Root finder solver for cubic equation

2021 3 8
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更新时间：2021 3 8
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I have the following cubic equation in beta, knowing gamma, theta and M1 how can I find the roots of beta using MATLAB?
for example, knowing the following:
M1 = 8:3:20;
gamma = 1.4;
theta = 8;

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David Goodmanson
David Goodmanson 2021-3-8
Hi Lujain. take a look at the roots function which requires four coefficients, all of which you have, so you can solve for tan(beta)
Jamie Al
Jamie Al 2021-3-8
Thanks David! I will try using roots and see how it goes.
Jamie Al
Jamie Al 2021-3-8
I tried something like this, but I am not sure it's giving back correct roots
M1 = 8:3:20;
gamma = 1.4;
theta = 8; %in degrees
a = 1+((gamma-1)/2).*M1.^2.*tan(theta);
b = 1-M1.^2;
c = (1+((gamma-1)/2).* M1.^2).*tan(theta);
d = 1;
P = [a b c d];
bb = roots(P);
beta = atand(bb); %take inv of tan

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Walter Roberson
Walter Roberson 2021-3-8
Ran in:

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Ran in:
syms M1 gamma theta_degrees tanbeta_degrees
eqn = 1 + (gamma - 1)/2 * M1.^2 * tand(theta_degrees).*tanbeta_degrees.^3 + (1-M1.^2).*tanbeta_degrees.^2 + (1+(gamma+1)/2.*M1.^2);
solbeta = solve(eqn, tanbeta_degrees);
beta_degrees = atand(solbeta)
beta_degrees = 
M1_ = 8:3:20;
gamma_ = 1.4;
theta_degrees_ = 8; %in degrees
beta_values = subs(beta_degrees, {M1, gamma, theta_degrees}, {M1_, gamma_, theta_degrees_})
beta_values = 
beta_numeric = double(beta_values)
beta_numeric = 3×5
-47.7576 -47.4871 -47.3704 -47.3097 -47.2741 48.6681 48.3829 48.2600 48.1960 48.1586 88.3627 88.3749 88.3801 88.3828 88.3844
whos beta_numeric
Name Size Bytes Class Attributes beta_numeric 3x5 120 double
Here, the different columns correspond to the different M1 values, and the different rows correspond to the different roots of the cubic. Results are in degrees.

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