02_roots_and_optimization

Repository for homework 2

Question 2

cat_cable = @(T) -35+T/10.*cosh(10./T*30)+30-T/10;
[root,fx,ea,iter]=falsepos(cat_cable,900,910,0.00001)

cat_cable = @(T) -35+T/10.*cosh(10./T*30)+30-T/10;
[root,fx,ea,iter]=bisect(cat_cable,908,909,0.00001)

cat_cable = @(T) -35+T/10.*cosh(10./T*30)+30-T/10;
[root,ea,iter]=mod_secant(cat_cable,900,910,0.00001)

| solver | initial guess | ea | number of iterations|
| --- | --- | --- | --- |
|falsepos   | 900, 910 |  0.00001| 4  |
|mod_secant | 900, 910 |  0.00001| 50 |
|bisect     | 900, 910 |  0.00001| 17 |

Question 3

f = @(x) (x-1)*exp(-(x-1)^2);
df = @(x) -(2*(x)^2-4*x+1)*exp(-(x-1)^2);
[root,ea,iter]=newtraph(f,df,3,.00001,5)

### divergence of Newton-Raphson method

| iteration | x_i | approx error |
| --- | --- | --- |
| 0 | 3 |    n/a    |
| 1 | 3 |   8.6957  |
| 2 | 3 |   6.8573  |
| 3 | 3 |   5.7348  |
| 4 | 3 |   4.9605  |
| 5 | 3 |   4.3873  |

### convergence of Newton-Raphson method

| iteration | x_i | approx error |
| --- | --- | --- |
| 0 |  1.2 |     n/a       |
| 1 |  1.2 |   22.1239     |
| 2 |  1.2 |    1.7402     |
| 3 |  1.2 |    0.0011     |
| 4 |  1.2 |    2.3315e-13 |
| 5 |  1.2 |    2.3315e-13 |

Question 4

a)

lj = @(x) 4*0.039*((2.394./x).^12-(2.394./x).^6);
[x,E,ea,its] = goldmin(lj,0,3)

x   = 2.6872
E   = -0.0390
ea  = 9.7092e-05
its = 27

b)

F = (0:0.0022/30:.0022)
lj = @(x) 4*0.039*((2.394./x).^12-(2.394./x).^6);
ET = @(dx) lj(dx)*(2.6872+dx)- F*dx;
[x,fx,ea,iter]=goldmin(ET,0,1)

x    =  1.0000
fx   =  1.0e+04
ea   =  8.6968e-05
iter =  27

c)

function SSE = sse_of_parabola(K,xdata,ydata)
    % calculate the sum of squares error for a parabola given a function, func, and xdata and ydata
    % output is SSE=sum of squares error
    K1=K(1);
    K2=K(2);
    y_function = K1*xdata+1/2*K2*xdata.^2;
    SSE = sum((ydata-y_function).^2);
end

d)

dx = zeros(1,50); % [in nm]
F_applied=linspace(0,0.0022,50)
[K,SSE_min]=fminsearch(@(K) sse_of_parabola(K,dx,F_applied),[1,1])

K =   1     1
SSE_min =   8.1490e-05

e)