7
$\begingroup$

The following code:

Series[x^2 Sqrt[1 + 1/x^4], {x, 0, 0}]

gives different results in Mathematica 11.0 and 11.2. In 11.0 I get the expected result

1+O[x]^2

while 11.2 gives

O[x]^2

Even worse, in Mathematica 11.2

Series[x^2 Sqrt[1 + 1/x^6], {x, 0, 0}]

also gives

O[x]^2

while in 11.0 I get

1/x + O[x]^1

Is this a known bug in 11.2? Does anyone have a workaround?

$\endgroup$
4
  • $\begingroup$ In v11.2, Limit[x^2 Sqrt[1 + 1/x^4], x -> 0] evaluates to 1 $\endgroup$
    – Bob Hanlon
    Nov 17, 2017 at 13:05
  • 1
    $\begingroup$ Despite 0 being a branch point I think this should probably not return an empty series. Provisionally treating it as a bug in 11.2. I should remark that in the absence of assumptions on x, 1+O[x]^2 is also not a correct result. $\endgroup$ Nov 17, 2017 at 16:50
  • $\begingroup$ @DanielLichtblau: sure, 1+O[x]^2 is not correct for all x. However, if I expand y Sqrt[1 + 1/y^2] v11.2 gives y Sqrt[1/y^2], so for the first example in my question x^2 Sqrt[1/x^4] seems a reasonable answer. Furthermore, even with Assuming [x \[Element] Reals, ... ] I get 0 and not 1. $\endgroup$
    – Olof
    Nov 18, 2017 at 4:25
  • 1
    $\begingroup$ What you propose as a reasonable answer is what I'm hoping to get into the next release. $\endgroup$ Nov 18, 2017 at 16:16

1 Answer 1

6
$\begingroup$

This works:

Assuming [x \[Element] Reals && x >= 0, Series[Sqrt[1 + 1/x^4] x^2, {x, 0, 0}]]
Assuming [x \[Element] Reals && x >= 0, Series[Sqrt[1 + 1/x^6] x^2, {x, 0, 0}]]
$\endgroup$
2
  • 1
    $\begingroup$ Thanks. Actually assuming x >= 0 is enough. $\endgroup$
    – Olof
    Nov 17, 2017 at 12:52
  • $\begingroup$ Or assume x <= 0 or x < 0 || x >= 0 $\endgroup$
    – Bob Hanlon
    Nov 17, 2017 at 13:10

Your Answer

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

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