%2multibyte Version: 5.50.0.2890 CodePage: 950
\documentclass[12pt,a4paper]{article}%
\usepackage{amsfonts}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{graphicx}%
\def \Def{{\bf {Definition}}}
\def \Thm{{\bf {Theorem}}}
\def \Cor{{\bf {Corollary}}}
\def \Lma{{\bf {Lemma}}}
\def \Rmk{{\bf {Remark}}}
\def \Hw{{\bf {Homework}}}
\def \Prop{{\bf {Proposition}}}
\def \Ex{{\bf {Example}}}
\def \Que{{\bf {Question}}}
\def \Ans{{\bf {Answer}}}
\def \Goal{{\bf {Goal}}}
\def \Facts{{\bf {Facts}}}
\def \pf{{\bf {Proof}}}
\def\x{t_i}
\input epsf
\begin{document}

\begin{center}
\textbf{\S 2.3 Properties of Gaussian quadrature rules}
\end{center}
The Gaussian quadrature rule, in addition to being optimal, has some
interesting and useful properities.
\begin{itemize}
\item (i) all nodes $t_{k}$ are real distinct, and contained in
$(a,b)$ (\Hw: 31)

\item (ii) all weights $w_{k}$ are positive

$0<\int_{a}^{b}l_{i}^{2}(t)w(t)dt=\sum \limits_{k=1}^{n}w_{k}l_{i}^{2}%
(t_{k})=w_{i}$, $l_{i}^{2}\in P_{2n-2}$, $2n-2<2n-1$

\item (iii) If $[a,b]$ is a finite interval,  then Gauss formula
converges  for any continuous function, i.e.
$E_{n}(f)\rightarrow0$ as $n\rightarrow \infty$%

\pf:\\
\\
$%
\begin{array}
[c]{l}%
\left \vert E_{n}(f)\right \vert =\left \vert E_{n}(f-\hat{p}_{2n-1})\right \vert
\\
=\left \vert \int_{a}^{b}(f-\hat{p}_{2n-1})w(t)dt-\sum \limits_{k=1}^{n}%
w_{k}\left[  f(t_{k})-\hat{p}_{2n-1}(t_{k})\right]  \right \vert \\
\leq \int_{a}^{b}\left \vert f(t)-\hat{p}_{2n-1}(f;t)\right \vert w(t)dt+\sum
\limits_{k=1}^{n}w_{k}\left \vert f(t_{k})-\hat{p}_{2n-1}(t_{k})\right \vert \\
\leq \left \Vert f-\hat{p}_{2n-1}(f)\right \Vert _{\infty}\left[  \int_{a}%
^{b}w(t)dt+\sum \limits_{k=1}^{n}w_{k}\right]
\end{array}
$

Notice that $\int_{a}^{b}w(t)dt=\sum \limits_{k=1}^{n}w_{k}=\mu_{0}$

$=2\mu_{0}\left \Vert f-\hat{p}_{2n-1}(f)\right \Vert
_{\infty}\rightarrow0$ as $n\rightarrow \infty$

\item (iv) The nodes in the n-pt Gauss formula separate those of the (n+1)-pt
formula. (\Hw: 32)

\end{itemize}

Prove that all zeros of $\pi_{n}(\cdot;w)$ are real, simple, and contained in
$\left(  a,b\right)$.
Let $a<x_{1}<\cdots<x_{l}<b$ are roots of $w_{n}$ in $\left(  a,b\right)  $
and of which $\pi_n$ changes sign.
Now let $q(x)=\prod \limits_{j=1}^{l}(x-x_{j})$, then $w_{n}(x)q(x)$ does
not change sign in $\left[  a,b\right]$.
So $(w_{n},q)=\int_{a}^{b}w(x)w_{n}(x)q(x)dx\neq0$
but if $l<n$ then $\rightarrow \leftarrow$ to Theorem 3.2.1
so $l=n$, i.e. all roots are distinct and with multiplicity 1.

\eject
The next property forms the basis of an efficient algorithm for
computing Gaussian quadrature formulae. Let
$$\pi_{k+1}(t)=(t-\alpha_{k})\cdot \pi_{k}(t)-\beta_{k}\pi_{k-1}(t)$$

$$\pi_{0}(t)=1, \pi_{-1}(t)=0, \beta_{0}\equiv\int_{a}^{b}w(x)dt=\mu_{0}$$
The nth-order Jacobi matrix for the weight function w is a
tridiagonal symmetric matrix defined by
\[
J_{n}=\left[
\begin{array}
[c]{cccc}%
\alpha_{0} & \sqrt{\beta_{1}} &  & 0\\
\sqrt{\beta_{1}} & \alpha_{1} & \ddots & 0\\
& \ddots & \ddots & \sqrt{\beta_{n-1}}\\
0 &  & \sqrt{\beta_{n-1}} & \alpha_{n-1}%
\end{array}
\right]
\]
%\[
%\left \vert \lambda I-J_{n}\right \vert =\left \vert
%\begin{array}
%[c]{cccc}%
%\lambda-\alpha_{0} & -\sqrt{\beta_{1}} &  & 0\\
%-\sqrt{\beta_{1}} & \lambda-\alpha_{1} & \ddots & 0\\
%& \ddots & \ddots & -\sqrt{\beta_{n-1}}\\
%0 &  & -\sqrt{\beta_{n-1}} & \lambda-\alpha_{n-1}%
%\end{array}
%\right \vert
%\]
%$$=(\lambda-\alpha_{n-1})\cdot \pi_{n-1}(\lambda)+(-1)\cdot(-\sqrt{\beta_{n-1}%
%})\cdot \left \vert
%\begin{array}
%[c]{cccc}%
%\lambda-\alpha_{0} & -\sqrt{\beta_{1}} &  & 0\\
%&  & \ddots & \vdots \\
%0 & \cdots & \lambda-\alpha_{n-3} & -\sqrt{\beta_{n-2}}\\
%0 & \cdots & \cdots & -\sqrt{\beta_{n-1}}
%\end{array}
%\right \vert $$
%$-\beta_{n-1}\cdot \pi_{n-2}(\lambda)$
\Prop:\\
  $J_{n}v^{(k)}=t_{k}v^{(k)}$, $v^{(k)T}v^{(k)}=1$ for some $v^{(k)}$. And
  the weight $w_k$ expressible in terms of the first component $v_1^{(k)}$
  of the corresponding (normalized) eigenvectors by
\[
w_k = \beta_0 (v_1^{(k)})^2, k=1,2\dots,n.
\]
Thus, to compute the Gaussian formula, we must solve an
eigenvalue/eigenvector problem for a symmetric tridiagonal matrix.




\pf


(1) claim: $\tilde{v}^{(i)}=\left[
\begin{array}
[c]{c}%
1\cdot \pi_{0}(t_{i})\\
\frac{1}{\sqrt{\beta_{1}}}\pi_{1}(t_{i})\\
\frac{1}{\sqrt{\beta_{1}\beta_{2}}}\pi_{2}(t_{i})\\
\vdots \\
\frac{1}{\sqrt{\beta_{1}\cdots \beta_{n-1}}}\pi_{n-1}(t_{i})
\end{array}
\right]  $ are eigenvectors.
\eject
(2)

$\left[
\begin{array}
[c]{cccc}%
\pi_{0}(t_{1}) & \pi_{0}(t_{2}) & \cdots & \pi_{0}(t_{n})\\
\frac{1}{\sqrt{\beta_{1}}}\pi_{1}(t_{1}) & \frac{1}{\sqrt{\beta_{1}}}\pi
_{1}(t_{2}) & \cdots & \frac{1}{\sqrt{\beta_{1}}}\pi_{1}(t_{n})\\
\vdots & \vdots & \cdots & \vdots \\
\frac{1}{\sqrt{\beta_{1}\cdots \beta_{n-1}}}\pi_{n-1}(t_{1}) & \frac{1}%
{\sqrt{\beta_{1}\cdots \beta_{n-1}}}\pi_{n-1}(t_{2}) & \cdots & \frac{1}%
{\sqrt{\beta_{1}\cdots \beta_{n-1}}}\pi_{n-1}(t_{n})
\end{array}
\right]  \cdot \left[
\begin{array}
[c]{c}%
w_{1}\\
w_{2}\\
\vdots \\
w_{n}%
\end{array}
\right]  $

$=\left[
\begin{array}
[c]{c}%
(\pi_{0},\pi_{0})\\
\frac{1}
{\sqrt{\beta_{1}}}(\pi_{1},\pi_{0})\\
\vdots \\
\frac{1}
{\sqrt{\beta_{1}\cdots \beta_{n-1}}}(\pi_{n-1},\pi_{0})
\end{array}
\right]  =\left[
\begin{array}
[c]{c}%
(\pi_{0},\pi_{0})\\
0\\
\vdots \\
0
\end{array}
\right] ,\mbox{ since }\pi_0 \equiv 1. $
\\
Eigenvectors of symmetric matrices for distinct eigenvalues are
orthogonal.
Therefore, multiplying by $\tilde{v}^{(i)}$ from the left yields
$$ \tilde{v}^{(i)T}\tilde{v}^{(i)}w_{i}=(\pi_{0},\pi_{0})\cdot
\tilde{v}_{1}^{(i)}$$

$$\Rightarrow(\tilde{v}^{(i)})^{T}\tilde{v}^{(i)}w_{i}=(\pi_{0},\pi_{0})$$

and $\tilde{v}^{(i)}=\frac{1}{v_{1}^{(i)}}v^{(i)}$, since
$\tilde{v}^{(i)}_1 = 1$, with $(v^{(i)})^{T}%
v^{(i)}=1$

Multiplying $(v^{(i)}_1)^{2}$ we have $v^{(i)T}v^{(i)}w_{i}=(v_1^{(i)}%
)^{2}(\pi_{0},\pi_{0})$

$\Rightarrow w_{i}=(v^{(i)}_1)^{2}(\pi_{0},\pi_{0})=(v^{(i)}_1)^{2}\beta_0 $%



\eject

Proof of claim:\\

$\left[
\begin{array}
[c]{cccc}%
\alpha_{0} & \sqrt{\beta_{1}} &  & 0\\
\sqrt{\beta_{1}} & \alpha_{1} & \ddots & 0\\
& \ddots & \ddots & \sqrt{\beta_{n-1}}\\
0 &  & \sqrt{\beta_{n-1}} & \alpha_{n-1}%
\end{array}
\right]  \left[
\begin{array}
[c]{c}%
\pi_{0}(t_{i})\\
\frac{1}{\sqrt{\beta_{1}}}\pi_{1}(t_{i})\\
\frac{1}{\sqrt{\beta_{1}\beta_{2}}}\pi_{2}(t_{i})\\
\vdots \\
\frac{1}{\sqrt{\beta_{1}\cdots \beta_{n-1}}}\pi_{n-1}(t_{i})
\end{array}
\right]  $
\\

(i) $\alpha_{0}\pi_{0}(\x)+\sqrt{\beta_{1}}\cdot
\frac{1}{\sqrt{\beta_{1}}}\pi
_{1}(\x)=\alpha_{0}\pi_{0}(\x)+\pi_{1}(\x)$

$=\alpha_{0}\pi_{0}(\x)+(\x-\alpha_{0})\pi_{0}(\x)=\x\cdot1\cdot
\pi_{0}(\x)$
\\

(ii) For $j=1,\cdots,n-2$

$\sqrt{\beta_{j}}\cdot \frac{1}{\sqrt{\beta_{1}\cdots
\beta_{j-1}}}\pi _{j-1}(\x)+\alpha_{j}\cdot
\frac{1}{\sqrt{\beta_{1}\cdots \beta_{j}}}\pi
_{j}(\x)+\sqrt{\beta_{j+1}}\cdot \frac{1}{\sqrt{\beta_{1}\cdots \beta_{j+1}}%
}\pi_{j+1}(\x)$

$=\frac{1}{\sqrt{\beta_{1}\cdots
\beta_{j}}}(\beta_{j}\pi_{j-1}(\x)+\alpha
_{j}\pi_{j}(\x)+(\x-\alpha_{j})\pi_{j}(\x)-\beta_{j}\pi_{j-1}(\x))$

$=\frac{1}{\sqrt{\beta_{1}\cdots \beta_{j}}}\cdot \x\cdot
\pi_{j}(\x)$
\\

(iii)

$\sqrt{\beta_{n-1}}\cdot \frac{1}{\sqrt{\beta_{1}\cdots
\beta_{n-2}}}\pi
_{n-2}(\x)+\alpha_{n-1}\cdot \frac{1}{\sqrt{\beta_{1}\cdots \beta_{n-1}}}%
\pi_{n-1}(\x)$

$=\frac{1}{\sqrt{\beta_{1}\cdots \beta_{n-1}}}\cdot(\beta_{n-1}\pi
_{n-2}(\x)+\alpha_{n-1}\pi_{n-1}(\x))$

$\because
0=\pi_{n}(\x)=(\x-\alpha_{n-1})\pi_{n-1}-\beta_{n-1}\pi_{n-2}(\x)$

$=\frac{1}{\sqrt{\beta_{1}\cdots \beta_{n-1}}}\cdot \x\cdot
\pi_{n-1}(\x)$

\begin{figure}
[h]
\begin{center}
\includegraphics[
height=6.9633cm,
width=9.5399cm
]%
{ch3-1.eps}%
\end{center}
\end{figure}


\newpage
%==============================================================================
Different approach: Let \\
$$\pi_{k+1}(t)=(t-\alpha_k)\pi_k(t)-\beta_k\pi_{k-1}(t)$$
$$\pi_{-1}(t) = 0, \quad\pi_0(t)=1$$
$$\alpha_k = \frac{(t\pi_k,\;\pi_k)}{(\pi_k,\;\pi_k)},\quad\beta_k
= \frac{(\pi_k,\;\pi_k)}{(\pi_{k-1},\;\pi_{k-1})}$$
We want to find $\pi_2(t)$ using equations defined above,
$$\pi_1(t) = (t-\alpha_0)\pi_0(t) - \beta_0\pi_{-1}(t)$$
$$\alpha_0 =
\frac{(t\pi_0,\;\pi_0)}{(\pi_0,\;\pi_0)}=\frac{\int_0^1t\cdot
t^{-\frac{1}{2}}dt}{\int_0^1 1\cdot1\cdot1\cdot
t^{-\frac{1}{2}}dt}=\frac{\frac{2}{3}}{2}=\frac{1}{3}$$\\
\begin{eqnarray*}
\pi_1(t)&=&(t-\frac{1}{3})\\
\pi_2(t)&=&(t-\alpha_1)\pi_1(t)-\beta_1\pi_0(t)\\
\alpha_1&=&\frac{(t\pi_1,\;\pi_1)}{(\pi_1,\;\pi_1)}=\frac{\int_0^1t\cdot(t-\frac{1}{3})\cdot(t-\frac{1}{3})\cdot
t^{-\frac{1}{2}}dt}{\int_0^1
(t-\frac{1}{3})\cdot(t-\frac{1}{3})\cdot t^{-\frac{1}{2}}dt}\\
&=&\frac{\int_0^1
t^{\frac{5}{2}}-\frac{2}{3}t^{\frac{3}{2}}+\frac{1}{9}t^{\frac{1}{2}}dt}{\int_0^1
t^{\frac{3}{2}}-\frac{2}{3}t^{\frac{1}{2}}+\frac{1}{9}t^{-\frac{1}{2}}dt}\\
&=&\frac{\frac{2}{7}-\frac{2}{3}\cdot\frac{2}{5}+\frac{1}{9}\cdot\frac{2}{3}}{\frac{2}{5}-\frac{2}{3}\cdot\frac{2}{3}+\frac{1}{9}\cdot2}=\frac{11}{21}\\
\beta_1&=&\frac{(\pi_1,\;\pi_1)}{(\pi_0,\;\pi_0)}=\frac{\int_0^1(t-\frac{1}{3})^2t^{-\frac{1}{2}}dt}{\int_0^1t^{-\frac{1}{2}}dt}
= \frac{4}{45}\\
\pi_2(t)&=&(t-\frac{11}{21})(t-\frac{1}{3})-\frac{4}{45}\cdot1\\
&=&t^2-\frac{6}{7}t+\frac{3}{35}\\
\end{eqnarray*}
This agrees with the previous compuations.
%%%Lee Please add the new tex after this 12/23
\begin{eqnarray*}
J_{2}&=&\left[
\begin{array}
[c]{cc}%
\frac{1}{3} & \sqrt{\frac{4}{45}} \\
\sqrt{\frac{4}{45}} & \frac{11}{21}\\
\end{array}
\right] =\left[
\begin{array}
[c]{cc}%
\alpha_{0} & \sqrt{\beta_{1}} \\
\sqrt{\beta_{1}} & \alpha_{1}  \\
\end{array}
\right]\\
|\lambda I- J_n| &=& \left|\left[
\begin{array}
[c]{cc}%
\lambda-\frac{1}{3}&\sqrt{\frac{4}{45}}\\
\sqrt{\frac{4}{45}}&\lambda-\frac{11}{21}\\
\end{array}
\right]\right|\\
&=&(\lambda-\frac{1}{3})(\lambda-\frac{11}{21})-\frac{4}{45}\\
\end{eqnarray*}
$t_1 =
\frac{1}{7}(3-2\sqrt{\frac{6}{5}}),\;t_2=\frac{1}{7}(3+2\sqrt{\frac{6}{5}})$\\
$\tilde{v}^{(1)} =\left(
\begin{array}
[c]{c}%
1\\
\sqrt{\frac{4}{45}}(t_1-\frac{1}{3})\\
\end{array}\right)
\quad \tilde{v}^{(2)} =\left(\begin{array}
[c]{c}%
1\\
\sqrt{\frac{4}{45}}(t_2-\frac{1}{3})\\
\end{array}\right)\\$
$ \tilde{v}^{(k)}
=\frac{\tilde{v}^{(k)}}{\|\tilde{v}^{(k)}\|}\\$  and
$w_k=\beta_0\cdot(v_1^{(k)})\\$
\\
\Rmk:\\
Which way is easier to compute?
%%%%%

\eject
\begin{center}
\textbf{\S 2.4 Some applications of Gauss quadrature rule}
\end{center}
We begin with the easy exercise of transforming the Gauss-Jacobi
quadrature rule from an arbitrary finite interval to the canonical
interval [-1,1]\\
(i) Gauss-Jacobi formula on $[-1,1]$

$$
{\displaystyle \int_{-1}^{1}}
 f(t)(1-t)^{\alpha}(1+t)^{\beta}dt,$$
where
$\omega(t)=(1-t)^{\alpha}(1+t)^{\beta}$, $\alpha>-1$, $\beta>-1$
Note that at $t=-1,1$ there are algebraic singularities.
\\
Goal: $$
{\displaystyle \int_{a}^{b}} (b-x)^{\alpha}(x-a)^{\beta}g(x)dx$$
A linear transformation of variables.
$$x=\frac{1}{2}(b-a)t+\frac{1}{2}(b+a),\qquad dx=\frac{1}{2}(b-a)dt$$
%
maps $[a,b]\rightarrow \lbrack-1,1]$ and the integral becomes

$$\int_{-1}^{1}(b-\frac{1}{2}(b-a)t-\frac{1}{2}(b+a))^{\alpha}(\frac{1}%
{2}(b-a)t+\frac{1}{2}(b+a)-a)^{\beta}$$

$$g(\frac{1}{2}(b-a)t+\frac{1}{2}(b+a))\frac{1}{2}(b-a)dt$$

$$=\int_{-1}^{1}(\frac{1}{2}(b-a)(1-t))^{\alpha}(\frac{1}{2}(b-a)(1+t))^{\beta
}f(t)\frac{1}{2}(b+a)dt$$

$$=[\frac{1}{2}(b-a)]^{\alpha+\beta+1}\int_{-1}^{1}(1-t)^{\alpha}(1+t)^{\beta
}f(t)dt,$$
%
where $f(t)=g(\frac{1}{2}(b-a)t+\frac{1}{2}(b+a))$ \  \ $-1\leq t\leq1$

$$
{\displaystyle \int_{a}^{b}} (b-x)^{\alpha}(x-a)^{\beta}g(x)dx$$

$$=[\frac{1}{2}(b-a)]^{\alpha+\beta+1}\sum \limits_{k=1}^{n}w_{K}^{J}g(\frac
{1}{2}(b-a)t_{K}^{J}+\frac{1}{2}(b+a))+E_{n}(g)$$

$t_{K}^{J}$, $w_{K}^{J}$ are (standard) Gauss-Jacobi nodes \& weights

and exact for $g\in P_{2n-1}$

\newpage

(ii) Iterated integrals
$$(Ig)(t):=%
%
{\displaystyle \int_{0}^{t}}
%
g(\tau)d\tau $$
And $I^p$ is the pth power. Then
%
$$
(I^{p+1}g)(1):=%
%
{\displaystyle \int_{0}^{1}}
%
(I^{p}g)(t)dt$$

From calculus that we have

$$(I^{p+1}g)(1)=%
%
{\displaystyle \int_{0}^{1}}
%
\frac{(1-t)^{p}}{p!}g(t)dt$$ For p =1, we have
$$
(I^2g)(1) = \int_0^1\left(\int_0^t g(\tau) d\tau\right)dt
=\int_0^1\int_\tau^1 g(\tau)dt\, d\tau = \int_0^1(1-\tau) g(\tau)
d\tau.
$$

When $p<<\infty$ we could use either trapezoidal or Simpson's rule.
But as $p\rightarrow \infty$, $(1-t)^{p}$ approaches a discontinuous
function. Use Jacobi weight on $[0,1],$ $\alpha=p$, $\beta=0$

$$(I^{p+1}g)(1)=\frac{1}{2^{p+1}p!}\sum \limits_{k=1}^{n}w_{K}^{J}g(\frac{1}%
{2}(1+t_{K}^{J}))+E_{n}(g)$$

(iii) Integration over $\Re$
$$
{\displaystyle \int_{-\infty}^{\infty}}
%
F(x)dx, \qquad F(x)\sim e^{-ax^{2}},\quad  x\rightarrow \pm
\infty,\quad a>0.$$

Let
$$x=\frac{t}{\sqrt{a}},\quad dx=\frac{1}{\sqrt{a}}dt,\quad f(t)=e^{t^{2}}%
F(\frac{t}{\sqrt{a}}),\quad f(\infty)=f(-\infty)=1$$

$$
%
{\displaystyle \int_{-\infty}^{\infty}}
%
F(x)dx=\frac{1}{\sqrt{a}}%
%
%
{\displaystyle \int_{-\infty}^{\infty}}
%
F(\frac{t}{\sqrt{a}})dt$$
%
$$=\frac{1}{\sqrt{a}}%
%
{\displaystyle \int_{-\infty}^{\infty}}
%
e^{-t^{2}}\cdot(e^{t^{2}}F(\frac{t}{\sqrt{a}}))dt$$
%
$$=\frac{1}{\sqrt{a}}%
%
{\displaystyle \int_{-\infty}^{\infty}}
%
e^{-t^{2}}f(t)dt.$$ Use Gauss Hermite formula
$$=\frac{1}{\sqrt{a}}\sum \limits_{k=1}^{n}w_{K}^{H}e^{(t_{K}^{h})^{2}}%
F(\frac{t_{K}^{H}}{\sqrt{a}})+E_{n}(f)$$

$$E_{n}(f)=0, \mbox{ if } e^{t^{2}}F(\frac{t}{\sqrt{a}})\in P_{2n-1}$$

i.e. $F(x)=e^{-t^{2}}\cdot p(x)=e^{-ax^{2}}\cdot p(x)$, $p(x)\in
P_{2n-1}$

\newpage

(iv)Integration over ${\mathcal{R}}_+$.\\
Compute
$$%
%
{\displaystyle \int_{0}^{\infty}}
%
F(x)dx$$
Assuming that
$$F(x)\sim \left \{
\begin{array}
[c]{l}%
x^{p}\text{ \  \ as }x\text{ }\downarrow0\text{ \ }(p>-1)\\
x^{-q}\text{ as }x\rightarrow \infty \text{ }(q<1)
\end{array}
\right.  $$
Similarly as in (iii), we define $f$ by
$$F(x)=\frac{x^{p}}{(1+x)^{p+q+\lambda}}f(x), \quad
\lambda=0,1,2,\cdots,2n-1.$$
So as to $f(x)\longrightarrow1$ as $x\downarrow0$ and $x\rightarrow \infty$.
The change of variables
$$x=\frac{1+t}{1-t},\quad (1-t)^{\lambda}=(\frac{2}{x+1})^{\lambda}
,\quad dx=\frac{2dt}{(1-t)^{2}}$$

$$\Rightarrow%
%
{\displaystyle \int_{0}^{\infty}}
%
F(x)dx=%
%
{\displaystyle \int_{-1}^{1}}
%
(\frac{1+t}{1-t})^{p}(\frac{2}{1-t})^{-(p+q)}f(\frac{1+t}{1-t})\frac
{2dt}{(1-t)^{2}}$$

$$=\frac{1}{2^{p+q-1}}%
%
{\displaystyle \int_{-1}^{1}}
%
(1-t)^{q-2}(1+t)^{p}g(t)dt,\quad \mbox{ when $\lambda=0$ }$$

where $g(t)=f(\frac{1+t}{1-t})$\\
Gauss-Jacobi quadrature , $\alpha=q-2$, $\beta=p$
$$%
%
{\displaystyle \int_{0}^{\infty}}
%
F(x)dx=\frac{1}{2^{p+q-1}}(\sum \limits_{k=1}^{n}w_{K}^{J}g(t_{K}^{J}%
)+E_{n}(g))$$

$$2^{-p-q+1}\sum \limits_{k=1}^{n}w_{K}^{J}\frac{2^{p+q}\frac{1}{(1-t_{K}%
^{J})^{p+q}}}{(\frac{1+t_{K}^{J}}{1-t_{K}^{J}})^{p}}\cdot F(\frac{1+t_{K}^{J}%
}{1-t_{K}^{J}})$$

$$=2\sum \limits_{k=1}^{n}w_{K}^{J}(1+t_{K}^{J})^{-P}(1-t_{K}^{J})^{-q}%
F(\frac{1+t_{K}^{J}}{1-t_{K}^{J}})+2^{-p-q+1}E_{n}(g)$$

\newpage

\begin{center}
\textbf{\S 2.4 }Approximation of linear functionals:
method of interpolation v.s method of undetermined coefficients
\end{center}
Up until now, we heavily relied on interpolation to obtain
approximations to derivatives, integrals, and the like. That is not
the only possible approach, however, to construct approximation
formulae, and indeed not usually the simplest one. Another is the
method of undetermined coefficients. Both approaches can be described
in a vastly more general setting, in which a given linear functional
is to be approximated by other linear functionals so as to have
exactness on a suitable finite-dimensional function space. Although
both approaches yield the same formulae, the mechanices involved are
quite different. Let's see two simple examples.\\
\Ex 1:\\
$$%
%
{\displaystyle \int_{0}^{1}}
%
f(x)dx\approx%
%
{\displaystyle \int_{0}^{1}}
%
p_{1}(x)dx=%
%
{\displaystyle \int_{0}^{1}}
%
((1-x)f(0)+xf(1))dx =\frac{1}{2}[f(0)+f(1)]$$

Method of undetermined coefficients :

$$%
%
{\displaystyle \int_{0}^{1}}
%
f(x)dx=a_{1}f(0)+a_{2}f(1),\quad \mbox{ if } f\in P_{1}(x)$$
This gives
$$1=a_{1}+a_{2}$$
$$\frac{1}{2}=a_{2}$$

$$\Rightarrow a_{1}=a_{2}=\frac{1}{2}$$

\Ex 2:\\
$$%
%
{\displaystyle \int_{0}^{1}}
%
\sqrt{x}f(x)dx\approx a_{1}f(0)+a_{2}%
%
{\displaystyle \int_{0}^{1}}
%
f(x)dx$$

Method of interpolation. ``Interpolation'' here means interpolation
to the given data -- the value of $f$ at $x=0$ and the integral of
$f$ from 0 and 1. We are thus seeking a polynomial $p\in P_{1}$ such
that

$$p(0)=f(0),\quad  %
%
{\displaystyle \int_{0}^{1}}
%
p(x)dx=%
%
{\displaystyle \int_{0}^{1}}
%
f(x)dx$$

\newpage

Let $p=c_{0}+c_{1}x$, we have

$$c_{0}=f(0),\quad c_{0}+\frac{1}{2}c_{1}=%
%
{\displaystyle \int_{0}^{1}}
%
f(x)dx$$

$$\Longrightarrow c_{0}=f(0),\quad  c_{1}=2\left \{
%
{\displaystyle \int_{0}^{1}}
%
f(x)dx-f(0)\right \}  $$

Therefore,

$$%
%
{\displaystyle \int_{0}^{1}}
%
\sqrt{x}f(x)dx\approx%
%
{\displaystyle \int_{0}^{1}}
%
\sqrt{x}p(x)dx=%
%
{\displaystyle \int_{0}^{1}}
%
\sqrt{x}(c_{0}+c_{1}x)dx$$

$$=\frac{2}{3}c_{0}+\frac{2}{5}c_{1}=\frac{2}{3}f(0)+\frac{2}{5}\cdot2\left \{
%
{\displaystyle \int_{0}^{1}}
%
f(x)dx-f(0)\right \}  $$

i.e. $$%
%
{\displaystyle \int_{0}^{1}}
%
\sqrt{x}f(x)dx\approx-\frac{2}{15}f(0)+\frac{4}{5}%
%
{\displaystyle \int_{0}^{1}}
%
f(x)dx$$

Method of undetermined Coefficients :\\
Equality of (2.40) for $f(x)=1$ and $f(x)=x$ immediately yields
$$\frac{2}{3}=%
{\displaystyle \int_{0}^{1}} \sqrt{x}\cdot1dx=a_{1}+a_{2}$$

$$\frac{2}{5}=\frac{1}{2}a_{2}$$

$$\Longrightarrow a_{1}=-\frac{2}{15},\quad a_{1}=\frac{4}{5}$$

\Rmk : In place of polynomials , we could have chosen other
classes of functions, as long as we make sure that their
dimension matches the number of " free parameters "%
\\
\eject
The essence of the two examples consists of showing how a centain
linear functional can be approximated in terms of other (presumably
simpler) linear functionals by forming a suitable linear combination
of the latter.\\
\noindent \Def: A linear functional $L$ on a linear space $F$ is a
map
$L:F\rightarrow\mathbb{R}$ which satisfies
$$L(f+g)=Lf+Lg~,~\forall~f,g\in F$$
$$L(cf)=cLf~,~\forall c\in\mathbb{R},~f\in F$$
\\
\Def: $F$ is a linear function space if
$$f,g\in F\Rightarrow f+g\in F$$
$$c\in\mathbb{R}\Rightarrow cf\in F$$
\\
\Ex: Linear functional:
\\
(i) $Lf=f(0);F=\{f:f~\mbox{is defined at}~x=0\}$
\\
(ii) $Lf=f''(\frac{1}{2});F=\{f:f~\mbox{has 2nd derivatives
at}~x=\frac{1}{2}\}$
\\
(iii) $Lf=\int^{1}_{0}fdx;F=C[0,1]$
\\
(iv) $Lf=\int^{1}_{0}fwdx;F=C[0,1]~w~\mbox{is intergrable}$
\\\\
\Ex: Nonlinear functional:
\\
$Kf=|f(0)|$
\\
$Kf=\int^{1}_{0}f^{2}dx$
\\
\newpage
\noindent General approximation problem:\\
\\
Given a linear functional
L on $\mathbb{F}$,  n special linear functionals
$L_{1},\ldots,L_{n}$ on $\mathbb{F}$ and
$\ell_{i}=L_{i}f~,i=1,\ldots,n,$ and a linear subspace
$\Phi\subset\mathbb{F}$ with dim $\Phi=n$
\\\\
Goal: We want to find an approximation formula,
so that $Lf\approx\sum_{i=1}^{n}a_{i}L_{i}f$ and is exact whenever
$f\in\Phi$
\\\\
Assumption: The "interpolation problem"
\begin{center}
Find $\varphi\in\Phi$ s.t. $L_{i}\varphi=s_{i},i=1,2,\ldots,n$ has
a unique solution, $\varphi(\cdot)=\varphi(s;\cdot)$
$\forall~s=[s_{1},\ldots,s_{n}]^{T}$
\end{center}
Let $\varphi=\sum_{j=1}^{n}c_{j}\varphi_{j},$ where
$\{\varphi_{1},\ldots,\varphi_{n}\}$ is a basis of $\Phi$
\\
and $L_{i}\varphi=\sum_{j=1}^{n}c_{j}L_{i}\varphi_{j}=s_{i}$
\\\\
$G=\left[\begin{array}{cccc}L_{1}\varphi_{1}&L_{1}\varphi_{2}
&\cdots&L_{1}\varphi_{n}
\\
L_{2}\varphi_{1}&L_{2}\varphi_{2}&\cdots&L_{2}\varphi_{n}
\\\cdots&\cdots&\cdots&\cdots\\L_{n}\varphi_{1}&L_{n}\varphi_{2}
&\cdots&L_{n}\varphi_{n}\end{array}\right]$
$\left[\begin{array}{cccc}c_{1}\\c_{2}\\\vdots\\c_{n}\end{array}\right]=
\left[\begin{array}{cccc}s_{1}\\s_{2}\\\vdots\\s_{n}\end{array}\right]$
\\\\$Gc=s$ is solvable iff $|G|\neq0$, the matrix G is called ``Gram matrix.''
\newpage
\noindent
Method of interpolation:
\\\\
Given
$$\ell=
\left[\begin{array}{ccc}\ell_{1}\\\vdots\\ \ell_{n}\end{array}\right]
,~~\ell_{i}=L_{i}f$$
\\
find $\varphi(\ell;\cdot)\in\Phi$ so that $Lf\approx L\varphi$ i.e.
$Gc=l$ and $\varphi=\sum c_{i}\varphi_{i},~Lf=L\varphi~
\\\mbox{if}~f\in\Phi$
\\
Writing$~\lambda_{j}=L\varphi_{j},~~\lambda=\left[\begin{array}{ccc}\lambda_{1}
\\\vdots\\\lambda_{n}\end{array}\right]$
\\
We have
$L\varphi=\sum_{j=1}^{n}c_{j}L\varphi_{j}=\left[\begin{array}{ccc}
\lambda_{1}&\cdots&\lambda_{n}\end{array}\right]
\left[\begin{array}{cccc}c_{1}\\\vdots\\c_{n}\end{array}\right]$
$=\lambda^{T}G^{-1}\ell=[(G^{T})^{-1}\lambda]^{T}\ell$
\\
i.e. $L\varphi=\sum_{i=1}^{n}a_{i}L_{i}f,~a=
\left[\begin{array}{ccc}a_{1}\\\vdots\\a_{n}\end{array}\right]=(G^{T})^{-1}\lambda$
\\\\\\\\
Method of undetermined coefficients:
\\\\
$L\varphi_{j}=\sum_{i=1}^{n}a_{i}L_{i}\varphi_{j}$
\\
$G^{T}=\left[\begin{array}{cccc}L_{1}\varphi_{1}&L_{2}\varphi_{1}&\cdots&L_{n}\varphi_{1}
\\L_{1}\varphi_{2}&L_{2}\varphi_{2}&\cdots&L_{n}\varphi_{2}
\\\cdots&\cdots&\cdots&\cdots\\L_{1}\varphi_{n}&L_{2}\varphi_{n}&\cdots&L_{n}\varphi_{n}
\end{array}\right]$
$\left[\begin{array}{cccc}a_{1}\\a_{2}\\\vdots\\a_{n}\end{array}\right]=
\left[\begin{array}{cccc}\lambda_{1}\\\lambda_{2}\\\vdots\\\lambda_{n}
\end{array}\right]$
\\
i.e. $\therefore~\left[\begin{array}{ccc}a_{1}\\\vdots\\a_{n}\end{array}\right]=(G^{T})^{-1}\lambda$
\\\\
so those two methods are mathematically equivalent.
\eject
\begin{center}
 {\large \bf  \S 2.6 Peano representation of linear functionals
 } \\
\end{center}
It may be argued that the method of interpolation, at least in the
case of polynomials is more powerful than the method of undetermined
coefficients because it also yields an expression for the error
term. The method of undetermined coefficients, in contrast, generates
only the coefficients in the approximation and gives no clue as to the
approximation error.\\
There is, however, a device due to Peano that allows us to discuss the
error after the approximation has been found. The point is that the error,
$$Ef=Lf-\sum^{n}_{i=1}a_{i}L_{i}f$$
is itself a linear functional, one that annihilates all polynomials,
say, of degree d,
$$Ep=0~\forall~p\in \mathbb{P}_{d}$$
Let $F=\mathbb{C}^{d+1}[a,b]$, then by Taylor's Theorem, we have
$$f(x)=f(a)+(x-a)\frac{f'(a)}{1!}+\cdots+(x-a)^{d}\frac{f^{(d)}(a)}{d!}
+\frac{1}{d!}\int^{x}_{a}(x-t)^{d}f^{(d+1)}(t)dt$$
Now, let
$$(x-t)_{+}=\left\{\begin{array}{ll}~x-t~~\mbox{if}~x-t\geq0~
\\~~~0~~~~~\mbox{if}~x-t<0~\end{array}
\right.$$
Thus
$$\int^{x}_{a}(x-t)^{d}f^{(d+1)}(t)dt=\int^{b}_{a}(x-t)^{d}_{+}f^{(d+1)}(t)dt$$
And
$$Ef=\frac{1}{d!}E\{\int^{b}_{a}(x-t)^{d}_{+}f^{(d+1)}(t)dt\}
=\frac{1}{d!}\int^{b}_{a}[E_{(x)}(x-t)^{d}_{+}]f^{(d+1)}(t)dt,$$
provided the interchange of E with the integral is legitimate. (For
most functionals it is.) The subscript $x$ is to indicate that E acts
on the variable $x$ (not t).
\newpage
\noindent
\Def:\\
$$K_{d}(t)=\frac{1}{d!}E_{(x)}(x-t)^{d}_{+},~t\in\mathbb{R}$$
Thus
$$Ef=\int^{b}_{a}K_{d}(t)f^{(d+1)}(t)dt$$
This is called
the Peano representation of the functional $E$, and $K_{d}$ the dth
Peano kernel for $E.$
\\\\
\Def:\\
 $E$ is definite of order $d$ if its Peano kernel $K_{d}$
doesn't change sign.
(We then say $K_{d}$ is definite).
\\
By Mean Value Theorem of integration we have
$$Ef=f^{((d+1))}(\tau)\int^{b}_{a}K_{d}(t) dt,\quad a<\tau< b$$
Moreover, if we let $$f=\frac{t^{d+1}}{(d+1)!}$$ then
$$E(\frac{t^{d+1}}{(d+1)!})=\int^{b}_{a}K_{d}(t) dt$$ denote it by
$e_{d+1}$, therefore $$Ef=e_{d+1}f^{d+1}(\tau) $$ hence
$$|Ef|\leq|e_{d+1}|\cdot||f^{(d+1)}||_{\infty}$$
\newpage
\noindent
\\
\Ex:
\\
$$Ef=\int^{1}_{0}\sqrt{x}f(x)dx-(-\frac{2}{15}f(0)+\frac{4}{5}\int^{1}_{0}f(x)dx)$$

$$Ef=\int^{1}_{0}K_{1}(x)f''(t)dt,~K_{1}=E_{(x)}(x-t)_{+}$$
\\
\begin{figure}[h]
\centerline{\epsfbox{figg.eps}}
\end{figure}
\\
Furthermore,
\\
$$K_{1}(t)=\int^{1}_{0}\sqrt{x}(x-t)_{+}dx
+\frac{2}{15}\cdot0-\frac{4}{5}\int^{1}_{0}(x-t)_{+}dx$$
%\\
%since$~x=0~\mbox{so}~x-t<0~\mbox{i.e.}~(0-t)_{+}\equiv0$
%\\
$$=\int^{1}_{t}\sqrt{x}(x-t)dx-\frac{4}{5}\int^{1}_{t}(x-t)dt=
\frac{2}{15}t(2t^{\frac{3}{2}}-3t+1)$$
Now the function in parenthese, say, $q(t)$, satisfies
$q(0)=1,\,q(1)=0,\,q^{'}(t)=-3(1-t^{1/2})<0$ for $0<t<1.$ Therefore
$K_{1}(t)$ is(positive) definite
\\
and
\\
$$e_{2}=E(\frac{t^{2}}{2!})=\int^{1}_{0}\sqrt{x}\frac{~x^{2}}{2}dx+\frac{2}{15}
\cdot0-\frac{4}{5}\int^{1}_{0}\frac{~x^{2}}{2}dx=\frac{1}{105}$$
\\$\mbox{i.e.}~Ef=\frac{1}{105}f''(\tau),0<\tau<1$
\end{document}
