# Interactive Real Analysis

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## 8.4. Taylor Series

### Theorem 8.4.6: Taylor's Theorem

Suppose f Cn+1([a, b]), i.e. f is (n+1)-times continuously differentiable on [a, b]. Then, for c [a,b] we have:
f(x) = where
Rn+1(x) = 1/n! (x-t)nf (n+1)(t) dt
In particular, 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.

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

f(x) = where
Rn+1(x) = 1/n! (x-t)nf (n+1)(t) dt

is true. Let's look at the term Rn+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

Rn+1(x) = 1/n! (x-t)nf (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) + Rn+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. Next | Previous | Glossary | Map