Taylor Series

Dividing by x4 gives us that x4 e 2 cosx2 1 m x4 m x

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Unformatted text preview: rom the Approximation Theorem 0 that we can ﬁnd a constant M1 such that for any u ∈ [−1, 1] −M1 u2 ≤ eu − (1 + u) ≤ M1 u2 DE Math 128 333 (B. Forrest)2 4.5. Introduction to Taylor Series CHAPTER 4. Sequences and Series 4 since 1 + u is the ﬁrst degree Taylor polynomial of eu . Now if x ∈ [−1, 1], then u = x ∈ 2 4 1 [−1, 1]. In fact, u ∈ [0, 2 ). It follows that if x ∈ [−1, 1] and we substitute u = x , then we 2 get x4 x4 −M1 x8 −M1 x8 ≤ e 2 − (1 + ) ≤ 4 2 4 We also can show that there exists a constant M2 such that for any v ∈ [−1, 1] −M2 v 4 ≤ cos(v ) − (1 − since 1 − v2 2 v2 ) ≤ M2 v 4 2 is the third degree Taylor polynomial for cos(v ). If x ∈ [−1, −1] then so is x2 . If we let v = x2 , then we see that −M2 x8 ≤ cos(x2 ) − (1 − x4 ) ≤ M2 x 8 2 The next step is to multiply each term in the previous inequality by −1 to get −M2 x8 ≤ (1 − x4 ) − cos(x2 ) ≤ M2 x8 . 2 (Remember, multiplying by a negative number reverses the inequality) We can now add ou...
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This document was uploaded on 03/23/2013.

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