Past exam of the approximation theory and finite element methods course of the University of Oxford / 2007 / A / 4

The endpoint chord is $l(x)=1-x$. Since the square-root function is concave, the gap $g(x)=\sqrt{1-x}-(1-x)$ is nonnegative. Put $t=\sqrt{1-x}$ to obtain $g=t-t^2$, whose maximum is $1/4$ at $t=1/2$, or $x=3/4$.

Shift the chord upward by half that maximum. ThenThe errors at $x=0,3/4,1$ are respectively $-1/8,+1/8,-1/8$. These are three alternating extrema, so equioscillation proves optimality and uniqueness for linear polynomials. This is minimax linear approximation of a square-root endpoint profile.

Use the Taylor expansion $\sqrt{1-x}=1-x/2-x^2/8+O(x^3)$. Expanding the rational form givesMatching through degree two yields $a_0=1$, $a_1-b_1=-1/2$, and $b_1^2-a_1{b}_1=-1/8$. The last equation becomes $b_1/2=-1/8$, soThis is the $[1/1]$ Padé approximant about zero. Its pole is at four, outside the interpolation interval; its order of matching is local and does not assert uniform minimax optimality.

Let $\beta (q)$ be the centered nondimensional cubic B-spline, even and supported on $[-2,2]$. The outer cubic must be $b(2-|q|{)}^3$ to make the function and its first two derivatives vanish at the support edge. Write its inner part as $A+B{q}^2+C|q{|}^3$. Matching the value and first two derivatives at $|q|=1$ gives $A=4b$, $B=-6b$, $C=3b$. Its nonzero knot values are then $4b,b,b$, so requiring their sum to be one fixes $b=1/6$. ThusIn particular $\beta (0)=2/3$, $\beta (\pm 1)=1/6$.

With $h=1/m$, represent the spline on $[0,1]$ as $s(x)={\sum }_{j=-1}^{m+1}{c}_j\beta (x/h-j)$. There are $m+3$ unknown coefficients. interpolation at $x_i=ih$, $0\le i\le m$, givesThe natural endpoint conditions, using ${\beta }^{\prime \prime }(0)=-2$ and ${\beta }^{\prime \prime }(\pm 1)=1$, giveThese $m+3$ equations are the required matrix system: the $m+1$ interpolation rows have coefficients $(1,4,1)$ at consecutive positions and right sides $6u(x_i)$; the two endpoint rows have $(1,-2,1)$ and zero right sides. No solution of that system is required. The scaling uses $\beta (x/h-j)$, without an extra $1/h$, because normalization is by knot values rather than by integral.