Pre-Calc Exam Notes 134

Pre-Calc Exam Notes 134 - ( x 1 x ) f ( x 1 ) f ( x 1 ) f (...

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134 Chapter 6 Additional Topics §6.2 1. Pick initial points x 0 and x 1 such that x 0 < x 1 and f ( x 0 ) f ( x 1 ) < 0 (i.e. the solution is somewhere between x 0 and x 1 ). 2. For n 2, defne the number x n by x n = x n 1 ( x n 1 x n 2 ) f ( x n 1 ) f ( x n 1 ) f ( x n 2 ) (6.2) as long as | x n 1 x n 2 |> ǫ error , where ǫ error > 0 is the maximum amount o± error desired (usually a very small number). 3. The numbers x 0 , x 1 , x 2 , x 3 , ... will approach the solution x as we go through more itera- tions, getting as close as desired. We will now show how to use this algorithm to solve the equation cos x = x . The solution to that equation is the root o± the ±unction f ( x ) = cos x x . And we saw that the solution is somewhere in the interval [0,1]. So pick x 0 = 0 and x 1 = 1. Then f (0) = 1 and f (1) =− 0.4597, so that f ( x 0 ) f ( x 1 ) < 0 (we are using radians, o± course). Then by defnition, x 2 = x 1
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Unformatted text preview: ( x 1 x ) f ( x 1 ) f ( x 1 ) f ( x ) = 1 (1 0) f (1) f (1) f (0) = 1 (1 0)( 0.4597) 0.4597 1 = 0.6851 , x 3 = x 2 ( x 2 x 1 ) f ( x 2 ) f ( x 2 ) f ( x 1 ) = 0.6851 (0.6851 1) f (0.6851) f (0.6851) f (1) = 0.6851 (0.6851 1)(0.0893) 0.0893 ( 0.4597) = 0.7363 , and so on. Using a calculator is not very efcient and will lead to rounding errors. A bet-ter way to implement the algorithm is with a computer. Listing 6.1 below shows the code (secant.java) or solving cos x = x with the secant method, using the Java programming lan-guage:...
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This note was uploaded on 01/21/2012 for the course MAC 1130 taught by Professor Dr.cheun during the Fall '11 term at FSU.

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