Taylor's theorem with Cauchy remainder

Ponder...

With , there is such that .

Theorem 3.23 (Taylor's Theorem: Cauchy Remainder)

Let as in Taylor’s Theorem: Lagrange Remainder 3.22 and define similarly. Then there exists such that

Proof

Again take . Let

so that . The function is continuous on and differentiable on with

For set ; then , so Rolle’s theorem provides with , that is

Substituting and rearranging,

Choosing gives the Cauchy remainder, while recovers the Lagrange form.

Take with , smooth on , and expand about : the Taylor polynomial is

and on .

The Lagrange form gives with . For and this is bounded by , but the argument fails on .

The Cauchy form instead writes

Since , the quotient is at most , and stays bounded independently of . Hence as for every : exponentially small, on the whole interval.