Question

If X_n = cos \(\frac{ π}{2^n}\) + i sin\(\frac{ π}{2^n}\) , then 

\[\prod_{n=1}^{\infty} x_n =\]

 

• A. 0
• B. 1
• C. -1
• D. i

Answer 1

The Correct Option is C

To solve the problem, we need to evaluate the infinite product of the complex numbers given by:

$X_n = \cos\left(\frac{\pi}{2^n}\right) + i \sin\left(\frac{\pi}{2^n}\right)$

1. Recognizing the Euler Form:
The expression is of the form:
$X_n = \cos\theta + i\sin\theta = e^{i\theta}$
So, we rewrite:
$X_n = e^{i\frac{\pi}{2^n}}$

2. Evaluate the Infinite Product:
We are asked to compute:
$\prod_{n=1}^{\infty} X_n = \prod_{n=1}^{\infty} e^{i\frac{\pi}{2^n}}$

3. Use Product of Exponentials:
$\prod_{n=1}^{\infty} e^{i\frac{\pi}{2^n}} = e^{i\sum_{n=1}^{\infty} \frac{\pi}{2^n}}$

4. Evaluate the Geometric Series:
$\sum_{n=1}^{\infty} \frac{\pi}{2^n} = \pi \sum_{n=1}^{\infty} \frac{1}{2^n} = \pi \cdot \left(\frac{1/2}{1 - 1/2}\right) = \pi \cdot 1 = \pi$

5. Final Expression:
$e^{i\sum \cdots} = e^{i\pi} = -1$

Final Answer:
The value of the infinite product is $ {-1} $.