4 problems found
Let \(\mathrm{f}(x)=\dfrac{\sin(n+\frac{1}{2})x}{\sin\frac{1}{2}x}\) for \(0 < x\leqslant\pi.\)
Solution:
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*}
For \(n=0,1,2,\ldots,\) the functions \(y_{n}\) satisfy the differential equation \[ \frac{\mathrm{d}^{2}y_{n}}{\mathrm{d}x^{2}}-\omega^{2}x^{2}y_{n}=-(2n+1)\omega y_{n}, \] where \(\omega\) is a positive constant, and \(y_{n}\rightarrow0\) and \(\mathrm{d}y_{n}/\mathrm{d}x\rightarrow0\) as \(x\rightarrow+\infty\) and as \(x\rightarrow-\infty.\) Verify that these conditions are satisfied, for \(n=0\) and \(n=1,\) by \[ y_{0}(x)=\mathrm{e}^{-\lambda x^{2}}\qquad\mbox{ and }\qquad y_{1}(x)=x\mathrm{e}^{-\lambda x^{2}} \] for some constant \(\lambda,\) to be determined. Show that \[ \frac{\mathrm{d}}{\mathrm{d}x}\left(y_{m}\frac{\mathrm{d}y_{n}}{\mathrm{d}x}-y_{n}\frac{\mathrm{d}y_{m}}{\mathrm{d}x}\right)=2(m-n)\omega y_{m}y_{n}, \] and deduce that, if \(m\neq n,\) \[ \int_{-\infty}^{\infty}y_{m}(x)y_{n}(x)\,\mathrm{d}x=0. \]
Solution: \begin{align*} && y_0(x) &= e^{-\lambda x^2} \\ && \lim_{x \to \pm \infty} y_0(x) &= 0 \Leftrightarrow \lambda > 0 \\ && \lim_{x \to \pm \infty} y'_0(x) &= \lim_{x \to \pm \infty} 2x\lambda e^{-\lambda x^2} \\ &&&= 0\Leftrightarrow \lambda > 0 \\ && y''_0(x) &= 4x^2 \lambda^2 e^{-\lambda x^2} + 2\lambda e^{-\lambda x^2} \\ \\ && y''_0 - \omega^2 x^2 y_0+(2\cdot 0 + 1) \omega y_0 &= e^{-\lambda x^2} \l 4x^2 \lambda^2 + 2 \lambda - \omega^2 x^2 + \omega\r \\ &&&=0 \Leftrightarrow \lambda = \pm \frac{\omega}{2} \end{align*} Therefore \(y_0\) satisfies if \(\lambda = \frac{\omega}{2}\) Similarly for \(y_1\), \begin{align*} && y_1(x) &= xe^{-\lambda x^2} \\ && \lim_{x \to \pm \infty} y_1(x) &= 0 \Leftrightarrow \lambda > 0 \\ && \lim_{x \to \pm \infty} y'_1(x) &= \lim_{x \to \pm \infty} \l -2x^2 \lambda e^{-\lambda x^2} + e^{-\lambda x^2} \r \\ &&&= 0\Leftrightarrow \lambda > 0 \\ && y''_0(x) &= e^{-\lambda x^2} \l 4x^3 \lambda^2-4x\lambda - 2x\lambda \r \\ &&&= e^{-\lambda x^2} \l 4x^3 \lambda^2-6x\lambda \r \\ && y''_1 - \omega^2 x^2 y_1+(2\cdot 1 + 1) \omega y_1 &= e^{-\lambda x^2} \l 4x^3\lambda^2-6x\lambda-\omega^2x^3+3\omega x\r \\ &&&=0 \Leftrightarrow \lambda = \pm \frac{\omega}{2} \end{align*} Therefore \(y_1\) satisfies if \(\lambda = \frac{\omega}{2}\) \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}\left(y_{m}\frac{\mathrm{d}y_{n}}{\mathrm{d}x}-y_{n}\frac{\mathrm{d}y_{m}}{\mathrm{d}x}\right) &= y'_my'_n+y_my''_n - y'_ny'_m-y_ny''_m \\ &= y_my''_n - y_ny''_m \\ &= y_m(\omega^2 x^2 y_n - (2n+1)\omega y_n) - y_n(\omega^2 x^2 y_m - (2m+1)\omega y_m) \\ &= y_my_n (2m-2n)\omega \\ &= 2(m-n) \omega y_my_n \end{align*} Therefore: \begin{align*} \int_{-\infty}^{\infty} y_m(x)y_n(x) \d x &= \int_{-\infty}^{\infty} \frac{1}{2(m-n)} \frac{\mathrm{d}}{\mathrm{d}x}\left(y_{m}\frac{\mathrm{d}y_{n}}{\mathrm{d}x}-y_{n}\frac{\mathrm{d}y_{m}}{\mathrm{d}x}\right) \d x \\ &= \frac{1}{2(m-n)} \left [ y_{m}\frac{\mathrm{d}y_{n}}{\mathrm{d}x}-y_{n}\frac{\mathrm{d}y_{m}}{\mathrm{d}x}\right]_{-\infty}^{\infty} \\ &\to 0 \end{align*} This condition is known as Orthogonality. In fact this question is talking about a Sturm-Liouville orthogonality condition, in particular for the quantum harmonic oscillator, and the eigenfunctions are related to Hermite polynomials.