by Martin Mendez, UASLP Copyright 2006 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
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1 Chapter 5 by Martin Mendez, 1
2 Roots of Equations Part Why? b m b a + b + c = 0 = aa 4ac But a b + c + d + e + f = 0 sin + = 0 =? =? by Martin Mendez,
3 Nonlinear Equation Solvers Bracketing Graphical Open Methods Bisection False Position Regula-Falsi Newton Raphson Secant All Iterative by Martin Mendez, 3
4 Eample 5.1 by Martin Mendez, 4
5 Bracketing Methods Or, two point methods for finding roots Two initial iti guesses for the root are required. These guesses must bracket or be on either side of the root. If one root of a real and continuous function, f=0, is bounded dby values = l, = u then f l. f u <0. The function changes sign on opposite sides of the root by Martin Mendez, 5
6 by Martin Mendez, 6
7 No answer No root Nice case one root Oops!! two roots!! Three roots Might work for a while!! by Martin Mendez, 7
8 Figure 5.3 Two roots Might work for a while!! Discontinuous function. Need special method by Martin Mendez, 8
9 MANY-MANY roots. What do we do? f=sin 10+cos 3 by Martin Mendez, 9
10 The Bisection Method For the arbitrary equation of one variable, f=0 1. Pick l and u such that they bound the root of interest, check if f l.f u <0.. Estimate the root by evaluating f[ l + u /]. 3. Find the pair If f l. f[ l + u /]<0, root lies in the lower interval, then u = l + u / and go to step. by Martin Mendez, 10
11 If f l. f[ l + u /]>0, root lies in the upper interval, then l = [ l + u /, go to step. If f l. f[ l + u /]=0, then root is l + u / and terminate. l l + + l u 4. Compare ε s with ε a l + u 5. If ε a < ε s, stop. Otherwise repeat the process. or u l + u u p100% p100% by Martin Mendez, 11
12 by Martin Mendez, 1
13 by Martin Mendez, 13
14 by Martin Mendez, 14
15 Evaluation of Method Pros Easy Always find root Number of iterations required to attain an absolute error can be computed a priori. Cons Slow Know a and b that bound root Multiple roots No account tis taken of f l and f u, if f l is closer to zero, it is likely l that root is closer to l. by Martin Mendez, 15
16 by Martin Mendez, 16
17 by Martin Mendez, 17
18 How Many Iterations will It Take? Length of the first Interval L o =b-a After 1 iteration L 1 =L o / After iterations L =L o /4 After k iterations L =L k k o / ε a L k 100% ε a ε s by Martin Mendez, 18
19 If the absolute magnitude of the error is ε s 100% = 10 4 and L o =, how many iterations will you have to do to get the required accuracy in the solution? 10 4 = k k = 10 4 k 14.3 = 15 by Martin Mendez, 19
20 by Martin Mendez, 0
21 The False-Position Method Regula-Falsi If a real root is bounded by l and u of f=0, Then we can approimate the solution by doing a linear interpolation between the points [ l, f l ] and [ u, f u ] to find the r value such that l r =0, l is the linear approimation of f. by Martin Mendez, 1
22 by Martin Mendez,
23 Procedure 1. Find a pair of values of, l and u such that f l =f l <0 and f u =f u >0.. Estimate the value of the root from the following formula Refer to Bo 5.1 and evaluate f r. by Martin Mendez, 3
24 by Martin Mendez, 4
25 by Martin Mendez, 5
26 by Martin Mendez, 6
27 by Martin Mendez, 7
28 3. Use the new point to replace one of the original points, keeping the two points on opposite sides of the ais. If f r <0 then l = r == > f l =f r If f r >0 then u = r == > f u =f r If f r =0 then you have found the root and need go no further! by Martin Mendez, 8
29 4. See if the new l and u are close enough for convergence to be declared. If they are not go back to step. Why this method? Faster Always converges for a single root. See Sec.5.3.1, Pitfalls of the False-Position Method Note: Always check by substituting estimated root in the original equation to determine whether f r 0. by Martin Mendez, 9
30 by Martin Mendez, 30
31 by Martin Mendez, 31
32 OPEN METHODS by Martin Mendez, 3
33 Simple Fied-point Iteration Rearrange the function so that is on the left side of the equation: f k = 0 g = = k o g given, k = 1, 1,... Bracketing methods are convergent. Fied-point methods may sometime diverge, depending on the stating point initial iti guess and how the function behaves. Chapter 6 33
34 Eample: f = f 0 g = or g = + or g = 1+ M Chapter 6 34
35 by Martin Mendez, 35
36 Almost linear by Martin Mendez, 36
37 Convergence Figure 6. =g can be epressed as a pair of equations: y 1 = y =g component equations Plot them separately. Chapter 6 37
38 by Martin Mendez, 38
39 Conclusion Fied-point iteration converges if g p 1 l slope of the line f = 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. Chapter 6 39
40 by Martin Mendez, 40
41 Newton Raphson Method Newton-Raphson Method Most widely used method. Based on Taylor series epansion:! 3 1 i i i i O f f f f Δ + Δ + Δ + = + 0 when f the value of The root is! 1 i 1 i 1 i i i i = Rearranging, 1 i i i i f f + = + Solve for 1 1 i i i i i i i f f f f = + + Newton-Raphson formula Chapter 6 41 i f
42 A convenient method for functions whose derivatives can be evaluated analytically. It may not be convenient for functions whose derivatives cannot be evaluated analytically. Fig. 6.5 Chapter 6 4
43 Fig. 6.6 Ej. f=ep-- Chapter 6 43
44 by Martin Mendez, 44
45 by Martin Mendez, 45
46 The Secant Method A slight variation of Newton s method for functions whose derivatives are difficult to evaluate. For these cases the derivative can be approimated by a backward finite divided difference. f 1 i i 1 i f i f i 1 i i 1 1 = i f i = 1,,3,K f i f i 1 i+ i Chapter 6 46
47 Fig. 6.7 Requires two initial estimates of, e.g, o, 1. However, because f is not required to change signs between estimates, it is not classified as a bracketing method. The secant method has the same properties as Newton s method. Convergence is not guaranteed for all o, f. Chapter 6 47
48 by Lale Yurttas, Teas A&M University Chapter 6 48
49 Fig. 6.8 Chapter 6 49
50 The Modified Secant Method The Modified Secant Method Rather than using two arbitrary values to estimate the derivative, an alternative approach involves a fractional perturbation of the independent variable to estimate f` f. ' i i i i f f + δ δ 1 i i i i i i i f f f + = + δ δ Where delta is a small fraction change Chapter 6 50
51 Chapter 6 51
52 Chapter 6 5
53 Roots of Polynomials Chapter 7 The roots of polynomials such as f = a + a + a + K+ a n o 1 n Follow these rules: 1. For an nth order equation, there are n real or comple roots.. If n is odd, there is at least one real root. 3. If comple root eist in conjugate pairs that is, λ+μiμ and λ-μi, μ where i=sqrt-1. Chapter 7 53 n
54 Conventional Methods The efficacy of bracketing and open methods depends on whether the problem being solved involves comple roots. If only real roots eist, these methods could be used. However, Finding good initial guesses complicates both the open and bracketing methods, also the open methods could be susceptible to divergence. Special methods have been developed to find the real and comple roots of polynomials Müller and Bairstow methods. Chapter 7 54
55 by Martin Mendez, 55
56 by Martin Mendez, 56
57 Müller Method Müller s method obtains a root estimate by projecting a parabola to the ais through three function values. Figure 7.3 Chapter 7 57
58 Müller Method The method consists of deriving the coefficients of parabola that goes through the three points: 1. Write the equation in a convenient form: f = a + b + c Chapter 7 58
59 The parabola should intersect the three points [. The parabola should intersect the three points [ o, f o ], [ 1, f 1 ], [, f ]. The coefficients of the polynomial can be estimated by substituting three polynomial can be estimated by substituting three points to give c b a f + + = c b a f c b a f o o o + + = Three equations can be solved for three unknowns, c b a f + + = 3. Three equations can be solved for three unknowns, a, b, c. Since two of the terms in the 3 rd equation are zero, it can be immediately solved for c=f. b a f f b a f f o o o + = + = Chapter b a f f + =
60 If - h - h If 1 1 o 1 o = = f f f f o o = = δ δ h h a h h b h h o o o o o + = + + δ δ Solved for a h a h b h = δ Solved for a and b 1 f h b o + δ δ δ f c ah b h h a o o = + = + = δ Chapter 7 60
61 Roots can be found by applying an alternative form of quadratic formula: 3 = + c c b ± b 4 ac The error can be calculated as ε a = % ±term yields two roots, the sign is chosen to agree with b. This will result in a largest denominator, and will give root estimate that is closest to. Chapter 7 61
62 Once 3 is determined, the process is repeated using the following guidelines: 1. If only real roots are being located, choose the two original points that are nearest the new root estimate, 3.. If both real and comple roots are estimated, t employ a sequential approach just like in secant method, 1,, and 3 to replace o, 1, and. Chapter 7 6
63 Chapter 7 63
64 Chapter 7 64
65 Chapter 7 65
66 Tarea Chapter 7 66
67 Polinomial evaluation and Differenciation Chapter 6 67
68 Evaluating the polinomial Evaluating the polinomial and its derivative Chapter 6 68
69 Polynomial deflation This is to eliminate the found root from the polynomial Forma factorizada Roots Si se divide el polinomio entre cualquiera de sus factores, el resultado serà un polinomio de grado 4 con un residuo igual a cero. 69
70 Seudocódigo r= an an=0 DOFOR i = n-1,0,-1 s=ai ai=r R=s+r*t ENDDO Ejemplo: f=-4+6=^+-44 Dividiento entre -4 Usando los parametros de suedocódigo: n=,a0=-4,a1=,a=1 y t=4 Entonces r=a=1 a=0 Divide un polinomio Iterando en el loop desde i=-1 hasta 0: de n-ésimo grado entre un monomial il Para i=1 -t Chapter 6 70
71 s=a1= a1=r=1 r=s+rt=+14=6 Para i = 0, s=a0=-4 a0=r=6 r=-4+64=0 Entonces, a0+a1=6+ Chapter 6 71
72 Chapter 6 7
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