600.0 B: 1530.0
polynomials composition of functions degree of polynomial functional equation simultaneous equations proof by contradiction quadratic

If \(\p(x)\) and \(\q(x)\) are polynomials of degree \(m\) and \(n\), respectively, what is the degree of \(\p(\q(x))\)?

  1. The polynomial \(\p(x)\) satisfies \[ \p(\p(\p(x)))- 3 \p(x)= -2x\, \] for all \(x\). Explain carefully why \(\p(x)\) must be of degree 1, and find all polynomials that satisfy this equation.
  2. Find all polynomials that satisfy \[ 2\p(\p(x)) +3 [\p(x)]^2 -4\p(x) =x^4 \] for all \(x\).

Solution: If \(\p(x)\) and \(\q(x)\) are polynomials of degree \(m\) and \(n\), \(\p(\q(x))\) has degree \(mn\).

  1. Suppose \(\p(\p(\p(x)))- 3 \p(x)= -2x\), and suppose \(p(x)\) has degree \(n = \geq 2\), then \(\p(\p(\p(x)))\) has degree \(n^3\) and so the left hand side has degree higher than \(1\) and the right hand side is degree \(1\). Therefore \(\p(x)\) is degree \(1\) or \(0\). If \(p(x) = c\) then \(c^3-3c = -2x\) but the LHS doesn't depend on \(x\) which is also a contradiction. Therefore \(\p(x)\) is degree \(1\). Suppose \(\p(x) = ax+b\) then: \begin{align*} && -2x &= \p(\p(\p(x))) - 3\p(x) \\ &&&= \p(\p(ax+b)) - 3(ax+b) \\ &&&= \p(a(ax+b)+b) - 3ax -3b \\ &&&= a(a^2x+ab+b) + b - 3ax - 3b \\ &&&= (a^3-3a)x + b(a^2+a-2) \\ \Rightarrow &&& \begin{cases} a^3-3a&=-2 \\ b(a^2+a-2) &= 0\end{cases} \\ \Rightarrow &&& \begin{cases} a^3-3a+2 = 0 \\ b = 0, a = 1, a = -2\end{cases} \\ \Rightarrow &&& \begin{cases} (a-1)(a^2+a-2) = 0 \\ b = 0, a = 1, a = -2\end{cases} \\ \Rightarrow && (a,b) &= (1, b), (-2,b) \end{align*}
  2. Suppose \(2\p(\p(x)) +3 [\p(x)]^2 -4\p(x) =x^4\) and let \(\deg \p(x) = n\), then LHS has degree \(\max(n^2,2n,n)\) and the right hand side has degree \(4\). Therefore \(\p(x)\) must have degree \(2\). Let \(\p(x) = ax^2 + bx + c\), then, considering the coefficient of \(x^4\) in \(2\p(\p(x)) + 3[\p(x)]^2 -4\p(x)\) we will have \(2a^3+3a^2=1 \Rightarrow 2a^3+3a^2-1 = (a+1)^2(2a-1) \Rightarrow a = -1, a=\frac12\). Consider the coefficient of \(x^3\) in \(2\p(\p(x)) + 3[\p(x)]^2 -4\p(x)\) we have \(4a^2b+6ab = 0 \Rightarrow 2ab(2a+3) = 0\) Since \(a = -1, \frac12\) this means \(b = 0\). Consider the constant coefficient in \(2\p(\p(x)) + 3[\p(x)]^2 -4\p(x)\) (using \(b = 0\)). \(2ac^2+c+3c^2-4c = 0 \Rightarrow c(2ac+3c-3) = 0\). Therefore \(c = 0\) or \(a = -1, c = 3, a = \frac12, c = \frac34\), so our possible polynomials are: \(\p(x) = -x^2, \frac12x^2, -x^2+3, \frac12x^2+\frac34\)

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1999 Paper 2 Q3
D: 1600.0 B: 1500.0
proof by induction differentiation polynomial Leibniz rule exponential function degree of polynomial leading coefficient inequality

Let $$ {\rm S}_n(x)=\mathrm{e}^{x^3}{{\d^n}\over{\d x^n}}{(\mathrm{e}^{-x^3})}.$$ Show that \({\rm S}_2(x)=9x^4-6x\) and find \({\rm S}_3(x)\). Prove by induction on \(n\) that \({\rm S}_n(x)\) is a polynomial. By means of your induction argument, determine the order of this polynomial and the coefficient of the highest power of \(x\). Show also that if \(\displaystyle \frac{\d S_n}{\d x}=0\) for some value \(a\) of \(x\), then \(S_n(a)S_{n+1}(a)\le0\).

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

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