Solution for x^3-x=y needed, is solution an approximation?

1. Oct 17, 2011

elginz

I have a formula to be solved for x that is y=x^3-x

I have a solution given x=((27y^2-4)^.5/23^2/3+y/2)^1/3 + 1/3((27y^2-4)^.5/23^2/3+y/2)^1/3 which seems to work

Is this solution an approxiamation?

2. Oct 17, 2011

LCKurtz

I would guess it probably is an approximation. But with your expressions full of "/" division signs and no parentheses, it is impossible to figure out what the formula actually is.

Either put in needed parentheses or, better, post it using the tex editor by using the $\sum$ button.

3. Oct 17, 2011

Matt Benesi

It's one of these 3 equations. Basically a variant of the http://en.wikipedia.org/wiki/Plastic_number" [Broken].

Note that $x_{i1}$ is one imaginary root, $x_{i2}$ is another, and lastly $x_{r}$ is the real root.

$$x_{i1}=\left( -\frac{\sqrt{3}\,i}{2}-\frac{1}{2}\right) \,{\left( \frac{\sqrt{27\,{y}^{2}-4}}{2\times{3}^{\frac{3}{2}}}-\frac{y}{2}\right) }^{\frac{1}{3}}+\frac{\frac{\sqrt{3}\,i}{2}-\frac{1}{2}}{3\,{\left( \frac{\sqrt{27\,{y}^{2}-4}}{2\times{3}^{\frac{3}{2}}}-\frac{y}{2}\right) }^{\frac{1}{3}}}$$
$$x_{i2}=\left( \frac{\sqrt{3}\,i}{2}-\frac{1}{2}\right) \,{\left( \frac{\sqrt{27\,{y}^{2}-4}}{2\times{3}^{\frac{3}{2}}}-\frac{y}{2}\right) }^{\frac{1}{3}}+\frac{-\frac{\sqrt{3}\,i}{2}-\frac{1}{2}}{3\,{\left( \frac{\sqrt{27\,{y}^{2}-4}}{2\times{3}^{\frac{3}{2}}}-\frac{y}{2}\right) }^{\frac{1}{3}}}$$
$$x_{r}={\left( \frac{\sqrt{27\,{y}^{2}-4}}{2\times{3}^{\frac{3}{2}}}-\frac{y}{2}\right) }^{\frac{1}{3}}+\frac{1}{3\,{\left( \frac{\sqrt{27\,{y}^{2}-4}}{2\times{3}^{\frac{3}{2}}}-\frac{y}{2}\right) }^{\frac{1}{3}}}$$

Corrected it a bit... looked like 233/2 instead of 2*33/2. Thus the 2x3....

Last edited by a moderator: May 5, 2017