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Cubic roots

  1. Jul 13, 2007 #1
    Hi! Who knows:
    can any cubic root like [itex]\sqrt[3]{x}[/itex] with x real be written as a form in which only square roots (real or complex) are involved?
     
  2. jcsd
  3. Jul 13, 2007 #2
    Sorry, not any but every!!!
     
  4. Jul 13, 2007 #3
    Yes it can, but sometimes the square root have to be of an irrational number like another cubic root so while it's the square root of a real number we haven't gotten rid of cubic roots. For example we have: [tex]\sqrt[3]{2}=\sqrt{\sqrt[3]{4}}[/tex] or [tex]\sqrt[3]{-2}=\sqrt{\sqrt[3]{4}i}[/tex].

    If x is nonnegative then we have a nonnegative real [tex]y=\sqrt[3]{x^2}[/tex]. Clearly [tex]x^2[/tex] exists for all real x and is itself a real nonnegative number. Similarly every nonnegative real number have a nonnegative real cubic root, so [tex]\sqrt[3]{x^2}[/tex] also exists and is a nonnegative real. The square root of y also exist since all real nonnegative numbers have a real nonnegative square root.
    [tex]\sqrt{y}=\sqrt[2*3]{x^2}=\sqrt[3]{x}[/tex]
    So the square root of y exists for all nonnegative x and it's equal to the cubic root of x.

    For negative x we have a positive real number z such that:
    x = -1z
    So:
    [tex]\sqrt[3]{x}=\sqrt[3]{-1z}=\sqrt[3]{-1}\sqrt{z}=-\sqrt[3]{z}[/tex]
    Since z is a nonnegative real number we know that we can write its cubic root as the square root of a real number. Let [tex]\sqrt{y}=\sqrt[3]{z}[/tex], then we have:
    [tex]-\sqrt[3]{z} = \sqrt{i} \sqrt{y} = \sqrt{yi}[/tex]
    So for all negative x we can write its cubic root as the square root of a pure imaginary number.
     
  5. Jul 14, 2007 #4

    HallsofIvy

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    Yes, obviously, if a is any real number, I can define y to be a2 and then say [itex]a= \sqrt{y}[/itex]. I don't think that was what edgo meant.

    In general, a cube root cannot be written in a form involving only square roots and rational numbers. That is because any number written in that form must be algebraic of order a power of two while any cuberoot is obviously algebraic of order 3.
     
  6. Jul 14, 2007 #5

    mathwonk

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    the proof halls is giving depends on the fact that there is a certain vector space associated to the root, and the dimension of that vector space equals the degree of the polynomial, namely 3. Moreover, if the expresion is writtena s a composition, the degrees are multiplicative, so if a square root were involved the product would be even, a contradiction. this is elementary field theory, and is based on the fact that the k vector space

    k[X]/(f(X)) where f is a polynomial of degree d, has k dimension d.
     
  7. Jul 14, 2007 #6
    Three times: Thank you.
    I'm not (yet?) familiar with the Galois theory. Then, I assumed that the cubic roots which are involved in the solution - via the Cardano or Harriot algorithms - of cubic equations were there because the algorithm introduced them. This not being so has the consequence that in the value of cos x etc. in general cubic roots are involved (since you can also solve most cubic equations with a goniometric algorithm). Am I right?
     
  8. Jul 15, 2007 #7
    There should not be such a thing as a quick reply in math. Rereading my last remarks urges me to give a correction on "you can solve most cubic equations with a goniometric algorithm". I meant cubic equations with 3 real roots and furthermore, most makes no sense at all as we are talking about the set of cubic equations and that set doesn't have a limited number of elements.
     
  9. Jul 15, 2007 #8

    HallsofIvy

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    It didn't occur to you that it might be because solving an equation is an "inverse" problem and that the inverse of cubing is the cube root?

    I don't know what you mean "goniometric algorithm". A google search for "goniometric algorithm" turn up nothing and a google search for "goniometric" led to one reference giving "trigonometric" as a secondary definition. All other references were to measurement devices.
     
  10. Jul 15, 2007 #9
    You know this goniometric algorithm for sure!
    We reduce any cubic equation (further: CE) with real coefficients to a form [itex]V=x^3+px+q=0[/itex]. Then V=0 also has real coefficients.
    When the discriminant [itex]D=\frac{q^2}{4}+\frac{p^3}{27}\prec0[/itex] we have the “casus irreducibilis”, in which case V=0 has three real roots.

    We set [itex]x=r\cos\alpha[/itex] and substitute that x into V=0: [itex]r^3{(\cos\alpha)}^3+pr\cos\alpha+q=0[/itex].
    There exists a goniometric equality [itex]\cos3\alpha=4{(\cos\alpha)}^3-3\cos\alpha[/itex].
    So, we multiply the terms of the left side of V=0 by 4 and replace [itex]4{(\cos\alpha)}^3[/itex] by [itex]\cos3\alpha+3\cos\alpha[/itex].
    It gives
    [itex]r^3(\cos3\alpha+3\cos\alpha)+4pr\cos\alpha+4q=0[/itex]
    then
    [itex]r^3\cos3\alpha+4q+(3r^2+4p)rcos\alpha=0[/itex]
    We can choose [itex]3r^2+4p=0[/itex] and this makes [itex]r^3\cos3\alpha+4q=0[/itex]

    We assume that [itex]\beta[/itex] meets the condition
    [itex]\alpha=\beta frac{2k\pi}{3}[/itex]

    The roots of [itex]V_1=0[/itex] can be written now as [itex]x_k=r\cos(\beta+ \frac{2k\pi}{3})[/itex] with k=1,2,3.

    I owe this solution to Dr P. Wijdenes; it’s dated from the times of the pre-calculator math (not to be confused with the pre-calculus math).

    No it didn't occur to me that solving a CE is the inverse of cubing. If I have a polynom of the 3rd power with real coefficients and I multiply this polynom
    with a factor, say, x-2, then I have a quartic equation with three cubic rootsand a root x=2.
    All I want to say is that it's not at the first sight clear: there are dark corners in the set of real numbers.
     
  11. Jul 15, 2007 #10
    Sorry, again a correction.
    False is: we assume that [itex]\beta[/itex] meets the condition etc.
    Correct is:
    We have to solve [itex]\alpha[/itex] in[itex]cos3\alpha=frac{-4p}{r^3}[/itex]

    We assume that [itex]\beta[/itex] is a solution for [itex]\alpha[/itex] in this equation.

    then [itex]\alpha=\beta+frac{2k\pi}{3}[/itex].

    Now the roots [itex]x_k[/itex] of V=0 are equal to [itex]rcos(\beta+frac{2k\pi}{3})[/itex] with k=1,2,3.
     
  12. Jul 15, 2007 #11
    Sorry, again a correction.Now I understand what went wrong!
    False is: we assume that [itex]\beta[/itex] meets the condition etc.
    Correct is:
    We have to solve [itex]\alpha[/itex] in [itex]cos3\alpha=\frac{-4p}{r^3}[/itex]

    We assume that [itex]\beta[/itex] is a solution for [itex]\alpha[/itex] in this equation.

    then [itex]\alpha=\beta+\frac{2k\pi}{3}[/itex].

    Now the roots [itex]x_k[/itex] of V=0 are equal to [itex]rcos(\beta+\frac{2k\pi}{3})[/itex] with k=1,2,3.
     
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