Exercise 6.7.8

Question

\begin{enumerate}[label=(\alph*)] 
    \item Fix $a \in[-1,1]$ and sketch
$$
h_{a}(x)=\frac{1}{2}(|x-a|+(x-a))
$$
over $[-1,1]$. Note that $h_{a}$ is polygonal and satisfies $h_{a}(x)=0$ for all $x \in[-1, a]$.
\item Explain why we know $h_{a}(x)$ can be uniformly approximated with a polynomial on $[-1,1]$.
\item Let $\phi$ be a polygonal function that is linear on each subinterval of the partition
$$
-1=a_{0}<a_{1}<a_{2}<\cdots<a_{n}=1 .
$$
Show there exist constants $b_{0}, b_{1}, \ldots, b_{n-1}$ so that
$$
\phi(x)=\phi(-1)+b_{0} h_{a_{0}}(x)+b_{1} h_{a_{1}}(x)+\cdots+b_{n-1} h_{a_{n-1}}(x)
$$
for all $x \in[-1,1]$.
\item Complete the proof of WAT for the interval $[-1,1]$, and then generalize to an arbitrary interval $[a, b]$.
     \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
    \item Left as an application for your favourite graphing calculator
    \item $|x-a|$ can be uniformly approximated, and multiplication by a constant and addition of polynomials preserves the ability to be uniform approximated.
    \item $b_0 = \frac{\phi(a_1) - \phi(a_0)}{a_1 - a_0}$, and for $n \geq 1 $, $b_n = \frac{\phi(a_{n+1}) - \phi(a_n)}{a_{n+1} - a_n} - b_{n-1}$
    \item Fix $\epsilon > 0$. For a function $f$ continuous over $[-1,1]$, apporixmate it uniformly within $\epsilon/2$ with a polygonal function $\phi(x)$, and approximate $\phi(x)$ uniformly within $\epsilon/2$ with a polynomial. The triangle inequality ensures that this polynomial uniformly approximates $f$ within $\epsilon$. To generalize over $[a,b]$ the same technique in Exercise 6.6.7 of scaling $x$ and $f$ can be used.
 \end{enumerate}

Exercise 6.7.8

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Exercise 6.7.8

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