How to solve

natski · Mar 25, 2007

Does anyone know how you would go about solving:

a + b*x + x^(-y) + c*x^(-2y) + d*x^(1-2y)=0

solving for x?

sutupidmath · Mar 26, 2007

can you give more information, on whad do a, b, c, d, x and y represent. Are some of them variables,functions, series, or merely constants?
maybe this information would be helpful to me, and to others also who have more expertise.

uart · Mar 26, 2007

natski said: ↑

Does anyone know how you would go about solving:

a + b*x + x^(-y) + c*x^(-2y) + d*x^(1-2y)=0

solving for x?

Sure. It's real easy.

a = -b*x - x^(-y) - c*x^(-2y) - d*x^(1-2y)

If that's not the answer you were expecting then how about specifying which variable you want to solve for.

jing · Mar 26, 2007

I think he did.

From OP

natski said: ↑

solving for x?

sutupidmath · Mar 26, 2007

natski said: ↑

Does anyone know how you would go about solving:

a + b*x + x^(-y) + c*x^(-2y) + d*x^(1-2y)=0

solving for x?

i guess natski already stated that he wants to solve for x.

uart · Mar 26, 2007

solving for x?

Whoops, that will teach me to read all of the post and not skip the last line.

Sorry, that'ss a transcendental equation in x, so there will be no closed form solution.

sutupidmath · Mar 26, 2007

ej uart how do u determine wheather it is transcendental equation in x?

uart · Mar 26, 2007

sutupidmath said: ↑

ej uart how do u determine wheather it is transcendental equation in x?

Oh yeah it's not transcendental in x. If y is an integer it's a polynomial in x.

natski · Mar 26, 2007

Um well y is a constant but a,b,c,d are functions of say z. So I want x=function of z or a function of a,b,c,d in this case. I tried to solve numerically for lots of different values of z, plot it and then fit a function to the plot. This worked except for my answer didn't agree with what it should. So I was hoping for a neater algebraic solution.

uart · Mar 27, 2007

So I was hoping for a neater algebraic solution.

Well there might be some partiucular values of "y" for which it can work, but in general I don't think it's going to be possible.