A Slide Show Demonstrating Newton s Method

Transcription

1 The AcroT E X Web Site, 1999 A Slide Show Demonstrating Newton s Method D. P. Story The Department of Mathematics and Computer Science The Universityof Akron, Akron, OH

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3 Now go up to the curve.

4 Now go up to the curve. Draw the tangent line.

5 Now go up to the curve. Draw the tangent line. Its equation is y = f( )+f ( )(x ).

6 x 1 Now go up to the curve. Draw the tangent line. Its equation is y = f( )+f ( )(x ). Let x 1 be in x-intercept of this tangent line.

7 x 1 Now go up to the curve. Draw the tangent line. Its equation is y = f( )+f ( )(x ). Let x 1 be in x-intercept of this tangent line. This intercept is given by the formula: x 1 = f() f ( ).

8 x 1 x 2 Now go up to the curve. Draw the tangent line. Its equation is y = f( )+f ( )(x ). Let x 1 be in x-intercept of this tangent line. This intercept is given by the formula: x 1 = f() f ( ). Now repeat using x 1 as the initial guess.

9 x 1 x 2 Now go up to the curve. Draw the tangent line. Its equation is y = f( )+f ( )(x ). Let x 1 be in x-intercept of this tangent line. This intercept is given by the formula: x 1 = f() f ( ). Now repeat using x 1 as the initial guess. The intercept x 2 is given by: x 2 = x 1 f(x 1) f (x 1 ).

10 2. Commentary The initial guess,, was close to the true root. From the picture, Frame 5, it appears our next estimate x 1, x 1 = f() f ( ) is a little closer to the unknown root than was. The next iterate, x 2, calculated from the formula x 2 = x 1 f(x 1) f (x 1 ) is closer still to the unknown root, see Frame 7. The process continues: Given that an estimate x n has already been calculated, the equation of the tangent line is calculated: y = f(x n )+f (x n )(x x n )

11 Section 2: Commentary The x-intercept is then calculated, f(x n )+f (x n )(x x n )=0 = x = x n f(x n) f (x n ) This intercept is labeled x n+1 and represents our next estimate of the unknown root. x n+1 = x n f(x n) f (1) (x n ) The initial guess, and the Newton Iteration formula, equation (1), together form an algorithm or a procedure of estimating the value of the root to the equation f(x) =0.

12 3. Examples Example 3.1. Find the positive root of the equation x 2 =2. Solution: The function is f(x) =x 2 2. Step 1: Compute derivative, f (x) =2x. Step 2: Construct the iteration formula: x n+1 = x n f(x n) f (x n ) = x n x2 n 2 2x n = x2 n +2 2x n Thus, for this problem, the iteration formula is x n+1 = x2 n +2 2x n This, together with an initial guess of =1.5 yields the following calculations.

13 Section 3: Examples Step 3: Construct a table of estimates. Initial guess of =1.5 and iteration formula of x n+1 = x2 n +2 2x n. Newton s Method f(x) =x 2 2, =1.5 n x n f(x n ) Thus, the positive root to the equation x 2 2=0is x , or, stated differently, Example 3.1.

14 Section 3: Examples Example 3.2. near =1.5. Find a solution to the equation x 3 = x + 1 that is Solution: The function is f(x) =x 3 x 1. The function f is always defined to make the given equation equivalent to an equation of the form f(x) = 0. (We just take everything on the right-hand side of the given equation to the left-hand side. The left-hand side is now an expression defining f(x). Step 1: Differentiate f (x) =3x 2 1. Step 2: Construct the Newton iteration formula: x n+1 = x n f(x n) f (x n ) = x n x3 n x n 1 3x 2 n 1 The iteration formula is x n+1 = x n x3 n x n 1 3x 2 n 1

15 Section 3: Examples Step 3: Construct the table of estimates. Newton s Method f(x) =x 3 x 1, =1.5 n x n f(x n ) Thus, the solution to the equation x 3 = x + 1 that is near =1.5 is x Example 3.2. AcroTEX: dpstory/acrotex.html D. P. Story: dpstory@uakron.edu