I am working through a paper on Random Walks in Random Environments at the moment (Solomon 1975), the introductory paper to the topic. I have come rather unstuck at proving an equality.
Prove $$e^u - \frac{e^u}{2}(\theta +1-[(\theta+1)^2 -4\theta e^{-2u}]^{\frac{1}{2}}) = \frac{2\theta}{\theta - 1} \cdot u + O(u^2)$$
Any help would be greatly appreciated!
Link to the paper: https://projecteuclid.org/euclid.aop/1176996444 It's on page 12!
I assume that $\theta>1$? (Otherwise, it is not the case, as we will see during the derivation.)
Use the Taylor expansions around $0$: \begin{align} e^u &= 1+u+O(u^2) \tag{1}\\ e^{-2u} &= 1-2u+O(u^2)\tag{2}\\ \sqrt{1+u} &= 1+\frac{u}{2}+O(u^2)\tag{3} \end{align} so that your LHS becomes, when $u\to 0$, \begin{align} \textrm{LHS} &= e^u - \frac{e^u}{2}(\theta +1-\sqrt{(\theta+1)^2 -4\theta (1-2u+O(u^2))}) \tag{From (2)}\\ &= e^u - \frac{e^u}{2}(\theta +1-\sqrt{1+\theta^2-2\theta +8\theta u+O(u^2)}) \\ &= e^u - \frac{e^u}{2}(\theta +1-\sqrt{(\theta-1)^2 +8\theta u+O(u^2)}) \\ &= e^u - \frac{e^u}{2}(\theta +1-\lvert \theta-1\rvert\sqrt{1 +\frac{8\theta}{(\theta-1)^2} u+O(u^2)}) \\ &= e^u - \frac{e^u}{2}(\theta +1-\lvert \theta-1\rvert\left(1 +\frac{4\theta}{(\theta-1)^2} u+O(u^2)\right)) \tag{From (3)}\\ &= e^u - \frac{e^u}{2}(\theta +1- (\theta-1) -\frac{4\theta}{\theta-1} u+O(u^2)) \tag{As $\theta > 1$}\\ &= e^u - e^u(1 -\frac{2\theta}{\theta-1} u+O(u^2))\\ &= e^u(\frac{2\theta}{\theta-1} u+O(u^2))\\ &= (1-u+O(u^2))\left(\frac{2\theta}{\theta-1} u+O(u^2)\right)\tag{From (1)}\\ &= \frac{2\theta}{\theta-1} u+O(u^2) \end{align} as claimed.
Note that if $\theta\in(0,1)$, this no longer holds (the step where we used $\lvert \theta-1\rvert = \theta-1$ no longer holds), and we would get instead that the LHS is $1-\theta+O(u)$.