Book contents
- Frontmatter
- Contents
- Preface
- Glossary of Selected Symbols
- 1 Introduction
- 2 Smoothness
- 3 Uniqueness
- 4 Identifying Functions and Directions
- 5 Polynomial Ridge Functions
- 6 Density and Representation
- 7 Closure
- 8 Existence and Characterization of Best Approximations
- 9 Approximation Algorithms
- 10 Integral Representations
- 11 Interpolation at Points
- 12 Interpolation on Lines
- References
- Supplemental References
- Author Index
- Subject Index
2 - Smoothness
Published online by Cambridge University Press: 05 August 2015
- Frontmatter
- Contents
- Preface
- Glossary of Selected Symbols
- 1 Introduction
- 2 Smoothness
- 3 Uniqueness
- 4 Identifying Functions and Directions
- 5 Polynomial Ridge Functions
- 6 Density and Representation
- 7 Closure
- 8 Existence and Characterization of Best Approximations
- 9 Approximation Algorithms
- 10 Integral Representations
- 11 Interpolation at Points
- 12 Interpolation on Lines
- References
- Supplemental References
- Author Index
- Subject Index
Summary
In this chapter we study one of the basic properties of ridge function decomposition, namely smoothness. In the first section we ask the following question. If
is smooth, does this imply that each of the fi is also smooth? In the second section we ask this same question with regard to generalized ridge functions, i.e., linear combinations of functions of the form f(Ax), where the A are d × n real matrices, and f : Rd → R.
Ridge Function Smoothness
Let Ck(Rn), k ∊ Z+, denote the usual class of real-valued functions with all derivatives of order up to and including k being continuous. Assume F ∊ Ck(Rn) is of the form
where r is finite, i.e., F ∊ M(a1, …, ar), and the ai are given pairwise linearly independent vectors in Rn. What can we say about the smoothness of the fi? Do the fi necessarily inherit all the smoothness properties of the F?
When r = 1 the answer is yes, and there is essentially nothing to prove. That is, if
F(x) = f1(a1 · x)
is in Ck(Rn) for some a1 ≠ 0, then for c ∊ Rn satisfying a1 · c = 1 and all t ∊ R we have that F(tc) = f1(t) is in Ck(R). This same result holds when r = 2. As the a1 and a2 are linearly independent, there exists a vector c ∊ Rn satisfying a1 · c = 0 and a2 · c = 1. Thus
F(tc) = f1(a1 · tc) + f2(a2 · tc) = f1(0) + f2(t).
Since F(tc) is in Ck(R), as a function of t, so is f2. The same result holds for f1.
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- Information
- Ridge Functions , pp. 12 - 18Publisher: Cambridge University PressPrint publication year: 2015