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I have been asked to find the wavelet expansion of $$ f(x)= x^2 \text{ for } 0\leq x<1, \quad 0 \text{ otherwise.} $$ I know I have to take the scalar product, and I should use the Haar wavelet. All I have to do is calculate the first couple of coefficients, the problem is I don't know how to start.

mike
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S.n
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The wavelet coefficient will be $$\int_{0}^{1}x^{2}\psi_{j,k}(x)\mathrm{d}x=\int_{0}^{1}x^{2}(\varphi_{j+1,2k}(x)-\varphi_{j+1,2k+1}(x))/\sqrt{2}\;\mathrm{d}x=2^{j/2}\left(\int_{k/2^{j}}^{(2k+1)/2^{j+1}}x^{2}\mathrm{d}x-\int_{(2k+1)/2^{j+1}}^{(k+1)/2^{j}}x^{2}\mathrm{d}x\right),$$ for $0\leq k\leq 2^{j}-1,$ at scales $j\geq 1.$ Now this is just a matter of integrating.

RideTheWavelet
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  • I think that $2^{(j-1)/2}$ should be $2^{(j/2)-1}$. – mike Aug 02 '17 at 00:08
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    I've made an edit. To make sure $\int_{0}^{1}\psi_{j,k}^{2}(x)\mathrm{d}x=1,$ we need to divide by $\sqrt{2}$ rather than $2,$ since both of the $\phi_{j+1,\cdot}$ should be normalized, and they are orthogonal. Then we have $2^{(j+1)/2}\cdot 2^{-1/2}=2^{j/2}.$ – RideTheWavelet Aug 03 '17 at 17:30
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    In hindsight, I should have caught this myself: the expression in parentheses is clearly the inner product between $x^{2}$ and the dilated, translated wavelet (without the renormalization), so the natural scaling factor should be $2^{j/2}.$ – RideTheWavelet Aug 03 '17 at 17:35