Math Problem Statement

find a positive root of the equation   2x=3+cosx   by Bisection method.  

Solution

To find the positive root of the equation 2x=3+cos(x)2x = 3 + \cos(x) using the Bisection Method, follow the steps below:

Steps:

  1. Rewrite the equation in the form f(x)=0f(x) = 0: f(x)=2x3cos(x)f(x) = 2x - 3 - \cos(x) The root of the equation corresponds to the value of xx where f(x)=0f(x) = 0.

  2. Choose an initial interval [a,b][a, b] such that f(a)×f(b)<0f(a) \times f(b) < 0. This ensures that there is a root in the interval.

  3. Apply the Bisection Method formula to find the midpoint cc: c=a+b2c = \frac{a + b}{2} Calculate f(c)f(c) and check the sign of f(c)f(c).

  4. Update the interval:

    • If f(a)×f(c)<0f(a) \times f(c) < 0, then the root lies in the interval [a,c][a, c], so update b=cb = c.
    • If f(b)×f(c)<0f(b) \times f(c) < 0, then the root lies in the interval [c,b][c, b], so update a=ca = c.
  5. Repeat the process until the interval is sufficiently small (i.e., until ba|b - a| is smaller than the desired tolerance).

Step-by-Step Calculation:

  1. Initial Interval: Let's find an interval [a,b][a, b] where f(x)f(x) changes sign.

    • For x=0x = 0: f(0)=2(0)3cos(0)=31=4f(0) = 2(0) - 3 - \cos(0) = -3 - 1 = -4
    • For x=2x = 2: f(2)=2(2)3cos(2)=43(0.416)=1+0.4161.416f(2) = 2(2) - 3 - \cos(2) = 4 - 3 - (-0.416) = 1 + 0.416 \approx 1.416

    Since f(0)=4f(0) = -4 and f(2)=1.416f(2) = 1.416, the function changes sign between x=0x = 0 and x=2x = 2. Therefore, the interval [0,2][0, 2] is a suitable starting point.

  2. First Iteration:

    • a=0a = 0, b=2b = 2
    • Midpoint c=0+22=1c = \frac{0 + 2}{2} = 1
    • f(1)f(1): f(1)=2(1)3cos(1)=230.540=1.540f(1) = 2(1) - 3 - \cos(1) = 2 - 3 - 0.540 = -1.540 Since f(1)<0f(1) < 0, the root lies in the interval [1,2][1, 2]. Update a=1a = 1.
  3. Second Iteration:

    • a=1a = 1, b=2b = 2
    • Midpoint c=1+22=1.5c = \frac{1 + 2}{2} = 1.5
    • f(1.5)f(1.5): f(1.5)=2(1.5)3cos(1.5)=330.0707=0.0707f(1.5) = 2(1.5) - 3 - \cos(1.5) = 3 - 3 - 0.0707 = -0.0707 Since f(1.5)<0f(1.5) < 0, the root lies in the interval [1.5,2][1.5, 2]. Update a=1.5a = 1.5.
  4. Third Iteration:

    • a=1.5a = 1.5, b=2b = 2
    • Midpoint c=1.5+22=1.75c = \frac{1.5 + 2}{2} = 1.75
    • f(1.75)f(1.75): f(1.75)=2(1.75)3cos(1.75)=3.53(0.178)=0.678f(1.75) = 2(1.75) - 3 - \cos(1.75) = 3.5 - 3 - (-0.178) = 0.678 Since f(1.75)>0f(1.75) > 0, the root lies in the interval [1.5,1.75][1.5, 1.75]. Update b=1.75b = 1.75.
  5. Repeat the process until the interval is sufficiently small. Continue bisecting until the root is accurate to the desired number of decimal places.

Final Answer:

The Bisection Method will give an increasingly accurate approximation of the root within the interval [1.5,1.75][1.5, 1.75]. After several iterations, you'll converge on the root.


Would you like me to complete more iterations, or explain anything further?

5 Related Questions:

  1. What are the advantages of the Bisection Method over other numerical methods?
  2. How does the choice of initial interval affect the convergence of the Bisection Method?
  3. Can the Bisection Method be applied to non-continuous functions?
  4. How do you determine the tolerance level for the Bisection Method?
  5. What is the convergence rate of the Bisection Method compared to Newton's Method?

Tip: Always ensure that the function changes sign within the interval before applying the Bisection Method.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Equations
Trigonometry
Numerical Methods

Formulas

Bisection Method formula

Theorems

-

Suitable Grade Level

Higher Education