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I claim, for $\operatorname{artanh}(\rho) = \frac{1}{2} \ln\left(\frac{1+\rho}{1-\rho}\right)$, i.e., the inverse hyperbolic tangent function, the following holds approximately under assumptions given below: \begin{align} y & = 1 + b_1\operatorname{artanh}(\rho) + b_2\operatorname{artanh}(\rho)^2 \\[5pt] {} & = 1+a_1\rho+a_2\rho^2+a_3\rho^3 +\cdots+a_j\rho^j+ \cdots \end{align}

for $a_j$ in $[0, 1]$. In other words, the given quadratic may be approximated as the sum of a power series. The power series converges to a non-negative constant, as proven by Goel and Ramalingam (1989, pp. 38-39).

Specifically, let $(X, Y)$ be a random sample of size $n$ from a bivariate normal population with correlation parameter $\rho$. Let $R(\cdot)$ be a ranking function that takes a real number sequence and replaces each real number with a positive integer indicating its position in the sorted sequence. Then G&R showed the expected number of fixed points between $R(X)$ and its permutation $R(Y)$ is the sum of the above power series.

The polynomial is a regression model fitted to data generated via Monte Carlo simulation. For my simulation parameters $(n = 30$ and $60, \rho = .1, .2, .3, .4, .5, .6, .7, .8, .9)$, fit is near perfect. It can be shown, however, that the approximation weakens for increasing $n$ and fails completely as $n$ goes to infinity, where all coefficients of the power series equal 1. [I assume the asymptotics are responsible for the inconsistencies identified by Carlo Beenakker.]

Now, given that context, my question is: can I relate $(b_1, b_2)$ to $a_j$, so that knowing the one allows me to compute the other? If this is not solvable for the given information, is it solvable in the special case of the power series being a geometric series, so that all $a_j$ are equal? I'm aware there is a relationship between Taylor series and power series that may be exploitable to express a polynomial as a power series, but not sure how or whether it relates here.

[Edit: I have corrected the post to say the expression is approximate and definitely fails under certain conditions. I've also added detail to the context to explain the distributional assumptions.]

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    $\begingroup$ I understand your question as asking for a relation between the coefficients $a_1,a_2,a_3,\ldots$ and $b_1,b_2$; I gave this relation in the answer box; I do not understand other statements in the post: the series does not terminate, there is an infinite series of $a$ coefficients, and the sign is not positive, it alternates. $\endgroup$
    – Carlo Beenakker
    Commented May 23 at 19:19
  • $\begingroup$ Thanks, @CarloBeenakker! I appreciate your proving the conjecture as originally stated to be false. I've corrected/revised the post to narrow the parameters/assumptions accordingly, and to correctly characterize the relationship as approximate and limited to certain parameter values. That the signs alternate when my empirical model holds approximately is a fascinating insight. It implies the true model looks very different! $\endgroup$
    – virtuolie
    Commented May 23 at 21:36
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    $\begingroup$ I do not see anything in the question that is defined as a quadratic polynomial. Also, the word "approximately" is not a technical term. $\endgroup$
    – Daniel Asimov
    Commented May 24 at 2:43
  • $\begingroup$ To preclude intuitive language in conjectures is to deny the role of intuition in the development of theorems. en.wikipedia.org/wiki/Proofs_and_Refutations $\endgroup$
    – virtuolie
    Commented May 24 at 20:09
  • $\begingroup$ express a polynomial as a power series, <--- a polynomial is a power series, though, one that only has finitely many non-zero coefficients. Also, all continuous functions are uniformly approximated on a compact interval by a polynomial, this is the Weierstrass approximation theorem. Finally, what's a reference for "G&R" more specific than a year? $\endgroup$
    – David Roberts ♦
    Commented Jun 1 at 17:35

1 Answer 1

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[This answers the question as originally stated. It has now been changed.]

Given the series expansion $$1+b_1\arctan x+b_2 \arctan^2 x=1+\sum_{k=1}^\infty a_k x^k$$ one has the relationships $$a_{2k+1}=(-1)^k\frac{b_1}{2k+1},$$ $$a_{2k}=(-1)^k b_2\sum_{i=0}^k \frac{1}{(2i+1)(2k-2i+1)}$$ $$\qquad=(-1)^k \frac{b_2}{4k+4}\left(H_{-k-\frac{3}{2}}+H_{k+\frac{1}{2}}+4\ln 2\right),$$ with $H_n$ the harmonic number.

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