Past exam of the approximation theory and finite element methods course of the University of Oxford / 2007 / A / 2

Use the usual approximation-space convention $P_n=\{p:\mathrm{deg}p\le n\}$. The source's phrase “degree $n$” must include lower degrees for its later space and interpolation statements; the consequence of a literal exact-degree reading is addressed in part (d).

Order the alternating points as $x_0<\cdots <x_{n+1}$. If $q$ had strictly smaller error than $E_n$, the valueswould have the alternating signs of $(u-p)(x_j)$: the second term has magnitude strictly less than $E_n$. The intermediate value theorem would then give at least $n+1$ distinct zeros of $q-p$, one in each gap. A nonzero polynomial of degree at most $n$ cannot have that many zeros. Thus no such $q$ exists, provingThe remainder of the equioscillation theorem says that a best uniform approximating polynomial exists, is unique, and is characterized by this $n+2$-point alternating extreme-error condition. For positive error, alternation is necessary as well as sufficient. If $u$ itself lies in $P_n$, the unique best polynomial is $u$ with zero error.

Let $\omega (t)={\prod }_{r=0}^n(t-x_r)$. For $x$ not one of the nodes, put $c=(u(x)-p_n(x))/\omega (x)$ and defineThis function has zeros at the $n+1$ interpolation nodes and at the additional point $x$. Repeated Rolle's theorem gives a point $\eta$ between the outermost of these points such that $F^{(n+1)}(\eta )=0$. Since $p_n^{(n+1)}=0$ and ${\omega }^{(n+1)}=(n+1)!$,Substitution yieldsAt an interpolation node both sides are zero, so any $\eta \in [-1,1]$ gives the identity there. The assumed $C^{n+1}$ regularity permits all applications of Rolle's theorem.

Choose the nodes as the $n+1$ roots of the Chebyshev polynomial $T_{n+1}$. Its leading coefficient is $2^n$, so the monic nodal product is $\omega =2^{-n}{T}_{n+1}$. Since $|T_{n+1}|\le 1$ on $[-1,1]$, part (b) bounds this particular interpolant byThe best error cannot exceed the error of a particular member of $P_n$. ThereforeThe choice of Chebyshev roots controls the maximum of the nodal product, not the size of any individual interpolation coefficient.

The polynomial $T_6$ isSubtract its monic version from $x^6$:This has degree four and hence belongs to the usual space $P_5$. The error is $T_6/32$, with norm $1/32$ and alternating extrema at the seven points $\cos (j\pi /6)$, $j=0,\dots ,6$. Part (a), followed by uniqueness, provesand that this is the unique best approximation of degree at most five.

If “degree five” is taken literally to exclude degree four, no best approximation exists in that nonclosed class: $p+ϵx^5$ approaches the error $1/32$ as $ϵ\to 0$, but uniqueness in $P_5$ excludes attainment by any degree-five polynomial. This is why the standard at-most-degree convention matters here.