# A nonlinear problem

• Sep 2nd 2012, 11:01 AM
markel
A nonlinear problem
I have been trying to find a solution to the following problem.

x(4/3) - x(-2/3) + c = 0 where -1 < x > 1

I think if I could solve,

s(-2) - s + c = 0,

I could use substitution to solve the first equation, but I have not had any luck with the second either.

Any help would be appreciated.
• Sep 2nd 2012, 01:26 PM
Soroban
Re: A nonlinear problem
Hello, markel!

Is there a typo?
You have a positive exponent and a negative exponent.
Even if the signs were alike, that "c" makes for a messy solution.

Quote:

$x^{\frac{4}{3}} -x^{-\frac{2}{3}} - c \:=\:0\;\;\;\text{ for }0 \,\le\,x\,<\,1$

I will wait for your clarification.

• Sep 2nd 2012, 01:56 PM
markel
Re: A nonlinear problem
Dear Soroban,

Thanks for responding. I'm afraid the negative exponent is correct. The limit however should be -1 < x < 1. I have used an Excel spreadsheet to calculate some examples by trial and error and the formula seems to be working. The task of simplifying the formula, however, has eluded me.
• Sep 2nd 2012, 08:50 PM
Prove It
Re: A nonlinear problem
Quote:

Originally Posted by markel
I have been trying to find a solution to the following problem.

x(4/3) - x(-2/3) + c = 0 where -1 < x > 1

I think if I could solve,

s(-2) - s + c = 0,

I could use substitution to solve the first equation, but I have not had any luck with the second either.

Any help would be appreciated.

\displaystyle \begin{align*} x^{\frac{4}{3}} - x^{-\frac{2}{3}} + c &= 0 \\ x^{\frac{2}{3}}\left( x^{\frac{4}{3}} - x^{-\frac{2}{3}} + c \right) &= 0x^{\frac{2}{3}} \\ x^{\frac{6}{3}} - 1 + c\,x^{\frac{2}{3}} &= 0 \\ X^3 + c\,X - 1 &= 0 \textrm{ if } X = x^{\frac{2}{3}} \end{align*}

While there is a solution to this cubic, it is NOT pretty. I suggest you read the attachment.