Numerical Methods for Root Finding: Bisection Method

1. Numerical Methods Solution of Equation

2. Root Finding Methods • Bisection Method • Fixed Point Method • Secant Method • Modified Secant • Successive approximation • Newton Raphson method • Berge Vieta

3. Motivation • Many problems can be re-written into a form such as: • f(x,y,z,…) = 0 • f(x,y,z,…) = g(s,q,…)

4. Motivation • A root, r, of function f occurs when f(r) = 0. • For example: • f(x) = x2 – 2x – 3 has tworoots at r = -1 and r = 3. • f(-1) = 1 + 2 – 3 = 0 • f(3) = 9 – 6 – 3 = 0 • We can also look at f in its factored form. f(x) = x2 – 2x – 3 = (x + 1)(x – 3)

5. Finding roots / solving equations • General solution exists for equations such as ax2 + bx + c = 0 • The quadratic formula provides a quick answer to all quadratic equations. • However, no exact general solution (formula) exists for equations with exponents greater than 4. • Transcendental equations: involving geometric functions (sin, cos), log, exp. These equations cannot be reduced to solution of a polynomial.

6. Examples

7. Problem-dependent decisions • Approximation: since we cannot have exactness, we specify our tolerance to error • Convergence: we also specify how long we are willing to wait for a solution • Method: we choose a method easy to implement and yet powerful enough and general • Put a human in the loop: since no general procedure can find roots of complex equations, we let a human specify a neighbourhood of a solution

8. Bisection Method • Based on the fact that the function will change signs as it passes thru the root. • Suppose we know a function has a root between a and b. (…and the function is continuous, … and there is only one root) • f(a)*f(b) < 0 • Once we have a root bracketed, we simply evaluate the mid-point and halve the interval.

9. Bisection Method • f(a) . F(b) < 0 • c=(a+b)/2 f(a)>0 f(c)>0 a c b f(b)<0

10. Bisection Method • Guaranteed to converge to a root if one exists within the bracket. a = c f(a)>0 b c a f(c)<0 f(b)<0

11. Bisection method… • Check if solution lies between aand b…F(a)*F(b) < 0 ? • Try the midpoint m: compute F(m) • If |F(m)| < ϵselectm as your approximate solution • Otherwise, if F(m) is of opposite sign to F(a) that is ifF(a)*F(m) < 0, then b = m. • Else a = m.

12. Stop Conditions • 1. number of iterations • 2. |f(x0)| ≤ϵ • 3. | x- x0| ≤ϵ

13. Bisection Method • Simple algorithm: Given: a and b, such that f(a)*f(b)<0 Given: error tolerance, err c=(a+b)/2.0; // Find the midpoint While( |f(c)| > err ) { if( f(a)*f(c) < 0 ) // root in the left half b = c; else // root in the right half a = c; c=(a+b)/2.0; // Find the new midpoint } return c;

14. Square root program • The (positive) square root function is continuous and has a single solution. c = x2 Example: F(x) = x2 - 4 F(x) = x2 - c

15. Example: bisection iteration

16. Remarks • Convergence • Guaranteed once a nontrivial interval is found • Convergence Rate • A quantitative measure of how fast the algorithm is • An important characteristics for comparing algorithms

17. we could compute exactly how many iterations we would need for a given amount of error. • The error is usually measured by looking at the width of the current interval you are considering (i.e. the distance between a and b). The width of the interval at each iteration can be found by dividing the width of the starting interval by 2 for each iteration. This is because each iteration cuts the interval in half. If we let the error we want to achieve ϵand n be the iterations we get the following:

18. Let: Length of initial interval L0 After k iterations, length of interval is Lk Lk=L0/2k Algorithm stops when Lk ϵ Plug in some values… Convergence Rate of Bisection This is quite slow, compared to the other methods… Meaning of eps

19. How to get initial (nontrivial) interval [a,b] ? • Hint from the physical problem • For polynomial equation, the following theorem is applicable: roots (real and complex) of the polynomial f(x) = anxn + an-1xn-1 +…+ a1x + aο satisfy the bound:

20. Roots are bounded by Hence, real roots are in [-10,10] Roots are –1.5251, 2.2626 ± 0.8844i Example complex

21. Problem with Bisection • Although it is guaranteed to converge under its assumptions, • Although we can predict in advance the number of iterations required for desired accuracy (b - a)/2n <e -> n > log((b - a ) / e ) • Too slow! Computer Graphics uses square roots to compute distances, can’t spend 15-30 iterations on every one! • We want more like 1 or 2, equivalent to ordinary math operation.

22. Examples • Locate the first nontrivial root of sin x = x3 using Bisection method with the initial interval from 0.5 to 1. perform computation until error 2% • Determine the real root of f(x)= 5x3-5x2+6x-2 using bisection method. Employ initial guesses of xl = 0 and xu = 1. iterate until the estimated error a falls below a level of s = 15%

23. Homework