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I'm confused about the difference between Series and Asymptotic. Is there a good rule of thumb of when to use which?

Observations:

  • Asymptotic of order $n+1$ is analogous to Series to order $n$
  • Series generates constant terms

Code below uses both to simplify the following expression

$$\text{expr}=\frac{1}{\frac{s}{2}+\frac{\sqrt{2} \sqrt{s}}{\pi -2 \tan ^{-1}\left(\frac{\sqrt{s}}{\sqrt{2}}\right)}}$$

ClearAll["Global`*"];
expr = 1/(s/2 + (Sqrt[2] Sqrt[s])/(\[Pi] - 2 ArcTan[Sqrt[s]/Sqrt[2]]));
SF = StringForm;
series[order_] := Normal@Series[expr, {s, 0, order}];
asymp[order_] := Asymptotic[expr, {s, 0, order}];
{approxSeries, approxAsymp} = 
  Table[{i, #[i]}, {i, 0, 3}] & /@ {series, asymp};

visualize[approx_, label_] := (
  Print[TableForm[approx, TableHeadings -> {{}, {"order", "expr"}}]];
  LogLogPlot @@ {{expr}~Join~(Last /@ approx), {s, 0., .2},
    PlotLegends -> {"true"}~Join~(SF["order ``", First@#] & /@ approx),
    PlotStyle -> {{Thick, Opacity[.5]}, Automatic, Dashed, Dotted, 
      DotDashed},
    PlotLabel -> label}
  )
Quiet[visualize[approxSeries, "Series"]]
Quiet[visualize[approxAsymp, "Asymptotic"]]

enter image description here

enter image description here

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  • $\begingroup$ Your code produces a warning "Asymptotic::aord: Approximation order specification 0 should be a positive integer or Infinity." omitted by you. $\endgroup$
    – user64494
    May 16 at 17:31
  • $\begingroup$ Also the question arises: what is the result without Quiets? $\endgroup$
    – user64494
    May 16 at 17:33
  • $\begingroup$ The constant in the result of Series follows from Limit[1/( s/2 + (Sqrt[2] Sqrt[s])/(\[Pi] - 2 ArcTan[Sqrt[s]/Sqrt[2]])) - \[Pi]/(Sqrt[2] Sqrt[s]), s -> 0 ] which results in -1 - \[Pi]^2/4. This term is not of importance for the asymptotic. $\endgroup$
    – user64494
    May 16 at 17:59
  • $\begingroup$ I've always wonder this myself too. I think 1. To avoid the SeriesData output and 2. Asymptotic generalizes Series in that it will give an expression asymptotic to the input. Series gives a (generalized) power series, and when this exists it's asymptotic to the input. Compare Asymptotic[PrimePi[x], {x, ∞, 3}] and Series[PrimePi[x], {x, ∞, 3}]. $\endgroup$
    – Greg Hurst
    May 17 at 15:29
  • 1
    $\begingroup$ Asymptotic is meant as a more powerful "superfunction". It can also handle, for example, inactivated operators (Asymptotic[Inactive[Integrate][Exp[x^2], x], x -> 0]). For a more detailed discussion about the peculiarities of implementation, I suggest watching Live CEOing Ep 279: Calculus & Algebra Features for WL 12.1. $\endgroup$
    – Domen
    May 17 at 16:16

1 Answer 1

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The definition of Asymptotic can be accessed by GeneralUtilities`PrintDefinitionsLocal, and it seems that part of its code actually depends on Series.

For example, under the context Asymptotics`AsymptoticDump` , the dependency is like

Asymptotic>ccAsymptotic>iAsymptotic>UnivariateAsymptotic>UnivariatePowerSeries

and

Asymptotics`AsymptoticDump`UnivariatePowerSeries//GeneralUtilities`PrintDefinitionsLocal

gives

...
resser = SeriesCoefficient[f, {x, x0, n}]
...

This can also be seen by

On[Series::esss]
Block[{Quiet},
Asymptotic[Exp[-1/x], {x, 0, 1}]
]

enter image description here

and as suggested by @MichaelE2,

Trace[Asymptotic[Exp[-1/x],{x,0,1}],_Series,TraceInternal->True]

(*{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{Series[x,{x,0,1}]}}}}}}}}}}}}}},{{{{{{{Series[E^(-1/x),{x,0,1}]}}}}}}}}}}}}}}}}}}}}}}}}}}*)
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9
  • $\begingroup$ How do you explain the difference between Asymptotic[Log[a*x], {x, Infinity, 2}, Assumptions -> a > 0] which results in Log[a x] and Series[Log[a*x], {x, Infinity, 2}, Assumptions -> a > 0] // Normal which outputs Log[a] + Log[x] (in 13.2.1)? $\endgroup$
    – user64494
    May 16 at 19:04
  • $\begingroup$ @Lacia, The series commands may also be seen in Trace[ Asymptotic[Exp[-1/x], {x, 0, 1}], _Series, TraceInternal -> True] $\endgroup$
    – Michael E2
    May 17 at 0:26
  • $\begingroup$ @user64494 maybe because there are lots of conditions and preprocess before SeriesCoefficient in the code of Asymptotic. $\endgroup$
    – Lacia
    May 17 at 5:44
  • $\begingroup$ @MichaelE2 yes, I agree. It surprises me that the source code of Asymptotic is not fully protected. $\endgroup$
    – Lacia
    May 17 at 5:45
  • $\begingroup$ @Lacia: Sorry, I dislike speculations. Trace[Asymptotic[Log[a*x], {x, 0, 1}, Assumptions -> a > 0], _Series, TraceInternal -> True] results in {{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{Series[x,{x,0,1}]}}}}}}}}}}}}}},{{{{{{{Series[Log[a x],{x,0,1}]}}}}}}}}}}}}}}}}}}}}}}}}}} suggesting that no assumption is taken into account. $\endgroup$
    – user64494
    May 17 at 6:33

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