REMARK ON A GENERAL ARITHMETIC FOURIER TRANSFORM

Krystyna Grytczuk

Abstract. In this note we prove a general result connected with arithmetic Fourier transforms from which follows the main result given by Walker [1].

1. Introduction. In the paper [1], W. J. Walker proved the following result concerning arithmetic Fourier transforms. Let f be an even function of period 2π which is normalized, so that

∫_{0}^{2π} f(θ)dθ = 0.

Suppose that the Fourier series

f(θ) = ∑_{n=1}^{∞} a_{n} cos nθ

is absolutely convergent to f. Moreover, let

δ_{j}^{q} = \{ \begin{matrix}0 & if j contains a prime factor greater than the qth prime \\ 1 & otherwise. \end{matrix}  (1)

Recall that the Möbius function is defined by

μ(n) = \{ \begin{matrix}1 & if n = 1 \\ (−1)^{r} & if n = p_{1}p_{2} ⋯ p_{r} \\ 0 & if p^{2} | n. \end{matrix}  (2)

In addition, define S(n) by

S(n) = \frac{1}{n} ∑_{m=0}^{n−1} f(\frac{2πm}{n}).  (3)

Then, it follows that

a_{n} = lim_{q→∞} ∑_{j=1}^{∞} δ_{j}^{q}μ(j)S(jn); n ≥ 1.  (4)

Moreover, Walker indicated some connections and applications to the field of signal processing and to artificial neural networks.

2. Result. In the present paper we prove a general result.

\underline{Theorem}. Let the function h satisfy the following condition.

\frac{1}{n} ∑_{m=0}^{n−1} h(\frac{wmj}{n} ) = \{ \begin{matrix}0 & if j ≠ kn \\ 1 & if j = kn, \end{matrix}  (C)

where w ≠ 0 is a fixed real constant and k is a positive integer and let the function f have the expansion on the Fourier series

f(x) = ∑_{n=1}^{∞} a_{n}h(nx),  (5)

which is absolutely convergent to f. Then

a_{n} = lim_{q→∞} δ_{j}^{q}μ(j)S(jn),  (6)

where

S(n) = \frac{1}{n} ∑_{m=0}^{n−1} f(\frac{wm}{n})  (7)

and δ_{j}^{q} and μ are defined by (1) and (2), respectively.

\underline{Proof}. First we prove that by the hypothesis it follows that

S(n) = ∑_{k=1}^{∞} a_{kn}.  (8)

Rewriting the series (5) in the form

f(x) = ∑_{j=1}^{∞} a_{j}h(jx)

and putting x = (wm)/n, where w ≠ 0 is a real constant we obtain

f(\frac{wm}{n}) = ∑_{j=1}^{∞} a_{j}h(\frac{wmj}{n} ).  (9)

From (7) and (9) we get

S(n) = ∑_{j=1}^{∞} a_{j} \frac{1}{n} ∑_{m=0}^{n−1} h(\frac{wmj}{n} ).  (10)

Applying to (10) condition (C), we obtain

S(n) = ∑_{j=1}^{∞} a_{j}

for j = kn, so

S(n) = ∑_{k=1}^{∞} a_{kn}

and we see that (8) is proved. On the other hand, we know the following summation formula for the Möbius function.

∑_{d|n} μ(d) = \{ \begin{matrix}1 & if n = 1 \\ 0 & if n > 1. \end{matrix}

Using this formula and the Möbius inversion formula, by an easy calculation, we obtain

a_{n} = ∑_{k=1}^{∞} μ(k)S(kn).  (11)

Now, assuming δ_{j}^{q} is defined as in (1), consider the expression

T_{q}(n) = ∑_{j=1}^{∞} δ_{j}^{q}μ(j)S(jn).  (12)

Then from (11) and (12) we can deduce that

T_{q}(n) = a_{n} + ∑_{k=2}^{∞} α_{k}^{q}a_{kn}  (13)

where α_{k}^{q} = 0, if k contains one of the first q primes and α_{k}^{q} = 1, otherwise. From (13) and by the assumption of the Theorem, it follows that

lim_{q→∞} ∑_{k=2}^{∞} α_{k}^{q}a_{kn} = 0.

Hence, from (12) and (13) we obtain

lim_{q→∞} T_{q}(n) = lim_{q→∞} ∑_{j=1}^{∞} δ_{j}^{q}μ(j)S(jn) = a_{n}.

The proof of the Theorem is complete.

3. Remark. Now, we observe that from our Theorem we can obtain Walker’s result as a particular case. Indeed, let h(x) = cos x; w = 2π. Then for x = (wmj)/n we have

∑_{m=0}^{n−1} h(\frac{wmj}{n} ) = ∑_{m=0}^{n−1} cos \frac{2πmj}{n} .  (14)

On the other hand, it is easy to see that

\{ \begin{matrix}cos \frac{2πmj}{n} + i sin \frac{2πmj}{n} = (cos \frac{2π}{n} + i sin \frac{2π}{n} )^{mj} = ϵ^{mj} \\ cos \frac{2πmj}{n} − i sin \frac{2πmj}{n} = (cos \frac{2π}{n} − i sin \frac{2π}{n} )^{mj} = ϵ^{−mj} & . \end{matrix}  (15)

By (15), it follows that

cos \frac{2πmj}{n} = \frac{1}{2}(ϵ^{mj} + ϵ^{−mj}).  (16)

Hence, from (14) and (16) we obtain

∑_{m=0}^{n−1} h(\frac{wmj}{n} ) = ∑_{m=0}^{n−1} cos \frac{2πmj}{n} = \frac{1}{2} ∑_{m=0}^{n−1}(ϵ^{mj} + ϵ^{−mj}).  (17)

But it is well–known that if ϵ is a root of unity of degree n, then

∑_{m=0}^{n−1} ϵ^{mj} = \{ \begin{matrix}0 & if j ≠ kn \\ n & if j = kn \end{matrix}  (18)

for some positive integer k. Therefore, by (17) and (18) it follows that

\frac{1}{n} ∑_{m=0}^{n−1} h(\frac{wmj}{n} ) = \{ \begin{matrix}0 & if j ≠ kn \\ 1 & if j = kn \end{matrix}

so that condition (C) is satisfied for the function h(x) = cos x. Thus, by the Theorem with w = 2π, we obtain Walker’s result, since,

S(n) = \frac{1}{n} ∑_{m=0}^{n−1} f(\frac{wm}{n}) = \frac{1}{n} ∑_{m=0}^{n−1} f(\frac{2πm}{n}).

\underline{Reference}

1. W. J. Walker, “The Arithmetic Fourier Transform and Real Neural Networks: Summability by Primes,” J. Math. Anal. Appl., 190 (1995), 211–219.

Krystyna Grytczuk Institute of Mathematics Technical University Zielona Góra, Poland