The formulae for the roots of a 3rd degree polynomial are given below Roots of a Polynomial: Root of a polynomial is the value of the independent variable at which the polynomial intersects the horizontal axis (the function has zero value) . The formulae for the roots of a 2nd degree polynomial are given below The formulae for the roots of a 3rd degree polynomial are given below First root (of three)

Example: Find the roots of the given polynomial. Roots of a Polynomial: The Matlab program can be used to calculate the roots of an n degree polynomial. Example: Find the roots of the given polynomial. ans = -1.0604 + 1.0863i -1.0604 - 1.0863i 0.5207 >>p=[5 8 6 -6]; roots(p) Example: Find the roots of the given polynomial. ans = 1.0043 + 2.7517i 1.0043 - 2.7517i -1.4940 + 0.3852i -1.4940 - 0.3852i 0.9793 >>p=[1 0 4 16 0 -20]; roots(p) All coefficients including those with zero must be specified. Otherwise the polynomial degree will be reduced.

NEWTON-RAPHSON ITERATION METHOD Solutions of Nonlinear Equations: Solutions of Nonlinear Equations: NEWTON-RAPHSON ITERATION METHOD f(x) x xi (İnitial value) f(xi)-0 Xi+1 Slope at this point is f'(xi) f(xi) Tangent line The Newton-Raphson method, or Newton’s method, is a powerful technique for solving equations numerically. Like so much of the differential calculus, it is based on the simple idea of linear approximation. This method is a method for finding successively better approximations to the real roots (or zeroes) of a real valued function. ε (error)

θ f f ' ε Newton-Raphson Example 1: Solutions of Nonlinear Equations: Newton-Raphson Example 1: Find one of the θ values, which satisfies the given equation. θ f f ' ε 1 -1.5858 2.3536 0.6738 1.6738 0.4368 3.6534 -0.1196 1.5542 0.0139 3.4213 -0.0041 1.5501 -0.00013 3.4134 3.95e-5

u f f ' ε Newton-Raphson Example 2: Solutions of Nonlinear Equations: Newton-Raphson Example 2: Find one of the u values, which satisfies the given equation. u f f ' ε 1 4.3899 5.4233 -0.8094 0.1905 -1.4883 6.6229 0.2247 0.4152 0.1569 7.8429 -0.0200 0.3952 0.00025 7.7801 -3.32e-5

Newton-Raphson Example 1: Newton-Raphson Example 2: clc, clear Solutions of Nonlinear Equations: MATLAB CODES The following changes are made in the program (nr1.m) to solve the problems. Newton-Raphson Example 1: Newton-Raphson Example 2: clc, clear x=1;xe=0.001*x; niter=20; %---------------------------------------------- for n=1:niter f=x^2-4+sqrt(x+1); df=2*x+0.5/(sqrt(x+1)); x1=x x=x1-f/df if abs(x-x1)<xe kerr=0;break end kerr,x clc, clear x=1;xe=0.001*x; niter=20; %---------------------------------------------- for n=1:niter f=5*x-cos(3*x)-1.6; df=5+3*sin(3*x); x1=x x=x1-f/df if abs(x-x1)<xe kerr=0;break end kerr,x x = fzero(@(x)x^2-4+sqrt(x+1),1) x = fzero(@(x)5*x-cos(3*x)-1.6,1)

f1(x1,x2)=0 f2(x1,x2)=0 Newton-Raphson Example 3: Solutions of Nonlinear Equations: Newton-Raphson iteration method is used to solve the nonlinear system of equations. Since there are more than one equations and unknown variables, partial derivatives of the equations with respect to each unkown variable are used in the solution procedure. f1(x1,x2)=0 f2(x1,x2)=0 Arbitrary initial values for x1 ve x2 are assigned and the iteration procedure is started by making the necessary changes in the computer program (newtonrn). The variables are stated as xb() in the program. Newton-Raphson Example 3: The equation of a circle with center coordinates (3,2) and a radius 5 is given on the right hand side. How do you find the intersection point of this circle and the parapola y=x2 ?

%---------------------------------------------- Solutions of Nonlinear Equations: The following changes are made in the program (nr.m) to solve the problem. clc, clear x=[1 4] ;xe=0.001*x; niter1=5;niter2=50; %---------------------------------------------- xe=transpose(abs(xe));kerr=1; for n=1:niter2 a(1,1)=2*(x(1)-3);a(1,2)=2*(x(2)-2); a(2,1)=-2*x(1);a(2,2)=1; b(1)=-((x(1)-3)^2+(x(2)-2)^2-25); b(2)=-(x(2)-x(1)^2); %---------------------------------------------- bb=transpose(b);eps=inv(a)*bb;x=x+transpose(eps) if n>niter1 if abs(eps)<xe kerr=0;break end x 1 2 3 4 9 x y (-1.82, 3.321) (2.643, 6.987) As shown in the figure, there are two valid solution sets. The output solution set is determined by the initial values of the unkown variables.