Find a two-term expansion for the root of 1+\sqrt(x^2+\epsilon)=e^x. I am trying to find a two term expansion for the root of $$1+\sqrt{(x^2+\epsilon)}=e^x$$. Since $$\epsilon << 1$$, I can tell that this equation behaves like $$1+x=e^x$$ which has a root close to zero. That gave me the idea that the root might look like $$x \approx x_0 + \epsilon^{\alpha} x_1 + \dots  $$. I used Taylor expansion for both functions $\sqrt{(x^2+\epsilon)}$ and $e^x$ and then plugged my guess into the Taylor expanded equation and balance of O(1) terms gave me $$x_0=0$$ as expected. My problem begins now. For the higher orders of $\epsilon$. Long story short this expansion ends up with some inconsistencies. So I thought maybe $x \approx x_0+\mu(\epsilon)$. But then again I get nowhere. I am getting frustrated and I would really appreciate some help.
Thank you in advance. 
 A: Here's a hint from my instructor. Rearranging the given equation as follows
\begin{align*}
x^2 + \varepsilon & = (e^x - 1)^2 \\
\varepsilon & = (e^x - 1)^2 - x^2 \\
& = (e^x - 1 - x)(e^x - 1 + x)
\end{align*}
You can then Taylor expand $e^x$ around 0, substitute the given asymptotic expansion and try to balance term. Note that for $\varepsilon=0$, $x=0$ is the solution to the equation $1 + \sqrt{x^2} = e^x$, so you may choose $x_0 = 0$. 
A: 
The first thing is to solve it numerically to vizualize how $x_\varepsilon$ varies in function of $\varepsilon$. 
Above is the graph of :$\displaystyle\quad\frac{\ln(\varepsilon)}{\ln(x_\epsilon)}$ when $\varepsilon=10^{-a}\quad a\in[1,10]$
So we can consider that $\varepsilon\ll 1\implies \varepsilon=O(x^3)$ or similarly $x=O(\sqrt[3]\varepsilon)$
Now we can compare the development of $\displaystyle \sqrt{1+\frac{\varepsilon}{x^2}}$ and the one of $e^x$ since $\displaystyle \frac{\varepsilon}{x^2}=O(x)$.


*

*$\displaystyle e^x=1+x+\frac 12 x^2+\frac 16 x^3+O(x^4)$

*$\displaystyle 1+\sqrt{x^2+\varepsilon}=1+x\left(1+\frac 12 \frac{\varepsilon}{x^2} -\frac 18 \frac{\varepsilon^2}{x^4}+O(x^3)\right)=1+x+\underbrace{\frac 12 \frac{\varepsilon}{x}}_{O(x^2)} -\underbrace{\frac 18 \frac{\varepsilon^2}{x^3}}_{O(x^3)}+O(x^4)$
If we consider the development up to $O(x^2)$ we get :
$\displaystyle \frac 12 x^2\sim\frac 12\frac{\varepsilon}x\iff \bbox[5px,border:2px solid]{x\sim\sqrt[3]{\varepsilon}}\quad$ this is more precise than just $x=O(\sqrt[3]{\varepsilon})$.


If we consider now the development up to $O(x^3)  $ we get :
$\displaystyle \frac 12 x^2+\frac 16x^3=\frac 12\frac{\varepsilon}x-\frac 18\frac{\varepsilon^2}{x^3}+o(\varepsilon)$
Let's have $x^3=\varepsilon+u\quad$ with $u=o(\varepsilon)\quad$ 
(we could have instead $x=\sqrt[3]{\varepsilon}+u$, but it complicates the calculation)
$\begin{array}{l}\require{cancel}
12x^5+4x^6=12x^2\varepsilon-3\varepsilon^2+o(\varepsilon^2)\\
12x^2(x^3-\varepsilon)+4(u+\varepsilon)^2+3\varepsilon^2=o(\varepsilon^2)\\
12x^2u+\cancel{4u^2}+\cancel{8u\varepsilon}+7\varepsilon^2=o(\varepsilon^2) & u^2,u\varepsilon=o(\varepsilon^2)\text{ so we cancel them  }
\end{array}$
In the same way $12x^2u\sim 12\varepsilon^\frac 23u$ other terms are negligible.
So we get $\quad\displaystyle 12\varepsilon^\frac 23u\sim-7\varepsilon^2\iff u\sim -\frac 7{12}\varepsilon^\frac 43$
Now reporting in $x^3=\varepsilon+u\quad$ we get $\quad \bbox[5px,border:2px solid]{x=\varepsilon^\frac 13-\frac 7{36}\varepsilon^\frac 23+o(\varepsilon^\frac 23)}$
We could continue the development, but it becomes delicate to know which terms we can keep and which ones we can dismiss. 
A: Starting from Chee Han's answer, we have 
$$\epsilon=(e^x - 1 - x)(e^x - 1 + x)$$ Now, using Taylor expansion around $x=0$
$$(e^x - 1 - x)=\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+O\left(x^6\right)$$
$$(e^x - 1 + x)=2 x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+O\left(x^6\right)$$ making
$$\epsilon=x^3+\frac{7 x^4}{12}+\frac{x^5}{4}+\frac{31 x^6}{360}+O\left(x^7\right)$$ Now, use series reversion to get
$$x= \epsilon ^{1/3}-\frac{7}{36} \epsilon ^{2/3}+\frac{13  }{432}\epsilon-\frac{733
   }{699840}\epsilon ^{4/3}+O\left(\epsilon ^{5/3}\right)$$ Let us try with $\epsilon=\frac 1 {27}$ (which is quite large); the above expansion will give $x=\frac{17733407}{56687040}\approx 0.312830$  while the "exact" solution would be $\approx 0.312826$
