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A math question

  1. Jul 19, 2008 #1
    Is there a easy way to solve x^3-x-1=0? I know there's newton's method and depressed method... but is there a easy way?
     
  2. jcsd
  3. Jul 19, 2008 #2

    mathman

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    Third degree and fourth degree polynomial equations have exact solutions in terms of radicals. With a little searching, you can find them.
     
  4. Jul 19, 2008 #3
    Hmm its not that simple.... you can't use long division or the |_____.....

    its a special polynomial equation..

    its from trying to solve (1+(1+(1+...)^1/3)^1/3)^1/3...
     
  5. Jul 19, 2008 #4
    It is that simple: there are solutions for cubics in terms of the coefficients and sums and radicals thereof. Just google for them.
     
  6. Jul 19, 2008 #5
  7. Jul 20, 2008 #6

    gel

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  8. Jul 20, 2008 #7
    can you explain how?
     
  9. Jul 20, 2008 #8

    gel

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    ok.
    Actually, because this cubic has only one real root, you should use the hyperbolic cosine instead.
    The triple angle formula is [itex]\cosh(3\theta)=4\cosh^3\theta-3\cosh\theta[/itex]. Try putting [itex]x=a\cosh\theta[/itex] into your formula for x.
    [tex]
    a^3\cosh^3(\theta)-a\cosh(\theta)=1.
    [/tex]
    notice if a2 = 4/3 then you can multiply both sides by 3/a,
    [tex]
    4\cosh^3(\theta)-3\cosh(\theta)=3/a=3\sqrt{3}/2
    [/tex]
    Putting this together, you have [itex]x = \frac{2}{\sqrt{3}}\cosh\left(\cosh^{-1}(3\sqrt{3}/2)/3\right)[/itex].
     
  10. Jul 20, 2008 #9
    Hmm? What's so hard about it? The constant term is -1, so chances are x - 1 or x +1 divides the polynomial. A little synthetic division magic and we see x-1 divides the cubic without remainder. Since x - 1 is a factor, we know:

    [tex](x^3 -x -1) = (x-1)(x^2 +x+1) = 0[/tex]

    So [tex]x-1 = 0[/tex] and or [tex]x^2 +x +1 = 0[/tex] (I assume you know how to solve quadratic equations)

    Little method I use for solving polynomials with an order > 2 is:

    1) Figure out which factors there might be of the equation. Usually in x^n +- x^(n-1)...+-C there is a factor x - c such that c is a factor of C. For instance, in [tex]x^4 - x^3 + x^2 - 8[/tex] possible factors could be x +- 2, +-4, or +-8.

    2) Divide the polynomial by the possible factors until you find one without a remainder.

    3) Repeat step 2 until you 'take out' all the factors.

    4) Try the answers, because some roots don't always work!
     
  11. Jul 21, 2008 #10

    HallsofIvy

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    No.

    Apparently this is harder than you think it is because x= 1 is obviously NOT a root. 13- 1- 1= -1, not 0.
    (x-1)(x2+ x+ 1)= x3+ x2+ x- x2- x- 1= x3- 1, not x3- x- 1.
     
  12. Jul 22, 2008 #11

    uart

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    Yes, unless you can just happen to guess one real solution (as per Daniel Y's attempt, it's always worth attempting this first --- although it appears Daniel made an arithmetic error or something in this case and got it wrong) then that "depressed cubic" method is really the best way to go.

    Personlly I like to do via a non-linear substitution. Fundamentally it's really the same thing as using those formulas you linked but I find it somehow more satisfying to use the substitution.

    Given a depressed cubic [itex]z^3 + pz + q = 0[/itex], if you make the non-linear substition [itex] z = w - p/(3w) [/itex] then it "magically" simplifies into a quadratic in [itex]w^3[/itex]. Give it a try, it's easy enough. Although be prepared to work in the complex domain for all the intermediate work, even for cases when all the final roots are real!
     
  13. Jul 22, 2008 #12
    ya my teacher showed me that one... I thought it was too complex... and I didn't understand it...
     
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