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

6. Fixed-point iterations
Math 685/CSI 700 Spring 08
Fixed-point iterations
George Mason University, Department of Mathematical Sciences

7. Example:

8. Start with a guess say x1=1, Generate
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 = 0.692
x4=e-x3= e-0.692=0.500
In general:
After a few more iteration we will get

9. Find a root near x=1.0 and x=2.0 Solution:
Problem
Find a root near x=1.0 and x=2.0
Solution:
Starting at x=1, x= 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 Math 685/CSI 700 Spring 08
George Mason University, Department of Mathematical Sciences

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 = (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 = (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
Math 685/CSI 700 Spring 08
Example: fixed point problems
George Mason University, Department of Mathematical Sciences

16. Examples: FPI Math 685/CSI 700 Spring 08
George Mason University, Department of Mathematical Sciences

17. Example: FPI Math 685/CSI 700 Spring 08
George Mason University, Department of Mathematical Sciences

18. Convergence of FPI Math 685/CSI 700 Spring 08
George Mason University, Department of Mathematical Sciences

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

24. Fixed Point Iteration xi+1=(10-xi2)/yi and yi+1=57-3xiyi2
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=
y=(57-3( )(3.5)2=
Second iteration:
x=( )/ =-0.209
y=57-3(-0.209)( )2=
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 -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 -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 =

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

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 =

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 Method Pros Cons Bisection Newton Secant
2/22/2019
Summary
Method
Pros
Cons
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 x0 such that f’(x0) is nonzero
Secant
- Fast (slower than Newton)
- One function evaluation per iteration
- Needs two initial points guess x0, x1 such that
f(x0)- f(x1) is nonzero