Math Problem Statement

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Find the Fourier series expansion of the periodic function f(x) = x on -π < x < π with f(x + 2π) = f(x).

Solution

To find the Fourier series expansion of the periodic function f(x)=xf(x) = x on the interval π<x<π-\pi < x < \pi, we proceed by calculating the Fourier coefficients.

Fourier Series:

The Fourier series of a function f(x)f(x) with period 2π2\pi is given by:

f(x)=a02+n=1(ancos(nx)+bnsin(nx))f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos(nx) + b_n \sin(nx) \right)

Where the coefficients a0a_0, ana_n, and bnb_n are defined as:

a0=1πππf(x)dxa_0 = \frac{1}{\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

Step 1: Find a0a_0

We first calculate a0a_0:

a0=1πππxdxa_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} x \, dx

This is a straightforward integral:

a0=1π[x22]ππ=1π(π22(π)22)=1π2π22=0a_0 = \frac{1}{\pi} \left[ \frac{x^2}{2} \right]_{-\pi}^{\pi} = \frac{1}{\pi} \left( \frac{\pi^2}{2} - \frac{(-\pi)^2}{2} \right) = \frac{1}{\pi} \cdot \frac{2\pi^2}{2} = 0

Thus, a0=0a_0 = 0.

Step 2: Find ana_n

Next, we calculate ana_n:

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

Since xcos(nx)x \cos(nx) is an odd function (because xx is odd and cos(nx)\cos(nx) is even), the integral of an odd function over a symmetric interval (like π-\pi to π\pi) is zero:

an=0for all na_n = 0 \quad \text{for all } n

Step 3: Find bnb_n

Now, we calculate bnb_n:

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

This is an integral of an odd function (both xx and sin(nx)\sin(nx) are odd, so their product is even), and we need to compute it:

bn=1π[xcos(nx)nππ+1nππcos(nx)dx]b_n = \frac{1}{\pi} \left[ -\frac{x \cos(nx)}{n} \Big|_{-\pi}^{\pi} + \frac{1}{n} \int_{-\pi}^{\pi} \cos(nx) \, dx \right]

Evaluating the boundary terms xcos(nx)ππx \cos(nx) \Big|_{-\pi}^{\pi}, we find that these terms vanish because cos(nπ)\cos(n\pi) alternates between ±1\pm 1, and the integrals of cos(nx)\cos(nx) over π-\pi to π\pi result in:

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

Final Fourier Series:

Thus, the Fourier series expansion of f(x)=xf(x) = x on π<x<π-\pi < x < \pi is given by:

f(x)=n=12(1)n+1nsin(nx)f(x) = \sum_{n=1}^{\infty} \frac{2 (-1)^{n+1}}{n} \sin(nx)

This is the Fourier series for the function f(x)=xf(x) = x, which consists only of sine terms (since it is an odd function).


Would you like more details on any of these steps or any related concept? Here are five questions you could explore further:

  1. How do we handle Fourier series for even and odd functions?
  2. What are the implications of the Fourier coefficients being zero for certain terms?
  3. How would the Fourier series change if the function were not odd?
  4. Can you derive a Fourier series for a piecewise function?
  5. How does the convergence of the Fourier series behave for discontinuous functions?

Tip: When calculating Fourier coefficients, always check the symmetry of the function to simplify the integrals (odd/even properties can eliminate terms).

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Math Problem Analysis

Mathematical Concepts

Fourier Series
Periodic Functions
Odd and Even Functions
Harmonic Analysis

Formulas

f(x) = a0/2 + Σ(an * cos(nx) + bn * sin(nx))
a0 = (1/π) ∫[-π, π] f(x) dx
an = (1/π) ∫[-π, π] f(x) cos(nx) dx
bn = (1/π) ∫[-π, π] f(x) sin(nx) dx

Theorems

Fourier Series Theorem
Orthogonality of Sine and Cosine Functions

Suitable Grade Level

Undergraduate Mathematics