\begin{frame} \frametitle{Continuously Compounded Interest} \vspace{-.5ex} \begin{block}{} If the interest is compounded $n$ times per year, then after $t$ years the value is\vspace{-2.5ex} \begin{talign} A_0 \cdot \left(1+\frac{r}{n}\right)^{nt} \end{talign} \end{block} \pause \begin{exampleblock}{} For instance, $1000\$$ with 6\% interest after $3$ years: \begin{itemize} \pause \item $1000\$ \cdot (1 + 0.06)^3 = 1191.02\$$ annual compounding \pause \item $1000\$ \cdot (1 + 0.03)^6 = 1194.05\$$ semiannual compounding \pause \item $1000\$ \cdot (1 + 0.015)^{12} = 1195.62\$$ quarterly compounding \pause \item $1000\$ \cdot (1 + 0.005)^{36} = 1196.68\$$ monthly compounding \pause \item $1000\$ \cdot (1 + 0.06/356)^{356\cdot 3} = 1197.20\$$ daily compounding \end{itemize} \end{exampleblock} \pause \begin{block}{} If we let $n \to \infty$, we get \emph{continuous compounding}:\vspace{-1ex} \begin{talign} A(t) = \lim_{n\to \infty} A_0 \cdot \left(1+\frac{r}{n}\right)^{nt} \mpause[1]{= A_0 \cdot \left( \lim_{n\to \infty} \left(1+\frac{r}{n}\right)^\frac{n}{r} \right)^{rt} } \mpause[2]{= A_0 \cdot e^{rt} } \end{talign} \end{block} \pause\pause\pause \begin{exampleblock}{} \begin{itemize} \item $1000\$ \cdot e^{0.06\cdot 3} = 1197.22\$$ continuous compounding \end{itemize} \end{exampleblock} \end{frame}