MATH 414 Lecture 34

« previous | Monday, April 14, 2014 | next »

Begin Exam 3 content

Properties of the $p_k$'s

Assume real-valued; if not, we'll need ${\bar{p}}_k$'s

Scaling function $\varphi (x)=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{p}_k\varphi (2x-k)$, where

$p_k=2\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\varphi (x)\varphi (2x-k)\mathrm{d}x$

Theorem 5.9:

1. $\sum _k{p}_{k-2\ell }{p}_{k-2m}=2{\delta }_{\ell ,m}$
2. $\sum _k{p}_{k-2\ell }^2=\sum _{k^{\prime }}{p}_{k^{\prime }}^2=2$

Theorem. $\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{p}_k=2$

Proof. Step 1: $\begin{array}{rl}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\varphi (x)\mathrm{d}x& =\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{p}_k\varphi (2x-k)\mathrm{d}x\\ & =\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{p}_k\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\varphi (2x-k)\mathrm{d}x\end{array}$

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

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\varphi (2x-k)=\frac{1}{2}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\varphi (u)\mathrm{d}u=\frac{M}{2}$

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

$M=\frac{1}{2}\sum _k{p}_k\cdot M$, where $M\ne 0$, so $1=\frac{1}{2}\sum _k{p}_k$, and $\sum _k{p}_k=2$

quod erat demonstrandum

1. $\begin{array}{rl}\sum _{\text{even}k}{p}_k& =1=\sum _{\ell =-\mathrm{\infty }}^{\mathrm{\infty }}{p}_{2\ell }\\ \sum _{\text{odd}k}{p}_k& =1=\sum _{\ell =-\mathrm{\infty }}^{\mathrm{\infty }}{p}_{2\ell +1}\end{array}$

Example: Haar

$\varphi (x)=\varphi (2x)+\varphi (2x+1)$, so $p_0=1,p_1=1$.

3. $p_0+p_1=2$

4. $p_0=1$ (even), $p_1=1$ (odd)

Example: Shannon

$\varphi (x)=\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}(x)=\{\begin{array}{ll}\frac{\sin \pi x}{\pi x}& x\ne 0\\ 1& x=0\end{array}$

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

Hence $p_k=\{\begin{array}{ll}1& k=0\\ \frac{2(-1{)}^{\frac{\ell -1}{2}}}{\pi k}& \text{odd}\ell =2k+1\\ 0& \text{otherwise}\end{array}$

Now $p_0=1$ is the only even term.

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

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

Daubechies $p=2$ MRA

$\varphi (x)=\sum _{k=0}^3{p}_k\varphi (2x-k)=p_0\varphi (2x)+p_1\varphi (2x-1)+p_2\varphi (2x-2)+p_3\varphi (2x-3)$, where

$p_k=\{\frac{1+\sqrt{3}}{4},\frac{3+\sqrt{3}}{4},\frac{3-\sqrt{3}}{4},\frac{1-\sqrt{3}}{4}\}$

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

Wavelets

$W_j=\{w\in V_{j+1}\mid w\perp V_j\}$

Looking for $\psi (x)$ such that ${\{\psi (x-k)\}}_{k\in \mathbb{\mathbb{Z}}}$ is an orthonormal basis for $W_0$.

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

find $q_k$'s such that $\psi (x)=\sum _{k\in \mathbb{\mathbb{Z}}}{q}_k\varphi (2x-k)$. We want

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

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

We have

$\begin{array}{rl}\psi (x)& =\sum _{k\in \mathbb{\mathbb{Z}}}{q}_k\varphi (2x-k)\\ \varphi (x)& =\sum _{\ell \in \mathbb{\mathbb{Z}}}{p}_{\ell }\varphi (2x-\ell )\end{array}$

Hence

$\begin{array}{rl}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\psi (x)\varphi (x)\mathrm{d}x& =\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\sum _k{q}_k\varphi (2x-k)\varphi (x)\mathrm{d}x\\ 0& =\sum _k\frac{1}{2}{q}_k\underset{p_k}{\underbrace{(2\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\varphi (x)\varphi (2x-k)\mathrm{d}x)}}& =\frac{1}{2}\sum _{k\in \mathbb{\mathbb{Z}}}{p}_k{q}_k\end{array}$

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

$\sum _{k\in \mathbb{\mathbb{Z}}}{p}_k{q}_k=\sum _{k=0}^3{p}_k{q}_k$

We want $q_0,q_1,q_2,q_3=q_{-2},q_{-1},q_0,q_1$?

We get $q_k={(-1)}^k{p}_{1-k}$, 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 $\psi (x)=\frac{1-\sqrt{3}}{4}\varphi (2x+2)+\frac{\sqrt{3}-3}{4}\varphi (2x+1)+\frac{3+\sqrt{3}}{4}\varphi (2x)-\frac{1-\sqrt{3}}{4}\varphi (2x-1)$, and $\varphi (x)\in V_1$.