Interactive Real Analysis - part of MathCS.org

• Real Analysis
• 1. Sets and Relations
• 2. Infinity and Induction
• 3. Sequences of Numbers
• 4. Series of Numbers
• 5. Topology
• 6. Limits, Continuity, and Differentiation
• 7. The Integral
• 8. Sequences of Functions
• 8.1. Pointwise Convergence
• 8.2. Uniform Convergence
• 8.3. Series and Power Series
• 8.4. Taylor Series
• 8.5. Approximation Theory
• 9. Historical Tidbits
• Java Tools

Theorem 8.4.6: Taylor's Theorem

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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

R_n+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.

Context

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) =

R_n+1(x) = ¹/_n! (x-t)ⁿf ⁽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)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) + 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.