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Derivative-window alias bounds including the P=2 knot branch

Statement

Use the data and aliases of Exact three-dimensional spline assignment aliases and FFT normalization. Write $h_{r} (0) = 1$ and, for $u_{r} = k_{r} / M_{r} \neq 0$ , $h_{r} (n) = ∣ u_{r} / (u_{r} + n) ∣^{P}$ for $n \neq 0$ ; when $u_{r} = 0$ , set $h_{r} (n) = 0$ for $n \neq 0$ . Put $a_{r} = \sum_{n \neq 0} h_{r} (n) .$ For $P \geq 3$ , define $v_{r} = \sum_{n \neq 0} ∣ k_{r} + n M_{r} ∣ h_{r} (n), J_{k, r} = (∣ k_{r} ∣ + v_{r}) \prod_{s \neq r} (1 + a_{s}) - ∣ k_{r} ∣,$ and $B_{k} = 2 π ∥ A^{- T} ∥_{2} {(\sum_{r = 1}^{3} J_{k, r}^{2})}^{1 / 2} .$ For $P \geq 3$ the physical derivative alias error satisfies $∥ \nabla_{x} b_{k} (t) - \nabla_{x} a_{k} (t) ∥_{2} \leq B_{k} .$ Each series has a finite p-series controller: for $0 < ∣ u_{r} ∣ < 1 / 2$ , $a_{r} \leq 2 \frac{∣ u_{r} ∣^{P}}{(1 - ∣ u_{r} ∣)^{P}} ζ (P), v_{r} \leq 2 M_{r} \frac{∣ u_{r} ∣^{P}}{(1 - ∣ u_{r} ∣)^{P - 1}} ζ (P - 1),$ where $ζ (s) = \sum_{m \geq 1} m^{- s}$ for $s > 1$ . For $J_{0} \geq 1$ , after summing exactly over $1 \leq ∣ n ∣ \leq J_{0}$ , the respective omitted tails are at most $2 \frac{∣ u_{r} ∣^{P}}{(1 - ∣ u_{r} ∣)^{P}} \frac{⌊ J_{0} ⌋^{1 - P}}{P - 1}, 2 M_{r} \frac{∣ u_{r} ∣^{P}}{(1 - ∣ u_{r} ∣)^{P - 1}} \frac{⌊ J_{0} ⌋^{2 - P}}{P - 2} .$

For the approved $P = 2$ case, the absolute derivative p-series above diverges when $u_{r} \neq 0$ . Instead use the finite derivative of the local piecewise window sum. At every open mesh box and each one-sided knot limit, put $C_{k, r} = \frac{2 M_{r}}{∣ D (k) ∣} + 2 π ∣ k_{r} ∣, B_{k} = ∥ A^{- T} ∥_{2} {(\sum_{r = 1}^{3} C_{k, r}^{2})}^{1 / 2} .$ This is a computable finite bound for the same physical derivative error. At a grid knot, use the declared $d_{k} (t)$ branch from Periodic cardinal-spline assignment and unitary mesh transform: it is a selected one-sided value or their average and obeys the bound, but is not asserted to be a classical derivative there. Thus for the full $P \geq 2$ range the bound is on $d_{k} (t) - \nabla_{x} a_{k} (t)$ , with $d_{k} = \nabla_{x} b_{k}$ wherever the classical derivative exists.

Facts & Assumptions

Given: A cell matrix $A$ , integers $M_{r} \geq 1$ , spline order $P \geq 2$ , signed mode $k$ with $∣ k_{r} ∣ < M_{r} / 2$ and $D (k) \neq 0$ , a fractional point $t \in [0, 1)^{3}$ , and the assignment factors $a_{k}, b_{k}$ of the periodic mesh definition.

[F1]

The exact signed alias expansion of $b_{k}$ has coefficients $r_{k, n}$ with $∣ r_{k, n} ∣ = \prod_{r} h_{r} (n_{r})$ , and $W_{P}$ is the causal $P$ -fold cardinal window (Exact three-dimensional spline assignment aliases and FFT normalization, Periodic cardinal-spline assignment and unitary mesh transform).

[F2]

The physical gradient is $\nabla_{x} = A^{- T} \nabla_{t}$ , and $\nabla_{x} a_{k} = - 2 π i A^{- T} k a_{k}$ (Neutral point sources and Fourier data on a three-dimensional torus).

[F3]

For $s > 1$ , $ζ (s) = \sum_{m \geq 1} m^{- s}$ converges (The p-series for a real exponent p converges exactly when p is greater than one).

[F4]

Uniform convergence of a series and its derivative series on a closed interval permits termwise differentiation, with real and imaginary parts treated separately (If continuously differentiable functions converge at one point and their derivatives converge uniformly on a closed interval, then the functions converge uniformly to a differentiable function whose derivative is the derivative limit).

Proof

technique · direct

1.1givenF3algebra

Suppose $P \geq 3$ and first take $u_{r} \neq 0$ . For $n \neq 0$ , $h_{r} (n) \leq \frac{∣ u_{r} ∣^{P}}{(1 - ∣ u_{r} ∣)^{P} ∣ n ∣^{P}}, ∣ k_{r} + n M_{r} ∣ h_{r} (n) \leq M_{r} \frac{∣ u_{r} ∣^{P}}{(1 - ∣ u_{r} ∣)^{P - 1} ∣ n ∣^{P - 1}},$ because $∣ u_{r} + n ∣ \geq (1 - ∣ u_{r} ∣) ∣ n ∣$ . Since $P > 1$ and $P - 1 > 1$ , [F3] proves absolute convergence and the stated $a_{r}, v_{r}$ bounds. To prove the remainders locally, put $L = ⌊ J_{0} ⌋ \geq 1$ . For $s > 1$ and each integer $m > L$ , monotonicity gives $m^{- s} \leq \int_{m - 1}^{m} x^{- s} d x$ . Summing and taking the limit yields $\sum_{m > J_{0}} m^{- s} \leq \int_{L}^{\infty} x^{- s} d x = L^{1 - s} / (s - 1)$ . Apply this with $s = P$ and $s = P - 1$ to the two majorants above. If $u_{r} = 0$ , all shifted factors and both series vanish.

1.2F1F2algebra

Now let $P = 2$ . The convolution $W_{2} (u) = \int_{0}^{1} 1_{[0, 1]} (u - s) d s$ is the length of $[0, 1] \cap [u - 1, u]$ , hence is $u$ on $[0, 1]$ , $2 - u$ on $[1, 2]$ , and $0$ outside $[0, 2]$ . Its derivative is $1$ on $(0, 1)$ , $- 1$ on $(1, 2)$ , and $0$ outside, with the indicated one-sided values at $0, 1, 2$ . On every open mesh box, only finitely many window translates are nonzero, so direct differentiation gives $\partial_{t_{r}} b_{k} (t) = \frac{M_{r}}{\overline{D (k)}} \sum_{ℓ \in Z^{3}} W_{2}^{'} (M_{r} t_{r} - ℓ_{r}) \prod_{s \neq r} W_{2} (M_{s} t_{s} - ℓ_{s}) e^{- 2 π i \sum_{s} k_{s} ℓ_{s} / M_{s}} .$ The periodic residue version is identical because $k_{s}$ are integers and the phase is unchanged when $ℓ_{s}$ is shifted by $M_{s}$ . For every non-knot $y$ , one translate lies in $(0, 1)$ and one in $(1, 2)$ , so $\sum_{ℓ_{r}} ∣ W_{2}^{'} (y - ℓ_{r}) ∣ = 2$ ; at a knot the same sum of absolute one-sided values is at most $2$ . Also $W_{2} \geq 0$ and $\sum_{ℓ} W_{2} (y - ℓ) = 1$ by the partition identity proved in [F1]. Therefore $∣ \partial_{t_{r}} b_{k} ∣ \leq 2 M_{r} / ∣ D (k) ∣$ . Since $∣ \partial_{t_{r}} a_{k} ∣ = 2 π ∣ k_{r} ∣$ , the fractional derivative error is at most $C_{k, r}$ .

2.1F1F4step 1.1algebra

For $P \geq 3$ , [F1] gives $b_{k} (t) = \sum_{n \in Z^{3}} r_{k, n} e^{- 2 π i (k + n ⊙ M) \cdot t} .$ The absolute derivative sum in coordinate $r$ is finite by step 1.1 and the product bounds $\sum_{n} \prod_{s} h_{s} (n_{s}) = \prod_{s} (1 + a_{s}) < \infty$ . Apply [F4] to finite partial sums on each coordinate interval; their values converge at one point by the absolute series and their derivative series converge uniformly by this bound. Thus the displayed series differentiates termwise. Subtracting the $n = 0$ term gives $∣ \partial_{t_{r}} b_{k} - \partial_{t_{r}} a_{k} ∣ \leq 2 π \sum_{n \neq 0} ∣ k_{r} + n_{r} M_{r} ∣ \prod_{s} h_{s} (n_{s}) = 2 π J_{k, r} .$ The last equality follows by summing the $n_{r} = 0$ term separately: the sum including $n = 0$ is $(∣ k_{r} ∣ + v_{r}) \prod_{s \neq r} (1 + a_{s})$ , from which the base $∣ k_{r} ∣$ is removed.

3.1F2step 2.1algebra

The physical gradient is $A^{- T}$ times the three fractional derivative components. Hence [F2] and step 2.1 give $∥ \nabla_{x} b_{k} - \nabla_{x} a_{k} ∥_{2} \leq ∥ A^{- T} ∥_{2} {(\sum_{r} (2 π J_{k, r})^{2})}^{1 / 2} = B_{k} .$ This proves the $P \geq 3$ branch.

4.1F2step 1.2algebra∎

Applying $A^{- T}$ to the three component bounds as in step 3.1 gives $∥ \nabla_{x} b_{k} - \nabla_{x} a_{k} ∥_{2} \leq ∥ A^{- T} ∥_{2} (\sum_{r} C_{k, r}^{2})^{1 / 2} = B_{k}$ . At a knot each one-sided limit satisfies the same coordinate bounds, and an average lies in their convex hull, so the declared branch value also satisfies them. No classical differentiability at a $P = 2$ knot is used.

Remarks

For $P = 2$ the absolute Fourier derivative majorant diverges at nonzero $u_{r}$ . The finite piecewise window derivative avoids that invalid rearrangement.
A tighter certified enclosure may subdivide the closed mesh boxes and bound the finite piecewise expression on each box. The explicit $C_{k, r}$ above is a valid initial enclosure and does not require a supremum oracle.

Depends on

Used by

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39 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Markus Deserno and Christian Holm, How to mesh up Ewald sums (I): A theoretical and numerical comparison of various particle mesh routines (1998) (standard reference, not scraped)
Multi-Level Ewald: A hybrid multigrid / Fast Fourier Transform approach to the electrostatic particle-mesh problem (full-text article, 2011) (standard reference, not scraped)