Numerical solutions of a nonlinear equation with different independent values in matlab

Question

For example, if I have this function: g = t^3 - 5*t^2 + 2

And g = [3 4 6 2 9 10 17 1]

I would like to solve the equation for each g[i] and obtain the resulting t vector.

Answer 1

This might guide you:

>> syms t g %// define symbolic variables
>> y = t^3 - 5*t^2 + 2 - g; %// define y so that equation is: y=0
>> g_data = [3 4 6 2 9 10 17 1]; %// define g values
>> n = 1; %// choose first value. Or use a loop: for n = 1:numel(g_data)
>> s = solve(subs(y, g, g_data(n))) %// substitute g value and solve equation y=0
s =
25/(9*((108^(1/2)*527^(1/2))/108 + 277/54)^(1/3)) + ((108^(1/2)*527^(1/2))/108 + 277/54)^(1/3) + 5/3
 5/3 - ((108^(1/2)*527^(1/2))/108 + 277/54)^(1/3)/2 - 25/(18*((108^(1/2)*527^(1/2))/108 + 277/54)^(1/3)) - (3^(1/2)*(25/(9*((108^(1/2)*527^(1/2))/108 + 277/54)^(1/3)) - ((108^(1/2)*527^(1/2))/108 + 277/54)^(1/3))*i)/2
 5/3 - ((108^(1/2)*527^(1/2))/108 + 277/54)^(1/3)/2 - 25/(18*((108^(1/2)*527^(1/2))/108 + 277/54)^(1/3)) + (3^(1/2)*(25/(9*((108^(1/2)*527^(1/2))/108 + 277/54)^(1/3)) - ((108^(1/2)*527^(1/2))/108 + 277/54)^(1/3))*i)/2
>> double(s) %// show solutions as floating point values
ans =
  5.039377328113847                     
 -0.019688664056924 + 0.445027607060817i
 -0.019688664056924 - 0.445027607060817i