Use of symmetry: sine and cosine series | 利用对称性：正弦与余弦级数

📚 Use of symmetry: sine and cosine series | 利用对称性：正弦与余弦级数

Fourier series allow us to express a periodic function as an infinite sum of sines and cosines. When the function possesses symmetry – in particular, when it is even or odd – many coefficients vanish automatically, saving significant computation. This article explores how to exploit parity to derive pure sine or cosine series, a core technique for IB Mathematics students working with trigonometric expansions.

傅里叶级数能将周期函数表示为无限多个正弦和余弦的和。当函数具有对称性时——尤其是当它是偶函数或奇函数时——许多系数会自然为零，从而大幅减少计算量。本文旨在探讨如何利用奇偶性导出纯粹的正弦或余弦级数，这是IB数学学生在处理三角展开时的核心技术之一。

1. Periodic Functions and Fourier Series | 周期函数与傅里叶级数

x+T)=f(x) for all x. For a 2π-periodic function, the Fourier series representation is f(x) ∼ a₀/2 + Σ_n=1^∞ (aₙ cos(nx) + bₙ sin(nx)). The symbol ∼ indicates that the series converges to f(x) under certain conditions.

如果一个函数满足 f(x+T)=f(x) 对所有 x 成立，则它是周期为 T 的周期函数。对于一个周期为 2π 的函数，其傅里叶级数表示为 f(x) ∼ a₀/2 + Σ_n=1^∞ (aₙ cos(nx) + bₙ sin(nx))。符号 ∼ 表示该级数在特定条件下收敛到 f(x)。

2. Euler's Formulas for Coefficients | 欧拉系数公式

d via integrals over one full period, typically [−π, π]. The constant term is a₀ = (1/π) ∫_−π^π f(x) dx. The cosine coefficients are aₙ = (1/π) ∫_−π^π f(x) cos(nx) dx for n ≥ 1. The sine coefficients are bₙ = (1/π) ∫_−π^π f(x) sin(nx) dx for n ≥ 1.

傅里叶系数通过整个周期上的积分计算，通常取区间 [−π, π]。常数项为 a₀ = (1/π) ∫_−π^π f(x) dx。余弦系数为 aₙ = (1/π) ∫_−π^π f(x) cos(nx) dx (n ≥ 1)。正弦系数为 bₙ = (1/π) ∫_−π^π f(x) sin(nx) dx (n ≥ 1)。

3. Even and Odd Functions | 奇函数与偶函数

nctions have graphs symmetric about the y-axis; odd functions are symmetric about the origin. Classical examples include cos(nx) (even) and sin(nx) (odd). The product of two even functions is even; product of two odd functions is even; and the product of an even and an odd function is odd.

如果对于所有 x 有 f(−x) = f(x)，则 f 为偶函数；若有 f(−x) = −f(x)，则为奇函数。偶函数图像关于 y 轴对称，奇函数图像关于原点对称。经典例子包括 cos(nx) 为偶函数，sin(nx) 为奇函数。偶×偶=偶，奇×奇=偶，偶×奇=奇。

4. Symmetry in Integration | 积分中的对称性

When integrating an odd function over a symmetric interval [−L, L], the result is zero: ∫_−L^L f_odd = 0. For an even function, the integral over [−L, L] is twice the integral over [0, L]: ∫_−L^L f_even(x) dx = 2∫₀^L f_even(x) dx. These properties directly eliminate many Fourier coefficients when f(x) itself has parity.

当在对称区间 [−L, L] 上积分奇函数时，结果为零：∫_−L^L f_奇(x) dx = 0。对于偶函数，在 [−L, L] 上的积分等于在 [0, L] 上积分的两倍：∫_−L^L f_偶(x) dx = 2∫₀^L f_偶(x) dx。当 f(x) 本身具有奇偶性时，这些性质能直接消除许多傅里叶系数。

5. Odd Functions and Sine Series | 奇函数与正弦级数

egrands f(x) cos(nx) are odd (odd × even = odd). The series reduces to a sine series: f(x) ∼ Σ_n=1^∞ bₙ sin(nx), where bₙ = (2/π) ∫₀^π f(x) sin(nx) dx. The symmetry halves the integration interval and eliminates half the coefficients.

如果 f(x) 在 [−π, π] 上是奇函数，则 a₀ = 0 且所有 aₙ = 0，因为被积函数 f(x) cos(nx) 是奇函数（奇×偶=奇）。级数退化为正弦级数：f(x) ∼ Σ_n=1^∞ bₙ sin(nx)，其中 bₙ = (2/π) ∫₀^π f(x) sin(nx) dx。对称性将积分区间减半并消去了一半系数。

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6. Even Functions and Cosine Series | 偶函数与余弦级数

use f(x) sin(nx) is odd (even × odd = odd). The series simplifies to a cosine series: f(x) ∼ a₀/2 + Σ_n=1^∞ aₙ cos(nx), with a₀ = (2/π) ∫₀^π f(x) dx and aₙ = (2/π) ∫₀^π f(x) cos(nx) dx. Again, the work is halved.

如果 f(x) 在 [−π, π] 上是偶函数，则所有 bₙ = 0，因为 f(x) sin(nx) 是奇函数（偶×奇=奇）。级数简化为余弦级数：f(x) ∼ a₀/2 + Σ_n=1^∞ aₙ cos(nx)，其中 a₀ = (2/π) ∫₀^π f(x) dx 且 aₙ = (2/π) ∫₀^π f(x) cos(nx) dx。同样，工作量减半。

7. Half-Range Expansions | 半幅展开

or evenly. An odd extension yields a sine series that converges to f(x) on (0, π) and to 0 at the endpoints. An even extension yields a cosine series matching f(x) on (0, π) and preserving continuity at the ends if f'(0)=f'(π)=0. The coefficients use the same half-range integrals as in sections 5 and 6.

通常函数只定义在 [0, π] 上，而我们希望将其展开为正弦或余弦级数。我们可以通过奇延拓或偶延拓将函数扩展到 [−π, π]。奇延拓得到正弦级数，在 (0, π) 上收敛到 f(x)，在端点处收敛到 0。偶延拓得到余弦级数，在 (0, π) 上与 f(x) 一致，并在端点处可能保持连续。系数计算与第5、第6节中的半区间积分相同。

8. Worked Example: Sawtooth Wave | 实例：锯齿波

= x on (−π, π). This is an odd function, so only sine terms survive. Compute bₙ = (2/π) ∫₀^π x sin(nx) dx. Integration by parts gives (2/π) [−x cos(nx)/n + sin(nx)/n²]₀^π = (2/π) [−π cos(nπ)/n] = 2(−1)ⁿ⁺¹/n. Hence the Fourier sine series is:

f(x) ∼ 2 Σ_n=1^∞ (−1)ⁿ⁺¹ sin(nx)/n

考虑 f(x) = x 在 (−π, π) 上。这是奇函数，因此只保留正弦项。计算 bₙ = (2/π) ∫₀^π x sin(nx) dx。分部积分得 (2/π) [−x cos(nx)/n + sin(nx)/n²]₀^π = (2/π) [−π cos(nπ)/n] = 2(−1)ⁿ⁺¹/n。因此傅里叶正弦级数为：

f(x) ∼ 2 Σ_n=1^∞ (−1)ⁿ⁺¹ sin(nx)/n

wtooth jumps, and the series converges to the average value 0, matching the odd extension.

在 x = ±π 处锯齿波发生跳跃，级数收敛到平均值 0，与奇延拓一致。

9. Worked Example: Absolute Value Function | 实例：绝对值函数

Take f(x) = |x| on [−π, π], an even function. All bₙ = 0. a₀ = (2/π) ∫₀^π x dx = π. For n ≥ 1, aₙ = (2/π) ∫₀^π x cos(nx) dx = (2/π) [ (x sin(nx))/n + cos(nx)/n² ]₀^π = (2/π) [cos(nπ)−1]/n² = 2[(−1)ⁿ−1]/(π n²). This vanishes for even n. For odd n = 2k+1, (−1)ⁿ = −1, so a_2k+1 = −4/(π (2k+1)²). The cosine series is:

|x| ∼ π/2 − (4/π) Σ_k=0^∞ cos((2k+1)x)/(2k+1)²

取 f(x) = |x| 在 [−π, π] 上，为偶函数。所有 bₙ = 0。a₀ = (2/π) ∫₀^π x dx = π。对于 n ≥ 1，aₙ = (2/π) ∫₀^π x cos(nx) dx = (2/π) [ (x sin(nx))/n + cos(nx)/

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