Math Problem Statement
Find the Fourier series of the periodic function f which period 2π where F(x)={0 -π<x<a 1 a<x<b 0 b<x<π
Solution
To find the Fourier series for the given periodic function , we proceed as follows:
1. Definition of the Function :
The function is periodic with period , and is defined as:
\begin{cases} 0, & -\pi < x < a, \\ 1, & a < x < b, \\ 0, & b < x < \pi.$$ ### 2. **Fourier Series Expansion:** The Fourier series for a function with period $$2\pi$$ is given by: $$f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty \left[ a_n \cos(nx) + b_n \sin(nx) \right],$$ where the coefficients are calculated as: - $$a_0 = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \, dx$$, - $$a_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \cos(nx) \, dx$$, - $$b_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \sin(nx) \, dx$$. ### 3. **Compute $$a_0$$:** $$a_0 = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \, dx.$$ The function $$f(x)$$ is nonzero only in the interval $$a < x < b$$: $$a_0 = \frac{1}{\pi} \int_a^b 1 \, dx = \frac{1}{\pi} \left[ x \right]_a^b = \frac{1}{\pi} (b - a).$$ ### 4. **Compute $$a_n$$:** $$a_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \cos(nx) \, dx.$$ Again, $$f(x)$$ is nonzero only in $$a < x < b$$: $$a_n = \frac{1}{\pi} \int_a^b \cos(nx) \, dx.$$ Using the integral of cosine: $$\int \cos(nx) \, dx = \frac{\sin(nx)}{n},$$ we get: \[ a_n = \frac{1}{\pi} \left[ \frac{\sin(nx)}{n} \right]_a^b = \frac{1}{\pi n} \left[ \sin(nb) - \sin(na) \right]. \] ### 5. **Compute $$b_n$$:** $$b_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \sin(nx) \, dx.$$ Similarly, $$f(x)$$ is nonzero only in $$a < x < b$$: $$b_n = \frac{1}{\pi} \int_a^b \sin(nx) \, dx.$$ Using the integral of sine: $$\int \sin(nx) \, dx = -\frac{\cos(nx)}{n},$$ we get: \[ b_n = \frac{1}{\pi} \left[ -\frac{\cos(nx)}{n} \right]_a^b = \frac{1}{\pi n} \left[ \cos(na) - \cos(nb) \right]. \] ### 6. **Final Fourier Series:** Substituting $$a_0$$, $$a_n$$, and $$b_n$$ into the Fourier series: $$f(x) = \frac{b-a}{2\pi} + \sum_{n=1}^\infty \left[ \frac{\sin(nb) - \sin(na)}{\pi n} \cos(nx) + \frac{\cos(na) - \cos(nb)}{\pi n} \sin(nx) \right].$$ This is the Fourier series representation of the function $$f(x)$$. --- ### Do you want detailed calculations for each step or have any specific questions? --- ### 5 Related Questions: 1. How can the Fourier coefficients be interpreted physically for this function? 2. What is the significance of the $$a_0$$ term in the Fourier series? 3. How would the Fourier series change if $$f(x)$$ were odd or even? 4. Can the convergence of the Fourier series be improved, and under what conditions? 5. How does the choice of $$a$$ and $$b$$ affect the frequency components of $$f(x)$$? ### Tip: To ensure accuracy in Fourier series problems, always check the piecewise definition of the function carefully, as coefficients depend only on the intervals where the function is non-zero.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Fourier Series
Periodic Functions
Trigonometric Integrals
Piecewise Functions
Formulas
Fourier Series: 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 Convergence
Orthogonality of Sine and Cosine Functions
Suitable Grade Level
Grades 11-12