Find the two term asymptotic expansion of the solution Find the two term asymptotic expansion of the solution of
$$
1 +\sqrt{x^2 + \epsilon} = e^x
$$
My approach: i tried solutions of the form $x = x_0+\epsilon ^ \alpha x_1$ and plug it in. Then using Taylor series expansion to expand $e^x\approx 1+x+\frac{x^2}{2}$. However, i only got $x_0=0$ and cannot determine the value of $\alpha$.
 A: I will assume that the two-term expansion reads $x=x_1 \epsilon^\alpha + x_2 \epsilon^\beta$ (usually in asymptotics we don't say that 0 is a "term of an approximation").
So the first difficulty is that the expansion of the square root takes a different form depending on whether $\alpha<1/2$ (in which case $x^2 \gg \epsilon$), $\alpha=1/2$ (in which case $x^2 \sim \epsilon$) or $\alpha>1/2$ (in which case $\epsilon \gg x^2$). So you can't be sure that you are correctly expanding the left side until you have some information about $\alpha$. (Of course you can always guess and then check for self-consistency.)
One way to get some information is to note that the leading term in the expansion of $e^x-1$ scales as $\epsilon^\alpha$. Thus it must be that the leading term in the expansion of $\sqrt{x^2+\epsilon}$ scales as $\epsilon^\alpha$, which means you can't have $\epsilon \gg x^2$. So $\alpha \leq 1/2$.
But that doesn't tell us whether $\alpha=1/2$ or $\alpha<1/2$, so it doesn't fully resolve the above ambiguity. To sort out this last ambiguity, try to solve the problem by assuming $\alpha=1/2$. You will find that there is no solution to the equation for the coefficient $x_1$, so you conclude that $\alpha<1/2$. (This is a quirk of the exact structure of this problem...lots of small perturbations of this problem result in $\alpha=1/2$.)
Now that you know that, you can start to expand the square root by assuming $x^2 \gg \epsilon$. You will find that the requirement that the $\epsilon^\alpha$ terms cancel does not even specify what $\alpha$ is; these terms are just $x_1 \epsilon^\alpha$ on both sides of the equation, which you can conclude only by assuming $\alpha<1/2$. So you must include another term in each expansion in order to pin down $\alpha$. It is safe to assume that $\beta$ is large enough that that term doesn't come into play yet, which yields the equation $1-\alpha=2\alpha$ so that $\alpha=1/3$. 
Now that you know the overall scaling behavior of $x$, you can write
$$\sqrt{x^2+\epsilon}=x \sqrt{1+\epsilon/x^2}.$$
Now you can expand that square root in powers of $\epsilon/x^2$. Similarly you can expand $e^x-1$ in powers of $x$. Now that you know the scaling of $x$, you know that $x$ and $\epsilon/x^2$ are of the same overall order in $\epsilon$, you should expand each side to have the same number of terms. Since the first term didn't specify anything about the parameters in the expansion of $x$, you will need at least three terms in order to get a two-term approximation. At this stage the problem finally becomes routine.
A: Let 
$$f(x) = \sqrt{x^2+\epsilon}-(e^x-1)$$
To find the asymptotic solution $f(x)=0$, expand the function,
$$0=x\left(1+\frac 12\frac{\epsilon}{x^2}-\frac18\frac{\epsilon^2}{x^4} \right) -\left( x+\frac12 x^2+\frac 16 x^3\right) $$
The terms of $x$ cancels and the next-order gives the leading-order solution $x=\epsilon^{1/3}$. So, assume the two-term solution has the form $x=\epsilon^{1/3}(1+c\epsilon^b)$ and plug into the above expansion to have,
$$\epsilon\epsilon^{2/3}(1+c\epsilon^b)^2 - \epsilon^{5/3}(1+c\epsilon^b)^5 -\frac 13\epsilon^{2}(1+c\epsilon^b)^6 = \frac{\epsilon^2}{4} $$
Then, expand the power terms and keep the two leading orders to solve for, 
$$ b= \frac13, \>\>\>\>\> c = -\frac{7}{36}$$
Thus, the two-term asymptotical expansion solution is,
$$x=\epsilon^{1/3}- \frac{7}{36} \epsilon^{2/3}$$
For illustration of accuracy, we list the solutions below for some values of $\epsilon$ and with their exact values provided in parentheses.
$$ \epsilon = 0.01, \>\>\> x=0.206\>(0.207)$$
$$ \epsilon = 0.02, \>\>\> x=0.257\>(0.258)$$
$$ \epsilon = 0.05, \>\>\> x=0.342\>(0.343)$$
$$ \epsilon = 0.10, \>\>\> x=0.422\>(0.425)$$
$$ \epsilon = 0.25, \>\>\> x=0.553\>(0.560)$$
The asymptotic solution is fairly accurate for small values of $\epsilon$. 
