📚 Fourier Series: Worked Examples | IB数学：傅里叶级数典型例题分析
Fourier series is a standard extension topic in IB Mathematics: Analysis and Approaches HL. It expresses a periodic function as an infinite sum of sine and cosine terms, and its examination questions typically ask for coefficient formulas, symmetry simplifications, and convergence values at discontinuities.
傅里叶级数是IB数学分析与方法HL的常见延伸考点。它把周期函数表示为正弦和余弦项的无穷和，考试题通常要求写出系数公式、利用对称性化简，以及计算间断点处的收敛值。
1. Basic Definitions and Euler Formulas | 基本定义与欧拉公式
For a 2π-periodic function f, the Fourier series is written as a constant term plus an infinite sum of cosine and sine terms.
对于周期为2π的函数f，傅里叶级数写成一个常数项加上余弦项和正弦项的无穷和。
f(x) = a₀/2 + ∑n=1∞ [aₙ cos(nx) + bₙ sin(nx)]
The coefficients are computed with Euler's formulas over one full period.
系数通过欧拉公式在一个完整周期上计算。
aₙ = (1/π) ∫−ππ f(x) cos(nx) dx, bₙ = (1/π) ∫−ππ f(x) sin(nx) dx
Notice that the constant term in the series is a₀/2, not a₀.
注意级数中的常数项是a₀/2，而不是a₀。
2. Example 1: Square Wave | 例1：方波
f(x) = 1 for 0 < x < π and f(x) = −1 for −π < x < 0, periodically repeated with period 2π.
设f为奇方波：当0 < x < π时f(x)=1，当−π < x < 0时f(x)=−1，并周期重复，周期为2π。
Since f is odd, every cosine coefficient is zero: aₙ = 0 for all n.
因为f是奇函数，所有余弦系数为零：对所有n都有aₙ=0。
For the sine coefficients, multiply by sin(nx). The product f(x)sin(nx) is even, so we double the integral over 0 to π.
对于正弦系数，乘以sin(nx)。乘积f(x)sin(nx)是偶函数，因此把0到π的积分加倍即可。
bₙ = (2/π) ∫0π sin(nx) dx = 2(1 − cos(nπ))/(π n)
This gives bₙ = 4/(π n) when n is odd and bₙ = 0 when n is even.
于是当n为奇数时bₙ=4/(π n)，当n为偶数时bₙ=0。
f(x) = (4/π) ∑k=0∞ sin((2k+1)x)/(2k+1)
The square wave shows how a discontinuous function can still be represented by smooth sine waves.
方波说明了不连续函数也能用光滑的正弦波来表示。
3. Odd and Even Symmetry | 奇偶性简化
Symmetry is the most powerful shortcut in Fourier coefficient questions.
对称性是傅里叶系数题中最有力的简化工具。
- If f is even, all sine coefficients vanish: bₙ = 0.
- 若f为偶函数，所有正弦系数为零：bₙ=0。
- If f is odd, all cosine coefficients vanish: aₙ = 0.
- 若f为奇函数，所有余弦系数为零：aₙ=0。
For even f, we also have aₙ = (2/π) ∫0π f(x) cos(nx) dx.
对于偶函数f，还有aₙ = (2/π) ∫0π f(x) cos(nx) dx。
For odd f, we have bₙ = (2/π) ∫0π f(x) sin(nx) dx.
对于奇函数f，有bₙ = (2/π) ∫0π f(x) sin(nx) dx。
Always check symmetry before writing integrals from −π to π.
在写从−π到π的积分之前，一定要先检查奇偶性。
4. Example 2: Sawtooth Wave | 例2：锯齿波
Let f(x) = x on (−π,π), repeated with period 2π. This is an odd function.
设f(x)=x，x在(−π,π)上，周期为2π。这是一个奇函数。
Because f is odd, aₙ = 0. The sine coefficients use integration by parts.
由于f是奇函数，aₙ=0。正弦系数需要分部积分。
bₙ = (2/π) ∫0π x sin(nx) dx = (−1)n+1 2/n
Therefore the Fourier series is:
因此傅里叶级数为：
x = 2 ∑n=1∞ (−1)n+1 sin(nx)/n
periodic extension jumps, and the series converges to the average of π and −π, which is 0.
该等式在开区间(−π,π)内成立。在x=π处，周期延拓发生跳跃，级数收敛于π和−π的平均值0。
5. Half-Range Sine and Cosine Series | 半幅正弦与余弦级数
When f is only given on [0,L], we often construct an odd or even extension to produce a half-range series.
当f只在[0,L]上给出时，我们通常构造奇延拓或偶延拓来得到半幅级数。
For the half-range sine series:
对于半幅正弦级数：
f(x) = ∑n=1∞ bₙ sin(nπx/L), bₙ = (2/L) ∫0L f(x) sin(nπx/L) dx
For the half-range cosine series:
对于半幅余弦级数：
f(x) = a₀/2 + ∑n=1∞ aₙ cos(nπx/L), aₙ = (2/L) ∫0L f(x) cos(nπx/L) dx
At the endpoints, the sine series may not equal f(L); it converges to the average of the left and right limits after odd extension.
在端点处，正弦级数不一定等于f(L)；它收敛于奇延拓后左右极限的平均值。
6. Piecewise Integrals in Coefficient Formulas | 分段函数积分技巧
are piecewise. Consider f(x) = x on 0 < x < π and f(x) = 0 on −π < x < 0, with period 2π.
许多考题给出的函数是分段的。设当0 < x < π时f(x)=x，当−π < x < 0时f(x)=0，周期为2π。
First find a₀:
先求a₀：
a₀ = (1/π) ∫0π x dx = π/2
So the constant term in the series is a₀/2 = π/4.
所以级数中的常数项是a₀/2=π/4。
Then calculate the cosine coefficients by integration by parts:
然后用分部积分求余弦系数：
aₙ = (1/π) ∫0π x cos(nx) dx = ((−1)n − 1)/(π n²)
The sine coefficients are:
正弦系数为：
bₙ = (1/π) ∫0π x sin(nx) dx = (−1)n+1/n
Write the final series by substituting these coefficients into the standard Fourier series formula.
把求得的系数代入标准傅里叶级数公式即可写出最终级数。
7. Convergence at Discontinuities | 间断点处的收敛值
By Dirichlet's theorem, at a point c where f has a finite jump, the Fourier series converges to the average of the left-hand limit and right-hand limit.
根据狄利克雷定理，在有限跳跃点c处，傅里叶级数收敛于左极限和右极限的平均值。
Fourier value at c = ½ [f(c⁺) + f(c⁻)]
e 1, at x = 0 the left limit is −1 and the right limit is 1, so the series converges to 0.
对于例1中的方波，在x=0处左极限为−1、右极限为1，因此级数收敛于0。
Do not substitute f(c) directly when c is a discontinuity.
当c是间断点时，不要直接把f(c)代进去。
8. Gibbs Phenomenon | 吉布斯现象
Near a jump discontinuity, the partial sums of a Fourier series overshoot the function values.
在跳跃间断点附近，傅里叶级数的部分和会超过原函数的值。
This overshoot is about 9% of the size of the jump, and it does not disappear as more terms are added.
这个超调量约为跳跃幅度的9%，而且不会随着项数增加而消失。
In IB questions, this is mainly a conceptual point; you may be asked to explain why the graph of a partial sum has "ripples" near a corner.
在IB考试中，这主要是概念性考点；你可能会被要求解释为什么部分和图像在拐角附近会出现"波纹"。
9. Parseval's Identity and a Famous Sum | 帕塞瓦尔等式与著名求和
Parseval's identity connects the average energy of f to the sum of the squares of the Fourier coefficients.
帕塞瓦尔等式将f的平均能量与傅里叶系数平方和联系起来。
(1/π) ∫−ππ [f(x)]² dx = a₀²/2 + ∑n=1∞ (aₙ² + bₙ²)
Apply this to the square wave with f(x)=±1. Since aₙ=0 and bₙ=4/(πn) for odd n, we get:
将等式用于f(x)=±1的方波。因为aₙ=0，奇数n时bₙ=4/(πn)，所以：
2 = (16/π²) ∑k=0∞ 1/(2k+1)²
Therefore the sum of reciprocal odd squares is π²/8, and from this we obtain the famous result:
因此奇数倒数平方和为π²/8，并由此得到著名结果：
∑n=1∞ 1/n² = π²/6
This is a classic IB extension question that combines Fourier series with series summation.
这是将傅里叶级数与级数求和结合的经典IB拓展题。
10. Common Mistakes and Exam Strategy | 常见错误与应试策略
The table below summarises the most frequent errors in Fourier series exam questions.
下表总结了傅里叶级数考题中最常见的错误。
| Mistake | 常见错误 | Correct Approach | 正确做法 |
| Writing a₀ instead of a₀/2 in the series | The constant term is always a₀/2 |
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