## Theorem 8.4.6: Taylor's Theorem

Suppose

*f C*, i.e.^{n+1}([a, b])*f*is*(n+1)*-times continuously differentiable on*[a, b]*. Then, for*c [a,b]*we have:wheref(x) =

In particular, the Taylor series for an infinitely often differentiable functionR_{n+1}(x) =^{1}/_{n!}(x-t)^{n}f^{(n+1)}(t) dt

*f*converges to*f*if and only if the remainder*R*converges to zero as_{(n+1)}(x)*n*goes to infinity.The statement involves "all integers" and therefore an induction proof might be in order. By the fundamental theorem of calculus we know that

f '(t) dt = f(x) - f(c)

f(x) = f(c) + f '(t) dt

That proves the base case *n=0* of our induction. Now assume that

wheref(x) =

R_{n+1}(x) =^{1}/_{n!}(x-t)^{n}f^{(n+1)}(t) dt

is true. Let's look at the term *R _{n+1}(x)* and integrate
by parts, where:

u'(t) = (x-t)^{n}u(t) = -^{1}/_{n+1}(x-t)^{n+1}v(t) = f^{(n+1)}(t)v'(t) = f^{(n+2)}(t)

Therefore

R_{n+1}(x) =^{1}/_{n!}(x-t)^{n}f^{(n+1)}(t) dt =

=^{1}/_{n!}u'(t) v(t) dt =

=^{1}/_{n!}(u(x)v(x) - u(c)v(c) - u(t) v'(t) dt) =

=^{1}/_{n!}^{1}/_{n+1}(x-c)^{n+1}f^{(n+1)}(c) +^{1}/_{n!}^{1}/_{n+1}(x-t)^{n+1}f^{(n+2)}(t) dt =

=^{1}/_{(n+1)!}(x-c)^{n+1}f^{(n+1)}(c) + R_{n+2}(x)

But then, putting that together with our assumption, we have:

f(x) = =

=

which finishes the induction.

It remains to show that the Taylor series for an infinitely often differentiable
function *f* converges to *f* if and only if the
remainder *R _{(n+1)}(x)* converges to zero as

*n*goes to infinity.

But what little there is to do to prove that is left to ... yup ... the reader.