Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Oct 17, 2011 #1
    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. jcsd
  3. Oct 17, 2011 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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 [itex]\sum[/itex] button.
  4. Oct 17, 2011 #3
    It's one of these 3 equations. Basically a variant of the http://en.wikipedia.org/wiki/Plastic_number" [Broken].

    Note that [itex]x_{i1}[/itex] is one imaginary root, [itex]x_{i2}[/itex] is another, and lastly [itex]x_{r}[/itex] is the real root.

    [tex] 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}}}[/tex]
    [tex]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}}}[/tex]
    [tex]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}}} [/tex]

    Corrected it a bit... looked like 233/2 instead of 2*33/2. Thus the 2x3....
    Last edited by a moderator: May 5, 2017
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook