# Section 3.8 – Newton’s Method. Solving Complicated Equations There is a well-known formula to solve all quadratic equations and there exist more complicated.

## Presentation on theme: "Section 3.8 – Newton’s Method. Solving Complicated Equations There is a well-known formula to solve all quadratic equations and there exist more complicated."— Presentation transcript:

Section 3.8 – Newton’s Method

Solving Complicated Equations There is a well-known formula to solve all quadratic equations and there exist more complicated equations to solve third- and fourth-degree polynomials. Unfortunately, a formula does not exist for any polynomial with a degree of five or higher. There are also plenty of other complicated equations that can not be solved with a formula. For instance: If we can not solve these equations algebraically, what other methods exist to approximate the answer?

A New Method to Solve Equations Suppose that a car dealer offers to sell you a car for \$18,000 or for payments of \$375 per month for five years. You would like to know what monthly interest rate the dealer is, in effect, charging you. To find the answer, you have to solve the equation: How would you solve such an equation?

A New Method to Solve Equations Use a graph to help solve: The solution is close to 0.011 The tangent line equation at x 1 = 0.011 is The x-intercept of the tangent line is closer to the solution:

The solution is closer to 0.00866 A New Method to Solve Equations Use a graph to help solve: The tangent line equation at x 2 = 0.00866 is The x-intercept of the tangent line is closer to the solution:

The solution is even closer to 0.00777. A New Method to Solve Equations Use a graph to help solve: Let’s Generalize the Algorithm.

The tangent line equation at x = x n is The solution is close to x n A New Method to Solve Equations Use a graph to help solve: The x-intercept of the tangent line is closer to the solution: A Closer Solution:

Newton’s Method Procedure to solve the equation f (x) = 0 : 1.Establish a small positive number Ɛ that determines the allowable tolerance for estimated solutions (typically given). 2.Compute f '(x). 3.With a graph or table, choose a number x 0 (with f '(x 0 )≠0 ) “close” to a solution of f (x)=0 as an initial estimate. 4.Compute a new approximation with the formula: 5.Repeat step 4 until ǀ x n+1 – x n ǀ < Ɛ. The estimate x n+1 then has the required accuracy. Each successive approximation is called an iteration. The Equation MUST equal 0.

Example 1 Establish the tolerance: Find the derivative : Use a graph to estimate the answer: Use Newton’s Method to solve. Continue the iterations until two successive approximations differ by less than 0.000001. f(x) = 0 :

Example 1 Use a graph to estimate the answer: Now graph the non-zero side: Since the equation equals 0, the x-intercept(s) are the solutions The solution is approximately 1. Now continue with Newton’s Method. Use Newton’s Method to solve. Continue the iterations until two successive approximations differ by less than 0.000001.

Example 1 Establish the tolerance: Find the derivative : Use a graph to estimate the answer: Compute the iterations: nxnxn f(xn)f(xn)f ' (x n ) x n – f(x n ) ÷f ' (x n ) 11-0.459697694-1.8414709850.7503638678 2 -0.018923073-1.6819049530.7391128909 3 -0.000046456-1.6736325440.7390851334 4 -2.8471x10 -10 -1.6736120290.7390851332 5 The solution of the equation is ~ 0.739085. The difference is less than 0.000001 A BC Always Store (->) these Results in the calculator Use Newton’s Method to solve. Continue the iterations until two successive approximations differ by less than 0.000001. f(x) = 0 : A – B÷C

Example 2 Use Newton’s Method to approximate the zeros of. Continue the iterations until two successive approximations differ by less than 0.0001. Establish the tolerance: Find the derivative : Use a graph to estimate the answer: Compute the iterations: nxnxn f(xn)f(xn)f ' (x n ) x n – f(x n ) ÷f ' (x n ) 12102.1 2 0.06111.232.094568121 3 0.000185723211.161646842.094551482 4 The root of the equation is ~ 2.0946. The difference is less than 0.0001 A BC Always Store (->) these Results in the calculator A – B÷C

Example 3 Use two iterations of Newton’s Method to approximate the x -coordinates of the intersection of y = x 5 and y = -x 2 – 2. Establish the tolerance: Find the derivative : Use a graph to estimate the answer: Compute the iterations: nxnxn f(xn)f(xn)f ' (x n ) x n – f(x n ) ÷f ' (x n ) 123-5/3 2 -8.08230452735.246913580-1.4373613543491 The x-coordinate of the intersection is ~ -1.437. End of the second iteration. A BC Always Store (->) these Results in the calculator Equation to solve: f(x) = 0 : A – B÷C

White Board Challenge Use Newton’s method to find the solution(s) to the following equation accurate to 3 decimal places:

White Board Challenge Use Newton’s method to approximate the following expression accurate to 3 decimal places:

Download ppt "Section 3.8 – Newton’s Method. Solving Complicated Equations There is a well-known formula to solve all quadratic equations and there exist more complicated."

Similar presentations