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I wonder if there is an explicit formula for the Maclaurin expansion of $\frac{x^2}{1 - x \cot x}$. We know an explicit formula for $1- x \cot x$.

Due to the continued fraction formula for $\tan x$, we know that all of the coefficients after the first are negative.

$\bf{Added:}$ I was lead to this question in trying to prove that in the Maclaurin expansion of $ \frac{x^2}{1 - x \cot x} + \frac{3}{5} ( 1 - x \cot x) - 2 $ all of the coefficients are positive. Since we have formulas for the expansion of the second term, we are interested in explicit formulas for the expansion of the first term.

$\bf{Added:}$ We have the continued fraction
$$\frac{x^2}{1 - x \cot x} = 3 - \frac{x^2}{ 5 - \frac{x^2}{ 7 - \cdots} } $$ following easily from the continued fraction for $\tan x$.

orangeskid
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    Wolfram gives this result. – John Wayland Bales May 11 '22 at 02:30
  • @John Wayland Bales: Thank you for your interest. Yes, I think we can calculate explicitly the coefficients up to a level-- but I wondered if there was an explicit formula like for $1- x \cot x$. – orangeskid May 11 '22 at 03:36
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    For $1 - x \cot x$ you have Bernoulli numbers in the coefficients. If you think Bernoulli numbers count as "explicit formula", then you may simply name the coefficients of $\frac{x^2}{1 - x\cot x}$: call them "orange numbers" and you get an explicit formula. The real thing is not whether a formula is explicit or not, but how many properties you can prove for these coefficients. – WhatsUp May 11 '22 at 05:53
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    There was an answer to this question in Section 6 of the paper: F. T. Howard, Properties of the van der Pol numbers and polynomials, J. Reine Angew. Math., vol. 260, pp. 35--46, 1973; available online at https://doi.org/10.1515/crll.1973.260.35. – qifeng618 May 27 '23 at 04:41
  • @WhatsUp At the site https://math.stackexchange.com/a/4254493, there are many “explicit” formulas for the Bernoulli numbers. I am sure that the Bernoulli numbers can be expressed by many closed-form expressions. – qifeng618 Jun 06 '23 at 22:10
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    There are more new results related to this question in the newly-published paper: Zhen-Ying Sun, Bai-Ni Guo, and Feng Qi, Determinantal expressions, identities, concavity, Maclaurin power series expansions for van der Pol numbers, Bernoulli numbers, and cotangent, Axioms 12 (2023), no. 7, Article 665, 14 pages; available online at https://doi.org/10.3390/axioms12070665. – qifeng618 Jul 06 '23 at 01:32
  • @qifeng618: Amazing results, it all seems so very interesting! Thanks! – orangeskid Jul 06 '23 at 06:08

1 Answers1

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Using Bessel functions, we find \begin{align*} \frac{{x^3 }}{{1 - x\cot x}} & = 3 - x\frac{{J_{5/2} (x)}}{{J_{3/2} (x)}} = \frac{3}{2} + x\frac{{J'_{3/2} (x)}}{{J_{3/2} (x)}} = 3 - 2x^2 \sum\limits_{k = 1}^\infty {\frac{1}{{1 - (x/j_{3/2,k}^2 )^2 }}} \\ & = 3 - 2x^2 \sum\limits_{k = 1}^\infty {\sum\limits_{n = 0}^\infty {\frac{1}{{j_{3/2,k}^{2n} }}x^{2n} } } = 3 - 2x^2 \sum\limits_{n = 0}^\infty {\left( {\sum\limits_{k = 1}^\infty {\frac{1}{{j_{3/2,k}^{2n} }}} } \right)x^{2n} } \\& = 3 - 2x^2 \sum\limits_{n = 0}^\infty {\sigma _{2n} \!\left( \tfrac{3}{2} \right)x^{2n} }, \end{align*} where $j_{3/2,k}$ denotes the $k$th positive zero of $J_{3/2}$ and $\sigma_n$ is the Rayleigh function of order $n$. If $n\geq 1$, we can write $$ \sigma _{2n} \!\left( {\tfrac{3}{2}} \right) = ( - 1)^{n - 1} 3 \cdot 2^{2n - 1} \frac{{V_{2n} }}{{(2n)!}}, $$ where $V_n$ is the $n$th van der Pol number. See this paper for properties of these numbers, including recurrence relations. In particular, your generating function is equation $(d)$ in Section $1$.

Gary
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    Thank you! Very informative. Great! – orangeskid May 13 '22 at 07:16
  • In Section 6 of the paper "F. T. Howard, Properties of the van der Pol numbers and polynomials, J. Reine Angew. Math., vol. 260, pp. 35--46, 1973; available online at https://doi.org/10.1515/crll.1973.260.35", the following series expansion was established: \begin{equation}\label{Howard-COT-Ser-1973} \frac{z^2}{1-z\cot z}=3-\frac{z^2}{5}+3\sum_{k=2}^{\infty}(-1)^k2^{2k}V_{2k}\frac{z^{2k}}{(2k)!}, \end{equation} where $V_{2k}$ denotes the van der Pol numbers. – qifeng618 Jun 05 '23 at 13:19
  • @qifeng618 Isn't it exactly what I am citing in my answer? – Gary Jun 05 '23 at 14:01
  • @Gary Did you mention “Section 6”? I think you didn’t. Is your answer same to the proof in “Section 6”? Not the same, I think. I am misled by your answer. – qifeng618 Jun 06 '23 at 21:27
  • @qifeng618 I came up with the answer myself and then found the paper by Howard. I am citing it so that people can learn more properties of these coefficients if they are interested. I wonder why this answer has been downvoted. – Gary Jun 06 '23 at 23:40
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    @Gary Several best answers given by me were also downvoted: Not everybody is an expert at the problem and its answers. As you wrote, your answer is slightly different from the one in “Section 6” of Howard’s paper. Due to your mention of Howard’s paper, I read and benefitted from the paper by Howard once again, thanks Gary. – qifeng618 Jun 08 '23 at 00:28
  • This answer and related comments are included in the paper: Zhen-Ying Sun, Bai-Ni Guo, and Feng Qi, Determinantal expressions, identities, concavity, Maclaurin power series expansions for van der Pol numbers, Bernoulli numbers, and cotangent, Axioms vol. 12 (2023), no. 7, Article 665, 14 pages; available online at https://doi.org/10.3390/axioms12070665. – qifeng618 Sep 01 '23 at 18:19