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The function to be expanded is: $$f(x) = \frac{x}{x-a\sqrt{1-e^{bx^2}}}$$ I would like to expand it up to second order terms, something like $c_0+c_1 x+c_2 x^2$. $x$ is not a relatively small quantity. $a$ and $b$ are real numbers (negative values).

Related information: this is a variant of getting formulae (14) (15) from (13) in Laskin's paper doi: 10.1117/12.2063388, and the author uploaded preprint here: https://www.researchgate.net/publication/266385491_Designing_refractive_beam_shapers_via_aberration_theory

I tried to get the Maclaurin term but the derivatives are cumbersome (mathDF.com may help with getting derivatives) and the square root part is annoying (didn't get through it by L'Hôpital's rule).

Benjamin Wang
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Jack2023
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  • This might help you manipulate the square root of the power series in the denominator: https://math.stackexchange.com/questions/2206822/finding-a-square-root-of-a-power-series-in-terms-of-another-power-series – Ethan Bolker Oct 12 '23 at 13:49
  • You can ask Wolfram Alpha. – Gary Oct 12 '23 at 13:52
  • Dear Ethan and Gary, thank you for your help of prompt edits as I am new here! And thank you for your references! I tried both but I still got stuck. Please take a look at the ELEGANT form of formulae (14)(15) in the preprint link. The authors didn't provide those by parabolic approximation (formula (13) is equal to a parabolic function at r=w0). It really draws me crazy how they went through the expansion... – Jack2023 Oct 12 '23 at 14:33
  • If $x$ isn't small then it doesn't make sense to expand in positive powers of $x$. – Ian Oct 12 '23 at 14:36
  • Dear Ian, thank you for your comment. This is related to an optics problem, in which x isn't small because rays are not paraxial. Honestly I feel like it's impossible to expand f(x) into Maclaurin terms. However, please take a look at the preprint link provided where the authors somehow made the expansion and the result is beautiful. – Jack2023 Oct 12 '23 at 15:05
  • @Ian@Gary@Ethan There is an answer on Quora by Jan link and I am verifying his proof at the moment. – Jack2023 Oct 12 '23 at 16:57

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To avoid sign problems, I think that it is easier to consider $$f(x) = \frac{x}{x+a\sqrt{1-e^{-bx^2}}}$$ where $a$ and $b$ are positive and above all that $x$ is positive.

Using steps $$1-e^{-bx^2}=b x^2-\frac{b^2 }{2}x^4+\frac{b^3 }{6}x^6+O\left(x^{8}\right)$$ Using binomial expansion $$\sqrt{1-e^{-bx^2}}=b^{1/2} x-\frac{ b^{3/2}}{4} x^3+\frac{5b^{5/2}}{96} x^5+O\left(x^6\right)$$ $$x+a\sqrt{1-e^{-bx^2}}= \left(1+a \sqrt{b}\right)x-\frac{a b^{3/2}}{4} x^3 +\frac{5 a b^{5/2}}{96} x^5+O\left(x^6\right)$$ Long division leads to $$f(x)=\frac{1}{1+a \sqrt{b}}+\frac{a b^{3/2} }{4 \left(a \sqrt{b}+1\right)^2}x^2+\frac{a b^{5/2} \left(a \sqrt{b}-5\right)}{96 \left(a \sqrt{b}+1\right)^3}x^4+O\left(x^5\right)$$

Claude Leibovici
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