1
$\begingroup$

There seems to be a bug for Mathematica when evaluating series expansion for expressions containing log functions. For example:

tmp1=-(1 - x)^2/(-1 + x^2)^2 - ((1 - x)^2 - ((-1 + x)^2*Log[1 - x])/
 2)/(-1 + x^2)^2;

When evaluating

Series[tmp1,{x,1,0}]//Normal

I get 0, which is obviously incorrect. To see this, we only need to evaluate

Series[tmp1//Expand,{x,1,0}]//Normal

Has anyone ever met similar problems?

p.s. The version of Mathematica on my PC is "11.0.1 for Linux x86 (64-bit) (September 21, 2016)".

$\endgroup$
  • $\begingroup$ I think the first result is correct since you ask Mathematica to perform a series expansion that is exact up to order 0, aren't you? $\endgroup$ – Henrik Schumacher Jan 23 '18 at 15:36
  • $\begingroup$ @ Henrik Schumacher I don't think so, because the first term in "tmp1" is obviously finite up to order 0 in x. $\endgroup$ – Wen Chern Jan 23 '18 at 15:38
  • 2
    $\begingroup$ I see. But Series[tmp1, {x, 1, 0}] returns SeriesData[x, 1, {}, 0, 0, 1] + Log[1 - x] SeriesData[x, 1, {}, 0, 0, 1] and that tells us that there is a singularity. No matter which constant you add. $\endgroup$ – Henrik Schumacher Jan 23 '18 at 15:42
  • 3
    $\begingroup$ On 11.1 I get 1/8 (-4 + Log[1 - x]) for both cases. $\endgroup$ – John Doty Jan 23 '18 at 15:49
  • 3
    $\begingroup$ @JohnDoty Ditto on V11.2. That makes it look like a bug that has been fixed. $\endgroup$ – Michael E2 Jan 23 '18 at 15:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.