Mathematics 36, section 5 -- AP Calculus

Discussion 2 -- Taylor Series

October 1, 1999

Background

Yesterday, we introduced the Taylor series of a function f(x) at x = a. As long as we can compute derivatives of f of all orders at x = a, the Taylor series is defined and looks like this:

Taylor series of f = sum_k=0^infinity f^(k)(a) (x - a)^k/k! For example, the Taylor series of f(x) = e^x at a = 0 looks like: sum_{k = 0}^infinity x^k / k! since f^(k)(0) = 1 for all k. Today we want to investigate the questions:

If we substitute x = c, does the Taylor series of f evaluated at x = c ``add up to f(c)''?
Does that work for all c? Some c? Which ones?

Recall that in class yesterday, we saw that the answer to question 1 for a particular x value is ``yes'' exactly when the error in the Taylor approximation with the nth degree polynomial:

E_n(x) = f(x) - p_n(x) = f(x) - sum_{k = 0}ⁿ f^(k)(a) (x - a)^k/k!

goes to zero as n -> infinity. For example, for f(x) = 1/(x + 1), at a = 0, the error is

E_n(x) = 1/(x + 1) - (1 - x + x^2 - ... +(-1)ⁿ xⁿ) =(-1)ⁿ⁺¹ xⁿ⁺¹ / (x + 1).

This goes to zero as n -> infinity only when |x| < 1.

Discussion Questions

A) Consider the Taylor series for f(x) = sin(x) at a = 0. As we know from Lab 3, by the Theoretical Error Bound for Taylor polynomial approximations,
|E_n(x)| = |sin(x) - p_n(x)| <= (1 / (n+1)!) |x|ⁿ⁺¹
(since all derivatives of sin(x) are bounded above by M = 1 for all x in R).
1. If |x| <= 1, then it is pretty clear that E_n(x) -> 0 as n -> infinity. Explain why. (What does the error bound do as n -> infinity?)
2. Now take an x > 1, like x = 4. What happens to the error bound 4ⁿ⁺¹ / (n+1)! as n -> infinity? (Try some experiments using a calculator -- evaluate the error bound for n = 2,3,4,5,6,10,20 at least and think about what that says for the size of the error.)
3. Repeat part 2 if x = 10. You might want to take n = 2,3,4,5,10,11,12,15,20 here.
4. Now let's try to find a general pattern here. Let x = c be any fixed number. Note that if we go from one n to the next integer n + 1, the corresponding error bounds are
  cⁿ⁺¹ / (n+1)! and cⁿ⁺² / (n+2)!
  The error bound for n+1 can be rewritten like this:
  
  cⁿ⁺² / (n+2)! = c cⁿ⁺¹ / ((n+2) (n+1)!) = (c / n+2) (cⁿ⁺¹ / (n+1)!),
  which shows something interesting. What will happen as n -> infinity here (recall c is fixed)? How does that explain what you saw in parts 2 and 3 with c = 4 and c = 10.
5. What is your conclusion here? For which x does the Taylor series of sin(x) ``add up to'' sin(x)?
B) Now consider the Taylor series for f(x) = e^x at a = 0. By the Theoretical Error Bound for Taylor polynomial approximations,
|E_n(x)| = |e^x - p_n(x)| <= (M / (n+1)!) |x|ⁿ⁺¹
if |f⁽ⁿ⁺¹⁾(z)| <= M for all z in [-x, x].
1. Take an x > 1, like x = 4. Explain why you could use M = 3⁴ = 81 in the error bound. What happens to the error bound 81 4ⁿ⁺¹ / (n+1)! as n -> infinity? (Try some experiments using a calculator -- evaluate the error bound for n = 2,3,4,5,6,10,20 at least and think about what that says for the size of the error.)
2. Repeat part 1 if x = 10. (What is one M you can use?) You might want to take n = 2,3,4,5,10,11,12,15,20 here.
3. Again, let's try to find a general pattern here. Let x = c be any fixed number. Note that if we go from one n to the next integer n + 1, what are the corresponding error bounds, and what is the ratio between them? What will happen as n -> infinity here (recall c is fixed)? How does that explain what you saw in parts 1 and 2 with c = 4 and c = 10.
4. What is your conclusion here? For which x does the Taylor series of e^x ``add up to'' e^x?

Assignment

Group write-ups due Tuesday, October 5.