SOLUTION OF NONLINEAR EQUATIONS Few examples of nonlinear

SOLUTION OF NONLINEAR EQUATIONS • Few examples of nonlinear equations follow:

• 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.

METHODS TO FIND ROOTS (we discuss only one root) 1. Bisection Method 2. False Position Method 3. Fixed Point Method 4. Newton’s Method 5. 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.

Bisection Algorithm 1. Select interval such that 2. Compute 3. If then else 4. When the interval is within a given tolerance or the n-th iteration is reached then stopped iteration, else repeat step 2

Application Consider the equation 1. Find an approximation value of the root within a tolerance of 2. Repeat problem above with ten iteration Answer: The problem is solved with matlab program

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

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

False Position Algorithm 1. Select interval such that 2. Compute 3. If then else 4. When the interval is within a given tolerance or the n-th iteration is reached then stopped iteration, else repeat step 2

Application Consider the equation 1. Find an approximation value of the root within a tolerance of 2. Repeat problem above with ten iteration Answer: The problem is solved with matlab program

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

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