If x_n = cos frac{pi}{2^n} + i sin frac{pi}{2 | vedclass

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(C) Given $x_n = \cos \frac{\pi}{2^n} + i \sin \frac{\pi}{2^n} = e^{i \frac{\pi}{2^n}}$.We need to find the product $P = \prod_{n=1}^{\infty} x_n$.$P

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$-1$

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(C) Given $x_n = \cos \frac{\pi}{2^n} + i \sin \frac{\pi}{2^n} = e^{i \frac{\pi}{2^n}}$.We need to find the product $P = \prod_{n=1}^{\infty} x_n$.$P = \prod_{n=1}^{\infty} e^{i \frac{\pi}{2^n}} = e^{i \pi \sum_{n=1}^{\infty} \frac{1}{2^n}}$.The sum of the geometric series is $\sum_{n=1}^{\infty} \frac{1}{2^n} = \frac{1/2}{1 - 1/2} = 1$.Therefore,$P = e^{i \pi (1)} = e^{i \pi}$.Using Euler's formula,$e^{i \pi} = \cos \pi + i \sin \pi = -1 + 0i = -1$.

If $x_n = \cos \frac{\pi}{2^n} + i \sin \frac{\pi}{2^n}$,then $\prod_{n=1}^{\infty} x_n =$

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