Editing Rectangular function (section)

==Rational approximation==
The pulse function may also be expressed as a limit of a [[rational function]]:

<math display="block">\Pi(t) = \lim_{n\rightarrow \infty, n\in \mathbb(Z)} \frac{1}{(2t)^{2n}+1}.</math>

===Demonstration of validity===
First, we consider the case where <math display=inline>|t|<\frac{1}{2}.</math> Notice that the term <math display=inline>(2t)^{2n}</math> is always positive for integer <math>n.</math> However, <math>2t<1</math> and hence <math display=inline>(2t)^{2n}</math> approaches zero for large <math>n.</math>

It follows that:
<math display="block">\lim_{n\rightarrow \infty, n\in \mathbb(Z)} \frac{1}{(2t)^{2n}+1} = \frac{1}{0+1} = 1, |t|<\tfrac{1}{2}.</math>

Second, we consider the case where <math display="inline">|t|>\frac{1}{2}.</math> Notice that the term <math display="inline">(2t)^{2n}</math> is always positive for integer <math>n.</math> However, <math>2t>1</math> and hence <math display="inline">(2t)^{2n}</math> grows very large for large <math>n.</math>

It follows that:
<math display="block">\lim_{n\rightarrow \infty, n\in \mathbb(Z)} \frac{1}{(2t)^{2n}+1} = \frac{1}{+\infty+1} = 0, |t|>\tfrac{1}{2}.</math>

Third, we consider the case where <math display="inline">|t| = \frac{1}{2}.</math> We may simply substitute in our equation:

<math display="block">\lim_{n\rightarrow \infty, n\in \mathbb(Z)} \frac{1}{(2t)^{2n}+1} = \lim_{n\rightarrow \infty, n\in \mathbb(Z)} \frac{1}{1^{2n}+1} = \frac{1}{1+1} = \tfrac{1}{2}.</math>

We see that it satisfies the definition of the pulse function. Therefore,

<math display="block">\operatorname{rect}(t) = \Pi(t) = \lim_{n\rightarrow \infty, n\in \mathbb(Z)} \frac{1}{(2t)^{2n}+1} = \begin{cases}
0 & \mbox{if } |t| > \frac{1}{2} \\
\frac{1}{2} & \mbox{if } |t| = \frac{1}{2} \\
1 & \mbox{if } |t| < \frac{1}{2}. \\
\end{cases}</math>