Open Methods Chapter 6 The Islamic University of Gaza Faculty of Engineering Civil Engineering Department Numerical Analysis ECIV 3306 Chapter 6 Open Methods

Open Methods Bracketing methods are based on assuming an interval of the function which brackets the root. The bracketing methods always converge to the root. Open methods are based on formulas that require only a single starting value of x or two starting values that do not necessarily bracket the root. These method sometimes diverge from the true root.

Open Methods- Convergence and Divergence Concepts f(x) f(x) x x xi xi+1 xi xi+1 Diverging increments Converging increments

1. Simple Fixed-Point Iteration Rearrange the function so that x is on the left side of the equation: Bracketing methods are “convergent”. Fixed-point methods may sometime “diverge”, depending on the stating point (initial guess) and how the function behaves.

Simple Fixed-Point Iteration Examples: 1. f(x) = x 2-2x+3  x = g(x)=(x2+3)/2 f(x) = sin x  x = g(x)= sin x + x f(x) = e-x- x  x = g(x)= e-x

Simple Fixed-Point Iteration Convergence x = g(x) can be expressed as a pair of equations: y1= x y2= g(x)…. (component equations) Plot them separately.

Simple Fixed-Point Iteration Convergence

Simple Fixed-Point Iteration Convergence Derivative mean value theorem: If g(x) are continuous in [a,b] then there exist at least one value of x= within the interval such that: i.e. there exist one point where the slope parallel to the line joining (a & b)

Simple Fixed-Point Iteration Convergence

Simple Fixed-Point Iteration Convergence Fixed-point iteration converges if : When the method converges, the error is roughly proportional to or less than the error of the previous step, therefore it is called “linearly convergent.”

Simple Fixed-Point Iteration-Convergence

Example: Simple Fixed-Point Iteration f(x) = e-x - x f(x) f(x)=e-x - x 1. f(x) is manipulated so that we get x=g(x) g(x) = e-x 2. Thus, the formula predicting the new value of x is: xi+1 = e-xi 3. Guess xo = 0 4. The iterations continues till the approx. error reaches a certain limiting value Root x f(x) f1(x) = x g(x) = e-x x

Example: Simple Fixed-Point Iteration i xi g(xi) ea% et% 0 0 1.0 1 1.0 0.367879 100 76.3 2 0.367879 0.692201 171.8 35.1 3 0.692201 0.500473 46.9 22.1 4 0.500473 0.606244 38.3 11.8 5 0.606244 0.545396 17.4 6.89 6 0.545396 0.579612 11.2 3.83 7 0.579612 0.560115 5.90 2.2 8 0.560115 0.571143 3.48 1.24 9 0.571143 0.564879 1.93 0.705 10 0.564879 1.11 0.399

Example: Simple Fixed-Point Iteration i xi g(xi) ea% et% 0 0 1.0 1 1.0 0.367879 100 76.3 2 0.367879 0.692201 171.8 35.1 3 0.692201 0.500473 46.9 22.1 4 0.500473 0.606244 38.3 11.8 5 0.606244 0.545396 17.4 6.89 6 0.545396 0.579612 11.2 3.83 7 0.579612 0.560115 5.90 2.2 8 0.560115 0.571143 3.48 1.24 9 0.571143 0.564879 1.93 0.705 10 0.564879 1.11 0.399

Flow Chart – Fixed Point Start Input: xo , s, maxi i=0 a=1.1s 1

1 False Stop True while a< s & i >maxi or xn=0 x0=xn Print: xo, f(xo) ,a , i False True

2. The Newton-Raphson Method Most widely used method. Based on Taylor series expansion: Solve for Newton-Raphson formula

The Newton-Raphson Method A tangent to f(x) at the initial point xi is extended till it meets the x-axis at the improved estimate of the root xi+1. The iterations continues till the approx. error reaches a certain limiting value. f(x) Root x xi xi+1 Slope f /(xi) f(xi)

Example: The Newton Raphson Method Use the Newton-Raphson method to find the root of e-x-x= 0  f(x) = e-x-x and f`(x)= -e-x-1; thus Iter. xi et% 0 0 100 1 0.5 11.8 2 0.566311003 0.147 3 0.567143165 0.00002 4 0.567143290 <10-8

Flow Chart – Newton Raphson Start Input: xo , s, maxi i=0 a=1.1s 1

1 False Stop True while a >s & i <maxi or xn=0 x0=xn Print: xo, f(xo) ,a , i False True

Pitfalls of The Newton Raphson Method

3. The Secant Method The derivative is replaced by a backward finite divided difference Thus, the formula predicting the xi+1 is:

The Secant Method Requires two initial estimates of x , e.g, xo, x1. However, because f(x) is not required to change signs between estimates, it is not classified as a “bracketing” method. The scant method has the same properties as Newton’s method. Convergence is not guaranteed for all xo, x1, f(x).

Secant Method: Example Use the Secant method to find the root of e-x-x=0; f(x) = e-x-x and xi-1=0, x0=1 to get x1 of the first iteration using: Iter xi-1 f(xi-1) xi f(xi) xi+1 et% 1 0 1.0 1.0 -0.632 0.613 8.0 2 1.0 -0.632 0.613 -0.0708 0.5638 0.58 3 0.613 -0.0708 0.5638 0.00518 0.5672 0.0048

Comparison of convergence of False Position and Secant Methods Use two estimate xl and xu Use two estimate xi and xi-1 f(x) must changes signs between xl and xu f(x) is not required to change signs between xi and xi-1 Xr replaces whichever of the original values yielded a function value with the same sign as f(xr) Xi+1 replace xi Xi replace xi-1 Always converge May be diverge

Comparison of convergence of False Position and Secant Methods Use the false-position and secant method to find the root of f(x)=lnx. Start computation with xl= xi-1=0.5, xu=xi = 5. False position method Secant method Iter xi-1 xi xi+1 1 0.5 5.0 1.8546 2 5 1.8546 -0.10438 Iter xl xu xr 0.5 5.0 1.8546 0.5 1.8546 1.2163 0.5 1.2163 1.0585

False Position and Secant Methods Although the secant method may be divergent, when it converges it usually does so at a quicker rate than the false position method See the next figure xl xi-1 xu xi

Comparison of the true percent relative Errors Et for the methods to the determine the root of f(x)=e-x-x

Flow Chart – Secant Method Start Input: x-1 , x0,s, maxi i=0 a=1.1s 1

1 False Stop True while a >s & i < maxi or Xi+1=0 Xi-1=xi Xi=xi+1 Print: xi , f(xi) ,a , i False True

Modified Secant Method Rather than using two initial values, an alternative approach is using a fractional perturbation of the independent variable to estimate  is a small perturbation fraction

Modified Secant Method: Example Use the modified secant method to find the root of f(x) = e-x-x and, x0=1 and =0.01

Multiple Roots f(x)= (x-3)(x-1)(x-1)(x-1) = x4- 6x3+ 125 x2- 10x+3 Double roots f(x) 1 3 triple roots x

Multiple Roots “Multiple root” corresponds to a point where a function is tangent to the x axis. Difficulties Function does not change sign with double (or even number of multiple root), therefore, cannot use bracketing methods. Both f(x) and f′(x)=0, division by zero with Newton’s and Secant methods which may diverge around this root.

4. The Modified Newton Raphson Method Another u(x) is introduced such that u(x)=f(x)/f /(x); Getting the roots of u(x) using Newton Raphson technique: This function has roots at all the same locations as the original function

Modified Newton Raphson Method: Example Using the Newton Raphson and Modified Newton Raphson evaluate the multiple roots of f(x)= x3-5x2+7x-3 with an initial guess of x0=0 Newton Raphson formula: Modified Newton Raphson formula:

Modified Newton Raphson Method: Example Newton Raphson Modified Newton-Raphson Iter xi et% iter xi et% 0 0 100 0 0 100 1 0.4286 57 1 1.10526 11 2 0.6857 31 2 1.00308 0.31 3 0.83286 17 3 1.000002 00024 4 0.91332 8.7 5 0.95578 4.4 6 0.97766 2.2 Newton Raphson technique is linearly converging towards the true value of 1.0 while the Modified Newton Raphson is quadratically converging. For simple roots, modified Newton Raphson is less efficient and requires more computational effort than the standard Newton Raphson method

Systems of Nonlinear Equations Roots of a set of simultaneous equations: f1(x1,x2,…….,xn)=0 f2 (x1,x2,…….,xn)=0 fn (x1,x2,…….,xn)=0 The solution is a set of x values that simultaneously get the equations to zero.

Systems of Nonlinear Equations Example: x2 + xy = 10 & y + 3xy2 = 57 u(x,y) = x2+ xy -10 = 0 v(x,y) = y+ 3xy2 -57 = 0 The solution will be the value of x and y which makes u(x,y)=0 and v(x,y)=0 These are x=2 and y=3 Numerical methods used are extension of the open methods for solving single equation; Fixed point iteration and Newton-Raphson.

Systems of Nonlinear Equations: 1.Fixed Point Iteration Use an initial guess x =1.5 and y =3.5 The iteration formulae: xi+1=(10-xi2)/yi and yi+1=57-3xiyi2 First iteration, x=(10-(1.5)2)/3.5=2.21429 y=(57-3(2.21429)(3.5)2=-24.37516 Second iteration: x=(10-2.214292)/-24.37516=-0.209 y=57-3(-0.209)(-24.37516)2=429.709 Solution is diverging so try another iteration formula

Systems of Nonlinear Equations: 1.Fixed Point Iteration Using iteration formula: xi+1=(10-xiyi)1/2 and yi+1=[(57-yi)/3xi]1/2 First guess: x=1.5 and y=3.5 1st iteration: x=(10-(1.5)(3.5))1/2=2.17945 y=((57-(3.5))/3(2.17945))1/2=2.86051 2nd iteration: x=(10-(2.17945)(2.86051))1/2=1.94053 y=((57-(2.86051))/3(1.94053))1/2=3.04955 The approach is converging to the true root, x=2 and y=3

Systems of Nonlinear Equations: 1.Fixed Point Iteration The sufficient condition for convergence for the two-equation case (u(x,y)=0 and v(x,y)=0) are:

Systems of Nonlinear Equations: 2. Newton Raphson Method Recall the standard Newton Raphson formula: which can be written as the following formula

Systems of Nonlinear Equations: 2. Newton Raphson Method By multi-equation version (in this section we deal only with two equation) the formula can be derived in an identical fashion: u(x,y)=0 and v(x,y)=0

Systems of Nonlinear Equations: 2. Newton Raphson Method And thus

Systems of Nonlinear Equations: 2. Newton Raphson Method x 2+ xy =10 and y + 3xy 2 = 57 are two nonlinear simultaneous equations with two unknown x and y they can be expressed in the form: