% Computer lab 2 & 3
% Johan Larsson
% Max Niia

% Function
f = @(x) 54*x^6 + 45*x^5 - 102*x^4 - 69*x^3 + 35*x^2 + 16*x - 4;

% Plotting
figure
fplot(f, [-2 2])
axis([-2 2 -40 40])
grid on

% From the drawn graph we can estimate intervals which contains one root
% [-1.5, -1.2], [-0.8, -0.5], [0.1, 0.3], [0.4, 0.65], [1.0, 1.3]

% Calculating the roots and keeping track of number of
% steps taken to calculate root.
% Store roots in r(x) and steps in xi(x)
format long
[r1, xi1] = secant(-1.5, -1.2);
[r2, xi2] = secant(-0.8, -0.5);
[r3, xi3] = secant(0.1, 0.3);
[r4, xi4] = secant(0.4, 0.65);
[r5, xi5] = secant(1.0, 1.3);

% To check if the roots are linear they have to satisfy the equation
% f'(r) = 0
l2 = vpa(secant_linearity(r2)) % f'(r) ~0

% Calculating the estimated value of alpha to check whether it's
% super linear or not (page 61)
[a1, alphas1] = super_linearity(xi1, r1)
[a2, alphas2] = super_linearity(xi2, r2)
[a3, alphas3] = super_linearity(xi3, r3)
[a4, alphas4] = super_linearity(xi4, r4)
[a5, alphas5] = super_linearity(xi5, r5)

% Roots 1, 3, 4 and 5 are superlinear since their alpha corresponds to
% similar values of 1.62

error1 = generate_errors(xi1, r1);
error2 = generate_errors(xi2, r2);
error3 = generate_errors(xi3, r3);
error4 = generate_errors(xi4, r4);
error5 = generate_errors(xi5, r5);

% Drawing error2
figure
plot(error2);

% Ignoring the first and the last few values since convergence might not
% take hold immidiately.
% The second root diverges at values
% xi(1) - x(5), for values at indexes >5 it converges
% at a rate of E_(i + 1) = 0.617*E_(i) which makes it linear

% Verification that linearity holds
factor = [];
for i = 7:size(error2, 2) - 10
    factor(i - 5) = error2(i + 1)/error2(i);
end

c = 0;
for i = 1:size(factor, 2)
    c = c + factor(i);
end
c = c/size(factor, 2)