Moreover all of the derivatives 2 of cosx are

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Unformatted text preview: 1, 1]. Then so is the function f (k+1) (x) g (x) =| | (k + 1)! DE Math 128 331 (B. Forrest)2 4.5. Introduction to Taylor Series CHAPTER 4. Sequences and Series The Extreme Value Theorem tells us that g (x) has a maximum on [−1, 1]. Therefore, there is an M such that f (k+1) (x) | |≤ M (k + 1)! for all x ∈ [−1, 1]. Let x ∈ [−1, 1]. Taylor’s Theorem shows that there is a c between x and 0 such that f (k+1) (c) k+1 | Rk,0 (x) |=| x | (k + 1)! Therefore, | f (x) − TK,0 (x) | = | Rk,0 (x) | =| f (k+1) (c) k+1 x | (k + 1)! ≤ M | x |k+1 since c is also in [−1, 1]. It follows that −M | x |k+1 ≤ f (x) − Tk,0 (x) ≤ M | x |k+1 . We summarize as follows: Theorem. [The Approximation Theorem] Assume that f (k+1) is continuous on [−1, −1]. Then there exists a constant M > 0 such that | f (x) − Tk,0 (x) |≤ M | x |k+1 or equivalently that −M | x |k+1 ≤ f (x) − Tk,0 (x) ≤ M | x |k+1 for each x ∈ [−1, 1]. This theorem is very helpful in calculating many limits. Example. cos(x)−1 . x2 x→0 Calculate lim 2 We know that for f (x) = cos(x) we have T2,0 = 1 − x . Moreover, all...
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