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Exponential sums and congruences

  1. Oct 27, 2008 #1
    let be the exponential sum

    [tex] S= \sum_{n=1}^{N}e( \frac{f(x)}{p}) [/tex]

    [tex] e(x)= exp( 2i \pi x) [/tex]

    my conjecture is that since the complex exponential takes its maximum value '1' when x is equal to an integer then

    [tex] Re(S)= \Pi (f,N) [/tex] with [tex]\Pi (f,N) [/tex] is the number of solutions on the interval (1,N) of the congruence

    [tex] f(x) =0 mod(p) [/tex] and f(x) is a Polynomial.
     
  2. jcsd
  3. Nov 4, 2008 #2
    Forgive me if this is a stupid question- but what's [itex]p[/tex]? Or did you mean [itex]n[/tex] instead or [itex]p[/tex] as the number of prime factors of [itex]n[/tex] or something?
     
  4. Nov 5, 2008 #3
    any prime
     
  5. Nov 5, 2008 #4
    I still don't get over what exactly that summation for [itex]S[/tex] is done. A clarification please?
     
  6. Nov 6, 2008 #5
    o sorry.. i should have written


    [tex] S= \sum_{n=1}^{N}e( \frac{f(n)}{p}) [/tex]


    the sum is taken over 'n' but if the prime 'p' divides f(n) then the complex exponential is equal to '1'
     
  7. Nov 6, 2008 #6
    Okay then [itex]p[/tex] is a prime of one's choosing.

    We have,

    [tex]S = \sum_{n = 1}^{N} \exp{\left(\frac{2\pi i}{p}f(n)\right)}[/tex]

    Then,
    [tex]\Re(S) = \sum_{n = 1}^{N} \cos{\left(\frac{2\pi}{p}f(n)\right)}[/tex]

    If [itex]f(n)[/tex] is a multiple of [itex]p[/tex], then the the real part of [itex]S[/tex] will 'count' each solution of that congruence, but what about certain [itex]f(n)[/tex] values that don't and hence give rise to non-zero real and imaginary components? They won't be 1 in a single go, but they can possibly accumulate to values greater than 1 I think. So some bounds for such a theorem also become necessary if I haven't missed anything.
     
    Last edited: Nov 6, 2008
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