4
$\begingroup$

Given $$l_0=\sqrt{\frac{\lambda_1(\lambda_0+\mu_0)}{\lambda_0(\lambda_1+\mu_1-1)}}$$

and

$$l_1=\sqrt{\frac{\lambda_0(\lambda_1+\mu_1)}{\lambda_1(\lambda_0+\mu_0-1)}}$$

l0 = Sqrt[(λ1*(λ0 + μ0))/(λ0*(λ1 + μ1 - 1))]    
l1 = Sqrt[(λ0*(λ1 + μ1))/(λ1*(λ0 + μ0 - 1))]

Is there any way to ask mathematica to determine if some functions $f_0$ and $f_1$ exist such that

$$f_0(l_0,l_1)=\frac{1}{\sqrt{\frac{\lambda_0+\mu_0}{\lambda_0}}}$$

and

$$f_1(l_0,l_1)=\frac{1}{\sqrt{\frac{\lambda_0+\mu_0-1}{\lambda_0}}}$$

f0[l0, l1] = 1/Sqrt[(λ0 + μ0)/λ0]
f1[l0, l1] = 1/Sqrt[(λ0 + μ0 - 1)/λ0]
$\endgroup$
3
  • $\begingroup$ You can try sol1 = Solve[{l0 == 1/Sqrt[(\[Lambda]0 + \[Mu]0)/\[Lambda]0], l1 == 1/Sqrt[(\[Lambda]0 + \[Mu]0 - 1)/\[Lambda]0]}, {\[Lambda]1, \[Mu]1}] and Simplify[{l0, l1} /. sol1] for a start. $\endgroup$ Aug 10, 2013 at 14:00
  • $\begingroup$ Can f0 and f1 contain lambdas and mus in addition to l0 and l1? $\endgroup$
    – David Park
    Aug 10, 2013 at 14:52
  • $\begingroup$ @DavidPark should contain only $l0$ and $l1$ $\endgroup$ Aug 10, 2013 at 20:39

1 Answer 1

3
$\begingroup$

From the wording of your post, you seem to ask whether there are functions $f_0$ and $f_1$, which have only two slots f0[slot1,slot2], such that when composed with the functions $l_0$ and $l_1$, which are functions with 4 variables {λ0,λ1,μ0,μ1}, return the given values.

Let us use Mathematica to write the rhs of $f_0$ and $f_1$ in terms of $l_0$, $l_1$, $μ_0$, and $μ_1$. That is, let us move from the space where {λ0,λ1,μ0,μ1} are the variables to the space where {l0,l1,μ0,μ1} are the variables.

fromλToL = Solve[{l0 == Sqrt[(λ1*(λ0 + μ0))/(λ0*(λ1 + μ1 - 1))],
 l1 == Sqrt[(λ0*(λ1 + μ1))/(λ1*(λ0 + μ0 - 1))]}, {λ0, λ1}][[1]];
FullSimplify[{f0[l0, l1] == 1/Sqrt[(λ0 + μ0)/λ0], 
 f1[l0, l1] == 1/Sqrt[(λ0 + λ0 - 1)/λ0]} /. 
   fromλToL, Assumptions -> l0 > 0 && l1 > 0]

enter image description here

It becomes obvious now that $f_0$ and $f_1$ cannot be functions of just $l_0$ and $l_1$.

Another reasoning that leads to the same answer: If there were those two functions $f_0$ and $f_1$ then the composition with the two functions $l_0$ and $l_1$ yield functions of {λ0,λ1,μ0,μ1}. Take derivatives of $f_0(l_0(λ0,λ1,μ0,μ1),l_1(λ0,λ1,μ0,μ1))=rhs0$ and $f_1(l_1(λ0,λ1,μ0,μ1))=rhs1$ respect to λ0,λ1,μ0, and μ1. You will get 8 differential equations. You can then prove (by solving for the derivatives and then replacing those derivatives in the remaining equations) that such a system of equations has no solution.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.