vanzzy

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6 years, 186 days

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These are replies submitted by vanzzy

@acer 

I see. But you have plotted both real and imaginary parts. How can I plot only the imaginary part of one root versus B, for different V's?

Also, I want to do this to V=0.8, V=0.9, V=0.99 for k=1 and k=5 in two different plots, but when I try to change your code, it doesn't work.

 

 

@Kitonum 

I see. Can I plot of one of the 4 roots for the same 'k' but different values of 'V', all in the same plot?

@Kitonum 

About the 'i', I used this, but it turned to 'I' when I copy/paste the code.

The function will have three roots, so how can I plot the imaginary part of 'w' as a function of 'B' for some values of 'V' and k'?

@Joe Riel 

Sorry! I will edit the post right now.

The equation is

f := proc (w, B, V, k) options operator, arrow; I*B*gm(V)^3*(w-k*V)^3+gm(V)^2*(w-k*V)^2-I*gm(V)^3*(w-k*V)*(w*V-k)^2-((1/3)*I)*B*gm(V)^3*(w*V-k)^2*(w-k*V)-(1/3)*gm(V)^2*(w*V-k)^2 end proc

where gm(V) is given by

gm := proc (V) options operator, arrow; 1/(1-V^2)^(1/2) end proc

I want to plot the imaginary part of 'w' as a function of 'B' for some discrete values of 'V' and 'k'. I tried using the tips you gave me:

F := proc (V, k) options operator, arrow; fsolve(f(w, B, V, k), complex) end proc

and used some regular 2d plot, but is says:

"Error, (in fsolve) {B, w} are in the equation, and are not solved for".

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