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Can a cubic polynomial be solved without arccos?

  1. Oct 12, 2015 #1
    I was reviewing the Cardano's method formula for a real cubic polynomial having 3 real roots. It seems that to do so, the arccos (or another arc*) of a term involving the p & q parameters of the reduced cubes must be done, and then followed by cos & sin of 1/3 of the result from that arccos - and AFAIK, the only way to do this is to use the series representation of both, which seems to me to be an imperfect method. Even using the complex formula for a fractional angle results in having to get deMoivre roots, which can only be done by going through the same arc* process.

    I wonder if this is so for the same reason that it is impossible to trisect an angle.
     
  2. jcsd
  3. Oct 12, 2015 #2

    SteamKing

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    There are algorithms for solving a cubic equation using root extraction alone. Cardano's method as formulated in the 16th century did not rely on inverse trig functions. This was a later refinement.

    https://en.wikipedia.org/wiki/Cubic_function

    This is the reason that an arbitrary angle cannot be trisected using a straightedge and compass alone. Certain angles, like a right angle, for instance, can be trisected using those tools.

    https://en.wikipedia.org/wiki/Angle_trisection
     
  4. Oct 13, 2015 #3
    I think that any operation that requires the cubic root of a complex number ends up having to do an inverse trig function to get the angle, followed by a trig function of the trisected angle to get the deMoivre root.
     
  5. Oct 14, 2015 #4
    Newtons method will work for complex equations as well. You'll need an initial guesss that is close enough to the root, but you can just use some conditions on (y/x) to makes this guess, instead of computing atn(y/x) to find the argument of the root.
     
  6. Oct 14, 2015 #5
    Well, Newton's method is also not direct, just like doing the arccos. I suppose that since cos & arccos are valid functions, it is still considered to be an analytical solution. Newton's method is not even analytical (i,e., it is numerical, which in theory could solve any polynomial.)
     
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