Past exam of the approximation theory and finite element methods course of the University of Oxford / 2008 / A / 1

Put $x=\cos \theta$. The trigonometric identitygives the Chebyshev polynomial recurrence $T_{n+1}=2x{T}_n-T_{n-1}$. Starting from $T_0=1$ and $T_1=x$, successive substitution yieldsThe recurrence also proves by induction that $T_n$ is a polynomial of degree $n$, with leading coefficient $2^{n-1}$ for $n\ge 1$.

For $n\ge 1$, set $p=2^{1-n}{T}_n$. It is monic by part (a), and its supremum norm on $[-1,1]$ is $E=2^{1-n}$. At the $n+1$ successive extreme points $x_j=\cos (j\pi /n)$ it alternates between $E$ and $-E$.

Suppose a monic degree-$n$ polynomial $q$ had $\Vert q{\Vert }_{\infty }<E$. At each $x_j$, the difference $p-q$ would have the sign of $p$. These signs alternate, so the intermediate value theorem forces a root in each of the $n$ successive gaps. But the leading monic terms cancel, making $\mathrm{deg}(p-q)\le n-1$, a contradiction. ThereforeThis is the monic monic Chebyshev extremal polynomial. The positive-degree restriction matters: at $n=0$ the printed scaling gives the constant two, while the only monic constant is one, so that version is false. The customary theorem uses $n\ge 1$.

Expand the fifth-degree Taylor polynomial $p_5$ in Chebyshev polynomials, using the identities for powers obtained from part (a):Delete the last two Chebyshev terms and convert back to powers. The economized cubic isSince $|T_j|\le 1$,Equality holds at $x=-1$, where both deleted terms have negative sign. Thus the economization changes the polynomial by at most $7/160$, exactly in supremum norm on $[-1,1]$.

For the target itself, $|\log (1+x)-p_3(x)|\le |\log (1+x)-p_5(x)|+7/160$ 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.