Question

Find the (real) Fourier series of period 2 for \(f(x)=\cosh x\) and \(g(x)=x^{2}\) in the range \(-1 \leq x \leq 1 .\) By integrating the series for \(f(x)\) twice, prove that $$ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{2} \pi^{2}\left(n^{2} \pi^{2}+1\right)}=\frac{1}{2}\left(\frac{1}{\sinh 1}-\frac{5}{6}\right) $$

Short answer

The given Fourier series can be evaluated and integrated, yielding the desired summation equality.

Step-by-step solution

Define the Fourier series

The Fourier series for a function with period 2 is given by \[ f(x) = a_0 + \sum_{n=1}^{\infty} \(a_n \cos(n\pi x) + b_n \sin(n\pi x)\) \]. In this case, the functions considered have period 2, thus we will use this form for expansion.

Compute the Fourier coefficients for \( f(x) = \cosh x \)

To find the coefficients, use the formulas:\[ a_0 = \frac{1}{2} \int_{-1}^{1} \cosh(x) dx \]\[ a_n = \int_{-1}^{1} \cosh(x) \cos(n\pi x) dx \]\[ b_n = \int_{-1}^{1} \cosh(x) \sin(n\pi x) dx \].Since \( \cosh x \) is even, \( b_n = 0 \) as the integrand of \( \cosh(x) \sin(nx) \) is odd and integrates to zero.

Calculate \( a_0 \) for \( f(x) = \cosh x \)

Compute \( a_0 \):\[ a_0 = \frac{1}{2} \int_{-1}^{1} \cosh(x) dx \].Using symmetry, \( a_0 = \int_{0}^{1} \cosh x dx \) which evaluates to:\[ a_0 = \sinh 1 \]

Calculate \( a_n \) for \( f(x) = \cosh x \)

Compute \( a_n \):\[ a_n = \int_{-1}^{1} \cosh(x) \cos(n\pi x) dx \].This evaluates to:\[ a_n = 2 \int_{0}^{1} \cosh(x) \cos(n\pi\ x) dx \].

Evaluate integrals for \( f(x) = \cosh x \)

Use integration by parts or tabulated integrals to evaluate \( a_n \) values:\[ a_n = \frac{2 (-1)^{n+1}}{n^2\pi^2 + 1} \frac{1}{\sinh 1} \].Thus the Fourier series for \( \cosh x \) is:\[ \cosh x = \sinh 1 + \sum_{n=1}^{\infty} \frac{2 (-1)^{n+1}}{n^2\pi^2 + 1} \cos(n\pi x) \]

Compute the Fourier coefficients for \( g(x) = x^2 \)

Use the formulas:\[ a_0 = \int_{-1}^{1} x^2 dx = \frac{2}{3} \]\[ a_n = 2 \int_{0}^{1} x^2 \cos(n\pi x) dx \]\[ b_n = 0 \] because \( x^2 \) is even.

Evaluate integrals for \( g(x) = x^2 \)

Use integration by parts to compute \( a_n \) which yields:\[ a_n = \frac{4 (-1)^n}{n^2\pi^2} \].Thus the Fourier series for \( x^2 \) is:\[ x^2 = \frac{2}{3} + \sum_{n=1}^{\infty} \frac{4 (-1)^n}{n^2\pi^2} \cos(n\pi x) \]

Integrate the series for \( f(x) = \cosh x \) twice.

Integrate Fourier series of \( \cosh x \) twice, matching the form of \( g(x) = x^2 \). The resulting expressions will help prove the summation equality.

Prove the given summation equality

Equate the integrated series of \( \cosh x \) with the series for \( x^2 \) and identify coefficients to match both sides. This process involves mathematical rearrangements that yield:\[ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{2} \pi^{2}(n^{2} \pi^{2}+1)}=\frac{1}{2}\left(\frac{1}{\sinh 1}-\frac{5}{6}\right) \]

Key concepts

Real Fourier Coefficients

The Fourier series decomposes a periodic function into a sum of sine and cosine terms. This is done using Fourier coefficients which are real numbers in the case of real-valued functions. For the given functions, the Fourier series is defined as:
\[ f(x) = a_0 + \frac{1}{2} \bigg[a_0 + \textstyle\big(\textstyle\frac{}{} \frac{2(-1)^{n+1}}{n^2 \pi^2 + 1} \bigg) \bigg] \] The coefficients \( a_0 \), \( a_n \), and \( b_n \) are calculated using integrals over one period. For even functions like \( \cosh x \), the Fourier series only includes cosine terms because \( b_n = 0 \). The same applies to \( x^2 \), which also results in \( b_n = 0 \).

Integration by Parts

Integration by parts is essential in computing some of the Fourier coefficients. It follows the principle:
\[ \textstyle \textstyle u dv = uv - \textstyle v du \] Choosing appropriate functions for \( u \) and \( dv \) is key.
For example, in finding coefficients for \( \cosh x \), set \( u = \cosh(x) \) and \( dv = \cos(n\bar{x}) dx \). Then compute:

• \( \textstyle u = \cosh(x), \textstyle du = \sinh(x) dx \)
• \( \ dv = \cos(n\ x) dx, \textstyle v = \frac{\sin(n\pi\ x)}{n\pi} \)

Using the integration by parts formula, evaluate:
\( \textstyle uv|_{0}^{1} - \textstyle \ v du \). Combining these calculations provides Fourier coefficients involving hyperbolic functions.

Hyperbolic Functions

Hyperbolic functions like \( \cosh(x) \) (hyperbolic cosine) and \( \sinh(x) \) (hyperbolic sine) arise in Fourier analysis.
\( \cosh x = \frac{e^x + e^{-x}}{2} \)
\( \sinh x = \frac{e^x - e^{-x}}{2} \)
These functions resemble trigonometric functions but relate to hyperbolas. This property simplifies integration and helps in expressing the coefficients, especially in the context of periodic functions. For instance, \( \cosh(x) \) is an even function, making the calculation of Fourier series straightforward due to symmetry.

Series Summation

To find specific sums of infinite series, Fourier series can be very powerful. After developing the Fourier series for a function, integrating those series repeatedly can help in evaluating complex sums. For instance, integrating the series for \( \cosh x \) twice matches the form of the series developed for \( x^2 \), providing a method to prove:

• \( \textstyle \ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2 \pi^2 \(n^2 \pi^2 + 1 \)} = \frac{1}{2} \left( \frac{1}{\sinh 1 } - \frac{5}{6} \right) \)

This highlights that sometimes complex summations can be tackled using Fourier series expansions and subsequent integrations, emphasizing their powerful mathematical applications.