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The problem provides the Fourier series expansion of a function $f(x)$ defined on the interval $(-L, L)$ with period $2L$: $f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left(a_n \cos\left(\frac{n\pi x}{L}\right) + b_n \sin\left(\frac{n\pi x}{L}\right)\right)$. The task is to prove the formulas for the Fourier coefficients $a_n$ and $b_n$: (i) $a_n = \frac{1}{L} \int_{-L}^{L} f(x) \cos\left(\frac{n\pi x}{L}\right) dx$ (ii) $b_n = \frac{1}{L} \int_{-L}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) dx$

AnalysisFourier SeriesOrthogonalityIntegration
2025/5/9

1. Problem Description

The problem provides the Fourier series expansion of a function f(x)f(x) defined on the interval (−L,L)(-L, L) with period 2L2L:
f(x)=a02+∑n=1∞(ancos⁡(nπxL)+bnsin⁡(nπxL))f(x) = \frac{a_0}{2} + \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).
The task is to prove the formulas for the Fourier coefficients ana_n and bnb_n:
(i) an=1L∫−LLf(x)cos⁡(nπxL)dxa_n = \frac{1}{L} \int_{-L}^{L} f(x) \cos\left(\frac{n\pi x}{L}\right) dx
(ii) bn=1L∫−LLf(x)sin⁡(nπxL)dxb_n = \frac{1}{L} \int_{-L}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) dx

2. Solution Steps

(i) To find ana_n, we multiply both sides of the Fourier series expansion by cos⁡(mπxL)\cos\left(\frac{m\pi x}{L}\right) and integrate from −L-L to LL, where mm is an integer.
∫−LLf(x)cos⁡(mπxL)dx=∫−LLa02cos⁡(mπxL)dx+∫−LL∑n=1∞(ancos⁡(nπxL)+bnsin⁡(nπxL))cos⁡(mπxL)dx\int_{-L}^{L} f(x) \cos\left(\frac{m\pi x}{L}\right) dx = \int_{-L}^{L} \frac{a_0}{2} \cos\left(\frac{m\pi x}{L}\right) dx + \int_{-L}^{L} \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) \cos\left(\frac{m\pi x}{L}\right) dx
We can interchange the integral and summation because of convergence properties.
∫−LLf(x)cos⁡(mπxL)dx=a02∫−LLcos⁡(mπxL)dx+∑n=1∞(an∫−LLcos⁡(nπxL)cos⁡(mπxL)dx+bn∫−LLsin⁡(nπxL)cos⁡(mπxL)dx)\int_{-L}^{L} f(x) \cos\left(\frac{m\pi x}{L}\right) dx = \frac{a_0}{2} \int_{-L}^{L} \cos\left(\frac{m\pi x}{L}\right) dx + \sum_{n=1}^{\infty} \left(a_n \int_{-L}^{L} \cos\left(\frac{n\pi x}{L}\right) \cos\left(\frac{m\pi x}{L}\right) dx + b_n \int_{-L}^{L} \sin\left(\frac{n\pi x}{L}\right) \cos\left(\frac{m\pi x}{L}\right) dx\right)
We will now use the orthogonality properties of sine and cosine functions:
∫−LLcos⁡(nπxL)cos⁡(mπxL)dx={0,n≠mL,n=m≠02L,n=m=0\int_{-L}^{L} \cos\left(\frac{n\pi x}{L}\right) \cos\left(\frac{m\pi x}{L}\right) dx = \begin{cases} 0, & n \neq m \\ L, & n = m \neq 0 \\ 2L, & n = m = 0 \end{cases}
∫−LLsin⁡(nπxL)cos⁡(mπxL)dx=0\int_{-L}^{L} \sin\left(\frac{n\pi x}{L}\right) \cos\left(\frac{m\pi x}{L}\right) dx = 0 for all nn and mm.
∫−LLsin⁡(nπxL)sin⁡(mπxL)dx={0,n≠mL,n=m≠00,n=m=0\int_{-L}^{L} \sin\left(\frac{n\pi x}{L}\right) \sin\left(\frac{m\pi x}{L}\right) dx = \begin{cases} 0, & n \neq m \\ L, & n = m \neq 0 \\ 0, & n = m = 0 \end{cases}
The first term on the right-hand side is:
∫−LLcos⁡(mπxL)dx=0\int_{-L}^{L} \cos\left(\frac{m\pi x}{L}\right) dx = 0 for m≠0m \neq 0.
If m=0m=0, ∫−LLa02cos⁡(0)dx=a02∫−LL1dx=a02(2L)=a0L\int_{-L}^{L} \frac{a_0}{2} \cos(0) dx = \frac{a_0}{2} \int_{-L}^{L} 1 dx = \frac{a_0}{2} (2L) = a_0L.
Since we are seeking ana_n (where n≠0n \neq 0), we have:
∫−LLf(x)cos⁡(nπxL)dx=an∫−LLcos⁡2(nπxL)dx=anL\int_{-L}^{L} f(x) \cos\left(\frac{n\pi x}{L}\right) dx = a_n \int_{-L}^{L} \cos^2\left(\frac{n\pi x}{L}\right) dx = a_n L
an=1L∫−LLf(x)cos⁡(nπxL)dxa_n = \frac{1}{L} \int_{-L}^{L} f(x) \cos\left(\frac{n\pi x}{L}\right) dx
(ii) To find bnb_n, we multiply both sides of the Fourier series expansion by sin⁡(mπxL)\sin\left(\frac{m\pi x}{L}\right) and integrate from −L-L to LL, where mm is an integer.
∫−LLf(x)sin⁡(mπxL)dx=∫−LLa02sin⁡(mπxL)dx+∫−LL∑n=1∞(ancos⁡(nπxL)+bnsin⁡(nπxL))sin⁡(mπxL)dx\int_{-L}^{L} f(x) \sin\left(\frac{m\pi x}{L}\right) dx = \int_{-L}^{L} \frac{a_0}{2} \sin\left(\frac{m\pi x}{L}\right) dx + \int_{-L}^{L} \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) \sin\left(\frac{m\pi x}{L}\right) dx
∫−LLf(x)sin⁡(mπxL)dx=a02∫−LLsin⁡(mπxL)dx+∑n=1∞(an∫−LLcos⁡(nπxL)sin⁡(mπxL)dx+bn∫−LLsin⁡(nπxL)sin⁡(mπxL)dx)\int_{-L}^{L} f(x) \sin\left(\frac{m\pi x}{L}\right) dx = \frac{a_0}{2} \int_{-L}^{L} \sin\left(\frac{m\pi x}{L}\right) dx + \sum_{n=1}^{\infty} \left(a_n \int_{-L}^{L} \cos\left(\frac{n\pi x}{L}\right) \sin\left(\frac{m\pi x}{L}\right) dx + b_n \int_{-L}^{L} \sin\left(\frac{n\pi x}{L}\right) \sin\left(\frac{m\pi x}{L}\right) dx\right)
∫−LLsin⁡(mπxL)dx=0\int_{-L}^{L} \sin\left(\frac{m\pi x}{L}\right) dx = 0 for all mm.
Thus, ∫−LLa02sin⁡(mπxL)dx=0\int_{-L}^{L} \frac{a_0}{2} \sin\left(\frac{m\pi x}{L}\right) dx = 0
Using orthogonality properties, we get:
∫−LLf(x)sin⁡(nπxL)dx=bn∫−LLsin⁡2(nπxL)dx=bnL\int_{-L}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) dx = b_n \int_{-L}^{L} \sin^2\left(\frac{n\pi x}{L}\right) dx = b_n L
Therefore,
bn=1L∫−LLf(x)sin⁡(nπxL)dxb_n = \frac{1}{L} \int_{-L}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) dx

3. Final Answer

(i) an=1L∫−LLf(x)cos⁡(nπxL)dxa_n = \frac{1}{L} \int_{-L}^{L} f(x) \cos\left(\frac{n\pi x}{L}\right) dx
(ii) bn=1L∫−LLf(x)sin⁡(nπxL)dxb_n = \frac{1}{L} \int_{-L}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) dx

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