Symbolic differentiation for very long equation
2 views (last 30 days)
Show older comments
Hi All,
I would like to do partial differentition for a relative long function. The function is give as following:
ni = 1;
nt = 1.4;
R_eff = Eff_Reflection(ni,nt); %Reff represents the fraction of photons that is internally diffusely reflected at the boundary.
syms f(ua,us,g,R,rho);
D = 1 / 3 / (ua + (1-g) * us);
z0 = 1 / (ua + (1 - g) * us);
zb = (1 + R)/(1 - R) * 2 * D;
u_eff = sqrt(3 * ua * (ua + (1 - g) * us));
r1 = sqrt(z0^2 + rho^2);
r2 = sqrt((z0 + 2*zb)^2 + rho^2);
f(ua,us,g,R,rho) = z0 * (u_eff + 1/r1) * exp(-u_eff * r1) / r1^2 + (z0 + 2*zb) * (u_eff + 1/r2) * exp(-u_eff * r2) / r2^2;
Df = diff(f,ua);
The answer for running the code is:
(exp(-3^(1/2)*(ua*(ua - us*(g - 1)))^(1/2)*((1/(ua - us*(g - 1)) - (4*(R + 1))/(3*(ua - us*(g - 1))*(R - 1)))^2 + rho^2)^(1/2))*(1/(ua - us*(g - 1)) - (4*(R + 1))/(3*(ua - us*(g - 1))*(R - 1)))*((3^(1/2)*(2*ua - us*(g - 1)))/(2*(ua*(ua - us*(g - 1)))^(1/2)) + ((1/(ua - us*(g - 1)) - (4*(R + 1))/(3*(ua - us*(g - 1))*(R - 1)))*(1/(ua - us*(g - 1))^2 - (4*(R + 1))/(3*(ua - us*(g - 1))^2*(R - 1))))/(rho^2 + (1/(ua - us*(g - 1)) - (4*R + 4)/(3*(ua - us*(g - 1))*(R - 1)))^2)^(3/2)))/((1/(ua - us*(g - 1)) - (4*(R + 1))/(3*(ua - us*(g - 1))*(R - 1)))^2 + rho^2) - (exp(-3^(1/2)*(ua*(ua - us*(g - 1)))^(1/2)*((1/(ua - us*(g - 1)) - (4*(R + 1))/(3*(ua - us*(g - 1))*(R - 1)))^2 + rho^2)^(1/2))*(1/(ua - us*(g - 1))^2 - (4*(R + 1))/(3*(ua - us*(g - 1))^2*(R - 1)))*(1/((1/(ua - us*(g - 1)) - (4*(R + 1))/(3*(ua - us*(g - 1))*(R - 1)))^2 + rho^2)^(1/2) + 3^(1/2)*(ua*(ua - us*(g - 1)))^(1/2)))/((1/(ua - us*(g - 1)) - (4*(R + 1))/(3*(ua - us*(g - 1))*(R - 1)))^2 + rho^2) + (exp(-3^(1/2)*(1/(ua - us*(g - 1))^2 + rho^2)^(1/2)*(ua*(ua - us*(g - 1)))^(1/2))*(1/((1/(ua - us*(g - 1))^2 + rho^2)^(3/2)*(ua - us*(g - 1))^3) + (3^(1/2)*(2*ua - us*(g - 1)))/(2*(ua*(ua - us*(g - 1)))^(1/2))))/((1/(ua - us*(g - 1))^2 + rho^2)*(ua - us*(g - 1))) - (exp(-3^(1/2)*(1/(ua - us*(g - 1))^2 + rho^2)^(1/2)*(ua*(ua - us*(g - 1)))^(1/2))*(1/(1/(ua - us*(g - 1))^2 + rho^2)^(1/2) + 3^(1/2)*(ua*(ua - us*(g - 1)))^(1/2)))/((1/(ua - us*(g - 1))^2 + rho^2)*(ua - us*(g - 1))^2) + (2*exp(-3^(1/2)*(1/(ua - us*(g - 1))^2 + rho^2)^(1/2)*(ua*(ua - us*(g - 1)))^(1/2))*(1/(1/(ua - us*(g - 1))^2 + rho^2)^(1/2) + 3^(1/2)*(ua*(ua - us*(g - 1)))^(1/2)))/((1/(ua - us*(g - 1))^2 + rho^2)^2*(ua - us*(g - 1))^4) - (exp(-3^(1/2)*(1/(ua - us*(g - 1))^2 + rho^2)^(1/2)*(ua*(ua - us*(g - 1)))^(1/2))*(1/(1/(ua - us*(g - 1))^2 + rho^2)^(1/2) + 3^(1/2)*(ua*(ua - us*(g - 1)))^(1/2))*((3^(1/2)*(1/(ua - us*(g - 1))^2 + rho^2)^(1/2)*(2*ua - us*(g - 1)))/(2*(ua*(ua - us*(g - 1)))^(1/2)) - (3^(1/2)*(ua*(ua - us*(g - 1)))^(1/2))/((1/(ua - us*(g - 1))^2 + rho^2)^(1/2)*(ua - us*(g - 1))^3)))/((1/(ua - us*(g - 1))^2 + rho^2)*(ua - us*(g - 1))) - (exp(-3^(1/2)*(ua*(ua - us*(g - 1)))^(1/2)*((1/(ua - us*(g - 1)) - (4*(R + 1))/(3*(ua - us*(g - 1))*(R - 1)))^2 + rho^2)^(1/2))*(1/(ua - us*(g - 1)) - (4*(R + 1))/(3*(ua - us*(g - 1))*(R - 1)))*((3^(1/2)*((1/(ua - us*(g - 1)) - (4*(R + 1))/(3*(ua - us*(g - 1))*(R - 1)))^2 + rho^2)^(1/2)*(2*ua - us*(g - 1)))/(2*(ua*(ua - us*(g - 1)))^(1/2)) - (3^(1/2)*(1/(ua - us*(g - 1)) - (4*(R + 1))/(3*(ua - us*(g - 1))*(R - 1)))*(1/(ua - us*(g - 1))^2 - (4*(R + 1))/(3*(ua - us*(g - 1))^2*(R - 1)))*(ua*(ua - us*(g - 1)))^(1/2))/(rho^2 + (1/(ua - us*(g - 1)) - (4*R + 4)/(3*(ua - us*(g - 1))*(R - 1)))^2)^(1/2))*(1/((1/(ua - us*(g - 1)) - (4*(R + 1))/(3*(ua - us*(g - 1))*(R - 1)))^2 + rho^2)^(1/2) + 3^(1/2)*(ua*(ua - us*(g - 1)))^(1/2)))/((1/(ua - us*(g - 1)) - (4*(R + 1))/(3*(ua - us*(g - 1))*(R - 1)))^2 + rho^2) + (2*exp(-3^(1/2)*(ua*(ua - us*(g - 1)))^(1/2)*((1/(ua - us*(g - 1)) - (4*(R + 1))/(3*(ua - us*(g - 1))*(R - 1)))^2 + rho^2)^(1/2))*(1/(ua - us*(g - 1)) - (4*(R + 1))/(3*(ua - us*(g - 1))*(R - 1)))^2*(1/(ua - us*(g - 1))^2 - (4*(R + 1))/(3*(ua - us*(g - 1))^2*(R - 1)))*(1/((1/(ua - us*(g - 1)) - (4*(R + 1))/(3*(ua - us*(g - 1))*(R - 1)))^2 + rho^2)^(1/2) + 3^(1/2)*(ua*(ua - us*(g - 1)))^(1/2)))/(rho^2 + (1/(ua - us*(g - 1)) - (4*R + 4)/(3*(ua - us*(g - 1))*(R - 1)))^2)^2
The answer that matlab give is very very long, it the results trustable?
0 Comments
Accepted Answer
Torsten
on 18 Mar 2022
Here is a code to test MATLAB's answer:
syms ua us g R rho
D = 1 / sym('3') / (ua + (1-g) * us);
z0 = 1 / (ua + (1 - g) * us);
zb = (1 + R)/(1 - R) * sym('2') * D;
u_eff = sqrt(sym('3') * ua * (ua + (1 - g) * us));
r1 = sqrt(z0^2 + rho^2);
r2 = sqrt((z0 + 2*zb)^2 + rho^2);
f = z0 * (u_eff + 1/r1) * exp(-u_eff * r1) / r1^2 + (z0 + 2*zb) * (u_eff + 1/r2) * exp(-u_eff * r2) / r2^2;
Df = diff(f,ua)
fnum = subs(f,[us,g,R,rho],[1 2 3 4]);
Dfnum = subs(Df,[us,g,R,rho],[1 2 3 4]);
ua = 2:0.1:10;
fnum = matlabFunction(fnum);
dfnum = (fnum(ua+1e-6)-fnum(ua))*1e6;
Dfnum = matlabFunction(Dfnum);
Dfnum = Dfnum(ua);
plot(ua,[dfnum;Dfnum])
More Answers (0)
See Also
Categories
Find more on Symbolic Math Toolbox in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!