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Pawel_Iks
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 Posted: Sat Jul 05, 2008 7:47 am Post subject: expression for n-th cyclotomic polynomial using Mobius funct |
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The n-th cyclotomic polynomial is defined as
Phi_n(x)=Product_{GCD(i,n)=1} (x-e_i), where e_i - squares of unit.
How to prove that:
Phi_n(x)=Product_{d divide n} (x^d-1)^mi(n/
d), (1)
where mi(n/d) is Mobius function?
I have an article where it is written that this formula is derived
from obious relation:
x^n-1=Product_{d divide n} Phi_d(x) (2)
by the exclusion inclusion rule ... but I have no idea what this rule
it is and how (1) could be derived from (2) such simply as is
described in this article ...
I have another question ... about some relations for the coefficients
of Phi_n(x) ... are there any? Is it true that they were considered as
a -1, 0 , 1 only in the past? Does anybody have some useful links
about history of this polynomials? |
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Tonico
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| Posted: Sat Jul 05, 2008 9:47 am Post subject: Re: expression for n-th cyclotomic polynomial using Mobius f |
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On Jul 5, 10:47 am, Pawel_Iks <pawel.labed...@gmail.com> wrote:
| Quote: |
The n-th cyclotomic polynomial is defined as
Phi_n(x)=Product_{GCD(i,n)=1} (x-e_i), where e_i - squares of unit..
How to prove that:
Phi_n(x)=Product_{d divide n} (x^d-1)^mi(n/
d), (1)
where mi(n/d) is Mobius function?
I have an article where it is written that this formula is derived
from obious relation:
x^n-1=Product_{d divide n} Phi_d(x) (2)
by the exclusion inclusion rule ... but I have no idea what this rule
it is and how (1) could be derived from (2) such simply as is
described in this article ...
I have another question ... about some relations for the coefficients
of Phi_n(x) ... are there any? Is it true that they were considered as
a -1, 0 , 1 only in the past? Does anybody have some useful links
about history of this polynomials?
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************************************************************
Google or Yahoo "Moebius Inversion": applying this inversion to the
above equation with x^n - 1 you get your result.
Regards
Tonio |
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