Nick Alger
|
Registered User
|
|
|
Dec 4 |
comment |
Approximation power of wavelets Same formula should hold for $d>1, p=\infty$ if I recall correctly. I think this is proved in the last chapter of Wojtaszczyk's book, [A Mathematical Introduction to Wavelets](books.google.com/books/about/…). |
|
Dec 3 |
comment |
Approximation power of wavelets One would solve $C \cdot 2^{-kd}|f|_{Lip(d)}=\epsilon$ for $k$. That gives the the resolution level $k$, from which the number of terms $N$ can be determined. In 1D on a periodic domain with Haar wavelets and $p=\infty$ Lipschitz smoothness with $d=1$, it simplifies. You get twice as many wavelets each time you increase the resolution, so $N \approx 2^k$. Altogether this yields $C/N = epsilon$ I believe. My personal knowledge is with Besov smoothness spaces and $p \neq \infty$ though, so you should probably double check it before using this result. |
|
Dec 3 |
revised |
Approximation power of wavelets added 138 characters in body |
|
Dec 3 |
answered | Approximation power of wavelets |
