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Z 2 l →{± 1 } be the mapping μ ( t )=( 1) c ,where c is the most significant
bit of t
Let μ :
Z 2 l ;inotherwordsitmaps0 , 1 , ..., 2 l− 1
1to+1and2 l− 1 , 2 l− 1 +
1 , ..., 2 l
1. Our goal is to express this map as a linear combination of
characters. Recall the Fourier transformation formula on
1to
Z
2 l :
2 l
2 l
1
1
1
2 l
μ =
μ j ψ j , where μ j =
μ ( x ) ψ j (
x ) .
(1)
j =0
x =0
Combining Lemma 4.1 and Corollary 7.4 of [5], we obtain
Lemma 3.1. Let l =3 . For the constants μ j =(1+ ω −j + ω 2 j + ω 3 j ) / 4
j =1 , 3 , 5 , 7 we have
μ = μ 1 ψ 1 + μ 3 ψ 3 + μ 5 ψ 5 + μ 7 ψ 7 ,
and μ j =0 , for even j .Furthermore
) 2 =2+ 2 .
(
|
μ 1 |
+
|
μ 3 |
+
|
μ 5 |
+
|
μ 7 |
Let q =2 l where l
4 .Then
q− 1
j =0 |
2
π ln( q )+1 .
μ j |
<
For all β
= 0 in the ring R = GR (2 l ,m ), we denote by Ψ β the additive character
,x
C
ω Tr( βx ) .
Ψ β : R
Note that for the defined above ψ k and Ψ β ,wehave:
ψ k (Tr( βx )) = Ψ βk ( x ) .
Let f ( X ) denote a polynomial in R [ X ]andlet
f ( X )= F 0 ( X )+2 F 1 ( X )+ ... +2 l− 1 F l− 1 ( X )
denote its 2-adic expansion. Let d i be the degree in x of F i . Let Ψ be an arbitrary
additive character of R ,andset D f to be the weighted degree of f , defined as
d 0 2 l− 1 ,d 1 2 l− 2 ,...,d l− 1 }
D f =max
{
.
With the above notation, and for any integers k, H such that 0 ≤ k<k + H− 1
2 m
2, we have, under mild technical conditions, the bound
2
+1 D f 2 m .
Ψ ( f ( ξ j ))
k + H− 1
π ln 4(2 m
1)
(2)
π
j = k
See [10] for details.
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