= Solution
Map $[a,b]$ to $[-1,1]$ by $z(r)=(2r-a-b)/(b-a)$. Since $a>0$, the normalization point $z(0)=-(a+b)/(b-a)$ lies outside that interval. The proposed <Chebyshev polynomial> ratio has degree $k$ and equals one at zero. On the spectral interval its maximum absolute value is
$$
\boxed{\frac{|T_k(1)|}{|T_k(-(a+b)/(b-a))|}.}
$$
The <Chebyshev minimax polynomial with exterior normalization> minimizes this maximum among degree-at-most-$k$ polynomials with $p(0)=1$. One may quote the minimax property. Its reason is the alternating $k+1$ extremal values: a strictly better competitor would force their difference to have at least $k$ roots in $[a,b]$ and the additional root at zero outside it, impossible for degree at most $k$.
The paper uses a monic normalization of $T_k$ for $k\geq1$, rather than the usual $T_k(1)=1$ convention. The common factor cancels in the ratio. With $\kappa=b/a>1$ and $q=(\sqrt\kappa-1)/(\sqrt\kappa+1)$, the hyperbolic-cosine formula gives the error factor $\boxed{2q^k/(1+q^{2k})\leq2q^k}$. This tends rapidly to zero when the spectral interval is well-conditioned. If $a=b$, use $p(r)=1-r/a$ for exact one-step termination instead of the displayed interval map.
Back to article page
