# Spatial and temporal periods and periodic functions

1. Apr 19, 2014

### Jhenrique

A periodic function is one that $f(\theta) = f(\theta + nT)$, by definition. However, the argument $\theta$ can be function of space and time ( $\theta(x, t)$ ), so exist 2 lines of development, one spatial and other temporal: $$f(\theta) = f(kx + \varphi) = f(2 \pi \xi x + \varphi) = f\left(\frac{2 \pi x}{\lambda} + \varphi \right)$$ $$f(\theta) = f(\omega t + \varphi) = f(2 \pi \nu t + \varphi) = f\left(\frac{2 \pi t}{T} + \varphi \right)$$ or the both together: $$f(\theta) = f(kx + \omega t + \varphi) = f(2 \pi \xi x + 2 \pi \nu t + \varphi) = f\left(\frac{2 \pi x}{\lambda} + \frac{2 \pi t}{T} + \varphi \right)$$ so, becomes obvius that $\lambda$ is the analogus of $T$, thus the correct wound't be say that a periodic function is one that $f(\theta) = f(\theta + nT + m\lambda)$ ?

2. Apr 19, 2014

### gopher_p

If you want to generalize the notion of a periodic function of a single variable to a multivariable situation, the most natural way to do that is to say that $f:\mathbb{R}^n\rightarrow \mathbb{R}^m$ is periodic with period $\vec{T}\in\mathbb{R}^n$ iff $f(\vec{x})=f(\vec{x}+n\vec{T})$ for all $\vec{x}\in\mathbb{R}^n$ and $n\in\mathbb{Z}$; i.e. the period is a vector/$n$-tuple rather than a scalar. Unlike the single variable case, the mutivariate case can have multiple linearly independent periods, although we wouldn't necessitate that to be true. There are some fun theorems that can be proven regarding the number of $\mathbb{Q}$-linearly independent periods that a continuous function can have.

3. Apr 20, 2014

### Jhenrique

happens that when I ploted in the geogebra the wave $f(\theta(x,t)) = \cos(kx - wt)$ happend that $f(\theta) = f(\theta + m \lambda)$ but $f(\theta) \neq f(\theta + n T)$ Why???

4. Apr 20, 2014

### gopher_p

That's weird. When I plot $f(\Phi(\sigma,\tau)) = \cos(\alpha\sigma - \beta\tau)$ in GnuPlot, I get both $f(\Phi)=f(\Phi+m\nu)$ and $f(\Phi)=f(\Phi+n\mu)$.