Can a cubic polynomial be solved without arccos?

In summary, the conversation discusses various methods for solving a cubic equation and the use of inverse trig functions in these methods. The Cardano's method is mentioned as a 16th century formulation that did not rely on inverse trig functions, but later refinements did. The conversation also touches on the impossibility of trisecting an angle and the use of Newton's method for solving complex equations.
  • #1
swampwiz
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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.
 
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  • #2
swampwiz said:
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.
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

I wonder if this is so for the same reason that it is impossible to trisect an angle.
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
 
  • #3
SteamKing said:
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

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.
 
  • #4
swampwiz said:
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.

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.
 
  • #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.)
 

1. Can a cubic polynomial always be solved without using arccos?

Yes, it is possible to solve a cubic polynomial without using arccos. There are multiple methods for solving cubic polynomials, such as factoring, using the cubic formula, or using the rational root theorem.

2. Why is arccos not always used to solve cubic polynomials?

Arccos is not always used to solve cubic polynomials because it is not always the most efficient method. Other methods, such as factoring or using the cubic formula, may be easier or more straightforward for certain polynomials.

3. Are there any advantages to using arccos to solve cubic polynomials?

Using arccos to solve cubic polynomials can be advantageous in certain cases, such as when the polynomial has complex roots. Arccos can also be used to find the exact solutions for some polynomials that cannot be solved using other methods.

4. Can arccos be used to solve any type of cubic polynomial?

No, arccos cannot be used to solve all cubic polynomials. There are some polynomials that do not have any real solutions and therefore cannot be solved using arccos or any other method.

5. Are there any limitations to using arccos to solve cubic polynomials?

Yes, there are limitations to using arccos to solve cubic polynomials. As mentioned before, there are some polynomials that do not have real solutions and therefore cannot be solved using arccos. Additionally, the use of arccos may result in complex or irrational solutions, which may not be desired in certain situations.

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