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

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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

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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

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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

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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.

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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

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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

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EE3561_Unit 2(c)AL-DHAIFALLAH14358

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EE3561_Unit 2(c)AL-DHAIFALLAH14359

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

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EE3561_Unit 2(c)AL-DHAIFALLAH Example

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

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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

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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

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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

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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.

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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.

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EE3561_Unit 2(c)AL-DHAIFALLAH Multiple Roots

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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

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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

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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

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

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

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

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

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EE3561_Unit 2(c)AL-DHAIFALLAH Example Solution

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

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

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EE3561_Unit 2(c)AL-DHAIFALLAH Secant Method

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EE3561_Unit 2(c)AL-DHAIFALLAH Secant Method

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EE3561_Unit 2(c)AL-DHAIFALLAH Secant Method

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

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EE3561_Unit 2(c)AL-DHAIFALLAH Modified Secant Method

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EE3561_Unit 2(c)AL-DHAIFALLAH Example

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

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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

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

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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

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EE3561_Unit 2(c)AL-DHAIFALLAH Example

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

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EE3561_Unit 2(c)AL-DHAIFALLAH Example

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

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EE3561_Unit 2(c)AL-DHAIFALLAH Example

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EE3561_Unit 2(c)AL-DHAIFALLAH Example

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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

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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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