Exponential decay approximation

Question

In many texts I see an approximation which I don't know where comes from.

How is $\displaystyle e^{-\frac{t}{a}}$ approximated to $(1 - \frac{t}{a})$ when $t\ll a$ ?

Answer 1

We know that the Maclaurin series of $e^x$ is $$e^x = 1+x +\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots\cdots$$ Hence if we let $x = \displaystyle\frac{-t}{a}$, we get $$\begin{align}e^{-\frac ta}&=1+\left(-\frac ta\right)+\frac{\left(-\frac{t}{a}\right)^2}{2!}+\frac{\left(-\frac{t}{a}\right)^3}{3!}+\frac{\left(-\frac{t}{a}\right)^4}{4!}+\cdots\cdots\\ &=1-\frac ta+\frac{t^2}{2a^2}-\frac{t^3}{6a^3}+\frac{t^4}{24a^4}-\cdots\cdots \end{align}$$ Now when $t\ll a$, we have $\displaystyle{\frac{t}{a}\ll 1}$. So the higher powers of $\displaystyle\frac ta$ $\displaystyle{\left(\text{i.e. }\frac{t^2}{a^2},\ \frac{t^3}{a^3},\ \frac{t^4}{a^4},\ \cdots\cdots\right)}$ are so small that they can be neglected. Hence in the above series all the terms with these powers of $\displaystyle\frac ta$ are neglected to get the approximation $$e^{-\frac ta}\approx1-\frac ta$$