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EE3561_Unit 2(c)AL-DHAIFALLAH14351 Lecture 5 Newton-Raphson Method Assumptions Interpretation Examples Convergence Analysis Reading Assignment: Sections.

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Presentation on theme: "EE3561_Unit 2(c)AL-DHAIFALLAH14351 Lecture 5 Newton-Raphson Method Assumptions Interpretation Examples Convergence Analysis Reading Assignment: Sections."— Presentation transcript:

1 EE3561_Unit 2(c)AL-DHAIFALLAH14351 Lecture 5 Newton-Raphson Method Assumptions Interpretation Examples Convergence Analysis Reading Assignment: Sections 6.2

2 EE3561_Unit 2(c)AL-DHAIFALLAH14352 Root finding Problems Many problems in Science and Engineering are expressed as These problems are called root finding problems

3 EE3561_Unit 2(c)AL-DHAIFALLAH14353 Solution Methods Several ways to solve nonlinear equations are possible.  Analytical Solutions possible for special equations only  Graphical Solutions Useful for providing initial guesses for other methods  Numerical Solutions Open methods Bracketing methods

4 EE3561_Unit 2(c)AL-DHAIFALLAH14354 Bracketing Methods  In bracketing methods, the method starts with an interval that contains the root and a procedure is used to obtain a smaller interval containing the root.  Examples of bracketing methods : Bisection method False position method

5 EE3561_Unit 2(c)AL-DHAIFALLAH14355 Open Methods  In the open methods, the method starts with one or more initial guess points. In each iteration a new guess of the root is obtained.  Open methods are usually more efficient than bracketing methods  They may not converge to the root.

6 EE3561_Unit 2(c)AL-DHAIFALLAH14356 Solution Methods Many methods are available to solve nonlinear equations  Bisection Method  Newton’s Method  Secant Method False position Method Muller’s Method Bairstow’s Method Fixed point iterations ………. These will be covered in EE3561

7 EE3561_Unit 2(c)AL-DHAIFALLAH14357 Newton-Raphson Method (also known as Newton’s Method) Given an initial guess of the root x 0, Newton-Raphson method uses information about the function and its derivative at that point to find a better guess of the root. Assumptions: f (x) is continuous and first derivative is known An initial guess x 0 such that f ’(x 0 ) ≠0 is given

8 EE3561_Unit 2(c)AL-DHAIFALLAH14358

9 EE3561_Unit 2(c)AL-DHAIFALLAH14359

10 EE3561_Unit 2(c)AL-DHAIFALLAH Newton’s Method F.m FP.m

11 EE3561_Unit 2(c)AL-DHAIFALLAH Example

12 EE3561_Unit 2(c)AL-DHAIFALLAH Example Iterationxkxk f(x k )f’(x k )x k+1 |x k+1 –x k |

13 Convergence analysis of the Newton’s Method  The Taylor series expansion can be represented as  An approximate solution can be obtained by  At the intersection with x axis f( x i+1 )=0, which gives EE3561_Unit 2(c)AL-DHAIFALLAH143513

14 Convergence analysis of the Newton’s Method  Substituting the true value x r into Eq (B6.2.1) yields  Subtracting Eq (B6.2.2) from Eq (B6.2.3) yields  Substituting and gives EE3561_Unit 2(c)AL-DHAIFALLAH143514

15 Convergence analysis of the Newton’s Method  If we assume convergence, both x i and should eventually approximated by the root x r, and Eq(B6.2.5) can be rearranged to yield  Which means that the current error is proportional to the square of the previous error which indicates quadratic convergence rate. EE3561_Unit 2(c)AL-DHAIFALLAH143515

16 EE3561_Unit 2(c)AL-DHAIFALLAH Convergence Analysis Remarks When the guess is close enough to a simple root of the function then Newton’s method is guaranteed to converge quadratically. Quadratic convergence means that the number of correct digits is nearly doubled at each iteration.

17 EE3561_Unit 2(c)AL-DHAIFALLAH Problems with Newton’s Method If the initial guess of the root is far from the root the method may not converge. Newton’s method converges linearly near multiple zeros { f(r) = f ’(r) =0 }. In such a case modified algorithms can be used to regain the quadratic convergence.

18 EE3561_Unit 2(c)AL-DHAIFALLAH Multiple Roots

19 EE3561_Unit 2(c)AL-DHAIFALLAH Problems with Newton’s Method Runaway The estimates of the root is going away from the root. x0x0 x1x1

20 EE3561_Unit 2(c)AL-DHAIFALLAH Problems with Newton’s Method Flat Spot The value of f’(x) is zero, the algorithm fails. If f ’(x) is very small then x 1 will be very far from x 0. x0x0

21 EE3561_Unit 2(c)AL-DHAIFALLAH Problems with Newton’s Method Cycle The algorithm cycles between two values x 0 and x 1 x 0 =x 2 =x 4 x 1 =x 3 =x 5

22 EE3561_Unit 2(c)AL-DHAIFALLAH Newton’s Method for Systems of nonlinear Equations

23 EE3561_Unit 2(c)AL-DHAIFALLAH Example  Solve the following system of equations

24 EE3561_Unit 2(c)AL-DHAIFALLAH Solution Using Newton’s Method

25 EE3561_Unit 2(c)AL-DHAIFALLAH Example Try this  Solve the following system of equations

26 EE3561_Unit 2(c)AL-DHAIFALLAH Example Solution

27 EE3561_Unit 2(c)AL-DHAIFALLAH Secant Method  Secant Method  Examples  Convergence Analysis

28 EE3561_Unit 2(c)AL-DHAIFALLAH Newton’s Method (Review)

29 EE3561_Unit 2(c)AL-DHAIFALLAH Secant Method

30 EE3561_Unit 2(c)AL-DHAIFALLAH Secant Method

31 EE3561_Unit 2(c)AL-DHAIFALLAH Secant Method

32 EE3561_Unit 2(c)AL-DHAIFALLAH Secant Method Stop NO Yes

33 EE3561_Unit 2(c)AL-DHAIFALLAH Modified Secant Method

34 EE3561_Unit 2(c)AL-DHAIFALLAH Example

35 EE3561_Unit 2(c)AL-DHAIFALLAH Example x(i)f( x(i) )x(i+1)|x(i+1)-x(i)|

36 EE3561_Unit 2(c)AL-DHAIFALLAH Convergence Analysis  The rate of convergence of the Secant method is super linear  It is better than Bisection method but not as good as Newton’s method

37 EE3561_Unit 2(c)AL-DHAIFALLAH Comparison of Root finding methods  Advantages/disadvantages  Examples

38 EE3561_Unit 2(c)AL-DHAIFALLAH Summary Bisection Reliable, Slow One function evaluation per iteration Needs an interval [a,b] containing the root, f(a) f(b)<0 No knowledge of derivative is needed Newton Fast (if near the root) but may diverge Two function evaluation per iteration Needs derivative and an initial guess x 0, f ’ (x 0 ) is nonzero Secant Fast (slower than Newton) but may diverge one function evaluation per iteration Needs two initial points guess x 0, x 1 such that f (x 0 )- f (x 1 ) is nonzero. No knowledge of derivative is needed

39 EE3561_Unit 2(c)AL-DHAIFALLAH Example

40 EE3561_Unit 2(c)AL-DHAIFALLAH Solution _______________________________ k x k f(x k ) ________________________________

41 EE3561_Unit 2(c)AL-DHAIFALLAH Example

42 EE3561_Unit 2(c)AL-DHAIFALLAH Five iterations of the solution  k x k f(x k ) f’(x k ) ERROR  ______________________________________      

43 EE3561_Unit 2(c)AL-DHAIFALLAH Example

44 EE3561_Unit 2(c)AL-DHAIFALLAH Example

45 EE3561_Unit 2(c)AL-DHAIFALLAH Example Estimates of the root of x-cos(x)= initial guess correct digit correct digits correct digits correct digits

46 EE3561_Unit 2(c)AL-DHAIFALLAH Example In estimating the root of x-cos(x)=0 To get more than 13 correct digits  4 iterations of Newton (x 0 =0.6)  43 iterations of Bisection method (initial interval [0.6,.8]  5 iterations of Secant method ( x 0 =0.6, x 1 =0.8)


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