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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?

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    If $f$ has a power series expansion around $x=a$, then $f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!} (x-a)^n$. So, differentiate and evaluate... Or, since the expansion is unique, use any technique (such as suggested by Harald below) to find an expansion. In this particular case, you could use the (generalized) binomial theorem.2012-12-18

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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 $f(\epsilon)=e^x-\frac12x^2e^x\epsilon +\frac 12(\frac23x^3+\frac14 x^4)e^x\epsilon^2+O(\epsilon^3)\\=e^x-\frac12x^2e^x\epsilon +(\frac13+\frac18 x)x^3e^x\epsilon^2+O(\epsilon^3)$

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    To answer the question a little more fully: In most treatments of calculus, one notes that there is quite a bit of difficulty defining $a^b$ when $b$ is irrational. The difficulty is typically resolvedd by introducing exp and its inverse the natural logarithm, and then using the exponential function to define $a^b=e^{b\ln a}$. This makes it quite natural to use the definition whenever doing anything non-trivial with powers.2012-12-19
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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 $u\to0$.

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    @MattE Absolutely. Thanks.2012-12-19
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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 $e^y=1+y+\cdots$