- #1

FrancescoMi

- 5

- 0

and psi' is the derivative of psi. I've written this code:

Matlab:

```
rho = 0.04; %In the image is lambda
c = 25;
eta = 0.02;
mu = 40;
sigma = 0.57;
alpha = -rho/eta;
syms p t
F = t^(-alpha-1)*exp(-t^2/2+((p-mu)/sigma*sqrt(2*eta))*t); %Funzione integranda non derivata
F_d = t^(-alpha-1)*(t*sqrt(2*eta))/sigma*exp(-t^2/2+((p-mu)/sigma*sqrt(2*eta))*t); %Funzione integranda derivata
D = (exp(((p-mu)/sigma*sqrt(2*eta))^2/4)/gamma(-alpha))*int(F,t,0,inf); %Funzione cilindrica non derivata
D_d = (exp((((p-mu)/sigma*sqrt(2*eta))^2)/4)/gamma(-alpha))*int(F_d,t,0,inf)+((eta*(p-mu))/(4*(sigma)^2))*(exp((((p-mu)/sigma*sqrt(2*eta))^2)/4)/gamma(-alpha))*int(F,t,0,inf); %Funzione cilindrica derivata
psi = exp((eta*(p-mu)^2)/(2*(sigma)^2))*D;
psi_d = (eta*(p-mu))/(sigma^2)*exp((eta*(p-mu)^2)/(2*(sigma)^2))*D + exp((eta*(p-mu)^2)/(2*(sigma)^2))*D_d; %Derivata di psi
Theta = (p-c)*psi_d-psi;
FreeBoundary = solve(Theta,p)
```

I've calculated the derivatives by myself, and I hope they are ok. What do you think about the code? It works only if rho>eta. Moreover for different values of sigma it give me some strange results.

Do you think the code is right and the derivatives are correct?