Is there a new solution for Cardano's formula?

[edgo]

Just asking: is it likely that there is any connection between roots as
\sqrt[3]{1} and \sqrt[7]{1}?

 

[arildno]

Just asking: is it likely that there is any connection between roots as
\sqrt[3]{1} and \sqrt[7]{1}?

Hmm, let's see:
Regarding the radix operations as functions from the reals to reals, i'd say the connection consists of an identity between your two expressions..

 

[qntty]

What exactly do you mean by a connection? They all (that is, the 3rd and 7th roots of unity) have a magnitude of 1 but I can't think of anything else.

 

[HallsofIvy]

If x= cuberoot of 1 then x³= 1.
If y= seventhroot of 1 then y⁷= 1.

Thus x³= y⁷. That's the "connection".

 

[edgo]

Thank you for the answers. The last one is the best one and I was that far myself. Only: the shoe doesn't fit. I don't want to be a headache with my eternal questions about cube roots but the truth is that I do have a bloody good reason to be superstitious about the combination Cardano/Galois/solutions of cubic equations. The problem is that it takes about 5 pages A4 loaded with formulas to get to the point where we can do business, so to say. And I am not familiar with the math. jargon in English, a jargon that tends to be very helpfull if one wants to explain new ideas. At the moment I am a member of a Dutch forum and have been busy to load their site with rather a lot formulas in LaTeX. We are now at the point where it can become a dialogue. May be you will never hear of me again, an indication that I was wrong. But if they don't can falsificate my theory then I am sure back here. I don't mean to say that this forum is my second choice, only native speech in math is my first choice, when it is new stuff. By the way, thanks for your immortal answer the other day that te solution of cubic equations is only: cube root (stuff) + cube root (other stuff). American eloquence.

 

[workmad3]

All real roots of unity are the same (i.e. 1). There are other roots of unity though, namely the complex ones :)

 

[matt grime]

-1, workmad3?

 

[workmad3]

oh yeah, there's that one too for some. Kinda forgotten that -1 was real for a second :) (or more likely forgotten that it was a root of unity... which is just as embarrassing)

They do all have a magnitude of 1 and lie on the unit circle in the complex plane though and some of the roots are shared (square roots are 1 and -1, while 4th roots are 1, -1, i and -i so have 2 roots in common).

Of course, that's assuming that you aren't defining the root operation as only the positive real roots of unity in which case they are all the same and equal to 1 (but that's a fairly limiting view and derives directly from the fact that 1 is a multiplicative identity number for the reals)

 

[edgo]

OK it's some time ago, but...
The connection I was referring to concerns the following. For the solutions of the equations x^7=1 and y^9=1 we can form the systems:
X_1=x+\frac{1}{x}
X_2=x^2+\frac{1}{x^2}
X_3=x^4+\frac{1}{x^4}
X_1=x^8+\frac{1}{x^8}=x+\frac{1}{x}
and
Y_1=y+\frac{1}{y}
Y_2=y^2+\frac{1}{y^2}
Y_3=y^4+\frac{1}{y^4}
Y_1=y^8+\frac{1}{y^8}=\frac{1}{y}+y.

We can form the equations X^3+X^2-2X-1=0 and Y^3-3Y+1=0; the solutions of these equations satisfy the relation of the form Z_i^2-p=Z_j with, in this case, p=2.
For any p we can always form 2 equations of which the solutions satisfy such a relation. I have named them cyclic equations with the cyclic constant p.
When 2 cyclic equations have the same constant, there always is a lot of isomorphy between the roots of those conjugated cyclic equations. I expect this to hold for the above mentioned solutions too. It would be helpful to get some input which is not originating from my work.

My question is: is this Forum open for a topic on a subject that takes about one full Forum-page to explain. It's all math, it's all proven (but the proofs are not included as to keep it short) and it's concerning what I believe to be the proof for the existence of an alternative solution for Cardano. I am looking for a kind of peer review. I didn't succeed in Holland due to a lack of time by the moderators