xy+y^5+3=0 My input is real x, from this implicit equation y will have both real and imag values for each x I want to plot a 3 D plot where z will represent imag y. Pleasehelp

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RAJAT 2023-5-23

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https://ww2.mathworks.cn/matlabcentral/answers/1971304-xy-y-5-3-0-my-input-is-real-x-from-this-implicit-equation-y-will-have-both-real-and-imag-values-for

评论： RAJAT 2023-12-15

Basically I want to plot a 3 D graph in which x axis will represent input values of x,, yaxis will represent real values of y and z axis wil represent imaginary values of y. As output Y will have both real and imaginary values

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Torsten 2023-5-23

编辑：Torsten 2023-5-23

You'll get 5 solutions for y given a value for x. Maybe 4 of these solutions for y have real and imaginary parts different from 0.

E.g. for x = x1 you may get

y1 = c1

y2 = c2r + 1i*c2i

y3 = c2r - 1i*c2i

y4 = c3r + 1i*c3i

y5 = c3r - 1i*c3i

where the c's are real numbers.

I can't understand what you want to plot here.

Dyuman Joshi 2023-5-23

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For each x, there will be 5 corresponding y values. How do you want to plot that?

x=rand;
roots([1 0 0 0 x 3])
ans = 
   1.0278 + 0.7884i
   1.0278 - 0.7884i
  -0.4376 + 1.1503i
  -0.4376 - 1.1503i
  -1.1805 + 0.0000i

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John D'Errico 2023-5-23

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编辑：John D'Errico 2023-5-23

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Simple enough. Just do exactly what you asked. Remember the roots of a 5th degree polynomial will in general have no algebraic solution, so we need to call roots.

I used a loop here, since roots is not vectorized, and although I could have written more sophisticated code that would do it all in one line, the code would be impossible to read.

nx = 100;

x = linspace(-10,10,nx)';

ysol = zeros(nx,5);

for n = 1:nx;

ysol(n,:) = roots([1 0 0 0 x(n) 3]).';

end

plot3(repmat(x,1,5),real(ysol),imag(ysol),'.')

grid on

box on

The color changes along those curves represents where two roots essentially switched branches. Remember that roots does not know in which order to return the roots, so we can see some confusion in the plot.