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Old 2020-04-11, 15:00   #1
wildrabbitt
 
Jul 2014

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Default need guidance with some maths

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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Old 2020-04-12, 09:44   #2
Nick
 
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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?
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Old 2020-04-13, 11:47   #3
wildrabbitt
 
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Thanks for your interest.
I wrote out what I understand of it.


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Old 2020-04-13, 13:54   #4
Nick
 
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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.

Last fiddled with by Nick on 2020-04-13 at 20:19 Reason: Corrected typo
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Old 2020-04-13, 15:44   #5
wildrabbitt
 
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Thanks very much. That's just what I needed. I can get on now with your help
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Old 2020-04-13, 16:11   #6
wildrabbitt
 
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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.

Last fiddled with by wildrabbitt on 2020-04-13 at 16:12
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Old 2020-04-13, 16:59   #7
Nick
 
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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.
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Old 2020-04-13, 18:49   #8
wildrabbitt
 
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Multiplicative Number Theory by Harold Davenport published by Springer (part of a series - Graduate Texts in Mathematics.


Thanks.
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Old 2020-04-13, 20:29   #9
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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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