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



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




f0[l0, l1] = 1/Sqrt[(λ0 + μ0)/λ0]
f1[l0, l1] = 1/Sqrt[(λ0 + μ0 - 1)/λ0]
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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. – b.gatessucks Aug 10 '13 at 14:00
Can f0 and f1 contain lambdas and mus in addition to l0 and l1? – David Park Aug 10 '13 at 14:52
@DavidPark should contain only $l0$ and $l1$ – Seyhmus Güngören Aug 10 '13 at 20:39
up vote 3 down vote accepted

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.

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