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.

Accepted Answer

Dyuman Joshi
Dyuman Joshi on 15 Aug 2022
Edited: Dyuman Joshi on 16 Aug 2022
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)
soltheta = 
0.17508061864338710498913163970348
Edit - Note that there are 2 solutions to the equation and only the first solution is obtained here (closer to 0).

More Answers (2)

Star Strider
Star Strider on 15 Aug 2022
When you delcared ‘a’ as symbolic, you cleared its numeric value.
Try this —
a = 2
a = 2
kepa = 3/13
kepa = 0.2308
lambda = 9
lambda = 9
b = -log(kepa)/lambda
b = 0.1629
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)
soltheta = 
format long
numtheta = double(soltheta)
numtheta = 2×1
0.175080618643387 1.824919381356613
figure
fplot(g, [-1 3])
yline(b)
grid
.

Torsten
Torsten on 15 Aug 2022
Change
syms theta a
to
syms theta

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