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Computed unitary radix-two FFT has the multiplicative stage error bound

Statement

Let N=2m, m≥0, and suppose a successful execution of Computed unitary radix-two FFT with certified twiddle phases has no range failures. If every twiddle certificate satisfies ϵℓ≤ϵtw≤1/4 and u≤1/16, put η=10u+ϵtw. Then

∥F^Nx−FNx∥2≤((1+η)m−1)∥x∥2.

The conjugated inverse routine has the same bound. If mη≤1, the right side is at most 2mη∥x∥2. Thus an O(ulog⁡N) shorthand requires both ϵtw=O(u) and ulog⁡N small; the theorem makes no such shorthand when twiddle errors are uncontrolled.

Facts & Assumptions

Given: N=2m, x∈CN, a successful execution, u≤1/16, and ϵℓ≤ϵtw≤1/4 for every twiddle certificate.

[F1]

For N=2M, split x into even and odd subsequences e,o; the unnormalised DFT obeys Xk(x)=Xk(e)+e−2πik/NXk(o) and Xk+M(x)=Xk(e)−e−2πik/NXk(o), while FN=N−1/2XN (The radix-two even/odd factorisation of the DFT, The unnormalised engineering DFT and its conversion to the unitary transform, The unitary discrete Fourier transform on Z/NZ). The recursive FFT computes XN exactly (The recursive radix-two fast Fourier transform, Correctness of the recursive radix-two FFT).

[F2]

Every computed normalized butterfly differs from its exact block by at most η times the pair input norm (A computed unitary radix-two butterfly has error at most (10u + εtw) times its input norm).

[F3]

The unitary DFT preserves the squared Euclidean norm; in particular the exact transforms of the even and odd subsequences have squared norms summing to ∥x∥22 (Finite Parseval and Plancherel identity for the unitary DFT).

[F4]

Complex conjugation preserves modulus and the Euclidean norm, and the inverse unitary DFT is obtained by conjugating input, applying the forward DFT, and conjugating output (Conjugation is an involutive real-field automorphism, zz‾=∣z∣2, and modulus is definite, multiplicative, and subadditive, Finite Fourier inversion for the unitary transform on Z/NZ).

[F5]

The real exponential is its power series (Real elementary functions and their domains).

[F6]

For z∈CN, ∥z∥22=∑j∣zj∣2 (The counting inner product on CZ/NZ).

Proof

1.1givenF1F7algebrabase

For m=0, N=1 and the algorithm returns the input exactly, so the error is 0=(1+η)0−1. For m≥1, [F1] gives the exact unnormalised top outputs Xk(e)+wXk(o) and Xk(e)−wXk(o), with w=e−2πik/N. The child unitary outputs are Ek=Xk(e)/N/2 and Ok=Xk(o)/N/2; dividing the parent unnormalised outputs by N=2N/2 gives exactly Uw(Ek,Ok)=2−1/2(Ek+wOk,Ek−wOk). By [F7], ∣w∣=1 and the scaling identity holds, so the columns of Uw have squared norm 1 and inner product 2−1(w−w)=0; hence Uw preserves the pair norm.

2.1givenF2F3F6step 1.1algebrainductionih

Suppose the recursive transform of length N/2 has error bound ρm−1=(1+η)m−1−1. By [F3] and [F6], stacking the exact even and odd transforms has norm ∥x∥2; stacking their computed errors has norm at most ρm−1∥x∥2, since the even and odd input coordinates partition x. Hence the stacked computed suboutputs have norm at most (1+ρm−1)∥x∥2. For each k, [F2] bounds the top butterfly error by η times the norm of its computed input pair. Squaring these bounds and summing over k shows that the total top-stage error norm is at most η(1+ρm−1)∥x∥2. The prior subtransform errors contribute at most ρm−1∥x∥2 through the direct sum of exact norm-preserving blocks in step 1.1. Therefore the total error is at most [ρm−1+η(1+ρm−1)]∥x∥2=((1+η)m−1)∥x∥2. Induction proves the forward bound for every m.

3.1F4step 2.1

By [F4], the inverse implementation is conjugation before and after the forward routine and conjugation preserves every coordinate modulus. Applying step 2.1 to x‾ gives the same error norm for the inverse direction.

4.1givenF5step 2.1algebradischarge-induction: step 2.1∎

If mη≤1, then 1+η≤eη by the power series [F5], so (1+η)m≤emη. For 0<t≤1, et−1=∑j≥1tj/j!≤t∑j≥11/j!<2t; at t=0, et−1=2t=0. Thus ((1+η)m−1)≤2mη. If ϵtw=O(u), this is O(mu)=O(ulog⁡2N), with the smallness condition already imposed.

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