Let our our interval be a b and split it into n subintervals Let h b a n Adding

# Let our our interval be a b and split it into n

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Let our our interval be [ a, b ] and split it into n subintervals. Let h = b - a n . Adding up the areas of the trapezoids gives us h f ( a ) + f ( a + h ) 2 + f ( a + h ) + f ( a + 2 h ) 2 + · · · + f ( b - h ) + f ( b ) 2 which we can simplify to arrive at the approximation Z b a f ( x ) dx h ( f ( a ) 2 + f ( a + h ) + · · · + f ( b - h ) + f ( b ) 2 ) := T n . (3.2) This is very simple to implement. Just break up your interval, plug in all of the end points, and make sure to take only half of the first and last values. Trying out the same integral from before, we get 1 2 π Z 1 0 e - x 2 / 2 dx 1 2 π 1 4 ( e 0 2 + e - ( 1 4 ) 2 / 2 + e - ( 1 2 ) 2 / 2 + e - ( 3 4 ) 2 / 2 + e - 1 / 2 2 ) 0 . 34008 The Trapezoidal Rule is much simpler to use than the midpoint rule, and it’s almost as accurate. Z b a f ( x ) dx - T n K ( b - a ) 3 12 n 2 (3.3)
3.2. TAYLOR SERIES APPROXIMATIONS 57 where K = max | f 00 ( x ) | for x [ a, b ]. This is only scratching the surface of a large subject. On tests, there will either be easy numbers to work with or you will be told to leave the approximation in the form of a sum. 3.2 Taylor Series Approximations What does it mean when we say 1 / 3 = ¯ . 3? Let’s think for a moment about what decimal notation means. 10 - 1 = 1 / 10 = . 1 10 - 2 = 1 / 100 = . 01 10 - 3 = 1 / 1000 = . 001 etc. So, 1 / 3 = ¯ . 3 means 1 3 = 3 10 + 3 100 + 3 1000 + · · · = X k =1 3 1 10 k which we can verify with our infinite geometric series formula: 3 / 10 1 - 1 / 10 = 1 / 3. Frequently, we use approximations like 1 / 3 . 33 in which we just round. From the series viewpoint, we’re taking an infinite series and simply lopping off all of the terms after the first two: 1 3 = 3 10 + 3 100 + 3 1000 + · · · = 3 10 + 3 100 . Wouldn’t it be nice if we could approximate complicated functions with relatively simple polynomials? The idea is that instead of using sums of pow- ers of 10 to represent numbers, we will use sums of powers of x to represent functions. For example, sin x x 1 - x 2 6 e x 3 - 1 x 2 x + x 5 2 1 - 4 x 1 - 2 x - 2 x 2 .
58 CHAPTER 3. APPROXIMATIONS WITH CALCULUS Those polynomials were not chosen at random! Try graphing the functions and polynomials on the same grid with x small. In order to do this accurately, we need a way to represent functions as infinite series and then take the first terms as our approximation. Recall linear approximations from 17A. Let f ( x ) be differentiable at x = a . Then we can make an approximation of f ( x ) for x close to a by looking at the tangent line: f ( x ) L ( x ) = f 0 ( a )( x - a ) + f ( a ) . All we did to create this rough approximation was to pick the simplest pos- sible L ( x ) that had the same value and derivative at a as f : L ( a ) = f ( a ) L 0 ( a ) = f 0 ( a ) . We have an easy initial value problem! Let’s recreate this linear approxima- tion L and then see how to improve upon it. L 0 ( a ) = f 0 ( a ) L ( x ) = Z f 0 ( a ) dx = f 0 ( a ) x + C = f 0 ( a ) x + C L ( a ) = f 0 ( a ) a + C = f ( a ) C = f ( a ) - f 0 ( a ) a L ( x ) = f 0 ( a )( x - a ) + f ( a ) . We could have guessed this result by realizing that holding the derivative fixed is the same as our approximation being a line, and finding the equation of a line once you know its slope and a point that it passes through should be familiar. However, this technique is generalizable, which makes it infinitely superior by mathematical standards.