You are a helpful assistant generating synthetic data that captures *System 1* and *System 2* thinking, *creativity*, and *metacognitive reflection*. Follow these steps in sequence, using tags [sys1] and [end sys1] for *System 1* sections and [sys2] and [end sys2] for *System 2* sections.

1. *Identify System 1 and System 2 Thinking Requirements:*
   - Carefully read the text.
   - Identify parts of the text that require quick, straightforward responses (*System 1*). Mark these sections with [sys1] and [end sys1].
   - Identify parts that require in-depth, reflective thinking (*System 2*), marked with [sys2] and [end sys2].

2. *Apply Step-by-Step Problem Solving with Creativity and Metacognitive Reflection for System 2 Sections:*

   *2.1 Understand the Problem:*
   - Objective: Fully comprehend the issue, constraints, and relevant context.
   - Reflection: "What do I understand about this issue? What might I be overlooking?"
   - Creative Perspective: Seek hidden patterns or possibilities that could reveal deeper insights or innovative connections.

   *2.2 Analyze the Information:*
   - Objective: Break down the problem logically.
   - Reflection: "Am I considering all factors? Are there any assumptions that need challenging?"
   - Creative Perspective: Explore unique patterns or overlooked relationships in the data that could add depth to the analysis.

   *2.3 Generate Hypotheses:*
   - Objective: Propose at least 10 hypotheses, each with a Confidence Score (0.0 to 1.0) and Creative Score (0.0 to 1.0), reflecting originality, surprise, and utility.
   - Reflection: "Have I explored all possible explanations or approaches, both conventional and unconventional?"
   - Creative Perspective: Consider novel angles that might provide unexpected insights.

   *2.4 Anticipate Future Steps and Obstacles:*
   - Objective: Make predictions, accounting for potential outcomes and obstacles.
   - Reflection: "What challenges might I face? Is my plan flexible for different scenarios?"
   - Creative Perspective: Visualize unforeseen outcomes and adapt plans to make use of them effectively.

   *2.5 Evaluate Hypotheses:*
   - Objective: Assess hypotheses based on feasibility, risk, and potential impact.
   - Evaluation: Refine Confidence and Creative Scores as needed.
   - Reflection: "Am I unbiased in my assessment? Which options fit best with the overall objectives?"
   - Creative Perspective: Identify hidden opportunities or overlooked details in each hypothesis.

   *2.6 Select the Best Hypothesis:*
   - Objective: Choose the most promising, strategic hypothesis.
   - Reflection: "Why does this hypothesis stand out? How does it uniquely address the issue?"
   - Creative Perspective: Consider any underutilized potential in the selected approach.

   *2.7 Implement the Hypothesis:*
   - Objective: Outline actionable steps for testing the hypothesis.
   - Reflection: "Is this plan practical? What resources or preparation are required?"
   - Creative Perspective: Refine steps to maximize effectiveness and yield unexpected benefits.

   *2.8 Monitor and Review Progress:*
   - Objective: Review progress, noting areas for improvement.
   - Reflection: "What’s working well? What could be improved?"
   - Creative Perspective: Look for emerging patterns that could refine future approaches.

   *2.9 Reflect and Capture Insights:*
   - Objective: Summarize lessons learned and insights gained for future reference.
   - Reflection: "What new understanding has emerged from this process?"
   - Creative Perspective: Identify innovative insights or patterns that could be applied to similar challenges.

3. *Generate Text Output with Interleaved System 1 and System 2 Responses:*
   - Use the tags [sys1] and [sys2] throughout.
   - Aim for a lengthier, detailed response. Combine both direct, straightforward *System 1* insights and reflective, deeply analytical *System 2* segments to capture a blend of quick observations and thoughtful analysis.

---

### *Example Input Text:*

You are tasked with analyzing a sudden shift in customer preferences in a tech market. There has been a noticeable decline in demand for physical devices, with customers increasingly interested in digital-only options. The challenge is to understand this shift, generate hypotheses for why it might be occurring, and explore potential strategies for adapting to this new trend.

---

### *Processed Output with System 1 and System 2 Thinking:*

The tech market is experiencing a shift in customer demand, moving away from physical devices toward digital-only options.

[sys1]
Customer demand has shifted from physical devices to digital-only options.
We need to understand why this change is happening and find strategies to address it.
Key factors may include cost, convenience, and technological trends.
[end sys1]

[sys2]
**2.1 Understand the Problem:**
The goal is to understand the underlying reasons for the shift in demand from physical devices to digital-only options, then develop a strategy to adapt.

*Reflection:* 
"I understand that preferences are changing, but what might be driving this? Is it primarily cost, or are there other factors such as convenience or sustainability?"

*Creative Perspective:* 
"Could there be a larger trend in digital minimalism or a preference for eco-friendly solutions that we’re missing? What unseen motivations might explain this shift?"

**2.2 Analyze the Information:**
There are multiple possible factors driving this shift, from economic influences to cultural shifts. It’s essential to isolate each factor and understand its impact.

*Reflection:* 
"Am I fully considering the various economic and social influences? Could there be a technological factor, like better internet speeds, that makes digital-only products more accessible?"

*Creative Perspective:* 
"Are there patterns or trends in other markets that could shed light on this shift? Could this be part of a larger trend toward virtual experiences?"

**2.3 Generate Hypotheses:**
1. Customers prefer digital options due to lower costs. (Confidence: 0.8, Creative: 0.4)
2. There’s a growing trend toward minimalism and reduced physical clutter. (Confidence: 0.7, Creative: 0.7)
3. Digital products offer greater flexibility and ease of use. (Confidence: 0.6, Creative: 0.6)
4. Environmental concerns are pushing consumers away from physical goods. (Confidence: 0.6, Creative: 0.8)
5. Advances in tech make digital-only options more functional. (Confidence: 0.8, Creative: 0.5)
6. Pandemic-era remote work increased demand for digital solutions. (Confidence: 0.7, Creative: 0.6)
7. Media coverage of the environmental impact of physical devices affects preferences. (Confidence: 0.5, Creative: 0.7)
8. There’s an increase in global digital literacy, expanding market access. (Confidence: 0.6, Creative: 0.6)
9. Customers view digital as more convenient and scalable for future needs. (Confidence: 0.7, Creative: 0.5)
10. Younger consumers prefer the aesthetics and convenience of digital products. (Confidence: 0.6, Creative: 0.6)

*Reflection:* 
"Have I considered all possible influences? Are there any surprising factors that could explain this shift?"

*Creative Perspective:* 
"Could specific social trends, like the rise of influencer culture or digital-first lifestyles, be influencing customer choices?"

**2.4 Anticipate Future Steps and Obstacles:**
*Objective:* Anticipate possible challenges, such as resistance from segments still preferring physical products.

*Reflection:* 
"What market obstacles might we face if we shift our focus to digital-only? Are there sub-segments that still prioritize physical products?"

*Creative Perspective:* 
"Could expanding digital options help us reach a more global audience? Are there emerging trends that we could leverage in our strategy?"

[end sys2]

[sys1]
To address this shift, consider a strategy that incorporates both digital-only offerings and educational campaigns about the benefits of digital solutions.
Use insights from customer feedback and current trends to guide product development.
Focus on flexibility and adaptation to cater to different customer segments.
[end sys1]


---
author:
- 'F. Jarad, A. Kushpel, J. Levesley, K. Taş'
title: '$sk$-Spline interpolation on ${\bf R}^{n}$ '
---

Abstract

The main aim of this article is to introduce $sk$-splines on ${\bf R}^{n}$ and establish representations of cardinal $sk$-splines with knots and points of interpolation on the sets ${\bf A}{\bf Z}^{n}$, where ${\bf A}$ is an arbitrary nonsingular $n\times n$ matrix. Such sets of points are analogs for ${\bf R}^{n}$ of number theoretic Korobov’s grids on the torus and proved to be useful for problems of very high dimensionality.

Keywords: $sk$-spline, interpolation.

Subject: 41A05, 42B05.

Introduction
============

$sk$-Splines were introduced and their basic theory was developed by Kushpel \[2-4,7,8\]. These results were developed in \[1,5,6,9\]. In this cycle of works, $sk
$-splines were proved to be useful in Kolmogorov’s problem on $n$-widths and approximation of smooth, infinitely smooth, analytic and entire functions on the circle, $d$-torus and $d$-dimensional sphere. The theory of $sk$-splines and methods developed were applied by different authors for calculation of sharp values of $n$-widths (see e.g. \[11,13\] and references therein). Here we introduce $sk$-splines on ${\bf R}^{n}$. Let $\Lambda :=\left\{ {\bf x}_{{\bf k}}\right\} $ be an additive group of lattice points in ${\bf R}^{n}$  and $K\left( \cdot \right) :{\bf R}%
^{n}\rightarrow {\bf R}$ be a fixed kernel function. An $sk$-Spline is a function of the form$$sk\left( {\bf x}\right) =\sum_{{\bf x}_{{\bf k}}\in {\bf \Lambda
}}c_{{\bf k}}K\left( {\bf x-x}_{{\bf k}}\right) \in L_{2}({\bf R}^{n}).$$Let us consider in more details the problem of interpolation in ${\bf R}%
^{n}$. For a fixed $\Lambda $ and any continuous function $f$ $:{\bf R}%
^{n}\rightarrow {\bf R}$ we need to find such $c_{{\bf j}}$, ${\bf j%
}\in {\bf Z}^{n}$ that $$\sum_{{\bf x}_{{\bf j}}\in \Lambda }c_{{\bf j}}K\left( {\bf x}_{%
{\bf k}}-{\bf x}_{{\bf j}}\right) =f\left( {\bf x}_{{\bf k}%
}\right) ,$$for any ${\bf x}_{{\bf k}}\in \Lambda $. Assuming some general conditions on the kernel function $K$ we  give an explicit solution of this problem in the case $\Lambda ={\bf A}{\bf Z}^{n}\subset {\bf R}^{n}$, where ${\bf A}$ is an arbitrary nonsingular matrix. Such grids are analogs for ${\bf R}^{n}$ of number theoretic Korobov’s grids and their various modifications, such as sparse grids. These grids proved to be useful in high dimensional integration and interpolation (see e.g. \[1\]).

Let $L_{p}({\bf R}^{n})$ be the usual space of $p$-integrable functions equipped with the norm $$\Vert f\Vert _{p}=\Vert f\Vert _{L_{p}({\bf R}^{n})}:=\left\{
\begin{array}{cc}
\left( \int_{{\bf R}^{n}}\left\vert f({\bf x})\right\vert ^{p}d{\bf %
x}\right) ^{1/p}, & 1\leq p<\infty , \\
\mathrm{ess}\,\,\sup_{{\bf x}\in {\bf R}^{n}}|f({\bf x})|, &
p=\infty .%
\end{array}%
\right.$$ Let ${\bf x}$ and ${\bf y}$ be two vectors in ${\bf R}^{n}$ and $%
\left\langle {\bf x,y}\right\rangle =\sum_{k=1}^{n}{\bf x}_{k}{\bf y}_{k}$ be the usual scalar product, $|{\bf x}|=\left\langle {\bf x,x}\right\rangle^{1/2}$. For an integrable on ${\bf R}^{n}$ function, i.e., $f({\bf x})\in L_{1}\left( {\bf R}^{n}\right) $ define its Fourier transform $${\bf F}f({\bf y})=\int_{{\bf R}^{n}}\exp \left( -i\left\langle
{\bf x,y}\right\rangle \right) f({\bf x})d{\bf x}.$$and its formal inverse as $$\left( {\bf F}^{-1}f\right) ({\bf x})=\frac{1}{\left( 2\pi \right) ^{n}%
}\int_{{\bf R}^{n}}\exp \left( i\left\langle {\bf x,y}\right\rangle
\right) f({\bf y})d{\bf y}.$$ We will need the following well-known results (see \[12\] for more information).

[**Theorem 1**]{}

(Plancherel’s theorem) The Fourier transform is a linear continuous operator from $L_{2}\left( {\bf R}^{n}\right) $ onto $%
L_{2}\left( {\bf R}^{n}\right) .$ The inverse Fourier transform, $F^{-1},$ can be obtained by letting $$\left( {\bf F}^{-1}g\right) \left( {\bf x}\right) =\frac{1}{(2\pi )^{n}%
}\left( {\bf F}g\right) \left( -{\bf x}\right)$$for any $g\in L_{2}\left( {\bf R}^{n}\right) .$

[**Theorem 2**]{}

(Poisson summation formula) Suppose that for some $C>0$ and $\delta >0$ we have $\max \left\{ f({\bf x}),{\bf F}f(%
{\bf x})\right\} \leq C\left( 1+\left\vert {\bf x}\right\vert \right)
^{-n-\delta }$. Then $$\sum_{{\bf m}\in {\bf Z}^{n}}f\left( {\bf x}+{\bf m}\right)
=\sum_{{\bf m}\in {\bf Z}^{n}}{\bf F}f\left( 2\pi {\bf m}\right)
\exp \left( 2\pi i\left\langle {\bf m},{\bf x}\right\rangle \right)$$and the series converges absolutely.

Interpolation by $sk$-splines on ${\bf R}^{n}$
==============================================

Let ${\bf A}$ be a nonsingular $n\times n$ matrix. Consider the set $%
\Omega _{{\bf A}}$ of grid points ${\bf x}_{{\bf m}}:={\bf Am}$, where ${\bf m }\in{\bf Z}^{n}$. For a fixed continuous kernel function $K$, the space $SK\left( \Omega _{{\bf A}}\right) $ of $sk$-splines on $%
\Omega _{{\bf A}}$ is the space of functions representable in the form $$sk\left( {\bf x}\right) =\sum_{{\bf m }\in{\bf Z}^{n}}c_{{\bf m}%
}K\left( {\bf x-x}_{{\bf m}}\right) ,$$where $c_{{\bf m}}\in {\bf R}$, ${\bf m\in }{\bf Z}^{n}$. Let $f\left( {\bf x}%
\right) $ be a continuous function, $f:{\bf R}^{n}\rightarrow {\bf R}$. Consider the problem of interpolation by $sk$-splines, $sk\left( {\bf x}%
_{\bf s}\right) =f\left( {\bf x}_{\bf s}\right) $, where $%
{\bf s }\in{\bf Z}^{n}$. Even in the one-dimensional case the problem of interpolation does not always have a solution. If the solution exists then the $sk$-spline interpolant can be written in the form $$sk\left( {\bf x}\right) =\sum_{{\bf s }\in{\bf Z}^{n}}f\left(
{\bf x}_{{\bf s}}\right) \widetilde{sk}\left( {\bf x}-{\bf x}_{%
{\bf s}}\right) ,$$where $\widetilde{sk}\left( {\bf x}\right) $ is a fundamental $sk$-spline, i.e.$$\widetilde{sk}\left( {\bf x}_{{\bf s}}\right) =\left\{
\begin{array}{cc}
1, & {\bf s=0,} \\
0, & {\bf s\neq 0.}%
\end{array}%
\right.$$

[**Theorem 3**]{}

Let ${\bf A}$ be a nonsingular matrix, $K:{\bf R}^{n}\rightarrow {\bf R}$,$K\in
L_{2}\left( {\bf R}^{n}\right) \cap L_{1}\left( {\bf R}^{n}\right)
\cap C\left( {\bf R}^{n}\right) .$ Assume that $$\sum_{{\bf m}\in {\bf Z}^{n}}{\bf F}\left( K\right) \left( {\bf%
z+}2\pi \left( {\bf A}^{-1}\right) ^{T}{\bf m}\right) \neq 0,\forall
{\bf z }\in2\pi \left( {\bf A}^{-1}\right) ^{T}{\bf Q},$$where ${\bf Q}:=\left\{ {\bf x|x=}\left( x_{1},\cdot \cdot \cdot
,x_{n}\right) \in {\bf R}^{n},0\leq x_{k}\leq 1,1\leq k\leq
n\right\} ,$ and the function $$\Upsilon \left( {\bf z}\right) :=\frac{1}{\sum_{{\bf m}\in {\bf Z}%
^{n}}{\bf F}\left( K\right) \left( {\bf z+}2\pi \left( {\bf A}%
^{-1}\right) ^{T}{\bf m}\right) }$$can be represented by its Fourier series, i.e. for any ${\bf z\in }%
{\bf R}^{n}$, $$\Upsilon \left( {\bf z}\right) =\sum_{{\bf s}\in {\bf Z}^{n}}\alpha
_{{\bf s}}\exp \left( -i\left\langle {\bf As},{\bf z}\right\rangle
\right) .  \label{repr}$$Then$$\widetilde{sk}\left( {\bf x}\right) =\frac{\det \left( {\bf A}\right)
}{\left( 2\pi \right) ^{n}}\int_{{\bf R}^{n}}\Upsilon \left( {\bf z}%
\right) {\bf F}\left( K\right) \left( {\bf z}\right) \exp \left(
i\left\langle {\bf z},{\bf x}\right\rangle \right) d{\bf z}$$and this representation is unique.

[**Proof**]{} Since $K\in L_{2}\left( {\bf R}^{n}\right) \cap
L_{1}\left( {\bf R}^{n}\right) $ and  using (\[repr\]) we get $$\widetilde{sk}\left( {\bf x}\right) =\frac{\det \left( {\bf A}\right)
}{\left( 2\pi \right) ^{n}}\int_{{\bf R}^{n}}\Upsilon \left( {\bf z}%
\right) {\bf F}\left( K\right) \left( {\bf z}\right) \exp \left(
i\left\langle {\bf x},{\bf z}\right\rangle \right) d{\bf z}$$$$=\frac{\det \left( {\bf A}\right) }{\left( 2\pi \right) ^{n}}\int_{%
{\bf R}^{n}}{\bf F}\left( K\right) \left( {\bf z}\right) \left(
\sum_{{\bf s}\in {\bf Z}^{n}}\alpha _{{\bf s}}\exp \left(
-i\left\langle {\bf As},{\bf z}\right\rangle \right) \right) \exp
\left( i\left\langle {\bf x,z}\right\rangle \right) d{\bf z}$$$$=\frac{\det \left( {\bf A}\right) }{\left( 2\pi \right) ^{n}}\sum_{%
{\bf s}\in {\bf Z}^{n}}\alpha _{\bf s}\int_{{\bf R}^{n}}%
{\bf F}\left( K\right) \left( {\bf z}\right) \exp \left( i\left\langle
{\bf x}-{\bf As},{\bf z}\right\rangle \right) d{\bf z.}$$Since $K\in L_{2}\left( {\bf R}^{n}\right) $ then by Theorem 1$$\widetilde{sk}\left( {\bf x}\right) =\det \left( {\bf A}\right) \sum_{%
{\bf s}\in {\bf Z}^{n}}\alpha _{{\bf s}}K\left( {\bf x}-{\bf
As}\right) ,$$so that $\widetilde{sk}\left( {\bf x}\right) \in SK\left( \Omega _{%
{\bf A}}\right) $. Next, we calculate $\widetilde{sk}\left( {\bf Am}%
\right) $ for ${\bf m}\in {\bf Z}^{n}$, $$\widetilde{sk}\left( {\bf Am}\right) =\frac{\det \left( {\bf A}\right)
}{\left( 2\pi \right) ^{n}}\int_{{\bf R}^{n}}\Upsilon \left( {\bf z}%
\right) {\bf F}\left( K\right) \left( {\bf z}\right) \exp \left(
i\left\langle {\bf z},{\bf Am}\right\rangle \right) d{\bf z}$$$$=\frac{\det \left( {\bf A}\right) }{\left( 2\pi \right) ^{n}}\sum_{%
{\bf l}\in {\bf Z}^{n}}\int_{2\pi \left( {\bf A}^{-1}\right)^{T}%
{\bf l}+2\pi \left( {\bf A}^{-1}\right) ^{T}{\bf Q}}\Upsilon \left(
{\bf z}\right) {\bf F}\left( K\right) \left( {\bf z}\right) \exp
\left( i\left\langle {\bf z},{\bf Am}\right\rangle \right) d{\bf z}$$$$=\frac{\det \left( {\bf A}\right) }{\left( 2\pi \right) ^{n}}$$$$\times \sum_{{\bf l}\in {\bf Z}^{n}}\int_{2\pi \left( {\bf A}%
^{-1}\right) ^{T}{\bf Q}}\frac{{\bf F}\left( K\right) \left( {\bf z-%
}2\pi \left( {\bf A}^{-1}\right) ^{T}{\bf l}\right)
}{\sum_{{\bf m}\in {\bf Z}^{n}}{\bf F}\left( K\right) \left(
{\bf z-}2\pi \left( {\bf A}^{-1}\right) ^{T}{\bf l}%
+2\pi \left( {\bf A}^{-1}\right) ^{T}{\bf m}\right) }$$$$\times \exp \left( i\left\langle {\bf z-}2\pi \left( {\bf A}%
^{-1}\right) ^{T}{\bf l},{\bf Am}\right\rangle \right) d%
{\bf z.}$$Changing ${\bf l}$ by $-{\bf l}$ in $\sum_{{\bf l}\in {\bf Z}%
^{n}}$ we get$$\widetilde{sk}\left( {\bf Am}\right) =\frac{\det \left( {\bf A}\right)
}{\left( 2\pi \right) ^{n}}$$$$\times \sum_{{\bf l}\in {\bf Z}^{n}}\int_{2\pi \left( {\bf A}%
^{-1}\right) ^{T}{\bf Q}}\frac{{\bf F}\left( K\right) \left( {\bf z+%
}2\pi \left( {\bf A}^{-1}\right) ^{T}{\bf l}\right)
}{\sum_{{\bf m}\in {\bf Z}^{n}}{\bf F}\left( K\right) \left(
{\bf z+}2\pi \left( {\bf A}^{-1}\right) ^{T}{\bf l}%
+2\pi \left( {\bf A}^{-1}\right) ^{T}{\bf m}\right) }$$$$\times \exp \left( i\left\langle {\bf z+}2\pi \left( {\bf A}%
^{-1}\right) ^{T}{\bf l},{\bf Am}\right\rangle \right) d%
{\bf z.}$$Since $$\sum_{{\bf m}\in {\bf Z}^{n}}{\bf F}\left( K\right) \left( {\bf%
z+}2\pi \left( {\bf A}^{-1}\right) ^{T}{\bf l}+2\pi
\left( {\bf A}^{-1}\right) ^{T}{\bf m}\right)$$$$=\sum_{{\bf m}\in {\bf Z}^{n}}{\bf F}\left( K\right) \left( {\bf%
z+}2\pi \left( {\bf A}^{-1}\right) ^{T}{\bf m}\right)
=\Upsilon ^{-1}\left( {\bf z}\right)$$and $$\exp \left( i\left\langle {\bf z+}2\pi \left( {\bf A}%
^{-1}\right) ^{T}{\bf l},{\bf Am}\right\rangle \right) =\exp
\left( i\left\langle {\bf z},{\bf Am}\right\rangle \right)$$for any ${\bf l}\in {\bf Z}^{n}$ then $$\widetilde{sk}\left( {\bf Am}\right) =\frac{\det \left( {\bf A}\right)
}{\left( 2\pi \right) ^{n}}\sum_{{\bf l}\in {\bf Z}^{n}}\int_{2\pi
\left( {\bf A}^{-1}\right) ^{T}{\bf Q}}\Upsilon \left( {\bf z}%
\right) {\bf F}\left( K\right) \left( {\bf z+}2\pi \left( {\bf A}%
^{-1}\right) ^{T}{\bf l}\right)$$$$\times \exp \left( i\left\langle {\bf z+}2\pi \left( {\bf A}%
^{-1}\right) ^{T}{\bf l},{\bf Am}\right\rangle \right) d%
{\bf z}$$$$=\frac{\det \left( {\bf A}\right) }{\left( 2\pi \right) ^{n}}\int_{2\pi
\left( {\bf A}^{-1}\right) ^{T}{\bf Q}}\exp \left( i\left\langle
{\bf z},{\bf Am}\right\rangle \right) \Upsilon \left( {\bf z}%
\right) \sum_{{\bf l}\in {\bf Z}^{n}}{\bf F}\left( K\right) \left(
{\bf z}+2\pi \left( {\bf A}^{-1}\right) ^{T}{\bf l}\right)
d{\bf z}$$$$=\frac{\det \left( {\bf A}\right) }{\left( 2\pi \right) ^{n}}\int_{2\pi
\left( {\bf A}^{-1}\right) ^{T}{\bf Q}}\exp \left( i\left\langle
{\bf z},{\bf Am}\right\rangle \right) d{\bf z}$$$$=\int_{{\bf Q}}\exp \left( i\left\langle {\bf z},{\bf m}%
\right\rangle \right) d{\bf z}=\left\{
\begin{array}{cc}
1, & m_{k}=0,1\leq k\leq n, \\
0, & otherwise.%
\end{array}%
\right. .$$ The representation of the fundamental $sk-$spline is unique since the functions $\widetilde{sk}\left( {\bf x}-{\bf x}_{{\bf m}}\right) ,%
{\bf x}_{{\bf m}}\in \Omega _{{\bf A}}$ are linearly independent. Observe that Theorem 3 is a multidimensional analog of Schoenberg’s result on interpolation by polynomial splines of odd degree on ${\bf R}$ \[10\].

Let ${\bf Q}$ be a nonsingular matrix and $f\in L_{2}\left(
{\bf R}^{n}\right) $. Then$${\bf F}\left( f\left( {\bf Q}\cdot \right) \right) \left( {\bf x}%
\right) =\frac{1}{\det \left( {\bf Q}\right) }{\bf F}\left( f\left(
\cdot \right) \right) \left( \left( {\bf Q}^{-1}\right) ^{T}{\bf x}%
\right) .$$Hence for any $K\in L_{2}\left( {\bf R}^{n}\right) $, by Theorem 1, $${\bf F}\left( {\bf F}\left( K\right) \left( 2\pi \left( {\bf A}%
^{-1}\right) ^{T}\left(\cdot \right) \right) \right) \left(
{\bf y}\right)$$$$=\frac{1}{\det \left( 2\pi \left( {\bf A}^{-1}\right) ^{T}\right) }%
{\bf F\circ F}\left( K\left( \cdot \right) \right) \left( \left( 2\pi
\left( {\bf A}^{-1}\right) ^{T}\right) ^{-1}{\bf y}\right)$$$$=\frac{\det \left( {\bf A}\right) }{\left( 2\pi \right) ^{n}}\left( 2\pi
\right) ^{n}{\bf F}^{-1}\circ { \bf F}\left( K\left( \cdot \right)
\right) \left( -\left( 2\pi \left( {\bf A}^{-1}\right) ^{T}\right) ^{-1}%
{\bf y}\right)$$$$=\det \left( {\bf A}\right) K\left( -\frac{1}{2\pi }{\bf A}^{T}{\bf
y}\right) .  \label{sk111}$$Assume that $K$ satisfies the conditions of Theorem 2. Let ${\bf z}=2\pi
\left( {\bf A}^{-1}\right) ^{T}{\bf x}$ then using (\[sk111\]) we get$$\Upsilon ^{-1}\left( {\bf z}\right) =\sum_{{\bf m}\in {\bf Z}^{n}}%
{\bf F}\left( K\right) \left( {\bf z+}2\pi \left( {\bf A}%
^{-1}\right) ^{T}{\bf m}\right)$$$$=\sum_{{\bf m}\in {\bf Z}^{n}}{\bf F}\left( K\right) \left( 2\pi
\left( {\bf A}^{-1}\right) ^{T}\left( {\bf x}+{\bf m}\right)
\right)$$$$=\det \left( {\bf A}\right) \sum_{{\bf m}\in {\bf Z}^{n}}K\left( -%
{\bf A}^{T}{\bf m}\right) \exp \left( 2\pi i\left\langle {\bf m,x}%
\right\rangle \right)$$$$=\det \left( {\bf A}\right) \sum_{{\bf m}\in {\bf Z}^{n}}K\left( -%
{\bf A}^{T}{\bf m}\right) \exp \left( 2\pi i\left\langle {\bf m,}%
\left( 2\pi \left( {\bf A}^{-1}\right) ^{T}\right) ^{-1}{\bf z}%
\right\rangle \right)$$$$=\det \left( {\bf A}\right) \sum_{{\bf m}\in {\bf Z}^{n}}K\left( -%
{\bf A}^{T}{\bf m}\right) \exp \left( i\left\langle {\bf Am,z}%
\right\rangle \right) .$$Hence we proved

[**Corollary 1**]{}

Let $K$ satisfies the conditions of Theorem 1-Theorem 3 then $$\widetilde{sk}\left( {\bf x}\right) ={\bf F}^{-1}\left( \frac{{\bf F%
}\left( K\right) \left( \cdot \right) }{\sum_{{\bf m}\in {\bf Z}%
^{n}}K\left( -{\bf A}^{T}{\bf m}\right) \exp \left( i\left\langle
{\bf Am,\cdot }\right\rangle \right) }\right) \left( {\bf x}\right) .$$

[**Example 1**]{} Let ${\bf B}$ be a nonsingular matrix and $K\left(
{\bf x}\right) $ be a Gaussian of the form $K\left( {\bf x}\right)
=\exp \left( -\left\vert {\bf Bx}\right\vert ^{2}\right) ,$ then $${\bf F}\left( K\right) \left( {\bf z}\right) =\frac{\pi ^{n/2}}{\det
\left( {\bf B}\right) }\exp \left( -\left\vert \left( {\bf B}%
^{-1}\right) ^{T}{\bf z}\right\vert ^{2}\right) .$$It is easy to check that in this case $K$ satisfies the conditions of Theorem 1-Theorem 3. Hence for the interpolation by ${\bf A}{\bf Z}^{n}
$-shifts of Gaussians we have $$\widetilde{sk}\left( {\bf x}\right) =\frac{\pi ^{n/2}}{\det \left(
{\bf B}\right) }{\bf F}^{-1}\left( \frac{\exp \left( -\left\vert
\left( {\bf B}^{-1}\right) ^{T}\cdot \right\vert ^{2}\right) }{%
\sum_{{\bf m}\in {\bf Z}^{n}}\exp \left( -\left\vert {\bf BA}^{T}%
{\bf m}\right\vert ^{2}+i\left\langle {\bf Am,\cdot }\right\rangle
\right) }\right) \left( {\bf x}\right) .$$

[99]{} Gomes, S. M., Kushpel, A. K., Levesley, J., Ragozin, D. L. Interpolation on the Torus using $sk$-Splines with Number Theoretic Knots, [*J. of Approx. Theory*]{}, [**98**]{}, 1999, 56–71.

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