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
a x bx c  0
2
 b  b2  4 a c
x1 
2a
 b  b2  4 a c
x2 
2a
The formulae for the roots of a 3rd degree polynomial are given below
a x3  b x 2  c x  d  0
First root (of three)
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.
5 x3  8 x 2  6 x  6  0
ans =
>>p=[5 8 6 -6]; roots(p)
-1.0604 + 1.0863i
-1.0604 - 1.0863i
0.5207
Example: Find the roots of the given polynomial.
x  4 x  16 x  20  0
5
3
2
>>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.
ans =
1.0043 + 2.7517i
1.0043 - 2.7517i
-1.4940 + 0.3852i
-1.4940 - 0.3852i
0.9793
Solutions of Nonlinear Equations:
NEWTON-RAPHSON ITERATION METHOD
f(x)
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.
f(xi)
Tangent line
f(xi)-0
0
Xi+1
xi (İnitial value)
x
x i  x i 1
f(x i )  0
f (x i ) 
x i  x i1
f(x i )
x i 1  x i 
f (x i )
Slope at this
point is f'(xi)
xi  xi1  f(xi )  f(xi )
ε (error)
f(x i )
x i 1  x i  
f (x i )
 f(xi )   f(xi )
Solutions of Nonlinear Equations:
Newton-Raphson Example 1:
2  4     1
f()  0
f  2  4    1
Find one of the θ values, which satisfies the given
equation.
f

f
1 1
f  2  
2 1
, x n1  x n  
(tet + 1)1/2 + tet2 - 4
40
f
f'
ε
1
-1.5858
2.3536
0.6738
1.6738
0.4368
3.6534
-0.1196
30
25
f(tet)
θ
35
20
15
10
1.5542
0.0139
3.4213
-0.0041
1.5501
-0.00013
3.4134
3.95e-5
5
0
-5
-1
0
1
1.55
2
3
tet
4
5
6
Solutions of Nonlinear Equations:
Newton-Raphson Example 2:
5u  cos(3u)  1.6
f(u)  0
f  5u  cos(3u)  1.6
f   5  3 sin(3u)
Find one of the u values, which satisfies the given
equation.
f

f
, x n1  x n  
5 u - cos(3 u) - 8/5
30
f
f'
ε
20
1
4.3899
5.4233
-0.8094
10
0.1905
-1.4883
6.6229
0.2247
0
0.4152
0.1569
7.8429
-0.0200
0.3952
0.00025
7.7801
-3.32e-5
f(u)
u
-10
-20
-30
-6
-4
-2
0
u
2
4
6
Solutions of Nonlinear Equations:
MATLAB CODES
The following changes are made in the program (nr1.m) to solve the problems.
Newton-Raphson Example 1:
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
end
kerr,x
x = fzero(@(x)x^2-4+sqrt(x+1),1)
Newton-Raphson Example 2:
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
end
kerr,x
x = fzero(@(x)5*x-cos(3*x)-1.6,1)
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
f    f
 f1
 x
 1
 f2
 x 1
f1 
x 2  1   f1 
  
f2  2   f2 
x 2 
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:
x  32  y  22  25
f1  x  32  y  22  25
f2  y  x 2
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 ?
f1
f1
 2x  3,
 2y  2
x
y
f2
f2
 2x ,
1
x
y
Solutions of Nonlinear Equations:
The following changes are made in the program (nr.m) to solve the problem.
clc, clear
y
x=[1 4] ; fe=[0.01 0.01];
niter1=5;niter2=50;
9
fe=transpose(abs(fe));kerr=1;
for n=1:niter2
(2.643, 6.987)
x
%Error Equations--------------------------a(1,1)=2*(x(1)-3);a(1,2)=2*(x(2)-2);
4
(-1.82, 3.321)
a(2,1)=-2*x(1);a(2,2)=1;
2
b(1)=-((x(1)-3)^2+(x(2)-2)^2-25);
1
b(2)=-(x(2)-x(1)^2);
%---------------------------------------------1 2 3
x
bb=transpose(b);eps=inv(a)*bb;x=x+transpose(eps);
if n>niter1
if abs(eps)<fe
kerr=0; break
else
display ('Roots are not found')
end
end
As shown in the figure, there are two valid solution sets. The output
end
solution set is determined by the initial values of the unkown variables.