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Unformatted text preview: Taylor Polynomials and Taylor Series Math 126 In many problems in science and engineering we have a function f ( x ) which is too complicated to answer the questions wed like to ask. In this chapter, we will use local information near a point x = b to find a simpler function g ( x ), and answer the questions using g instead of f . How useful the answers will be depends upon how closely the function g approximates f , so we also need to estimate, or bound, the error in this approximation: f g . 1. Tangent Line Error Bound. Ken is at work and his car is located at his home twenty miles north. Fifteen minutes from now, in the absence of any other information, his best guess is that the car is still at home. How accurate is this guess? If Ken knows his son will drive the car no faster than 40 miles per hour in the city, how far away can the car be in 15 minutes? A moments reflection should give you the estimate (or bound) that his son can drive the car no further than 10 miles ( 1 4 hour at 40 mph) so that the car will be within 10 miles of home, or no more than 30 miles from Ken. We can also see this by using the Fundamental Theorem of Calculus. Suppose Kens son is driving the car in a straight line headed north and suppose x ( t ) is the distance of the car from Ken at time t then x ( t ) x (0) = integraldisplay t x prime ( u ) du. Here x prime ( u ) is the velocity at time u . In this problem we know  x prime ( u )  40 for all u between 0 and t , so that  x ( t ) x (0)  = vextendsingle vextendsingle vextendsingle vextendsingle integraldisplay t x prime ( u ) du vextendsingle vextendsingle vextendsingle vextendsingle integraldisplay t  x prime ( u )  du (1 . 1) integraldisplay t 40 du = 40 t. 1 There are a couple of steps above that need justification. First we used the inequality vextendsingle vextendsingle vextendsingle vextendsingle integraldisplay x prime ( u ) du vextendsingle vextendsingle vextendsingle vextendsingle integraldisplay  x prime ( u )  du. The integral integraltext x prime ( u ) du represents the area which is below the curve y = x prime ( u ) and above the uaxis minus the area which is above the curve and below the u axis. Whereas the righthand side is equal to the total area between the curve and the u axis, and so the righthand side is at least as big as the left. Secondly we replaced the function  x prime ( u )  by the larger constant 40 and the area under the curve y = 40 is at least as big as the area under the curve y =  x prime ( u )  . This was a rather longwinded way to get the same bound, but it works in general: if  f prime ( t )  M for all t between b and x , then  f ( x ) f ( b )  M  x b  ....
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 Fall '07
 Smith
 Polynomials, Taylor Series

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