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This is a very powerful result [link] , [link] . It not only ties the number of zero moments to the regularity but also to the degree ofpolynomials that can be exactly represented by a sum of weighted and shifted scaling functions.

Theorem 21 If ψ ( t ) is K -times differentiable and decays fast enough, then the first K - 1 wavelet moments vanish [link] ; i.e.,

d k d t k ψ ( t ) < , 0 k K


m 1 ( k ) = 0 . 0 k K

Unfortunately, the converse of this theorem is not true. However, we can relate the differentiability of ψ ( t ) to vanishing moments by

Theorem 22 There exists a finite positive integer L such that if m 1 ( k ) = 0 for 0 k K - 1 then

d P d t P ψ ( t ) <

for L P > K .

For example, a three-times differentiable ψ ( t ) must have three vanishing moments, but three vanishing moments results in only one-timedifferentiability.

These theorems show the close relationship among the moments of h 1 ( n ) , ψ ( t ) , the smoothness of H ( ω ) at ω = 0 and π and to polynomial representation. It also states a loose relationship with thesmoothness of φ ( t ) and ψ ( t ) themselves.

Daubechies' method for zero wavelet moment design

Daubechies used the above relationships to show the following important result which constructs orthonormal wavelets with compact support with themaximum number of vanishing moments.

Theorem 23 The discrete-time Fourier transform of h ( n ) having K zeros at ω = π of the form

H ( ω ) = 1 + e i ω 2 K L ( ω )


| H ( ω ) | 2 + | H ( ω + π ) | 2 = 2

if and only if L ( ω ) = | L ( ω ) | 2 can be written

L ( ω ) = P ( sin 2 ( ω / 2 ) )

with K N / 2 where

P ( y ) = k = 0 K - 1 K - 1 + k k y k + y K R ( 1 2 - y )

and R ( y ) is an odd polynomial chosen so that P ( y ) 0 for 0 y 1 .

If R = 0 , the length N is minimum for a given regularity K = N / 2 . If N > 2 K , the second term containing R has terms with higher powers of y whose coefficients can be used for purposes other than regularity.

The proof and a discussion are found in Daubechies [link] , [link] . Recall from [link] that H ( ω ) always has at least one zero at ω = π as a result of h ( n ) satisfying the necessary conditions for φ ( t ) to exist and have orthogonal integer translates. We are now placing restrictions on h ( n ) to have as high an order zero at ω = π as possible. That accounts for the form of [link] . Requiring orthogonality in [link] gives [link] .

Because the frequency domain requirements in [link] are in terms of the square of the magnitudes of the frequency response, spectralfactorization is used to determine H ( ω ) and therefore h ( n ) from | H ( ω ) | 2 . [link] becomes

| H ( ω ) | 2 = 1 + e i ω 2 2 K | L ( ω ) | 2 .

If we use the functional notation:

M ( ω ) = | H ( ω ) | 2 and L ( ω ) = | L ( ω ) | 2

then [link] becomes

M ( ω ) = | cos 2 ( ω / 2 ) | K L ( ω ) .

Since M ( ω ) and L ( ω ) are even functions of ω they can be written as polynomials in cos ( ω ) and, using cos ( ω ) = 1 - 2 sin 2 ( ω / 2 ) , [link] becomes

M ˜ ( sin 2 ( ω / 2 ) ) = | cos 2 ( ω / 2 ) | K P ( sin 2 ( ω / 2 ) )

which, after a change of variables of y = sin 2 ( ω / 2 ) = 1 - cos 2 ( ω / 2 ) , becomes

M ˜ ( y ) = ( 1 - y ) K P ( y )

where P ( y ) is an ( N - K ) order polynomial which must be positive since it will have to be factored to find H ( ω ) from [link] . This now gives [link] in terms of new variables which are easier to use.

In order that this description supports an orthonormal wavelet basis, we now require that [link] satisfies [link]

| H ( ω ) | 2 + | H ( ω + π ) | 2 = 2

which using [link] and [link] becomes

M ( ω ) + M ( ω + π ) = ( 1 - y ) K P ( y ) + y K P ( 1 - y ) = 2 .

Equations of this form have an explicit solution found by using Bezout's theorem. The details are developed by Daubechies [link] . If all the ( N / 2 - 1 ) degrees of freedom are used to set wavelet moments to zero, we set K = N / 2 and the solution to [link] is given by

Questions & Answers

can someone help me with some logarithmic and exponential equations.
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it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
I'm not sure why it wrote it the other way
I got X =-6
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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Differences Between Laspeyres and Paasche Indices
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No. 7x -4y is simplified from 4x + (3y + 3x) -7y
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
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Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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Source:  OpenStax, Wavelets and wavelet transforms. OpenStax CNX. Aug 06, 2015 Download for free at https://legacy.cnx.org/content/col11454/1.6
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