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Math Help - Real algebraic Expressions for cubic root?

  1. #1
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    Real algebraic Expressions for cubic root?


    We know the square roots of a complex number a + ib \quad (a,b \in \mathbf{R}) can be written as
    x + iy = \pm \frac {1}{\sqrt {2}} (\sqrt {a + \sqrt {a^2 + b^2}} + i sgn(b + ) \sqrt { - a + \sqrt {a^2 + b^2}})
    where sgn(b + ) = 1 when b \geq 0 otherwise sgn(b + ) = - 1

    Thus we have real algebraic expressions for the real and imaginary parts of the square root, without trigonometric functions involved.

    Does this still the case for cubic roots? I tried with no success. Thought that's not possible but cannot prove my guess
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by elim View Post
    We know the square roots of a complex number a + ib \quad (a,b \in \mathbf{R}) can be written as
    x + iy = \pm \frac {1}{\sqrt {2}} (\sqrt {a + \sqrt {a^2 + b^2}} + i sgn(b + ) \sqrt { - a + \sqrt {a^2 + b^2}})
    where sgn(b + ) = 1 when b \geq 0 otherwise sgn(b + ) = - 1

    Thus we have real algebraic expressions for the real and imaginary parts of the square root, without trigonometric functions involved.

    Does this still the case for cubic roots? I tried with no success. Thought that's not possible but cannot prove my guess
    You want to find  p,q\in\mathbb{R} such that  \sqrt[3]{a+bi} = p+qi \implies a+bi = (p+qi)^3 \implies p^3-3pq^2+(3p^2q-q^3)i = a+bi

    Therefore we have two equations and two unknowns:

     \begin{cases} p^3-3pq^2 = a\\3p^2q-q^3 = b \end{cases}

    Looks rather messy to solve, but I imagine it's doable. I might attempt it later.

    Edit: I'm starting to think this system isn't solvable in general...
    Last edited by chiph588@; March 31st 2010 at 12:51 PM.
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    It is solvable, I just cannot prove that in general the resulting expressions cannot get rid of trigonometric functions.
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by elim View Post
    It is solvable, I just cannot prove that in general the resulting expressions cannot get rid of trigonometric functions.
    Right, when I said it wasn't solvable, I meant expressing it without trig functions involved.
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by chiph588@ View Post
    You want to find  p,q\in\mathbb{R} such that  \sqrt[3]{a+bi} = p+qi \implies a+bi = (p+qi)^3 \implies p^3-3pq^2+(3p^2q-q^3)i = a+bi

    Therefore we have two equations and two unknowns:

     \begin{cases} p^3-3pq^2 = a\\3p^2q-q^3 = b \end{cases}

    Looks rather messy to solve, but I imagine it's doable. I might attempt it later.

    Edit: I'm starting to think this system isn't solvable in general...
    Ok, I'll show you in general you can't express  p and  q in terms of radicals.

    From the two equations we can get  3p\sqrt{p^3-a}-\left(\frac{\sqrt{p^3-a}}{p}\right)^3 = b (solve for  q in equation 1 and substitute into equation 2).

    Now square both sides:
     -\frac{a^3}{p^6}+\frac{3a^2}{p^3}-\frac{6a^2}{p^2}-9ap^2+12ap-3a+9p^5-6p^4+p^3=b^2

    Simplify:
     9p^{11}-6p^{10}+p^9-9ap^8+12ap^7-(3a+b^2)p^6-6a^2p^4+3a^2p^3-a^3=0

    Since there is no formula that can solve for a polynomial of degree greater than  4 in terms of radicals, we see that this equation is unsolvable in general.

    Applying this method to  q yields the same results.
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  6. #6
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    Let (a,b),(x,y) \in \mathbf{R}^2 such that a+ib=(x+iy)^3 = x^3-3xy^2+i(3x^2y-y^3)
    Then when a=0 we have (x,y)=(0,-\sqrt[3]{b}) as a solution.
    If a \neq 0, then x \neq 0 and we have
    b^2 = (3x^2y-y^3)^2=\left(\frac{x^3-a}{3x}\right)\left(3x^2-\frac{x^3-a}{3x}\right)^2 hence
    64x^9-48ax^6-(15a^2+27b^2)x^3-a^3=0
    Let x^3=z+a/4, \quad p=-\frac{27}{64}(a^2+b^2),\quad q=-\frac{27a}{256}(a^2+b^2), then we have
    z^3+pz+q = 0 and so z can be expressed in terms of radicals and so can x and y.

    But the problem is that the radicals themselves(if using Cardano's method or known methods for me) are involving complex numbers and we get cyclic situations!

    Seems we can express the problem as this:
    Does real cubic equation z^3+pz+q = 0 solvable in real field?
    Or how to disprove the equation is solvable in real field?
    Last edited by elim; April 2nd 2010 at 10:46 AM.
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  7. #7
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by elim View Post
    Let (a,b),(x,y) \in \mathbf{R}^2 such that a+ib=(x+iy)^3 = x^3-3xy^2+i(3x^2y-y^3)
    Then when a=0 we have (x,y)=(0,-\sqrt[3]{b}) as a solution.
    If a \neq 0, then x \neq 0 and we have
    b^2 = (3x^2y-y^3)^2=\left(\frac{x^3-a}{3x}\right)\left(3x^2-\frac{x^3-a}{3x}\right)^2 hence
    64x^9-48ax^6-(15a^2+27b^2)x^3-a^3=0
    Let x^3=z+a/4, \quad p=-\frac{27}{64}(a^2+b^2),\quad q=-\frac{27a}{256}(a^2+b^2), then we have
    z^3+pz+q = 0 and so z can be expressed in terms of radicals and so can x and y.

    But the problem is that the radicals themselves(if using Cardano's method or known methods for me) are involving complex numbers and we get cyclic situations!

    Seems we can express the problem as this:
    Does real cubic equation z^3+pz+q = 0 solvable in real field?
    Or how to disprove the equation is solvable in real field?
    Wow, didn't think of it this way!
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