It follows that for any x b b x e n0 f n 0 n x

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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 x0 |≤ . (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 x 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) = k=0 x2k+1 . (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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