At work we use the fault tree analysis tool FaultTree+ from Isograph. The user guide version v15.0 presents a section with different failure models, one of them being the so-called dormant model type.
The following is stated there at p.455:
Section Dormant Model Type:
The expression for the mean unavailability of an individual dormant event is given below:
\begin{align*} \color{blue}{Q_{\mathrm{MEAN}}} \color{blue}{=\frac{\lambda \tau-\left(1-e^{-\lambda \tau}\right)+\lambda\,\mathrm{MTTR}\left(1-e^{-\lambda\tau}\right)}{\lambda \tau+\lambda\,\mathrm{MTTR}\left(1-e^{-\lambda\tau}\right)}}\tag{1} \end{align*} In (1) we have \begin{align*} Q_{\mathrm{MEAN}}&=\ \text{Mean unavailability value}\\ \lambda&=\ \text{Constant failure rate}\\ \mathrm{MTTR}&=\ \text{Mean time to repair}\\ \tau&=\ \text{Test interval} \end{align*} The following is valid provided $\lambda\,\mathrm{MTTR}\ll 1$ and $\tau\,\mathrm{MTTR}\ll 1$: \begin{align*} \color{blue}{Q_{\mathrm{MEAN}}\approx\frac{\lambda\tau}{2}+\lambda\,\mathrm{MTTR}}\tag{2} \end{align*}
Considerations:
At first I thought we should see rather at a glance how the approximation formula (2) can be derived from (1). But it was not that obvious to me. Finally I managed to derive it via a series expansion and noted that an expansion up to second order was necessary to show the validity. This was astonishing as I would have expected some hints in the user guide if a rather elaborated derivation is necessary to derive the approximation formula.
Question:
Do I miss something obvious which admits a simpler derivation than the solution given below?
Solution:
Currently I have the following solution: It is convenient to use in the following the repair rate $\mu:=\frac{1}{\mathrm{MTTR}}$. We consider $Q_{\lambda}(\tau):=Q_{\mathrm{MEAN}}$ as function in $\tau$ depending on a parameter $\lambda$ and write (1) in the form \begin{align*} \color{blue}{Q_{\lambda}(\tau)}&=\frac{\lambda \tau-\left(1-e^{-\lambda \tau}\right)+\frac{\lambda}{\mu}\left(1-e^{-\lambda\tau}\right)}{\lambda \tau+\frac{\lambda}{\mu}\left(1-e^{-\lambda\tau}\right)}\\ &\color{blue}{=1-\frac{1-e^{-\lambda\,\tau}}{\lambda \tau +\frac{\lambda}{\mu}\left(1-e^{-\lambda\,\tau}\right)}}\tag{3} \end{align*} In order to show (2) we need to expand the exponential series $e^{-\lambda\tau}$ up to the order of two: \begin{align*} e^{-\lambda\tau}=1-\lambda\tau+\frac{1}{2}\lambda^2\tau^2+\mathcal{O}\left(\tau^3\right)\tag{4} \end{align*} We obtain from (3) and (4) \begin{align*} \color{blue}{Q_{\lambda}(\tau)}&=1-\frac{\lambda\tau-\frac{1}{2}\lambda^2\tau^2+\mathcal{O}\left(\tau^3\right)} {\lambda\tau+\frac{\lambda}{\mu}\left(\lambda \tau-\frac{1}{2}\lambda^2\tau^2+\mathcal{O}\left(\tau^3\right)\right)}\tag{5.1}\\ &=1-\frac{1-\frac{1}{2}\lambda\tau+\mathcal{O}\left(\tau^2\right)} {1+\frac{\lambda}{\mu}\left(1-\frac{1}{2}\lambda\tau+\mathcal{O}\left(\tau^2\right)\right)}\tag{5.2}\\ &=1-\frac{1-\frac{1}{2}\lambda\tau+\mathcal{O}\left(\tau^2\right)} {1+\frac{\lambda}{\mu}-\frac{\lambda^2}{2\mu}\tau+\mathcal{O}\left(\tau^2\right)}\\ &=1-\frac{\mu}{\lambda+\mu}\cdot\frac{1-\frac{1}{2}\lambda\tau+\mathcal{O}\left(\tau^2\right)} {1-\frac{\lambda^2}{2\left(\lambda+\mu\right)}\tau+\mathcal{O}\left(\tau^2\right)}\tag{5.3}\\ &=1-\frac{\mu}{\lambda+\mu}\left(1-\frac{1}{2}\lambda\tau+\mathcal{O}\left(\tau^2\right)\right) \left(1+\frac{\lambda^2}{2\left(\lambda+\mu\right)}\tau+\mathcal{O}\left(\tau^2\right)\right)\tag{5.4}\\ &=1-\frac{\mu}{\lambda+\mu}\left(1-\left(\frac{1}{2}\lambda-\frac{\lambda^2}{2\left(\lambda+\mu\right)}\right)\tau+\mathcal{O}\left(\tau^2\right)\right)\tag{5.5}\\ &=\frac{\lambda}{\lambda+\mu}+\frac{\lambda\mu^2}{2\left(\lambda+\mu\right)^2}\tau+\mathcal{O}\left(\tau^2\right)\tag{5.6}\\ &\,\,\color{blue}{\approx \frac{\lambda}{\mu}+\frac{\lambda}{2}\tau}\tag{5.7}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\Box \end{align*} and the claim (2) follows. Note in the step from (5.6) to (5.7) we use that $\frac{\lambda}{\mu}=\lambda\,\mathrm{MTTR}\ll 1$. In the series expansion (5.4) we use both assumptions, $\frac{\lambda}{\mu}\ll 1$ and $\frac{\tau}{\mu}\ll 1$.
Comment:
In (5.1) we use the series expansion (3) in the representation from (2).
In (5.2) we cancel $\lambda\tau$.
In (5.3) we factor out $1+\frac{\lambda}{\mu}=\frac{\lambda+\mu}{\mu}$ from the denominator.
In (5.4) we do a geometric series expansion $\frac{1}{1-z}=1+z+\mathcal{O}\left(z^2\right)$.
In (5.5) we multiply out putting all terms with order of $\tau$ greater equal $2$ into $\mathcal{O}\left(\tau^2\right)$.
In (5.6) we do some simplifications.
You could have done the same work without any approximation in a first step since, in fact, you assume that $\lambda \tau$ is small. $$\begin{align*} Q_{\mathrm{MEAN}} =\frac{\lambda \tau-\left(1-e^{-\lambda \tau}\right)+\lambda\,M\left(1-e^{-\lambda\tau}\right)}{\lambda \tau+\lambda\,M\left(1-e^{-\lambda\tau}\right)}\tag{1} \end{align*}$$ It is normal that you need to go to the second order if you want $\tau$ to appear in the formula because the expansion of the numerator is $$\lambda ^2 M \color{red}{\tau} +\frac{1}{2}\lambda ^2 \left(1-\lambda M\right)\tau ^2-\frac{1}{6} \lambda ^3 (1-\lambda M)\tau ^3+O\left(\tau ^4\right)$$ and for the denominator $$ \lambda \left(1 +\lambda M\right)\color{red}{\tau}-\frac{1}{2} \lambda ^3 M\tau ^2+\frac{1}{6} \lambda ^4 M \tau ^3+O\left(\tau ^4\right)$$ Then, long division to get $$Q_{\mathrm{MEAN}}=\frac{\lambda M}{1+\lambda M}+\frac{\lambda }{2 (1+\lambda M)^2}\tau -\frac{ \lambda^2\left(2-\lambda M \right)}{12 (1+\lambda M)^3}\tau ^2+O\left(\tau ^3\right)\tag 2$$
Now, if you want to consider that $\lambda M \ll 1$, then what you wrote,
If you want a better aproximation which will be $O\left(\tau ^3\right)$ with only first power of $\tau$ $$Q_{\mathrm{MEAN}}=\frac {\frac{\lambda M}{1+\lambda M}+\frac{\lambda (3-\lambda M)}{6 (1+\lambda M)}\tau } {1+\frac{\lambda (2-\lambda M)}{6 (1+\lambda M)}\tau }\tag 3$$ for which the asymptotic error is $\frac{\lambda ^3 }{72 (1+\lambda M)^2}\tau ^3$.
Edit
It could be interesting to write $$Q_{\mathrm{MEAN}}=\frac{\lambda M}{1+\lambda M}\Bigg[1+\frac 1M x -\frac{\lambda (2-\lambda M)}{3 M}x^2\Bigg]+O(x^3) \quad \text{with} \quad x=\frac{\tau }{2 (1+\lambda M)}$$