Share The Formula You Found

+Label +Description

Recently Referred Formulae

a_{n+1}-\alpha=p(a_n-\alpha)=p^{n+1}(a_0-\alpha)
\alpha=p\alpha+q
a_{n+1}=-a_n+1, a_1=1
a_{n+1}=2a_n
\left(\begin{array}{cc}x_{n+1}\\y_{n+1}\end{array}\right)=\begin{array}{cc}A\end{array}\left(\begin{array}{cc}x_n\\y_n\end{array}\right)=\begin{array}{cc}A^n\end{array}\left(\begin{array}{cc}x_1\\y_1\end{array}\right)