1 problem found
If \(\mathrm{Q}\) is a polynomial, \(m\) is an integer, \(m\geqslant1\) and \(\mathrm{P}(x)=(x-a)^{m}\mathrm{Q}(x),\) show that \[ \mathrm{P}'(x)=(x-a)^{m-1}\mathrm{R}(x) \] where \(\mathrm{R}\) is a polynomial. Explain why \(\mathrm{P}^{(r)}(a)=0\) whenever \(1\leqslant r\leqslant m-1\). (\(\mathrm{P}^{(r)}\) is the \(r\)th derivative of \(\mathrm{P}.\)) If \[ \mathrm{P}_{n}(x)=\frac{\mathrm{d}^{n}}{\mathrm{d}x^{n}}(x^{2}-1)^{n} \] for \(n\geqslant1\) show that \(\mathrm{P}_{n}\) is a polynomial of degree \(n\). By repeated integration by parts, or otherwise, show that, if \(n-1\geqslant m\geqslant0,\) \[ \int_{-1}^{1}x^{m}\mathrm{P}_{n}(x)\,\mathrm{d}x=0 \] and find the value of \[ \int_{-1}^{1}x^{n}\mathrm{P}_{n}(x)\,\mathrm{d}x. \] {[}Hint. \textit{You may use the formula \[ \int_{0}^{\frac{\pi}{2}}\cos^{2n+1}t\,\mathrm{d}t=\frac{(2^{2n})(n!)^{2}}{(2n+1)!} \] without proof if you need it. However some ways of doing this question do not use this formula.}{]}
Solution: \begin{align*} && P(x) &= (x-a)^mQ(x) \\ \Rightarrow && P'(x) &= m(x-a)^{m-1}Q(x) + (x-a)^mQ'(x) \\ &&&= (x-a)^{m-1}(\underbrace{mQ(x) + (x-a)Q'(x)}_{\text{a polynomial}}) \\ &&&= (x-a)^{m-1}R(x) \end{align*} Therefore \(P^{(r)}(a) = 0\) for \(1 \leq r \leq m-1\) since each time we differentiate we will have a factor of \((x-a)^{m-r}\) which is zero when we evaluate at \(x = a\). If \(P_n(x) = \frac{\d^n}{\d x^n}(x^2-1)^n\) then we are differentiating a degree \(2n\) polynomial \(n\) times. Each time we differentiate we reduce the degree by \(1\), therefore the degree of \(P_n\) is \(n\). \begin{align*} && \int_{-1}^1 x^mP_n(x) \d x &= \left [x^m \underbrace{\frac{\d^{n-1}}{\d x^{n-1}}\left ( (x-1)^{n} (x+1)^{n} \right)}_{\text{has a factor of }x-1\text{ and }x+1}\right]_{-1}^1 - \int_{-1}^1 mx^{m-1}\frac{\d^{n-1}}{\d x^{n-1}}\left ( (x-1)^{n} (x+1)^{n} \right) \d x\\ &&&= 0 - \int_{-1}^1 mx^{m-1}\frac{\d^{n-1}}{\d x^{n-1}}\left ( (x-1)^{n} (x+1)^{n} \right) \d x\\ &&&= -\left [mx^{m-1} \underbrace{\frac{\d^{n-2}}{\d x^{n-2}}\left ( (x-1)^{n} (x+1)^{n} \right)}_{\text{has a factor of }x-1\text{ and }x+1}\right]_{-1}^1+ \int_{-1}^1 m(m-1)x^{m-2}\frac{\d^{n-2}}{\d x^{n-2}}\left ( (x-1)^{n} (x+1)^{n} \right) \d x\\ &&&= m(m-1)\int_{-1}^1 x^{m-2}\frac{\d^{n-2}}{\d x^{n-2}}\left ( (x-1)^{n} (x+1)^{n} \right) \d x\\ &&& \cdots \\ &&&= (-1)^m m!\int_{-1}^1 \frac{\d^{n-m}}{\d x^{n-m}} \left ( (x-1)^{n} (x+1)^{n} \right) \d x\\ &&&= 0 \end{align*} If \(n = m\), we have \begin{align*} && \int_{-1}^1 x^n P_n(x) \d x&= (-1)^nn! \int_{-1}^1 (x^2-1)^n \d x \\ && &= (-1)^{2n}n! \cdot 2\int_{0}^1 (1-x^2)^n \d x \\ x = \sin \theta, \d x = \cos \theta \d \theta: &&&= 2 \cdot n!\int_{0}^{\pi/2} \cos^{2n} \theta \cdot \cos \theta \d \theta \\ &&&= 2 \cdot n!\int_{0}^{\pi/2} \cos^{2n+1} \theta \d \theta \\ &&&= 2 \cdot n!\frac{(2^{2n})(n!)^{2}}{(2n+1)!} \\ &&&= \frac{(2^{2n+1})(n!)^{3}}{(2n+1)!} \\ \end{align*}