Simple Power Series Expansion for Problems similar to $f = (1 + \epsilon \,x)^{1/\epsilon}$ I was flicking through a book on perturbation methods and saw a simple question asking the reader to expand the following expression for $f$ in a power series (up to the first 2 terms):
$f = (1 + \epsilon \,x)^{1/\epsilon}$, where $\epsilon$ is a small parameter. I'm sure this is very simple, but I wasn't certain about the best way to approach this. A quick look at mathematica tells me the solution is $e^x - \frac{1}{2} (e^x x^2) \,\epsilon + ...$. How would I go about getting this answer - and more importantly, how would I systematically find series expansions for problems similar to this one?
 A: Since $\epsilon^{-1}\log(1+\epsilon x)=\epsilon^{-1}\left(\epsilon x-\frac12\epsilon^2x^2+o(\epsilon^2)\right)=x-\frac12\epsilon x^2+o(\epsilon)$, one gets $f(\epsilon)=\exp(\epsilon^{-1}\log(1+\epsilon x))=\mathrm e^x\exp\left(-\frac12\epsilon x^2+o(\epsilon)\right)$, that is,
$$
f(\epsilon)=\mathrm e^x\left(1-\tfrac12\epsilon x^2\right)+o(\epsilon).
$$
To sum up, one uses $\log(1+u)=u-\frac12u^2+o(u^2)$ and $\mathrm e^u=1+u+o(u)$ when $u\to0$.
A: The short answer: Use the identity $a^b=e^{b\ln a}$ to get $$(1+\epsilon x)^{1/\epsilon}=\exp\Bigl(\frac{\ln(1+\epsilon x)}{\epsilon}\Bigr),$$
then use known series expansions of the logarithm and exponential functions.
In more detail, $\ln(1+x)=x-\frac12x^2+\cdots$, so $\ln(1+\epsilon x)/\epsilon=x-\frac12\epsilon x^2+\cdots$.
Substitute that into the exponential function, using the identity $e^{x+y}=e^xe^y$ to get $$(1+\epsilon x)^{1/\epsilon}=e^x\exp(-\tfrac12\epsilon x^2+\cdots),$$
and finally substitue $-\tfrac12\epsilon x^2+\cdots$ for $y$ in $e^y=1+y+\cdots$
A: Note that $f(\epsilon)=(1+\epsilon x)^{1/\epsilon}= \exp(\frac1\epsilon\ln(1+\epsilon x))$.
The expansion of $\ln(1+t)$ is $t-\frac 12t^2+\frac13t^3\pm\ldots$ (should be well-known), hence the expansion of $g(\epsilon):=\frac1\epsilon\ln(1+\epsilon x)$ is $x-\frac12x^2\epsilon+\frac13x^3\epsilon^2\pm\ldots$ (which is just a nice encoding of the facts $g(0)=x$, $g'(0)=-\frac12x^2$, $g''(0)=\frac23x^3$ etc.) and thus after applying exponentiation, we obtain
$f(0)=e^x$, $f'(0)=g'(0)e^x$, $f''(0)=(g''(0)+g'(0)^2)e^x$, hence if I'm not mistaken
$$f(\epsilon)=e^x-\frac12x^2e^x\epsilon +\frac 12(\frac23x^3+\frac14 x^4)e^x\epsilon^2+O(\epsilon^3)\\=e^x-\frac12x^2e^x\epsilon +(\frac13+\frac18 x)x^3e^x\epsilon^2+O(\epsilon^3)$$
