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Chapter 14: Problem 6

Sum the trigonometric series $$ \sum_{n=1}^{\infty}(-1)^{n+1} \frac{\sin n x}{n} $$ and show that it equals \(x / 2\).

Short Answer

Expert verified
The given series sums to \(\frac{x}{2}\).

Step by step solution

01

Recognize the Series as a Known Expansion

The series given is a known expansion related to the Fourier series of the function defined on the interval (-蟺, 蟺). Specifically, it resembles the Fourier series expansion of the function f(x) = x, which gives rise to the series expansion in terms of sine functions.
02

Recall the Fourier Series

Recall that the Fourier series representation of a sawtooth wave function, f(x) = x on (-蟺, 蟺), is given by: \[f(x) = x = 2\sum_{n=1}^{\infty} (-1)^{n+1} \frac{\sin(nx)}{n}\] This is valid on the interval (-蟺, 蟺), where the function is periodic with period 2蟺.
03

Interpret the Given Series

Notice that the given series is essentially the same but multiplied by \(\frac{1}{2}\): \[\sum_{n=1}^{\infty} (-1)^{n+1} \frac{\sin(nx)}{n} = \frac{1}{2} \left(x\right)\] This means the given series represents exactly one-half of the function f(x) = x.
04

Show Equivalence on the Interval

Therefore, on the interval (-蟺, 蟺), the sum of the series simplifies to \(\frac{x}{2}\). This demonstrates the equality:\[\sum_{n=1}^{\infty} (-1)^{n+1} \frac{\sin(nx)}{n} = \frac{x}{2}\] This is known due to the properties of Fourier series representing periodic functions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometric Series
The concept of a trigonometric series encompasses expansions made using trigonometric functions like sine and cosine. These series are pivotal in representing complex periodic functions. A typical trigonometric series may look like a sum of sine and cosine terms as seen here:
  • Sine and cosine functions inherently relate to angles and periodic oscillations.
  • Each function in the series has coefficients that adjust its amplitude and phase.
  • These series particularly stand out in fields like signal processing and acoustics, where wave-like data is prevalent.
By solving such a series, you essentially decompose a signal or a function into its harmonic components. In our exercise, the given series resembles the well-known Fourier series, underpinning yet another connection between trigonometric series and harmonic decomposition.
Sawtooth Wave
A sawtooth wave is a specific type of signal or periodic function that looks like the teeth on a saw blade. This shape makes it linear with sharp discontinuities between intervals. Here are some interesting features:
  • When plotted, it has a distinct linear rise over time followed by a sudden drop.
  • The mathematical representation is often through a Fourier series.
  • The function is defined as: \( f(x) = x \) within a certain range, often (-蟺, 蟺).
One key characteristic of a sawtooth wave is its spacious bandwidth due to its sharp changes (harmonic richness). This quality makes it invaluable in synthesizers and electronic music as it鈥檚 capable of generating rich, edgy sounds. In our solution, we utilize its Fourier series expansion to sum up to the expression provided, linking the sawtooth wave to the series result.
Periodic Function
In mathematics, a periodic function repeats its values at regular intervals, known as the period. If you think of it in everyday terms, it鈥檚 much like a clock ticking. Consider the following:
  • Its value remains unchanged when shifted by a full period, like how moving a clock hand 360 degrees brings it back to 12.
  • Common examples include sine and cosine functions.
  • These functions can be described by Fourier series, transforming them into sums of trigonometric terms.
The periodic function depicted by the Fourier series in our example functions optimally within the interval (-蟺, 蟺), confirming its periodicity with period 2蟺. This property ensures smooth, continuous, and reliable behavior when repeating, critical in both theoretical and applied mathematics.

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Most popular questions from this chapter

Assuming that \(\int_{-\pi}^{\pi}[f(x)]^{2} d x\) is finite, show that $$ \lim _{m \rightarrow \infty} a_{m}=0, \quad \lim _{m \rightarrow \infty} b_{m}=0 . $$ Hint. Integrate \(\left[f(x)-s_{n}(x)\right]^{2}\), where \(s_{n}(x)\) is the \(n\) th partial sum and use Bessel's inequality, Section \(9.4\). For our finite interval the assumption that \(f(x)\) is square integrable \(\left(\int_{-\pi}^{\pi}|f(x)|^{2} d x\right.\) is finite) implies that \(\int_{-\pi}^{\pi}|f(x)| d x\) is also finite, The converse does not hold.

(a) Show that the Dirac delta function \(\delta(x-a)\), expanded in a Fourier sine series in the half interval \((0, L),(0

Expand $$ f(x)=\left\\{\begin{array}{ll} 1, & x^{2}x_{0}^{2} \end{array}\right. $$ in the interval \([-\pi, \pi]\). Note. This variable width square wave is of some importance in electronic music.

Let \(f(z)=\ln (1+z)=\sum_{n=1}^{\infty}(-1)^{n+1} z^{n} / n .\) (This series converges to \(\ln (1+z)\) for \(|z| \leq 1\), except at the point \(z=-1 .\) ) (a) From the imaginary parts show that $$ \ln \left(2 \cos \frac{\theta}{2}\right)=\sum_{n=1}^{\infty}(-1)^{n+1} \frac{\cos n \theta}{n}, \quad-\pi<\theta<\pi $$ (b) Using a change of variable, transform part (a) into $$ -\ln \left(2 \sin \frac{\theta}{2}\right)=\sum_{n=1}^{\infty}(-1)^{n+1} \frac{\cos n \theta}{n}, \quad 0<\theta<2 \pi $$

(a) Find the Fourier series representation of $$ f(x)=\left\\{\begin{array}{lr} 0, & -\pi
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