Bracketing Methods Chapter 5 The Islamic University of Gaza Faculty of Engineering Civil Engineering Department Numerical Analysis ECIV 3306 Chapter 5 Bracketing Methods

PART II ROOTS OF EQUATIONS Bracketing Methods Bisection method False Position Method Open Methods Simple fixed point iteration Newton Raphson Secant Modified Newton Raphson System of Nonlinear Equations Roots of polynomials Muller Method

Study Objectives for Part Two

ROOTS OF EQUATIONS Root of an equation: is the value of the equation variable which make the equations = 0.0 But

ROOTS OF EQUATIONS Non-computer methods: - Closed form solution (not always available) - Graphical solution (inaccurate) Numerical systematic methods suitable for computers

The roots exist where f(x) crosses the x-axis. Graphical Solution Plot the function f(x) f(x) roots The roots exist where f(x) crosses the x-axis. x f(x)=0 f(x)=0

Graphical Solution: Example The parachutist velocity is What is the drag coefficient c needed to reach a velocity of 40 m/s if m=68.1 kg, t =10 s, g= 9.8 m/s2 c f(c) c=14.75 Check: F (14.75) = 0.059 ~ 0.0 v (c=14.75) = 40.06 ~ 40 m/s

Numerical Systematic Methods I. Bracketing Methods f(x) f(x) No roots or even number of roots Odd number of roots f(xl)=+ve f(xl)=+ve roots roots f(xu)=+ve x x xl xu f(xu)=-ve xl xu

Bracketing Methods (cont.) Two initial guesses (xl and xu) are required for the root which bracket the root (s). If one root of a real and continuous function, f(x)=0, is bounded by values xl , xu then f(xl).f(xu) <0. (The function changes sign on opposite sides of the root)

Bracketing Methods 1. Bisection Method Generally, if f(x) is real and continuous in the interval xl to xu and f (xl).f(xu)<0, then there is at least one real root between xl and xu to this function. The interval at which the function changes sign is located. Then the interval is divided in half with the root lies in the midpoint of the subinterval. This process is repeated to obtained refined estimates.

Step 1: Choose lower xl and upper xu guesses for the root such that: f(x) Step 1: Choose lower xl and upper xu guesses for the root such that: f(xl).f(xu)<0 Step 2: The root estimate is: xr = ( xl + xu )/2 Step 3: Subdivide the interval according to: If (f(xl).f(xr)<0) the root lies in the lower subinterval; xu = xr and go to step 2. If (f(xl).f(xr)>0) the root lies in the upper subinterval; xl = xr and go to step 2. If (f(xl).f(xr)=0) the root is xr and stop xr = ( xl + xu )/2 f(xu) xl xr1 xu x f(xu) f(xr1) f(x) (f(xl).f(xr)<0): xu = xr xr = ( xl + xu )/2 f(xu) f(xr2) xl xu x xr2 f(xu)

Bisection Method - Termination Criteria For the Bisection Method ea > et The computation is terminated when ea becomes less than a certain criterion (ea < es)

Bisection method: Example The parachutist velocity is What is the drag coefficient c needed to reach a velocity of 40 m/s if m = 68.1 kg, t = 10 s, g= 9.8 m/s2 f(c) c

and so on…... f(x) Assume xl =12 and xu=16 f(xl)=6.067 and f(xu)=-2.269 The root: xr=(xl+xu)/2= 14 Check f(12).f(14) = 6.067•1.569=9.517 >0; the root lies between 14 and 16. Set xl = 14 and xu=16, thus the new root xr=(14+ 16)/2= 15 Check f(14).f(15) = 1.569•-0.425= -0.666 <0; the root lies bet. 14 and 15. Set xl = 14 and xu=15, thus the new root xr=(14+ 15)/2= 14.5 and so on…... 6.067 1.569 x 12 14 16 -2.269 f(x) (f(12).f(14)>0): xl = 14 1.569 15 x 14 16 -0.425 -2.269

Bisection method: Example In the previous example, if the stopping criterion is et = 0.5%; what is the root? Iter. Xl Xu Xr ea% et% 1 12 16 14 5.279 -- 2 14 16 15 6.667 1.487 3 14 15 14.5 3.448 1.896 4 14.5 15 14.75 1.695 1.204 5 14.75 15 14.875 0.84 0.641 6 14.74 14.875 14.813 0.422 0.291

Bisection method

Flow Chart –Bisection Start False Stop Input: xl , xu , s, maxi f(xl). f(xu)<0 i=0 a=1.1s False while a> s & i <maxi Stop Print: xr , f(xr ) ,a , i

True Test=0 Test<0 False xu+xl =0 Test=f(xl). f(xr) a=0.0 xu=xr xl=xr False

Bracketing Methods 2. False-position Method The bisection method divides the interval xl to xu in half not accounting for the magnitudes of f(xl) and f(xu). For example if f(xl) is closer to zero than f(xu), then it is more likely that the root will be closer to f(xl). False position method is an alternative approach where f(xl) and f(xu) are joined by a straight line; the intersection of which with the x-axis represents and improved estimate of the root.

2. False-position Method False position method is an alternative approach where f(xl) and f(xu) are joined by a straight line; the intersection of which with the x-axis represents and improved estimate of the root.

False-position Method -Procedure f(x) f(xu) xl xr xu x f(xl) f(xr)

False-position Method -Procedure Step 1: Choose lower xl and upper xu guesses for the root such that: f(xl).f(xu)<0 Step 2: The root estimate is: Step 3: Subdivide the interval according to: If (f(xl).f(xr)<0) the root lies in the lower subinterval; xu = xr and go to step 2. If (f(xl).f(xr)>0) the root lies in the upper subinterval; xl = xr and go to step 2. If (f(xl).f(xr)=0) the root is xr and stop

False position method: Example The parachutist velocity is What is the drag coefficient c needed to reach a velocity of 40 m/s if m =68.1 kg, t =10 s, g= 9.8 m/s2 f(c) c

False position method: Example Assume xl = 12 and xu=16 f(xl)= 6.067 and f(xu)= -2.269 The root: xr=14.9113 f(12) . f(14.9113) = -1.5426 < 0; The root lies bet. 12 and 14.9113. Assume xl = 12 and xu=14.9113, f(xl)=6.067 and f(xu)=-0.2543 The new root xr= 14.7942 This has an approximate error of 0.79% f(x) 6.067 14.91 x 12 16 -2.269

False position method: Example

Flow Chart –False Position Start Input: xl , x0 , s, maxi f(xl). f(xu)<0 i=0 a=1.1s False while a> s & i <maxi Stop Print: xr , f(xr ) ,a , i

True Test=0 Test<0 False i=1 or xr=0 Test=f(xl). f(xr) a=0.0 xu=xr xr0=xr Test<0 xl=xr False

False Position Method-Example 2

False Position Method - Example 2

Roots of Polynomials: Using Software Packages MS Excel:Goal seek f(x)=x-cos x

MS Excel: Solver u(x,y)= x2+xy-10 =0 v(x,y)=y+3xy2-57=0