如何证明关于直线x=a与x=b对称的函数是周期函数？

Hey there! Let's break down this proof clearly— it's all about turning symmetry rules into equations and manipulating them to spot the repeating pattern (the period).

First, Recall What Symmetry Means #

When a function $f(x)$ is symmetric about the line $x=a$, it means the value of the function at $a + x$ is the same as at $a - x$ for any $x$. In math terms:
f(a + x) = f(a - x)

Similarly, symmetry about $x=b$ gives us:
f(b + x) = f(b - x)

These two equations are our starting point.

Step-by-Step Derivation of the Period #

Our goal is to find a non-zero constant $T$ such that f(x + T) = f(x) for all $x$. Let's do this with simple substitutions:

1. Rewrite $f(x)$ using the $x=a$ symmetry:
   Let $t = x - a$, so $x = a + t$. Plugging into the symmetry rule:
   f(x) = f(a + t) = f(a - t) = f(a - (x - a)) = f(2a - x)

2. Now apply the $x=b$ symmetry to the result from step 1:
   Treat $2a - x$ as a new variable $y$. Using the $x=b$ rule, f(y) = f(2b - y). Substitute $y = 2a - x$:
   f(2a - x) = f(2b - (2a - x)) = f(x + 2(b - a))

3. Combine the two steps:
   We just showed f(x) = f(x + 2(b - a)).

The constant $T = 2|b - a|$ is the function's period (we take the absolute value to make it positive, since periods are positive by convention). And since $a ≠ b$, $T$ isn't zero— exactly what we need for a periodic function!

Quick Example to Verify #

Take $\sin(x)$: it's symmetric about $x=\frac{\pi}{2}$ and $x=\frac{3\pi}{2}$. Calculating the period using our formula: $2|\frac{3\pi}{2} - \frac{\pi}{2}| = 2\pi$, which is exactly the known period of sine. Perfect, that checks out.

Key Takeaways #

• Start by translating symmetry into concrete function equations— that's the foundation.
• The period is twice the distance between the two symmetry lines— easy to remember for quick checks.

内容的提问来源于stack exchange，提问作者alans