A periodic function repeats its values at regular intervals. Periodicity appears throughout mathematics and the sciences: in the oscillation of springs, the motion of tides, the rotation of planets, and alternating electrical current.
Quick Reference
| Concept | Meaning |
|---|---|
| Periodic with period $T$ | $f(x + T) = f(x)$ for all $x$ in the domain |
| Period not unique | If $T$ is a period, so are $2T$, $3T$, $-T$, etc. |
| Fundamental period | The smallest positive period of $f$ (if it exists) |
| Constant functions | Periodic with every $T$; no fundamental period |
Definition
A function $f$ is periodic with period $T$ if:
- Whenever $x$ lies in the domain of $f$, so does $x + T$, and
- $f(x + T) = f(x)$ for every $x$ in the domain of $f$.
The first condition ensures the domain is itself "periodic," allowing the shift to make sense. A function need not be defined for all real numbers to be periodic.
Multiple Periods and the Fundamental Period
The period of a periodic function is not unique. If $T$ is a period of $f$, then $2T$, $3T$, $-T$, etc., are also periods:
$f(x + 2T) = f((x+T) + T) = f(x+T) = f(x),$$f(x - T) = f(x - T + T) = f(x).$In general:
If $f$ is periodic with period $T$, then $f(x + nT) = f(x)$ for every integer $n$.
The smallest positive period of a periodic function (if it exists) is called the fundamental period of the function.
A constant function $f(x) = c$ is periodic with every positive $T$ as a period, since $f(x + T) = c = f(x)$. Because there is no smallest positive number, a constant function has no fundamental period.
Examples
Find the fundamental period of $f(x) = 2x - \lfloor 2x \rfloor$.
Solution
Suppose $T$ is a period of $f$. Then $f(x + T) = f(x)$ for all $x$: $2(x + T) - \lfloor 2(x+T) \rfloor = 2x - \lfloor 2x \rfloor.$ Expanding: $2x + 2T - \lfloor 2(x+T) \rfloor = 2x - \lfloor 2x \rfloor.$ Canceling $2x$ from both sides: $2T = \lfloor 2(x+T) \rfloor - \lfloor 2x \rfloor.$ The right side is always an integer (the difference of two integers), so $2T$ must be an integer. To find the fundamental period, we choose $2T$ to be the smallest positive integer, which is $1$: $2T = 1 \implies T = \frac{1}{2}.$ The fundamental period of $f$ is $\dfrac{1}{2}$.
Show that $f(x) = x^2 - \lfloor x^2 \rfloor$ is not periodic.
Solution
Assume for contradiction that $f$ has period $T$. Then $f(x + T) = f(x)$ for all $x$, which means: $(x + T)^2 - \lfloor (x+T)^2 \rfloor = x^2 - \lfloor x^2 \rfloor.$ Expanding $(x+T)^2 = x^2 + 2xT + T^2$: $x^2 + 2xT + T^2 - \lfloor (x+T)^2 \rfloor = x^2 - \lfloor x^2 \rfloor.$ Canceling $x^2$: $2xT + T^2 = \lfloor (x+T)^2 \rfloor - \lfloor x^2 \rfloor.$ The right side is always an integer (difference of two floor values). So the left side $2xT + T^2$ must be an integer for every real number $x$. But $2xT + T^2$ is a linear function of $x$ with slope $2T$. If $T \neq 0$, this linear function takes all real values as $x$ varies, not just integer values. This is a contradiction. Therefore no such $T$ exists, and $f$ is not periodic.
