MATH 414 Lecture 38

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Scaling Function

$\varphi (x)=\sum _k{p}_k\varphi (2x-k)$

$\hat{\varphi }(\xi )=P\left({\mathrm{e}}^{-\frac{i\xi }{2}}\right)\hat{\varphi }\left(\frac{\xi }{2}\right)$, where $P(z)=\frac{1}{2}\sum _k{p}_k{z}^k$

Hence $\hat{\varphi }(\xi )=\frac{1}{\sqrt{2\pi }}\prod _{r=1}^{\mathrm{\infty }}P\left({\mathrm{e}}^{-\frac{i\xi }{2^r}}\right)$

• This means we can start with $p_k$'s and get $\hat{\varphi }(\xi )$ and $\varphi (x)={{ℱ}}^{-1}\left[\hat{\varphi }\right]$
• $p_k$'s satisfy at least the following properties:
  • $\sum _k{p}_{k-2\ell }{p}_k=2{\delta }_{\ell ,0}$
  • $\sum _k{p}_k=2$
  • $\sum _k{p}_{2k}=1$ and $\sum _k{p}_{2k+1}=1$.

Wavelet

$\hat{\psi }(\xi )=-{\mathrm{e}}^{-\frac{i\xi }{2}}P(-{\mathrm{e}}^{\frac{i\xi }{2}})\hat{\varphi }\left(\frac{\xi }{2}\right)$

$\hat{\psi }$ is known!

Haar Example

$P(z)=\frac{1}{2}(1+z)$

$\begin{array}{rl}{v}_N& =\prod _{r=1}^N\frac{1}{2}(1+{\mathrm{e}}^{-\frac{i\xi }{2^r}})\\ & =\left(\frac{1+{\mathrm{e}}^{-\frac{i\xi }{2^N}}}{2}\right)(1+{\mathrm{e}}^{-\frac{i\xi }{2^{N-1}}})\dots (1+{\mathrm{e}}^{-\frac{i\xi }{2}})\end{array}$

Let $z={\mathrm{e}}^{-\frac{i\xi }{2^N}}$. Then

$\begin{array}{rl}{v}_N& =\left(\frac{1+z}{2}\right)\left(\frac{1+z}{2}\right)\left(\frac{1+z^4}{2}\right)\dots \left(\frac{1+z^{2^{N-1}}}{2}\right)\\ (1-z)v_N& =\frac{1}{2^N}(1-z^{2n})\\ & =\frac{1}{2^N}(1-{\mathrm{e}}^{-i\xi })\\ v_N& =\frac{1}{1-{\mathrm{e}}^{-\frac{i\xi }{2^N}}}(1-{\mathrm{e}}^{-i\xi })\frac{1}{2^N}\end{array}$

We take a power series expansion of ${\mathrm{e}}^x=1+x+\frac{x^2}{2}+\dots$

$\begin{array}{rl}1-{\mathrm{e}}^{-\frac{i\xi }{2^N}}& =\left(\frac{i\xi }{2^N}\right)+\frac{(\dots )}{2^{2N}}+\dots \\ 2^N(1-{\mathrm{e}}^{-\frac{i\xi }{2^N}})& =i\xi +O(2^{-N})\end{array}$

The limit as $N\to \mathrm{\infty }$ gives

$V_N=\dots =\frac{2\sin \left(\frac{\xi }{2}\right)}{\xi }$

Hence $\hat{\varphi }(\xi )=\frac{1}{\sqrt{2\pi }}\frac{2}{\xi }\sin \left(\frac{\xi }{2}\right)={ℱ}\left[{\varphi }_{haar}\right]$

$\hat{\varphi }(\xi )=\frac{1}{\sqrt{2\pi }}\prod _{r=1}^{\mathrm{\infty }}P\left({\mathrm{e}}^{-\frac{i\xi }{2^r}}\right)$, where ${\hat{\varphi }}_N$ is a partial product.

When $\hat{\varphi }$ satisfies an iterative relation:

• ${\varphi }_0(x)={\varphi }_{haar}(x)$
• ${\varphi }_N(x)=\sum _k{p}_k{\varphi }_{N-1}(2x-k)$

Theorem. Suppose $P(z)=\frac{1}{2}\sum _k{p}_k{z}^k$, where the $p_k$'s are given. If:

1. ${|P(z)|}^2+{|P(-z)|}^2=1$, $\left|z\right|=1$, where $z={\mathrm{e}}^{it}$
2. $P(1)=1$
3. $\left|P\left({\mathrm{e}}^{it}\right)\right|>0$ for $\left|t\right|\le \frac{\pi }{2}$

Then $\frac{1}{\sqrt{2\pi }}\prod _{r=1}^{N}P\left({\mathrm{e}}^{-\frac{i\xi }{2}}\right)$ converges in $L^2$ to a function $\hat{\varphi }(\xi )$.

Proof.

quod erat demonstrandum

Point:

1. $\varphi (x)={{ℱ}}^{-1}[\hat{\varphi }(\xi )]$ is a scaling function
2. ...

Moments of Wavelet

A moment of a function $\psi (x)$ is an integral of the form:

$M_k=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^k\psi (x)\mathrm{d}x$, where $k=0,1,\dots$

Recall that if $M_0=0$ and $M_1=0$, then for the Daubechies ( $N=2$ ) wavelet (projection of $f(x)$ onto ${\psi }_{jk}$ component of $W_j$ space):

$\begin{array}{rl}{b}_k^{j-1}& =⟨f,{\psi }_{jk}⟩=⟨f(x),\psi (2^{j}x-k)⟩\\ & =\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(x)\psi (2^{j}x-k)\mathrm{d}x\\ & =\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(2^{-j}k+2^{-j}u)\psi (u)\mathrm{d}u\end{array}$

We can take the Taylor series expansion of $f$ centered at $2^{-j}u$

$f(2^{-j}k+2^{-j}u)=f\left(2^{-j}k\right)+f^{\prime }\left(2^{-j}k\right)2^{-j}u+\frac{f^{″}\left(2^{-j}k\right)}{2}{\left(2^{-j}u\right)}^2+\dots$

... and plug it into the inner product projection for $b_k^{j-1}$ above:

$\begin{array}{rl}{b}_k^{j-1}& =f\left(2^{-j}k\right)\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\psi (x)\mathrm{d}x+f^{\prime }\left(2^{-j}k\right)2^{-j}u\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}u\psi (x)\mathrm{d}x+f^{″}\left(2^{-j}k\right)2^{-2j}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{u}^2\psi (x)\mathrm{d}x+\dots \\ & \approx \frac{f^{″}\left(2^{-j}k\right)2^{-2j}}{2}{M}_2\end{array}$

If $j$ is much greater than 1 and $f$ is smooth, then $b_k^{j-1}$ is small.

If $f^{\prime }$ is discontinuous in $k\le u<k+1$, then coefficients can become large.

Hence we can detect singularities (discontinuities) in $f^{\prime }$ because $b_k^{j-1}$ will be relatively large near $x_{sing}=2^{-j}k$

If $M_k=0$ for $k=0,\dots ,N-1$, then for smooth $f$,

$b_k^{j-1}\approx \frac{f^{(N)}\left(2^{-j}k\right)}{N!}{2}^{-Nj}{M}_N+\dots$

$\psi (\xi )=-{\mathrm{e}}^{-\frac{i\xi }{2}}P(-{\mathrm{e}}^{\frac{i\xi }{2}})\hat{\varphi }\left(\frac{\xi }{2}\right)$

$M_k=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^k\psi (x)\mathrm{d}x=\sqrt{2\pi }{i}^k\frac{{\mathrm{d}}^k\hat{\psi }}{\mathrm{d}{\xi }^k}$

Vanishing Moments

Demand: $P(z)={\left(\frac{1+z}{2}\right)}^N\tilde{P}(z)$, $\tilde{P}(-1)\ne 0$

$\hat{\psi }(\xi )={\sin }^N\left(\frac{\xi }{4}\right)F\left({\mathrm{e}}^{\frac{i\xi }{2}}\right)\sim {\xi }^{N}F(\dots )$

${\frac{{\mathrm{d}}^k\hat{\psi }}{\mathrm{d}{\xi }^k}|}_{\xi =0}=0$ implies $M_k=0$

Require: $P(z)=\frac{1}{2}(p_0+p_{1}z+p_2{z}^2+p_3{z}^3)$

$P(z)=\frac{{(1+z)}^2}{2}(az+b)$, where $\tilde{P}(z)=az+b$

Failed to parse (unknown function "\i"): {\displaystyle \begin{align} \hat{\psi}(\xi) &= -\mathrm{e}^{-\frac{i\,\xi}{2}} \, \left( \frac{1-\mathrm{e}^{ \frac{i\,\xi}{2} }}{2} \right)^2 \, \hat{\phi} \left( \frac{\xi}{2} \right) \\ &= (-i)^2 \, \sin^2 \left( \frac{\xi}{4} \right) \, \hat{\phi} \left( \frac{\xi}{2} \right) \\ &= \sin^2 \left( \frac{\xi}{4} \right) \, \tilde{P} \left( -\mathrm{e}^{ \frac{i \, \i}{2}} \right) \, \hat{\phi} \left( \frac{\xi}{2} \right) \\ &= \dots \end{align}}

Going back to conditions on $P(z)$...

$P(1)={\left(\frac{2}{2}\right)}^2(a+b)=a+b=1$, hence $P(z)={\left(\frac{1+z}{2}\right)}^2(bz+1-b)$

Satisfy: ${|P(z)|}^2+{|P(-z)|}^2=1$, where $z={\mathrm{e}}^{i\theta }$.

${\cos }^4\left(\frac{\theta }{2}\right){|b{\mathrm{e}}^{i\theta }+1-b|}^2+{\sin }^4\left(\frac{\theta }{2}\right){|-b{\mathrm{e}}^{i\theta }+1-b|}^2=1$

Use trig identities until you're sick in the face... What you get when all is said and done is:

$\begin{array}{rl}{a}^2+{(1-a)}^2& =4\\ 2a^2-2a+1& =4\\ a^2-a-3& =0\\ a& =\frac{1\pm \sqrt{3}}{4}\end{array}$

$a$ and $b$ form the $p_k$'s for the Daubechies wavelet