Interesting Graphs: Wave

Interestingly, there are mathematical equations that describe waves.

We all know that sine and cosine functions are wave-like functions. One fact you may not know is that a rational function can look like a single pulse wave! [Edited on 26.1: Just discovered another function that resembles a wave: $e^{-x^2}$!]

In general, $y=\dfrac{a}{x^2+b}$ looks like a pulse centered at $x=0$, where $a,b$ are positive real numbers $\neq 0$.

Question: Why does it have to be "+b"?

Having learnt transformation of graphs, we know that $f(x-a)$ is formed by translating $f(x)$ to the right a units; $f(x+a)$ to the left a units, where $a>0$.

For waves,

$y(x, t) = f(x-vt)$ $\Rightarrow$ wave travelling to the right with speed v,

$y(x, t) = f(x+vt)$ $\Rightarrow$ wave travelling to the left with speed v.

Example:

$f(x) = \dfrac{5}{x^2+1}$
"Purple wave": $f(x-vt)$
"Orange wave": $-f(x+vt)$

The "purple wave" is moving to the right, where the "orange wave" is moving to the left. In this case, $v=5$, $t=1$ or $v=1$, $t=5$.

Demonstration of Constructive Interference

For simplicity, choose $v=1$.

At $t=1$,

"Green wave": moving to the right
$y=\dfrac{5}{(x-1+10)^2+1}$
There has to be $\pm$constant, in this case, +10, because otherwise we won't have constructive interference.

"Blue wave": moving to the left

At $t=5$, we have constructive interference: superposition of two waves (represented by the green curve).

$\dfrac{10}{(x+5)^2+1} = \dfrac{5}{(x-5+10)^2+1} + \dfrac{5}{(x+5)^2+1}$

At $t=10$,

"Green wave": moving to the right

"Blue wave": moving to the left

Demonstration of Destructive Interference

Destructive interference: The two waves cancel out each other.

For your information:

Cauchy distribution

The simplest Cauchy distribution is called the standard Cauchy distribution. It is the distribution of a random variable that is the ratio of two independent standard normal variables and has the probability density function $$f(x;0,1)=\dfrac{1}{\pi(1+x^2)}.$$

Its cumulative distribution function has the shape of $\arctan x$:

$$F(x;0,1)=\frac{1}{\pi}\arctan x+\frac{1}{2}.$$