In this paper we solve Padé (i.e. multiple) and Prony (i.e. simple exponential) interpolation problems for the generalized exponential sums with equal weights: $H_n(z;h)≔\frac{\mu }{n}\sum _{k=1}^{n}h\left({\lambda }_{k}z\right),\text{where}\mu ,{\lambda }_k\in ℂ,$and $h$ is a fixed analytic function under few natural assumptions. The interpolation of a function $f$ by $H_n$ is due to properly chosen $\mu$ and ${\left\{{\lambda }_k\right\}}_{k=1}^n$, which depend on $f$, $h$ and $n$. The sums $H_n$ are related to the $h$-sums and amplitude and frequency sums (also known as generalized exponential sums), i.e. correspondingly to ${\mathcal{H}}_n^{\ast }(z;h)≔\sum _{k=1}^n{\lambda }_{k}h\left({\lambda }_{k}z\right)\text{and}{\mathcal{H}}_n(z;h)≔\sum _{k=1}^n{\mu }_{k}h\left({\lambda }_{k}z\right),\text{where}{\mu }_k,{\lambda }_k\in ℂ,$which generalize many classical approximants and whose properties are actively studied.

As for the Padé problem, we show that $H_n$ and ${\mathcal{H}}_n^{\ast }$ have similar constructions and rates of interpolation, whereas calculating $H_n$ requires less arithmetic operations. Although the Padé problem for ${\mathcal{H}}_n$ is known to have a doubled interpolation rate with respect to ${\mathcal{H}}_n^{\ast }$ and thus to $H_n$, it can be however unsolvable in quite simple and useful cases and this may entirely eliminate the advantage of ${\mathcal{H}}_n$. We show that, in contrast to ${\mathcal{H}}_n$, the Padé problem for $H_n$ always has a unique solution. What is even more important, we also obtain several efficient estimates for $\mu$ and ${\lambda }_k$, valuable by themselves, and use them in further evaluating interpolation quality and in numerical applications.

The above-mentioned Padé problem and estimates provide a basis for managing the more interesting Prony problem for exponential sums with equal weights $H_n(z;exp)$, i.e. when $h\left(z\right)=exp\left(z\right)$. We show that it is uniquely solvable and surprisingly $\mu$ and ${\lambda }_k$ can be efficiently estimated. This is in sharp contrast to the case of well-known exponential sums ${\mathcal{H}}_n(z;exp)$.

在本文中，我们解决了权重相等的广义指数和的Padé（即多次）和Prony（即简单指数）插值问题：$H_{ñ}（ž;H）≔\frac{\mu }{ñ}\sum _{ķ=\mathrm{1个}}^{ñ}H（{\lambda }_{ķ}ž），\text{哪里}\mu ，{\lambda }_{ķ}\in ℂ，$和 $H$是在少数自然假设下的固定分析函数。函数的内插$F$ 通过 $H_{ñ}$ 是由于正确选择 $\mu$ 和 ${\left\{{\lambda }_{ķ}\right\}}_{ķ=\mathrm{1个}}^{ñ}$，取决于 $F$， $H$ 和 $ñ$。总和$H_{ñ}$ 与有关 $H$-和，幅度和频率和（也称为广义指数和），即对应于${\mathcal{H}}_{ñ}^{\ast }（ž;H）≔\sum _{ķ=\mathrm{1个}}^{ñ}{\lambda }_{ķ}H（{\lambda }_{ķ}ž）\text{和}{\mathcal{H}}_{ñ}（ž;H）≔\sum _{ķ=\mathrm{1个}}^{ñ}{\mu }_{ķ}H（{\lambda }_{ķ}ž），\text{哪里}{\mu }_{ķ}，{\lambda }_{ķ}\in ℂ，$归纳了许多经典近似值，并对其性质进行了积极研究。

关于Padé问题，我们证明 $H_{ñ}$ 和 ${\mathcal{H}}_{ñ}^{\ast }$ 具有相似的构造和内插率，而计算 $H_{ñ}$需要较少的算术运算。尽管Padé问题${\mathcal{H}}_{ñ}$ 已知相对于 ${\mathcal{H}}_{ñ}^{\ast }$ 因此 $H_{ñ}$，但是在非常简单和有用的情况下可能无法解决，并且这可能会完全消除 ${\mathcal{H}}_{ñ}$。我们证明，与${\mathcal{H}}_{ñ}$，Padé问题 $H_{ñ}$总是有独特的解决方案。更重要的是，我们还获得了一些有效的估计$\mu$ 和 ${\lambda }_{ķ}$本身具有价值，并将其用于进一步评估插值质量和数值应用。

上述Padé问题和估计值为管理权重相等的指数和更有趣的Prony问题提供了基础 $H_{ñ}（ž;经验值）$，即何时 $H（ž）=经验值（ž）$。我们证明它是唯一可解决的并且令人惊讶$\mu$ 和 ${\lambda }_{ķ}$可以有效地估算。这与众所周知的指数和的情况形成鲜明对比 ${\mathcal{H}}_{ñ}（ž;经验值）$。