How to solve V=B+CT(V^N) for V?

  • #1

Main Question or Discussion Point

Does anyone know how to solve V=B+CT(V^N) for V?

The answer is V=[B^(1-N)+C(N-1)T]^(1/(1-N))
I don't understand how to get there.
Please help me out. Thank you
 

Answers and Replies

  • #2
chiro
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Hey thiago.omena and welcome to the forums.

Are these matrices or just plain old numbers?
 
  • #3
Hey thiago.omena and welcome to the forums.

Are these matrices or just plain old numbers?
In the question? They are just variables. I am trying to isolate the V.
 
  • #4
haruspex
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Does anyone know how to solve V=B+CT(V^N) for V?
The answer is V=[B^(1-N)+C(N-1)T]^(1/(1-N))
I don't understand how to get there.
What makes you think that's the answer? It doesn't look right to me. In fact, I do not believe there is a solution in closed form.

Btw, it's better to use the formatting this site provides. If you click Go Advanced you'll see a toolbar with e.g. X2 for superscript. Select the text to be superscripted and click the icon:
V=B+CT(VN)
 
  • #5
HallsofIvy
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[itex]V= B+ CTV^N[/itex] is the same as the polynomial equation [itex]CTV^N- V+ B= 0[/itex]. There is NO general formula for solving a polynomial equation of degree greater than 5. In fact, it was proved about a hundred years ago that there cannot be a formula (involving only roots of numbers) for such equations.

And it is easy to show that your [itex]V=[B^{1-N}+C(N-1)T]^(1/(1-N))[/itex] is NOT a general solution to the equation. In the case that C= T= 1, B= -2 and N= 2, the equation becomes [itex]V= -2+ V^2[/itex] which is the same as [itex]V^2- V- 2= (V- 2)(V+ 1)= 0[/itex] which has roots 2 and -1. But your formula gives [itex]V= [(-2)^{-1}+ (1)(-1)(1)]^{1/-1}= [-1/2+ -1]^{-1}= [-3/2]^{-1}= -2/3[/itex] which is NOT a solution.
 
  • #6
Mute
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Does anyone know how to solve V=B+CT(V^N) for V?

The answer is V=[B^(1-N)+C(N-1)T]^(1/(1-N))
I don't understand how to get there.
Please help me out. Thank you
Is this a thermodynamic equation? If it is, you're likely looking for an asymptotic solution for V >> 1 and N >> 1. As has been said, solving the equation exactly is not possible in general, but an approximate solution in the large V and/or N limits may be possible, and is perhaps the answer you claim to have.
 

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