1. SOLUTION OF NONLINEAR EQUATIONS
Few examples of nonlinear equations follow:

2. The primary reason why we solve nonlinear equations by using computer method is that nonlinear equations have no closed-form solution except for very few problem.
Many researcher had been found the analytic solution of polynomial equations for fourth order, but there are no closed-form solutions for higher order.
Roots of the nonlinear equations are found by computer methods based on iterative procedures.

3. METHODS TO FIND ROOTS (we discuss only one root)
course-3/numerical method
METHODS TO FIND ROOTS (we discuss only one root)
Bisection Method
False Position Method
Fixed Point Method
Newton’s Method
Secant Method
Note:
The first two methods require a preliminary effort to estimate an appropriate interval that contains the desired root.
The last three methods need an initial guess to find the root.

5. Bisection Algorithm Select interval such that Compute If then else
When the interval is within a given tolerance or the n-th iteration is reached then stopped iteration, else repeat step 2

6. course-3/numerical method
Application
Consider the equation
Find an approximation value of the root within a tolerance of
Repeat problem above with ten iteration
Answer:
The problem is solved with matlab program
catatan

7. Error Analysis The interval size after n iteration steps becomes
This also represents the maximum error bound
Hence, the number of iteration steps required for the given error tolerance by is the smallest integer satisfying
or equivalently

8. Scheme Analysis Advantage:
The bisection method is the simplest, safest, and most robust scheme for finding one root in a given interval
Disadvantage:
Slow to converge for a large interval
Cannot find a pair of double roots
Does not recognize the difference between root and singularity

10. False Position Algorithm
Select interval such that
Compute
If then else
When the interval is within a given tolerance or the n-th iteration is reached then stopped iteration, else repeat step 2

11. course-3/numerical method
Application
Consider the equation
Find an approximation value of the root within a tolerance of
Repeat problem above with ten iteration
Answer:
The problem is solved with matlab program
catatan

12. Error Analysis The interval size after n iteration steps becomes
This also represents the maximum error bound
Hence, the number of iteration steps required for the given error tolerance by is the smallest integer satisfying
or equivalently

13. Scheme Analysis Advantage:
The bisection method is the simplest, safest, and most robust scheme for finding one root in a given interval
Disadvantage:
Slow to converge for a large interval
Cannot find a pair of double roots
Does not recognize the difference between root and singularity
example