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Moment of wavelet

  1. Jan 10, 2013 #1
    I don't understand how to get the equation [itex]\sum_k (-1)^kk^pc_k=0[/itex] from [itex]∫t^pψ(t)dt=0[/itex] from here on page 80. Can somebody explain it?
     
  2. jcsd
  3. Jan 11, 2013 #2

    Simon Bridge

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    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?
     
  4. Jan 11, 2013 #3
    i can see 5.14 but i can not obtain 5.15 and 5.16.
    Actually i get [itex]\sum_k (-1)^kc_k.0=0[/itex] instead of [itex]\sum_k (-1)^k k^p c_k=0[/itex] from the integral [itex]∫t^p [\sum_k (-1)^kc_k \phi(2t+k-N+1)]dt=0[/itex].
     
  5. Jan 12, 2013 #4

    Simon Bridge

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    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?
     
  6. Jan 13, 2013 #5
    I substituted

    [itex]\psi(t)=[∑_k(−1)^kc_k\phi(2t+k−N+1)][/itex]

    into

    [itex]∫t\psi(t)dt[/itex].

    Then i got

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

    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 [itex]∑_k(−1)^kkc_k[/itex].
     
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