3
$\begingroup$

The following code

Series[Log[1 - zz] Log[2 zz^3] - PolyLog[2, 1/zz], {zz, 0, 0}, 
 Assumptions -> zz > 0]
% // Normal

returns

(log^2(zz)/2-I \[Pi] log(zz)-\[Pi]^2/3)+O(zz^1)

log^2(zz)/2-I \[Pi] log(zz)-\[Pi]^2/3

on Mathematica versions 8, 9, 10.3, 11.0 and 11.1. So I assume that intrinsically Mathematica has no issues with this expansion. For some reason version 11.2 not only gives me one order more than I requested, but also returns an expression with an infinity

(-zz+O(zz^2)) log(O(zz^3))+((log^2(zz)/2-I \[Pi] log(zz)-\[Pi]^2/3)+zz+O(zz^2))

zz \[Infinity]+zz+log^2(zz)/2-I \[Pi] log(zz)-\[Pi]^2/3

I would say at least that this behavior is inconsistent with the output of the previous versions. Is it a bug or is there something I'm missing about Series in version 11.2?

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7
  • 1
    $\begingroup$ Not yet a definitive answer, but i suspect it is a bug. Will investigate tomorrow or so. $\endgroup$ Commented Nov 12, 2017 at 17:29
  • $\begingroup$ @DanielLichtblau Notice that SeriesCoefficient[ Log[1 - zz] Log[2 zz^3] - PolyLog[2, 1/zz], {zz, 0, 0}, Assumptions -> zz > 0] does not evaluate on version 11.2 but works fine with 11.1, 10.3 and so on. $\endgroup$
    – vsht
    Commented Nov 13, 2017 at 2:52
  • $\begingroup$ @DanielLichtblau Any news on this issue? $\endgroup$
    – vsht
    Commented Nov 15, 2017 at 7:18
  • $\begingroup$ It's an issue with handling of empty coefficient lists in SeriesData objects. Will be testing a possible fix, not yet sure what the outcome will be. $\endgroup$ Commented Nov 15, 2017 at 16:25
  • $\begingroup$ Since this got bumped to the homepage, I added the 'bugs' tag. $\endgroup$ Commented Feb 11, 2018 at 15:34

2 Answers 2

1
$\begingroup$

A workaround is to use ReplaceAll as follows:

Assuming[
    zz>0,
    Log[1-zz] Log[2 zz^3] - PolyLog[2, 1/zz] /. zz->zz+O[zz]^3
] //TeXForm

$\left(\frac{\log ^2(\operatorname{zz})}{2}-i \pi \log (\operatorname{zz})-\frac{\pi ^2}{3}\right)+O\left(\operatorname{zz}^1\right)$

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0
$\begingroup$

You take too small order in {zz, 0, 0}. The following works well in version 11.2.

Series[Log[1 - zz] Log[2 zz^3] - PolyLog[2, 1/zz], {zz, 0, 4}, Assumptions -> zz > 0]

$$\left(\frac{\log ^2(\text{zz})}{2}-i \pi \log (\text{zz})-\frac{\pi ^2}{3}\right)+\text{zz} (-3 \log (\text{zz})+1-\log (2))+\frac{1}{4} \text{zz}^2 (-6 \log (\text{zz})+1-2 \log (2))+\text{zz}^3 \left(-\log (\text{zz})+\frac{1}{9}-\frac{\log (2)}{3}\right)+\frac{1}{16} \text{zz}^4 (-12 \log (\text{zz})+1-4 \log (2))+O\left(\text{zz}^5\right) $$

% // Normal

$$ \frac{1}{16} \text{zz}^4 (-12 \log (\text{zz})+1-4 \log (2))+\text{zz}^3 \left(-\log (\text{zz})+\frac{1}{9}-\frac{\log (2)}{3}\right)+\frac{1}{4} \text{zz}^2 (-6 \log (\text{zz})+1-2 \log (2))+\frac{\log ^2(\text{zz})}{2}+\text{zz} (-3 \log (\text{zz})+1-\log (2))-i \pi \log (\text{zz})+\frac{\pi ^2}{3} $$

The order can be decreased to 3 from 4, but no to 2.

Addition. The above shows the answer

(log^2(zz)/2-I \[Pi] log(zz)-\[Pi]^2/3)+O(zz^1)

in some previous versions is not correct because $x\log x \neq O(x),\, x \to 0$.

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5
  • $\begingroup$ Sorry, but I do not quite understand your answer. The point of looking at the 0th order is to extract the leading order behavior of the function around the expansion point. I'm really not interested in higher orders here. If you look at ex1 = SeriesCoefficient[ Log[1 - zz] Log[2 zz^3] - PolyLog[2, 1/zz], {zz, 0, 0}, Assumptions -> zz > 0], ex2 = SeriesCoefficient[ Log[1 - zz] Log[2 zz^3] - PolyLog[2, 1/zz], {zz, 0, 1}, Assumptions -> zz > 0] and Plot[{Re[ex1], zz Re[ex2]}, {zz, -1, 1}] in version 11.0 or any older version, then it is clear that $\endgroup$
    – vsht
    Commented Nov 13, 2017 at 2:50
  • 1
    $\begingroup$ the largest contribution around zz=0 comes from the logs, not logs times zz. And this is exactly what I want to know. So in my view the result given by all the previous versions of Mathematica is correct. A simpler example which shares the same features would be Normal[Series[(a Log[x] + b Log[x]^2) Sqrt[1 + x], {x, 0, 0}]] Here the 0th order expansion correctly returns only the logs both on Mathematica 11.2 and older versions. $\endgroup$
    – vsht
    Commented Nov 13, 2017 at 2:50
  • $\begingroup$ @vsht: The expansion (log^2(zz)/2-I [Pi] log(zz)-[Pi]^2/3)+O(zz^1) is incorrect because of O(zz^1). It should be (log^2(zz)/2-I [Pi] log(zz)-[Pi]^2/3)+O(zz^1*log(zz)).Don't hesitate to ask for further explanation in need. $\endgroup$
    – user64494
    Commented Nov 13, 2017 at 15:32
  • $\begingroup$ AFAIK it is a notation convention of Mathematica to show only the expansion parameter in the Landau symbol. You may not like it for various reasons, but this is how it works across many versions. Moreover, this behavior is not limited to my example. Normal[Series[(a Log[x] + b Log[x]^2) Sqrt[1 + x], {x, 0, 0}]] behaves in exactly the same way. Anyway, my original question was about Series returning naked singularities and more orders than requested, not about the big O notation. So I still fail to see how your answer is related to the original question. $\endgroup$
    – vsht
    Commented Nov 14, 2017 at 3:10
  • $\begingroup$ @vsht: I submitted that bug in Series. $\endgroup$
    – user64494
    Commented Feb 11, 2018 at 19:08

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