Introduction

Transformations on the real number line of the type

$$\begin{aligned} \phi (x) =x-\lambda - \frac{b}{x- \mu }, \end{aligned}$$
(1)

with \(\lambda , \mu , b\) real numbers with \(b>0\), called doubling transformations, have the interesting property that for any absolutely integrable function \(F(x)\) on \(\mathbb R\), we have

$$\begin{aligned} \int _{-\infty }^{\infty } F(\phi (x)) \ dx = \int _{- \infty }^{\infty } F(x) \ dx. \end{aligned}$$
(2)

This result can be found for example in Wilson’s Advanced Calculus [4, p. 386]; a proof is provided in Section “Proof of the Integral Formula” for the reader’s convenience. The name doubling comes from the fact that \(\phi (x)\) is a two-to-one function on \(\mathbb R -\{\mu \}\), and thus the graph of \(F(\phi (x))\) is a doubling of the graph of \(F(x)\). By taking compositions of doubling functions we generate new rational functions satisfying (2). Our interest in this paper is to characterize all such rational functions. This extends work started by Hagler in [1, 2] and [3] where doubling transformation were used in the study of orthogonal systems of polynomials such as those of Jacobi, Hermite and Laguerre.

We shall establish criteria for recognizing when a given rational function is a composition of \(n\) doubling transformations. For example, the following are all examples of compositions of two doubling transformations,

$$\begin{aligned}&\frac{(x-a)(x+a^2)(x-a^3)(x+a^4)}{x(x-a^2)(x+a^3)}, \quad (a>1),\end{aligned}$$
(3)
$$\begin{aligned}&\frac{(x^2-a^2)(x^2-a^8)}{x(x+a^2)(x-a^3)}, \quad (a>1),\end{aligned}$$
(4)
$$\begin{aligned}&\frac{(x^2-a^2)(x^2-b^2)}{x(x^2-ab)}, \quad (0<a<b); \end{aligned}$$
(5)

see Examples 9, 10 and 11. Thus, by (2) we obtain integral formulae such as

$$\begin{aligned} \sqrt{\pi } = \int _{-\infty }^\infty e^{-x^2} \ dx = \int _{-\infty }^{\infty } e^{-\left[ \frac{(x-a)(x+a^2)(x-a^3)(x+a^4)}{x(x-a^2)(x+a^3)}\right] ^2} \ dx, \end{aligned}$$

for \(a>1\) and

$$\begin{aligned} \pi = \int _{-\infty }^{\infty } \frac{\ dx}{1+x^2} = \int _{-\infty }^{\infty }\frac{x^2(x^2-ab)^2 \ dx}{x^2(x^2-ab)^2 + (x^2-a^2)^2(x^2-b^2)^2}, \end{aligned}$$

for \(0<a<b\).

In general, for compositions of two doubling transformations we establish the following characterization.

Theorem 1

A rational function is a composition of two doubling transformations if and only if it is of the form \(f_2(x- \mu )\) for some real number \(\mu \) and rational function

$$\begin{aligned} f_2(x)=\frac{\prod _{i=1}^4 (x- \theta _i^{(2)})}{x\prod _{i=1}^2(x-\theta _i^{(1)})}, \end{aligned}$$

where the \(\theta _i^{(j)}\) are real numbers satisfying

$$\begin{aligned}&\theta _4^{(2)}<\theta _2^{(1)}< \theta _2^{(2)}<0<\theta _3^{(2)}<\theta _1^{(1)}< \theta _1^{(2)},\\&\theta _{1}^{(1)}\theta _{2}^{(1)}= \theta _{1}^{(2)} \theta _{2}^{(2)}= \theta _{3}^{(2)} \theta _{4}^{(2)}. \end{aligned}$$

Theorem 1 is an immediate consequence of Theorem 7, which establishes both existence and uniqueness criteria for such decompositions. Similar criteria are established for \(n=3\) in Section “The case \(n=3\)” and for general \(n\) in Section “The General Case”. For \(n=3\) we give an explicit family of such functions in Example 12. Writing down explicit families for \(n>3\) is more challenging and we have not endeavored to do so at this point.

Compositions of Doubling Transformations

Suppose that for \(1 \le k \le n\), \(\phi _k\) is a doubling transformation,

$$\begin{aligned} \phi _k(x) = x- \lambda _k - \frac{b_k}{x-\mu _k}, \end{aligned}$$

for some \(\lambda _k, b_k,\mu _k \in \mathbb R\), \(b_k>0\). For any real number \(\mu \), let \(t_{\mu }\) denote the translation \(t_{\mu }(x):= x- \mu \), and \(t_{\mu }^{-1}\) denote its inverse, \(t_{\mu }^{-1}(x) = x+ \mu \). Then we have

$$\begin{aligned} \phi _n \circ \phi _{n-1} \circ \cdots \circ \phi _1&= (\phi _n \circ t_{\mu _n}^{-1})\circ (t_{\mu _n}\circ \phi _{n-1} \circ t_{\mu _{n-1}}^{-1}) \circ \cdots \circ \left( t_{\mu _{2}} \circ \phi _1 \circ t_{\mu _1}^{-1}\right) \circ t_{\mu _1}\\&= \rho _n \circ \rho _{n-1} \circ \cdots \circ \rho _1 \circ t_{\mu _1} \end{aligned}$$

say, where \(\rho _k\) is a doubling transformation with pole at 0, \(1 \le k \le n\). Thus we have the following lemma.

Lemma 2

Any function that can be expressed as a composition of doubling transformations can be expressed in the form \(f(x-\mu )\) for some \(\mu \in \mathbb R\), where \(f\) is a composition of doubling transformations having poles at 0.

Henceforth, we shall assume that the poles \(\mu _k\) in all of our doubling transformations are zero. The \(\phi _k\) then take the form

$$\begin{aligned} \phi _k(x):= x- \lambda _{k}- \frac{b_{k}}{x}= \frac{(x-\alpha _k)(x- \beta _k)}{x}, \end{aligned}$$
(6)

where \(\alpha _k>0,\beta _k<0\) are real numbers given by

$$\begin{aligned} \alpha _k,\beta _k = \frac{\lambda _{k}\pm \sqrt{\lambda _{k}^2 +4b_{k}}}{2}, \end{aligned}$$

and satisfy the basic relations

$$\begin{aligned}&\alpha _k+\beta _k = \lambda _{k},\end{aligned}$$
(7)
$$\begin{aligned}&\alpha _k\beta _k= -b_{k}. \end{aligned}$$
(8)

Set

$$\begin{aligned} f_n(x)= \phi _n \circ \phi _{n-1} \circ \cdots \circ \phi _1(x). \end{aligned}$$
(9)

Each \(\phi _k\) is strictly increasing on \((-\infty ,0)\) and on \((0, \infty )\), and is a 2-to-1 mapping on the extended real number line \(\mathbb R \cup \{\infty \}\), where we define \(\phi _k(0) = \infty \), \(\phi _k(\infty ) = \infty \). Thus \(f_n\) is a \(2^n\)-to-1 mapping on the extended real number line, strictly increasing on the open intervals where it is defined, having \(2^n-1\) distinct real simple poles, and can be expressed in reduced form as a rational function

$$\begin{aligned} f_n(x) = \frac{p_n(x)}{q_n(x)}, \end{aligned}$$

for some monic polynomials \(p_n(x),q_n(x)\) of degrees \(2^n\), \(2^n-1\) respectively. The graph of \(f_n(x)\) consists of \(2^n\) connected components \(\mathcal C_r^{(n)}\), \(1 \le r \le 2^n\), each containing a unique zero \(\theta _r^{(n)}\) of \(f_n(x)\). We call \(\mathcal C_r^{(n)}\) the component of the graph corresponding to \(\theta _r^{(n)}\). Thus we have,

$$\begin{aligned} p_n(x) = \prod _{r=1}^{2^n} (x - \theta _r^{(n)}). \end{aligned}$$

Since

$$\begin{aligned} f_n(x) = \phi _n(f_{n-1}(x)) = f_{n-1}(x)-\lambda _n -b_n/f_{n-1}(x), \end{aligned}$$

we see that the poles of \(f_n\) are just the zeros of \(f_{n-1}\) together with the poles of \(f_{n-1}\). From this observation and the fact that \(p_n(x), q_n(x)\) are monic, we see that \(f_1(x)= \frac{p_1(x)}{x}\), \(f_2(x)= \frac{p_2(x)}{xp_1(x)}\), and by induction that,

$$\begin{aligned} f_n(x) = \frac{p_n(x)}{xp_1(x) p_2(x) \cdots p_{n-1}(x)}. \end{aligned}$$

The zeros of \(f_n(x)\) can be put into positive-negative pairs inductively as follows. For \(n=1\) we already have the positive-negative pair \((\theta _1^{(1)}, \theta _2^{(1)})=(\alpha _1, \beta _1)\). We will use odd subcripts for the positive zeros and even subscripts for the negative zeros. Suppose that we are given a pairing of the zeros of \(f_{n-1}(x)\), say \(( \theta _{2k-1}^{(n-1)}, \theta _{2k}^{(n-1)})\), \(1 \le k \le 2^{n-2}\). Let \(\mathcal C_{2k-1}^{(n-1)}, \mathcal C_{2k}^{(n-1)}\) be the corresponding components of the graph of \(f_{n-1}(x)\). Note that \(\mathcal C_{2k-1}^{(n-1)}\) lies in the right half-plane (\(x>0\)), while \(\mathcal C_{2k}^{(n-1)}\) lies in the left half-plane (\(x<0\)). Since \(f_n(x)=\phi _n(f_{n-1}(x))\), the zeros of \(f_n(x)\) are just the solutions of the equations \(f_{n-1}(x)= \alpha _n\), \(f_{n-1}(x) = \beta _n\). Let \(\theta _{2k-1}^{(n)}\) be the solution of \(f_{n-1}(x) = \alpha _n\) with \((x,\alpha _n)\) on \(\mathcal C_{2k-1}^{(n-1)}\) and \(\theta _{2k}^{(n)}\) the solution of \(f_{n-1}(x) = \alpha _n\) with \((x, \alpha _n)\) on \(\mathcal C_{2k}^{(n-1)}\). Then \((\theta _{2k-1}^{(n)}, \theta _{2k}^{(n)})\) is a uniquely defined pair of zeros of \(f_n(x)\). Similarly, let \((\theta _{2^{n-1}+2k-1},\theta _{2^{n-1}+2k})\) be the positive negative pair corresponding to the solutions of \(f_{n-1}(x) = \beta _n\).

In particular, we see that for \(n=2\),

$$\begin{aligned} \theta _4^{(2)}<\theta _2^{(1)}<\theta _2^{(2)} <0<\theta _3^{(2)}< \theta _1^{(1)}< \theta _1^{(2)}. \end{aligned}$$
(10)

Thus, between any two consecutive zeros of \(q_2(x)=xp_1(x)\) there is a zero of \(p_2(x)\). Also, to the right of the largest zero of \(q_2(x)\) and to the left of the smallest zero of \(q_2(x)\) there is a zero of \(p_2(x)\). Similarly, for \(n=3\) one obtains from the construction above that,

$$\begin{aligned} \theta _8^{(3)}&<\theta _4^{(2)}<\theta _4^{(3)}<\theta _2^{(1)}< \theta _6^{(3)}<\theta _2^{(2)}<\theta _2^{(3)}<0\nonumber \\ 0&<\theta _7^{(3)}<\theta _3^{(2)}<\theta _3^{(3)}<\theta _1^{(1)}<\theta _5^{(3)} <\theta _1^{(2)}<\theta _1^{(3)}, \end{aligned}$$
(11)

and we see the same type of splicing of zeros. Continuing this process we have the following lemma.

Lemma 3

The Splicing Principle. For \(n \ge 2\), between any two consecutive zeros of \(q_n(x)\) (with respect to the standard ordering on \(\mathbb R\)), as well as to the right of the largest zero and to the left of the smallest zero of \(q_n(x)\), there is a unique zero of \(p_{n}(x)\). Moreover, we have the following consecutive triples of zeros of \(q_n(x)\):

$$\begin{aligned} \theta _{2^{n-1}+2k-1}^{(n)}<\theta _{2k-1}^{(n-1)}< \theta _{2k-1}^{(n)}, \quad \quad \theta _{2^{n-1}+2k}^{(n)}<\theta _{2k}^{(n-1)}<\theta _{2k}^{(n)}, \quad 1 \le k \le 2^{n-2}. \end{aligned}$$
(12)

The positive-negative pairs can be defined in a purely algebraic manner as follows. They correspond to the choice of positive-negative signs in the successive applications of the quadratic formula that one would use for calculating the zeros. For instance,

$$\begin{aligned} \theta _1^{(2)}&=\left( \lambda _1 + \alpha _2+ \sqrt{(\lambda _1+\alpha _2)^2 +4b_1}\right) /2\\ \theta _2^{(2)}&=\left( \lambda _1 + \alpha _2- \sqrt{(\lambda _1+\alpha _2)^2 +4b_1}\right) /2\\ \theta _3^{(2)}&=\left( \lambda _1 + \beta _2+ \sqrt{(\lambda _1+\beta _2)^2 +4b_1}\right) /2\\ \theta _4^{(2)}&=\left( \lambda _1 + \beta _2- \sqrt{(\lambda _1+\beta _2)^2 +4b_1}\right) /2. \end{aligned}$$

Fundamental Relations

As noted above, the zeros of \(f_n(x)\) are the solutions of the equations \(f_{n-1}(x)= \alpha _n\), \(f_{n-1}(x) = \beta _n\). We call the former the \(\alpha \) zeros of \(f_n(x)\) and the latter the \(\beta \) zeros of \(f_n(x)\). Let

$$\begin{aligned} f_{n-1}^*(x)&: =\phi _{n-1}^*\circ \phi _{n-2} \circ \cdots \circ \phi _1(x), \\ f_{n-1}^{**}(x)&:= \phi _{n-1}^{**}(x)\circ \phi _{n-2} \circ \cdots \circ \phi _1(x), \end{aligned}$$

where

$$\begin{aligned} \phi _{n-1}^*(x)&:= x- ( \lambda _{n-1}+\alpha _n) - b_{n-1}/x,\\ \phi _{n-1}^{**}(x)&:= x-( \lambda _{n-1}+\beta _n)- b_{n-1}/x. \end{aligned}$$

Let \(\theta _j^{*(n-1)}\), \(\theta _j^{**(n-1)}\) denote the zeros of \(f_{n-1}^*, f_{n-1}^{**}\) respectively. Then the alpha zeros of \(f_n\) are just the zeros of \(f_{n-1}^*\), while the beta zeros of \(f_n\) are the zeros of \(f_{n-1}^{**}\), that is,

$$\begin{aligned} \theta _{2k-1}^{(n)}&= \theta _{2k-1}^{*(n-1)}, \quad \theta _{2k}^{(n)} = \theta _{2k}^{*(n-1)}, \end{aligned}$$
(13)
$$\begin{aligned} \theta _{2^{n-1}+2k-1}^{(n)}&= \theta _{2k-1}^{**(n-1)},\quad \theta _{2^{n-1}+2k}^{(n)} = \theta _{2k}^{**(n-1)}. \end{aligned}$$
(14)

Because \(f_{n-1}^*(x)\) and \(f_{n-1}^{**}(x)\) are functions of type (9), identical to \(f_{n-1}(x)\) except for the value of \(\lambda _{n-1}\), we have

Lemma 4

The Correspondence Principle. Any relationship satisfied by the zeros of \(f_{n-1}(x)\) that has no dependence on \(\lambda _{n-1}\), will be satisfied by the corresponding \(\alpha \)-zeros and \(\beta \)-zeros of \(f_n(x)\).

As an example of this phenomena we give the most basic such relationship.

Lemma 5

The Basic Relationship. For any \(n \ge 1\) and any positive-negative pair \((\theta _{2k-1}^{(n)}, \theta _{2k}^{(n)})\) of zeros of \(f_n(x)\) we have

$$\begin{aligned} \theta _{2k-1}^{(n)}\theta _{2k}^{(n)} = -b_1. \end{aligned}$$

Proof

The relationship holds for \(n=1\) by (8). Since it has no dependence on the \(\lambda _n\) the same relationship holds by induction and the Correspondence Principle for all \(n\). \(\square \)

The next lemma provides a more general class of relationships.

Lemma 6

For any positive integer \(n\), we have

$$\begin{aligned} \textit{(i)} \quad&\sum _{i=1}^{2^n} \theta _i^{(n)} = \lambda _n+2 \lambda _{n-1}+2^2 \lambda _{n-2} +\dots + 2^{n-1} \lambda _{1}.\end{aligned}$$
(15)
$$\begin{aligned} \textit{(ii)} \quad&\sum _{i=1}^{2^{n-1}} \theta _i^{(n)}- \sum _{i=1}^{2^{n-1}} \theta _i^{(n-1)} = \alpha _n \end{aligned}$$
(16)
$$\begin{aligned} \textit{(iii)} \quad&\sum _{i=1}^{2^{n-1}} \theta _{2^{n-1}+i}^{(n)}- \sum _{i=1}^{2^{n-1}} \theta _i^{(n-1)}= \beta _n. \end{aligned}$$
(17)
$$\begin{aligned} \textit{(iv)} \quad&-b_n = \left( \sum _{i=1}^{2^{n-1}} \theta _i^{(n)}- \sum _{i=1}^{2^{n-1}} \theta _i^{(n-1)}\right) \left( \sum _{i=1}^{2^{n-1}} \theta _{2^{n-1}+i}^{(n)}- \sum _{i=1}^{2^{n-1}} \theta _i^{(n-1)}\right) . \end{aligned}$$
(18)

When \(n=1\), the relationship in (iv) is just the basic relationship of Lemma 5.

Proof

The proof of \((i)\) is by induction on \(n\) the case \(n=1\) being the identity \(\alpha _1+\beta _1= \lambda _1\). Suppose the statement is true for \(n\). Then for \(n+1\) the \(\alpha \)-zeros of \(f_{n+1}(x)\) are the solutions of \(f_n(x)= \alpha _{n+1}\), that is, the zeros of \(f_n^*(x)\) where \(f_n^*(x)\) is the same as \(f_n(x)\) with \(\lambda _n\) replaced by \(\lambda _n + \alpha _{n+1}\). Thus by the induction assumption the sum of the \(\alpha \)-zeros is \(\lambda _n+ \alpha _{n+1}+\sum _{i=1}^{n-1} 2^i \lambda _{n-i}.\) Similarly, the sum of the \(\beta \)-zeros of \(f_{n+1}(x)\) is \(\lambda _n+ \beta _{n+1} + \sum _{i=1}^{n-1} 2^i \lambda _{n-i} \). Thus the full sum of zeros of \(f_{n+1}(x)\) is

$$\begin{aligned} \alpha _{n+1}+\beta _{n+1} + 2\lambda _n + 2 \sum _{i=1}^{n-1} 2^i \lambda _{n-i} = \lambda _{n+1} +2 \lambda _n +2^2 \lambda _{n-1} + \dots + 2^{n} \lambda _1. \end{aligned}$$

The sum in (ii) is just the sum of the \(\alpha \)-zeros of \(f_n(x)\) minus the sum of all the zeros of \(f_{n-1}(x)\), which by \((i)\) equals \(\sum _{i=0}^{n-1} 2^i \lambda _{n-i} + \alpha _n - \sum _{i=0}^{n-1} 2^i \lambda _{n-i}= \alpha _n.\) The proof of (iii) is identical, using the \(\beta \)-zeros in place of the \(\alpha \)-zeros. The identity in (iv) follows from (ii), (iii) and the fact that \(\alpha _n\beta _n = -b_n\). \(\square \)

Since the identity in part (iv) of the lemma has no dependence on the \(\lambda _i\), one can invoke the correspondence lemma to produce further relations. We do this in Section “The General Case”, but first we explore the cases \(n=2\) and \(n=3\) in detail.

The case \(n=2\)

Compositions of two doubling transformations are quadrupling transformations, that is, 4-to-1 mappings on the extended real number line. In the notation of the previous section, any such transformation has the form

$$\begin{aligned} f_2(x)=\phi _2 \circ \phi _1(x) = \frac{p_2(x)}{xp_1(x)}=\frac{\prod _{i=1}^4 (x- \theta _i^{(2)})}{x\prod _{i=1}^2(x-\theta _i^{(1)})}, \end{aligned}$$
(19)

where, by the Splicing Principle (Lemma 3), Basic Relationship (Lemma 5) and Lemma 6 (iv) and (i), we have

$$\begin{aligned} \theta _4^{(2)}&<\theta _2^{(1)}< \theta _2^{(2)}<0<\theta _3^{(2)}<\theta _1^{(1)}< \theta _1^{(2)}, \end{aligned}$$
(20)
$$\begin{aligned} -b_1&=\theta _{1}^{(1)}\theta _{2}^{(1)}= \theta _{1}^{(2)} \theta _{2}^{(2)}= \theta _{3}^{(2)} \theta _{4}^{(2)}, \end{aligned}$$
(21)
$$\begin{aligned} -b_2&=\left( \theta _{1}^{(2)} + \theta _{2}^{(2)}-\theta _1^{(1)}- \theta _2^{(1)}\right) \left( \theta _{3}^{(2)} + \theta _{4}^{(2)}-\theta _1^{(1)}- \theta _2^{(1)}\right) , \end{aligned}$$
(22)
$$\begin{aligned} \lambda _1&= \theta _1^{(1)} + \theta _2^{(1)}, \end{aligned}$$
(23)
$$\begin{aligned} \lambda _2&= \theta _1^{(2)}+ \theta _2^{(2)} + \theta _3^{(2)} +\theta _4^{(2)} -2 \theta _1^{(1)}-2\theta _2^{(1)}. \end{aligned}$$
(24)

Thus, the values \(\lambda _1,\lambda _2,b_1,b_2\) are uniquely determined by \(f_2(x)\), and we have established one direction of the following theorem.

Theorem 7

Let \(f(x)=\frac{p(x)}{q(x)}\) be a rational function over \(\mathbb R\), expressed in reduced form with \(q(x)\) monic. Then \(f(x)\) is a composition of two doubling transformations with poles at zero if and only if \(p(x)\) is a monic fourth degree polynomial with distinct real zeros \(\theta _i^{(2)}\), \(1 \le i \le 4\), \(q(x)\) is a third degree polynomial with distinct real zeros \(0, \theta _1^{(1)}, \theta _2^{(1)}\), and the zeros of \(p(x),q(x)\) can be ordered in such a manner that the relations in (20) and (21) hold for some \(b_1 \in \mathbb R\).

To establish the converse part of the theorem we need the following uniqueness lemma.

Lemma 8

For a given set of values \(b_1,b_2,\lambda _1,\lambda _2 \in \mathbb R\), with \(b_1>0, b_2>0\), the system of equations (21), (22), (23), (24) has a unique solution \(\theta _1^{(1)}, \theta _2^{(1)}, \theta _i^{(2)}\), \(1 \le i \le 4\), satisfying (20).

Proof

The values \(\theta _1^{(1)}, \theta _2^{(1)}\) are roots of the quadratic equation \(x^2 -\lambda _1 x-b_1=0\), with \(\theta _1^{(1)}>0>\theta _2^{(1)}\). Put \(X=\theta _1^{(2)}+ \theta _2^{(2)}\), \(Y= \theta _3^{(2)}+ \theta _4^{(2)}\). Then we have

$$\begin{aligned} (X-\lambda _1)(Y- \lambda _1) = -b_2, \quad X+Y-2\lambda _1 = \lambda _2, \quad X>Y. \end{aligned}$$

The first two equations give \(X,Y\) as zeros of a quadratic equation, and the inequality \(X>Y\) then uniquely determines \(X,Y\). Next, the system

$$\begin{aligned} \theta _1^{(2)}\theta _2^{(2)} = -b_1, \quad \theta _1^{(2)}+\theta _1^{(2)} = X, \quad \theta _1^{(2)}> \theta _2^{(2)}. \end{aligned}$$

uniquely determines \(\theta _1^{(2)}, \theta _2^{(2)}\), while the system

$$\begin{aligned} \theta _3^{(2)}\theta _4^{(2)}= - b_1, \quad \theta _3^{(2)} + \theta _4^{(2)} =Y, \quad \theta _3^{(2)}>\theta _4^{(2)}, \end{aligned}$$

uniquely determines \(\theta _3^{(2)}, \theta _4^{(2)}\). \(\square \)

Proof of Theorem 7

Suppose that \(f(x)= \frac{p(x)}{q(x)}\) is a given rational function with \(p(x)\) monic and having zeros \(\theta _1^{(2)}\), \(\theta _2^{(2)}\), \(\theta _3^{(2)}\), \(\theta _4^{(2)}\), and \(q(x)\) monic with zeros 0, \(\theta _1^{(1)}\), \(\theta _2^{(1)}\). Suppose also that the relations in (20) and (21) hold for some \(b_1\in \mathbb R\). Define \(b_2,\lambda _1\) and \(\lambda _2\) as in (22), (23) and (24), and set

$$\begin{aligned} \phi _1(x):=x-\lambda _1- \frac{b_1}{x}, \quad \quad \phi _2(x) :=x -\lambda _2- \frac{b_2}{x}. \end{aligned}$$

By (20) we see that \(b_1>0\) and \(b_2>0\) and so \(\phi _1, \phi _2\) are doubling transformations. The zeros and poles of \(\phi _2 \circ \phi _1\) satisfy the relations (20), (21), (22), (23) and (24). Thus, by Lemma 8, the zeros and poles of \(\phi _2 \circ \phi _1\) must be the values \(\theta _1^{(1)}, \theta _2^{(1)}, \theta _i^{(2)}\), \(1 \le i \le 4\). Since \(\phi _2 \circ \phi _1\) is a ratio of monic polynomials, we have \(\phi _2 \circ \phi _1 (x)= f(x)\). \(\square \)

In the next three examples, we give families of quadrupling transformations having particularly nice sets of zeros and poles. The strategy for constructing such examples is to find a set of real numbers satisfying (20) and (21).

Example 9

Let \(a>1\). Observing that \(-a^4<-a^3<-a^2<0<a<a^2<a^3\) we put

$$\begin{aligned} \theta _1^{(2)}=a^3,\quad \theta _2^{(2)}=-a^2,\quad \theta _3^{(2)} = a,\quad \theta _4^{(2)} = -a^4,\quad \theta _1^{(1)}= a^2,\quad \theta _2^{(1)} = -a^3. \end{aligned}$$

Then, with \(b_1= a^5\), \(b_2= -2a^3(a-1)^2(a^2+1)\), relations (21) and (22) hold. Defining \(\lambda _1=a^2-a^3\), \(\lambda _2= -a(a-1)^3\), we have

$$\begin{aligned} f_2(x) = \phi _2\circ \phi _1(x) = \frac{(x-a)(x+a^2)(x-a^3)(x+a^4)}{x(x-a^2)(x+a^3)}. \end{aligned}$$

Example 10

Let \(a>1\). Observing that \(-a^4<-a^2<-a<0<a<a^3<a^4\), we put

$$\begin{aligned} \theta _4^{(2)}=-a^4,\quad \theta _3^{(2)}= a,\quad \theta _2^{(2)} = -a,\quad \theta _1^{(2)} = a^4,\quad \theta _2^{(1)}= -a^2,\quad \theta _1^{(1)} = a^3. \end{aligned}$$

Then we have \(b_1= a^5\), \(b_2= a^2(a^4-1)(a^2-1)\), \(\lambda _1=a^3-a^2\), \(\lambda _2=-2(a^3-a^2)\), and

$$\begin{aligned} f_2(x) = \phi _2 \circ \phi _1 (x) = \frac{(x^2-a^2)(x^2-a^8)}{x(x+a^2)(x-a^3)}. \end{aligned}$$

Example 11

Let \(0<a<b\), so that \(-b<-\sqrt{ab}<-a<0<a<\sqrt{ab}<b\). Then with \(b_1=ab\), \(b_2 =(a-b)^2\), \(\lambda _1=\lambda _2=0\) we have

$$\begin{aligned} f_2(x)= \phi _2 \circ \phi _1 (x) = \frac{(x^2-a^2)(x^2-b^2)}{x(x^2-ab)}. \end{aligned}$$

The Case \(n=3\).

Let

$$\begin{aligned} f_3(x) = \phi _3 \circ \phi _2 \circ \phi _1 (x)= \frac{\prod _{i=1}^8 (x- \theta _i^{(3)})}{ x \prod _{i=1}^2 (x- \theta _i^{(1)}) \prod _{i=1}^4 (x- \theta _i^{(2)})}. \end{aligned}$$
(25)

By the Splicing principle,

$$\begin{aligned} \theta _8^{(3)}&<\theta _4^{(2)}<\theta _4^{(3)}<\theta _2^{(1)}< \theta _6^{(3)}<\theta _2^{(2)}<\theta _2^{(3)}<0 \nonumber \\ 0&<\theta _7^{(3)}<\theta _3^{(2)}<\theta _3^{(3)}<\theta _1^{(1)} <\theta _5^{(3)}<\theta _1^{(2)}<\theta _1^{(3)}. \end{aligned}$$
(26)

By the basic relationship,

$$\begin{aligned} -b_1=\theta _{1}^{(3)} \theta _{2}^{(3)}=\theta _{3}^{(3)} \theta _{4}^{(3)}= \theta _{5}^{(3)} \theta _{6}^{(3)}=\theta _{7}^{(3)} \theta _{8}^{(3)}= \theta _{1}^{(2)} \theta _2^{(2)} = \theta _3^{(2)} \theta _4^{(2)} = \theta _1^{(1)}\theta _2^{(1)}. \end{aligned}$$
(27)

By Lemma 6 (iv) together with the Correspondence Principle,

$$\begin{aligned} -b_2&= \left( \theta _1^{(3)}+\theta _2^{(3)} - \theta _1^{(1)}-\theta _2^{(1)}\right) \left( \theta _3^{(3)}+ \theta _4^{(3)} - \theta _1^{(1)} - \theta _2^{(1)}\right) \nonumber \\&=\left( \theta _5^{(3)}+\theta _6^{(3)} - \theta _1^{(1)}-\theta _2^{(1)}\right) \left( \theta _7^{(3)}+ \theta _8^{(3)} - \theta _1^{(1)} - \theta _2^{(1)}\right) \nonumber \\&= \left( \theta _{1}^{(2)} + \theta _{2}^{(2)}-\theta _1^{(1)}- \theta _2^{(1)}\right) \left( \theta _{3}^{(2)} + \theta _{4}^{(2)}-\theta _1^{(1)}- \theta _2^{(1)}\right) , \end{aligned}$$
(28)

and again by Lemma 6 (iv),

$$\begin{aligned} -b_3&=\left( \theta _1^{(3)}+\theta _2^{(3)}+\theta _3^{(3)} +\theta _4^{(3)} - \theta _1^{(2)}- \theta _2^{(2)} - \theta _3^{(2)} - \theta _4^{(2)}\right) \nonumber \\&\quad \quad \cdot \left( \theta _5^{(3)}+\theta _6^{(3)}+\theta _7^{(3)} +\theta _8^{(3)} - \theta _1^{(2)}- \theta _2^{(2)} - \theta _3^{(2)} - \theta _4^{(2)}\right) . \end{aligned}$$
(29)

By Lemma 6 (\(i\)) the values \(\lambda _1,\lambda _2, \lambda _3\) satisfy

$$\begin{aligned} \lambda _1&= \sum _{i=1}^2 \theta _i^{(1)},\\ \lambda _2&= \sum _{i=1}^4 \theta _i^{(2)} -2 \sum _{i=1}^2 \theta _i^{(1)},\\ \lambda _3&= \sum _{i=1}^8 \theta _i^{(3)} -2 \sum _{i=1}^4 \theta _i^{(2)}- 4 \sum _{i=1}^2 \theta _i^{(1)}. \end{aligned}$$

Again, given any \(\theta _i^{(k)}\) satisfying the relations in (26), (27) and (28), by defining \(b_1,b_2,b_3, \lambda _1, \lambda _2, \lambda _3\) as above we obtain an \(f_3(x)\) of the form (25). Thus we have the analogue of Theorem 7 for the case \(n=3\); see Theorem 13.

Example 12

Let \(a_0=1.839...\) be the real zero of \(x^3-x^2-x-1=0\). Let \(a\) be any real with \(a>a_0\), and set \(\theta _1^{(1)}=1, \theta _2^{(1)}=-1\),

$$\begin{aligned} \theta _1^{(2)}&=a, \quad \theta _2^{(2)}= -\frac{1}{a},\quad \theta _3^{(2)} = \frac{a-1}{a+1}, \quad \theta _4^{(2)} = -\frac{a+1}{a-1}\\ \theta _1^{(3)}&= a^2,\quad \theta _2^{(3)}= -\frac{1}{a^2},\quad \theta _3^{(3)} =\frac{a^2-1}{a^2+1},\quad \theta _4^{(3)} = -\frac{a^2+1}{a^2-1}, \\ \theta _5^{(3)}&= \frac{a^2+1}{a^2-1},\quad \theta _6^{(3)}= - \frac{a^2-1}{a^2+1},\quad \theta _7^{(3)}= \frac{1}{a^2},\quad \theta _8^{(3)}= -a^2. \end{aligned}$$

It is easy to see that (26) holds since, by assumption, \(a^3-a^2-a-1>0\), and that (27) holds with \(b_1=1\). Next, we have

$$\begin{aligned}&\left( \theta _1^{(3)}+\theta _2^{(3)} - \theta _1^{(1)}-\theta _2^{(1)}\right) \left( \theta _3^{(3)}+ \theta _4^{(3)} - \theta _1^{(1)} - \theta _2^{(1)}\right) \\&\quad = \left( a^2-\frac{1}{a^2}\right) \left( \frac{a^2-1}{a^2+1}- \frac{a^2+1}{a^2-1}\right) =-4,\\&\left( \theta _5^{(3)}+\theta _6^{(3)} - \theta _1^{(1)}-\theta _2^{(1)}\right) \left( \theta _7^{(3)}+ \theta _8^{(3)} - \theta _1^{(1)} - \theta _2^{(1)}\right) \\&\quad = \left( \frac{a^2+1}{a^2-1}-\frac{a^2-1}{a^2+1}\right) \left( \frac{1}{a^2} - a^2\right) =-4,\\&\left( \theta _{1}^{(2)} + \theta _{2}^{(2)}-\theta _1^{(1)}- \theta _2^{(1)}\right) \left( \theta _{3}^{(2)} + \theta _{4}^{(2)}-\theta _1^{(1)}- \theta _2^{(1)}\right) \\&\quad = \left( a-\frac{1}{a}\right) \left( \frac{a-1}{a+1}- \frac{a+1}{a-1}\right) =-4, \end{aligned}$$

and so (28) holds with \(b_2=4\). Thus there exist \(\phi _1,\phi _2, \phi _3\) such that

$$\begin{aligned} f_3(x): =\phi _3\circ \phi _2 \circ \phi _1 (x)= \frac{\left( x^2- \frac{1}{a^4}\right) \left( x^2-a^4\right) \left( x^2 - \left( \frac{a^2+1}{a^2-1}\right) ^2\right) \left( x^2 - \left( \frac{a^2-1}{a^2-1}\right) ^2 \right) }{x(x^2-1)\left( x+\frac{1}{a}\right) (x-a)\left( x+\frac{a+1}{a-1}\right) \left( x-\frac{a-1}{a+1}\right) }. \end{aligned}$$

The General Case

For general \(n\), we start with the relationship of Lemma 6,

$$\begin{aligned} -b_m = \left( \sum _{i=1}^{2^{m-1}} \theta _i^{(m)}- \sum _{i=1}^{2^{m-1}} \theta _i^{(m-1)}\right) \left( \sum _{i=1}^{2^{m-1}} \theta _{2^{m-1}+i}^{(m)}- \sum _{i=1}^{2^{m-1}} \theta _i^{(m-1)}\right) , \end{aligned}$$

for \(m \le n\). For a fixed \(m\) we apply the Correspondence Principle to obtain \(2^{n-m+1}-1\) companion relationships:

$$\begin{aligned} -b_m&= \left( \sum _{i=1}^{2^{m-1}} \theta _{2^{m}\ell +i}^{(m+j)}- \sum _{i=1}^{2^{m-1}} \theta _i^{(m-1)}\right) \left( \sum _{i=1}^{2^{m-1}} \theta _{2^{m-1}(2\ell +1)+i}^{(m+j)}- \sum _{i=1}^{2^{m-1}} \theta _i^{(m-1)}\right) ,\end{aligned}$$
(30)
$$\begin{aligned}&0 \le j \le n-m, \quad 0 \le \ell \le 2^j-1, \end{aligned}$$
(31)

where \(\theta _1^{(0)}:=0\). Thus for \(m=1,2,3\) and \(j,\ell \) satisfying (31), we have

$$\begin{aligned} -b_1&= \theta _{2 \ell +1}^{(1+j)}\theta _{2\ell +2}^{(1+j)}, \end{aligned}$$
(32)
$$\begin{aligned} -b_2&= \left( \theta _{4\ell +1}^{(2+j)} + \theta _{4\ell +2}^{(2+j)} - \theta _{1}^{(1)}-\theta _2^{(1)}\right) \left( \theta _{4\ell +3}^{(2+j)} + \theta _{4\ell +4}^{(2+j)} - \theta _{1}^{(1)}-\theta _2^{(1)}\right) , \end{aligned}$$
(33)
$$\begin{aligned} -b_3&= \left( \theta _{8\ell +1}^{(3+j)} + \theta _{8\ell +2}^{(3+j)}+ \theta _{8\ell +3}^{(3+j)} + \theta _{8\ell +4}^{(3+j)}- \theta _1^{(2)}- \theta _2^{(2)} - \theta _3^{(2)} -\theta _4^{(2)} \right) \nonumber \\&\quad \quad \cdot \left( \theta _{8\ell +5}^{(3+j)} + \theta _{8\ell +6}^{(3+j)}+ \theta _{8\ell +7}^{(3+j)} + \theta _{8\ell +8}^{(3+j)}- \theta _1^{(2)}- \theta _2^{(2)} - \theta _3^{(2)} -\theta _4^{(2)} \right) . \end{aligned}$$
(34)

There are \(2^n-1\), \(2^{n-1}-1\) and \(2^{n-2}-1\) equations in (32), (33) and (34) respectively. For \(m=n\) we have a single relationship,

$$\begin{aligned} -b_n&= \left( \theta _1^{(n)} + \cdots + \theta _{2^{n-1}}^{(n)}-\theta _1^{(n-1)} - \cdots - \theta _{2^{n-1}}^{(n-1)}\right) \nonumber \\&\quad \quad \cdot \left( \theta _{2^{n-1}+1}^{(n)} + \cdots + \theta _{2^{n}}^{(n)}-\theta _1^{(n-1)} - \cdots - \theta _{2^{n-1}}^{(n-1)} \right) . \end{aligned}$$
(35)

Altogether, there are \(\sum _{m=1}^n (2^{n-m+1}-1) = 2^{n+1}-2-n\) equations in \(2^{n+1}-2\) variables \(\theta _i^{(j)}\). Insisting that the \(b_i\) be positive gives \(n\) more soft conditions.

Theorem 13

Existence-Uniqueness property. Let \(n\) be a positive integer. Given a collection of \(2^{n+1}-2\) real numbers \(\theta _i^{(j)}\), \(1 \le j \le n\), \(1 \le i \le 2^j\) satisfying the \(2^{n+1}-2-n\) equations in (30) for some positive real numbers \(b_m\), \(m \le n\), and ordered in accordance with the Splicing principle (12), there exists a unique sequence of doubling transformations \(\phi _i(x)\), \(1 \le i \le n\), with poles at zero, such that with \(f_n(x): = \phi _n \circ \phi _{n-1} \circ \cdots \circ \phi _1(x) \), we have

$$\begin{aligned} f_n(x) = \frac{\prod _{i=1}^{2^n} (x- \theta _i^{(n)})}{ x \prod _{j=1}^{n-1}\prod _{i=1}^{2^j} (x- \theta _i^{(j)})}. \end{aligned}$$

Proof

The proof is an extension of the proof of Theorem 7 and so we just give a sketch. To define the \(\phi _i\), we simply define the \(b_i\) as in (30), and the \(\lambda _i\) by the relations in Lemma 6 (i), to wit,

$$\begin{aligned} \lambda _1&= \sum _{i=1}^2 \theta _i^{(1)}, \end{aligned}$$
(36)
$$\begin{aligned} \lambda _2&= \sum _{i=1}^4 \theta _i^{(2)} -2 \sum _{i=1}^2 \theta _i^{(1)} \end{aligned}$$
(37)
$$\begin{aligned} \lambda _3&=\sum _{i=1}^8 \theta _i^{(3)} -2 \sum _{i=1}^4 \theta _i^{(2)}-4 \sum _{i=1}^2 \theta _i^{(1)}\\ \qquad \qquad \qquad \qquad \qquad \qquad&\,\,\vdots \nonumber \end{aligned}$$
(38)
$$\begin{aligned} \lambda _n&=\sum _{i=1}^{2^n} \theta _i^{(n)} - 2 \sum _{i=1}^{2^{n-1}}\theta _{i}^{(n-1)}- \cdots - 2^{n-1}\sum _{i=1}^2 \theta _i^{(1)}. \end{aligned}$$
(39)

The composition \(\phi _n \circ \phi _{n-1} \circ \cdots \circ \phi _1\) then has zeros and poles satisfying (30) and the \(\lambda _i\) equations (36), (37), \(\dots \), (39). The theorem then follows by the analogue of the uniqueness lemma, Lemma 8, which can be proven by induction following the line of argument in the proof of Lemma 8. \(\square \)

Proof of the Integral Formula

Theorem 14

Let \(\phi (x)=x - \lambda - \frac{b}{x-\mu }\) with \(\lambda ,b \in \mathbb R\), \(b>0\). Then for any continuous, absolutely integrable function \(F(x)\) on \((-\infty , \infty )\), we have

$$\begin{aligned} \int _{-\infty }^{\infty } F(\phi (x)) \ dx = \int _{- \infty }^{\infty } F(x) \ dx. \end{aligned}$$

Proof

Let \(I=\int _{- \infty }^{\infty } F(x) \ dx.\) Replacing \(x\) by \(x+\mu \), we may assume that \(\mu =0\) and

$$\begin{aligned} \phi (x) = x - \lambda - \frac{b}{x}. \end{aligned}$$

Since \(\phi (x)\) is continuously differentiable on \((0, \infty )\) with image \((-\infty , \infty )\), and \(\phi '(x) = 1+ \frac{b}{x^2}\), we have, substituting \(x= \phi (u)\),

$$\begin{aligned} I= \int _{0}^\infty F( \phi (u)) \phi '(u) \ du = \int _{0}^\infty F(\phi (u)) + F(\phi (u)) \frac{b}{u^2} \ du. \end{aligned}$$
(40)

Now, since \(|F(\phi (u))| \le |F(\phi (u))||\phi '(u)| \), and \(|F(\phi (u)) \frac{b}{u^2}|\le |F(\phi (u))||\phi '(u)|\) for \(u \in (0, \infty )\), and \(F(\phi (u))\phi '(u)\) is absolutely integrable on \((0, \infty )\), the functions \(F(\phi (u))\) and \(F(\phi (u))\frac{b}{u^2}\) are integrable on \((0, \infty )\), and so we can break up the integral in (40) to get,

$$\begin{aligned} I= \int _{0}^\infty F(\phi (u)) \ du + \int _{0}^{\infty } F(\phi (u)) \frac{b}{u^2} \ du. \end{aligned}$$

Next, we substitute \(u=-b/t\), and note that \(\phi \left( \frac{-b}{t}\right) = \phi (t)\), \(dt= \frac{b}{ u^2 }\ du\), to obtain

$$\begin{aligned} I&= \int _{0}^\infty F(\phi (u)) \ du + \int _{-\infty }^0 F(\phi (-b/t)) \ dt \\&= \int _{0}^\infty F(\phi (u)) \ du + \int _{-\infty }^0 F(\phi (t)) \ dt\\&= \int _{-\infty }^\infty F(\phi (u)) \ du . \end{aligned}$$

\(\square \)