Calculating the Fourier Coefficients | 计算傅里叶系数

📚 Calculating the Fourier Coefficients | 计算傅里叶系数

Fourier series allow us to represent a periodic function as an infinite sum of sines and cosines. In IB Mathematics, understanding how to calculate the Fourier coefficients – a₀, aₙ, and bₙ – is essential for both paper-based problems and conceptual understanding of signal decomposition, heat equations, and wave analysis. This article provides a step-by-step guide to deriving and computing these coefficients, explores simplifications for even and odd functions, and works through typical IB-style examples.

傅里叶级数让我们能够将一个周期函数表示为无限个正弦与余弦项的和。在 IB 数学中，掌握如何计算傅里叶系数——a₀、aₙ 和 bₙ——对于考试中的计算题以及理解信号分解、热传导方程和波动分析都至关重要。本文逐步指导你推导并计算这些系数、探讨偶函数与奇函数的简化方法，并通过典型的 IB 例题进行演示。

1. Introduction to Fourier Series | 傅里叶级数简介

A Fourier series expresses a periodic function f(x) with period 2π as a linear combination of cos(nx) and sin(nx) for integer n. The standard form is:

傅里叶级数将周期为 2π 的函数 f(x) 表示为 cos(nx) 与 sin(nx)（n 为整数）的线性组合。标准形式为：

f(x) = a₀/2 + Σⁿ₌₁∞ (aₙ cos(nx) + bₙ sin(nx))

The constant term a₀/2 is used to keep the coefficient formula for a₀ consistent with aₙ. For a function with period 2L, the argument is scaled to nπx/L.

常数项用 a₀/2 是为了使 a₀ 的公式与 aₙ 保持一致。对于周期为 2L 的函数，只需将自变量缩放为 nπx/L。

If the function satisfies Dirichlet conditions (piecewise continuous, finite number of maxima/minima, finite number of discontinuities), the series converges to f(x) at points of continuity and to the average of the left and right limits at jump discontinuities.

如果函数满足狄利克雷条件（分段连续、极大极小值数目有限、间断点数目有限），该级数在连续点处收敛于 f(x)，在跳跃间断点处收敛于左右极限的平均值。

---

2. Orthogonality of Sine and Cosine Functions | 正弦与余弦函数的正交性

The key to isolating each coefficient is the orthogonality of the trigonometric functions over the interval [–π, π]. The relevant integrals are summarised below:

分离各个系数的关键在于三角函数在区间 [–π, π] 上的正交性。相关积分总结如下：

Integral	Value for m ≠ n	Value for m = n
∫–ππ sin(mx) sin(nx) dx	0	π (m,n ≥ 1)
∫–ππ cos(mx) cos(nx) dx	0	π (m,n ≥ 1); 2π if m=n=0
∫–ππ sin(mx) cos(nx) dx	0 (always)	0

These orthogonality relations allow us to 'pick out' a single coefficient when we multiply the Fourier series by a specific sine or cosine and integrate term by term.

这些正交关系使得我们在将傅里叶级数乘以某个特定的正弦或余弦并逐项积分时，可以“挑出”单个系数。

---

3. Derivation of the Fourier Coefficients | 傅里叶系数的推导

Assume f(x) has a Fourier expansion. To find a₀, integrate the series over [–π, π]:

假设 f(x) 可以展开为傅里叶级数。为求 a₀，在 [–π, π] 上对级数积分：

∫_–π^π f(x) dx = ∫_–π^π (a₀/2) dx + Σ ∫ (aₙ cos(nx) + bₙ sin(nx)) dx

in(nx) dx = 0 over a full period, only the constant term survives. This yields a₀ = (1/π)∫_–π^π f(x) dx.

由于 cos(nx) 和 sin(nx) 在一个完整周期上的积分为零，只有常数项保留下来。由此得到 a₀ = (1/π)∫_–π^π f(x) dx。

and integrate. By orthogonality, only the term aₙ cos²(nx) contributes, giving aₙ = (1/π)∫_–π^π f(x) cos(nx) dx. Similarly, multiplying by sin(nx) isolates bₙ.

为求 aₙ (n ≥ 1)，将级数乘以 cos(nx) 并积分。根据正交性，只有 aₙ cos²(nx) 项有贡献，得到 aₙ = (1/π)∫_–π^π f(x) cos(nx) dx。类似地，乘以 sin(nx) 可分离出 bₙ。

---

4. The Formulas for a₀, aₙ, and bₙ | a₀、aₙ 和 bₙ 的公式

For a 2π-periodic function f(x), the Euler–Fourier formulas are:

对于以 2π 为周期的函数 f(x)，欧拉-傅里叶公式为：

• a₀ = (1/π) ∫_–π^π f(x) dx

  a₀ = (1/π) ∫_–π^π f(x) dx

• aₙ = (1/π) ∫_–π^π f(x) cos(nx) dx, n ≥ 1

  aₙ = (1/π) ∫_–π^π f(x) cos(nx) dx, n ≥ 1

• bₙ = (1/π) ∫_–π^π f(x) sin(nx) dx, n ≥ 1

  bₙ = (1/π) ∫_–π^π f(x) sin(nx) dx, n ≥ 1

If the period is 2L, the interval becomes [–L, L] and the coefficients are scaled: aₙ = (1/L)∫_–L^L f(x) cos(nπx/L) dx, bₙ = (1/L)∫_–L^L f(x) sin(nπx/L) dx. The constant term is a₀/2 with a₀ = (1/L)∫_–L^L f(x) dx.

如果周期为 2L，积分区间变为 [–L, L]，系数需相应缩放：aₙ = (1/L)∫_–L^L f(x) cos(nπx/L) dx，bₙ = (1/L)∫_–L^L f(x) sin(nπx/L) dx。常数项为 a₀/2，其中 a₀ = (1/L)∫_–L^L f(x) dx。

Many IB problems focus on period 2π, but you must be comfortable with the general 2L case as well.

许多 IB 考题集中在周期 2π 的情形，但你也必须熟悉一般的 2L 情况。

---

5. Fourier Series for Even and Odd Functions | 偶函数和奇函数的傅里叶级数

Exploiting symmetry drastically reduces the computation. Recall f is even if f(–x) = f(x) and odd if f(–x) = –f(x).

利用对称性可以极大地减少计算量。回忆：若 f(–x) = f(x)，则 f 为偶函数；若 f(–x) = –f(x)，则 f 为奇函数。

• If f(x) is even on [– /2 + Σ aₙ cos(nx). Moreover, aₙ can be computed using half-interval integrals: a₀ = (2/π)∫₀^π f(x) dx, aₙ = (2/π)∫₀^π f(x) cos(nx) dx.

  若 f(x) 在 [–π, π] 上为偶函数，则所有 bₙ = 0。级数化为傅里叶余弦级数：f(x) = a₀/2 + Σ aₙ cos(nx)。此外，可利用半区间积分计算 aₙ：a₀ = (2/π)∫₀^π f(x) dx，aₙ = (2/π)∫₀^π f(x) cos(nx) dx。

• aₙ = 0. The series reduces to a Fourier sine series: f(x) = Σ bₙ sin(nx), with bₙ = (2/π)∫₀^π f(x) sin(nx) dx.

  若 f(x) 为奇函数，则 a₀ = 0 且所有 aₙ = 0。级数简化为傅里叶正弦级数：f(x) = Σ bₙ sin(nx)，其中 bₙ = (2/π)∫₀^π f(x) sin(nx) dx。

Always check for symmetry before computing all three coefficients – it saves time and reduces errors.

在计算全部三个系数之前，务必检查对称性——这能节省时间并减少错误。

---

6. Half-Range Expansions | 半幅展开

When a function is defined only on [0, L], we can extend it to [–L, L] as an even or odd function to obtain a cosine or sine series. This technique is known as half-range expansion.

当函数仅定义在 [0, L] 上时，我们可以将其偶延拓或奇延拓到 [–L, L]，从而得到余弦或正弦级数。这种技巧称为半幅展开。

• Half-range cosine series: extend f as even, then aₙ = (2/L)∫₀^L f(x) cos(nπx/L) dx, bₙ = 0.

  半幅余弦级数：将 f 偶延拓，则 aₙ = (2/L)∫₀^L f(x) cos(nπx/L) dx，bₙ = 0。

• Half-range sine series: extend f as odd, then aₙ = 0, bₙ = (2/L)∫₀^L f(x) sin(nπx/L) dx.

  半幅正弦级数：将 f 奇延拓，则 aₙ = 0，bₙ = (2/L)∫₀^L f(x) sin(nπx/L) dx。

This is particularly useful in IB when a function is given only on a half period and you are asked for a sine or cosine expansion.

这在 IB 中尤其有用，当函数只给出半个周期，并要求你展开为正弦或余弦级数时。

---

广告 / Sponsored

7. Complex Exponential Form of Fourier Series | 傅里叶级数的复指数形式

Using Euler's formula e^inx = cos(nx) + i sin(nx), we can write the Fourier series in a compact exponential form:

利用欧拉公式 e^inx = cos(nx) + i sin(nx)，我们可以将傅里叶级数写成紧凑的指数形式：

f(x) = Σⁿ₌⁺∞_n=–∞ cₙ e^inx

where the complex coefficients are given by cₙ = (1/(2π)) ∫_–π^π f(x) e^–inx . For real f(x), the relationship to the trigonometric coefficients is c₀ = a₀/2, cₙ = (aₙ – i bₙ)/2 for n > 0, and c_–n = (aₙ + i bₙ)/2.

其中复系数由 cₙ = (1/(2π)) ∫_–π^π f(x) e^–inx dx 给出。对于实函数 f(x)，它与三角系数的关系为 c₀ = a₀/2，对 n > 0 有 cₙ = (aₙ – i bₙ)/2，c_–n = (aₙ + i bₙ)/2。

This form is elegant and often appears in more advanced IB HL topics or calculus options. It also simplifies the derivation of coefficients via a single formula.

这种形式非常优雅，常出现在较高级的 IB HL 专题或微积分选修中。它也用单一公式简化了系数的推导。

---

8. Worked Example: Square Wave | 示例：方波的傅里叶系数

Consider the odd, 2π-periodic square wave defined by:

考虑以下奇函数、以 2π 为周期的方波：

f(x) = –1 for –π < x < 0, f(x) = 1 for 0 < x < π, and f(0)=f(π)=0.

Because f is odd, a₀ = 0, aₙ = 0. We only need bₙ = (2/π)∫₀^π (1) sin(nx) dx.

因为 f 是奇函数，a₀ = 0，aₙ = 0。我们只需计算 bₙ = (2/π)∫₀^π (1) sin(nx) dx。

∫₀^π sin(nx) dx = [–cos(nx)/n]₀^π = (1 – cos(nπ))/n. Since cos(nπ) = (–1)ⁿ, we obtain bₙ = (2/(nπ))(1 – (–1)ⁿ).

∫₀^π sin(nx) dx = [–cos(nx)/n]₀^π = (1 – cos(nπ))/n。由于 cos(nπ) = (–1)ⁿ，得到 bₙ = (2/(nπ))(1 – (–1)ⁿ)。

Thus bₙ = 0 for even n, and bₙ = 4/(nπ) for odd n. The Fourier series is f(x) = (4/π) Σ_k=0^∞ sin((2k+1)x) / (2k+1).

因此，当 n 为偶数时 bₙ = 0，当 n 为奇数时 bₙ = 4/(nπ)。傅里叶级数为 f(x) = (4/π) Σ_k=0^∞ sin((2k+1)x) / (2k+1)。

This classic result shows how a square wave contains only odd harmonics.

这个经典结果展示了方波仅包含奇次谐波。

---

9. Worked Example: Sawtooth Wave | 示例：锯齿波

Take the odd, 2π-periodic function f(x) = x for –π < x < π, extended periodically. Since it is odd, a₀=0 and aₙ=0. Compute bₙ = (1/π)∫_–π^π x sin(nx) dx.

取以 2π 为周期的奇函数 f(x) = x，x ∈ (–π, π)，并周期延拓。由于是奇函数，a₀=0，aₙ=0。计算 bₙ = (1/π)∫_–π^π x sin(nx) dx。

sing integration by parts or noting the integrand is even (x sin(nx) is even), bₙ = (2/π)∫₀^π x sin(nx) dx. Let u = x, dv = sin(nx) dx → du = dx, v = –cos(nx)/n.

运用分部积分或注意到被积函数为偶函数（x sin(nx) 为偶函数），bₙ = (2/π)∫₀^π x sin(nx) dx。令 u = x，dv = sin(nx) dx → du = dx，v = –cos(nx)/n。

∫ x sin(nx) dx = –x cos(nx)/n + ∫ cos(nx)/n dx = –x cos(nx)/n + sin(nx)/n². Evaluating from 0 to π yields –π cos(nπ)/n. Since cos(nπ) = (–1)ⁿ, we get bₙ = (2/π) * (–π(–1)ⁿ / n) = –2(–1)ⁿ/n = 2(–1)ⁿ⁺¹/n.

∫ x sin(nx) dx = –x cos(nx)/n + ∫ cos(nx)/n dx = –x cos(nx)/n + sin(nx)/n²。从 0 到 π 求值得 –π cos(nπ)/n。因为 cos(nπ) = (–1)ⁿ，得到 bₙ = (2/π) * (–π(–1)ⁿ / n) = –2(–1)ⁿ/n = 2(–1)ⁿ⁺¹/n。

Thus the Fourier sine series is f(x) = 2 Σ_n=1^∞ ((–1)ⁿ⁺¹/n) sin(nx).

因此傅里叶正弦级数为 f(x) = 2 Σ_n=1^∞ ((–1)ⁿ⁺¹/n) sin(nx)。

This series converges to x for x ∈ (–π, π), but at x = ±π it converges to 0, the average of the jump.

该级数在 (–π, π) 内收敛于 x，但在 x = ±π 处收敛到 0，即跳跃处的平均值。

---

10. Gibbs Phenomenon and Convergence | 吉布斯现象与收敛性

When a Fourier series approximates a function with a jump discontinuity, the partial sums overshoot the function near the jump. This overshoot does not vanish as more terms are added – it approaches about 9% of the jump. This is the Gibbs phenomenon.

当傅里叶级数逼近一个带有跳跃间断的函数时，部分和在跳跃附近会出现超调。随着项数增加，这种超调并不会消失，而是趋近于跳跃幅度的约 9%。这就是吉布斯现象。

IB students are not usually required to quantify the overshoot, but they should recognise that near a jump the partial sum oscillates and the maximum error remains finite. This is important when interpreting Fourier series approximations.

IB 学生通常不需要量化超调，但应认识到在跳跃点附近部分和会产生振荡，且最大误差保持有限。这在解释傅里叶级数近似时很重要。

For smooth, continuous periodic functions, the Fourier series converges uniformly and the coefficients decay rapidly; for functions with discontinuities, the coefficients decay like 1/n.

对于光滑、连续的周期函数，傅里叶级数一致收敛且系数迅速衰减；而对于有间断的函数，系数以 1/n 的速度衰减。

---

11. Applications in IB Mathematics | IB数学中的应用

In the IB syllabus, Fourier coefficients appear in the Calculus option (HL). Typical exam questions ask you to:

在 IB 大纲中，傅里叶系数出现在 HL 微积分选修部分。典型考题要求你：

• Derive the coefficients for a given periodic function.

  推导给定周期函数的系数。

• Determine the Fourier series up to a certain harmonic.

  求出傅里叶级数直某项谐波。

• Use the series to evaluate infinite sums, e.g. Σ 1/(2k+1)² = π²/8.

  利用级数计算无穷和，例如 Σ 1/(2k+1)² = π²/8。

• Identify the convergence value at a specific point.

  确定特定点处的收敛值。

Mastery of integration techniques (integration by parts, trigonometric identities) is crucial, as coefficient calculations often reduce to routine integrals.

熟练掌握积分技巧（分部积分、三角恒等式）至关重要，因为系数计算常常归结为常规的积分。

---

12. Summary and Tips for Exams | 总结与考试提示

To succeed with Fourier coefficients in IB:

要在 IB 中成功掌握傅里叶系数：

• Memorise the standard formulas for both 2π and 2L periods.

  记住 2π 和 2L 周期下的标准公式。

• Always check whether the function is even, odd, or neither – this can halve your work.

  始终检查函数是偶函数、奇函数还是非奇非偶——这可以省去一半的工作量。

• Use orthogonality to verify your integrals; many vanish by symmetry.

  利用正交性检验你的积分；许多积分因对称性而为零。

• When computing bₙ for odd functions, integrate over half the interval and double the result.

  为奇函数计算 bₙ 时，在半区间上积分并将结果加倍。

• Watch out for n = 0 separately in the aₙ formula to avoid division by zero.

  注意在 aₙ 公式中单独处理 n = 0，避免除以零。

• Present your final series clearly, showing the first few terms to demonstrate the pattern.

  清晰地写出最终的级数，展示前几项以体现规律。

With careful practice, evaluating Fourier coefficients becomes a systematic process. Understanding the underlying orthogonality and symmetry not only helps in exams but also builds intuition for advanced topics like Fourier transforms.

经过认真练习，计算傅里叶系数会成为一个系统的流程。理解背后的正交性与对称性不仅有助于考试，还能为傅里叶变换等高级主题建立直觉。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）