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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?

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  • $\begingroup$ 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
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    $\begingroup$ 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. $\endgroup$
    – Daniel Lichtblau
    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
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    $\begingroup$ What you propose as a reasonable answer is what I'm hoping to get into the next release. $\endgroup$
    – Daniel Lichtblau
    Nov 18, 2017 at 16:16

1 Answer 1

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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}]]
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    $\begingroup$ 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

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