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Mathematics Mathematica Equation Solving Wolfram Language

Why Mathematica Can't Solve this Equation

Songklod Rastapana

Posted 10 years ago

I have tries to solve an equation by using Solve[] to find pl as a function of other variables. However, it generates an Error! Can anyone help? In[780]:= ClearAll["Global`"] Clear[temp] temp = Solve[(((1 - pi)(pl - ph)/(1 - ph))^((1 - pi)\[Lambda]l))((pi(pl - ph)/(pl - 1))^(pi)) == ((pi(pl - ph)R/(ph(pl - 1)))^(pi(1 - \[Lambda]h)))(((pl - ph)(pi\[Lambda]h + \[Lambda]l - pi\[Lambda]l)/(\[Lambda]h(pl - 1) + \[Lambda]l(1 - ph)))^(pi\[Lambda]h + \[Lambda]l - pi\[Lambda]l)),{pl}] During evaluation of In[780]:= Solve::nsmet: This system cannot be solved with the methods available to Solve. >> Out[782]= Solve[(((1 - pi) (-ph + pl))/(1 - ph))^((1 - pi) \[Lambda]l) ((pi (-ph + pl))/(-1 + pl))^pi == ((pi (-ph + pl) R)/(ph (-1 + pl)))^(pi (1 - \[Lambda]h)) (((-ph + pl) (pi \[Lambda]h + \[Lambda]l - pi \[Lambda]l))/((-1 + pl) \[Lambda]h + (1 - ph) \[Lambda]l))^(pi \[Lambda]h + \[Lambda]l - pi \[Lambda]l), {pl}] Attachments:*

I have tries to solve an equation by using Solve[] to find pl as a function of other variables. However, it generates an Error! Can anyone help?

In[780]:= ClearAll["Global`*"]
Clear[temp]
temp = Solve[(((1 - pi)*(pl - ph)/(1 - ph))^((1 - pi)*\[Lambda]l))*((pi*(pl - ph)/(pl - 1))^(pi)) == ((pi*(pl - ph)*R/(ph*(pl - 1)))^(pi*(1 - \[Lambda]h)))*(((pl - ph)*(pi*\[Lambda]h + \[Lambda]l - pi*\[Lambda]l)/(\[Lambda]h*(pl - 1) + \[Lambda]l*(1 - ph)))^(pi*\[Lambda]h + \[Lambda]l - pi*\[Lambda]l)),{pl}]

During evaluation of In[780]:= Solve::nsmet: This system cannot be solved with the methods available to Solve. >>

Out[782]= Solve[(((1 - pi) (-ph + pl))/(1 - ph))^((1 - pi) \[Lambda]l) ((pi (-ph + pl))/(-1 + pl))^pi == ((pi (-ph + pl) R)/(ph (-1 + pl)))^(pi (1 - \[Lambda]h)) (((-ph + pl) (pi \[Lambda]h + \[Lambda]l - pi \[Lambda]l))/((-1 + pl) \[Lambda]h + (1 - ph) \[Lambda]l))^(pi \[Lambda]h + \[Lambda]l - pi \[Lambda]l), {pl}]

POSTED BY: Songklod Rastapana

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Bill Simpson

Posted 10 years ago

I doubt it.

POSTED BY: Bill Simpson

Songklod Rastapana

Posted 10 years ago

Thank you very much Bill. I am just wondering would it be possible to get closed-form solution in this case?

POSTED BY: Songklod Rastapana

Bill Simpson

Posted 10 years ago

Sometimes NMinimize of the Norm of an expression with some constraints NMinimize[{Norm[ (((1 - pi)(pl - ph)/(1 - ph))^((1 - pi)\[Lambda]l))((pi(pl - ph)/(pl - 1))^(pi)) - ((pi(pl - ph) R/(ph(pl - 1)))^(pi(1 - \[Lambda]h)))(((pl - ph)(pi\[Lambda]h + \[Lambda]l - pi\[Lambda]l)/ (\[Lambda]h(pl - 1) + \[Lambda]l(1 - ph)))^(pi\[Lambda]h + \[Lambda]l - pi\[Lambda]l))], 0<=pi<=1 && 0<=\[Lambda]h<=1 && 0<=\[Lambda]l<=1 && 1<=R<=2 && 0<=ph<=1 && 1<=pl<=2}, {pi, \[Lambda]h, \[Lambda]l, R, ph, pl}] can find solutions. In this case it also complains about some things that you might think carefully about.

Sometimes NMinimize of the Norm of an expression with some constraints

NMinimize[{Norm[
   (((1 - pi)*(pl - ph)/(1 - ph))^((1 - pi)*\[Lambda]l))*((pi*(pl - ph)/(pl - 1))^(pi)) - ((pi*(pl - ph)*
   R/(ph*(pl - 1)))^(pi*(1 - \[Lambda]h)))*(((pl - ph)*(pi*\[Lambda]h + \[Lambda]l - pi*\[Lambda]l)/
   (\[Lambda]h*(pl - 1) + \[Lambda]l*(1 - ph)))^(pi*\[Lambda]h + \[Lambda]l - pi*\[Lambda]l))], 
   0<=pi<=1 && 0<=\[Lambda]h<=1 && 0<=\[Lambda]l<=1 && 1<=R<=2 && 0<=ph<=1 && 1<=pl<=2},
   {pi, \[Lambda]h, \[Lambda]l, R, ph, pl}]

can find solutions.

In this case it also complains about some things that you might think carefully about.

POSTED BY: Bill Simpson

Songklod Rastapana

Posted 10 years ago

Sorry, I should also mention that; 0 <= pi <= 1 && 0 <= \[Lambda]h <= 1 && 0 <= \[Lambda]l <= 1 && 1 <= R <= 2 && 0 <= ph <= 1 && 1 <= pl <= 2 Thanks!!!

POSTED BY: Songklod Rastapana

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