Is there an easy method to solve x³ - x - 1 = 0?

[glueball8]

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?

 

[mathman]

Third degree and fourth degree polynomial equations have exact solutions in terms of radicals. With a little searching, you can find them.

 

[glueball8]

Third degree and fourth degree polynomial equations have exact solutions in terms of radicals. With a little searching, you can find them.

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...

 

[n_bourbaki]

It is that simple: there are solutions for cubics in terms of the coefficients and sums and radicals thereof. Just google for them.

 

[glueball8]

you mean this? http://www.josechu.com/ecuaciones_polinomicas/cubica_solucion.htm

 

[gel]

You can also express the solution as a sine or cosine, using the http://en.wikipedia.org/wiki/List_o...ouble-.2C_triple-.2C_and_half-angle_formulae".

 

[glueball8]

You can also express the solution as a sine or cosine, using the http://en.wikipedia.org/wiki/List_o...ouble-.2C_triple-.2C_and_half-angle_formulae".

can you explain how?

 

[gel]

ok.
Actually, because this cubic has only one real root, you should use the hyperbolic cosine instead.
The triple angle formula is $\cosh(3\theta)=4\cosh^3\theta-3\cosh\theta$. Try putting $x=a\cosh\theta$ into your formula for x.
$$a^3\cosh^3(\theta)-a\cosh(\theta)=1.$$
notice if a² = 4/3 then you can multiply both sides by 3/a,
$$4\cosh^3(\theta)-3\cosh(\theta)=3/a=3\sqrt{3}/2$$
Putting this together, you have $x = \frac{2}{\sqrt{3}}\cosh\left(\cosh^{-1}(3\sqrt{3}/2)/3\right)$.

 

[Daniel Y.]

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:

$$(x^3 -x -1) = (x-1)(x^2 +x+1) = 0$$

So $$x-1 = 0$$ and or $$x^2 +x +1 = 0$$ (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 $$x^4 - x^3 + x^2 - 8$$ 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!

 

[HallsofIvy]

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:

$$(x^3 -x -1) = (x-1)(x^2 +x+1) = 0$$

No.

So $$x-1 = 0$$ and or $$x^2 +x +1 = 0$$ (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 $$x^4 - x^3 + x^2 - 8$$ 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!

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.

 

[uart]

you mean this? http://www.josechu.com/ecuaciones_polinomicas/cubica_solucion.htm

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 $z^3 + pz + q = 0$, if you make the non-linear substition $z = w - p/(3w)$ then it "magically" simplifies into a quadratic in $w^3$. 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!

 

[glueball8]

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 $z^3 + pz + q = 0$, if you make the non-linear substition $z = w - p/(3w)$ then it "magically" simplifies into a quadratic in $w^3$. 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!

ya my teacher showed me that one... I thought it was too complex... and I didn't understand it...