Basis of Mathematical Modeling

LECTURE 3

Numerical Analysis with MATLAB

Dr. N.K. Sakhnenko, PhD, Professor Associate

Outline

Finding roots with MATLAB Interpolation Numerical Integration Optimization Problems: Minimizing Functions of One

Variable; Minimizing Functions of Several Variables; Linear Programming

Root finding

Computing roots of the polynomials

Polynomials are represented in MATLAB by their coefficients in the descending order of powers.For instance, the cubic polynomial p(x) = 3x3 + 2x2 + 1 is represented asp = [3 2 0 1]. Its roots can be found using function roots.

>> format short>> r = roots(p)r = -1.0000 0.1667 + 0.5528i 0.1667 - 0.5528i

Let now f be an one variable function f=f(x). MATLAB function fzero computes a zero of the function f using user supplied initial guess.In the following example let f(x) = cos(x) – x. First we define a function y = f1(x)

function y = f1(x)y = cos(x) - x;

To compute its zero we use MATLAB function fzero>> format long>>r = fzero('f1', 0.5)r =0.73908513321516

Name of the function whose zero is computed is entered as a string. Second argument of function fzero is the initial approximation of r.

Finding zeros using MATLAB function fzero

f=0

Function fzeroA user can enter a two-element vector that designates a starting interval. In our example we choose [ 0 1] as a starting interval to obtainr = fzero('f1', [0 1])r =0.73908513321516

One can check last result using function fevalerr = feval('f1', r)err =0

By adding the third input parameter tol you can force MATLAB to compute the zero of a function with the relative error tolerance tol. In our example we let tol = 10-3 to obtainrt = fzero('f1', .5, 1e-3)rt =0.73886572291538

Function fzero

0)cos()sin( 2 xxx

function f = myf(x)f = sin(x) - х.^2.*cos(x);

>> x1=fzero ('myf', -5)x1 = -4.7566

>> x2=fzero ('myf', 0.1)x2 = 7.9161e-019

>> x3=fzero ('myf', 5)x3 = 4.6665

f=0f=0 f=0

Systems of nonlinear equations

1 1 2

2 1 2

1 2

( , ,... ) 0

( , ,... ) 0

...

( , ,... ) 0

n

n

n n

F x x x

F x x x

F x x x

MATLAB function fsolve computes solution of this system as well.

For more information help fsolve

Systems of nonlinear equations1 2 1

1 2 1

(2 ) cos

2 cos

x

x

x x x e

x x x e

function F=mysys(x)F(1)=x(1)*(2-x(2))-cos(x(1))*exp(x(2));F(2)=2+x(1)-x(2)-cos(x(1))-exp(x(2));

>> [x,f]=fsolve('mysys', [0 0])

x = 0.7391 0.4429

f = 1.0e-011 *

-0.4702 -0.6404

InterpolationGiven set of n+1 points xk , yk, k=(0…n), with x0 < x1 < … < xn, find a function f(x) whose graph interpolates the data points, i.e., f(xk) = yk, for k = 0, 1, …, n.The general form of the function interp1 is yi = interp1(x, y, xi, method), where the vectors x and y are the vectors holding the x- and the y- coordinates of points to be interpolated, respectively, xi is a vector holding points of evaluation, i.e., yi = f(xi) and method is an optional string specifying an interpolation method. The following methods work with the function interp1:

Nearest neighbor interpolation, method = 'nearest'. Produces a locally piecewise constant interpolant.

Linear interpolation method = 'linear'. Produces a piecewise linear interpolant.

Cubic spline interpolation, method = 'spline'. Produces a cubic spline interpolant.

Cubic interpolation, method = 'cubic'. Produces a piecewise cubic polynomial.

InterpolationIn this example, the following points (xk, yk) = (k/5, sin(2k)), k = 0, 1, … , 5,x = 0:pi/5:pi;y = sin(2.*x);are interpolated using the 'cubic' method of interpolation. The interpolant isevaluated at the following pointsxi = 0:pi/100:pi;yi = interp1(x, y, xi, 'cubic');Points of interpolation together with the resulting interpolant are displayed belowplot(x, y, 'o', xi, yi), title('Cubic interpolant of y = sin(2x)')

Numeric Integration

22

0

0

sin cos 1xdx x

Two MATLAB functions quad('f ', a, b, tol) and quad8('f ', a, b, tol) are designed for numerical integration of the univariate functions. The input parameter 'f' is a string containing the name of the function to be integrated from a to b. The fourth input parameter tol is optional and specifies user's chosen relative error in the computed integral.

>> quad('sin',0,pi/2)ans = 0.99999999787312>> quad('sin',0,pi/2,0.0000001)ans = 0.99999999946896

The default accuracy is 0.001

function f=fint(x) f=exp(-x.^2);

>> quad('fint',0,1)

ans =

0.74682418072642

21

0

xe dx

Numeric Integration of the bivariate functions

Function dblquad computes a numerical approximation of the double integral

( , )D

f x y dxdy

where D = {(x, y): a < x< b, c< y <d} is the domain of integration. Syntax of the function is dblquad (fun, a, b, c, d, tol), where the parameter tol has the same meaning as in the function quad.

Numeric Integration of the bivariate functions

The M-file esin is used to evaluate function f:

function z = esin(x,y);z = exp(-x.*y).*sin(x.*y);

>>result = dblquad('esin', -1, 1, 0, 1)result =

-0.22176646183245

Let ( , ) sin( ), 1 1, 0 1.xyf x y e xy x y

Optimization

The problem of optimality criterion finding is in a base of most managing protocols in modern telecom technologies.

The optimization problem is the problem of finding the best solution from all feasible solutions (from Wikipedia, the free encyclopedia)

This may correspond to maximizing profit in a company, or minimizing loss in a conflict. For Telecom, for example, it is necessary to minimize middle delivery of packages in a network; to maximize the productivity of telecom system; to minimize the cost of the using of resources of networks; to maximize a profit of the sale of telecommunications services.

Minimizing Functions of One Variable

For a given mathematical function of a single variable coded in an M-file, you can use the fminbnd function to find a local minimum of the function in a given interval. If one needs a maximum he should to put minus before the function.

[x,fval] = fminbnd(@myfun,x1,x2) returns value of x in which the function myfun(x) has a minimum on the interval x1<x<x2 and returns fval, that is the value of the function in the minimum.

function f = myfun(x)f = ... % the function under consideration

For minimizing one can use the following functions[x,fval] = fminsearch(@myfun,x0)[x,fval] = fminunc(@myfun, x0)where x0 is initial guess.

Minimizing Functions of One Variable

Find the minimum of the function f(x)=x3-2x-5 on the interval from 0 to 2

clear all;fplot('fun1',[0,2]); grid on; title('fun1'); xlabel('x'); ylabel('f(x)');[x,fval]=fminbnd(@fun1,0,2)% fun1.mfunction f = fun1(x)f=x^3-2*x-5

Solution: x=0.8165; fval=-6.0887

Minimizing Functions of One VariableFor minimizing problem one can use the following functions[x,fval] = fminsearch(@myfun,x0)[x,fval] = fminunc(@myfun, x0)where x0 is initial guess. E.g., Find the minimum of the function f(x)=arctg(x3-2x-5) on the interval from -2 to 2

clear all;fplot('fun2',[-2,2]); grid on; title('fun2'); xlabel('x'); ylabel('f(x)');

[x,fval]=fminunc(@fun2,0)Solution: x=0.8165; fval=-1.4080[x,fval]=fminunc(@fun2,-1)Solution: x =-2; fval=-1.5708% fun2.mfunction f=fun2(x)f=atan(x^3-2*x-5);

Minimizing Functions of Several Variables

The fminsearch function minimizes a function of several variables.

Syntaxx = fminsearch(fun,x0)x = fminsearch(fun,x0,options)[x,fval] = fminsearch(...)

Description fminsearch finds the minimum of a scalar function of several variables, starting at an initial estimate. This is generally referred to as unconstrained nonlinear optimization. x = fminsearch(fun,x0) starts at the point x0 and finds a local minimum x of the function described in fun. x0 can be a scalar, vector, or matrix.

Minimizing Functions of Several Variables

Find the minimum of the function f(x)= 3x2 + 2xy + y2+5 near the point [20, -20]

clear all;x0 = [20,-20];[x,fval] = fminunc(@fun3,x0)% fun3.mfunction f = fun3(x)f = 3*x(1)^2 + 2*x(1)*x(2) + x(2)^2+5;

Linear Programming

1 1 2 2

11 1 12 2 1 1

1 1 2 2

1 2

minimize ( ) ...

subject to

...

...

... ,

and 0, 0,..., 0.

n n

n n

m m mn n m

n

f x c x c x c x

a x a x a x b

a x a x a x b

x x x

A linear programming problem (or LP in brief) is any decision problem of the form

Linear Programming

First, enter the coefficients f = [-5; -4; -6]A = [1 -1 1 3 2 4 3 2 0];b = [20; 42; 30];lb = zeros(3,1);Next, call a linear programming routine:

[x,fval] = linprog(f,A,b,[],[],lb)Optimization terminated successfully.x = 0.0000 15.0000 3.0000fval = -78.0000

1 2

1 2 3

1 2 3

1 2

1 2 3

Find that minimize ( ) 5 4 6

subject to

20

3 2 4 42

3 2 30,

and 0, 0, 0.

nx f x x x x

x x x

x x x

x x

x x x