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2003 Paper 3 Q3
If \(m\) is a positive integer, show that \(\l 1+x \r^m + \l 1-x \r^m \ne 0\) for any real \(x\,\). The function \(\f\) is defined by \[ \f (x) = \frac{ (1+x )^m - ( 1-x )^m}{ (1+x )^m + (1-x )^m} \;. \] Find and simplify an expression for \(\f'(x)\). In the case \(m=5\,\), sketch the curves \(y = \f (x)\) and \(\displaystyle y = \frac1 { \f (x )}\;\).
Solution: If \(m\) is even, clearly that expression is positive since it's the sum of two (different) squares. If \(m\) is odd, then we can expand it as a sum of powers of \(x^2\) with a leading coefficient of \(1\) so it is also positive. \begin{align*} && f (x) = \frac{ (1+x )^m - ( 1-x )^m}{ (1+x )^m + (1-x )^m} \\ && f'(x) &= \frac{(m(1+x )^{m-1} + m( 1-x )^{m-1})((1+x)^m + (1-x)^m ) - ((1+x )^m - ( 1-x )^m)(m(1+x)^{m-1} - m(1-x)^{m-1} )}{\l (1+x)^m + (1-x)^m \r^2} \\ &&&= \frac{2m(1+x)^m(1-x)^{m-1}+2m(1+x)^{m-1}(1-x)^m}{\l (1+x)^m + (1-x)^m \r^2} \\ &&&= \frac{2m(1+x)^{m-1}(1-x)^{m-1}(1+x+1-x)}{\l (1+x)^m + (1-x)^m \r^2} \\ &&&= \frac{4m(1+x)^{m-1}(1-x)^{m-1}}{\l (1+x)^m + (1-x)^m \r^2} \\ \end{align*}
1989 Paper 3 Q5
Given that \(y=\cosh(n\cosh^{-1}x),\) for \(x\geqslant1,\) prove that \[ y=\frac{(x+\sqrt{x^{2}-1})^{n}+(x-\sqrt{x^{2}-1})^{n}}{2}. \] Explain why, when \(n=2k+1\) and \(k\in\mathbb{Z}^{+},\) \(y\) can also be expressed as the polynomial \[ a_{0}x+a_{1}x^{3}+a_{2}x^{5}+\cdots+a_{k}x^{2k+1}. \] Find \(a_{0},\) and show that
- \(a_{1}=(-1)^{k-1}2k(k+1)(2k+1)/3\);
- \(a_{2}=(-1)^{k}2(k-1)k(k+2)(2k+1)/15.\)
Solution: Recall, \(\cosh^{-1} x = \ln (x + \sqrt{x^2-1})\) \begin{align*} \cosh(n \cosh^{-1} x) &= \frac12 \left ( \exp(n \cosh^{-1} x) + \exp(-n\cosh^{-1}x) \right) \\ &= \frac12 \left ((x + \sqrt{x^2-1})^n + (x + \sqrt{x^2-1})^{-n} \right) \\ &= \frac12 \left ((x + \sqrt{x^2-1})^n + (x - \sqrt{x^2-1})^{n} \right) \\ \end{align*} When \(n = 2k+1\) \begin{align*} \cosh(n \cosh^{-1} x)&= \frac12 \left ((x + \sqrt{x^2-1})^n + (x - \sqrt{x^2-1})^{n} \right) \\ &= \frac12 \left (\sum_{i=0}^{2k+1}\binom{2k+1}{i}x^{2k+1-i}\left ( (\sqrt{x^2-1}^{i} + (-\sqrt{x^2-1})^{i} \right) \right) \\ &=\sum_{i=0}^{k} \binom{2k+1}{2i}x^{2k+1-2i}(x^2-1)^i \\ &=\sum_{i=0}^{k} \binom{2k+1}{2i}x^{2(k-i)+1}(x^2-1)^i \\ \end{align*} Which is clearly a polynomial with only odd degree terms. \begin{align*} a_0 &= \frac{\d y}{\d x} \vert_{x=0} \\ &= \sum_{i=0}^k\binom{2k+1}{2i} \left ( (2(k-i)+1)x^{2(k-i)}(x^2-1)^i + 2i\cdot x^{2(k-i)+2}(x^2-1) \right) \\ &= \binom{2k+1}{2k} (-1)^{k} \\ &= (-1)^k(2k+1) \end{align*}
- \begin{align*} a_1 &= \binom{2k+1}{2k}\binom{k}{1}(-1)^{k-1}+\binom{2k+1}{2(k-1)}(-1)^{k-1} \\ &=(-1)^{k-1}\cdot ( (2k+1)k + \frac{(2k+1)\cdot 2k \cdot (2k-1)}{3!}) \\ &= (-1)^{k-1}(2k+1)k\frac{3 + 2k-1}{3} \\ &= (-1)^{k-1}2(2k+1)k (k+1) \end{align*}
- \begin{align*} a_2 &= \binom{2k+1}{2k} \binom{k}{2}(-1)^{k-2} + \binom{2k+1}{2(k-1)} \binom{k-1}{1} (-1)^{k-2}+\binom{2k+1}{2(k-2)} (-1)^{k-2} \\ &= \binom{2k+1}{1} \binom{k}{2}(-1)^{k-2} + \binom{2k+1}{3} \binom{k-1}{1} (-1)^{k-2}+\binom{2k+1}{5} (-1)^{k-2} \\ &= (-1)^{k} \left (\binom{2k+1}{1} \frac{k(k-1)}{2} + \binom{2k+1}{3}(k-1)+\binom{2k+1}{5} \right) \\ &= (-1)^{k} \left ( \frac{(2k+1)k(k-1)}{2} + \frac{(2k+1)k(2k-1)}{3} + \frac{(2k+1)k(2k-1)(k-1)(2k-3)}{5\cdot2\cdot3} \right) \\ &= (-1)^k (2k+1)k\frac{1}{30} \left ( 15(k-1) + 10(2k-1)+(2k-1)(k-1)(2k-3) \right) \end{align*}