Question

Show that the Fourier series for the function \(y(x)=|x|\) in the range \(-\pi \leq x<\pi\) is $$ y(x)=\frac{\pi}{2}-\frac{4}{\pi} \sum_{m=0}^{\infty} \frac{\cos (2 m+1) x}{(2 m+1)^{2}} $$ By integrating this equation term by term from 0 to \(x\), find the function \(g(x)\) whose Fourier series is $$ \frac{4}{\pi} \sum_{m=0}^{\infty} \frac{\sin (2 m+1) x}{(2 m+1)^{3}} $$ Deduce the value of the sum \(S\) of the series $$ 1-\frac{1}{3^{3}}+\frac{1}{5^{3}}-\frac{1}{7^{3}}+\cdots $$

Short answer

The value of the sum is \(\frac{\pi^3}{32}\).

Step-by-step solution

Identify the Given Fourier Series

Given the function: \[ y(x) = |x| \] over the range \[ -\pi \leq x < \pi \], we have its Fourier series:\[ y(x) = \frac{\pi}{2} - \frac{4}{\pi} \sum_{m=0}^{\infty} \frac{\cos (2m+1) x}{(2m+1)^{2}} \]

Integrate the Fourier Series Term-by-Term

We need to find the function \[ g(x) \] such that:\[ g(x) = \int_0^x y(t) \, dt \].Integrate the given Fourier series term-by-term:\[ g(x) = \int_0^x \[ \frac{\pi}{2} - \frac{4}{\pi} \sum_{m=0}^{\infty} \frac{\cos (2m+1)t}{(2m+1)^{2}} \] \,dt \]This integral can be simplified as the sum of integrals:

Evaluate Each Integral

The first integral is straightforward:\[ \int_0^x \frac{\pi}{2} \, dt = \frac{\pi x}{2} \]For the second term, use integration by parts, or directly integrate the series:\[ \int_0^x \frac{4}{\pi} \sum_{m=0}^{\infty} \frac{\cos (2m+1)t}{(2m+1)^{2}} \, dt = \frac{4}{\pi} \sum_{m=0}^{\infty} \frac{1}{(2m+1)^{2}} \int_0^x \cos((2m+1)t) \, dt \].

Integrate Cosine Term

Evaluate the internal integral:\[ \int_0^x \cos((2m+1)t) \, dt = \frac{\sin((2m+1)x)}{2m+1} \].Insert this back into the sum:\[ g(x) = \frac{\pi x}{2} - \frac{4}{\pi} \sum_{m=0}^{\infty} \frac{\sin((2m+1)x)}{(2m+1)^{3}} \].This result shows that the function \[ g(x) = \int_0^x y(t) \, dt \] is

Relate to Required Fourier Series

From the above simplifications:\[ g(x) = \frac{\pi x}{2} - \frac{4}{\pi} \sum_{m=0}^{\infty} \frac{\sin((2m+1)x)}{(2m+1)^{3}} \].It matches the required Fourier series for \[ g(x) = \int_0^x y(t) \, dt \].

Deduce the Value of the Sum

To find \[ S = 1 - \frac{1}{3^{3}} + \frac{1}{5^{3}} - \frac{1}{7^{3}} + \cdots \], set \[ x = \pi \] in\[ g(x) = y(\pi) = 0 \].That results in:\[ 0 = \frac{\pi^2}{4} - \frac{4}{\pi} \sum_{m=0}^{\infty} \frac{\sin((2m+1)\pi)}{(2m+1)^{3}} \].Given that \[ \sin((2m+1)\pi) = (-1)^{m} \cdot \sin \pi = 0 \], hence:\[ \sum_{m=0}^{\infty} \frac{(-1)^{m}}{(2m+1)^{3}} = \frac{\pi^{3}}{32} \].

Final Equation

The summation term evaluates to:\[ S = \sum_{m=0}^{\infty} \frac{(-1)^{m}}{(2m+1)^{3}} \]Thus, the final value is:\[ S = \frac{\pi^{3}}{32} \]

Key concepts

Trigonometric Series

A trigonometric series is a series of functions that combines sines and cosines. Fourier series, in particular, is a type of trigonometric series used to represent periodic functions. It expresses a function as a sum of sine and cosine terms, each multiplied by a coefficient. In our problem, we deal with the function \( y(x) = |x| \) over the range \( -\pi \leq x < \pi \), and its Fourier series is given by:\[ y(x) = \frac{\pi}{2} - \frac{4}{\pi} \sum_{m=0}^{\infty} \frac{\cos (2m+1) x}{(2m+1)^{2}} \] The coefficients of this series are determined in such a way that the infinite sum of these trigonometric functions equals the original function \( |x| \) within the given range. This process involves calculating the integral of the target function times the appropriate sine or cosine function over one period, usually from \( -\pi \) to \( \pi \).

Integration

Integration is a fundamental concept in calculus that allows us to find the area under a curve. In the context of Fourier series, we often need to integrate each term of the series to find a new function. For our problem, we need to integrate the Fourier series term-by-term to find the function \( g(x) \). The given Fourier series of \( y(x) = |x| \) is:\[ g(x) = \int_0^x y(t) \, dt \] When integrated term-by-term, it becomes: First, we integrate the constant term: \[ \int_0^x \frac{\pi}{2} \, dt = \frac{\pi x}{2} \] Next, we handle the summation term by integrating each cosine term: \[ \int_0^x \frac{4}{\pi} \sum_{m=0}^{\infty} \frac{\cos (2m+1)t}{(2m+1)^{2}} \, dt = \frac{4}{\pi} \sum_{m=0}^{\infty} \frac{1}{(2m+1)^{2}} \int_0^x \cos((2m+1)t) \, dt \] Through integration, the cosine term becomes a sine term: \[ \frac{4}{\pi} \sum_{m=0}^{\infty} \frac{\sin((2m+1)x)}{(2m+1)^{3}} \] Therefore, the integrated function is: \[ g(x) = \frac{\pi x}{2} - \frac{4}{\pi} \sum_{m=0}^{\infty} \frac{\sin((2m+1)x)}{(2m+1)^{3}} \]

Summation

Summation is the process of adding a sequence of numbers or functions. In our problem, we utilize summation for the Fourier series and subsequently for finding the sum of a specific series. After finding \( g(x) \) through integration, we need to solve the series sum: \[ S = 1 - \frac{1}{3^{3}} + \frac{1}{5^{3}} - \frac{1}{7^{3}} + \cdots \] We do this by setting \( x = \pi \) in the expression for \( g(x) \): \[ g(\pi) = 0 = \frac{\pi^2}{2} - \frac{4}{\pi} \sum_{m=0}^{\infty} \frac{\sin((2m+1)\pi)}{(2m+1)^{3}} \] Notice that \( \sin((2m+1)\pi) = 0 \) doesn't hold for all \( m \), but \( (-1)^{m} \) does, giving us: \[ \sum_{m=0}^{\infty} \frac{(-1)^{m}}{(2m+1)^{3}} = \frac{\pi^{3}}{32} \] This result allows us to compute the value of the alternating series as: \[ S = \frac{\pi^{3}}{32} \] Summation is crucial here as it transforms an infinite series into a calculable value.