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Polynomial

Introduction to Matlab 9

Omed Ghareb AbdullahSulaimani UniversitySulaimani UniversityCollege of SciencesCollege of SciencesPhysics DepartmentPhysics Department

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Polynomials• The polynomials are represented by their coefficients in

MATLAB.• Consider the following polynomial:

Y(x) = x3 + 3x2 + 3x + 1• For x is scalar: use scalar operations

Y = x^3 + 3*x^2 + 3*x + 1;

• For x is a vector or a matrix: use array or element by element operation

Y = x.^3 + 3*x.^2 + 3.*x + 1;• function polyval(p,x): evaluates a polynomial with

coefficients in vector p for the values in x.

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Example:Example:

Y(x) = x3 + 3x2 + 3x + 1

Polynomials

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100

150

200

250

Y(x

)

Y(x)=x3+3x2+3x+1

Y(x) x + 3x + 3x + 1

clear all;clcx = linspace (-5, 5, 100);

Y = x.^3+3.*x.^2+3.*x+1;

-5 -4 -3 -2 -1 0 1 2 3 4 5-100

-50

0

x

plot (x, Y, 'r.-')xlabel ('x')ylabel ('Y(x)')title('Y(x)=x^3+3x^2+3x+1');

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Example:Example:

Y(x) = x3 + 3x2 + 3x + 1

Polynomials

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100

150

200

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Y(x

)

Y(x)=x3+3x2+3x+1

Y(x) x + 3x + 3x + 1

clear all;clcx = linspace (-5, 5, 100);p = [ 1 3 3 1];Y = polyval (p, x);

-5 -4 -3 -2 -1 0 1 2 3 4 5-100

-50

0

x

plot (x, Y, 'r.-')xlabel ('x')ylabel ('Y(x)')title('Y(x)=x^3+3x^2+3x+1');

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Polyval: Evaluate polynomialy = polyval (p,x), when p is a vector of length n+1 whose

elements are the coefficients of a polynomial, is the value of the polynomial evaluated at x.

If x is a matrix or vector the polynomial is evaluated atIf x is a matrix or vector, the polynomial is evaluated at all points in x, using polyvalm for evaluation in a matrix sense.

Order n+1 (= number of coefficients)Degree n

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Polynomials AdditionLet we have two polynomials:

152.15.23 1231 ++−= xxxy

The sum of polynomial y1 and y2 can be calculated by sum of the coefficients of two polynomials:the coefficient vectors must be the same length.

181.28.15 1232

1

−++= xxxy

y

p1=[3 -2 5 1 2 15];p1=[3, 2.5, 1.2, 15];

p2=[5, 1.8, 2.1, -18];

sum=p1+p2sum =

8.0000 -0.7000 3.3000 -3.00006

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Polynomials MultiplicationLet we have two polynomials:

152.15.23 1231 ++−= xxxy

The product of polynomial y1 and y2 can be calculated by using 'conv' commend..

181.28.15 1232

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−++= xxxy

y

p1=[3, -2.5, 1.2, 15];

p2=[5, 1.8, 2.1, -18];

prod= conv (p1,p2)prod=

15.0000 -7.1000 7.8000 17.9100 74.5200 9.9000 -270.00007

Polynomials DivisionLet we have two polynomials:

152.15.23 1231 ++−= xxxy

The division of polynomial y1 and y2 can be calculated by using 'deconv' commend..

181.28.15 1232

1

−++= xxxy

y

p1=[3 -2 5 1 2 15];p1=[3, 2.5, 1.2, 15];

p2=[5, 1.8, 2.1, -18];

[q,r]= deconv (p1,p2)q= r=

0.6000 -0.0000 -3.5800 -0.0600 25.8000

q is the quotient polynomial coefficient, and r is the remainder polynomial coefficient.

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Polynomials DerivativesLet we have two polynomials:

152.15.23 1231 ++−= xxxy

The derivation of polynomial y1 or y2 can be calculated by using 'polyder' commend..

181.28.15 1232

1

−++= xxxy

y

p1=[3 -2 5 1 2 15];p1=[3, 2.5, 1.2, 15];

p2=[5, 1.8, 2.1, -18];

d= polyder(p1)d=

9.0000 -5.0000 1.20009

Roots of Nonlinear equation

Solutions of the nonlinear equations, can be found using Newton’s methodcan be found using Newton s method

)(xf)f(xx=x n

n+n '1 −)(xf n

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Roots of Nonlinear equation

Find the solution of the following non‐linear i i h i h l ( 8)equation, with in the tolerance (108)

)f(x

)3sin(5)( 2304

−+− xe=xf x

)(xf)f(xx=x

n

nn+n '1 − Tolxx nn ≤−+1

)3cos(104)(' 2303 4

−+− xxex=xf x

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Newton Method% Newton's method for solving equationsclc;clear allclc;clear allxo = 2;tol = 1e-3;iter = 0; done = 0;while ~ done,iter = iter + 1;f=exp(xo^4-30)+5*sin(xo^2-3);f=exp(xo^4-30)+5*sin(xo^2-3);fd=4*xo^3*exp(xo^4-30)+10*xo*cos(xo^2-3);x = xo - f / fd;done=( abs(x-xo)<tol );xo=x

end

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5y=exp(x.4-30)+5*sin(x.2-3)

Newton Method

% ploting function y=exp(x.^4-30)+5*sin(x.^2-3)

x=linspace(-2,2);

y=exp(x.^4-30)+5*sin(x.^2-3);

plot(x,y,'r.-')-1

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1

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y-ax

ies

-1.7321 1.7321

xlabel('x-axies');

ylabel('y-axies');

title('y=exp(x.^4-30)+5*sin(x.^2-3)');

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2-5

-4

-3

-2

x-axies

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function [x,iter] = newton(xo,tol,maxit)% [x,iter] = NEWTON(xo,tol,maxit) % Newton's method for solving equations

newton.mNewton Method

% Newton s method for solving equations% x: the root, n: iteration number% xo:first guess for the root% tol: tolerance, maxit: maximum iterationiter = 0; done=0;while ~done,iter = iter + 1;f=exp(xo^4-30)+5*sin(xo^2-3);fd ( ) ( )

>> [x,iter]=newton(2,1e-3,10)

fd=4*xo^3*exp(xo^4-30)+10*xo*cos(xo^2-3);x = xo - f / fd;done=(iter>=maxit) | ( abs(x-xo)<tol );xo=x;

end

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Find the root of:

f(x)= x3 – 3x2 + 3x – 1 = 0

Polynomial Roots: Newton’s Method

f’(x)=3x2 – 6x + 3

Answer (root) found after n iterations15

% Newton's method for solving equations

clc;clear all

xo 3;

Matlab Implementation: Newton’s Method

xo = 3;

tol = 1e-3;

iter = 0;

done = 0;

while ~ done,

iter = iter + 1;

f=xo^3 3*xo^2+3*xo 1;f=xo^3-3*xo^2+3*xo-1;

fd=3*xo^2-6*xo+3;

x = xo - f / fd;

done=( abs(x-xo)<tol );

xo=x

end16

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x=linspace(‐2,4)  

y=(x ^3‐3 *x ^2+3*x‐1);

Polynomial Roots

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y=(x. 3‐3. x. 2+3 x‐1);

plot(x,y,'.r‐','LineWidth',2);

roots([1 ‐3 3 ‐1])

-2 -1 0 1 2 3 4-30

-20

-10

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Polynomial RootsDraw graph of polynomial:

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p = [1 0 -9]; x = -10:0.01:10; y=polyval(p,x);plot(x,y)

id

-10 -8 -6 -4 -2 0 2 4 6 8 10-20

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Note that the Roots are +3 and - 3

grid on

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Polynomial RootsThe roots function can be used to calculates the roots of a polynomial.

p = [1 0 -9]; r=roots(p)

r = 3 - 3

By convention, MATLAB stores roots in column vectors. The function polyreturns the polynomial coefficients given the roots.

p2 = poly(r)p p y( )p2 = 1 0 -9

poly and roots are inverse functions, up to ordering, scaling, and roundofferror.

Find Roots roots(p) returns the roots of the polynomial p in column vector form.

Find Polynomials poly(r) returns the coefficient vector of the polynomial having roots r.19

Systems of Linear Equations• Recall solution of simple Systems of Linear

Equations.

• MATLAB’s built-in matrix capabilities allow for easy solutions as well.

General Approach

1. Rearrange the equation with all unknown quantities on the left-hand side and all known on the right-hand side.g

2. Write the equation in matrix form.

3. Solve the resulting equation.

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Systems of Linear EquationsGaussian Substitution:

3x + 2y – 1z = 10

3x + 2y – 1z = 10

-3x + 9y + 6z = 15

+ _______________

11y + 5z = 25 11y + 5z 25

-3x + 9y + 6z = 15

x - y - z = -1

11y + 5z = 25

-x + 3y + 2z = 5

x - y - z = -1

+ _______________

2y + z = 4

11y + 5z = 25

-10y - 5z = -20

+ _____________

y = 5

Etc…21

Systems of Linear Equations

• Create a matrix of the coefficients

• Create a column vector of the

clear all;clc

A=[3,2,-1;-1,3,2;1,-1,-1];

• Create a column vector of the values

• Use the inverse command to tell MATLAB to solve the system of equations.

B=[10;5;-1];

S1=inv(A)*BS1 =

‐2.0000

5.0000

‐6.0000 22

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Systems of Linear Equations

• Create a matrix of the coefficients

• Create a column vector of the

clear all;clc

A=[3,2,-1;-1,3,2;1,-1,-1];

• Create a column vector of the values

• Use the left division ( \ ) operator to tell MATLAB to solve the system of equations.

B=[10;5;-1];

S2=A\BS2 =

‐2.0000

5.0000

‐6.0000 23

Systems of Linear Equations

• Create a matrix of the coefficients

• Create a column vector of the

clear all;clc

A=[3,2,-1;-1,3,2;1,-1,-1];

• Create a column vector of the values

• Use rref (reduced row echelon form) function to solve.

B=[10;5;-1];

C=[A B];

S3=rref(C)

S3 =

1     0     0    ‐2

0     1     0     5

0     0     1    ‐6 24

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RREF   Reduced row echelon formR = RREF(A) produces the reduced row echelon form of A.

[R,jb] = RREF(A,TOL) uses the given tolerance in the rank tests.

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