---

The Bisection Method

---

• Introduction
  • Bisection Method:

    Bisection Method = a numerical method in Mathematics to find a root of a given function

    ---

  • Root of a function:

    Root of a function f(x) = a value a such that: f(a) = 0

    Example:

    Function: f(x) = x2 - 4 Roots: x = -2, x = 2 Because: f(-2) = (-2)2 - 4 = 4 - 4 = 0 f(2) = (2)2 - 4 = 4 - 4 = 0

  ---

  ---

• A Mathematical Property
  • Well-known Mathematical Property:

    If a function f(x) is continuous on the interval [a..b] and sign of f(a) ≠ sign of f(b), then: There is a value c ∈ [a..b] such that: f(c) = 0 I.e., there is a root c in the interval [a..b] Example:

  ---

  ---

• The Bisection Method
  • The Bisection Method is a successive approximation method that narrows down an interval that contains a root of the function f(x)

  • The Bisection Method is given an initial interval [a..b] that contains a root

    (We can use the property sign of f(a) ≠ sign of f(b) to find such an initial interval)

  • The Bisection Method will cut the interval into 2 halves and check which half interval contains a root of the function

  • The Bisection Method will keep cut the interval in halves until the resulting interval is extremely small

    The root is then approximately equal to any value in the final (very small) interval.

    ---

  • Example:

    Suppose the interval [a..b] is as follows: We cut the interval [a..b] in the middle: m = (a+b)/2 Because sign of f(m) ≠ sign of f(a) , we proceed with the search in the new interval [a..b]: We can use this statement to change to the new interval: b = m;

    ---

  • In the above example, we have changed the end point b to obtain a smaller interval that still contains a root

    In other cases, we may need to changed the end point b to obtain a smaller interval that still contains a root

    Here is an example where you have to change the end point a:

    Initial interval [a..b]: After cutting the interval in half, the root is contained in the right-half, so we have to change the end point a:

    ---

  • Example: find the root of f(x) = x² − 5 between [0, 4]

    There is a root between [0,4] because:: f(0) = 02 − 5 = −5 f(4) = 42 − 5 = 11 Start: a = 0; f(a) = -5 b = 4; f(b) = 11 Iteration 1: m = (a + b)/2 = 2 f(m) = 22 − 5 = -1 Because f(m) < 0, we replace a with m a = 2; f(a) = -1 b = 4; f(b) = 11 Iteration 2: m = (a + b)/2 = 3 f(m) = 32 − 5 = 4 Because f(m) > 0, we replace b with m a = 2; f(a) = -1 b = 3; f(b) = 4 Iteration 3: m = (a + b)/2 = 2.5 f(m) = 2.52 − 5 = 1.25 Because f(m) > 0, we replace b with m a = 2; f(a) = -1 b = 2.5; f(b) = 1.25 And so on....

    ---

  • Rough description (pseudo code) of the Bisection Method:

    Given: interval [a..b] such that: sign of f(a) ≠ sign of f(b) repeat (until the interval [a..b] is "very small") { a+b m = -----; // m = midpoint of interval [a..b] 2 if ( sign of f(m) ≠ sign of f(b) ) { use interval [m..b] in the next iteration (i.e.: replace a with m) } else { use interval [a..m] in the next iteration (i.e.: replace b with m) } } Approximate root = (a+b)/2; (any point between [a..b] will do because the interval [a..b] is very small)

    Re-writing the pseudo code in term of a while-statement:

    Given: interval [a..b] such that: sign of f(a) ≠ sign of f(b) while (interval [a..b] is NOT "very small") { a+b m = -----; // m = midpoint of interval [a..b] 2 if ( sign of f(m) ≠ sign of f(b) ) { use interval [m..b] in the next iteration (i.e.: replace a with m) } else { use interval [a..m] in the next iteration (i.e.: replace b with m) } } Approximate root = (a+b)/2; (any point between [a..b] will do because the interval [a..b] is very small)

    ---

  • Structure Diagram of the Bisection Algorithm:

    ---

    ---

  • Example execution:

    We will use a simple function to illustrate the execution of the Bisection Method Function used: f(x) = x2 - 3 Roots: √3 = 1.7320508... and −√3 = −1.7320508... We will use the starting interval [0..4] since: f(0) = 02 − 3 = −3 f(4) = 42 − 3 = 13 The interval [0..4] contains a root because: sign of f(0) ≠ sign of f(4)

    Steps taken by the Bisection Method:

    Iteration 1: New interval: [0..2] (it contains √3 = 1.7320508.. !!!) Iteration 2: New interval: [1..2] (it contains √3 = 1.7320508.. !!!) Iteration 3: New interval: [1.5 .. 2] (it contains √3 = 1.7320508.. !!!) And so on !! Result: The interval gets smaller and smaller But it will always contain the root √3 When the interval is smaller than 0.000001, the while-loop will exit At that moment, the end points of the interval will be very close to root √3

    ---

    ---

  • Java program:

    // Bisection Method - Solves: x^2 - 3 = 0 public class Bisection01 { public static void main(String[] args) { final double epsilon = 0.00001; double a, b, m, y_m, y_a; a = 0; b = 4; while ( (b-a) > epsilon ) { m = (a+b)/2; // Mid point y_m = m*m - 3.0; // y_m = f(m) y_a = a*a - 3.0; // y_a = f(a) if ( (y_m > 0 && y_a < 0) || (y_m < 0 && y_a > 0) ) { // f(a) and f(m) have different signs: move b b = m; } else { // f(a) and f(m) have same signs: move a a = m; } System.out.println("New interval: [" + a + " .. " + b + "]"); // Print progress } System.out.println("Approximate solution = " + (a+b)/2 ); } }

    Output:

    Initial interval: [0.0 .. 4.0] New interval: [0.0 .. 2.0] (We did the first 3 by hand above) New interval: [1.0 .. 2.0] New interval: [1.5 .. 2.0] New interval: [1.5 .. 1.75] New interval: [1.625 .. 1.75] New interval: [1.6875 .. 1.75] New interval: [1.71875 .. 1.75] New interval: [1.71875 .. 1.734375] New interval: [1.7265625 .. 1.734375] New interval: [1.73046875 .. 1.734375] New interval: [1.73046875 .. 1.732421875] New interval: [1.7314453125 .. 1.732421875] New interval: [1.73193359375 .. 1.732421875] New interval: [1.73193359375 .. 1.732177734375] New interval: [1.73193359375 .. 1.7320556640625] New interval: [1.73199462890625 .. 1.7320556640625] New interval: [1.732025146484375 .. 1.7320556640625] New interval: [1.7320404052734375 .. 1.7320556640625] New interval: [1.7320480346679688 .. 1.7320556640625] Approximate solution = 1.7320518493652344

    ---

  • Example Program: (Demo above code)                                                
    • Prog file: click here

    How to run the program:

    Right click on link and save in a scratch directory To compile:   javac Bisection01.java To run:          java Bisection01

  ---

  ---

  ---

  ---

  ---

  ---

• Finding the roots of a different function
  • Suppose we want to find the root of

    f(x) = x3 + x - 3

    in the interval [0..4]

    ---

  • Changes to the Java program:

    // Solves: x^3 + x - 3 = 0 public class Bisection02 { public static void main(String[] args) { final double epsilon = 0.00001; double a, b, m, y_m, y_a; a = 0; b = 4; while ( (b-a) > epsilon ) { m = (a+b)/2; // Mid point y_m = m*m*m + m - 3.0; // y_m = f(m) y_a = a*a*a + a - 3.0; // y_a = f(a) if ( (y_m > 0 && y_a < 0) || (y_m < 0 && y_a > 0) ) { // f(a) and f(m) have different signs: move b b = m; } else { // f(a) and f(m) have same signs: move a a = m; } System.out.println("New interval: [" + a + " .. " + b + "]"); } System.out.println("Approximate solution = " + (a+b)/2 ); } }

    ---

  • Example Program: (Demo above code)                                                
    • Prog file: click here

    How to run the program:

    Right click on link and save in a scratch directory To compile:   javac Bisection02.java To run:          java Bisection02

    ---

  • We had to change the body of the Java program to find the root of a different function

    This is very inconvenient

    ---

  • It would be more convenient to if we can write:

    y_m = f(m); y_a = f(a);

    When we solve a different function, we make changes to the function definition f()

    This is indeed possible in Java...

    However, we have not study user-defined methods yet.

    But we will get there very soon.