How to convert this function of one variable into MATLAB code?
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Hi I'm trying to convert this function
into MATLAB code.
into MATLAB code.I'm fairly new to understanding the syntax so maybe there's an obvious error to someone experienced.
I think I'm close with this but I'm expecting the function to pass through the x-axis.
This is my current code.
Many thanks!
a1 = 0.5;
n = 4;
S = 0.5;
K = S^n;
x1 = linspace(0,3);
x2 = ((K^2.*(1-(x1))+K*(1+(a1)).*(x1).^4-K.*(x1).^5)./(K.*(x1)+(x1).^5-(a1).*(x1).^4)).^(1/4);
plot(x1,x2, 'g')
Antworten (2)
You have done a good job, but there are a couple of small (imperative) points to consider - to take out real and imaginary parts of x2.
clc; clearvars
a1 = 0.5;
n = 4;
S = 0.5;
K = S^n;
x1 = linspace(0,3);
x2 = ((K^2*(1-(x1))+K*(1+(a1)).*x1.^4-K*x1.^5)./(K*x1+x1.^5-a1.*x1.^4)).^(1/4);
plot(x1,real(x2), 'r-x', x1, imag(x2), 'b'), grid on; legend('RE(x_2)', 'IM(x_2)', 'location', 'best')
xlabel('x_1'); ylabel('x_2')
1 Kommentar
Your syntax looks good.
I am concerned that the difference in starting value, 1.4 vs 1.2. Let's check,
a1 = 0.5;
n = 4;
S = 0.5;
K = S^n;
syms x1
x2 = ((K^2*(1-(x1))+K*(1+(a1)).*x1.^4-K*x1.^5)./(K*x1+x1.^5-a1.*x1.^4)).^(1/4);
limit(x2, x1, 0)
vpa(solve(x2==1.4,x1))
vpa(solve(x2==1.2,x1))
So possibly the other one just plotted more densely.
fplot([real(x2),imag(x2)], [0 3])
ax = gca;
ax.XAxis.MinorTick = 'on';
ax.YAxis.MinorTick = 'on';
ax.XAxis.MinorTickValues = 0:.5/5:3;
ax.YAxis.MinorTickValues = 0:.2/5:1.4;
grid on
ax.XMinorGrid = 'on';
ax.YMinorGrid = 'on';
Looks okay. I think you were just seeing an artifact of not plotting densely enough.
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