Plotting a polynomial problem

4 次查看(过去 30 天)
Trying to plot this polynomial for different values of A but no luck so far.
M=linspace(-10,10,100);
g=10;
for A=0.1:0.1:0.9
D=-(A^2-A+1/3)*(((A-1/2)*M*g-0.25*A^2)/(2*M*((A-1/2)^2+1)+0.5*(A^2-A+1/3)))-0.5*A^2
plot(M,D)
hold all
end
legend('A=0.1','A=0.2','A=0.3','A=0.4','A=0.5','A=0.6','A=0.7','A=0.8','A=0.9')

采纳的回答

Image Analyst
Image Analyst 2014-7-31
You forgot to use .* instead of *.
Corrected code:
workspace;
clc;
M=linspace(-10,10,100);
g=10;
lineColors = lines(9);
counter = 1;
for A=0.1:0.1:0.9
D = -(A^2-A+1/3) .* (((A-1/2)*M*g-0.25.*A^2) ./ (2*M*((A-1/2)^2+1)+0.5*(A^2-A+1/3)))-0.5*A^2
plot(M, D, 'LineWidth', 2, 'Color', lineColors(counter, :));
hold on
counter = counter + 1;
end
legend('A=0.1','A=0.2','A=0.3','A=0.4','A=0.5','A=0.6','A=0.7','A=0.8','A=0.9')
grid on;
% Enlarge figure to full screen.
set(gcf, 'Units', 'Normalized', 'OuterPosition', [0 0 1 1]);

更多回答(1 个)

Star Strider
Star Strider 2014-7-31
I don’t understand what ‘no luck so far’ means (I can only assume that a plot of several parallel lines of dots wasn’t your desired result), but after ‘vectorising’ your code (the line that assigns ‘D’) it produces a much different plot than the one your code originally produced:
M=linspace(-10,10,100);
g=10;
for A=0.1:0.1:0.9
D=-(A^2-A+1/3).*(((A-1/2).*M.*g-0.25.*A.^2)./(2*M.*((A-1/2).^2+1)+0.5.*(A.^2-A+1/3)))-0.5*A.^2;
plot(M,D)
hold all
end
legend('A=0.1','A=0.2','A=0.3','A=0.4','A=0.5','A=0.6','A=0.7','A=0.8','A=0.9')
‘Vectorising’ produces element-by-element operations, rather than MATLAB’s default matrix operations. It requires using the ‘dot operator’ in front of multiplication, division, and exponentiation operators: (.*), (./), (.^).
(The only exception is the transpose operator ('), where (.') produces a simple transpose rather than the default complex conjugate transpose. I mention that for completeness. It doesn’t relate to your code.)

类别

Help CenterFile Exchange 中查找有关 Loops and Conditional Statements 的更多信息

Community Treasure Hunt

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

Start Hunting!

Translated by