0 votes 0 votes For the function $f(t) = e^{-t}/\tau$, the Taylor series approximation for $t\ll$$\tau$ is $1+\frac{t}{\tau}$ $1-\frac{t}{\tau}$ $1-\frac{t^2}{2\tau^2}$ $1+t$ Calculus gate2012 calculus taylor-series + – Milicevic3306 asked Mar 25, 2018 • edited Mar 15, 2021 by soujanyareddy13 Milicevic3306 7.9k points answer See all 0 reply