deletions | additions       diff --git a/$n^{th}$-order decay.tex b/$n^{th}$-order decay.tex  index 01de591..9a60b04 100644  --- a/$n^{th}$-order decay.tex  +++ b/$n^{th}$-order decay.tex 
...
We now shift our attention to $n^{th}$ order reactions. These $n^{th}$ order reactions are of the form of one reactant turning into product. The inequality between the statistical distance and Fisher divergence can also be derived for these irreversible decay reactions.  The time dependent rate coefficient is  \begin{equation}  k(t)\equiv(\frac{d\frac{1}{(S(t)^{n-1)}}}{dt})\equiv(n-1)\omega[A_0]^{n-1} k_n(t) \equiv (\frac{d\frac{1}{(S(t)^{n-1)}}}{dt})\equiv(n-1)\omega[A_0]^{n-1}  \end{equation}  As shown in equation 4, the statistical length is the integral of the cumulative time dependent rate coefficient over a period of time $\Delta{t}$. The statistical length is   \begin{equation}  \mathcal{L}(\Delta t)^2=\left[\int_{t_i}^{t_f}(n-1)\omega([A_0]^{n-1})dt\right]^2 \mathcal{L}_n(\Delta t)^2 = \left[\int_{t_i}^{t_f}(n-1)\omega([A_0]^{n-1})dt\right]^2  \end{equation}  Following length, the Fisher divergence is the integral of the cumulative time dependent rate coefficient squared over a period of time $\Delta{t}$. The Fisher divergence is  \begin{equation}  \frac{\mathcal{J}( \Delta \frac{\mathcal{J}_n(\Delta  t)}{\Delta t}=\int_{t_i}^{t_f}{(n-1)^2\omega^2}([A_0]^{n-1})^{2} t} = \int_{t_i}^{t_f}{(n-1)^2\omega^2}([A_0]^{n-1})^{2}  dt \end{equation}  The inequality becomes  \begin{equation}