Exponential sums and congruences

[mhill]

let be the exponential sum

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

e(x)= exp( 2i \pi x)

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

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

f(x) =0 mod(p) and f(x) is a Polynomial.

 

[yasiru89]

Forgive me if this is a stupid question- but what's p[/tex]? Or did you mean n[/tex] instead or p[/tex] as the number of prime factors of n[/tex] or something?

 

[soandos]

any prime

 

[yasiru89]

any prime

I still don't get over what exactly that summation for S[/tex] is done. A clarification please?

 

[mhill]

o sorry.. i should have written


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


the sum is taken over 'n' but if the prime 'p' divides f(n) then the complex exponential is equal to '1'

 

[yasiru89]

Okay then p[/tex] is a prime of one&#039;s choosing.<br /> <br /> We have,<br /> <br /> S = \sum_{n = 1}^{N} \exp{\left(\frac{2\pi i}{p}f(n)\right)}<br /> <br /> Then,<br /> \Re(S) = \sum_{n = 1}^{N} \cos{\left(\frac{2\pi}{p}f(n)\right)}<br /> <br /> If f(n)[/tex] is a multiple of p[/tex], then the the real part of S[/tex] will &amp;amp;amp;#039;count&amp;amp;amp;#039; each solution of that congruence, but what about certain f(n)[/tex] values that don&amp;amp;amp;amp;#039;t and hence give rise to non-zero real and imaginary components? They won&amp;amp;amp;amp;#039;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&amp;amp;amp;amp;#039;t missed anything.