121616949-math.284 - x N 1 N 1 fl fl fl fl< 005 Since...

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270 Chapter 10 Sequences and Series - 5 - 4 - 3 - 2 - 1 0 1 1 2 3 4 5 ....................................................................................................................................................................................... . . . . . . . . . . . . . . . . . . . . . . . . . . ................................................................................................................................................................................................ Figure 10.5 sin x and a polynomial approximation. Every derivative of sin x is ± sin x or ± cos x , so | f ( N +1) ( z ) | ≤ 1. The factor ( x - a ) N +1 is a bit more difficult, since x - a could be quite large. Let’s pick a = 0 and | x | ≤ π/ 2; if we can compute sin x for x [ - π/ 2 , π/ 2], we can of course compute sin x for all x . We need to pick N so that
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Unformatted text preview: x N +1 ( N + 1)! fl fl fl fl < . 005 . Since we have limited x to [-π/ 2 , π/ 2], fl fl fl fl x N +1 ( N + 1)! fl fl fl fl < 2 N +1 ( N + 1)! . The quantity on the right decreases with increasing N , so all we need to do is find an N so that 2 N +1 ( N + 1)! < . 005 . A little trial and error shows that N = 8 works, and in fact 2 9 / 9! < . 0015, so sin x = 8 X n =0 f ( n ) (0) n ! x n ± . 0015 = x-x 3 6 + x 5 120-x 7 5040 ± . 0015 . Figure 10.5 shows the graphs of sin x and and the approximation on [0 , 3 π/ 2]. As x gets larger, the approximation heads to negative infinity very quickly, since it is essentially acting like-x 7 ....
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