Math Problem Statement

f(x) = e^x - cos(x) - 2 using bisection method

Solution

To apply the bisection method to find the root of the function f(x)=excos(x)2f(x) = e^x - \cos(x) - 2, we need to follow these steps:

  1. Identify an interval [a,b][a, b] where the function changes sign, i.e., f(a)f(b)<0f(a) \cdot f(b) < 0.
  2. Calculate the midpoint c=a+b2c = \frac{a + b}{2}.
  3. Evaluate f(c)f(c).
  4. Determine the sub-interval: If f(a)f(c)<0f(a) \cdot f(c) < 0, then set b=cb = c; otherwise, set a=ca = c.
  5. Repeat steps 2-4 until the desired accuracy is achieved.

Step 1: Find an Interval

Let's evaluate f(x)f(x) at some values to find an appropriate interval:

  • f(0)=e0cos(0)2=112=2f(0) = e^0 - \cos(0) - 2 = 1 - 1 - 2 = -2
  • f(1)=e1cos(1)22.7180.54020.178f(1) = e^1 - \cos(1) - 2 \approx 2.718 - 0.540 - 2 \approx 0.178

Since f(0)<0f(0) < 0 and f(1)>0f(1) > 0, we can choose the interval [0,1][0, 1].

Step 2: Bisection Process

Let's perform the bisection method for a few iterations:

  1. First Iteration:

    • c=0+12=0.5c = \frac{0 + 1}{2} = 0.5
    • f(0.5)=e0.5cos(0.5)21.6490.87721.228f(0.5) = e^{0.5} - \cos(0.5) - 2 \approx 1.649 - 0.877 - 2 \approx -1.228
    • Since f(0)<0f(0) < 0 and f(0.5)<0f(0.5) < 0, update a=0.5a = 0.5.
  2. Second Iteration:

    • c=0.5+12=0.75c = \frac{0.5 + 1}{2} = 0.75
    • f(0.75)=e0.75cos(0.75)22.1170.73120.614f(0.75) = e^{0.75} - \cos(0.75) - 2 \approx 2.117 - 0.731 - 2 \approx -0.614
    • Update a=0.75a = 0.75.
  3. Third Iteration:

    • c=0.75+12=0.875c = \frac{0.75 + 1}{2} = 0.875
    • f(0.875)=e0.875cos(0.875)22.3980.63920.241f(0.875) = e^{0.875} - \cos(0.875) - 2 \approx 2.398 - 0.639 - 2 \approx -0.241
    • Update a=0.875a = 0.875.
  4. Fourth Iteration:

    • c=0.875+12=0.9375c = \frac{0.875 + 1}{2} = 0.9375
    • f(0.9375)=e0.9375cos(0.9375)22.5510.58620.034f(0.9375) = e^{0.9375} - \cos(0.9375) - 2 \approx 2.551 - 0.586 - 2 \approx 0.034
    • Since f(0.875)<0f(0.875) < 0 and f(0.9375)>0f(0.9375) > 0, update b=0.9375b = 0.9375.
  5. Fifth Iteration:

    • c=0.875+0.93752=0.90625c = \frac{0.875 + 0.9375}{2} = 0.90625
    • f(0.90625)2.4600.61620.156f(0.90625) \approx 2.460 - 0.616 - 2 \approx -0.156
    • Update a=0.90625a = 0.90625.

Continue this process until the interval is sufficiently small (e.g., when ba<ϵ|b - a| < \epsilon for a chosen ϵ\epsilon).

Conclusion

You can continue this iteration to achieve your desired precision. The root lies within the interval and can be approximated as cc converges.

Would you like further details on any specific part of this process, or do you have any questions? Here are some related questions you might consider:

  1. How does the bisection method compare to other root-finding methods?
  2. What is the significance of choosing the initial interval correctly?
  3. Can the bisection method be applied to complex functions?
  4. What are some limitations of the bisection method?
  5. How would you implement the bisection method in a programming language?

Tip: Always verify that f(a)f(a) and f(b)f(b) have opposite signs before starting the bisection process!

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Math Problem Analysis

Mathematical Concepts

Numerical Methods
Root Finding
Functions

Formulas

f(c) = e^c - cos(c) - 2
c = (a + b) / 2

Theorems

Intermediate Value Theorem

Suitable Grade Level

Grades 11-12