Math Problem Statement
If π is the Fourier series of π(π₯)={3,16βπ₯2,β4<π₯<00β€π₯<4 , then π(π₯)=
Solution
To determine the Fourier series for the given piecewise function , let's break down the process step by step.
Given:
\begin{cases} 3, & -4 < x < 0 \\ 16 - x^2, & 0 \leq x < 4 \\ 0, & \text{periodically extended outside these intervals.} \end{cases}$$ The Fourier series of $$g(x)$$ is expressed as: $$f(x) = a_0 + \sum_{n=1}^\infty \left( a_n \cos\left(\frac{n\pi x}{L}\right) + b_n \sin\left(\frac{n\pi x}{L}\right) \right),$$ where $$L$$ is half the period, $$a_0$$ is the average value (DC component), $$a_n$$ are the coefficients for the cosine terms, and $$b_n$$ are the coefficients for the sine terms. Assume the function is periodic with period $$T = 8$$, so $$L = 4$$. --- ### Step 1: Calculate $$a_0$$ The average value $$a_0$$ is: $$a_0 = \frac{1}{2L} \int_{-L}^{L} g(x) \, dx = \frac{1}{8} \int_{-4}^{4} g(x) \, dx.$$ Split the integral according to $$g(x)$$: $$a_0 = \frac{1}{8} \left( \int_{-4}^{0} 3 \, dx + \int_{0}^{4} (16 - x^2) \, dx \right).$$ - For $$\int_{-4}^{0} 3 \, dx$$: $$\int_{-4}^{0} 3 \, dx = 3[x]_{-4}^{0} = 3(0 - (-4)) = 12.$$ - For $$\int_{0}^{4} (16 - x^2) \, dx$$: $$\int_{0}^{4} (16 - x^2) \, dx = \int_{0}^{4} 16 \, dx - \int_{0}^{4} x^2 \, dx.$$ $$\int_{0}^{4} 16 \, dx = 16[x]_{0}^{4} = 16(4 - 0) = 64,$$ $$\int_{0}^{4} x^2 \, dx = \frac{x^3}{3} \bigg|_{0}^{4} = \frac{4^3}{3} - \frac{0^3}{3} = \frac{64}{3}.$$ $$\int_{0}^{4} (16 - x^2) \, dx = 64 - \frac{64}{3} = \frac{192}{3} - \frac{64}{3} = \frac{128}{3}.$$ Thus: $$a_0 = \frac{1}{8} \left( 12 + \frac{128}{3} \right) = \frac{1}{8} \cdot \frac{36 + 128}{3} = \frac{164}{24} = \frac{41}{6}.$$ --- ### Step 2: Calculate $$a_n$$ The cosine coefficients $$a_n$$ are given by: $$a_n = \frac{1}{L} \int_{-L}^{L} g(x) \cos\left(\frac{n\pi x}{L}\right) \, dx.$$ Split the integral as before: $$a_n = \frac{1}{4} \left( \int_{-4}^{0} 3 \cos\left(\frac{n\pi x}{4}\right) \, dx + \int_{0}^{4} (16 - x^2) \cos\left(\frac{n\pi x}{4}\right) \, dx \right).$$ You can compute these terms for specific values of $$n$$, but they will involve integration by parts for $$\int x^2 \cos\left(\frac{n\pi x}{4}\right) dx$$. --- ### Step 3: Calculate $$b_n$$ The sine coefficients $$b_n$$ are given by: $$b_n = \frac{1}{L} \int_{-L}^{L} g(x) \sin\left(\frac{n\pi x}{L}\right) \, dx.$$ Similarly: $$b_n = \frac{1}{4} \left( \int_{-4}^{0} 3 \sin\left(\frac{n\pi x}{4}\right) \, dx + \int_{0}^{4} (16 - x^2) \sin\left(\frac{n\pi x}{4}\right) \, dx \right).$$ Due to symmetry: - The sine terms $$b_n$$ may cancel depending on the odd/even nature of $$g(x)$$. --- ### Final Expression Combining $$a_0$$, $$a_n$$, and $$b_n$$, you get the full Fourier series $$f(x)$$. Would you like me to compute $$a_n$$ and $$b_n$$ for specific $$n$$ values, or summarize the process further?Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Fourier Series
Piecewise Functions
Integration
Formulas
Fourier series: f(x) = a0 + Ξ£ (an cos(nΟx/L) + bn sin(nΟx/L))
Average coefficient: a0 = (1/2L) β« from -L to L g(x) dx
Cosine coefficient: an = (1/L) β« from -L to L g(x) cos(nΟx/L) dx
Sine coefficient: bn = (1/L) β« from -L to L g(x) sin(nΟx/L) dx
Theorems
Fourier Series Expansion
Orthogonality of Trigonometric Functions
Suitable Grade Level
Undergraduate (Calculus and Differential Equations)
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