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Mathematics
General Math
Regarding the Continous Wavelet Transform 'a' parameter
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[QUOTE="chiro, post: 5472396, member: 157252"] Hey tomizzo. The wavelet transform doesn't exist on a subset of the real line - it exists across it. The Fourier transform exists for some finite length (according to the length of the first harmonic) where-as the wavelet transform has the same "intuition" (something harmonic analysis studies) but it exists across L^2(R). This difference in covering the whole real line (as opposed to the length of the fundamental harmonic) complicates things quite a bit mathematically and technically but the result is that you have the wavelet and other transforms that exist across the entire real line. The inner product intuition is a good one to have. Also - wavelet's often have to be derived from their relations (this is the case for the Daubechies wavelet) because of these complications. There is a lot of interesting theory (I took a class in wavelet's a very long time ago) but the difference between a finite interval and the entire real line being consistent with the inner product axioms (which are a function of Hilbert spaces) changes things a lot. It's (analogously) a lot like the difference between infinities and finite quantities - making it consistent requires one to add new constraints that didn't exist when the subset of this new space was considered. [/QUOTE]
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Regarding the Continous Wavelet Transform 'a' parameter
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