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Section 3.8 – Newton’s Method

Section 3.8 – Newton’s Method. Solving Complicated Equations.

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Section 3.8 – Newton’s Method

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  1. Section 3.8 – Newton’s Method

  2. 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?

  3. 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?

  4. A New Method to Solve Equations Use a graph to help solve: The solution is close to 0.011 The tangent line equation at x1 = 0.011 is The x-intercept of the tangent line is closer to the solution:

  5. A New Method to Solve Equations Use a graph to help solve: The solution is closer to 0.00866 The tangent line equation at x2 = 0.00866 is The x-intercept of the tangent line is closer to the solution:

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

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

  8. Newton’s Method The Equation MUST equal 0. Procedure to solve the equation f (x) = 0: • Establish a small positive number Ɛ that determines the allowable tolerance for estimated solutions (typically given). • Compute f '(x). • With a graph or table, choose a number x0 (with f '(x0)≠0)“close” to a solution of f (x)=0 as an initial estimate. • Compute a new approximation with the formula: • Repeat step 4 until ǀ xn+1 – xn ǀ < Ɛ. The estimate xn+1 then has the required accuracy. Each successive approximation is called an iteration.

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

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

  11. Example 1 Use Newton’s Method to solve . Continue the iterations until two successive approximations differ by less than 0.000001. f(x) = 0 : Establish the tolerance: Find the derivative : Always Store (->) these Results in the calculator Use a graph to estimate the answer: Compute the iterations: A – B÷C A B C 1 1 -0.459697694 -1.841470985 0.7503638678 2 0.7503638678 -0.018923073 -1.681904953 0.7391128909 3 0.7391128909 -0.000046456 -1.673632544 0.7390851334 4 0.7390851334 -2.8471x10-10 -1.673612029 0.7390851332 5 0.7390851332 The difference is less than 0.000001 The solution of the equation is ~0.739085.

  12. 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 : Always Store (->) these Results in the calculator Use a graph to estimate the answer: Compute the iterations: A – B÷C A B C 1 2 -1 10 2.1 2 2.1 0.061 11.23 2.094568121 3 2.094568121 0.0001857232 11.16164684 2.094551482 4 2.094551482 The difference is less than 0.0001 The root of the equation is ~2.0946.

  13. Example 3 Use two iterations of Newton’s Method to approximate the x-coordinates of the intersection of y = x5and y = -x2 – 2. Equation to solve: f(x) = 0 : Establish the tolerance: Find the derivative : Always Store (->) these Results in the calculator Use a graph to estimate the answer: A – B÷C Compute the iterations: A B C 1 -1 2 3 -5/3 2 -5/3 -8.082304527 35.246913580 -1.4373613543491 End of the second iteration. The x-coordinate of the intersection is ~ -1.437.

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

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

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