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

[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?

 

[LCKurtz]

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?

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.

 

[Matt Benesi]

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

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 23^3/2 instead of 2*3^3/2. Thus the 2x3...