I am having a bit of difficulties with expanding perturbation series. I need to find a first and second order perturbative approximation to the root of $1+(x^2+\epsilon)^{1/2}=e^x$.
I first tried the following: $$1+(x^2+\epsilon)^{1/2}=e^x$$ Set $\epsilon = 0$ $$1+(x^2)^{1/2}=e^x$$ $$1+x=e^x$$ $$x=0$$
But the derivative of $1+x-e^x$ evaluated at $x=0$ is $1-e^0=0$, so it is singular, and we can't use Taylor expansion.
So, I tried,
$$1+(x^2+\epsilon)^{1/2}=e^x$$ $$(x^2+\epsilon)^{1/2}=e^x-1$$ $$x^2+\epsilon=(e^x-1)^2$$ $$x^2=(e^x-1)^2-\epsilon$$ Let $y=e^x-1$ $$e^x=y+1$$ So, $x=ln(y+1)$
Plugging it back to the above equation, $$(ln(y+1))^2=y^2-\epsilon$$
Then I think I need to use the method of dominant balance and then go back to the very first equation written in terms of $x$ and $\epsilon$, but I am not quite sure how or if this is correct. Can someone please help? Thanks!
As mentioned in my comment, your equation isn't $1+x-e^x$ when $\epsilon=0$, it's $1+|x|-e^x$ so the derivative isn't defined there.
You naive expansion found the $O(1)$ term as $0$, correctly, but this wasn't the right expansion to try. You should do dominant balance here.
Assume that the solution scales with $\epsilon^\alpha$, for some $\alpha>0$, i.e. $x\sim\epsilon^\alpha x_0$.
So $x$ is small (this is reasonable, $e^x$ should grow faster than $\sqrt{x^2+\epsilon}$), which lets me express $e^x$ as its Taylor series, so now I am trying to solve $$x^2+\epsilon=(e^x-1)^2=(x+x^2/2+x^3/6+\ldots)^2=x^2+x^3+7x^4/12+\ldots,$$ and hence $\epsilon=x^3+7x^4/12+\ldots$. From here the solution should be easy enough to find, dominant balance gives the expansion as powers of $1/3$, and the first two terms give $$ x = \epsilon^{1/3}-\frac{7}{36}\epsilon^{2/3}. $$
This approximation seems to do a pretty good job.