Advanced Search

We establish two truncations of Gauss’ square exponent theorem and a finite extension of Euler’s identity. For instance, we prove that for any positive integer $n$, $$\sum _{k=0}^n{(-1)}^k\left[\begin{array}{c}2n-k\\ k\end{array}\right]{(q;q^2)}_{n-k}{q}^{\left(\genfrac{}{}{0pt}{}{k+1}{2}\right)}=\sum _{k=-n}^n{(-1)}^k{q}^{k^2},$$ where $$\left[\begin{array}{c}n\\ m\end{array}\right]=\prod _{k=1}^m\frac{1-q^{n-k+1}}{1-q^k}\text{and}{(a;q)}_n=\prod _{k=0}^{n-1}(1-a{q}^k).$$