Past exam of the approximation theory and finite element methods course of the University of Oxford / 2009 / A / 2

Let $t_1,\dots ,t_m$ be the distinct roots in $(0,1)$ at which ${\varphi }_r$ changes sign, and put $q(x)={\prod }_{j=1}^m(x-t_j)$. Both ${\varphi }_r$ and $q$ change sign at each such point, while roots of even multiplicity do not change sign. Consequently ${\varphi }_{r}q$ has one constant sign on $(0,1)$ except at its zeros, and is not identically zero. Its integral cannot vanish.

If $m<r$, however, orthogonality of the monic orthogonal polynomial ${\varphi }_r$ to every polynomial of lower degree gives $({\varphi }_r,q)=0$, a contradiction. Hence $m\ge r$. Since $\mathrm{deg}{\varphi }_r=r$, there must be exactly $r$ such roots, each of multiplicity one. They exhaust the degree, leaving no complex, exterior, or endpoint roots. Thus all roots are simple and lie strictly in $(0,1)$, which proves the requested weaker inclusion in $[0,1]$. This is the sign-change proof of the zeros of orthogonal polynomials theorem.

The monic orthogonal polynomials ${\varphi }_0,\dots ,{\varphi }_{r+1}$ form a basis of $P_{r+1}$, ordered by degree. Since $x{\varphi }_r$ is monic of degree $r+1$,For $j\le r-2$, symmetry of multiplication by $x$ gives $(x{\varphi }_r,{\varphi }_j)=({\varphi }_r,x{\varphi }_j)=0$, because $\mathrm{deg}(x{\varphi }_j)<r$. Only $a_r$ and $a_{r-1}$ remain. Taking inner products with ${\varphi }_r$ yields $a_r=({\varphi }_r,x{\varphi }_r)/({\varphi }_r,{\varphi }_r)$.

Similarly,since $x{\varphi }_{r-1}-{\varphi }_r$ has degree at most $r-1$. Rearranging proves the three-term recurrence for monic orthogonal polynomials:

The monic degree-zero polynomial is ${\varphi }_0=1$. Orthogonality of $x-a$ to one gives $a={\int }_0^{1}xdx=1/2$, so ${\varphi }_1=x-1/2$.

The relevant integrals areInsert these into the three-term recurrence for monic orthogonal polynomials:ThusTheir mutual orthogonality can also be checked directly by integrating on $[0,1]$.

The best approximation in the $L^2$ norm is the orthogonal projection onto $\mathrm{span}\{{\varphi }_0,{\varphi }_1\}$. ThereforeUsing ${\int }_0^1{x}^{k}dx=1/(k+1)$,It follows thatThe error is orthogonal to every linear polynomial, so the Pythagorean identity proves optimality and uniqueness. At $k=0$, this gives $p_1=1$; at $k=1$, it gives $p_1=x$. Both are exact because the target itself then belongs to $P_1$.

Put $x=1-t$. For $0<k\le 1$,whereas the constrained rational interpolant expands asMatching the constant term gives $a=1+b$, and matching the coefficient of $t$ gives $1/(1+b)=k$. Thus $a=1/k$, $b=(1-k)/k$, andThis is an endpoint-constrained rational approximation to a power: it matches value and derivative at $x=1$ and also enforces $r(0)=0$ for $k>0$.

There is a terminology qualification. A standard degree-$(1,1)$ Padé approximant about $t=0$ matches through $t^2$, but the displayed constrained expression has quadratic coefficient $-k(1-k)$ rather than $-k(1-k)/2$. The unconstrained Padé approximant is insteadThe source's requested formula is obtained by the two stated endpoint matching conditions within its restricted family, rather than by full Padé matching. At $k=1$ both reduce to $x$. At $k=0$, the requested expression is one for $x>0$ with a removable value at zero; finite $a,b$ in the original restricted form do not represent that constant function. The endpoint is understood by this limit.