if x r=cos(pi/2^r)+i sin(pi/2^r),then x1*x2*x3*x4....infinity=?

To tackle the problem of finding the product of the series \( x_1, x_2, x_3, \ldots \) where \( x_r = \cos\left(\frac{\pi}{2^r}\right) + i \sin\left(\frac{\pi}{2^r}\right) \), we can start by recognizing that this is a complex exponential form. According to Euler's formula, we know that \( e^{i\theta} = \cos(\theta) + i \sin(\theta) \). Therefore, we can express each term \( x_r \) as:

Transforming the Terms

Using Euler's formula, we write:

x_r = e^{i \frac{\pi}{2^r}}

Understanding the Infinite Product

We want to find the infinite product:

x_1 \cdot x_2 \cdot x_3 \cdots

Substituting our expression for \( x_r \) into the product gives:

x_1 \cdot x_2 \cdot x_3 \cdots = e^{i \frac{\pi}{2^1}} \cdot e^{i \frac{\pi}{2^2}} \cdot e^{i \frac{\pi}{2^3}} \cdots

Combining Exponents

When multiplying exponentials, we can add the exponents:

e^{i \left( \frac{\pi}{2^1} + \frac{\pi}{2^2} + \frac{\pi}{2^3} + \cdots \right)}

Calculating the Series

Next, we need to evaluate the series:

\(\frac{\pi}{2^1} + \frac{\pi}{2^2} + \frac{\pi}{2^3} + \cdots\)

This is a geometric series where the first term \( a = \frac{\pi}{2} \) and the common ratio \( r = \frac{1}{2} \). The sum \( S \) of an infinite geometric series can be calculated using the formula:

S = \frac{a}{1 - r}

Plugging in our values:

S = \frac{\frac{\pi}{2}}{1 - \frac{1}{2}} = \frac{\frac{\pi}{2}}{\frac{1}{2}} = \pi

Final Result

Thus, we find that:

x_1 \cdot x_2 \cdot x_3 \cdots = e^{i \pi}

According to Euler's identity, \( e^{i \pi} = -1 \). Therefore, the product of the series is:

-1

Summary

In conclusion, the infinite product \( x_1 \cdot x_2 \cdot x_3 \cdots \) evaluates to \(-1\). This demonstrates not only the power of series and exponential functions but also the beauty of how complex numbers can simplify certain mathematical expressions.