How to get the positive root of the cubic a*x^3+3*b*x^2+3*c*x+d=0 where a=1,c=b+2, d=-3 and b=.1:.1:2.
If rr is the root, find rr/(rr+1) for each b.
MATLAB: How to get the positive root of the cubic a*x^3+3*b*x^2+3*c*x+d=0
rootzeros
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You can probably improve on this and in terms of actual efficiency I am not sure it's better than what you have, but I am assuming you wanted something that makes it so you don't run a loop on all the matrix multiplies:
%%%% OLD %%%%
clear all;close all;%% Import the data file
M = importdata('6x300_input_data.txt');%% u,v,w are inputs taken from the data file that are each their own column
u = M.data(:,3);v = M.data(:,4);w = M.data(:,5);%% Time variable in seconds that is pulled from data
t_in = M.data(:,6);lambda = -60; %longitude in degrees
omega = 0.004; %Rate
phi = 30; %measurement in degrees
UVW = [u,v,w];%% Array to populate the series of A matrices that will be multiplied by
%each instance of UVW
A = arrayfun(@(t_in)[ -sind(lambda - omega*t_in) cosd(lambda - omega*t_in) 0; -sind(phi)*cosd(lambda-omega*t_in) -sind(phi)*sind(lambda - omega*t_in) cosd(phi); cosd(phi)*cosd(lambda - omega*t_in) cosd(phi)*sind(lambda - omega*t_in) sind(phi)],t_in,'UniformOutput',false);%% proper outputs that are desired but want to find a better way...
C = A{1,1}*UVW(1,:)'D = A{2,1}*UVW(2,:)'E = A{3,1}*UVW(3,:)'F = A{4,1}*UVW(4,:)'G = A{5,1}*UVW(5,:)'H = A{6,1}*UVW(6,:)'disp(C:H);%%%% NEW %%%%
%% General Idea:
a = A{2,1}(:);b = repmat(UVW(2,:),3,1);b = b(:);p = a.*b; % individual products
%% Implement:
A_hat = [ -sind(lambda - omega*t_in), ... cosd(lambda - omega*t_in), ... zeros(size(t_in)), ... -sind(phi)*cosd(lambda-omega*t_in), ... -sind(phi)*sind(lambda - omega*t_in), ... ones(size(t_in))*cosd(phi), ... cosd(phi)*cosd(lambda - omega*t_in), ... cosd(phi)*sind(lambda - omega*t_in), ... ones(size(t_in))*sind(phi)];B_hat = repmat(UVW,1,3);P = A_hat .* B_hat;%% I think differences may just be finite-math issues:
output = [sum(P(:,1:3),2), sum(P(:,4:6),2), sum(P(:,7:9),2)].';disp('output(:,1) - C:');disp(output(:,1) - C);disp('output(:,2) - D:');disp(output(:,2) - D);disp('output(:,3) - E:');disp(output(:,3) - E);disp('output(:,4) - F:');disp(output(:,4) - F);disp('output(:,5) - G:');disp(output(:,5) - G);disp('output(:,6) - H:');disp(output(:,6) - H); The eigenvalues of the matrix that are being calculated are correct and real. They might look like complex numbers due to the fact that the complex unit 'I' appears, but in fact they are not. This can be seen e.g. by plugging in real values for 'a', 'b' and 'c'.
Of course, the problem of realizing a complex number is a bit more complicated, since there are nested roots involved in the eigenvalues (implicitly contributing complex numbers that, when plugging in real values for 'a', 'b' and 'c', cancel with the more obvious appearances of the complex unit 'I' in the output and finally giving a real result).
In general, the MuPAD (MATLAB symbolic engine) library for Linear Algebra does not react to assumptions like 'assume(x, Type::Real)'. Such assumptions will not have any effect in the intermediate computational steps MuPAD performs. The reason why it does not react to properties is due to avoiding performance issues.
In general, you can try simplifying the results obtained from linear algebra computations E.g. use 'simplify' or 'Simplify' to make the results more elementary. Unfortunately, in this particular matrix, the simplifier cannot be used to completely remove the complex unit 'I' from the result.
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