Exercise 6.7.7

Question

\begin{enumerate}[label=(\alph*)] 
    \item Use the fact that $|a|=\sqrt{a^{2}}$ to prove that, given $\epsilon>0$, there exists a polynomial $q(x)$ satisfying
$$
|| x|-q(x)|<\epsilon
$$
for all $x \in[-1,1]$.
\item Generalize this conclusion to an arbitrary interval $[a, b]$.
     \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
    \item Let the polynomial $p(x)$ be the partial sum of the Taylor series of $\sqrt{1-x}$ which satisfies $\left|p(x) - \sqrt{1-x}\right| < \epsilon$, and let $q(x) = p(1-x^2)$. We then have
        $$\left|q(x) - \sqrt{1-(1-x^2)}\right| = \left||x| - q(x)\right| < \epsilon$$ as desired.
    \item Let $c = \max\{|a|,|b|\}$, and let $p(x)$ satisfy $\left||x| - p(x)\right| < \epsilon / c $. Then
    $$\left|\left|\frac{x}{c}\right| - p\left(\frac{x}{c}\right)\right| < \frac{\epsilon}{c}$$
    $$\left||x| - c \cdot p\left(\frac{x}{c}\right)\right| < \epsilon$$
     for $x \in [-c,c] \supseteq [a,b]$, so we can use the polynomial $c \cdot p(\frac{x}{c})$.
 \end{enumerate}

Exercise 6.7.7

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Exercise 6.7.7

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