Homework #11

13.7

For this problem I first graphed he equation to see if there was a maximum for that range of x, and there was so I implemented the Golden-Section Search and reached the convergence of the true value of 0.8408 at x= -0.3472

Matlab:

%Graph that shows theres a maximum between -2:1

x = (-2:0.1:1);

f = @(x) -x.^4 - 2*x.^3 - 8*x.^2 - 5*x;

plot(x,f(x));

%plot(x,f_x(x));

xl = -2;

xu = 1;

 d = ((sqrt(5) - 1)/2)*(xu-xl);

 x1 = xl + d;

 x2 = xu - d;

 sol= [xl f(xl) x2 f(x2) x1 f(x1) xu f(xu) d]

 for iter = 1:13

 if f(x2) > f(x1)

    xu = x1;

    x1 = x2;

    d = ((sqrt(5) - 1)/2)*(xu-xl);

    x2 = xu - d;

 end

 if f(x1) > f(x2)

    xl = x2;

    x2 = x1;

    d = ((sqrt(5) - 1)/2)*(xu-xl);

    x1 = xl + d; 

 end

 sol= [xl f(xl) x2 f(x2) x1 f(x1) xu f(xu) ]

 end

 hold on;

 plot(x2,f(x2),'*r')

13.8

(a) for this problem I used the some code you provided us and altered it to work with this problem. 

Output:

GoldenRatio = -0.3472    0.8408

Parabolic = -0.3473    0.8408

Newton =  -0.3473    0.8408

Matlab:

theta0 = 50;

x = [-2:0.01:1];

y = @(x) -x.^4 - 2*x.^3 - 8*x.^2 - 5*x;

figure( 1);

plot(x,y(x));

hold on;

% Golden Ration Search

xl= -2;

xu = 1;

for ii = 0:26,

d = (xu-xl)*(sqrt(5)-1)/2;

x1 = xl+d;

x2 = xu-d;

if(y(x1)>y(x2))

xl = x2;

else

xu = x1;

end;

plot(xu,y(xu),'*r')

plot(xl,y(xl),'*r')

end

plot(xu,y(xu),'*g')

GoldenRatio = [xu y(xu)]

figure(2)

x = [0:0.01:20];

plot(x,y(x));

hold on;

x0 = -2;

x1 = -1;

x2 = 1;

% parabolic interpolation

for ii = 0:6,

x3 = y(x0)*(x1^2-x2^2) + y(x1)*(x2^2-x0^2)+y(x2)*(x0^2-x1^2);

x3 = x3/(2*y(x0)*(x1-x2)+2*y(x1)*(x2-x0)+2*y(x2)*(x0-x1));

if(abs(x3-x1)<1.0e-6) break; end

if(y(x3)>y(x1))

x0 = x1;

x1 = x3;

% x2 = x2;

else

x2 = x1;

x1 = x3;

% x0 = x0; 

end;

plot(x3,y(x3),'*r')

end

plot(x3,y(x3),'*g')

Parabolic = [x3 y(x3)]

% Newtons

figure(3)

y = @(x) -x.^4 - 2*x.^3 - 8*x.^2 - 5*x;

yp = @(x) -4*x.^3 -6*x.^2 -16*x -5;

ypp = @(x) -4*(3*x.^2 + 3*x +4);

x = [0:0.01:20];

plot(x,y(x),x,yp(x))

hold

x = 0;

for ii = 0:6,

x = x-yp(x)/ypp(x);

plot(x,y(x),'*r')

end

Newton = [x y(x)]

plot(x,y(x),'*g')

13.21

For this problem I implemented the golden section search to calculate the time and altitude of peak elevation

Output:

GoldenRatio = 3.8699(time)  192.8537(elevation)

Matlab:

g = 9.81;

z0 = 100;

v0 = 55;

m = 80;

c= 15;

t = (0:0.01:10);

f = @(t) z0 + (m/c)*(v0+((m*g)/c))*(1-exp(-((c/m)*t))) - ((m*g)/c)*t;

figure(1);

plot(t,f(t));

hold on;

% Golden Ration Search

xl= 0;

xu = 10;

for ii = 0:10,

d = (xu-xl)*(sqrt(5)-1)/2;

x1 = xl+d;

x2 = xu-d;

if(f(x1) > f(x2))

xl = x2;

else

xu = x1;

end;

plot(xu,f(xu),'*r')

plot(xl,f(xl),'*r')

end

plot(xu,f(xu),'*g')

GoldenRatio = [xu f(xu)]

13.22