Math Problem Statement

Find the Fourier Series of the function f(x) = 1, defined on the interval (-л, л]

Solution

To find the Fourier series of the function f(x)=1f(x) = 1 defined on the interval (π,π](- \pi, \pi], we follow these steps:

  1. Determine the Fourier coefficients a0a_0, ana_n, and bnb_n.

The Fourier series of a function f(x)f(x) defined on (π,π](- \pi, \pi] is given by: f(x)=a0+n=1(ancos(nx)+bnsin(nx))f(x) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos(nx) + b_n \sin(nx) \right)

The coefficients are calculated as follows:

a0=12πππf(x)dxa_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) \, dx an=1πππf(x)cos(nx)dxa_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx bn=1πππf(x)sin(nx)dxb_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx

  1. Calculate a0a_0:

a0=12πππ1dxa_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} 1 \, dx

Since f(x)=1f(x) = 1 over the entire interval: a0=12π[x]ππ=12π(π(π))=12π2π=1a_0 = \frac{1}{2\pi} \cdot \left[ x \right]_{-\pi}^{\pi} = \frac{1}{2\pi} \cdot ( \pi - (-\pi)) = \frac{1}{2\pi} \cdot 2\pi = 1

  1. Calculate ana_n:

an=1πππ1cos(nx)dxa_n = \frac{1}{\pi} \int_{-\pi}^{\pi} 1 \cos(nx) \, dx

Since cos(nx)\cos(nx) is an odd function when n0n \neq 0 and we are integrating over a symmetric interval: an=1π0=0forn1a_n = \frac{1}{\pi} \cdot 0 = 0 \quad \text{for} \quad n \geq 1

  1. Calculate bnb_n:

bn=1πππ1sin(nx)dxb_n = \frac{1}{\pi} \int_{-\pi}^{\pi} 1 \sin(nx) \, dx

Since sin(nx)\sin(nx) is an odd function and we are integrating over a symmetric interval: bn=1π0=0forn1b_n = \frac{1}{\pi} \cdot 0 = 0 \quad \text{for} \quad n \geq 1

Therefore, the Fourier coefficients ana_n and bnb_n for n1n \geq 1 are all zero.

  1. Construct the Fourier series:

Given that a0=1a_0 = 1 and an=bn=0a_n = b_n = 0 for n1n \geq 1, the Fourier series for f(x)=1f(x) = 1 is: f(x)=1+n=1(0cos(nx)+0sin(nx))=1f(x) = 1 + \sum_{n=1}^{\infty} \left( 0 \cdot \cos(nx) + 0 \cdot \sin(nx) \right) = 1

Thus, the Fourier series of the function f(x)=1f(x) = 1 defined on the interval (π,π](- \pi, \pi] is simply:

f(x)=1f(x) = 1

Summary:

The Fourier series of the function f(x)=1f(x) = 1 on the interval (π,π](- \pi, \pi] is:

f(x)=1f(x) = 1

Would you like any further details or have any questions?

Here are some related questions to expand on the topic:

  1. How do you find the Fourier series of a function defined on an interval other than (π,π](- \pi, \pi]?
  2. What is the physical significance of the Fourier coefficients?
  3. How does the Fourier series converge for different types of functions?
  4. Can you provide an example of a function with a more complex Fourier series?
  5. How are Fourier series used in signal processing?
  6. What is the relationship between the Fourier series and the Fourier transform?
  7. How do Gibbs phenomena appear in Fourier series?
  8. What are the conditions for a function to be represented by a Fourier series?

Tip: When calculating Fourier coefficients, taking advantage of symmetry (odd/even functions) can simplify the integrals significantly.

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Math Problem Analysis

Mathematical Concepts

Fourier Series
Trigonometric Functions
Integral Calculus

Formulas

Fourier series coefficients
Integral formulas for Fourier coefficients

Theorems

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Suitable Grade Level

Undergraduate Level