Fourier Series Practice Problems

Fourier series is a mathematical tool used to decompose periodic functions into a sum of simpler sine and cosine waves. Understanding how to solve Fourier series practice problems is crucial for anyone studying signal processing, differential equations, or any field involving periodic functions. This guide will walk you through various problems, explaining each step in detail to enhance your comprehension.

Practice Problems and Solutions

Problem 1: Find the Fourier series for f(x) = x on the interval [-π, π].

Solution:

Given

f(x) = x on -π, π

Since f(x) is an odd function

f(-x) =-f(x) , its Fourier series contains only sine terms.

Step 1: Calculate a₀:

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

dx = \frac{1}{\pi} \left[ \frac{x^2}{2} \right]_{-\pi}^{\pi}

= \frac{1}{\pi} \left( \frac{\pi^2}{2} - \frac{\pi^2}{2} \right) = 0

Step 2: Calculate a_n:

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

dx = 0 (because the integrand is an odd function over symmetric limits, the integral is zero)

Step 3:

Calculate b_n:

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

dx = \frac{1}{\pi} \left[ -\frac{x \cos(nx)}{n} + \frac{\sin(nx)}{n^2} \right]_{-\pi}^{\pi}

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

Therefore, the Fourier series for f(x) is: f(x) = \sum_{n=1}^\infty \frac{2}{n} (-1)^{n+1} \sin(nx)

Problem 2: Find the Fourier series for f(x) = |x| on the interval [-π, π].

Solution:

Given f(x) = |x| on [-π, π] .

Since f(x) is an even function f(-x) = f(x)

its Fourier series contains only cosine terms.

Step 1: Calculate a₀:

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

dx = \frac{2}{\pi} \int_0^{\pi} x

dx = \frac{2}{\pi} \cdot \frac{\pi^2}{2} =\pi

Step 2: Calculate a_n:

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

dx = \frac{2}{\pi} \int_0^{\pi} x \cos(nx)dx

Integration by parts gives

int_0^{\pi} x \cos(nx),

dx = \frac{1}{n^2} \left( (-1)^n - 1 \right)
Thus,a_n = \frac{2}{\pi n^2} \left( (-1)^n - 1 \right)))

Step 3: Calculate b_n:

b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} |x| \sin(nx)
dx = 0 (since the integrand is an odd function)

Therefore, the Fourier series is: f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos(nx) = \frac{\pi}{2} - \frac{4}{\pi} \sum_{n=1}^\infty \frac{\cos\big((2n - 1)x\big)}{(2n - 1)^2}\\

Problem 3: Find the Fourier series for f(x) = x² on the interval [-π, π].

Solution:

Given f(x) = x² on [-π, π]

Since f(x) is an even function f(-x) = f(x) , its Fourier series contains only cosine terms.

Step 1: Calculate a₀ :

a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi}

x^2 \, dx = \frac{2}{\pi} \int_0^{\pi} x^2

= \frac{2}{\pi} \cdot \frac{\pi^3}{3}
= \frac{2 \pi^2}{3}

Step 2: Calculate a_n:

a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \cos(nx)

dx = \frac{2}{\pi} \int_0^{\pi} x^2 \cos(nx)

Using integration by parts twice, the integral evaluates to:

\int_0^\pi x^2 \cos(nx)

dx = \frac{2\pi (-1)^n}{n^2} - \frac{2}{n^3} \sin(n\pi)

= \frac{2\pi (-1)^n}{n^2} {since \sin(n\pi) = 0)

Therefore, a_n = \frac{2}{\pi} \times \frac{2\pi (-1)^n}{n^2} = \frac{4 (-1)^n}{n^2}

Step 3: Calculate b_n:

b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \sin(nx)

dx = 0 (because x²sin(nx) is an odd function over [- П, П] )

Therefore, the Fourier series is f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos(nx) = \frac{\pi^2}{3} + 4 \sum_{n=1}^\infty \frac{(-1)^n}{n^2} \cos(nx)

Problem 4: Find the Fourier series for f(x) = x³ on the interval [-π, π].

Solution:

Step 1: Determine a₀

a₀ = (1/π) ∫ x³ dx from -π to π = 0

Step 2: Determine aₙ

aₙ = (1/π) ∫ x³ cos(nx) dx from -π to π

= (6/n²) (-1)ⁿ⁺¹ - (6π/n³) (-1)ⁿ = 0

Step 3: Determine bₙ

bₙ = (1/π) ∫ x³ sin(nx) dx from -π to π
= -6π²/n³ + 24/n³

Therefore, the Fourier series is:

f(x) = Σ ((6/n²) (-1)ⁿ⁺¹ + (-6π²/n³ + 24/n³) sin(nx)

Problem 5: Find the Fourier series for f(x) = x on the interval [0, 2π].

Solution:

The given function f(x) = x on [0, 2П] is an odd function if extended periodically with period.:

Step 1: Calculate a₀:

a₀ = \frac{1}{\pi} \int_{0}^{2\pi} x \, dx = \frac{1}{\pi} \left[ \frac{x^2}{2} \right]_0^{2\pi} = 2\pi

Step 2: Calculate a_n:

a_n = \frac{1}{\pi} \int_{0}^{2\pi} x \cos(nx) \, dx = 0, \quad \text{for all } n

Step 3: Calculate b_n:

b_n = \frac{1}{\pi} \int_{0}^{2\pi} x \sin(nx) \, dx = -\frac{2}{n}

Therefore, the Fourier series is:

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

= \pi - \sum_{n=1}^{\infty} \frac{2}{n} \sin(nx)

Problem 6: Find the Fourier series for f(x) = |sin(x)| on the interval [-π, π].

Solution:

f(x)=∣sinx∣, −π ≤ x ≤ π.

Since ∣sin⁡x∣ is even, the Fourier series contains only cosine terms:

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

Step 1: Constant term a₀:

a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} |\sin x| \, dx= \frac{2}{\pi} \int_0^{\pi} \sin x \, dx= \frac{2}{\pi} \cdot 2= \frac{4}{\pi}

\frac{a_0}{2} = \frac{2}{\pi}

Step 2: Cosine coefficients a_n

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

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

\int_0^{\pi} \sin(kx)\, dx = \frac{1 - \cos(k\pi)}{k}
a_n = \frac{1}{\pi} \left( \frac{1 - \cos((n+1)\pi)}{n+1} + \frac{1 - \cos((1-n)\pi)}{1-n} \right)

Simplify using:

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

|\sin x| = \frac{2}{\pi} - \frac{4}{\pi} \sum_{m=1}^\infty \frac{\cos(2mx)}{4m^2 - 1}, \quad -\pi \le x \le \pi

Unsolved Practice Problems on Fourier Series

Problem 1: Find the Fourier series for f(x) = sin(x) on the interval [-π, π].

Problem 2: Find the Fourier series for f(x) = x² - π² on the interval [-π, π].

Problem 3: Find the Fourier series for f(x) = 1 for 0 ≤ x < π and f(x) = -1 for π ≤ x < 2π.

Problem 4: Find the Fourier series for f(x) = e^x on the interval [-π, π].

Problem 5: Find the Fourier series for f(x) = cos⁡²(x) on the interval [−π, π].

Problem 6: Find the Fourier series for f(x) = x⁴ on the interval [−π, π].