Re: Coefficients of the Cyclotomic Polynomials

On Mar 14, 7:27 pm, Gottfried Helms <h...@xxxxxxxxxxxxx> wrote:
Am 14.03.2007 17:05 schrieb Gerry:



Hi all,

Say we offset the roots of the unit circle by a complex number (a+ib)

I know that for a<>0 and b=0 the coefficients are polynomials built by
Pascal's Triangle.

For example the Cyclotomic Polynomial of degree 6
gives the all one polynomial AOP of degree 5 as in

x^6-1=(x-1)(1+x+x^2+x^3+x^4+x^5)

Defining the coefficients of the degree 5 polynomial using Pascal's
triangle
(which can be defined by the Stirling numbers of the first kind) as in

c0=1+a+a^2+a^3+a^4+a^5
c1=1+2a+3a^2+4a^3+5a^4
c2=1+3a+6a^2+10a^3
c3=1+4a+10a^2
c4=1+5a
c5=1

we get the general polynomial of degree 5

c0+c1x+c2x^2+c3x^3+c4x^4+c5x^5

representing the roots of the unit circle shifted accross the real
axis.

Questions:

1) The polynomial coefficients c become complex as soon as b<>0.
How can we represent the complex coefficients?

2) Is there a definition for the coefficients of the cyclotomic
polynomials if one rotates the roots say clockwise or
counterclockwise?
It looks like the binomial coefficients are not very helpfull in this
case.
I suspect that the coefficient are related to the gamma function.

Any comments are welcome.

Gerry

Hmm, I don't know, whether I understood it right.
What I think is, that you want to replace in

x^6 - 1 = 0

x by (y- (a+bi)) to get a shifting of the cyclotomial
roots into the complex plane.
But then it would only mean to assign the binomial-coefficients
according to

(y - (a+bi))^6 - 1 = 0

and expanding the parenthese.

Or did you mean a completely different thing?

Gottfried Helms- Hide quoted text -

- Show quoted text -

Hi Gottfried,

Your answering question 1, if this gives the results
for a=0 i get the AOP.
for a=1 i get the polynomial : 6+15x+20x^2+15x^3+6^4+x^5
for a=2 i get the polynomial : 63+129x+111x^2+49x^3+11x^4+x^5
Yes
If b<>0 you get the complex part also.

Thanks
Gerry

.

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