# 0.7 Generalizations of the basic multiresolution wavelet system  (Page 24/28)

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## Local cosine and sine bases

Recall the four orthonormal trigonometric bases for ${L}^{2}\left(\left(0,1\right)\right)$ we described earlier.

1. $\left\{{\Phi }_{n},\left(t\right)\right\}=\left\{\sqrt{2},cos,\left(\pi \left(n+\frac{1}{2}\right)t\right)\right\}$ , $n\in \left\{0,,,1,,,2,,,...\right\}$ ;
2. $\left\{{\Phi }_{n},\left(t\right)\right\}=\left\{\sqrt{2},sin,\left(\pi \left(n+\frac{1}{2}\right)t\right)\right\}$ , $n\in \left\{0,,,1,,,2,,,...\right\}$ ;
3. $\left\{{\Phi }_{n},\left(t\right)\right\}=\left\{1,,,\sqrt{2},cos,\left(\pi nt\right)\right\}$ , $n\in \left\{1,,,2,,,...\right\}$ ;
4. $\left\{{\Phi }_{n},\left(t\right)\right\}=\left\{\sqrt{2},sin,\left(\pi nt\right)\right\}$ , $n\in \left\{0,,,1,,,2,,,...\right\}$ .

The bases functions have discontinuities at $t=0$ and $t=1$ because they are restrictions of the cosines and sines to the unit interval by rectangular windowing.The natural extensions of these basis functions to $t\in \text{ℝ}$ (i.e., unwindowed cosines and sines) are either even (say “+”) or odd (say “-”)symmetric (locally) about the endpoints $t=0$ and $t=1$ . Indeed the basis functions for the four cases are $\left(+,-\right)$ , $\left(-,+\right)$ , $\left(+,+\right)$ and $\left(-,-\right)$ symmetric, respectively, at $\left(0,1\right)$ . From the preceding analysis, this means that unfolding these basis functionscorresponds to windowing if the unfolding operator has the right polarity. Also observe that the basis functions are discontinuous atthe endpoints. Moreover, depending on the symmetry at each endpoint all odd derivatives (for “+” symmetry) or evenderivatives (for “ $-$ ” symmetry) are zero. By choosing unfolding operators of appropriate polarity at the endpoints (with non overlapping actionregions) for the four bases, we get smooth basis functions of compact support. For example, for (+, $-$ ) symmetry, the basis function ${U}_{+}\left({r}_{0},0,{ϵ}_{0}\right){U}_{+}\left({r}_{1},1,{ϵ}_{1}\right){\psi }_{n}\left(t\right)$ is supported in $\left(-{ϵ}_{0},1+{ϵ}_{1}\right)$ and is as many times continuously differentiable as ${r}_{0}$ and ${r}_{1}$ are.

Let $\left\{{t}_{j}\right\}$ be an ordered set of points in $\text{ℝ}$ defining a partition into disjoint intervals ${I}_{j}=\left[{t}_{j},{t}_{j+1}\right]$ . Now choose one of the four bases above for each intervalsuch that at ${t}_{j}$ the basis functions for ${I}_{j-1}$ and that for ${I}_{j}$ have opposite symmetries. We say the polarity at ${t}_{j}$ is positive if the symmetry is $-\right)\left(+$ and negative if it is $+\right)\left(-$ . At each ${t}_{j}$ choose a smooth cutoff function ${r}_{j}\left(t\right)$ and action radius ${ϵ}_{j}$ so that the action intervals do not overlap. Let $p\left(j\right)$ be the polarity of ${t}_{j}$ and define the unitary operator

${U}^{☆}=\prod _{j}{U}_{p\left(j\right)}^{☆}\left({r}_{j},{t}_{j},{ϵ}_{j}\right).$

Let $\left\{{\psi }_{n},\left(t\right)\right\}$ denote all the basis functions for all the intervals put together. Then $\left\{{\psi }_{n},\left(t\right)\right\}$ forms a nonsmooth orthonormal basis for ${L}^{2}\left(\text{ℝ}\right)$ . Simultaneously $\left\{{U}^{☆},{\psi }_{n},\left(t\right)\right\}$ also forms a smooth

orthonormal basis for ${L}^{2}\left(\text{ℝ}\right)$ . To find the expansion coefficients of a function $f\left(t\right)$ in this basis we use

$〈f,,,{U}^{☆},{\psi }_{n}〉=〈U,f,,,{\psi }_{n}〉.$

In other words, to compute the expansion coefficients of $f$ in the new (smooth) basis, one merely folds $f$ to $Uf$ and finds its expansion coefficients with respect to the originalbasis. This allows one to exploit fast algorithms available for coefficient computation in the original basis.

So for an arbitrary choice of polarities at the end points ${t}_{j}$ we have smooth local trigonometric bases. In particular by choosing the polarityto be positive for all ${t}_{j}$ (consistent with the choice of the first basis in all intervals) we get local cosine bases.If the polarity is negative for all ${t}_{j}$ (consistent with the choice of the second basis for all intervals), we get localsine bases. Alternating choice of polarity (consistent with the alternating choice of the third and fourth bases in the intervals) thusleads to alternating cosine/sine bases.

can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
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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Sherica
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Sherica
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Tamia
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a perfect square v²+2v+_
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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yes
Asali
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Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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China
Cied
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I start with an easy one. carbon nanotubes woven into a long filament like a string
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Porter
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Stotaw
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
Azam
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Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
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Prasenjit
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silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
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