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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    $\begingroup$ 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. $\endgroup$
    – copper.hat
    Dec 18 '12 at 22:18

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 $u\to0$.

  • $\begingroup$ Dear did, A minor comment/question: since $\epsilon$ is stated to be a small parameter, would it be better to write $o(\epsilon^2)$ rather than $o(x^2)$? Regards, $\endgroup$
    – Matt E
    Dec 19 '12 at 21:12
  • $\begingroup$ @MattE Absolutely. Thanks. $\endgroup$
    – Did
    Dec 19 '12 at 21:14

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 $e^y=1+y+\cdots$


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

  • $\begingroup$ Thanks! You and Harald have both been very helpful. How is it that you knew to make the initial change (from $(1 + \epsilon x)^{1/\epsilon}$ to $exp(\frac{1}{\epsilon} ln(1 + \epsilon x))$)? What is the general form of the problem that would require this treatment? $\endgroup$
    – covertbob
    Dec 18 '12 at 22:34
  • $\begingroup$ Taking the derivative of $f(x)^{g(x)}$ is usually done by reading this as $\exp(g(x)\ln f(x))$ (or rather: This is the definiton of the suggestive notation $f(x)^{g(x)}$) $\endgroup$ Dec 18 '12 at 22:45
  • $\begingroup$ 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. $\endgroup$ Dec 19 '12 at 21:55

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