Question

Math

Posted 6 months ago

`$\sum_{n=1}^{\infty} \frac{\sin (n x)}{n}$`

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Answer from Sia

Posted 6 months ago

Solution by Steps

step 1

The sum of the series is given by the formula derived from the asksia-ll calculator: $\sum_{n=1}^{\infty} \frac{\sin(nx)}{n} = \frac{1}{2} i \left(\log(1 - e^{ix}) - \log(e^{-ix}(-1 + e^{ix}))\right)$ when $x$ is a real number

step 2

The formula can be simplified using properties of logarithms: $\frac{1}{2} i \left(\log(1 - e^{ix}) - \log(e^{-ix}) + \log(-1 + e^{ix})\right)$

step 3

Recognize that the complex logarithm can have multiple values due to the periodicity of the exponential function, but the principal value is taken here

step 4

The approximation of the sum is given by substituting $e \approx 2.71828$ and $i$ as the imaginary unit: $\approx \frac{1}{2} i \left(\log(1 - 2.71828^{ix}) - \log(2.71828^{-ix}(-1 + 2.71828^{ix}))\right)$ when $x$ is a real number

Answer

$\sum_{n=1}^{\infty} \frac{\sin(nx)}{n} = \frac{1}{2} i \left(\log(1 - e^{ix}) - \log(e^{-ix}(-1 + e^{ix}))\right)$

Key Concept

Summation of a trigonometric series involving sine and natural numbers

Explanation

The series is a special case of the Fourier series and can be expressed in terms of complex logarithms. The asksia-ll calculator provides a formula for the sum of the series, which involves complex numbers and the natural logarithm.

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