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

1. Sep 1, 2012

### thiago.omena

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

I don't understand how to get there.

2. Sep 1, 2012

### chiro

Hey thiago.omena and welcome to the forums.

Are these matrices or just plain old numbers?

3. Sep 1, 2012

### thiago.omena

In the question? They are just variables. I am trying to isolate the V.

4. Sep 2, 2012

### haruspex

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. Sep 2, 2012

### HallsofIvy

Staff Emeritus
$V= B+ CTV^N$ is the same as the polynomial equation $CTV^N- V+ B= 0$. 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 $V=[B^{1-N}+C(N-1)T]^(1/(1-N))$ is NOT a general solution to the equation. In the case that C= T= 1, B= -2 and N= 2, the equation becomes $V= -2+ V^2$ which is the same as $V^2- V- 2= (V- 2)(V+ 1)= 0$ which has roots 2 and -1. But your formula gives $V= [(-2)^{-1}+ (1)(-1)(1)]^{1/-1}= [-1/2+ -1]^{-1}= [-3/2]^{-1}= -2/3$ which is NOT a solution.

6. Sep 3, 2012

### Mute

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.