Problem with for loop plotting a graph

3 views (last 30 days)
Hi I'm really new to Matlab so you'll have to bare with me. I'm trying to plot a graph using the for function but to no avail.
I'm using the following code:
>> for a=2.0:-0.02:0.0
p=[f(a),g(a),h(a),r(a),t(a)] (where f,g,h,r,t are complicated functions of a I don't fancy writing down.)
R=roots(p)
C=imag(R)
c=C(1,1)
....
end
I'm looking to plot the complex part of the first root of each polynomial against each value of a or if its possible plotting the complex part of every root against a for every a on the same graph with different colours representing each root.
Thanks, sorry if the problem is ill described.
  2 Comments
madhan ravi
madhan ravi on 6 Dec 2018
Edited: madhan ravi on 6 Dec 2018
attach the script file and the relevent function files
Alexander Kimbley
Alexander Kimbley on 6 Dec 2018
Thanks, this is the entire polynomial.

Sign in to comment.

Accepted Answer

Star Strider
Star Strider on 6 Dec 2018
The Symbolic Math Toolbox is inefficient for this sort of problem.
Try this:
U = 1;
B = 0.1;
H = 0.5;
pfcn = @(a) [2*exp(2*a*(1-H)).*(1 - tanh(a*H)), -6*U*(1 + tanh(a*H)).*exp(2*a*(1 - H)) + 2*U*(1 - tanh(a*H)),(7*U^2 *(1 + tanh(a*H)) - 2*B^2 *tanh(a*H)).*exp(2*a*(1 - H)) - 5*U^2 *(1 - tanh(a*H)), exp(2*a*(1 - H)).*(2*U*B^2 *tanh(a*H) - 4*U^3 *(1 + tanh(a*H))) - (2*U*B^2 *tanh(a*H) - 4*U^3 *(1 + tanh(a*H))), U^2 *(U^2 *(1 + tanh(a*H)) - B^2 *tanh(a*H)).*exp(2*a*(1 - H)) - U^2 *(U^2 *(1 - tanh(a*H)) - B^2 *tanh(a*H))];
a=2.0:-0.02:0.0;
for k1 = 1:numel(a)
p = pfcn(a);
R = roots(p);
C(:,k1) = imag(R);
end
figure
plot(a, C, '.')
grid
figure
meshc(C)
grid on
That should tell you all you need to know about the imaginary roots of your function at each ‘a’ value.
  14 Comments
Alexander Kimbley
Alexander Kimbley on 7 Feb 2019
Thanks a lot! You've been such a godsend!
Star Strider
Star Strider on 7 Feb 2019
As always, my pleasure! Thank you!

Sign in to comment.

More Answers (0)

Categories

Find more on Linear Algebra in Help Center and File Exchange

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!