aa r X i v : . [ m a t h . HO ] S e p A Remark on Fourier Transform
Guangwu Xu ∗ Abstract
In this note, we describe an interpretation of the (continuous)Fourier transform from the perspective of the Chinese Remainder The-orem. Some related issues are discussed.
Keywords: Chinese Remainder Theorem; Fourier Transform.
The usual Chinese Remainder Theorem (CRT) concerns reconstructing aninteger given its remainders with respect to a set of coprime moduli. Fora set of pairwise coprime positive integers { m , m , . . . , m µ − } , we let Γ = Q µ − j =0 m j . Then there is an isomorphism Z / (Γ) → µ − Y j =0 Z / ( m j ) (1) n (cid:18) n (mod m ) , n (mod m ) , · · · , n (mod m µ − ) (cid:19) . The crucial part is to verify the inverse (of the isomorphism), which is givenby the CRT: finding (the unique) solution n <
Γ of the system of congruence ∗ Department of EE & CS, University of Wisconsin-Milwaukee, Milwaukee, WI 53211,USA; e-mail: [email protected] . x ≡ r (mod m ) x ≡ r (mod m ) · · · x ≡ r µ − (mod m µ − ) (2)CRT provides a formula (reported by Jiushao Qin (aka Chin Chiu-shao) in1247 [4]) for the solution: n = µ − X j =0 r j u j Γ m j (mod Γ) (3)where u j = (cid:0) Γ m j (cid:1) − (mod m j ). It is remarked that in [4], Qin explained abeautiful algorithm for computing (the positive value) u j (under the nameof ‘Dayan deriving one’). As discussed in [6, 7], Qin’s algorithm is somehowmore concise and efficient than the modern Extended Euclidean algorithm.We would like to point out that Qin also noted the following relation µ − X j =0 u j Γ m j = 1 + ℓ Γ , (4)and gave it the names ‘positive use’ (for ℓ = 1) and ‘universal use’ (for ℓ > Z / ( m j )) of smaller size and arithmetic operations on thosesmaller rings are closed and independent, so computation task of large scalecan be decomposed into that of smaller scales and performed in parallel, see[1, 2]. Other applications of CRT in fast computation can be found in publickey cryptography.It is well-known that the Chinese remainder theorem has a general for-mulation to decompose a certain ring to be a product of ‘smaller’ quotientrings. The correctness of the decomposition in essence is (3). As an example,replacing Z / (Γ) with R [ x ] / ( x n −
1) and Z / ( m j ) with R [ x ] / ( x − e − jπin ), weget the (finite) discrete Fourier transform: for a polynomial f ∈ R [ x ] with2egree less than n , f (cid:18) f (1) , f ( e − πin ) , f ( e − πin ) , · · · , f ( e − n − πin ) (cid:19) . The inverse of the transform is exactly in the same format like (3) and it isnow commonly called the Lagrange interpolation formula. For the computa-tional significance of the discrete Fourier transform, besides being paralleliz-able, the symmetry of the roots of x n − Now we consider a discretization of the continuous Fourier transform underthe framework of Chinese remainder theorem. We shall assume functionsinvolved are all good enough (e.g., in Schwartz space [5]) so the convergencewill not be an issue.Let f : R → C be a rapidly decreasing smooth function, its Fouriertransform is given by ˆ f ( y ) = Z R f ( x ) e − πixy dx (5)The inverse of the transform is proved to be f ( x ) = Z R ˆ f ( y ) e πixy dy. (6)We will use carefully formulated finite sums to approximate these twoabsolutely convergent integrals. Let N, M be large even integers (depend on f ). 3or the spacial domain R , we choose interval [ − N , N ] and partition itwith each unit interval being divided into M equal subintervals. Take thefollowing partition points x j = − N + jM for j = 0 , , · · · , M N − R , we choose interval [ − M , M ] and partitionit with each unit interval being divided into N equal subintervals. Take thefollowing partition points y k = − M + kN for k = 0 , , · · · , M N -1.To approximately evaluate the transform at partition points y k , we seethat ˆ f ( y k ) ≈ Z N − N f ( x ) e − πixy k dx ≈ MN − X j =0 f ( x j ) e − πix j y k M = 1 M MN − X j =0 f ( x j ) e − πi ( Mxj )( Nyk ) MN . (7)Similarly, we can get f ( x j ) ≈ N MN − X k =0 ˆ f ( y k ) e πi ( Nyk )( Mxj ) MN (8)We should note that N y k (resp. M x j ) runs over all integers from − MN to MN − k = 0 , , · · · , M N − j = 0 , , · · · , M N − w = NM and and ω = e − πi MN . Denote P ( X ) = 1 M (cid:0) f ( x w )+ f ( x w +1 ) X + · · · + f ( x w − ) X w − + f ( x ) X w + f ( x ) X w +1 + · · · + f ( x w − ) X w − (cid:1) . Then (7) says that P ( X ) (mod X − ω Ny k ) ≈ ˆ f ( y k ) = ˆ f ( y Ny k + MN ) , and hence ˆ f can be approximated by the Chinese remainder representationof P ( X ) ( treated as an element of C [ X ] / ( X MN −
1) ).With this interpretation, we can illustrate a way of recovering f given thevalues of ˆ f ( y k ) , k = 0 , , · · · , N M − X MN − m j = X − ω j . We also note that (cid:16) Γ m j ( ω j ) (cid:17) − is (cid:16) Γ m j (cid:17) − m j ), i.e., u j = Q NM − ℓ =0 ,ℓ = j ( ω j − ω ℓ ) = ω j MN . We have heuristically that P ( X ) ≈ NM − X k =0 ˆ f ( y ( k + MN ) (mod NM ) ) u k Γ m k . We note that M f ( x j ) is the coefficient of X q in P ( X ) with q = ( j + MN )(mod M N ), in order to get its expression, we take q th derivatives of bothsides and evaluate them at X = 0: f ( x j ) ≈ N MN − X k =0 ˆ f ( y k ) e πi ( Nyk )( Mxj ) MN . As mentioned above, this derivation of the approximation formula forinverse Fourier transform we just illustrated is rather heuristic. A morerigorous treatment inside this framework can be done by using the Dirichletkernel which is defined as D ( x ) = ( sin( πMx + πxN )sin( πxN ) if xN / ∈ Z M N + 1 if xN ∈ Z We shall simply explain this for the case of x j = 0, i.e., j = MN .1 N MN − X k =0 ˆ f ( y k ) = 1 N MN − X k =0 Z R f ( x ) e − πixy k dx = 1 N Z R f ( x ) MN − X k =0 e − πixy k dx = 1 N Z R f ( x ) MN X k =0 e − πi xN ( − MN + k ) dx − N Z R f ( x ) e − πiMx dx = 1 N Z R f ( x ) D ( x ) dx + O (1) N ≈ f (0) . We also want to remark that parallelization and FFT are thus possiblefor numerical evaluations of the transform. It is interesting to note that we can omit the (mod ) operation here since the degreeof both sides are less than NM. We also note that P NM − j =0 u j Γ m j = P NM − j =0 Γ mj Γ mj ( ω j ) = 1. .1 Poisson summation formula The classical 1-dimensional Poisson summation formula states that for arapidly decreasing smooth function f , one has X n ∈ Z f ( n ) = X n ∈ Z ˆ f ( n ) . (9)It is noted that from our previous formula (5),ˆ f ( − M f ( − M · · · + ˆ f ( M − f ( y ) + ˆ f ( y N ) + ˆ f ( y N ) + · · · + ˆ f ( y ( M − N ) ≈ M M − X ℓ =0 MN − X j =0 f ( x j ) e − πi ( − MN ℓN ) MxjMN = 1 M MN − X j =0 f ( x j ) M − X ℓ =0 (cid:16) e − πi jM (cid:17) ℓ = f ( − N f ( − N · · · + f ( N − Acknowledgement
The author thanks Dr. Shangbin Cui for useful discussions.
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