Introduction to Refinable Functions

During the summers of 2008 and 2009 I participated in a NSF funded REU at Texas A&M. Under the guidance of Dr. David Larson I researched, among other things, refinable functions. Here and in following posts I will present some of the results obtained during these mathematically stimulating summers.

Definition: A function $f : \mathbb{R} \rightarrow \mathbb{R}$ is finitely m-refinable if there exist integers $N_1, N_2$ with $N_1 < N_2$ and $c_{N_1} , \ldots , c_{N_2} \in \mathbb{R}$ with $c_{N_1} \not = 0$ and $c_{N_2} \not = 0$ such that $f(x) = \sum_{k = N_1}^{N_2} c_{k} f(m x - k)$ for all $x \in \mathbb{R}$ .

The set $\{ c_i \}_{i=N_1}^{N_2}$ is called the scaling (or refinement) sequence of $f$ and $\sum_{k = N_1}^{N_2} c_{k} f(m x - k)$ is called the scaling equation.

For the sake of simplicity we will consider only finitely 2-refinable functions. So when we say a function is refinable, unless otherwise noted, we mean a finitely 2-refinable function.

Examples:

Our first example is the characteristic function of $[0,1)$ also known as the scaling function $\phi$ of the Haar Wavelet defined so that

$\phi (x) =\begin{cases} 1 & x \in [0,1)\\ 0 & x \not \in [0,1) \\ \end{cases}$ and graphed below.

phi

It has the refinement sequence $c_0=c_1=1$ so that $\phi(x) = \phi(2x) + \phi(2x-1)$ . The first term is graphed in red the and the second term graphed in green below.

phirefinement

The second example is the hat function defined so that

$Hat(x) = \begin{cases} x & x \in [0,1)\\ 2-x & x \in [1,2)\\ 0 & x \not \in [0,2)\\ \end{cases}$

and so called because of it’s shape shown below.

Hat It has the refinement sequence $c_0=c_2 = \frac{1}{2}$ and $c_1 = 1$ so that $Hat(x) = \frac{1}{2} Hat(2x) + Hat(2x - 1) + \frac{1}{2} Hat(2x - 2)$ . Each term in the refinement sequence is graphed below.

hatrefinement

The final example of a refinable function is actually a whole class of functions that are refinable: polynomials. While this might not be obvious at first (it wasn’t for me) take a few minutes and work it out on your own. It’s a fun little puzzle and provides you with a better understanding of refinable functions and their interesting properties. In my next post I will prove some basic properties of refinable functions.

Mathematica: The images used in this post were created using Mathematica and the following code:

Clear[\[Phi]]
\[Phi][x_] := \[Piecewise] {
 {1, 0 <= x < 1},
 {0, True}
 }
Plot[\[Phi][x], {x, -.5, 1.5}, PlotStyle -> Thick,
 PlotRange -> {0, 1.5}]
Plot[{\[Phi][2 x], \[Phi][2 x - 1]}, {x, -.5, 1.5},
 PlotStyle -> {{Thick, Dashed, Red}, {Thick, Dashed, Green}},
 PlotRange -> {0, 1.5}]

Clear[Hat]
Hat[x_] := \[Piecewise] {
 {x, 0 <= x < 1},
 {2 - x, 1 <= x < 2},
 {0, True}
 }
Hat[x_] := If[0 <= x < 1, x, If[1 <= x < 2, 2 - x, 0]]
Plot[Hat[x], {x, -.5, 2.5}, PlotStyle -> Thick,
 PlotRange -> {0, 1.2}]
Plot[{1/2 Hat[2 x], Hat[2 x - 1], 1/2 Hat[2 x - 2]}, {x, -.5, 2.5},
 PlotStyle -> Thick, PlotRange -> {0, 1.2}]
Plot[.5 Hat[2 x] + Hat[2 x - 1] + .5 Hat[2 x - 2], {x, -.5, 2.5},
 PlotStyle -> Thick]

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This entry was posted on Saturday, July 18th, 2009 at 10:07 am and is filed under Refinable Functions. Tagged: Refinable Functions. You can feed this entry. You can leave a response, or trackback from your own site.

5 Responses

John, good work! I am positively thrilled you are finally blogging and it is even more splendid the topic is about something you are passionate about–Maths. I am now going to set my calculator to “stun”.

I look forward to learning more about this intriguing subject-maths-as the world follows your blog.

Take care,

Morgan

Morgan - July 21, 2009 at 12:08 am

Reply
- Thanks Morgan. I’m going to take some Garry gum and then some anti-Garry gum in the hopes that I might create some interesting maths.
  
  jmeuser - July 21, 2009 at 11:35 pm
  
  Reply
John, I have some questions. I have tried to find scaling equations for a couple of polynomials. For polynomials, will the scaling sequence always be between 0 and 1? For arbitrary m, are all polynomials m-refinable? Given a scaling equation, is there a way to solve for f in closed form, or at least know something about f? Are there initial conditions required? I found a scaling equation for f(x)=x^2+1. I noticed that when I found the second derivative, the sum of the scaling sequence was 1; is this perhaps associated with being a constant? Is this perhaps some sign of polynomial behavior, that if the nth derivative of a scaling equation yields a scaling sequence whose sum is 1, is f an nth degree polynomial?

P.S.: How do I post LaTeX symbols?

Ted - July 21, 2009 at 3:37 pm

Reply
- To use LaTeX in wordpress (which is the reason I put this blog on wordpress) simply write the dollar sign followed by latex and then enclose it in a dollar sign at the end. You can also follow this link to find out more about how WordPress supports LaTeX: http://support.wordpress.com/latex/
  
  jmeuser - July 21, 2009 at 5:29 pm
  
  Reply
- I know that all polynomials are finitely 2-refinable, and I think with a few additions it can easily be shown to be m-refinable. If you come with a proof I would be happy to let you post it here. I was thinking of writing a post on refinable polynomials sometime in the future.
  
  Sadly if you are given a scaling sequence and even some initial conditions on $f$ I don’t think you are guaranteed to find a reconstruction formula. More on this later though.
  
  You are right about the constant functions. Let $f: \mathbb{R} \rightarrow \mathbb{R}$ so that $f(x) = a \in \mathbb{R}$ for all $x \in \mathbb{R}$ and choose some $N_1 < N_2$ then any refinement sequence must satisfy
  
  $f(x) = a = \sum_{k = N_1}^{N_2} c_k f(m x - k) = \sum_{k=N_1}^{N_2} c_k a$
  By dividing both sides by a you see that this implies that
  $\sum_{k = N_1}^{N_2} c_k = 1$ .
  
  jmeuser - July 21, 2009 at 6:33 pm
  
  Reply

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