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Newton’s Iteration Method for Finding the Zero of a Function Introduction: To estimate a zero for a function f x we begin by choosing a point x0 ”close” to the actual zero, calculating the equation of the tangent line thru the point x0 , f x0 then finding where this tangent line intersects the x-axis. If we call this point x1 , it should be a better approximation to the zero we’re looking for. Repeat. Example: Find a zero for the function f x x3 x 2 3 slope = f x0 x , f x 0 0 x0 x1 To find the zero of f x do the following. 1. Let x0 be an estimate for the zero. 2. Then x0 , f x0 is a point on the curve. From the point on the curve x0 , f x0 , if we sight down the line tangent, where the line crosses the x-axis at x1 is closer to the real zero than x0 . 3. The equation of the tangent line (slope = mtan f x0 ) thru the point x0 , f x0 is y f x0 ( x x0 ) f x0 . It crosses the x-axis when y 0 . Solving for x x1 we obtain x1 x0 4. Repeat. x2 x1 f x1 f x1 , x3 x2 f x2 f x2 , … Convergence is often very quick! f x0 f x0 Back to our example: From f x x3 x 2 3 and f x 3x 2 2 x we have the formula xn 1 xn xn3 xn2 3 3xn2 2 xn Set this up on a TI-84 y1 x 3 x 2 3 y2 3 x 2 2 x 2x x y1 x / y2 x x We generate the following table of values n 0 1 2 3 4 xn 2 1.875 1.863793103 1.863706533 1.863706528 Therefore 1.863706528 is the zero of the equation f x x3 x 2 3 . This method always works provided that the initial guess is sufficiently close to the solution and f xn is never zero (why?). Convergence should be fast. Example: Solve x 3 3 x 2 1 0 Let x0 3 be an initial guess and iterative compute xn1 xn n xn 0 1 2 3 4 3 2.888888888 2.879451567 2.879385242 2.879385242 xn3 3xn2 1 3xn2 6 xn Exercises: Using Newton’s Formula locate the smallest positive root for each of the following to within the accuracy of the TI 83/84 calculator. f x0 for that function f x0 a. Obtain the derivative and write down Newton’s formula x1 x0 b. Graph the function to find an integer value close to the root to use as your initial value c. Fill in the table below for each function f x x2 5 n 0 1 2 3 4 5 xn f x x2 5 n 0 1 2 3 4 5 xn f x x3 n 0 1 2 3 4 5 xn Newton’s Formulas f x x3 3x 17 n 0 1 2 3 4 5 xn Newton’s Formulas f x x 5 17 f x x2 x 1 n 0 1 2 3 4 5 xn n 0 1 2 3 4 5 xn