= Solution
{c}
Expand the fifth-degree <Taylor polynomial> $p_5$ in <Chebyshev polynomials>, using the identities for powers obtained from part (a):
$$
p_5=-\frac{11}{32}T_0+\frac{11}{8}T_1-\frac38T_2
+\frac7{48}T_3-\frac1{32}T_4+\frac1{80}T_5.
$$
Delete the last two Chebyshev terms and convert back to powers. The economized cubic is
$$
\boxed{p_3(x)=\frac1{32}+\frac{15}{16}x-\frac34x^2+\frac7{12}x^3.}
$$
Since $|T_j|\leq1$,
$$
\|p_5-p_3\|_\infty
\leq\frac1{32}+\frac1{80}=\frac7{160}.
$$
Equality holds at $x=-1$, where both deleted terms have negative sign. Thus \b[the economization changes the <polynomial> by at most $7/160$], exactly in supremum <norm> on $[-1,1]$.
For the target itself, $|\log(1+x)-p_3(x)|\leq|\log(1+x)-p_5(x)|+7/160$ wherever $x>-1$. The logarithm is singular at $-1$, so its total uniform error on all of $(-1,1]$ is infinite; the finite estimate concerns the additional <polynomial> change, or the increase on any smaller interval where the original error is bounded.
\Image[/past-exam-of-the-approximation-theory-and-finite-element-methods-course-of-the-university-of-oxford/2008-log-economization.png]
{title=Chebyshev economization removes two bounded high-degree components}
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