三本书中任意周期傅里叶系数的不同形式为何等价？

Hey there, let's unpack why these two sets of Fourier series coefficient formulas look different but are actually equivalent—it all comes down to a simple difference in how each book defines the period of the function and tweaks the notation to match. Let's break this down step by step:

1. First, Clarify the Period Definition in Each Book #

• Book 1: Here, the function (f(t)) has a period of (2L). That's why the integral runs from (-L) to (L)—that's one full period (length (2L)). The fundamental angular frequency here is (\omega_0 = \frac{2\pi}{\text{period}} = \frac{2\pi}{2L} = \frac{\pi}{L}), which is why the cosine/sine terms use (\frac{\pi n}{L}t).
  The coefficients (a_n) and (b_n) use a normalization factor of (\frac{1}{L}), which aligns with the standard Fourier series formula for a function with period (2L).

• Book 2: In this case, the function (f(t)) has a period of (L) directly. The integral runs from (-L/2) to (L/2)—that's one full period of length (L). The fundamental angular frequency here is (\omega_0' = \frac{2\pi}{\text{period}} = \frac{2\pi}{L}), so the cosine/sine terms use (\frac{2\pi n}{L}t).
  The normalization factor here is (\frac{2}{L}), which matches the standard formula for a function with period (L).

2. Proving They're Equivalent #

Let's map the notation between the two books to show they produce the same result:

• Let’s say in Book 1, we rename the period (2L) to (T) (so (L = T/2)). Substitute this into Book 1's formula for (a_n):
  [
  a_n = \frac{1}{T/2}\int_{-T/2}^{T/2}f(t)\cos\left(\frac{\pi n}{T/2}t\right)dt = \frac{2}{T}\int_{-T/2}^{T/2}f(t)\cos\left(\frac{2\pi n}{T}t\right)dt
  ]
• Now look at Book 2's formula: if we let Book 2's (L) equal (T) (since Book 2's (L) is the full period), Book 2's (a_n) becomes:
  [
  a_n = \frac{2}{T}\int_{-T/2}^{T/2}f(t)\cos\left(\frac{2\pi n}{T}t\right)dt
  ]
  They're identical! The only difference was what each book called "L"—Book 1 used (L) for half the period, Book 2 used (L) for the full period.

3. A Quick Example to Make It Concrete #

Suppose (f(t)) has a period of 4:

• Book 1 would set (L=2) (half the period), so:
  [
  a_n = \frac{1}{2}\int_{-2}^{2}f(t)\cos\left(\frac{\pi n}{2}t\right)dt
  ]
• Book 2 would set (L=4) (full period), so:
  [
  a_n = \frac{2}{4}\int_{-2}^{2}f(t)\cos\left(\frac{2\pi n}{4}t\right)dt = \frac{1}{2}\int_{-2}^{2}f(t)\cos\left(\frac{\pi n}{2}t\right)dt
  ]
  Same exact calculation, same result.

At the end of the day, it's just a notation mix-up—no mathematical inconsistency here!

内容的提问来源于stack exchange，提问作者JDoeDoe