Interpolation by generalized exponential sums with equal weights

preprint2020arXivOpen access

Abstract

Here we solve Padé and Prony interpolation problems for the generalized exponential sums with equal weights: Hn(z;h)=μn∑k=1nh(λkz),whereμ,λk∈C,H_n(z; h)=\fracμ{n}\sum_{k=1}^n h(λ_k z),\quad \text{where}\quad μ,λ_k\in \mathbb{C}, and hh is a fixed analytic function under few natural assumptions. The interpolation of a function ff by HnH_n is due to properly chosen μμ and {λk}k=1n\{λ_k\}_{k=1}^n, which depend on ff, hh and nn. The sums HnH_n are related to the hh-sums and generalized exponential sums, i.e. to Hn∗(z;h)=∑k=1nλkh(λkz)andHn(z;h):=∑k=1nμkh(λkz),whereμk,λk∈C,\mathcal{H}^*_n(z; h)=\sum_{k=1}^n λ_k h(λ_k z)\quad \text{and}\quad\mathcal{H}_n(z; h):=\sum_{k=1}^n μ_k h(λ_k z),\quad \text{where}\quad μ_k,λ_k\in \mathbb{C}, which generalize many classical approximants and whose properties are actively studied. As for the Padé problem, we show that HnH_n and Hn∗\mathcal{H}_n^* have similar constructions and rates of interpolation, whereas calculating HnH_n requires less arithmetic operations. Although the Padé problem for Hn\mathcal{H}_n is known to have a doubled interpolation rate with respect to Hn∗\mathcal{H}_n^* and thus to HnH_n, it can be however unsolvable in many useful cases and this may entirely eliminate the advantage of Hn\mathcal{H}_n. We show that, in contrast to Hn\mathcal{H}_n, the Padé problem for HnH_n always has a unique solution. More importantly, we also obtain efficient estimates for μμ and λkλ_k, valuable by themselves, and use them in further evaluating interpolation quality and in applications. The Padé problem and estimates provide a basis for managing the more interesting Prony problem for exponential sums with equal weights Hn(z;exp⁡)H_n(z;\exp), i.e. when h(z)=exp⁡(z)h(z)=\exp(z). We show that it is uniquely solvable and surprisingly μμ and λkλ_k can be efficiently estimated. This is in sharp contrast to the case of well-known exponential sums Hn(z;exp⁡)\mathcal{H}_n(z;\exp).

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