For some reason Mathematica will not evaluate this asymptotic series to the requested order.


1/x ( x ((1 - x)^2) + E^x);
Series[%, {x, \[Infinity], 0}]


$ \left( x^2 + \frac{1}{\mathcal{O} \left( \frac{1}{x} \right)} \right) + e^x \left( \frac{1}{x} + \mathcal{O} \left( \frac{1}{x} \right)^2 \right)$

The problem being that in the first term it doesn't give me the $\mathcal{O} (x^1)$ and $\mathcal{O} (x^0)$ that I requested.

Mathematica does not make this error for simpler expressions, but I haven't been able to nail down what exactly about this expression is giving me problems.

A couple things I have tried:

(1) I could simply go to a higher order in x in order to get the 'missing' terms, but this is computationally expensive on more complicated expressions.

(2) If I Expand[%] the expression prior to the series expansion that gives me the correct terms, the only thing is, I would never have guessed that I have to put that in there. Plus I often deal with expressions much more complicated than this and it is hard to visually inspect all the $\mathcal{O}(1/\Lambda)$'s all over the place

Is there a simple reason I should have anticipated this being a problem for Mathematica? And is there a simple command to use that doesn't slow things down? - sometimes using Expand[%] prior to doing a computation can slow down when dealing with large expressions.

  • 1
    $\begingroup$ Interesting. Not sure if it's a bug or a feature. I'll take a look. $\endgroup$ Jul 28, 2014 at 23:00
  • $\begingroup$ @DanielLichtblau, Any News on this? $\endgroup$
    – user9660
    Jun 26, 2016 at 8:34
  • $\begingroup$ @Louis I don't understand the question. Have you tried the computation in a recent version of Mathematica? Do you obtain an unexpected result? Generally speaking, if you ask someone to revisit a thread from ~2 years ago, it would be a good idea to state specifically what you have checked, in what version, and what is the (presumably unexpected and/or undesired) status you encounter. $\endgroup$ Jun 27, 2016 at 16:49


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Browse other questions tagged or ask your own question.