Ejercicios En Matlab

1. ENCONTRAR LAS RAICES DEL POLINOMIO EN MATLAB:

A.
>> pol=[1 -6 -12]

pol =

1 -6 -12

>> roots(pol)

ans =

7.5826
-1.5826

Donde se desprende: x1=7.5826 yx2=-1.5826

B.
>> pol=[3 4 10]

pol =

3 4 10

>> roots(pol)

ans =

-0.6667 + 1.6997i
-0.6667 - 1.6997i

Es decir: x1=-0.6667 + 1.6997i y x2=-0.6667 - 1.6997i

2. GRAFIQUELA SIGUIENTE TRAYECTORIA TEMPORAL EN MATLAB:

>> t=0:0.001:12
>> y=exp(-0.7*t).*(cos(3*t)+4*sin(4*t))+6
>> plot(t,y)

¿Es convergente o divergente?

ES CONVERGENTE A LA TRAYECTORIA

3.MINIMIZAR EN MATLAB:

4. HALLAR LAS SIGUIENTES INTEGRALES IMPROPIAS EN MATLAB:

a. >> syms x

>> int(x^(2),0,1)ans =

1/3

b. >> syms t

>> (1/(2*pi)^0.5)*int(exp(-0.5*t^2),-inf,0.85)

ans =7186705221432913/36028797018963968*2^(1/2)*pi^(1/2)+7186705221432913/36028797018963968*2^(1/2)*pi^(1/2)*erf(17/40*2^(1/2))

5. REPRESENTE GRÁFICAMENTE EN MATLAB:

>> x0=[1,1,1];
>> [x,fval]=fminunc(@myfun,x0)Warning: Gradient must be provided for trust-region method;
using line-search method instead.
> In fminunc at 247

Optimization terminated: relative infinity-norm of gradient less thanoptions.TolFun.

x =

1.0e-006 *

0.4212 -0.1979 0.0624

fval =

2.0000

Por condicion de primer orden:

Por lo tanto: x1=0.4212 ; x2=-0.1979 ; x3=0.0624

El valor de z=2

Porcondicion de primer orden:

exitflag =

1

output =

iterations: 10
funcCount: 44
stepsize: 1
firstorderopt: 1.5795e-006
algorithm: 'medium-scale:Quasi-Newton line search'
message: 'Optimization terminated: relative infinity-norm of gradient less than options.TolFun.'

grad =

1.0e-005 *

0.1580
-0.1222
0.0566...