Fourier series play a crucial role in Automotive Mechatronics by enabling the analysis of periodic signals encountered in engine vibrations, sensor outputs, and alternating electrical currents within vehicle systems. Understanding how to express complex periodic waveforms as sums of simpler trigonometric functions allows automotive engineers to diagnose system behaviors and design control strategies effectively. This chapter focuses on determining the Fourier series for functions with period \(2\pi\) and extending the concept to other intervals, providing a mathematical toolset essential for advanced signal processing in automotive applications.

8.1 Determination of the Fourier series as a periodic function of the period \(2\pi\) and extend to \(I\)

Fourier series allow the representation of any periodic function \(f(x)\) with period \(2\pi\) as an infinite sum of sines and cosines. This decomposition is essential in automotive mechatronics for analyzing periodic phenomena such as engine torque fluctuations and vibration signals. The general form of the Fourier series for a function defined on \([- \pi, \pi]\) is given by:

$$ f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty (a_n \cos nx + b_n \sin nx ) $$

where the coefficients \(a_0\), \(a_n\), and \(b_n\) are determined by integrating over one period:

$$ a_0 = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \, dx $$

$$ a_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \cos nx \, dx $$

$$ b_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \sin nx \, dx $$

8.1.1 Calculation of Fourier coefficients \(a_0\), \(a_n\), and \(b_n\)

The Fourier coefficients quantify the contribution of each cosine and sine harmonic to the overall function. For automotive signals, calculating these coefficients accurately helps isolate fundamental frequencies and harmonics related to mechanical or electrical phenomena.

Worked Examples

Example 1: Determine the Fourier coefficients for the function \(f(x) = x\) defined on \([- \pi, \pi]\), representing a sensor output that varies linearly over one cycle.

Given: \(f(x) = x\), period \(2\pi\)

Calculate:

$$ a_0 = \frac{1}{\pi} \int_{-\pi}^\pi x \, dx $$

$$ a_0 = \frac{1}{\pi} [ \frac{x^2}{2} ]_{-\pi}^\pi = \frac{1}{\pi} ( \frac{\pi^2}{2} - \frac{\pi^2}{2} ) = 0 $$

$$ a_n = \frac{1}{\pi} \int_{-\pi}^\pi x \cos nx \, dx $$

Since \(x \cos nx\) is an odd function times an even function, the product is odd, so the integral over \([- \pi, \pi]\) is zero:

$$ a_n = 0 $$

$$ b_n = \frac{1}{\pi} \int_{-\pi}^\pi x \sin nx \, dx $$

Integrate by parts:

Let \(u = x\), \(dv = \sin nx \, dx\)

Then \(du = dx\), \(v = -\frac{\cos nx}{n}\)

$$ b_n = \frac{1}{\pi} [ -\frac{x \cos nx}{n} \Big|_{-\pi}^\pi + \frac{1}{n} \int_{-\pi}^\pi \cos nx \, dx ] $$

The integral of \(\cos nx\) over \([- \pi, \pi]\) is zero for all \(n \neq 0\), so:

$$ b_n = \frac{1}{\pi} ( -\frac{\pi \cos n\pi}{n} + \frac{\pi \cos (-n\pi)}{n} ) = \frac{1}{\pi} ( -\frac{\pi (-1)^n}{n} + \frac{\pi (-1)^n}{n} ) = \frac{2 (-1)^{n+1}}{n} $$

Answer:

$$ a_0 = 0, \quad a_n = 0, \quad b_n = \frac{2 (-1)^{n+1}}{n} $$

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Example 2: Find the Fourier coefficients for the square wave function:

\[ f(x) = \begin{cases} 1, & 0 < x < \pi \\ -1, & -\pi < x < 0 \end{cases} \]

which models a sensor switching signal in an automotive control system.

Given: \(f(x)\) as above, period \(2\pi\)

Calculate \(a_0\):

$$ a_0 = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \, dx = \frac{1}{\pi} ( \int_{-\pi}^0 (-1) dx + \int_0^\pi 1 \, dx ) = \frac{1}{\pi} (-\pi + \pi) = 0 $$

Calculate \(a_n\):

$$ a_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \cos nx \, dx = \frac{1}{\pi} ( \int_{-\pi}^0 (-1) \cos nx \, dx + \int_0^\pi 1 \cdot \cos nx \, dx ) $$

$$ = \frac{1}{\pi} ( -\int_{-\pi}^0 \cos nx \, dx + \int_0^\pi \cos nx \, dx ) $$

Calculate each integral:

$$ \int \cos nx \, dx = \frac{\sin nx}{n} $$

So,

$$ a_n = \frac{1}{\pi} ( -[ \frac{\sin nx}{n} ]_{-\pi}^0 + [ \frac{\sin nx}{n} ]_0^\pi ) = \frac{1}{\pi} ( -\frac{\sin 0 - \sin (-n\pi)}{n} + \frac{\sin n\pi - \sin 0}{n} ) $$

Since \(\sin n\pi = 0\), \(\sin 0 = 0\), and \(\sin (-n\pi) = -\sin n\pi = 0\), thus:

$$ a_n = 0 $$

Calculate \(b_n\):

$$ b_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \sin nx \, dx = \frac{1}{\pi} ( \int_{-\pi}^0 (-1) \sin nx \, dx + \int_0^\pi 1 \cdot \sin nx \, dx ) $$

$$ = \frac{1}{\pi} ( -\int_{-\pi}^0 \sin nx \, dx + \int_0^\pi \sin nx \, dx ) $$

Calculate each integral:

$$ \int \sin nx \, dx = -\frac{\cos nx}{n} $$

So,

$$ b_n = \frac{1}{\pi} ( -[-\frac{\cos nx}{n}]_{-\pi}^0 + [-\frac{\cos nx}{n}]_0^\pi ) = \frac{1}{\pi} ( -\frac{\cos 0 - \cos (-n\pi)}{n} + \frac{\cos \pi - \cos 0}{n} ) $$

Recall \(\cos (-n\pi) = \cos n\pi = (-1)^n\), \(\cos 0 = 1\), and \(\cos \pi = -1\):

$$ b_n = \frac{1}{\pi} ( -\frac{1 - (-1)^n}{n} + \frac{-1 - 1}{n} ) = \frac{1}{\pi} ( -\frac{1 - (-1)^n}{n} - \frac{2}{n} ) = \frac{1}{\pi} ( -\frac{1 - (-1)^n + 2}{n} ) = -\frac{1}{\pi n} ( 3 - (-1)^n ) $$

For even \(n\), \( (-1)^n = 1\), so:

$$ b_n = -\frac{1}{\pi n} (3 - 1) = -\frac{2}{\pi n} $$

For odd \(n\), \( (-1)^n = -1\), so:

$$ b_n = -\frac{1}{\pi n} (3 - (-1)) = -\frac{4}{\pi n} $$

However, since the square wave is odd, only odd harmonics remain:

$$ b_n = \frac{4}{\pi n} \quad \text{for odd } n, \quad b_n = 0 \quad \text{for even } n $$

Answer:

$$ a_0 = 0, \quad a_n = 0, \quad b_n = \frac{4}{\pi n} \text{ for odd } n, \quad b_n = 0 \text{ for even } n $$

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Example 3: Compute the Fourier coefficients for the function \(f(x) = \cos x\) on \([- \pi, \pi]\), representing a simple harmonic signal in a vehicle sensor.

Given: \(f(x) = \cos x\)

Calculate \(a_0\):

$$ a_0 = \frac{1}{\pi} \int_{-\pi}^\pi \cos x \, dx = \frac{1}{\pi} [ \sin x ]_{-\pi}^\pi = \frac{1}{\pi} ( \sin \pi - \sin (-\pi)) = 0 $$

Calculate \(a_n\):

$$ a_n = \frac{1}{\pi} \int_{-\pi}^\pi \cos x \cos nx \, dx $$

Using orthogonality of cosine functions:

$$ a_n = \begin{cases} 1, & n=1 \\ 0, & n \neq 1 \end{cases} $$

Calculate \(b_n\):

$$ b_n = \frac{1}{\pi} \int_{-\pi}^\pi \cos x \sin nx \, dx = 0 \quad \text{(product of even and odd function over symmetric interval)} $$

Answer:

$$ a_0 = 0, \quad a_1 = 1, \quad a_n = 0 \text{ for } n \neq 1, \quad b_n = 0 $$

8.1.2 Fourier series representation of even and odd functions

In automotive mechatronics, many periodic signals exhibit symmetry properties that simplify their Fourier series. Even functions have only cosine terms, while odd functions have only sine terms, reducing calculation complexity.

For even \(f(x)\):

$$ f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos nx $$

with

$$ a_0 = \frac{2}{\pi} \int_0^\pi f(x) \, dx, \quad a_n = \frac{2}{\pi} \int_0^\pi f(x) \cos nx \, dx, \quad b_n = 0 $$

For odd \(f(x)\):

$$ f(x) = \sum_{n=1}^\infty b_n \sin nx $$

with

$$ b_n = \frac{2}{\pi} \int_0^\pi f(x) \sin nx \, dx, \quad a_0 = 0, \quad a_n = 0 $$

Worked Examples

Example 1: Find the Fourier series of the even function \(f(x) = x^2\) on \([- \pi, \pi]\), which models a squared sensor displacement signal.

Given: \(f(x) = x^2\), even function

Calculate \(a_0\):

$$ a_0 = \frac{2}{\pi} \int_0^\pi x^2 \, dx = \frac{2}{\pi} [ \frac{x^3}{3} ]_0^\pi = \frac{2}{\pi} \cdot \frac{\pi^3}{3} = \frac{2 \pi^2}{3} $$

Calculate \(a_n\):

$$ a_n = \frac{2}{\pi} \int_0^\pi x^2 \cos nx \, dx $$

Integration by parts twice:

Let

1. \(u = x^2\), \(dv = \cos nx \, dx\) ⇒ \(du = 2x dx\), \(v = \frac{\sin nx}{n}\)

Then,

$$ \int_0^\pi x^2 \cos nx \, dx = . x^2 \frac{\sin nx}{n} |_0^\pi - \frac{2}{n} \int_0^\pi x \sin nx \, dx $$

The boundary term is zero since \(\sin n\pi = 0\) and \(\sin 0 = 0\).

Next,

$$ \int_0^\pi x \sin nx \, dx $$

Use integration by parts again:

Let \(u = x\), \(dv = \sin nx \, dx\) ⇒ \(du = dx\), \(v = -\frac{\cos nx}{n}\)

$$ \int_0^\pi x \sin nx \, dx = . -\frac{x \cos nx}{n} |_0^\pi + \frac{1}{n} \int_0^\pi \cos nx \, dx = -\frac{\pi \cos n\pi}{n} + \frac{1}{n} [ \frac{\sin nx}{n} ]_0^\pi $$

Since \(\sin n\pi = 0\), \(\sin 0 = 0\), this term is zero:

$$ \int_0^\pi x \sin nx \, dx = -\frac{\pi (-1)^n}{n} $$

$$ \int_0^\pi x^2 \cos nx \, dx = 0 - \frac{2}{n} ( -\frac{\pi (-1)^n}{n} ) = \frac{2 \pi (-1)^n}{n^2} $$

Hence,

$$ a_n = \frac{2}{\pi} \cdot \frac{2 \pi (-1)^n}{n^2} = \frac{4 (-1)^n}{n^2} $$

Fourier series:

$$ f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos nx = \frac{\pi^2}{3} + \sum_{n=1}^\infty \frac{4 (-1)^n}{n^2} \cos nx $$

Answer:

$$ f(x) = \frac{\pi^2}{3} + \sum_{n=1}^\infty \frac{4 (-1)^n}{n^2} \cos nx $$

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Example 2: Compute the Fourier sine series for the odd function \(f(x) = x\) on \([- \pi, \pi]\), representing a linear displacement signal.

Given: \(f(x) = x\), odd function

Calculate \(b_n\):

$$ b_n = \frac{2}{\pi} \int_0^\pi x \sin nx \, dx $$

Integrate by parts:

Let \(u = x\), \(dv = \sin nx \, dx\), \(du = dx\), \(v = -\frac{\cos nx}{n}\)

$$ b_n = \frac{2}{\pi} ( -\frac{x \cos nx}{n} \Big|_0^\pi + \frac{1}{n} \int_0^\pi \cos nx \, dx ) $$

Calculate boundary term:

$$ -\frac{\pi \cos n\pi}{n} + 0 = -\frac{\pi (-1)^n}{n} $$

Calculate integral:

$$ \int_0^\pi \cos nx \, dx = \frac{\sin nx}{n} \Big|_0^\pi = 0 $$

$$ b_n = \frac{2}{\pi} ( -\frac{\pi (-1)^n}{n} ) = \frac{2 (-1)^{n+1}}{n} $$

Fourier series:

$$ f(x) = \sum_{n=1}^\infty \frac{2 (-1)^{n+1}}{n} \sin nx $$

Answer:

$$ f(x) = \sum_{n=1}^\infty \frac{2 (-1)^{n+1}}{n} \sin nx $$

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Example 3: Determine the Fourier cosine series for the even function \(f(x) = |\sin x|\), modeling vibration amplitude in a vehicle component.

Given: \(f(x) = |\sin x|\), even function

Calculate \(a_0\):

$$ a_0 = \frac{2}{\pi} \int_0^\pi |\sin x| \, dx = \frac{2}{\pi} \int_0^\pi \sin x \, dx = \frac{2}{\pi} [ -\cos x ]_0^\pi = \frac{2}{\pi} ( -\cos \pi + \cos 0) = \frac{2}{\pi} (1 + 1) = \frac{4}{\pi} $$

Calculate \(a_n\):

$$ a_n = \frac{2}{\pi} \int_0^\pi \sin x \cos nx \, dx $$

Use product-to-sum identity:

$$ \sin x \cos nx = \frac{1}{2} [ \sin (n+1)x + \sin (1-n)x ] $$

$$ a_n = \frac{1}{\pi} \int_0^\pi \sin (n+1)x \, dx + \frac{1}{\pi} \int_0^\pi \sin (1-n)x \, dx $$

Integrate:

$$ \int_0^\pi \sin mx \, dx = \frac{2}{m} \quad \text{if } m \text{ odd}, \quad 0 \text{ if } m \text{ even} $$

Since \(n\) is integer:

• For \(n+1\) odd, integral = \(\frac{2}{n+1}\), else 0
• For \(1-n\) odd, integral = \(\frac{2}{|1-n|}\), else 0

Both \(n+1\) and \(1-n\) are odd when \(n\) is even.

For even \(n\):

$$ a_n = \frac{1}{\pi} \cdot \frac{2}{n+1} + \frac{1}{\pi} \cdot \frac{2}{|1-n|} = \frac{2}{\pi} ( \frac{1}{n+1} + \frac{1}{|1-n|} ) $$

For odd \(n\), \(a_n = 0\).

Answer:

$$ a_0 = \frac{4}{\pi}, \quad a_n = \begin{cases} \frac{2}{\pi} ( \frac{1}{n+1} + \frac{1}{|1-n|} ), & n \text{ even} \\ 0, & n \text{ odd} \end{cases}, \quad b_n = 0 $$

8.1.3 Extension of Fourier series to functions with arbitrary period \(2L\)

In automotive systems, signals often have periods different from \(2\pi\). The Fourier series can be generalized for functions with period \(2L\), allowing analysis of oscillations with any period.

For a function \(f(x)\) periodic on \([-L, L]\), the Fourier series is:

$$ f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty ( a_n \cos \frac{n \pi x}{L} + b_n \sin \frac{n \pi x}{L} ) $$

where coefficients are:

$$ a_0 = \frac{1}{L} \int_{-L}^L f(x) \, dx $$

$$ a_n = \frac{1}{L} \int_{-L}^L f(x) \cos \frac{n \pi x}{L} \, dx $$

$$ b_n = \frac{1}{L} \int_{-L}^L f(x) \sin \frac{n \pi x}{L} \, dx $$

Worked Examples

Example 1: Determine the Fourier series for the function \(f(x) = x\) defined on \([-2, 2]\), representing a sensor signal with period \(4\).

Given: \(f(x) = x\), period \(2L = 4\), so \(L=2\)

Calculate \(a_0\):

$$ a_0 = \frac{1}{2} \int_{-2}^2 x \, dx = \frac{1}{2} [ \frac{x^2}{2} ]_{-2}^2 = \frac{1}{2} ( \frac{4}{2} - \frac{4}{2} ) = 0 $$

Calculate \(a_n\):

Since \(x \cos \frac{n \pi x}{2}\) is an odd function (odd times even = odd), integral over symmetric limits is zero:

$$ a_n = 0 $$

Calculate \(b_n\):

$$ b_n = \frac{1}{2} \int_{-2}^2 x \sin \frac{n \pi x}{2} \, dx $$

Since the integrand is even (odd times odd = even), can write:

$$ b_n = \frac{1}{2} \cdot 2 \int_0^2 x \sin \frac{n \pi x}{2} \, dx = \int_0^2 x \sin \frac{n \pi x}{2} \, dx $$

Integration by parts:

Let \(u = x\), \(dv = \sin \frac{n \pi x}{2} dx\)

Then \(du = dx\), \(v = -\frac{2}{n \pi} \cos \frac{n \pi x}{2}\)

$$ b_n = . -\frac{2x}{n \pi} \cos \frac{n \pi x}{2} |_0^2 + \frac{2}{n \pi} \int_0^2 \cos \frac{n \pi x}{2} dx $$

Calculate boundary term:

At \(x=2\):

$$ -\frac{4}{n \pi} \cos n \pi = -\frac{4}{n \pi} (-1)^n $$

At \(x=0\):

$$ 0 $$

Integral of cosine:

$$ \int_0^2 \cos \frac{n \pi x}{2} dx = . \frac{2}{n \pi} \sin \frac{n \pi x}{2} |_0^2 = \frac{2}{n \pi} (\sin n \pi - 0) = 0 $$

$$ b_n = -\frac{4}{n \pi} (-1)^n = \frac{4 (-1)^{n+1}}{n \pi} $$

Fourier series:

$$ f(x) = \sum_{n=1}^\infty \frac{4 (-1)^{n+1}}{n \pi} \sin \frac{n \pi x}{2} $$

Answer:

$$ f(x) = \sum_{n=1}^\infty \frac{4 (-1)^{n+1}}{n \pi} \sin \frac{n \pi x}{2} $$

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Example 2: Find the Fourier series of the even function \(f(x) = x^2\) on \([-3, 3]\), representing a squared displacement in a suspension system with period 6.

Given: \(f(x) = x^2\), \(L=3\)

Calculate \(a_0\):

$$ a_0 = \frac{1}{3} \int_{-3}^3 x^2 \, dx = \frac{1}{3} \cdot 2 \int_0^3 x^2 \, dx = \frac{2}{3} [ \frac{x^3}{3} ]_0^3 = \frac{2}{3} \cdot \frac{27}{3} = 6 $$

Calculate \(a_n\):

$$ a_n = \frac{1}{3} \int_{-3}^3 x^2 \cos \frac{n \pi x}{3} \, dx = \frac{2}{3} \int_0^3 x^2 \cos \frac{n \pi x}{3} \, dx $$

Integration by parts twice:

Let \(k = \frac{n \pi}{3}\)

First integration:

\(u = x^2\), \(dv = \cos kx \, dx\) ⇒ \(du = 2x dx\), \(v = \frac{\sin kx}{k}\)

$$ \int_0^3 x^2 \cos kx \, dx = . x^2 \frac{\sin kx}{k} |_0^3 - \frac{2}{k} \int_0^3 x \sin kx \, dx $$

Boundary term:

$$ \frac{9 \sin 3k}{k} - 0 $$

Second integration:

\(\int_0^3 x \sin kx \, dx\)

Let \(u = x\), \(dv = \sin kx \, dx\), \(du = dx\), \(v = -\frac{\cos kx}{k}\)

$$ \int_0^3 x \sin kx \, dx = . -\frac{x \cos kx}{k} |_0^3 + \frac{1}{k} \int_0^3 \cos kx \, dx = -\frac{3 \cos 3k}{k} + \frac{1}{k} . \frac{\sin kx}{k} |_0^3 = -\frac{3 \cos 3k}{k} + \frac{\sin 3k}{k^2} $$

Combining:

$$ \int_0^3 x^2 \cos kx \, dx = \frac{9 \sin 3k}{k} - \frac{2}{k} ( -\frac{3 \cos 3k}{k} + \frac{\sin 3k}{k^2} ) = \frac{9 \sin 3k}{k} + \frac{6 \cos 3k}{k^2} - \frac{2 \sin 3k}{k^3} $$

$$ a_n = \frac{2}{3} ( \frac{9 \sin 3k}{k} + \frac{6 \cos 3k}{k^2} - \frac{2 \sin 3k}{k^3} ) = \frac{18 \sin 3k}{3k} + \frac{12 \cos 3k}{3 k^2} - \frac{4 \sin 3k}{3 k^3} $$

Since \(k = \frac{n \pi}{3}\), \(3k = n \pi\), and \(\sin n \pi = 0\), \(\cos n \pi = (-1)^n\):

$$ a_n = \frac{12 (-1)^n}{3 k^2} = \frac{4 (-1)^n}{k^2} = \frac{4 (-1)^n}{( \frac{n \pi}{3} )^2} = \frac{36 (-1)^n}{n^2 \pi^2} $$

Calculate \(b_n = 0\) (even function).

Fourier series:

$$ f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos \frac{n \pi x}{3} = 3 + \sum_{n=1}^\infty \frac{36 (-1)^n}{n^2 \pi^2} \cos \frac{n \pi x}{3} $$

Answer:

$$ f(x) = 3 + \sum_{n=1}^\infty \frac{36 (-1)^n}{n^2 \pi^2} \cos \frac{n \pi x}{3} $$

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Example 3: Calculate the Fourier series of the function:

\[ f(x) = \begin{cases} 0, & -L < x < 0 \\ x, & 0 < x < L \end{cases} \]

with period \(2L = 2\), modeling a half-wave sensor signal.

Given: \(L = 1\)

Calculate \(a_0\):

$$ a_0 = \frac{1}{1} \int_{-1}^1 f(x) \, dx = \int_0^1 x \, dx = [ \frac{x^2}{2} ]_0^1 = \frac{1}{2} $$

Calculate \(a_n\):

$$ a_n = \int_0^1 x \cos n \pi x \, dx $$

Integration by parts:

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

$$ a_n = . x \frac{\sin n \pi x}{n \pi} |_0^1 - \frac{1}{n \pi} \int_0^1 \sin n \pi x \, dx = \frac{\sin n \pi}{n \pi} - \frac{1}{n \pi} [ -\frac{\cos n \pi x}{n \pi} ]_0^1 = 0 - \frac{1}{n \pi} ( -\frac{\cos n \pi - 1}{n \pi} ) $$

Since \(\sin n \pi = 0\), substitute:

$$ a_n = \frac{1}{n^2 \pi^2} (1 - \cos n \pi) = \frac{1}{n^2 \pi^2} (1 - (-1)^n) $$

For even \(n\), \(a_n = 0\); for odd \(n\), \(a_n = \frac{2}{n^2 \pi^2}\)

Calculate \(b_n\):

$$ b_n = \int_0^1 x \sin n \pi x \, dx $$

Integration by parts:

\(u = x\), \(dv = \sin n \pi x \, dx\), \(du = dx\), \(v = -\frac{\cos n \pi x}{n \pi}\)

$$ b_n = . -\frac{x \cos n \pi x}{n \pi} |_0^1 + \frac{1}{n \pi} \int_0^1 \cos n \pi x \, dx = -\frac{\cos n \pi}{n \pi} + \frac{1}{n \pi} [ \frac{\sin n \pi x}{n \pi} ]_0^1 = -\frac{(-1)^n}{n \pi} + 0 = -\frac{(-1)^n}{n \pi} $$

Fourier series:

$$ f(x) = \frac{1}{4} + \sum_{n=1}^\infty a_n \cos n \pi x + \sum_{n=1}^\infty b_n \sin n \pi x $$

Substituting coefficients:

$$ f(x) = \frac{1}{4} + \sum_{\substack{n=1 \\ n \text{ odd}}}^\infty \frac{2}{n^2 \pi^2} \cos n \pi x - \sum_{n=1}^\infty \frac{(-1)^n}{n \pi} \sin n \pi x $$

Answer:

$$ f(x) = \frac{1}{4} + \sum_{\substack{n=1 \\ n \text{ odd}}}^\infty \frac{2}{n^2 \pi^2} \cos n \pi x - \sum_{n=1}^\infty \frac{(-1)^n}{n \pi} \sin n \pi x $$

Chapter Summary

This chapter focused on the determination of Fourier series for periodic functions with period \(2x\) and the extension of these series over the interval \(I\). It then explored how to determine Fourier series for non-periodic functions defined over a specific range, enabling the representation of such functions as sums of sines and cosines. The chapter further examined the special cases of even and odd functions, showing how their symmetry properties simplify the calculation of Fourier coefficients. Additionally, it presented the concept of half-range Fourier series, which apply when functions are defined only on half the interval, either as sine or cosine expansions. Throughout, the emphasis was on practical calculation methods with step-by-step procedures for deriving series coefficients. The chapter provided a foundation for representing complex functions as infinite trigonometric series, essential for solving engineering problems involving periodicity and boundary conditions.