Numerical Solution of Nonlinear Equations Lecture 4 Chapter 2 (continued) Numerical Solution of Nonlinear Equations

Numerical methods for the solution of Nonlinear equations Single Nonlinear Equations Fixed-Point Iteration. bisection method Newton–Raphson method Secant method System of Nonlinear Equations Newton Method

3- Newton–Raphson (tangents) method The Newton–Raphson formula can be derived from the Taylor series expansion of f (x) about x:

Geometric interpretation The equation of the tangent to the curve y = f(x) at P(x(0), y(0)) which is x f(x) x* x0 y0 x3 x1 y1 x2

4- Secant method The secant method can be regarded as a modification of the Newton method in the sense that the derivative is replaced by a difference approximation based on the successive estimates By this approximation, the iterative formula of becomes

We observe that the Secant method will take iterations more than required when using Newton-Raphson method.

Newton Method for a System of Nonlinear Equations Consider the system of two equations by taking the Taylor series expansion up to first-order about some estimate point (x0,y0) This can be arranged into a matrix–vector form as

we solve for (x, y) to get the updated vector estimate let, Jacobian matrix

For 2x2 system we prefer to apply Cramer’s rule.

Steps of Newton–Raphson method for simultaneous, nonlinear equations:

The Jacobian matrix is defined as From the plot we can get a rough estimate of the coordinates of an intersection point: (x0,y0 ) =(0.5, 1.5), which we use as the starting values.

the coordinates of the four intersection points are ±(0.618 03, 1.618 03) and ± (1.618 03, 0.618 03)

Apply Bisection Method Matlab Functions Apply Bisection Method fzero Find root of continuous function of one variable Syntax x = fzero(fun,x0) >> fzero(@(x)exp(x)-exp(-x)-1,4) ans = 0.481211825059603

Apply Newton-Raphson Method fsolve Solve system of nonlinear equations Equation Solves a problem specified by F(x)=0 for x where x is a vector and F(x) is a function that returns a vector value. Syntax x = fsolve(fun,x0) Apply Newton-Raphson Method >> x=fsolve(@(x)x.^3+cos(x),-1) Equation solved. fsolve completed because the vector of function values is near zero as measured by the default value of the function tolerance, and the problem appears regular as measured by the gradient. <stopping criteria details> x = -0.865474075950360

% For system of nonlinear equations >> x=fsolve(@(x)[x(1).^2+x(2).^2-3;x(1).*x(2)-1],[0.5,1.5]) Equation solved. fsolve completed because the vector of function values is near zero as measured by the default value of the function tolerance, and the problem appears regular as measured by the gradient. <stopping criteria details> x = 0.618033988957653 1.618033988958327

End of Chapter 2