1. Numerical Methods Root Finding 4

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(x k )=0, hence x k+1 =x k 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, x i+1 =g(x i ) The approximate error is given by:

5. Fixed-point iterations

6. Example:

7. Iterative Solution 1.Start with a guess say x 1 =1, 2.Generate a)x 2 =e -x1 = e -1 = 0.368 b)x 3 =e -x2 = e -0.368 = 0.692 c)x 4 =e -x3 = e -0.692 =0.500 In general: After a few more iteration we will get Find the root of f(x) = e -x – x

8. Problem Find a root near x=1.0 and x=2.0 Solution:  Starting at x=1, x=0.292893 at 15 th 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

9. Examples

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

11. The rearrangement x = (x 3 + 3)/7 leads to the iteration To find the middle root , let initial approximation x 0 = 2. Fixed Point Iteration The iteration slowly converges to give  = 0.441 (to 3 s.f.) etc.

12. The rearrangement x = (x 3 + 3)/7 leads to the iteration For x 0 = 2 the iteration will converge on the middle root , since g’(  ) < 1. Fixed Point Iteration  = 0.441 (to 3 s.f.)  x0x0 x2x2 x1x1 x3x3 y = (x 3 + 3)/7 y = x

13. Fixed Point Iteration - breakdown The rearrangement x = (x 3 + 3)/7 leads to the iteration For x 0 = 3 the iteration will diverge from the upper root . The iteration diverges because g’(  ) > 1.  x0x0 x1x1

14. Example: fixed point problems

15. Examples: FPI

16. Example: FPI

17. Convergence of FPI

18. 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).

19. b 1 =a 1 +p 0 b 0 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 p 0 = 1. Synthetic Division will be performed as below: Let f (x) = a 0 x 3 + a 1 x 2 + a 2 x + a 3 p0p0 a0a0 a1a1 a2a2 a3a3 b0b0 b1b1 b2b2 b3b3 c0c0 c1c1 c2c2 c3c3 p0b0p0b0 p1b1p1b1 p2b2p2b2 p0p0 s i m i l a r l y p 1 = p 0 – b 3 /c 2 Repeat synthetic division using p 1

20. 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

21. 120-51 22-3-2 241 1 22-3 241 Iteration No. 1: 320-51 261340 21249187 3 61839 636147 Iteration No. 2: p 1 = p 0 – b 3 /c 2 = 1 – (-2)/1 = 3 Not required p 2 = p 1 – b 3 /c 2 = 3 – 40/49 = 2.1837

22. 1.518520-51 23.037-0.38830.4104 26.0748.8351 1.5185 3.0374.6117-0.5896 3.0379.2234 Iteration No. 5: 1.472120-51 22.9442-0.66580.01986 25.88848.0025 1.4721 2.94424.3342-0.9801 2.94428.6683 Iteration No. 6: p 5 = p 4 – b 3 /c 2 = 1.5185 – 0.4104/8.8351 = 1.4721 p 6 = p 5 – b 3 /c 2 = 1.4721 – 0.01986/8.0025 = 1.469624

23. the equation x 3 +x 2 -3x-3 using Birge-Vieta Method where x 0 = 2. Using the synthetic division, 2|1 1 -3 -3 | 2 6 6 |1 3 3 3 ¬ f(x 0 ) | 2 10 |1 5 13 ¬ f ’ (x 0 ) Now, x 1 = 2 – 3/13 = 1.7692

24. Examples Determine the lowest positive root of: f(x) = 8 e -x sin (x) - 1 Using the Newton-Raphson method (three iterations, x 0 = 0.3) and Using the secant method (four iterations, x -1 = 0.5 and x 0 = 0.4). Using the modified secant method (three iterations, x 0 = 0.3, d = 0.01).

25. Summary MethodProsCons Bisection - Easy, Reliable, Convergent - One function evaluation per iteration - No knowledge of derivative is needed - Slow - Needs an interval [a,b] containing the root, i.e., f(a)f(b)<0 Newton - Fast (if near the root) - Two function evaluations per iteration - May diverge - Needs derivative and an initial guess x 0 such that f’(x 0 ) is nonzero Secant - Fast (slower than Newton) - One function evaluation per iteration - No knowledge of derivative is needed - May diverge - Needs two initial points guess x 0, x 1 such that f(x 0 )- f(x 1 ) is nonzero