# 0.3 Approximation and processing in bases  (Page 2/5)

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## Sampling theorems

Let us consider finite energy signals $\parallel \phantom{\rule{0.166667em}{0ex}}\overline{f}{\parallel }^{2}=\int {|\phantom{\rule{0.166667em}{0ex}}\overline{f}\left(x\right)|}^{2}\phantom{\rule{0.166667em}{0ex}}dx$ of finite support, which is normalized to $\left[0,1\right]$ or ${\left[0,1\right]}^{2}$ for images. A sampling process implementsa filtering of $\phantom{\rule{0.166667em}{0ex}}\overline{f}\left(x\right)$ with a low-pass impulse response ${\overline{\phi }}_{s}\left(x\right)$ and a uniform sampling to output a discrete signal:

$f\left[n\right]=\phantom{\rule{0.166667em}{0ex}}\overline{f}☆{\overline{\phi }}_{s}\left(ns\right)\phantom{\rule{1.em}{0ex}}\text{for}\phantom{\rule{1.em}{0ex}}0\le n

In two dimensions, $n=\left({n}_{1},{n}_{2}\right)$ and $x=\left({x}_{1},{x}_{2}\right)$ . These filtered samples can also be written as inner products:

$\overline{f}☆{\overline{\phi }}_{s}\left(ns\right)=\int f\left(u\right)\phantom{\rule{0.166667em}{0ex}}{\overline{\phi }}_{s}\left(ns-u\right)\phantom{\rule{0.166667em}{0ex}}du=⟨\phantom{\rule{0.166667em}{0ex}}f\left(x\right),{\phi }_{s}\left(x-ns\right)⟩$

with ${\phi }_{s}\left(x\right)={\overline{\phi }}_{s}\left(-x\right)$ . Chapter 3 explains that φ s is chosen, like in the classic Shannon–Whittaker sampling theorem, so that a family of functions ${\left\{{\phi }_{s}\left(x-ns\right)\right\}}_{1\le n\le N}$ is a basis of an appropriate approximation space U N . The best linear approximation of $\phantom{\rule{0.166667em}{0ex}}\overline{f}$ in U N recovered from these samples is the orthogonal projection  $\phantom{\rule{0.166667em}{0ex}}{\overline{f}}_{N}$ of $\phantom{\rule{0.166667em}{0ex}}f$ in U N , and if the basis is orthonormal, then

$\begin{array}{c}\hfill {\overline{f}}_{N}\left(x\right)=\sum _{n=0}^{N-1}f\left[n\right]\phantom{\rule{0.166667em}{0ex}}{\phi }_{s}\left(x-ns\right).\end{array}$

A sampling theorem states that if $\phantom{\rule{0.166667em}{0ex}}\overline{f}\in {\mathbf{U}}_{N}$ then $\phantom{\rule{0.166667em}{0ex}}\overline{f}={\overline{f}}_{N}$ so recovers $\phantom{\rule{0.166667em}{0ex}}\overline{f}\left(x\right)$ from the measured samples. Most often, $\phantom{\rule{0.166667em}{0ex}}\overline{f}$ does not belong to this approximation space. It is called aliasing in the context of Shannon–Whittaker sampling, where U N is the space of functions having a frequency support restricted to the N lower frequencies. The approximation error $\parallel \phantom{\rule{0.166667em}{0ex}}\overline{f}-{\overline{f}}_{N}{\parallel }^{2}$ must then be controlled.

## Linear approximation error

The approximation error is computed by finding an orthogonal basis $\mathcal{B}={\left\{{\overline{g}}_{m}\left(x\right)\right\}}_{0\le m<+\infty }$ of the whole analog signal space ${\mathbf{L}}^{\mathbf{2}}\left(\mathbb{R}\right){\left[0,1\right]}^{2}$ , with the first N vector ${\left\{{\overline{g}}_{m}\left(x\right)\right\}}_{0\le m that defines an orthogonal basis of U N . Thus, the orthogonal projection on U N can be rewritten as

${\overline{f}}_{N}\left(x\right)=\sum _{m=0}^{N-1}⟨\phantom{\rule{0.166667em}{0ex}}\overline{f},{\overline{g}}_{m}⟩\phantom{\rule{0.166667em}{0ex}}{\overline{g}}_{m}\left(x\right).$

Since $\phantom{\rule{0.166667em}{0ex}}\overline{f}={\sum }_{m=0}^{+\infty }⟨\phantom{\rule{0.166667em}{0ex}}\overline{f},{\overline{g}}_{m}⟩\phantom{\rule{0.166667em}{0ex}}{\overline{g}}_{m}$ , the approximation error is the energy of the removed inner products:

${ϵ}_{l}\left(N,f\right)=\parallel \phantom{\rule{0.166667em}{0ex}}\overline{f}-{\overline{f}}_{N}{\parallel }^{2}=\sum _{m=N}^{+\infty }{|⟨\phantom{\rule{0.166667em}{0ex}}\overline{f},{\overline{g}}_{m}⟩|}^{2}.$

This error decreases quickly when N increases if the coefficient amplitudes $|⟨\phantom{\rule{0.166667em}{0ex}}\overline{f},{\overline{g}}_{m}⟩|$ have a fast decay when the index m increases. The dimension N is adjusted to the desired approximation error.

Figure (a) shows a discrete image $\phantom{\rule{0.166667em}{0ex}}f\left[n\right]$ approximated with $N={256}^{2}$ pixels. Figure(c) displays a lower-resolution image $\phantom{\rule{0.166667em}{0ex}}{f}_{N/16}$ projected on a space ${\mathbf{U}}_{N/16}$ of dimension $N/16$ , generated by $N/16$ large-scale wavelets. It is calculated by setting all thewavelet coefficients to zero at the first two smaller scales. The approximation error is $\parallel \phantom{\rule{0.166667em}{0ex}}f-{f}_{N/16}{\parallel }^{2}/{\parallel \phantom{\rule{0.166667em}{0ex}}f\parallel }^{2}=14\phantom{\rule{0.45pt}{0ex}}×{10}^{-3}$ . Reducing the resolution introduces more blur and errors.A linear approximation space U N corresponds to a uniform grid that approximates precisely uniform regular signals.Since images $\phantom{\rule{0.166667em}{0ex}}\overline{f}$ are often not uniformly regular, it is necessary to measure it at a high-resolution N . This is why digital cameras have a resolution that increases as technology improves.

## Sparse nonlinear approximations

Linear approximations reduce the space dimensionality but can introduce important errors when reducing the resolutionif the signal is not uniformly regular, as shown by Figure(c). To improve such approximations, more coefficients should be kept where needed—notin regular regions but near sharp transitions and edges.This requires defining an irregular sampling adapted to the local signal regularity. This optimized irregular sampling has a simple equivalent solutionthrough nonlinear approximations in wavelet 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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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
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is it 3×y ?
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Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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Cied
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Porter
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Porter
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Stotaw
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Azam
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Prasenjit
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Azam
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Prasenjit
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Damian
silver nanoparticles could handle the job?
Damian
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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.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
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