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Chapter 2: Problem 157

If \(x_{n}=\cos \left(\pi / 2^{n}\right)+i \sin \left(\pi / 2^{n}\right)\) then \(x_{1} x_{2} x_{3} \ldots \ldots \infty=\) (a) - i (b) \(-1\) (c) (d) 1

Short Answer

Expert verified
The product of the infinite sequence \(x_1x_2x_3\ldots\) is equal to \(-1\). The correct answer is (b) \(-1\).

Step by step solution

01

Identify Euler's formula

Euler's formula is given as: \[e^{ix} = \cos x + i\sin x\]
02

Rewrite the expression using Euler's formula

Using Euler's formula, we can rewrite the expression for \(x_n\) as: \[x_n = e^{i\left(\pi / 2^{n}\right)}\]
03

Find the product of the sequence

We are looking for the product of the infinite sequence \(x_1x_2x_3\ldots\), which can be represented as the following: \[x_1x_2x_3\ldots = e^{i\left(\frac{\pi}{2^1}\right)} \cdot e^{i\left(\frac{\pi}{2^2}\right)} \cdot e^{i\left(\frac{\pi}{2^3}\right)}\ldots\]
04

Combine exponents with the same base

As the base of the exponential function is consistent, we can combine the exponents: \[x_1x_2x_3\ldots = e^{i\left(\frac{\pi}{2} + \frac{\pi}{4} + \frac{\pi}{8} + \ldots\right)}\]
05

Find the sum of the geometric series

The exponent is a geometric series with a first term of \(\frac{\pi}{2}\) and a common ratio of \(\frac{1}{2}\). The sum of this series can be calculated using the formula: \[\frac{\text{first term}}{1 - \text{common ratio}}\] So, the sum of the series is given by: \[\frac{\frac{\pi}{2}}{1 - \frac{1}{2}} = \frac{\frac{\pi}{2}}{\frac{1}{2}} = \pi\]
06

Substitute the sum back into the expression and simplify

Now, substitute the sum of the geometric series back into the expression for the product of the infinite sequence: \[x_1x_2x_3\ldots = e^{i\pi}\] Using Euler's formula, we know that \(e^{i\pi} = \cos\pi + i\sin\pi\), and as \(\cos\pi = -1\) and \(\sin\pi = 0\), we have: \[x_1x_2x_3\ldots = -1\] Therefore, the product of the infinite sequence is equal to: \[\boxed{-1}\] The correct answer is (b) \(-1\).

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Most popular questions from this chapter

It \(\mathrm{z}^{2}+\mathrm{z}+1=0\) where \(\mathrm{z}\) is a complex number, then the value of \([z+(1 / z)]^{2}+\left[z^{2}+\left(1 / z^{2}\right)\right]^{2}+\left[z^{3}+\left(1 / z^{3}\right)\right]^{2}+\ldots\) \(+\left[z^{6}+\left(1 / z^{6}\right)\right]^{2}\) is (a) 18 (b) 54 (c) 6 (d) 12

If \(\mathrm{w}\) is one of the cube root of 1 other then 1 then $$ \left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & -1-\mathrm{w}^{2} & \mathrm{w}^{2} \\ 1 & \mathrm{w}^{2} & \mathrm{w}^{4} \end{array}\right|= $$ (a) \(3 \mathrm{w}\) (b) \(3 \mathrm{w}(\mathrm{w}-1)\) (c) \(3 \mathrm{w}^{2}\) (d) \(3 \mathrm{w}(1-\mathrm{w})\)

For any integer \(\mathrm{n}, \arg \left[(\sqrt{3}+\mathrm{i})^{4 n+1} /(1-\mathrm{i} \sqrt{3})^{4 \mathrm{n}}\right]=\) (a) \((\pi / 3)\) (b) \((\pi / 6)\) (c) \((2 \pi / 3)\) (d) \((5 \pi / 6)\)

\([\\{1+\cos (\pi / 12)+i \sin (\pi / 12)\\} /\\{1+\cos (\pi / 12)\) \(-i \sin (\pi / 12)\\}]^{36}=\) (a) \(-1\) (b) 1 (c) 0 (d) \((1 / 2)\)

If \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are integers, not all equal, and \(\mathrm{w}\) is a cube root of unity \((\mathrm{w} \neq 1)\) Then the minimum value of \(\left|\mathrm{a}+\mathrm{bw}+\mathrm{cw}^{2}\right|\) is (a) 0 (b) 1 (c) \((\sqrt{3} / 2)\) (d) \((1 / 2)\)
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