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  {\large \textbf{Mathematical Tripos Part IB: Easter 2001}}

  {\Large \textbf{Numerical Analysis -- Detailed Example
      10\footnote{Corrections and suggestions to these notes should
        be emailed to \texttt{a.shadrin@damtp.cam.ac.uk}. All handouts
        are available on the WWW at the URL 
        \texttt{http://www.damtp.cam.ac.uk/user/na/PartIB/Handouts.html}.}}}

  \vspace{8pt}
  \textbf{\textit{\LARGE Orthogonal polynomials}}
\end{center}

Let
\begin{displaymath}
  \langle f,g\rangle =\int_{-1}^1 f(x)g(x) \CC{d}x.
\end{displaymath}
We use the three-term recurrence relation to form the orthogonal
polynomials: 
\begin{Eqnarray*}
  &&p_{-1}(x)\equiv0,\qquad p_0(x)\equiv1,\\
  n=0:&\quad& \alpha_0=\frac{\int_{-1}^1 x\CC{d}x}{\int_{-1}^1
    \CC{d}x}=0,\qquad \beta_0=0 \mbox{\ (by definition)}\\
  &&\qquad\Rightarrow\qquad p_1(x)=x\cdot 1-0=x,\\
  n=1:&& \alpha_1=\frac{\int_{-1}^1 x^3\CC{d}x}{\int_{-1}^1
    x^2\CC{d}x}=0,\qquad \beta_1=\frac{\int_{-1}^1
    x^2\CC{d}x}{\int_{-1}^1 \CC{d}x}=\Frac13\\
  &&\qquad\Rightarrow\qquad p_2(x)=x\cdot x-\Frac13 \cdot
  1=x^2-\Frac13,\\
  n=2:&& \alpha_2=\frac{\int_{-1}^1 (x^2-\frac13)^2x\CC{d}x}
  {\int_{-1}^1 (x^2-\frac13)^2\CC{d}x}=0,\qquad
  \beta_2=\frac{\int_{-1}^1 (x^2-\frac13)^2\CC{d}x} {\int_{-1}^1
    x^2\CC{d}x} =\Frac{4}{15}\\
  &&\qquad\Rightarrow\qquad p_3(x)=x\cdot
  (x^2-\Frac13)-\Frac{4}{15}x=x^3-\Frac35 x,\\
  n=3:&&\alpha_3=\frac{\int_{-1}^1 (x^3-\frac35 x)^2 x\CC{d}x}
  {\int_{-1}^1 (x^3-\frac35x)^2\CC{d}x}=0,\qquad
  \beta_3=\frac{\int_{-1}^1 (x^3-\frac35 x)^2\CC{d}x} {\int_{-1}^1
    (x^2-\frac13)^2 \CC{d}x} =\Frac{9}{35}\\
  &&\qquad\Rightarrow\qquad p_4(x)=x\cdot (x^3-\Frac35 x)-\Frac{9}{35}
  (x^2-\Frac13) =x^4-\Frac67 x^2+\Frac{3}{35}.
\end{Eqnarray*}

Left to the reader:
\begin{enumerate}
\item Prove that $\alpha_n=0$ for all $n\geq0$;
\item Prove that each $p_{2n}$ is an even polynomial and each
  $p_{2n+1}$ is an odd polynomial.
\end{enumerate}


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