Calculating roots of an equation in Matlab.
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I am trying to calculate switching points. To do this I need to calculate the root of this equation (theta). To do this I have tried the code below.
a = 2
kepa = 3/13
lambda = 9
b = -log(kepa)/lambda
syms theta a
g = 2*(normcdf(a) + (theta - 1) .* normcdf(a*(1-theta)) + 1/(a*sqrt(2*pi)) * (exp(-(a.^2)/2) - ...
exp(-((a.*(1-theta)).^2)/2))) - theta == b;
soltheta = solve(g, theta)
This outputs:
soltheta =
Empty sym: 0-by-1.
I am not sure why this is the case? I know a solution exists and is around 0.175. How do I get this to output a solution for theta? Any help will be greatly appreciated, thank you.
채택된 답변
Dyuman Joshi
2022년 8월 15일
편집: Dyuman Joshi
2022년 8월 16일
Some slight tweaks
I used vpasolve cause symbolic solver will give an error and will return the answer using vpasolve only.
Defining a variable and then declaring it as a syms variable will overwrite it's value.
a = 2;
kepa = 3/13;
lambda = 9;
b = -log(kepa)/lambda;
syms theta
g = 2*(normcdf(a) + (theta - 1) .* normcdf(a*(1-theta)) + 1/(a*sqrt(2*pi)) * (exp(-(a.^2)/2) - ...
exp(-((a.*(1-theta)).^2)/2))) - theta == b;
soltheta = vpasolve(g, theta)
Edit - Note that there are 2 solutions to the equation and only the first solution is obtained here (closer to 0).
추가 답변 (2개)
Star Strider
2022년 8월 15일
When you delcared ‘a’ as symbolic, you cleared its numeric value.
Try this —
a = 2
kepa = 3/13
lambda = 9
b = -log(kepa)/lambda
syms theta
g = 2*(normcdf(a) + (theta - 1) .* normcdf(a*(1-theta)) + 1/(a*sqrt(2*pi)) * (exp(-(a.^2)/2) - ...
exp(-((a.*(1-theta)).^2)/2))) - theta;
soltheta(1,:) = vpasolve(g == b, -1);
soltheta(2,:) = vpasolve(g == b, 2)
format long
numtheta = double(soltheta)
figure
fplot(g, [-1 3])
yline(b)
grid
.
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