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Unformatted text preview: Taylor Series and Polynomials 1 Motivations The purpose of Taylor series is to approximate a function with a polynomial; not only we want to be able to approximate, but we also want to know how good the approximation is. We are already familiar with first order approximations. The limit: lim x → sin x x = 1 tells us that when x is very close to zero sin x can be approximated with x . Using this result we are able to conclude that the integral: Z 1 sin x x dx converges if and only if Z 1 1 dx is convergent; and we can also say that the series: + ∞ X n =1 n sin 1 n is convergent if and only if: + ∞ X i =1 1 is convergent. The first case is a convergent integral and the second case is a divergent series. In both these two cases the precision provided by the first order approximation of sine suffices. Let’s analyze the following example: Z 1 sin x x 5 1 x 4 dx We want to determine if the improper integral is convergent or divergent. As in the previous example we can try to use limit comparison and we obtain the following: sin x x 5 1 x 4 = 1 x 4 sin x x 1 ≈ x → 1 x 4 · 0 = 0 When the limit is zero, limit comparison implies convergence if the integral of the function we are comparing with is convergent, but it doesn’t imply divergence if the integral we compare with is divergent. In this specific case the limit is zero and the integral we are comparing with is: Z 1 1 x 4 dx 1 which is divergent. Limit comparison in this case is inconclusive. We have used that sin x x is approximately zero when x is small, and because of this ap proximation limit comparison is inconclusive. Is there a more precise way of approximating this function? Since we want to approximate with a polynomial we can try to find out if there is a monomial ax b such that: lim x → sin x x ax b = 1 where a is a real number and b is a natural number.This limit is a 0 / 0 case, so we can apply de l’Hˆ opital: lim x → sin x x ax b = lim x → cos x 1 abx b 1 Remember now the fundamental limit: lim x → 1 cos x x 2 = 1 2 By comparing the two limits we can conclude that b = 3 and a = 1 6 , so finally: lim x → sin x x x 3 6 = 1 or equivalently: sin x ≈ x → x x 3 6 We can now use this more refined approximation to solve the improper integral: sin x x 5 1 x 4 = x 2 6 x 4 sin x x x 3 6 ≈ x → 1 x · 1 6 We are now allowed to use limit comparison in both the two directions; since the integral of 1 x is divergent in [0 , 1] we can conclude that the integral we are studying is actually divergent. We could iterate this method and have better approximations of sin. The function: sin x x + x 3 6 approaches to zero as x goes to zero. We can determine how fast by comparing with a monomial cx d . We want to find c,d such that the following limit is 1: lim x → sin x x + x 3 6 cx d = 1 We can apply de l’Hˆ opital three times and find out that d = 5 and c = 1 5!...
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This note was uploaded on 11/03/2011 for the course MATH 1090 taught by Professor Greenwood during the Spring '08 term at MIT.
 Spring '08
 greenwood
 Calculus, Polynomials, Approximation, Taylor Series

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