# How to verify this equation $\prod_{j=1}^\infty{1+\exp(i2^{-j}\omega)\over2}={1-\exp(i\omega)\over i\omega}$?

This equation comes from A Wavelet Tour of Signal Processing, 3rd Ed. page 288. $$\prod_{j=1}^\infty{1+\exp(i2^{-j}\omega)\over2}={1-\exp(i\omega)\over i\omega}$$

I don't think it is right because when $\omega\to0$, the RHS tends toward $-1$, but the LHS is $1$. I have checked the errata of this book and found nothing. Can anyone tell me if this equation is right? If it is, how to prove it? Thanks.

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You're right: I get

$$\prod_{j=1}^\infty{1+\exp(i2^{-j}\omega)\over2}=\frac{e^{i \omega}-1}{i \omega}$$

The way to see this is to observe that

$$\frac{1+e^{i \omega/2^j}}{2} = e^{i \omega/2^{j+1}} \cos{\frac{\omega}{2^{j+1}}}$$

The products of the exponential and the cosines may be evaluated separately (assuming they each converge, which they do). The product of the exponentials is easily shown to be

$$\prod_{j=1}^{\infty} e^{i \omega/2^{j+1}} = e^{i \omega/2}$$

The product of the cosines is a well-known result which follows from

\begin{align}\sin{\omega} &= 2 \sin{\frac{\omega}{2}}\cos{\frac{\omega}{2}}\\ &= 2^2 \sin{\frac{\omega}{2^2}}\cos{\frac{\omega}{2}} \cos{\frac{\omega}{2^2}}\\ &= 2^N \sin{\frac{\omega}{2^N}} \prod_{j=1}^{N} \cos{\frac{\omega}{2^{j}}}\end{align}

Therefore, in the limit as $N \rightarrow \infty$:

$$\prod_{j=1}^{\infty} \cos{\frac{\omega}{2^{j+1}}} = \frac{\sin{\omega}}{\omega \cos{(\omega/2)}}$$

The net product is then

$$2 e^{i \omega/2} \frac{\sin{(\omega/2)}}{\omega}$$

The result follows.

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