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 wildrabbitt 2020-04-11 15:00

need guidance with some maths

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Hi, I've been trying to understand some maths in one of my text books for a long time now. I've written out all the relevant material and then explained why I can't understand it. If anyone can point out the right way to understand it, as usual I'd be most appreciative.

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 Nick 2020-04-12 09:44

I'm having trouble understanding what you have written.
You start with q, which I assume is a fixed prime number, and define zeta as a primitive qth root of unity in the complex numbers.
You then choose a polynomial F and suppose that $$F(\zeta^m)=F(\zeta)$$ whenever $$v(m)\equiv 0\pmod{e}$$.
Presumably this e is not the constant 2.718... that you used earlier! Is e a fixed integer here?
In explaining what v(n) means, you then refer to both e and f - what is f?

 wildrabbitt 2020-04-13 11:47

2 Attachment(s)
I wrote out what I understand of it.

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 Nick 2020-04-13 13:54

Let's take your example of q=7. If

$f_0\zeta^0+f_1\zeta^1+f_2\zeta^2+f_3\zeta^3+f_4\zeta^4+f_5\zeta^5= g_0\zeta^0+g_1\zeta^1+g_2\zeta^2+g_3\zeta^3+g_4\zeta^4+g_5\zeta^5$
where the $$f_i$$ and $$g_i$$ are polynomials with integer (or rational) coefficients
then $$f_0=g_0$$, $$f_1=g_1$$,...,$$f_6=g_6$$.
So it's not a problem if the polynomial F has a constant term.
Replacing $$\zeta$$ with $$\zeta^m$$ permutes the coefficients of $$\zeta^1$$ up to $$\zeta^6$$ and leaves $$\zeta^0$$ unaltered.

 wildrabbitt 2020-04-13 15:44

Thanks very much. That's just what I needed. I can get on now with your help :smile:

 wildrabbitt 2020-04-13 16:11

Except I still don't know how it can be

that if $F(x) = \prod_{R}(x - \zeta^{R})$ has a constant term,

then $F(\zeta) = A_{0}(x)\eta_{0} + A_{1}(x)\eta_{1}$

as it seems to me there can't be a constant term on the right hand side of the latter.

 Nick 2020-04-13 16:59

On your 1st sheet, you have
$F(\zeta)=\sum_{r=1}^{q-1}A_r\zeta^r$
so there is no term with $$\zeta^0$$ (or $$\zeta^q$$ which is also 1).
On your 2nd sheet, you apply the idea to a polynomial which does have a term with $$\zeta^0$$.
Is the book you are using a well known one? If one of has it, perhaps we can help you pinpoint where the misunderstanding arises.

 wildrabbitt 2020-04-13 18:49

Multiplicative Number Theory by Harold Davenport published by Springer (part of a series - Graduate Texts in Mathematics.

Thanks.

 Nick 2020-04-13 20:29

OK, I think I understand the confusion.
Let's use your example of q=7 with $$\zeta=e^{2\pi i/7}$$.
Then we have
$\zeta^0=1,\zeta^1=\zeta,\zeta^2,\zeta^3,\zeta^4,\zeta^5,\zeta^6$
all distinct and $$\zeta^7=1$$ again.
We also have the equation
$\zeta^6+\zeta^5+\zeta^4+\zeta^3+\zeta^2+\zeta+1=0$
so it follows that
$-1=\zeta^6+\zeta^5+\zeta^4+\zeta^3+\zeta^2+\zeta$
Thus you can use $$\zeta^0$$ to $$\zeta^5$$ inclusive OR $$\zeta^1$$ to $$\zeta^6$$ inclusive and still equate coefficients (they are linearly independent).

I hope this helps!

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