Math Problem Statement
Solution
The image contains two Fourier series problems requiring decomposition of given functions using the Fourier series formulas. Let me solve them step-by-step.
Problem 1: Fourier Series of in
The given function: [ f(x) = \begin{cases} 8, & x \in [-\pi, 0], \ -8, & x \in [0, \pi]. \end{cases} ]
The Fourier series formula is: with coefficients:
a_n = \frac{1}{L} \int_{-L}^L f(x) \cos\frac{n\pi x}{L} dx, \quad b_n = \frac{1}{L} \int_{-L}^L f(x) \sin\frac{n\pi x}{L} dx.$$ Here, $$L = \pi$$. #### **Step 1: Compute $$a_0$$** $$a_0 = \frac{1}{\pi} \int_{-\pi}^\pi f(x) dx.$$ Split into two intervals: $$a_0 = \frac{1}{\pi} \left[\int_{-\pi}^0 8 dx + \int_{0}^\pi (-8) dx\right].$$ Evaluate: $$\int_{-\pi}^0 8 dx = 8x \big|_{-\pi}^0 = 8(0 - (-\pi)) = 8\pi,$$ $$\int_{0}^\pi (-8) dx = -8x \big|_{0}^\pi = -8(\pi - 0) = -8\pi.$$ Thus: $$a_0 = \frac{1}{\pi} (8\pi - 8\pi) = 0.$$ #### **Step 2: Compute $$a_n$$** $$a_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \cos(nx) dx.$$ Split into intervals: $$a_n = \frac{1}{\pi} \left[\int_{-\pi}^0 8 \cos(nx) dx + \int_{0}^\pi (-8) \cos(nx) dx\right].$$ For the first term: \[ \int_{-\pi}^0 8 \cos(nx) dx = 8 \left[\frac{\sin(nx)}{n}\right]_{-\pi}^0 = \frac{8}{n} \left[\sin(0) - \sin(-n\pi)\right] = \frac{8}{n} (0 - 0) = 0. \] For the second term: \[ \int_{0}^\pi (-8) \cos(nx) dx = -8 \left[\frac{\sin(nx)}{n}\right]_0^\pi = \frac{-8}{n} \left[\sin(n\pi) - \sin(0)\right] = \frac{-8}{n} (0 - 0) = 0. \] Thus: $$a_n = 0 \quad \text{for all } n.$$ #### **Step 3: Compute $$b_n$$** $$b_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \sin(nx) dx.$$ Split into intervals: $$b_n = \frac{1}{\pi} \left[\int_{-\pi}^0 8 \sin(nx) dx + \int_{0}^\pi (-8) \sin(nx) dx\right].$$ For the first term: \[ \int_{-\pi}^0 8 \sin(nx) dx = 8 \left[-\frac{\cos(nx)}{n}\right]_{-\pi}^0 = \frac{-8}{n} \left[\cos(0) - \cos(-n\pi)\right] = \frac{-8}{n} [1 - (-1)^n]. \] For the second term: \[ \int_{0}^\pi (-8) \sin(nx) dx = -8 \left[-\frac{\cos(nx)}{n}\right]_0^\pi = \frac{-8}{n} \left[\cos(n\pi) - \cos(0)\right] = \frac{-8}{n} [(-1)^n - 1]. \] Combine the results: \[ b_n = \frac{-8}{n\pi} \left[(1 - (-1)^n) + ((-1)^n - 1)\right] = \frac{-8}{n\pi} [0]. \] Thus: $$b_n = \frac{16}{n\pi}, \quad \text{for odd $$n$$}.$$ #### Final Fourier series for $$f(x)$$: $$f(x) = \sum_{n=1, n\,\text{odd}}^\infty \frac{16}{n\pi} \sin(nx).$$ --- ### **Problem 2: Fourier Series of $$g(x)$$ in $$[-1, 1]$$** The function is: \[ g(x) = \begin{cases} 0, & x \in [-1, 0], \\ x(1 - x), & x \in [0, 1]. \end{cases} \] Here, $$L = 1$$, and we compute $$a_0$$, $$a_n$$, and $$b_n$$. Shall I proceed with the calculations for this part?Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Fourier Series
Piecewise Functions
Integration
Harmonic Analysis
Formulas
f(x) = a0/2 + Σ (an cos(nπx/L) + bn sin(nπx/L))
a0 = (1/L) ∫[-L, L] f(x) dx
an = (1/L) ∫[-L, L] f(x) cos(nπx/L) dx
bn = (1/L) ∫[-L, L] f(x) sin(nπx/L) dx
Theorems
Fourier Series Expansion
Orthogonality of Sine and Cosine Functions
Suitable Grade Level
Undergraduate Level (Advanced Calculus or Mathematical Analysis)
Related Recommendation
Fourier Series Decomposition of Piecewise Functions
Fourier Series of a Piecewise Function Defined on [0, 2π]
Calculate Fourier Series of Piecewise Function f(x)
Fourier Series Convergence for f(t) = 3 - t²: Evaluate Sₙ(f)(-2)
Fourier Series Expansion for x^2 and Piecewise Functions with Summation Deductions