- #1

- 32

- 0

[tex]\sqrt[3]{1}[/tex] and [tex]\sqrt[7]{1}[/tex]?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter edgo
- Start date

- #1

- 32

- 0

[tex]\sqrt[3]{1}[/tex] and [tex]\sqrt[7]{1}[/tex]?

- #2

arildno

Science Advisor

Homework Helper

Gold Member

Dearly Missed

- 9,970

- 134

Hmm, let's see:

[tex]\sqrt[3]{1}[/tex] and [tex]\sqrt[7]{1}[/tex]?

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

- #3

- 290

- 2

- #4

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 963

If y= seventhroot of 1 then y

Thus x

- #5

- 32

- 0

- #6

- 52

- 0

- #7

matt grime

Science Advisor

Homework Helper

- 9,395

- 4

-1, workmad3?

- #8

- 52

- 0

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)

- #9

- 32

- 0

The connection I was referring to concerns the following. For the solutions of the equations [tex]x^7=1[/tex] and [tex]y^9=1[/tex] we can form the systems:

[tex]X_1=x+\frac{1}{x}[/tex]

[tex]X_2=x^2+\frac{1}{x^2}[/tex]

[tex]X_3=x^4+\frac{1}{x^4}[/tex]

[tex]X_1=x^8+\frac{1}{x^8}=x+\frac{1}{x}[/tex]

and

[tex]Y_1=y+\frac{1}{y}[/tex]

[tex]Y_2=y^2+\frac{1}{y^2}[/tex]

[tex]Y_3=y^4+\frac{1}{y^4}[/tex]

[tex]Y_1=y^8+\frac{1}{y^8}=\frac{1}{y}+y[/tex].

We can form the equations [tex]X^3+X^2-2X-1=0[/tex] and [tex]Y^3-3Y+1=0[/tex]; the solutions of these equations satisfy the relation of the form [tex]Z_i^2-p=Z_j[/tex] 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

Share:

- Replies
- 1

- Views
- 3K