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I am trying to solve

$\dfrac{\rho\ ((1-b)\ \lambda _f\ \lambda _s\ P _A)^{\rho }}{b-1}+\dfrac{\rho \left(\dfrac{b\ \text{ps}\ x}{1-\text{ps}}\right)^{\rho }}{b}=0$

in terms of $b$,

I entered

   Solve[((((b ps x)/(1 - ps))^ρ ρ)/
     b + ((-(-1 + b) PA λf λs)^ρ ρ)/(-1 + b)) == 0, b] // Simplify

but Solve function results to this message

!Solve::nsmet: This system cannot be solved with the methods available to Solve. >>{1]

How can I solve this equation?

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    $\begingroup$ Please post the actual Mathematica code you have used. You can use the "edit" link below your question to update it. $\endgroup$ Commented Sep 22, 2012 at 7:01
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    $\begingroup$ >Welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Read the FAQs! 3) When you see good Q&A, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. ALSO, remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign. Please consider changing your user name to something more rememberable. $\endgroup$ Commented Sep 22, 2012 at 7:03
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    $\begingroup$ If you simplify by hand a bit your equation (removing irrelevant bits relative to b): eqn = (b ((ps x)/(1 - ps))^[Rho] + (1 - b) (P[CapitalAlpha] [Lambda]f [Lambda]s)^[Rho]) == 0; Solve[eqn, b] works; $\endgroup$
    – chris
    Commented Sep 22, 2012 at 7:30
  • $\begingroup$ Thank you all very much, I am new to this kind of site, I registered today because I am having problems with mathematica, your advices were really helpful. I will try again $\endgroup$
    – Jones
    Commented Sep 22, 2012 at 7:56
  • $\begingroup$ @Chris, you could give that as an answer... $\endgroup$ Commented Sep 22, 2012 at 23:22

1 Answer 1

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Your equation needs such a massive amounts of assumptions that I can imagine Mathematica can't find the solution. Here is my rough manual derivation in Mathematica's TraditionalForm side-stepping many of the intricacies that will prevent Mathematica from solving this (for instance, I assumed $\rho\neq0$). A numerical test in Mathematica follows the derivation.

Mathematica graphics

Table[((((b ps x)/(1 - ps))^\[Rho] \[Rho])/
        b + ((-(-1 + b) PA \[Lambda]f \[Lambda]s)^\[Rho] \[Rho])/(-1 + b)) 
  /. b -> 1/(1 + ((\[Lambda]f*\[Lambda]s*PA (1 - ps))/(ps*x))^(\[Rho]/(1 - \[Rho]))) 
  /. {ps -> Random[],x -> Random[], \[Rho] -> Random[], 
      \[Lambda]f -> Random[], \[Lambda]s -> Random[], PA -> Random[]}, 
  {100}
] // Chop // Quiet

{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, Indeterminate, 0, Indeterminate, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, Indeterminate, 0, 0, -6.9383157*10^-9, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1.196618005*10^-8, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1.205482381*10^-9, Indeterminate, 0, Indeterminate, 0, 0, 0, 0, Indeterminate, 0, Indeterminate, 0, 0, 0, Indeterminate, 0, 0, 0}

The Indeterminates stem from division by (near) zero, where the Random function generates a pole.

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  • $\begingroup$ You also assumed $\rho\neq1$ obviously. $\endgroup$
    – chris
    Commented Sep 23, 2012 at 14:18
  • $\begingroup$ @chris Yeah, I wrote "for instance, I assumed $\rho\neq0$" as an obvious example of what to exclude. There are more if you like: \[Rho] > 0 && b > 0 && \[Rho] != 1 && b != 0 && b != 1 && b > 0 && ps != 1 && x != 1 && ps != 0 && ps > 0 && x > 0 && PA > 0 && \[Lambda]f > 0 && \[Lambda]s > 0. I used these in Assuming, but that didn't help much. $\endgroup$ Commented Sep 23, 2012 at 14:22
  • $\begingroup$ Quite a few indeed :-) $\endgroup$
    – chris
    Commented Sep 23, 2012 at 14:23

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