Chebyshev Expansions - Society for Industrial and Applied
... This result is easy to prove by noticing that fs = ( + I)s f0 , s = 0, 1, . . . , n, and by
expanding the binomial of commuting operators and I (I being the identity, Ifi = fi ).
The formula for the remainder in Theorem 3.2 resembles that for the Taylor formula
of degree n (Lagrange form), except ...