Math Problem Statement
Solution
To find the Fourier expansion of the given function , we need to express the function as a Fourier series on the interval .
The given piecewise function is:
\begin{cases} -x, & -1 < x < 0 \\ x, & 0 < x < 1 \end{cases}$$ Since the function is odd (because $$f(-x) = -f(x)$$), we only need to find the sine terms of the Fourier series (Fourier sine series). The general Fourier sine series for an odd function is given by: $$f(x) = \sum_{n=1}^{\infty} b_n \sin\left(n\pi x\right)$$ where the Fourier coefficients $$b_n$$ are computed as: $$b_n = \frac{2}{L} \int_0^L f(x) \sin\left(\frac{n\pi x}{L}\right) dx$$ Here, $$L = 1$$ because the interval is $$[-1, 1]$$. ### Step 1: Compute the Fourier Coefficients $$b_n$$ Since the function is odd, we only need to calculate the sine coefficients: $$b_n = 2 \int_0^1 f(x) \sin(n\pi x) dx$$ For $$f(x) = x$$ on $$[0, 1]$$, we compute $$b_n$$ as: $$b_n = 2 \int_0^1 x \sin(n\pi x) dx$$ This integral can be solved using integration by parts. Let: - $$u = x$$, so $$du = dx$$ - $$dv = \sin(n\pi x) dx$$, so $$v = -\frac{1}{n\pi} \cos(n\pi x)$$ Now applying integration by parts: $$b_n = 2 \left[ -\frac{x \cos(n\pi x)}{n\pi} \Big|_0^1 + \frac{1}{n\pi} \int_0^1 \cos(n\pi x) dx \right]$$ Evaluating the first term at the limits: $$-\frac{x \cos(n\pi x)}{n\pi} \Big|_0^1 = -\frac{1 \cdot \cos(n\pi)}{n\pi} + 0 = -\frac{(-1)^n}{n\pi}$$ Now for the second term: $$\frac{1}{n\pi} \int_0^1 \cos(n\pi x) dx = \frac{1}{n\pi} \left[ \frac{\sin(n\pi x)}{n\pi} \Big|_0^1 \right] = 0$$ Thus, we have: $$b_n = \frac{2(-1)^{n+1}}{n\pi}$$ ### Step 2: Write the Fourier Series Now that we have the coefficients $$b_n = \frac{2(-1)^{n+1}}{n\pi}$$, the Fourier sine series expansion for $$f(x)$$ is: $$f(x) = \sum_{n=1}^{\infty} \frac{2(-1)^{n+1}}{n\pi} \sin(n\pi x)$$ This is the Fourier series for the given piecewise function. --- Would you like a detailed breakdown of each calculation step or need further clarifications? Here are some related questions for exploration: 1. How does the parity of a function affect its Fourier series expansion? 2. What are the conditions for using only the sine terms in the Fourier series? 3. How can you verify that a function is odd or even? 4. How do you apply integration by parts for solving Fourier coefficients? 5. What is the physical significance of Fourier series in signal processing? **Tip:** For odd functions defined on symmetric intervals like $$[-L, L]$$, the Fourier series will only involve sine terms, simplifying the expansion process.Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Fourier Series
Piecewise Function
Odd Functions
Formulas
Fourier sine series: f(x) = Σ b_n * sin(nπx)
Fourier coefficients: b_n = (2/L) * ∫ f(x) sin(nπx/L) dx
Theorems
Fourier Series Theorem
Integration by Parts
Suitable Grade Level
Undergraduate level (Calculus II or III)
Related Recommendation
Fourier Series Expansion of a Piecewise Function f(x) = π + x and π - x
Fourier Series Expansion for Piecewise Functions: Compute up to n=5
Fourier Series of Piecewise Periodic Function with Odd Symmetry
Fourier Series Expansion of a Piecewise Function f(x) = {-π, x - π}
Fourier Sine Series for Piecewise Function f(x) on [0, π]