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

  {\Large \textbf{Numerical Analysis -- Detailed Example
      13\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 The Peano Kernel Theorem}}
\end{center}

Given $f\in\CC{C}^3[0,\infty)$, we integrate
\begin{displaymath}
  \int_0^\infty \CC{e}^{-x}f(x)\CC{d}x\approx b_1 f(c_1) +b_2 f(c_2).
\end{displaymath}
The quadratic monic polynomial orthogonal w.r.t.\ $\langle g,h\rangle
=\int_0^\infty \CC{e}^{-x}g(x)h(x)\CC{d}$ is $x^2-4x+2$, with zeros
\begin{displaymath}
  c_1=2-\sqrt{2},\qquad c_2=2+\sqrt{2},
\end{displaymath}
which we choose as the quadrature nodes. The weights are
\begin{displaymath}
  b_1=\Frac12+\Frac{\sqrt{2}}{4},\qquad
  b_2=\Frac12-\Frac{\sqrt{2}}{4}, 
\end{displaymath}
whence the quadrature is exact for all $f\in\BB{P}_3[x]$
(verify!). The error functional is
\begin{displaymath}
  L[f]=\int_0^\infty \CC{e}^{-x}f(x)\CC{d}x
  -(\Frac12+\Frac{\sqrt{2}}{4}) f(2-\sqrt{2})
  -(\Frac12-\Frac{\sqrt{2}}{4}) f(2+\sqrt{2}).
\end{displaymath}
We evaluate the Peano kernel $K(\theta)=L[(x-\theta)_+^3]$:\\
{\bf 1.} $0\leq\theta\leq 2-\sqrt{2}$:
\begin{Eqnarray*}
  K(\theta)&=&\int_\theta^\infty
  \CC{e}^{-x} (x-\theta)^3 \CC{d}x- (\Frac12+\Frac{\sqrt{2}}{4})
  (2-\sqrt{2}-\theta)^3 -(\Frac12-\Frac{\sqrt{2}}{4})
  (2+\sqrt{2}-\theta)^3\\
  &=&\theta^3-3\theta^2+6\theta-6+6\CC{e}^{-\theta},
\end{Eqnarray*}
{\bf 2.} $2-\sqrt{2}\leq\theta\leq2+\sqrt{2}$:
\begin{Eqnarray*}
  K(\theta)&=&\int_\theta^\infty
  \CC{e}^{-x} (x-\theta)^3 \CC{d}x-(\Frac12-\Frac{\sqrt{2}}{4})
  (2+\sqrt{2}-\theta)^3\\
  &=&(\Frac12-\Frac{\sqrt{2}}{4})\theta^3-\Frac32\theta^2
  +(3+\Frac32\sqrt{2})\theta-(3+2\sqrt{2})+6\CC{e}^{-\theta},
\end{Eqnarray*}
{\bf 3.} $2+\sqrt{2}\leq \theta$:
\begin{Eqnarray*}
  K(\theta)&=&\int_\theta^\infty   \CC{e}^{-x} (x-\theta)^3 \CC{d}x
  =6\CC{e}^{-\theta}. 
\end{Eqnarray*}
It is possible to verify that $K(\theta)\geq0$ (sketch the curve to
persuade yourself!) and long calculation (or few seconds with symbolic
algebra) show that
\begin{displaymath}
  \|K\|_1=\int_0^\infty K(\theta)\CC{d}\theta=1\qquad\Rightarrow\qquad
  |L[f]|\leq \Frac16 \|f^{\CC{(iv)}}\|_\infty.
\end{displaymath}
(Remember that we need to divide the bound by $3!=6$.) Likewise,
\begin{Eqnarray*}
  \|K\|_2=\left\{\int_0^\infty [K(\theta)]^2\CC{d}\theta\right\}^{1/2}
  \approx 0.527664    &\qquad\Rightarrow\qquad& |L[f]| \leq  0.087944 \cdot
  \|f^{\CC{(iv)}}\|_2,\\
  \|K\|_\infty=\max_{\theta\geq0} |K(\theta)|\approx 0.403021
  &\qquad\Rightarrow\qquad& |L[f]| \leq  0.067170 \cdot \|f^{\CC{(iv)}}\|_1.
\end{Eqnarray*}



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