m8l02 - S ECTION 1.2 1.2. T HE R EMAINDER T HEOREM 9 T HE R...

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SECTION 1.2 THE REMAINDER THEOREM 9 1.2. T HE R EMAINDER T HEOREM Our goal in this section is to explain why Taylor polynomials make such nice approxima- tions to certain functions. The object of interest for us is therefore the remainder term , R n x f x T n x . While you may be accustomed to “remainder” being a notion from division, here we are using its more general meaning as “the part left over.” Three remainder terms for sin x (centered at 0 ) are shown below. As can be seen from these plots, the Taylor polynomials for sin x give better and better approximations near x 0 . 1 1 π π R 1 x 1 1 π π R 3 x 1 1 π π R 5 x If we can show that the remainder term is close to 0 , then that means that the Taylor polynomial is a good approximation to the function. For example, consider the remainder terms plotted above; T 5 x is a better approximation to sin x than T 3 x because R 5 x is close to 0 for more values of x than R 3 x . Our main tool in bounding the remainder term is, somewhat surprisingly, a result from Calculus I. Rolle’s Theorem. Suppose that f is a differentiable function. If f a f b for some a b then there is at least one number c between a and b at which f c 0 . Rolle’s Theorem is named after the French mathematician Michael Rolle (1652–1719), although it also appears (without proof) in the 12th century work of the Indian astronomer Bh¯askara II (circa 1114–1185). In order to show how Rolle’s Theorem can possibly tell us anything about Taylor poly- nomials, we begin with the following problem. Example 1. Determine how close the Taylor polynomial of degree 3 for sin x centered at 0 , T 3 x x x 3 6 , is to sin x at the point x 1 2 . Of course, there is an easy way to cheat: just compute the difference between T 3 1 2 and sin 1 2 . In fact, we did this in the last section. But let’s pretend for a moment that we don’t have access to calculators or computers, so we have to estimate the error, rather than simply compute the error. This is a valuable exercise because error estimates are useful theoretically.
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10 CHAPTER 1T AYLOR POLYNOMIALS Solution. There is certainly some value of K for which sin 1 2 T 3 1 2 K 1 2 4 , because we could just solve for K (if we did so, we would Fnd that K 0 . 00414 , but remember that we are pretending not to have calculators!). Let K be this value. Now deFne the function F x by F x sin x T 3 x Kx 4 . This function is 0 at two points: 0 (which is, not coincidentally, the center), and 1 2 , because that’s how we deFned K . Therefore we can apply Rolle’s Theorem to see that F x sin x T 3 x 4 3 is 0 at some point c 1 between 0 and 1 2 . But F x is also 0 at x 0 , because sin x T 3 x at x 0 and 4 3 is 0 at x 0 . Using Rolle’s Theorem again, we see that F x sin x T 3 x 12 2 is 0 at some point c 2 between 0 and c 1 . Again, F 0 0 , so Rolle’s Theorem says that F x sin x T 3 x 24 is 0 at some point c 3 between 0 and c 2 . Still, F 0 0 , so we can apply Rolle’s Theorem once again: F 4 sin 4 x T 4 3 x 24 K is 0 at some point c 4 between 0 and c 3 . This is the point we are really interested in, so let’s set c c 4 . Notice that T 4 3 x is 0 , because T 3 x
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This note was uploaded on 12/27/2011 for the course MAC 1114 taught by Professor Gentimis during the Spring '11 term at University of Florida.

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m8l02 - S ECTION 1.2 1.2. T HE R EMAINDER T HEOREM 9 T HE R...

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