Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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. – copper.hat Dec 18 '12 at 22:18
up vote 2 down vote accepted

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

share|cite|improve this answer
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? – covertbob Dec 18 '12 at 22:34
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 $f(x)^{g(x)}$) – Hagen von Eitzen Dec 18 '12 at 22:45
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. – Harald Hanche-Olsen Dec 19 '12 at 21:55

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 $u\to0$.

share|cite|improve this answer
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, – Matt E Dec 19 '12 at 21:12
@MattE Absolutely. Thanks. – 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 $e^y=1+y+\cdots$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.