Asymptotic behavior of $-gTt-gT^2e^{\frac{-t}{T}}$ for small $t$

Question

I want to solve this using Taylor series expansion of $e^{f(x)}$

$$\begin{align}x=-gTt-gT^2e^{\frac{-t}{T}}+gT^2+x_0\end{align}$$

Show that for small values of t $(t\ll T)$, the equation for x is approximately $x\approx{x_0-\frac{1}{2}gt^2}$ , where T is a constant

Please help.

Answer 1

$$\begin{align}x=-gTt-gT^2e^{\frac{-t}{T}}+gT^2+x_0\end{align}$$

Note: $e^{-t/T} = 1 - \frac{t}{T} + \frac{t^2}{2T^2}+....$

Therefore $e^{-t/T} \approx 1 - \frac{t}{T} + \frac{t^2}{2T^2}$

$x\approx gTt- gT^2(1 - \frac{t}{T} + \frac{t^2}{2T^2})+gT^2+x_0 $

$x\approx gTt- gT^2- gTt +\frac{1}{2}gt^2+gT^2+x_0 $

$x\approx \frac{1}{2}gt^2+x_0 $

Answer 2

Hint: $e^{-t/T} \approx 1 - t/T + 0.5 t^2 / T^2$

Answer 3

Using Taylor Series You get:

$e^{-t/T} \approx 1 - t/T + 0.5 t^2 / T^2$