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Unformatted text preview: is, we will have
| f (k+1) (ck) |≤ 1
From this it follows immediately that
| Rk,a (x0 ) |=|
| x0 |k
k→∞ k! However, we know that lim f (k+1) (ck ) k+1
| x0 |k+1
(k + 1)!
(k + 1)! = 0, so the Squeeze Theorem shows that
lim Rk,a (x0 ) = 0 k→∞ Therefore, since x0 was chosen at random, for f (x) = cos(x) and any x ∈ R, we have
∞ f (x) =
n=0 DE Math 128 f (n) (0) n
n! 336 (B. Forrest)2 CHAPTER 4. Sequences and Series 4.5. Introduction to Taylor Series In particular, for any x ∈ R ∞ kx (−1) cos(x) =
k=0 2k 2k ! . A nearly identical argument applies to sin(x) as it did above for cos(x) to show that for
any x ∈ R, sin(x) agrees with the value of its Taylor series. That is
∞ (−1)k sin(x) =
(2k + 1)! These two examples suggest the following very useful theorem:
Theorem. [Convergence Theorem for Taylor Series]
Assume that f (x) has derivatives of all orders on an interval I containing x = a. Assume
also that there exists an M such that
| f (k) (x) |≤ M
for all k and for all x ∈ I ....
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