Math Problem Statement
Hitunglah nilai ao,an,bn
Hitunglah nilai ao, an, bn
Solution
The function given in the image is defined piecewise as:
1 & \text{for } 0 < x < \frac{1}{2}, \\ -1 & \text{for } \frac{1}{2} < x < 1. \end{cases}$$ You are likely asking for the Fourier coefficients $$a_0$$, $$a_n$$, and $$b_n$$. To find these, we need to follow the process for calculating the Fourier series coefficients. ### General Fourier Series Form: For a periodic function $$f(x)$$ defined on $$[0, 1]$$, the Fourier series is given by: $$f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos(2\pi n x) + b_n \sin(2\pi n x) \right).$$ #### 1. Calculation of $$a_0$$: The coefficient $$a_0$$ represents the average value of the function over one period. It is given by: $$a_0 = 2 \int_0^1 f(x) \, dx.$$ Using the piecewise definition of $$f(x)$$, we split the integral: $$a_0 = 2 \left( \int_0^{1/2} 1 \, dx + \int_{1/2}^1 (-1) \, dx \right).$$ #### 2. Calculation of $$a_n$$: The cosine coefficients $$a_n$$ are calculated as: $$a_n = 2 \int_0^1 f(x) \cos(2\pi n x) \, dx.$$ This also needs to be split into two parts due to the piecewise nature of $$f(x)$$: $$a_n = 2 \left( \int_0^{1/2} \cos(2\pi n x) \, dx + \int_{1/2}^1 (-\cos(2\pi n x)) \, dx \right).$$ #### 3. Calculation of $$b_n$$: The sine coefficients $$b_n$$ are given by: $$b_n = 2 \int_0^1 f(x) \sin(2\pi n x) \, dx.$$ Again, we need to split the integral: $$b_n = 2 \left( \int_0^{1/2} \sin(2\pi n x) \, dx + \int_{1/2}^1 (-\sin(2\pi n x)) \, dx \right).$$ Let me calculate these integrals step by step for you.The calculations for the Fourier coefficients yield the following results: 1. **$$a_0$$**: $$a_0 = 0$$ This implies that the function has no constant (average) component. 2. **$$a_n$$**: $$a_n = \begin{cases} \frac{2 \sin(\pi n) - \sin(2 \pi n)}{\pi n}, & n \neq 0 \\ 0, & n = 0 \end{cases}$$ Using the fact that $$\sin(k \pi) = 0$$ for any integer $$k$$, this simplifies further for integer values of $$n$$, so all the $$a_n$$ coefficients will be zero. 3. **$$b_n$$**: $$b_n = \begin{cases} \frac{-2 \cos(\pi n) + \cos(2 \pi n) + 1}{\pi n}, & n \neq 0 \\ 0, & n = 0 \end{cases}$$ Again, simplifying for integer $$n$$, this will yield the non-zero sine coefficients. Thus, the Fourier series only has sine terms, and the cosine terms vanish. Would you like to see more details or have any questions about the process? Here are five related questions for further exploration: 1. How does the function $$f(x)$$ being piecewise affect the Fourier series? 2. Why does the Fourier series have only sine terms in this case? 3. How would the Fourier coefficients change if the limits of the intervals were different? 4. Can we represent any piecewise function using Fourier series? 5. What is the significance of the $$a_0$$ term in Fourier series? **Tip**: When dealing with periodic functions, always check for symmetries (even or odd functions) to simplify the Fourier coefficient calculations.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Piecewise functions
Fourier series
Integration
Formulas
a_0 = 2 ∫_0^1 f(x) dx
a_n = 2 ∫_0^1 f(x) cos(2πnx) dx
b_n = 2 ∫_0^1 f(x) sin(2πnx) dx
Theorems
Fourier series
Suitable Grade Level
College/University level