2 b g x dx a g x dx a for any function g x and

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Unformatted text preview: (u − 0)2 |≤ . 2! 2 328 (B. Forrest)2 CHAPTER 4. Sequences and Series 4.5. Introduction to Taylor Series Observe that the integral is over the interval [0, 0.1]. If 0 ≤ x ≤ 0.1, then squaring gives us that 02 ≤ x2 ≤ (0.1)2 , or equivalently, that 0 ≤ x2 ≤ 0.01 Multiplying each term in the above inequality by −1 reverses the inequality to show that if 0 ≤ x ≤ 0.1, then −0.01 ≤ −x2 ≤ 0 This is helpful because if we choose any x ∈ [0, .1] and then let u = −x2 , the above inequality shows that −0.01 ≤ u = −x2 ≤ 0. However, for such a u, we know that | eu − (1 + u) |≤ u2 . 2 Substituting −x2 for u in the above inequality, we see that if x ∈ [0, 0.1], then 2 | e−x − (1 + (−x2 )) |≤ or 2 | e−x − (1 − x2 ) |≤ Recall that b | (−x2 )2 2 x4 . 2 b g (x) dx |≤ a | g (x) | dx a for any function g (x) and that if f (x) ≤ g (x), then for a ≤ b b b f (x) dx ≤ a g (x) dx. a Using these rules of integration and the above inequality, we get 0 .1 0.1 2 e−x dx − | 0 0 .1 2 e−x − (1 − x2 ) dx | (1 − x2 ) dx | = | 0 0 0.1 2 | e−x − (1 −...
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This document was uploaded on 03/23/2013.

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