Numerical Methods Root Finding

1. 4 Numerical Methods Root Finding

2. Fixed-Point Iteration---- Successive Approximation • Many problems also take on the specialized form: g(x)=x, where we seek, x, that satisfies this equation. • In the limit, f(xk)=0, hence xk+1=xk f(x)=x g(x)

3. Fractals • Images result when we deal with 2-dimensions. • Such as complex numbers. • Color indicates how quickly it converges or diverges.

4. Simple Fixed-Point Iteration • Rearrange the function f(x)=0 so that x is on the left-hand side of the equation: x=g(x) • Use the new function g to predict a new value of x - that is, xi+1=g(xi) • The approximate error is given by:

5. Begin Compute or choose initial object Compute next object No Object sufficient? End Yes Successive Approximations

6. Fixed-point iterations

7. Example:

8. Iterative Solution • Find the root of f(x) = e-x – x • Start with a guess say x1=1, • Generate • x2=e-x1= e-1= 0.368 • x3=e-x2= e-0.368 = 0.692 • x4=e-x3= e-0.692=0.500 In general: After a few more iteration we will get

9. Problem • Find a root near x=1.0 and x=2.0 • Solution: • Starting at x=1, x=0.292893 at 15th iteration • Starting at x=2, it will not converge • Why? Relate to g'(x)=x. for convergence g'(x) < 1 • Starting at x=1, x=1.707 at iteration 19 • Starting at x=2, x=1.707 at iteration 12 • Why? Relate to

10. Examples

11. Fixed Point Iteration The equation f(x) = 0, where f(x) = x3- 7x + 3, may be re-arranged to give x = (x3 + 3)/7. Intersection of the graphs of y = x and y = (x3 + 3)/7 represent roots of the original equation x3- 7x + 3 = 0. y = (x3 + 3)/7 y = x

12. Fixed Point Iteration The rearrangement x = (x3 + 3)/7 leads to the iteration To find the middle root a, let initial approximation x0 = 2. etc. The iteration slowly converges to give a = 0.441 (to 3 s.f.)

13. Fixed Point Iteration The rearrangement x = (x3 + 3)/7 leads to the iteration For x0 = 2 the iteration will converge on the middle root a, since g’(a) < 1. y = x y = (x3 + 3)/7 a x3 x2 x1 x0 a = 0.441 (to 3 s.f.)

14. Fixed Point Iteration - breakdown The rearrangement x = (x3 + 3)/7 leads to the iteration For x0 = 3 the iteration will diverge from the upper root a. a x0 x1 The iteration diverges because g’(a) > 1.

15. Example: fixed point problems

16. Examples: FPI

17. Example: FPI

18. Convergence of FPI

19. Simple Fixed-Point Iteration Convergence

20. 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.”

21. Simple Fixed-Point Iteration-Convergence

22. More on Convergence Graphically the solution is at the intersection of the two curves. We identify the point on y2 corresponding to the initial guess and the next guess corresponds to the value of the argument x where y1 (x) = y2 (x). Convergence of the simple fixed-point iteration method requires that the derivative of g(x) near the root has a magnitude less than 1. Convergent, 0≤g’<1 Convergent, -1<g’≤0 Divergent, g’>1 Divergent, g’<-1

23. 1 while a< s & i>maxi • False Print: xo, f(xo) ,a , i i=1 or xn=0 Stop • True x0=xn

24. 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

25. Birge – Vieta Method • Used for finding roots of polynomial functions. • Uses “synthetic division” of polynomial to extract factor of the given polynomial in the form of (x – p).

26. Problem: Find roots of f (x) = 2x³ – 5x + 1 using Birge – Vieta Method. Solution: Assume that x = 1 is root of the equation. Hence initial approximation of the solution is p0 = 1. Synthetic Division will be performed as below: Let f (x) = a0x3 + a1x2 + a2x + a3 p0 a0 a1 a2 a3 p0b0 p1b1 p2b2 p0 b0 b1=a1+p0b0 b1 b2 b3 p1 = p0 – b3/c2 s i m i l a r l y Repeat synthetic division using p1 c0 c1 c2 c3

27. Birge-Vieta Method • NR method with f(x) and f'(x) evaluated using Horner’s method • Once a root is found, reduce order of polynomial

28. Iteration No. 1: 1 2 0 -5 1 2 2 -3 1 2 2 -3 -2 2 4 1 2 4 1 -1 p1 = p0 – b3/c2 = 1 – (-2)/1 = 3 Iteration No. 2: 3 2 0 -5 1 6 18 39 Not required 3 2 6 13 40 6 36 147 2 12 49 187 p2 = p1 – b3/c2 = 3 – 40/49 = 2.1837

29. Iteration No. 5: 1.5185 2 0 -5 1 3.037 4.6117 -0.5896 1.5185 2 3.037 -0.3883 0.4104 3.037 9.2234 2 6.074 8.8351 p5 = p4 – b3/c2 = 1.5185 – 0.4104/8.8351 = 1.4721 Iteration No. 6: 1.4721 2 0 -5 1 2.9442 4.3342 -0.9801 1.4721 2 2.9442 -0.6658 0.01986 2.9442 8.6683 2 5.8884 8.0025 p6 = p5 – b3/c2 = 1.4721 – 0.01986/8.0025 = 1.469624

30. the equation x3+x2-3x-3 using Birge-Vieta Method where x0= 2. Using the synthetic division, 2|1 1 -3 -3 | 2 6 6 |1 3 3 3¬f(x0) | 2 10 |1 5 13¬f ’(x0) Now, x1= 2 – 3/13 = 1.7692

31. Examples • Determine the lowest positive root of: f(x) = 8 e-x sin (x) - 1 Using the Newton-Raphson method (three iterations, x0 = 0.3) and Using the secant method (four iterations, x-1 = 0.5 and x0 = 0.4). Using the modified secant method (three iterations, x0 = 0.3, d = 0.01).

32. Summary