Therefore r10 01 sinc 012 2 1 012 2

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Unformatted text preview: os(x) and f (x) = − sin(x). It follows that there exists some c betwewn 0 DE Math 128 326 (B. Forrest)2 CHAPTER 4. Sequences and Series 4.5. Introduction to Taylor Series and .01 such that the error in our linear approximation is given by | R1,0 (.01) |=| − sin(c) f (c) (.01 − 0)2 |=| (.01)2 | 2 2 It is important to note that the theorem does not tell us the value of c but rather just that it exists. This may seem to make it impossible to say anything significant about the error, but this is actually not the case. The key observation is that no matter what point c is, | − sin(c) |≤ 1. Therefore, | R1,0 (.01) | = | − sin(c) (.01)2 | 2 1 (.01)2 2 < 10−4 ≤ This simple process seems to be remarkably accurate. In fact, it turns out that this estimate is actually much better than even the above calculation suggests. This is true because not only does T1,0 (x) = x, but we also have that T2,0 (x) = T1,0 (x) = x. This means that there is a new number c between 0 and .01 such that | sin(.01) −...
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