If x_n = cos left(frac{pi}{4^n}right) + i sin | vedclass

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(A) Given $x_n = \cos \left(\frac{\pi}{4^n}\right) + i \sin \left(\frac{\pi}{4^n}\right) = e^{i \frac{\pi}{4^n}}$.The product $P = x_1 x_2 x_3 \ldots

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$\frac{1 + i \sqrt{3}}{2}$

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(A) Given $x_n = \cos \left(\frac{\pi}{4^n}\right) + i \sin \left(\frac{\pi}{4^n}\right) = e^{i \frac{\pi}{4^n}}$.The product $P = x_1 x_2 x_3 \ldots \infty$ is given by:$P = e^{i \frac{\pi}{4^1}} \cdot e^{i \frac{\pi}{4^2}} \cdot e^{i \frac{\pi}{4^3}} \ldots = e^{i \pi \left(\frac{1}{4} + \frac{1}{4^2} + \frac{1}{4^3} + \ldots \right)}$.The exponent is a geometric series with first term $a = \frac{1}{4}$ and common ratio $r = \frac{1}{4}$.The sum of the infinite geometric series is $S = \frac{a}{1 - r} = \frac{1/4}{1 - 1/4} = \frac{1/4}{3/4} = \frac{1}{3}$.Therefore,$P = e^{i \pi \left(\frac{1}{3}\right)} = e^{i \frac{\pi}{3}}$.Using Euler's formula,$e^{i \frac{\pi}{3}} = \cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right) = \frac{1}{2} + i \frac{\sqrt{3}}{2} = \frac{1 + i \sqrt{3}}{2}$.

If $x_n = \cos \left(\frac{\pi}{4^n}\right) + i \sin \left(\frac{\pi}{4^n}\right)$,then the product $x_1 x_2 x_3 \ldots \infty$ is equal to

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