Taylor's theorem with Lagrange remainder
With , there is such that .
Let be continuous, and assume its first derivatives are continuous as well, and that it is -times differentiable on . [This is all satisfied if where .]
Let the Taylor remainder be
Then there exists such that
Proof
By translation invariance it suffices to take ; otherwise apply the argument to .
Let be chosen so that , where
Then and for all . Applying Rolle’s theorem to between and gives with ; applying it again to between and gives ; after steps there is with
Hence , so
A second proof, applying Rolle’s theorem to a family of auxiliary functions , yields the Cauchy form of the remainder as well.
Order bound
If with , then is continuous and hence bounded on the compact interval: for
the Lagrange form gives
so is as . This does not say that as : even when , nothing controls how grows with . Nothing is special about : applying the theorem to covers increments with .
Related
Stated in
- Theorem 3.22 (Taylor's Theorem: Lagrange Remainder)§3.3 Higher Derivatives and Taylor’s Theorem