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

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

I don't understand how to get there.

chiro
Hey thiago.omena and welcome to the forums.

Are these matrices or just plain old numbers?

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.

haruspex
Homework Helper
Gold Member
2020 Award
Does anyone know how to solve V=B+CT(V^N) for V?
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)

HallsofIvy
Homework Helper
$V= B+ CTV^N$ 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.

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