# 0.16 Nonlinear approximation and wavelet analysis  (Page 2/2)

 Page 2 / 2

Suppose $f$ is piecewise Lipschitz and ${f}_{k}$ ia a piecewise constant.

$|f\left(t\right)-{f}_{k}\left(t\right)|\approx \Delta$

where $\Delta$ is a constant equal to average of $f$ on right and left side of discontinuity in this interval.

$⇒||f-{f}_{k}{||}_{{L}_{2}}^{2}=O\left({k}^{-1}\right)$

where ${k}^{-1}$ is the width of the interval. Notice this rate is quite slow.

This problem naturally suggests the following remedy: use very small intervals near discontinuities and larger intervals insmooth regions. Specifically, suppose we use intervals of width ${k}^{-2\alpha }$ to contain the discontinuities and the intervals ofwidth ${k}^{-1}$ elsewhere. Then accordingly piecewise polynomial approximation ${\stackrel{˜}{f}}_{k}$ satisfies

$||f-{\stackrel{˜}{f}}_{k}{||}_{{L}_{2}}^{2}=O\left({k}^{-2\alpha }\right).$

We can accomplish this need for "adaptive resolution" or "multiresolution" using recursive partitions and trees.

We discussed this idea already in our examination of classification trees. Here is the basic idea again, graphically.

Consider a function $f\in {B}^{\alpha }\left({C}_{\alpha }\right)$ that contains no more than m points of discontinuity, and is ${H}^{\alpha }\left({C}_{\alpha }\right)$ away from these points.

Lemma

Consider a complete RDP with n intervals, then there exists anassociated pruned RDP with $O\left(klogn\right)$ intervals, such that an associated piecewise degree $⌈\alpha ⌉$ polynomial approximation $\stackrel{˜}{\left(}{f\right)}_{k}$ , has a squared approximation error of $O\left(min\left({k}^{-2\alpha },{n}^{-1}\right)\right)$ .

Assume $n>k>m$ . Divide $\left[0,1\right]$ into $k$ intervals. If $f$ is smooth on a particular interval $I$ , then

$|f\left(t\right)-{\stackrel{˜}{f}}_{k}\left(t\right)|=O\left({k}^{-2\alpha }\right)\forall t\in I.$

In intervals that contain a discontinuity, recursively subdivide into two until the discontinuity is contained in an interval ofwidth ${n}^{-1}$ . This process results in at most $lo{g}_{2}n$ addition subintervals per discontinuity, and the squared approximationerror is $O\left(k-2\alpha \right)$ on all of them accept the $m$ intervals of width ${n}^{-1}$ containing the discontinuities where the error is $O\left(1\right)$ at each point.

Thus, the overall squared ${L}_{2}$ norm is

$||f-{\stackrel{˜}{f}}_{k}{||}_{{L}_{2}}^{2}=O\left(min\left({k}^{-2\alpha },{n}^{-1}\right)\right)$

and there are at most $k+lo{g}_{2}n$ intervals in the partition. Since k>m, we can upperbound the number of intervals by $2klo{g}_{2}n$ .

Note that if the initial complete RDP has $n\approx {k}^{2\alpha }$ intervals, then the squared error is $O\left({k}^{-2\alpha }\right)$ .

Thus, we only incur a factor of $2\alpha logk$ additional leafs and achieve the same overall approximation error as in the ${H}^{\alpha }\left({C}_{\alpha }\right)$ case. We will see that this is a small price to pay in order to handle not only smooth functions, but alsopiecewise smooth functions.

## Wavelet approximations

Let $f\in {L}^{2}\left(\left[0,1\right]\right)$ ; $\int {f}^{2}\left(t\right)dt<\infty$ .

A wavelet approximation is a series of the form

$f={c}_{o}+\sum _{j\ge 0}\sum _{k=1}^{{2}^{j}}{\psi }_{j,k}$

where ${c}_{o}$ is a constant $\left({c}_{o}={\int }_{0}^{1}f\left(t\right)dt\right)$ ,

$={\int }_{0}^{1}f\left(t\right){\psi }_{j,k}\left(t\right)dt$

and the basis functions ${\psi }_{j,k}$ are orthonormal, oscillatory signals, each with an associated scale ${2}^{-j}$ and position $k{2}^{-j}$ . ${\psi }_{j,k}$ is called the wavelet at scale ${2}^{-j}$ and position $k{2}^{-j}$ .

## Haar wavelets

${\psi }_{j,k}\left(t\right)={2}^{j/2}\left({\mathbf{1}}_{\left\{t\in \left[{2}^{-j}\left(k-1\right),{2}^{-j}\left(k-1/2\right)\right]\right\}}-{\mathbf{1}}_{\left\{t\in \left[{2}^{-j}\left(k-1/2\right),{2}^{-j}k\right]\right\}}\right)$
${\int }_{0}^{1}{\psi }_{j,k}\left(t\right)dt=0$
${\int }_{0}^{1}{\psi }_{j,k}^{2}\left(t\right)dt={\int }_{\left(k-1\right){2}^{-j}}^{k{2}^{-j}}{2}^{j}dt=1$
${\int }_{0}^{1}{\psi }_{j,k}\left(t\right){\psi }_{l,m}\left(t\right)dt={\delta }_{j,l}.{\delta }_{k,m}$
If $f$ is constant on $\left[{2}^{-j}\left(k-1\right),{2}^{-j}k\right]$ , then
$\int f{\psi }_{j,k}\left(t\right)=0.$

Suppose $f$ is piecewise constant with at most $m$ discontinuities. Let

${f}_{J}={c}_{o}+\sum _{j=0}^{J-1}\sum _{k=1}^{{2}^{j}}{\psi }_{j,k}.$

Then, ${f}_{J}$ has at most $mJ$ non-zero wavelet coefficients; i.e., $=0$ for all but $mJ$ terms, since at most one Haar Wavelet at each scale senses each point of discontinuity. Said another way, allbut at most $m$ of the wavelets at each scale have support over constant regions of $f$ .

${f}_{J}$ itself will be piecewise constant with discontinuities only possible occurring at end points of the intervals $\left[{2}^{-J}\left(k-1\right),{2}^{-J}k\right]$ . Therefore, in this case

$||f-{f}_{J}{||}_{{L}_{2}}^{2}=O\left({2}^{-J}\right).$

Daubechies wavelets are the extension of the Haar wavelet idea. Haar wavelets have one "vanishing moment":

${\int }_{0}^{1}{\psi }_{j,k}=0.$

Daubechies wavelets are "smoother" basis functions with extra vanishing moments. The Daubechies- $N$ wavelet has $N$ vanishing moments.

${\int }_{0}^{1}{t}^{l}{\psi }_{j,k}dt=0forl=0,1,...,N-1.$

The Daubechies-1 wavelet is just the Haar case.

If $f$ is a piecewise degree $\le N$ polynomial with at most m pieces, then using the Daubechies- $N$ wavelet system.

$||f-{f}_{J}{||}_{{L}_{2}}^{2}=O\left({2}^{-J}\right);$

and

${f}_{J}\left(t\right)={c}_{o}+\sum _{j=0}^{J-1}\sum _{k=1}^{{2}^{j}}{\psi }_{j,k}\left(t\right)$

has at most $O\left(mJ\right)$ non-zero wavelet coefficients. ${f}_{J}$ is called the Discrete Wavelet Transform (DWT) approximation of $f$ . The key idea is the same as we saw with trees.

## Sampled data

We can also use DWT's to analyze and represent discrete, sampled functions. Suppose,

$\underline{f}=\left[f\left(1/n\right),f\left(2/n\right),...,f\left(n/n\right)\right]$

then we can write $\underline{f}$ as

$\underline{f}={c}_{o}+\sum _{j=0}^{lo{g}_{2}n-1}\sum _{k=1}^{{2}^{j}}<\underline{f},{\underline{\psi }}_{j,k}>{\underline{\psi }}_{j,k}$

where

${\underline{\psi }}_{j,k}=\left[{\psi }_{j,k}\left(1\right),{\psi }_{j,k}\left(2\right),...,{\psi }_{j,k}\left(n\right)\right]$

is a discrete time analog of the continuous time wavelets we considered before. In particular,

$\sum _{i=1}^{n}{i}^{l}{\psi }_{j,k}\left(i\right)=0,l=0,1,...,N-1$

for the Daubechies- $N$ discrete wavelets.

$<\underline{f},{\underline{\psi }}_{j,k}>={\underline{f}}^{T}{\underline{\psi }}_{j,k}$

Thus, we also have an analogous approximation result: If $\underline{f}$ are samples from a piecewise degree $\le N$ polynomial function with a finite number $m$ of discontinuities, then $\underline{f}$ has $O\left(mJ\right)$ non-zero wavelet coefficients.

## ApproximatingFunctions with wavelets

Suppose $f\in {B}^{\alpha }\left({C}_{\alpha }\right)$ and has a finite number of discontinuities. Let ${f}_{p}$ denote piecewise degree- $N\left(N=⌈\alpha ⌉\right)$ polynomial approximation to $f$ with $O\left(k\right)$ pieces; a uniform partition into $k$ equal length intervals followed by addition splits at the points of discontinuity.

Then

$|f\left(t\right)-{f}_{p}\left(t\right){|}^{2}=O\left({k}^{\left(}-2\alpha \right)\right)\forall t\in \left[0,1\right]$
$⇒|f\left(i/n\right)-{f}_{p}{\left(i/n\right)|}^{2}=O\left({k}^{-2\alpha }\right)i=1,...,n$
$⇒1/n||\underline{f}-{\underline{f}}_{p}{||}_{{L}_{2}}^{2}=O\left({k}^{-2\alpha }\right)\right)$

and ${\underline{f}}_{p}$ has $O\left(klo{g}_{2}n\right)$ non-zero coefficients according to our previous analysis.

## Wavelets in 2-d

Suppose $f$ is a 2-D image that is piecewise polynomial:

A pruned RDP of $k$ squares decorated with polyfits gives

$||f-{f}_{k}{||}_{{L}_{2}}^{2}=O\left({k}^{-1}\right).$

Let $\underline{f}{=\left[f\left(i/k,j/k\right)}_{i,j=1}^{n}$ sample range.

${f}_{n}\left(t\right)=\sum _{i,j=1}^{k}f\left(i/k,j/kk\right){\mathbf{1}}_{\left\{t\in \left[i-1/k,i/k\right)x\left[j-1/k,j/k\right)\right\}}$

then

$||f-{f}_{n}{||}_{{L}_{2}}^{2}=O\left({k}^{-1}\right)$

$O\left(1\right)$ error on $k$ of the ${k}^{2}$ pixels, near zero elsewhere. The DWT of $\underline{f}$ has $O\left(k\right)$ non-zero wavelet coefficients. $O\left({2}^{j}\right)$ at scale ${2}^{-j},j=0,1,...,logn.$

show that the set of all natural number form semi group under the composition of addition
what is the meaning
Dominic
explain and give four Example hyperbolic function
_3_2_1
felecia
⅗ ⅔½
felecia
_½+⅔-¾
felecia
The denominator of a certain fraction is 9 more than the numerator. If 6 is added to both terms of the fraction, the value of the fraction becomes 2/3. Find the original fraction. 2. The sum of the least and greatest of 3 consecutive integers is 60. What are the valu
1. x + 6 2 -------------- = _ x + 9 + 6 3 x + 6 3 ----------- x -- (cross multiply) x + 15 2 3(x + 6) = 2(x + 15) 3x + 18 = 2x + 30 (-2x from both) x + 18 = 30 (-18 from both) x = 12 Test: 12 + 6 18 2 -------------- = --- = --- 12 + 9 + 6 27 3
Pawel
2. (x) + (x + 2) = 60 2x + 2 = 60 2x = 58 x = 29 29, 30, & 31
Pawel
ok
Ifeanyi
on number 2 question How did you got 2x +2
Ifeanyi
combine like terms. x + x + 2 is same as 2x + 2
Pawel
x*x=2
felecia
2+2x=
felecia
×/×+9+6/1
Debbie
Q2 x+(x+2)+(x+4)=60 3x+6=60 3x+6-6=60-6 3x=54 3x/3=54/3 x=18 :. The numbers are 18,20 and 22
Naagmenkoma
Mark and Don are planning to sell each of their marble collections at a garage sale. If Don has 1 more than 3 times the number of marbles Mark has, how many does each boy have to sell if the total number of marbles is 113?
Mark = x,. Don = 3x + 1 x + 3x + 1 = 113 4x = 112, x = 28 Mark = 28, Don = 85, 28 + 85 = 113
Pawel
how do I set up the problem?
what is a solution set?
Harshika
find the subring of gaussian integers?
Rofiqul
hello, I am happy to help!
Abdullahi
hi mam
Mark
find the value of 2x=32
divide by 2 on each side of the equal sign to solve for x
corri
X=16
Michael
Want to review on complex number 1.What are complex number 2.How to solve complex number problems.
Beyan
yes i wantt to review
Mark
16
Makan
x=16
Makan
use the y -intercept and slope to sketch the graph of the equation y=6x
how do we prove the quadratic formular
Darius
hello, if you have a question about Algebra 2. I may be able to help. I am an Algebra 2 Teacher
thank you help me with how to prove the quadratic equation
Seidu
may God blessed u for that. Please I want u to help me in sets.
Opoku
what is math number
4
Trista
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
can you teacch how to solve that🙏
Mark
Solve for the first variable in one of the equations, then substitute the result into the other equation. Point For: (6111,4111,−411)(6111,4111,-411) Equation Form: x=6111,y=4111,z=−411x=6111,y=4111,z=-411
Brenna
(61/11,41/11,−4/11)
Brenna
x=61/11 y=41/11 z=−4/11 x=61/11 y=41/11 z=-4/11
Brenna
Need help solving this problem (2/7)^-2
x+2y-z=7
Sidiki
what is the coefficient of -4×
-1
Shedrak
A soccer field is a rectangle 130 meters wide and 110 meters long. The coach asks players to run from one corner to the other corner diagonally across. What is that distance, to the nearest tenths place.
Jeannette has $5 and$10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives.
What is the expressiin for seven less than four times the number of nickels
How do i figure this problem out.
how do you translate this in Algebraic Expressions
why surface tension is zero at critical temperature
Shanjida
I think if critical temperature denote high temperature then a liquid stats boils that time the water stats to evaporate so some moles of h2o to up and due to high temp the bonding break they have low density so it can be a reason
s.
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!