| course |
course_year |
question_number |
tags |
title |
year |
Methods |
IB |
49 |
|
Paper 1, Section II, B |
2009 |
Find a power series solution about $x=0$ of the equation
$$x y^{\prime \prime}+(1-x) y^{\prime}+\lambda y=0,$$
with $y(0)=1$, and show that $y$ is a polynomial if and only if $\lambda$ is a non-negative integer $n$. Let $y_{n}$ be the solution for $\lambda=n$. Establish an orthogonality relation between $y_{m}$ and $y_{n}(m \neq n)$.
Show that $y_{m} y_{n}$ is a polynomial of degree $m+n$, and hence that
$$y_{m} y_{n}=\sum_{p=0}^{m+n} a_{p} y_{p}$$
for appropriate choices of the coefficients $a_{p}$ and with $a_{m+n} \neq 0$.
For given $n>0$, show that the functions
$$\left{y_{m}, y_{m} y_{n}: m=0,1,2, \ldots, n-1\right}$$
are linearly independent.
Let $f(x)$ be a polynomial of degree 3. Explain why the expansion
$$f(x)=a_{0} y_{0}(x)+a_{1} y_{1}(x)+a_{2} y_{2}(x)+a_{3} y_{1}(x) y_{2}(x)$$
holds for appropriate choices of $a_{p}, p=0,1,2,3$. Hence show that
$$\int_{0}^{\infty} e^{-x} f(x) d x=w_{1} f\left(\alpha_{1}\right)+w_{2} f\left(\alpha_{2}\right)$$
where
$$w_{1}=\frac{y_{1}\left(\alpha_{2}\right)}{y_{1}\left(\alpha_{2}\right)-y_{1}\left(\alpha_{1}\right)}, \quad w_{2}=\frac{-y_{1}\left(\alpha_{1}\right)}{y_{1}\left(\alpha_{2}\right)-y_{1}\left(\alpha_{1}\right)},$$
and $\alpha_{1}, \alpha_{2}$ are the zeros of $y_{2}$. You need not construct the polynomials $y_{1}(x), y_{2}(x)$ explicitly.