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I have a function f($x$) given by the expression

$$f (x) = \frac{\left(1+x\left[1-\sqrt{1+x^2}\right]\right)^2-x+x^3\left[1-\sqrt{1+x^2}\right]^2}{1+x^2\left(1-\sqrt{1+x^2}\right)^2}$$

and would like to expand it for two limits of $x$: $x \gg 1$ and $x \ll 1$. From the posts I've read here, there seems to be a simple 'command' to do so in Mathematica:

$$x \gg 1 \longrightarrow \text{Series}\left[f(x),\{x,\text{Infinity},4\}\right]$$

$$x \ll 1 \longrightarrow \text{Series}\left[f(x),\{x,0,4\}\right]$$

Since I am quite new to this, I am a bit confused about how this works. Why is it that for $x \gg 1$ we consider Infinity, and for the other limit we consider $x = 0$? From the documentation, Series 'generates a power series expansion for f about the point x=x0', and I cannot understand why this is the same as our situation.

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  • $\begingroup$ Terms like expand, limit, infinity, $0$ etc. are all vague. Mathematics is the art of using well defined (perfectly clear) concepts. $\endgroup$
    – Artes
    Commented Jun 2, 2021 at 14:33
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    $\begingroup$ People here generally like users to post code as Mathematica code instead of just images or TeX, so they can copy-paste it. It makes it convenient for them and more likely you will get someone to help you. You may find the meta Q&A, How to copy code from Mathematica so it looks good on this site, helpful $\endgroup$
    – Michael E2
    Commented Jun 2, 2021 at 14:38
  • $\begingroup$ @MichaelE2 Thank you very much for the tip! $\endgroup$
    – miniplanck
    Commented Jun 2, 2021 at 14:40
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    $\begingroup$ How are $x \gg 1$ and $x \ll 1$ defined in your field or problem? I believe translating that to the language of Series[] or Asymptotic[] is the key to the choice of the point for expansion. In terms of asymptotic expansions, Wikipedia is typically opaque, but for "the point x=0," take in the Wikipedia article $L = x_0$ and for finite $x_0$, $\varphi_n(x) = (x-x_0)^n$ 'in the limit $x \rightarrow x_0$'; and in the limit $x_0\rightarrow\infty$, $\varphi_n(x)=x^{-n}$. $\endgroup$
    – Michael E2
    Commented Jun 2, 2021 at 15:13
  • $\begingroup$ This is better suited for math.SE (it's about the math really, not the software). $\endgroup$
    – Daniel Lichtblau
    Commented Jun 2, 2021 at 22:27

2 Answers 2

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Maybe this post answers your question: https://math.stackexchange.com/questions/51770/taylor-expansion-at-infinity Expansion at large x is equivalent to expansion at x-->1/x for small x.

f[x_] := ((1 + x (1 - Sqrt[1 + x^2]))^2 - x + 
    x^3 (1 - Sqrt[1 + x^2])^2)/(1 + x^2 (1 - Sqrt[1 + x^2])^2)

Normal[Series[f[x], {x, Infinity, 4}]]

Normal[Series[f[1/x], {x, 0, 4}]] /. x -> 1/x

(* Both give you the output: 1 - 5/x^4 - 4/x^3 - 2/x^2 + x *)
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  • $\begingroup$ I think it does answer my question. Much appreciated! $\endgroup$
    – miniplanck
    Commented Jun 2, 2021 at 14:41
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$\begingroup$ 
$Version

(* "12.3.0 for Mac OS X x86 (64-bit) (May 10, 2021)" *)

Clear["Global`*"]

x << 1 and x >> 1 represent Asymptotic (v12.1 or later) behavior.

For x << 1

Asymptotic[f[x], {x, 0, 4}]

(* 1 - x - x^3 *)

In earlier versions

Series[f[x], {x, 0, 4}] // Normal

(* 1 - x - x^3 *)

For x >> 1

Asymptotic[f[x], {x, Infinity, 4}]

(* 1 - 5/x^4 - 4/x^3 - 2/x^2 + x *)

In earlier versions

Series[f[x], {x, Infinity, 4}] // Normal

(* 1 - 5/x^4 - 4/x^3 - 2/x^2 + x *)
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