MATH 414 Lecture 34

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Begin Exam 3 content

Properties of the $p_{k}$'s

Assume real-valued; if not, we'll need ${\overline {p}}_{k}$'s

Scaling function $\phi (x)=\sum _{k=-\infty }^{\infty }p_{k}\,\phi (2x-k)$, where

$p_{k}=2\int _{-\infty }^{\infty }\phi (x)\,\phi (2x-k)\,\mathrm {d} x$

Theorem 5.9:

1. $\sum _{k}p_{k-2\ell }p_{k-2m}=2\delta _{\ell ,m}$
2. $\sum _{k}p_{k-2\ell }^{2}=\sum _{k'}p_{k'}^{2}=2$

Theorem. $\sum _{k=-\infty }^{\infty }p_{k}=2$

Proof. Step 1: ${\begin{aligned}\int _{-\infty }^{\infty }\phi (x)\,\mathrm {d} x&=\int _{-\infty }^{\infty }\sum _{k=-\infty }^{\infty }p_{k}\phi (2x-k)\,\mathrm {d} x\\&=\sum _{k=-\infty }^{\infty }p_{k}\,\int _{-\infty }^{\infty }\phi (2x-k)\,\mathrm {d} x\end{aligned}}$

Step 2: change variables $u=2x-k$:

$\int _{-\infty }^{\infty }\phi (2x-k)={\frac {1}{2}}\int _{-\infty }^{\infty }\phi (u)\,\mathrm {d} u={\frac {M}{2}}$

Step 3: It takes work to show this, but we're left with

$M={\frac {1}{2}}\,\sum _{k}p_{k}\cdot M$, where $M\neq 0$, so $1={\frac {1}{2}}\,\sum _{k}p_{k}$, and $\sum _{k}p_{k}=2$

quod erat demonstrandum

1. ${\begin{aligned}\sum _{{\mbox{even}}\ k}p_{k}&=1=\sum _{\ell =-\infty }^{\infty }p_{2\ell }\\\sum _{{\mbox{odd}}\ k}p_{k}&=1=\sum _{\ell =-\infty }^{\infty }p_{2\ell +1}\end{aligned}}$

Example: Haar

$\phi (x)=\phi (2x)+\phi (2x+1)$, so $p_{0}=1,p_{1}=1$.

3. $p_{0}+p_{1}=2$

4. $p_{0}=1$ (even), $p_{1}=1$ (odd)

Example: Shannon

$\phi (x)=\mathrm {sinc} (x)={\begin{cases}{\frac {\sin {\pi \,x}}{\pi \,x}}&x\neq 0\\1&x=0\end{cases}}$

Scaling: $\phi (x)=\phi (2x)+\sum _{k=-\infty }^{\infty }{\frac {2(-1)^{k}}{(2k+1)\,\pi }}\,\phi (2x-1-2k)$

Hence $p_{k}={\begin{cases}1&k=0\\{\frac {2(-1)^{\frac {\ell -1}{2}}}{\pi \,k}}&{\mbox{odd}}\ \ell =2k+1\\0&{\mbox{otherwise}}\end{cases}}$

Now $p_{0}=1$ is the only even term.

Let $f(t)=\sum _{k\in \mathbb {Z} }{\frac {-2i}{(2k+1)\pi }}\,\mathrm {e} ^{(2k+1)\,i\,t}$ be the fourier series for the alternating function

Then $f\left({\frac {\pi }{2}}\right)=1=\sum _{k\in \mathbb {Z} }{\frac {-2i}{(2k+1)\pi }}\,\mathrm {e} ^{i\,k\,\pi }\cdot \mathrm {e} ^{i\,{\frac {\pi }{2}}}=\sum _{k\in \mathbb {Z} }{\frac {2(-1)^{k}}{(2k+1)\pi }}$

Daubechies $p=2$ MRA

$\phi (x)=\sum _{k=0}^{3}p_{k}\,\phi (2x-k)=p_{0}\,\phi (2x)+p_{1}\,\phi (2x-1)+p_{2}\,\phi (2x-2)+p_{3}\,\phi (2x-3)$, where

$p_{k}=\left\{{\frac {1+{\sqrt {3}}}{4}},{\frac {3+{\sqrt {3}}}{4}},{\frac {3-{\sqrt {3}}}{4}},{\frac {1-{\sqrt {3}}}{4}}\right\}$

These were derived from the properties, and not from a given $\phi$.

Wavelets

$W_{j}=\left\{w\in V_{j+1}~\mid ~w\perp V_{j}\right\}$

Looking for $\psi (x)$ such that $\left\{\psi (x-k)\right\}_{k\in \mathbb {Z} }$ is an orthonormal basis for $W_{0}$.

From this we get $\left\{2^{\frac {j}{2}}\,\psi \left(2^{\frac {j}{2}}\,x-k\right)\right\}_{k\in \mathbb {Z} }$ is an orthonormal basis for $W_{j}$.

find $q_{k}$'s such that $\psi (x)=\sum _{k\in \mathbb {Z} }q_{k}\,\phi (2x-k)$. We want

1. $\int \psi (x-k)\,\phi (x-\ell )\,\mathrm {d} x=0$ for all $k,\ell$
2. $\int \psi (x-k)\,\psi (x-\ell )\,\mathrm {d} x=\delta _{k,\ell }$

If $k=\ell$, we get $\int \psi (x)\,\phi (x)\,\mathrm {d} x=0$.

We have

${\begin{aligned}\psi (x)&=\sum _{k\in \mathbb {Z} }q_{k}\,\phi (2x-k)\\\phi (x)&=\sum _{\ell \in \mathbb {Z} }p_{\ell }\,\phi (2x-\ell )\end{aligned}}$

Hence

${\begin{aligned}\int _{-\infty }^{\infty }\psi (x)\,\phi (x)\,\mathrm {d} x&=\int _{-\infty }^{\infty }\sum _{k}q_{k}\,\phi (2x-k)\,\phi (x)\,\mathrm {d} x\\0&=\sum _{k}{\frac {1}{2}}\,q_{k}\underbrace {\left(2\int _{-\infty }^{\infty }\phi (x)\phi (2x-k)\,\mathrm {d} x\right)} _{p_{k}}&={\frac {1}{2}}\,\sum _{k\in \mathbb {Z} }p_{k}\,q_{k}\end{aligned}}$

(since we are deriving the Daubechies wavelet, we will set $k\in \left[0,3\right]$, but this could be arbitrary.)

$\sum _{k\in \mathbb {Z} }p_{k}\,q_{k}=\sum _{k=0}^{3}p_{k}\,q_{k}$

We want $q_{0},q_{1},q_{2},q_{3}=q_{-2},q_{-1},q_{0},q_{1}$?

We get $q_{k}=\left(-1\right)^{k}\,p_{1-k}$, and we check that

• $\sum _{k}q_{k}\,q_{k-2\ell }=2\delta _{0,\ell }$
• $\sum _{k}p_{k}=2$
• $\sum _{k}q_{k}=0$

Note: In the Haar wavelet, we have $p_{0}{=}1$, $p_{1}{=}1$, $q_{0}{=}1$, and $q_{1}{=}-1$

$q_{-2}={\frac {1-{\sqrt {3}}}{2}}$, $q_{-1}={\frac {{\sqrt {3}}-3}{4}}$, $q_{0}={\frac {3+{\sqrt {3}}}{4}}$, $q_{1}={\frac {{\sqrt {3}}-1}{4}}$

So for $W_{0}$, we get $\psi (x)={\frac {1-{\sqrt {3}}}{4}}\,\phi (2x+2)+{\frac {{\sqrt {3}}-3}{4}}\,\phi (2x+1)+{\frac {3+{\sqrt {3}}}{4}}\,\phi (2x)-{\frac {1-{\sqrt {3}}}{4}}\,\phi (2x-1)$, and $\phi (x)\in V_{1}$.