- #1
fluidistic
Gold Member
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I am clueless. Here's the model I tried to follow:
With output
Which is 1)Define a function in terms of some variable(s).
2)Define another function in terms of some variable(s) and in terms of the function defined in 1).
3)Evaluate the function in 2) to see the explicit and eventually simplified form of that function.
I tried this with a more complicated function but the output is troublesome:
1)
2)
3)
with output
which reduces to
which is way too simple.
On my draft I've got a more "complex" function than this ultra reduction and there's absolutely no way to get what Maxima got. It's almost as if Maxima did not divide by Z(beta, etc) when evaluating 3).
Am I missing something? Or Maxima is bugged?
Code:
g(a):=a^2-1;
Code:
f(a,b):=a+g(a)-b;
Code:
f(a,b);
Code:
−b+a^2+a−1
2)Define another function in terms of some variable(s) and in terms of the function defined in 1).
3)Evaluate the function in 2) to see the explicit and eventually simplified form of that function.
I tried this with a more complicated function but the output is troublesome:
1)
Code:
Z(beta, delta, hbar, omega):=exp(-beta*hbar*omega)*((exp(-beta*delta)-1)/(1-exp(-2*beta*hbar*omega))+1/(1-exp(-beta*hbar*omega))^2);
Code:
P_n(beta, delta, hbar, omega,n):=(exp(-beta*(hbar*omega*(1+2*n)+delta)))/Z(beta, delta, hbar, omega);
Code:
P_n(beta, delta, hbar, omega,n);
Code:
e^(beta*hbar*omega−beta*(hbar*(2*n+1)*omega+delta))
Code:
e^(−2*beta*hbar*n*omega−beta*delta)
On my draft I've got a more "complex" function than this ultra reduction and there's absolutely no way to get what Maxima got. It's almost as if Maxima did not divide by Z(beta, etc) when evaluating 3).
Am I missing something? Or Maxima is bugged?