Wavelet Theory: Solving ∫t^pψ(t)dt=0

[omer21]

I don't understand how to get the equation \sum_k (-1)^kk^pc_k=0 from ∫t^pψ(t)dt=0 from here on page 80. Can somebody explain it?

 

[Simon Bridge]

It's an approximation method - follows from the definitions in eqs. 5.12-13 on p79.
Can you see how 5.14 and 5.15 are obtained?

 

[omer21]

i can see 5.14 but i can not obtain 5.15 and 5.16.
Actually i get \sum_k (-1)^kc_k.0=0 instead of \sum_k (-1)^k k^p c_k=0 from the integral ∫t^p [\sum_k (-1)^kc_k \phi(2t+k-N+1)]dt=0.

 

[Simon Bridge]

Then you should go back over the derivations and make sure you understand what they are saying ... starting from the latest that you appear to be able to follow:

5.14 was the case for p=0.
i.e. $\phi(t)$ is capable of expressing only the zeroth monomial.
therefore, the zeroth-order moment of the wavelet function must be zero.
Which is to say...

$$\int_{-\infty}^\infty \psi(t)dt=0$$ ... which, to my mind, means that $\psi$ must be odd... which places limits on the expansion coefficients as the book says.

5.15 was the case for p=1 ... so $\phi(t)$ is capable of expressing up to the first monomial.
therefore, the zeroth and first-order moment of the wavelet function must be zero.
the zeroth order moment is above. The first order is given by...

$$\int t\psi(t)dt =0$$ ... so what steps did you follow from here?

 

[omer21]

I substituted

\psi(t)=[∑_k(−1)^kc_k\phi(2t+k−N+1)]

into

∫t\psi(t)dt.

Then i got

∫t[∑_k(−1)^kc_k\phi(2t+k−N+1)]dt=∑_k(−1)^kc_k∫t\phi(2t+k−N+1)]dt=0.

Actually i am not sure how to find the above integral but i did some change of variables and used integration by parts however i could not reach ∑_k(−1)^kkc_k.