Calculating roots of an equation in Matlab.

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.

[Star Strider]

There are two roots.

Answer 1

[Dyuman Joshi]

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).

[Charles]

Thank you!!!

Answer 2

[Star Strider]

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

.

Answer 3

[Torsten]

Change

syms theta a

to

syms theta