Skopje, Makedonija ABOUT CHARACTERS ON VILENKIN ...

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$Fractional and operational calculus with generalized fractional ...$

$Numerical Methods Course Notes Version 0.1 (UCSD Math 174, Fall ...$

42 MEDO PEPIĆ e) For n ∈ N given by (5) and x ∈ G given by (3) holds ⎛ ( ) ∞∑ k j j∑ ⎞ χ n (x) =exp⎝2πi · x s m s ⎠ . m j+1 j=0 s=0 Proof. (Lemma 1). a) By the facts that r n ∈ Γ n+1 Γ n and Γ n+1 Γ n is a cyclic group of prime order p n+1 follows r p n+1 n ∈ Γ n Γ n−1 . Therefrom and using the fact that Γ n Γ n−1 is a cyclic group of prime order p n follows r p n+1.p n n ∈ Γ n−1 Γ n−2 . By continuing reasoning describe above we conclude that for each 0 ≤ s ≤ n holds rn pn+1.pn....ps+1 ∈ Γ s Γ s−1 . This proves assertion a) since obviously holds (∀s ∈{0, 1, 2,...,n})p n+1 .p n ...p s+1 = m n+1 m s . Let us notice, that from (21) follows r p n+1.p n ...p s+1 n = r p s+1 s = r s−1 . b) If s>n,then r n ∈ Γ s and therefrom r n (x s g s )=1sincex s g s ∈ Γ s Γ s+1 .Let us arbitrarily choose s ∈{0, 1, 2,...,n}. Then x s g s ∈ Γ s Γ s+1 , for all 1 ≤ m n+1 ms x s
ABOUT CHARACTERS ON VILENKIN GROUPS 43 χ k (x) = n∏ j=0 e) By (5) ∧ (18 ∗ ) ∧ c) wehave χ n (x) = r kj j n (x) = ∏ k j · exp ⎜ ⎝ 2πi j=0 j=0 ⎛ ⎞ j∑ x s m s ⎟ ⎠ s=0 m j+1 ⎛ ( ) n∑ k j j∑ ⎞ = exp⎝2πi · x s m s ⎠ . m j+1 ∞∏ j=0 r kj j s=0 ∞ (x) = ∏ k j · exp ⎜ ⎝ 2πi j=0 j=0 ⎛ ⎞ j∑ x s m s ⎟ ⎠ s=0 m j+1 ⎛ ( ) ∞∑ k j j∑ ⎞ = exp⎝2πi · x s m s ⎠ . m j+1 s=0 □ Proof. (Theorem 1). Let G be given Vilenkin group. Let k ∈ [m n ,m n+1 )be natural number given by k = n∑ k j m j , 1 ≤ k n
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