structural:scholar_edition, information:extraction, structural:summarization, quality:correction
Demonstration of validity

First, we consider the case where | t | < 1 2. {\textstyle |t|<{\frac {1}{2}}.} Notice that the term (2 t) 2 n {\textstyle (2t)^{2n}} is always positive for integer n. {\displaystyle n.} However, 2 t < 1 {\displaystyle 2t<1} and hence (2 t) 2 n {\textstyle (2t)^{2n}} approaches zero for large n. {\displaystyle n.}

It follows that: lim n → ∞, n ∈ (Z) 1 (2 t) 2 n + 1 = 1 0 + 1 = 1, | t | < 1 2. {\displaystyle \lim _{n\rightarrow \infty,n\in \mathbb {(} Z)}{\frac {1}{(2t)^{2n}+1}}={\frac {1}{0+1}}=1,|t|<{\tfrac {1}{2}}.}

Second, we consider the case where | t | > 1 2. {\textstyle |t|>{\frac {1}{2}}.} Notice that the term (2 t) 2 n {\textstyle (2t)^{2n}} is always positive for integer n. {\displaystyle n.} However, 2 t > 1 {\displaystyle 2t>1} and hence (2 t) 2 n {\textstyle (2t)^{2n}} grows very large for large n. {\displaystyle n.}

It follows that: lim n → ∞, n ∈ (Z) 1 (2 t) 2 n + 1 = 1 + ∞ + 1 = 0, | t | > 1 2. {\displaystyle \lim _{n\rightarrow \infty,n\in \mathbb {(} Z)}{\frac {1}{(2t)^{2n}+1}}={\frac {1}{+\infty +1}}=0,|t|>{\tfrac {1}{2}}.}

Third, we consider the case where | t | = 1 2. {\textstyle |t|={\frac {1}{2}}.} We may simply substitute in our equation:

lim n → ∞, n ∈ (Z) 1 (2 t) 2 n + 1 = lim n → ∞, n ∈ (Z) 1 1 2 n + 1 = 1 1 + 1 = 1 2. {\displaystyle \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}}.}

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

rect (t) = Π (t) = lim n → ∞, n ∈ (Z) 1 (2 t) 2 n + 1 = { 0 if | t | > 1 2 1 2 if | t | = 1 2 1 if | t | < 1 2. {\displaystyle \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}}}