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

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.