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I. Basic math.
II. Pricing and Hedging.
III. Explicit techniques.
IV. Data Analysis.
V. Implementation tools.
VI. Basic Math II.
VII. Implementation tools II.
1. Calculational Linear Algebra.
2. Wavelet Analysis.
A. Elementary definitions of wavelet analysis.
B. Haar functions.
C. Multiresolution analysis.
D. Orthonormal wavelet bases.
E. Discrete wavelet transform.
F. Construction of MRA from scaling filter or auxiliary function.
G. Consequences and conditions for vanishing moments of wavelets.
a. Vanishing moments vs decay at infinity.
b. Vanishing moments vs approximation.
c. Sufficient conditions for vanishing moments.
d. Reproduction of polynomials.
e. Smoothness of compactly supported wavelets with vanishing moments.
H. Existence of smooth compactly supported wavelets. Daubechies polynomials.
I. Semi-orthogonal wavelet bases.
J. Construction of (G)MRA and wavelets on an interval.
3. Finite element method.
4. Construction of approximation spaces.
5. Time discretization.
6. Variational inequalities.
VIII. Bibliography
Notation. Index. Contents.

Sufficient conditions for vanishing moments.


roposition

(Sufficient conditions for vanishing moments) Let MATH be an MRA, scaling function and wavelet (see the definition ( Multiresolution analysis ) and propositions ( Scaling equation ),( Existence of orthonormal wavelet bases 2 )). Assume that

1. $\phi$ is compactly supported and (therefore) MATH is finite, where the sequence MATH is defined in the proposition ( Scaling equation ).

Then the following statements are equivalent:

(a) MATH , $k=0,1,...,n-1$ , for some MATH ,

(b) MATH , $k=0,1,...,n-1$ ,

(c) there is a representation MATH MATH for some finite sequence MATH ,

(d) MATH , $k=0,1,...,n-1$ .

Proof

(a) MATH (b). According to the proposition ( Scaling equation 2 ) MATH and according to the proposition ( Integral of scaling function ), MATH Therefore MATH can be zero if and only if MATH .

We continue MATH MATH We use the previous result MATH : MATH and conclude MATH .

We continue similarly for any $k$ : for any derivative MATH there will be a term MATH and the rest of the terms are zero at $z=0$ by results of the previous steps.

Proof

(b) $\Leftrightarrow$ (c) Set MATH By condition (1) the $A\left( z\right) $ is defined for all $y$ except, possibly, $y=0$ . We have MATH MATH MATH Hence, if (b) takes place then MATH We do Taylor decomposition, (see the proposition ( Taylor decomposition in Schlomilch, Lagrange and Cauchy forms )): MATH Thus, by (b) and $\left( \#\right) $ , MATH The MATH then must be of the form MATH for a finite sequence MATH . This is (c).

The calculation may be performed in opposite direction: assume (c), then the $A\left( y\right) $ has the form MATH and we do argument in reverse arriving to (b).

Proof

(b) $\Leftrightarrow$ (d) The proof is a direct verification. We use finiteness of MATH to differentiate MATH termwise and substitute into (b).





Notation. Index. Contents.


















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