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Actually one can resort to the two-scale equation in multiresolution analysis. Perform Fourier transformation on both side of $\phi(t)=\phi(2t)+\phi(2t-1)$, it turns out that

$$\hat\phi(\omega)=\frac{1}{2}\hat\phi(\frac{1}{2}\omega)+\frac{1}{2}\hat\phi(\frac{1}{2}\omega)e^{-\frac{i\omega}{2}}$$ , that is, $$\frac{\hat{\phi}(2\omega)}{\hat{\phi}(\omega)}=\frac{1+e^{-i\omega}}{2}=m(\omega)$$ Therefore $$\hat\phi(\omega)=\prod_{k=1}^{\infty}m\left(\frac{\omega}{2^k}\right)=\prod_{k=1}^{\infty}\frac{1+e^{-i2^{-k}\omega}}{2}\hat\phi(0)$$

My question is, how to calculate this limit? Thank you~

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OK...I found the answer pages later in the textbook...but this is still a good question so I decide not to spoil others. Or should I delete it? – ziyuang Jun 23 '11 at 14:01
You can add your own answer and accept it. – Aryabhata Jun 23 '11 at 15:47
up vote 2 down vote accepted

If $\omega=0$, then the product is $1$. Let $z=e^{-i\omega}$. Observe that $$ \prod_{k=1}^{K}\frac{1+z^{2^{-k}}}{2}=\frac{1}{2^K}\frac{1-z}{1-z^{2^{-K}}}.$$ Treating $2^K(1-z^{2^{-K}})$ as a derivative as $K\to\infty$ for $-e^{i\omega} t$ at $t=0$, we get $$\prod_{k=1}^{\infty}\frac{1+e^{-i2^{-k}\omega}}{2}=\frac{1-e^{-i\omega}}{i\omega}.$$ In the limit as $\omega\to 0$, we also get $1$.

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