Math Problem Statement
solve this
Solution
To find the Fourier series expansion of the periodic function on the interval , we proceed by calculating the Fourier coefficients.
Fourier Series:
The Fourier series of a function with period is given by:
Where the coefficients , , and are defined as:
Step 1: Find
We first calculate :
This is a straightforward integral:
Thus, .
Step 2: Find
Next, we calculate :
Since is an odd function (because is odd and is even), the integral of an odd function over a symmetric interval (like to ) is zero:
Step 3: Find
Now, we calculate :
This is an integral of an odd function (both and are odd, so their product is even), and we need to compute it:
Evaluating the boundary terms , we find that these terms vanish because alternates between , and the integrals of over to result in:
Final Fourier Series:
Thus, the Fourier series expansion of on is given by:
This is the Fourier series for the function , which consists only of sine terms (since it is an odd function).
Would you like more details on any of these steps or any related concept? Here are five questions you could explore further:
- How do we handle Fourier series for even and odd functions?
- What are the implications of the Fourier coefficients being zero for certain terms?
- How would the Fourier series change if the function were not odd?
- Can you derive a Fourier series for a piecewise function?
- How does the convergence of the Fourier series behave for discontinuous functions?
Tip: When calculating Fourier coefficients, always check the symmetry of the function to simplify the integrals (odd/even properties can eliminate terms).
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Math Problem Analysis
Mathematical Concepts
Fourier Series
Periodic Functions
Odd and Even Functions
Harmonic Analysis
Formulas
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 Theorem
Orthogonality of Sine and Cosine Functions
Suitable Grade Level
Undergraduate Mathematics
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