N-Tupling Transformations and Invariant Deﬁnite · N-Tupling Transformations and Invariant...

Int. J. Appl. Comput. Math (2015) 1:527–541DOI 10.1007/s40819-015-0025-y

ORIGINAL PAPER

N-Tupling Transformations and Invariant DefiniteIntegrals

Todd Cochrane · Lee Goldstein

Published online: 23 January 2015© Springer India Pvt. Ltd. 2015

Abstract Functions on the real number line of the type ψ(x) = c + x − bx−μ

, with b > 0,have the interesting property that for any continuous, absolutely integrable function F on R,the graph of F(ψ(x)) is a “doubling” of the graph of F(x), while the integral overR remainsinvariant,

∫ ∞−∞ F(ψ(x)) dx = ∫ ∞

−∞ F(x) dx . In this paper, we discover new families ofn-to-1 mappings on R that have the same invariance property.

Keywords Doubling transformations · Invariant integrals

Mathematics Subject Classification 26A42 · 26A33 · 26A48 · 26C15

Introduction

Functions on the real number line of the type

ψ(x) = c + x − b

x − μ, (1)

with b, c, μ real numbers, b > 0, called doubling transformations, have the interestingproperty that for any continuous, absolutely integrable function F on R, we have

∞∫

−∞F(ψ(x)) dx =

∞∫

−∞F(x) dx . (2)

T. Cochrane (B)Department of Mathematics, Kansas State University, Manhattan, KS 66506, USAe-mail: [email protected]

L. GoldsteinLawrence, KS, USAe-mail: [email protected]

123

http://crossmark.crossref.org/dialog/?doi=10.1007/s40819-015-0025-y&domain=pdf

528 Int. J. Appl. Comput. Math (2015) 1:527–541

0

0.4

0.8

422–4–

Fig. 1 Graph of y = e−(x−2/x)2

Thus, for example,∞∫

−∞e−

(x− b

x

)2

dx =∞∫

−∞e−x2 dx = √

π. (3)

Note that the integrand on the left-hand side can be made into a continuous function on R

by defining its value at x = 0 to be 0. This result can be found for example in Wilson’sAdvanced Calculus [7, p. 386], and makes a nice exercise for a student of advanced calculus(Fig. 1).

The name doubling comes from the fact that ψ is a 2-to-1 function on R − {μ}, and thusthe graph of F ◦ ψ is a doubling of the graph of F , as one can see in the figure above for thecase F(x) = e−x2 . The graph is doubled, while the area bounded by the graph remains thesame. Hagler [4,5] and [6] made use of doubling transformations in the study of orthogonalsystems of polynomials such as those of Jacobi, Hermite and Laguerre.

Our interest here is in obtaining further examples of functions satisfying the invarianceproperty in (2). Specifically, we seek n-to-1 mappings ψ on R such that the identity in(2) holds. We call n the tupling number of ψ , and call ψ an n-tupling transformation. Thisinvestigation has led us to an interesting blend of calculus, linear algebra and abstract algebra.

For any n we succeed in obtaining a class of n-tupling transformations satisfying (2),specifically, the class of rational functions of the type

ψ(x) = c + x − λ2n−1∑

k=1

sin2( jπ/n)

sin2(kπ/n)

x − μ − λsin((k− j)π/n)

sin(kπ/n)

, (4)

where c, λ, μ ∈ R, λ �= 0, and j is an integer with 1 ≤ j ≤ n/2, gcd( j, n) = 1; see “Explicitformula for ψ(x)” section. When n = 1, the ψ in (4) is just a translation, ψ(x) = x + c,while for n = 2, ψ is just the doubling transformation (1) given above. It is not hard toshow that compositions of functions of type (4) also satisfy the invariance property (2). Moreinterestingly, functions of type (4) admit composition-factorizations for certain composite n.For example, if n = 6 and ψ6 denotes a 6-tupling transformation of type (4), then we canwrite ψ6 = ψ2 ◦ ψ3 for some doubling transformation ψ2 and tripling transformation ψ3,each of type (4); see Examples 6, 7, 8.

To state our main result, let L1(R) denote the set of real valued functions F onR such that|F | is integrable on R, and C0(R) denote the set of continuous functions on R that vanish atinfinity, that is, limx→±∞ F(x) = 0.

Theorem 1 For any F ∈ C0(R) ∩ L1(R), and any ψ of the type (4), we have F ◦ ψ ∈C0(R) ∩ L1(R), and the equality in (2) holds. Moreover, the same conclusion follows if ψ isreplaced by any composition of functions of type (4).

It is easy to see that translations ψ(x) = x + c satisfy (2) for any F ∈ L1(R). In Lemma11 we show in fact that they are the only 1-to-1 rational functions satisfying (2) for allF ∈ C0(R) ∩ L1(R). For values of n ≥ 2, we do not know if the functions of type (4),

123

Int. J. Appl. Comput. Math (2015) 1:527–541 529

0.4

0.8

422–4–

Fig. 2 Graph of y = F ◦ ψ1 ◦ ψ2(x), a = 1, b = 2

0

0.4

0.8

422–4–

Fig. 3 Graph of y = F ◦ ψ1 ◦ ψ2(x), a = 1.7, b = 2

or compositions thereof, are the only n-to-1 rational functions satisfying (2). Theorem 1is an immediate consequence of Theorem 4, and the fact that the set �, defined in (5), isclosed under composition. A number of open problems for further investigation are given in“Factorization in �” section.

What makes the functions in (4) special is that they satisfy their own algebraic invarianceproperty, ψ(φ(x)) = ψ(x), for some rational function φ(x) of finite order (see (9)). Forexample, if ψ is the doubling function ψ(x) = x − 1

x and φ(x) = −1x , then φ is of order 2,

that is φ(φ(x)) = x , andψ(φ(x)) = ψ(x). Before getting into the discussion of constructingsuch functions lets look at another example.

Example 2 Let a, b be distinct positive real numbers andψ1, ψ2 be doubling transformations

given by ψ1(x) = x − (a−b)2

x , ψ2(x) = x − abx . The composition ψ1 ◦ ψ2 is a quadrupling

(4-tupling) transformation given by

ψ(x) := ψ1 ◦ ψ2(x) = (x2 − a2)(x2 − b2)

x(x2 − ab),

an odd functionwith poles at 0,±√ab and zeros at±a,±b. Let F(x) = 1

1+x2. Then F(ψ(x))

is an even function with local maxima at the zeros of ψ and zeros at the poles of ψ , and, byTheorem 1, we have that

π =∞∫

−∞

dx

1 + x2=

∞∫

−∞F(ψ(x)) dx =

∞∫

−∞

x2(x2 − ab)2 dx

x2(x2 − ab)2 + (x2 − a2)2(x2 − b2)2.

It is interesting to note that the value of the integral is independent of the choice of a and b.Below is a plot of F(ψ(x)) with a = 1, b = 2, a quadrupling of the graph of f (x) = 1

1+x2.

As a approaches b, the valley between the peaks at x = a and x = b narrows, until wereach a = b whence the graph is just a doubling of the graph of F(x). Indeed, when a = b,ψ1(x) = x and ψ(x) = ψ2(x). Fixing b = 2 and letting a increase, one can animate thegraph and show the peak at a passing through the peak at b, all the while conserving the totalarea bounded below the graph (Figs. 2, 3).

123


Setting the Stage for the Investigation

In seeking functionsψ satisfying the integral invariance property (2), we restrict our attentionto rational functions ψ that are strictly increasing on R, that is, strictly increasing on eachopen interval where they are defined, and satisfy ψ(x) → ∞ as x → ∞, (and consequentlyψ(x) → −∞ as x → −∞). Any such ψ has a tupling number n equal to one more than thenumber of its real poles, where, as usual, the poles of ψ are the zeros of the denominator ofψ (written in reduced form). Strictly speaking, we can view ψ as either an n-to-1 mappingon its domain, or as an n-to-1 mapping on the extended real number line R∪{∞}, taking thevalue∞ at each of its poles, including the pole at∞. Further, we restrict F to be a function inC0(R)∩ L1(R). For any such F andψ , we have that F ◦ψ ∈ C0(R), the continuity achievedby defining the value of F ◦ ψ at any pole of ψ to be zero, and that F ◦ ψ ∈ L1(R) (Lemma14). In particular, both integrals in (2) exist.

Thus our set of interest is,

� :={

ψ(x) ∈ R(x) : ψ(x) is strictly increasing, limx→∞ ψ(x) = ∞ and(1.2) holds for any F ∈ C0(R) ∩ L1(R)

}

, (5)

where R(x) is the field of rational functions. We may view the elements of � as eitherformal expressions in an indeterminate symbol x or as functions of x , the context making itclear which vantage point we are taking. As formal expressions, the field R(x) is a monoidwith respect to function composition (closed under compositions, associative, possessing anidentity) with identity element x . Thus, we will also use the symbol x to denote the identityfunction on R.

We say that an element f ∈ R(x) is invertible (in R(x)) and has inverse g if f ◦ g =x = g ◦ f for some g ∈ R(x). It is not hard to see (Lemma 10) that an element φ ∈ R(x) isinvertible if and only if it is an invertible linear fractional transformation,

φ(x) = ax + b

cx + d, a, b, c, d ∈ R, ad − bc �= 0.

Note that φ is increasing if ad − bc > 0 and decreasing if ad − bc < 0. As functionson the complex plane, such transformations are called Möbius transformations, and play animportant role in the study of analytic mappings on the complex plane; see eg. [3, Chapter3]or [1, Chapter3].

Any invertible rational function is certainly 1-to-1 on its domain, but the converse (oddly)is not true. For instance the polynomial ψ(x) = x3 is 1-to-1 on R but it is not invertible as anelement of R(x). If we restrict our attention to rational functions in � this subtle distinctiongoes away. In Lemma 11 we show that the only 1-to-1 functions in� are just the translationsψ(x) = x + r , r ∈ R.

Note that the set � is closed under function composition. Indeed, if ψ1, ψ2 ∈ �, thenψ1 ◦ ψ2 is increasing, ψ1 ◦ ψ2(x) → ∞ as x → ∞, and for any F ∈ C0(R) ∩ L1(R), wehave F ◦ ψ1, F ◦ ψ1 ◦ ψ2 ∈ C0(R) ∩ L1(R) (Lemma 14), and so applying (2) to ψ2 with Freplaced by F ◦ ψ1, we see that

∞∫

−∞F(ψ1 ◦ ψ2(x)) dx =

∞∫

−∞(F ◦ ψ1)(ψ2(x)) dx =

∞∫

−∞F(ψ1(x)) dx =

∫ ∞

−∞F(x) dx .

Thus, � is a submonoid of (R(x), ◦).If ψ1 is an n-to-1 mapping in � and ψ2 is an m-to-1 mapping in �, then ψ1 ◦ ψ2 is an

mn-to-1 mapping in �. In this manner, one can construct n-to-1 mappings in � by factoring

123


n, say n = n1 · n2 · · · nr , and expressing ψ as a composition of functions in � with tuplingnumbers n1, . . . , nr .

We noted above that anyψ ∈ � must have n−1 distinct real poles r1, . . . , rn−1, for somen. In Lemma 12, we show in addition that ψ(x) ∼ x as x → ∞, that is, limx→∞ ψ(x)

x = 1.Thus, it is reasonable to seek transformations ψ having partial fraction expansions of thetype

ψ(x) = c + x − b1x − r1

− · · · − bn−1

x − rn−1, (6)

for some c, bi ∈ R with bi > 0, 1 ≤ i ≤ n − 1 (although we do not know if ψ must havethis form). The bi must be positive to make ψ increasing on R. We shall construct functionsof this type in � by building them out of elements of finite order.

For any rational function φ and positive integer k, let φk = φ ◦φ ◦ · · ·◦φ, the compositionof φ with itself k-times. We say that φ has finite order if φn = x (the identity function) forsome positive integer n, and the minimal such n is called the order of φ. If φ has order n, thenφ is invertible (φ−1 = φn−1) and so, as noted above, it must be an invertible linear fractionaltransformation.

Let G = G(R) denote the group of invertible linear fractional transformations over R, thegroup operation being composition. It is easy to see (and well known) that G is isomorphicto GL2(R)/�, the multiplicative group of invertible 2 by 2 matrices over the reals modulothe subgroup � of nonzero multiples of the identity matrix, via the correspondence

φ(x) := ax + b

cx + d→ Aφ :=

[a bc d

]

.

Using this correspondence we can readily characterize all rational functions φ of order n.

Theorem 3 Let φ be an increasing rational function over R. Then φ is of order n ≥ 2 if andonly if φ is a linear fractional transformation of the form

φ(x) = 2λ cos( jπ/n) + μ − λ2

x − μ, (7)

for some λ,μ ∈ R, λ �= 0, and integer j with 1 ≤ j ≤ n/2, gcd( j, n) = 1.

Given any increasing linear fractional transformation φ of order n ≥ 2 we define

ψ(x) = c + x + φ(x) + φ2(x) + · · · + φn−1(x), (8)

with c ∈ R. It is plain that ψ(x) is increasing on R, satisfies ψ(x) → ∞ as x → ∞, andthat ψ is invariant under composition by φ, that is,

ψ(φ(x)) = ψ(x). (9)

This invariance property is the key ingredient needed to prove that any such ψ satisfies theintegral invariance property (2).

Theorem 4 If φ(x) is an increasing linear fractional transformation of order n ≥ 2 andψ(x) is defined by (8), then ψ ∈ � and ψ(x) is an n-to-1 mapping on the extended realnumber line.

The proofs of Theorem 3 and Theorem 4 are given in “Proof of Theorem 3” and “Proofof Theorem 4” sections respectively. Putting the two theorems together yields the explicitclass of n-to-1 mappings ψ ∈ �, given in (4). The details of this derivation are provided in“Explicit formula for ψ(x)” section.

123


Example 5 In this example we make explicit the formula in (4) for n = 2, 3 and 4. If n = 2,

then j = 1, φ(x) = μ − λ2

x−μ, and

ψ(x) = c + x − λ2

x − μ, λ �= 0, c, μ ∈ R. (10)

This recaptures the original transformation in (1). If n = 3, then j = 1, φ(x) = λ+μ− λ2

x−μ,

and

ψ(x) = c + x − λ2

x − μ− λ2

x − μ − λ, λ �= 0, c, μ ∈ R. (11)

If n = 4, then j = 1, φ(x) = √2λ + μ − λ2

x−μ, and

ψ(x) = c + x − λ2

x − μ− λ2/2

x − μ − λ√22

− λ2

x − μ − √2λ

, λ �= 0, c, μ ∈ R. (12)

Factorization in �

Recall, every transformation ψ in � is a strictly increasing n-to-1 mapping on R for some n,called the tupling number of ψ . Also, the set of invertible transformations in � is just the setof translations x + a, a ∈ R. These are the only elements of � with tupling number 1. Forα, β ∈ �, we say that α is a factor of β if β = α ◦ γ or β = γ ◦ α for some γ ∈ �. We calla noninvertible transformation ψ ∈ � irreducible if it cannot be expressed as a compositionof two non-invertible transformations in �. Otherwise it is called composite. In the lattercase, ψ = ψ1 ◦ ψ2 for some non-invertible ψ1, ψ2 ∈ �. Certainly, if ψ has a prime tuplingnumber, then ψ is irreducible. For composite tupling numbers, transformations of type (4)need not be irreducible as the following examples show for the cases n = 4, 6 and 8.

Before stating the examples let us note that the set � is invariant under translations,ψ(x) → ψ(x+a) andψ(x) → ψ(x)+b, and invariant under the dilationψ(x) → λψ(x/λ).Indeed, for λ > 0, F ∈ C0(R) ∩ L1(R), ψ ∈ �,

∞∫

−∞F(λψ(x/λ)) dx = λ

∞∫

−∞F(λψ(u)) du = λ

∞∫

−∞F(λu) du =

∞∫

−∞F(v) dv,

where in the second to last step we applied (2) to the function F(λx). A similar argumentholds for λ < 0. Thus, in the study of transformations ψ of type (4), we may assume thatc = 0, μ = 0 and λ = 1, that is, ψ takes on a generic form,

ψ(x) = x −n−1∑

k=1

sin2( jπ/n)

sin2(kπ/n)

x − sin((k− j)π/n)sin(kπ/n)

, (13)

for some j with 1 ≤ j ≤ n/2, gcd( j, n) = 1. We omit the details in the examples since,for the most part, they were obtained by rather brute force methods, and it would be nice toestablish instead a theoretical framework for the existence of such factorizations. The guidingprinciple we used was to match the poles and residues of the two sides.

123


Example 6 Factoring a 4-tupling transformation. Let ψ4(x) be a generic 4-tupling transfor-mation as in (13) (n = 4, j = 1), so that,

ψ4(x +√22 ) = x +

√22 − 1

x +√22

− 1

2x− 1

x −√22

.

Put

α(x) := x +√22 − 2

x, β(x) := x − 1

2x.

Then α, β are doubling transformations of type (10) and

α(β(x)) = x +√22 − 1

2x− 2

x − 12x

= x +√22 − 1

2x−

(1

x −√22

+ 1

x +√22

)

= ψ4

(x +

√22

),

that is, ψ4(x) = α ◦ β(x −√22 ).

Example 7 Factoring a 6-tupling transformation. Suppose that n = 6 and that ψ6(x) is thegeneric 6-tupling transformation (13) ( j = 1). Then

ψ6(x) = x − 1

x− 1

3(x −

√33

) − 1

4(x −

√32

) − 1

3(x − 2

√3

3

) − 1

x − √3.

Let ψ2, ψ3 be doubling and tripling transformations of types (10), (11), given by

ψ2(x) := x +√3

2− 9

4x, ψ3(x) := x −

√3

6− 1

3x− 1

3(x −

√33

) .

Then

ψ6(x) = ψ2 ◦ ψ3

(x −

√33

).

Example 8 Factoring an 8-tupling transformation. Suppose that n = 8 and that ψ8(x) is thegeneric 8-tupling transformation with j = 1,

ψ8(x) = x −7∑

k=1

sin2(π/8)sin2(kπ/8)

x − sin((k−1)π/8)sin(kπ/8)

.

Let α, β, γ be doubling transformations given by

α(x) := x − (2 − √2)/4

x,

β(x) := x − 2 − √2

x,

γ (x) := x + 12

√2 + √

2 − 4(2 − √2)

x.

Then,

ψ8(x) = γ ◦ β ◦ α

(

x − 12

√

2 + √2

)

.

123


The next example gives families of composite 4-tupling transformations in � that are notof type (12).

Example 9 In [2] the authors give criteria for determining when a rational function is acomposition of two doubling transformations, from which one can readily create exampleshaving particularly nice sets of zeros and poles, such as Example 2 and the following; see [2,Examples 4.1, 4.2, 4.3].

α ◦ β(x) = (x − a)(x + a2)(x − a3)(x + a4)

x(x − a2)(x + a3), (a > 1); (14)

γ ◦ δ(x) = (x2 − a2)(x2 − a8)

x(x + a2)(x − a3), (a > 1); (15)

where

α(x) = x + a(a − 1)3 − 2a3(a − 1)2(a2 + 1)

x, β(x) = x + (a3 − a2) − a5

x,

γ (x) = x + 2(a3 − a2) − a2(a4 − 1)(a2 − 1)

x, δ(x) = x + (a2 − a3) − a5

x.

It is plain that these examples are not transformations of the type (12). Indeed, the poles oftransformations of type (12) are always in arithmetic progression.

There remain a number of open problems:Open Problem 1. Characterize the set of all irreducible transformations in �. In particular,which elements of type (4) are irreducible? The examples above show that any ψ of type (4)with tupling number 4, 6 or 8 is composite. Can further examples of this type be found?Open Problem 2. Does every transformation in � have an essentially unique decomposi-tion as a composition of irreducible transformations? By essentially unique, we mean up tomultiples of invertible factors. For instance if ψ = α1 ◦ α2 ◦ · · · ◦ αk then we also haveψ = (α1 ◦ φ) ◦ (φ−1 ◦ α2 ◦ φ) · · · (φ−1 ◦ αk) for any translation φ. In [2] the authors provedthat such is the case if one restricts � to compositions of doubling transformations.OpenProblem3.Let�0 be the semigroup of transformations generated (under composition)by the transformations of type (4). Since � is closed under compositions, �0 ⊆ �. Does�0 = �?

Open Problem 4.Does every element of � have a partial fraction expansion of the type (6)?In Theorem 6 we show that every element of �0 has such an expansion.Open Problem 5. Are there any rational functions satisfying the invariance property (2) forany F ∈ C0(R) ∩ L1(R), that are not in � or −�?

Lemmas

Lemma 10 If φ(x) ∈ R(x) is an invertible rational function with respect to composition,then φ(x) is an invertible linear fractional transformation.

Proof Say φ(x) = p(x)q(x) , with p(x), q(x) relatively prime polynomials over R. Since φ(x) is

invertible (in R(x)), it is 1-to-1 on C and so the equation φ(x) = 0 has at most one solutioninC. Thus p(x) has at most one complex zero, and its multiplicity is at most one since, beingan invertible function, φ′(x) is nonzero onC and consequently the system p(x) = p′(x) = 0

123


has no solution. Thus p(x) is of degree at most 1. Since 1/φ(x) is also invertible, we see thatq(x) also is of degree at most one. ��Lemma 11 If ψ is a 1-to-1 rational function on R such that the integral invariance property(2) holds for any F ∈ C0(R) ∩ L1(R), then ψ is a translation.

Proof Letψ ∈ � be a 1-to-1 rational function onR. Let β denote its inverse (as a function onthe reals). In particular, β is continuous onR, wherever it is defined. Let [a, b] be any intervalcontained in the domain of ψ . On this interval, ψ has no pole and must be either increasingor decreasing, so we’ll assume the former. Let 1[a,b] denote the characteristic function on theinterval [a, b], and for any ε > 0, let F be a continuous function onR agreeing with 1[a,b] on[a, b], supported on (a − ε, b + ε), and satisfying 0 ≤ F(x) ≤ 1 for all x . Such a functionis called a majorizing function for 1[a,b]. We assume that ε is sufficiently small so that nopole of ψ is in (a − ε, b+ ε). Plainly, F ∈ C0(R) ∩ L1(R). Write F(x) = 1[a,b](x) + E(x),where E(x) is a function supported on (a − ε, a) ∪ (b, b + ε) with 0 ≤ E(x) ≤ 1 for all x .Then by (2) we have

b − a =∞∫

−∞1[a,b](x) dx ≤

∞∫

−∞F(x) dx =

∞∫

−∞F(ψ(x)) dx

=∞∫

−∞1[a,b](ψ(x)) + E(ψ(x)) dx

≤∞∫

−∞1[a,b](ψ(x)) + 1[a−ε,a](ψ(x)) + 1[b,b+ε](ψ(x)) dx

= (β(b) − β(a)) + (β(a) − β(a − ε)) + (β(b + ε) − β(b)),

and sob − a ≤ β(b + ε) − β(a − ε).

Since this inequality holds for any ε > 0 we conclude, by the continuity of β, that b − a ≤β(b) − β(a). Using a minorizing function, one obtains in a similar manner the oppositeinequality, and so b − a = β(b) − β(a). Thus, for any such a < b we have

β(b) − β(a)

b − a= 1.

Taking the limit as b → a we see that β ′(a) = 1 for any real number a where ψ is defined.Thus β(x) = x + c for some constant c on any open interval on which it is defined, theconstant c a priori depending on the interval, and ψ(x) = x − c on a corresponding interval.Being a rational function, we must then have ψ(x) = x − c on R. ��Lemma 12 If ψ ∈ � then ψ(x) ∼ x as x → ∞, that is, limx→∞ ψ(x)

x = 1.

Proof Let ψ ∈ � be an n-to-1 increasing function with poles r1 < r2 < · · · < rn−1. Putr0 = −∞, rn = ∞. Let βi denote the inverse function of ψ |(ri ,ri+1), the restriction of ψ tothe interval (ri , ri+1), 0 ≤ i ≤ n − 1. Note, βi is continuously differentiable on R. Then,arguing as in the previous lemma, we obtain

b − a =n−1∑

i=0

(βi (b) − βi (a)).

123


Dividing by b − a and taking the limit as b → a, we see that∑n−1

i=0 β ′i (x) = 1 for all

x , and thus∑n−1

i=0 βi (x) = x + c, for some constant c. Inserting ψ(x) for x , we have∑n−1

i=0 βi (ψ(x)) = ψ(x) + c. Now for any x ∈ (rn−1,∞), we have βn−1(ψ(x)) = x , andso for any such x , we have

x +n−2∑

i=0

βi (ψ(x)) = ψ(x) + c.

Observing that limx→∞ βi (ψ(x)) = ri+1, for 0 ≤ i ≤ n − 2, we obtain

limx→∞

ψ(x)

x= lim

x→∞x − c + ∑n−2

i=0 βi (ψ(x))

x= 1.

��Lemma 13 For any ψ ∈ � there exists a positive constant B such that ψ ′(x) ≥ B for all xin the domain of ψ .

Proof Since ψ(x) ∼ x as x → ∞, we have

ψ(x) = x + c + r(x)

g(x), ψ ′(x) = 1 + g(x)r ′(x) − r(x)g′(x)

g(x)2,

for some polynomials r(x), g(x)with either r(x) identically zero, or deg(r(x)) < deg(g(x)).Thus, ψ ′(x) → 1 as x → ±∞. In particular, ψ ′(x) > 1

2 on some neighborhood of ±∞, say(−∞, s) ∪ (t,∞), s, t ∈ R. Also, at each real pole of ψ the slope goes to ∞ from the leftand the right. Thus there exists an open interval about each pole on which ψ ′(x) > 1

2 . Thecomplement of the union of these open intervals is a compact set. Since, ψ ′ is continuouson this compact set and strictly positive, its minimum value on this set must be a positivenumber. ��Lemma 14 Let F ∈ C0(R) ∩ L1(R), and ψ ∈ �. Then F ◦ ψ ∈ C0(R) ∩ L1(R).

Proof We noted earlier that for any such F and ψ , F ◦ ψ ∈ C0(R), the continuity achievedby defining F ◦ ψ to be zero at any pole of ψ . We turn now to the integrability of F ◦ ψ .Let r1 < r2 < · · · < rn−1 be the distinct poles of ψ and put r0 = −∞, rn = ∞. For0 ≤ i ≤ n − 1, since ψ is an increasing, continuously differentiable function on the interval(ri , ri+1) with image (−∞,∞), we see, upon making a substitution, that

∞∫

−∞|F(x)| dx =

ri+1∫

ri

|F(ψ(u))|ψ ′(u) du. (16)

To be precise, one can break up the integral on the left-hand side into two pieces, say over(−∞, 0] and over [0,∞). For 0 ≤ i ≤ n−1, let zi be the zero of ψ in the interval (ri , ri+1).Then, for zi < t < ri+1, we have

ψ(t)∫

0

|F(x)| dx =t∫

zi

|F(ψ(u))|ψ ′(u) du.

Taking the limit as t → ri+1, yields∫ ∞0 |F(x)| dx = ∫ ri+1

zi|F(ψ(u))|ψ ′(u) du. Doing the

same thing for the second piece, and then putting the two pieces together, yields the equalityin (16). Since (16) holds for i = 0, . . . , n − 1, we obtain

123


∞∫

−∞|F(ψ(u))|ψ ′(u) du =

n−1∑

i=0

ri+1∫

ri

|F(ψ(u))|ψ ′(u) du = n

∞∫

−∞|F(x)| dx .

In particular, since ψ ′(u) is always positive and F ∈ L1(R), we see that F(ψ(x))ψ ′(x)∈ L1(R). Now, by Lemma 13, there exists a B > 0 such that ψ ′(x) ≥ B for all x inthe domain of ψ . Thus |F(ψ(u))| ≤ 1

B |F(ψ(u))|ψ ′(u) for all u in the domain of ψ , andtherefore F ◦ ψ ∈ L1(R). ��

Proof of Theorem 3

Let φ be an increasing rational function on R of order n ≥ 2. Then, as noted earlier, φ

is invertible and so by Lemma 10, φ is a linear fractional transformation. If φ is a linearpolynomial then it is plain that φ has finite order if and only if φ is the identity function.Thus, since n > 1, φ is not a polynomial and so it must have a pole. Note that, since φ hasorder n, then so does any conjugate α−1 ◦ φ ◦ α of φ. In particular, letting α(x) = x − μ,μ ∈ R, we see that φ(x − μ) + μ has order n. Thus, by making a translation if necessary,we may assume that φ has a pole at 0. Any increasing, linear fractional transformation withpole at 0 has the form φ(x) = a − b

x with a, b ∈ R, b > 0, and corresponds to a matrix Aφ

as given below,

φ(x) := a − b

x= ax − b

x→ Aφ :=

[a −b1 0

]

.

Since φn(x) = x , we have Anφ = C I for some nonzero real number C , where I is

the identity matrix. The minimal polynomial for Aφ is just the characteristic polynomialdet (Aφ − t I ) = t2 − at + b, and so this polynomial must be a divisor of tn − C . Thust2 − at + b = (t − r1)(t − r2) for some distinct n-th roots r1, r2 of C . If r1, r2 are realthen we must have n even, r1 = −r2 and b = −r22 , contradicting our assumption that b ispositive. Therefore r1 and r2 must be complex conjugates. If C > 0, say C = λn for some

λ > 0, then r1 = λe2 jπ in for some integer j , while if C < 0, say C = −λn with λ > 0, then

r1 = λe(2 j+1)π i

n , for some integer j . Putting the two cases together, we can say that

r1 = λejπ in , r2 = λe

− jπ in , a = 2λ cos( jπ/n), b = λ2, (17)

for some j with 1 ≤ j ≤ n − 1, and some λ > 0. Replacing j with n − j has the effect ofchanging the sign of λ (and interchanging r1 and r2). Since we are assuming that φ has ordern, we must have gcd( j, n) = 1. Thus φ has the form

φ(x) = 2λ cos( jπ/n) − λ2

x, 1 ≤ j ≤ n/2, gcd( j, n) = 1, λ �= 0. (18)

To complete the proof of Theorem 3,we simply note that ifφ has a pole atμ thenφ(x+μ)−μ

must have the form above.

Explicit formula for ψ(x)

Suppose now that ψ(x) = c + x + φ(x) + · · · + φn−1(x), with φ(x) as in (18). The valuesr1 and r2 in (17) are eigenvalues of Aφ with corresponding eigenvectors (r1, 1), (r2, 1), andso,

123


Aφ = PDP−1, with D =[r1 00 r2

]

, and P =[r1 r21 1

]

.

It follows that for any positive integer k, the transformation φk corresponds to the matrix

Akφ = PDk P−1 = 1

r1−r2

[r1 r21 1

] [rk1 00 rk2

] [1 −r2

−1 r1

]

= 1r1−r2

[rk+11 − rk+1

2 λ2(rk2 − rk1 )

rk1 − rk2 λ2(rk−12 − rk−1

1 )

]

.

Thus, for any positive integer k,

φk(x) = λ sin( j (k + 1)π/n)x − λ2 sin( jkπ/n)

sin( jkπ/n)x − λ sin( j (k − 1)π/n), (19)

an increasing function with pole at

αk := λsin( j (k − 1)π/n)

sin( jkπ/n), 1 ≤ k ≤ n − 1. (20)

Doing a long division and then using the trigonometric identity,

1 − sin( j (k + 1)π/n) sin( j (k − 1)π/n)

sin2( jkπ/n)= sin2( jπ/n)

sin2( jkπ/n), (21)

we obtain

φk(x) = λsin( j (k + 1)π/n)

sin( jkπ/n)−

λ2(1 − sin( j (k+1)π/n) sin( j (k−1)π/n)

sin2( jkπ/n)

)

x − λsin( j (k−1)π/n)

sin( jkπ/n)

= λsin( j (k + 1)π/n)

sin( jkπ/n)− λ2

sin2( jπ/n)

sin2( jkπ/n)


sin( jkπ/n)

.

After absorbing constants into the value c one obtains from the definition of ψ ,

ψ(x) := c + x +n−1∑

k=1

φk(x) = c′ + x − λ2n−1∑

k=1

sin2( jπ/n)

sin2( jkπ/n)


sin( jkπ/n)

, (22)

for some constant c′. Since gcd( j, n) = 1, as k runs from 1 to n − 1, jk also runs through acomplete set of nonzero residues (mod n), and so making a change of variables we arriveat the formula in (4) (with μ = 0).

Proof of Theorem 4

Let φ be an increasing linear fractional transformation of order n ≥ 2 and ψ be defined byψ(x) = c+ x +φ(x)+φ2(x)+ · · · +φn−1(x), with c ∈ R. As before, we may assume thatthe pole of φ is 0, and thus φ is as given (18), and ψ as given in (22). Plainly, ψ is a strictlyincreasing function on R with n − 1 distinct real poles, α1, . . . , αn−1 as given in (20), andso it is an n-to-1 mapping on R. Moreover, by (22), ψ(x) → ∞ as x → ∞.

Put φ0(x) = x , and for 1 ≤ k ≤ n − 1, φk(x) = φk(x). Then, since φn(x) = x , we have

ψ(φk(x)) = ψ(x), for 1 ≤ k ≤ n − 1. (23)

123


Note also that

ψ(x) = c + φ0(x) + φ1(x) + · · · + φn−1(x),

and that ψ is continuously differentiable on its domain, with

ψ ′(x) = φ′0(x) + φ′

1(x) + · · · + φ′n−1(x).

We order the poles,

αi1 < αi2 < · · · < αin−1 ,

and put αi0 := −∞, αin := ∞.Let F ∈ C0(R)∩L1(R). In particular, F is integrable onR and, by Lemma 14, so is F ◦ψ .

Noting that for 0 ≤ ≤ n − 1, ψ is an increasing, continuously differentiable function onthe interval (αi , αi +1), with image (−∞,∞) we see upon making a u-substitution, that

∞∫

−∞F(x) dx =

αi +1∫

αi

F(ψ(u))ψ ′(u) du.

Thus, we obtain

n

∞∫

−∞F(x) dx =

n−1∑

=0

αi +1∫

αi

F(ψ(u))ψ ′(u) du

=n−1∑

=0

αi +1∫

αi

F (ψ(u))

n−1∑

k=0

φ′k(u) du

=n−1∑

=0

αi +1∫

αi

n−1∑

k=0

F (ψ(u)) φ′k(u) du.

At this point we would like to express the integral of the sum as a sum of the integrals, but thisrequires each piece to be integrable.We claim, in fact, that each piece is absolutely integrable.Indeed, since 0 < φ′

k(u) ≤ ψ ′(u) for all u (where defined) we have |F(ψ(u))φ′k(u)| ≤

|F(ψ(u))ψ ′(u)|, but the latter function is integrable on (αi , αi +1), since F ◦ ψ ∈ L1(R).Thus, we have

n

∞∫

−∞F(x) dx =

n−1∑

k=0

n−1∑

=0

αi +1∫

αi

F(ψ(u))φ′k(u) du

=n−1∑

k=0

n−1∑

=0

αi +1∫

αi

F(ψ(φk(u)))φ′k(u) du, (since ψ(φk(x)) = ψ(x)),

123


=n−1∑

k=0

∞∫

−∞F(ψ(φk(u))φ′

k(u) du

=n−1∑

k=0

∫ ∞

−∞F(ψ(t)) dt, (letting t = φk(u))

= n

∞∫

−∞F(ψ(t)) dt.

Partial Fraction Expansions

Let �0 be the semigroup generated by elements of type (4), under composition.

Theorem 15 Any transformation ψ ∈ �0 is an increasing n-tupling transformation (forsome n) with partial fraction expansion of the type,

ψ(x) = c + x − b1x − r1

− b2x − r2

− · · · − bn−1

x − rn−1, (24)

for some real number c, positive real numbers bi , and distinct real numbers ri , 1 ≤ i ≤ n−1.

Proof By definition, the generators of �0 have partial fraction expansions of type (24), andso it suffices to show that ifψ1,ψ2 are transformations type (24), then so is their composition.Clearly, any function of type (24) is a strictly increasing n-to-1 function on the extended realnumber line. Thus ifψ1 ism-to-1 andψ2 is n-to-1 thenψ1◦ψ2 is a strictly increasingmn-to-1function. In particular, it has mn − 1 distinct real poles, say w1, . . . , wmn−1. Moreover, thepoles must be simple, and the residue of each pole wi (that is, the coefficient of 1

x−wi,) must

be negative (in order for ψ1 ◦ ψ2 to be increasing on a neighborhood of the pole.) Also notethat, written in reduced form, ψ1 = f1

g1, ψ2 = f2

g2, for some polynomials f1, g1, f2, g2 of

degreesm,m−1, n, n−1 respectively. Thus the composition has the formψ1 ◦ψ2 = f3g3

forsome polynomials f3, g3 of degrees mn,mn − 1 respectively, and so ψ1 ◦ ψ2 has a partialfraction expansion of the type

ψ1 ◦ ψ2(x) = c + bx − b1x − w1

− · · · − bmn−1

x − wmn−1,

for some real numbers c, b and positive real numbers b1, . . . , bmn−1. Since ψ1 ◦ ψ2(x) ∼ xas x → ∞ (by Lemma 12 or as seen directly), we must have b = 1. ��

References

1. Ahlfors, L.: Complex Analysis, 2nd edn. McGraw-Hill, New York (1966)2. Cochrane, T., Goldstein, L.: Doubling transformations and invariance of definite integrals. Int. J. Appl.

Comp. Math. 1–12 (2014)3. Conway, J.B.: Functions of One Complex Variable. Graduate Texts in Mathematics, vol. 11, 2nd edn.

Springer, New York (1978)4. Hagler, B.A.: A transformation of orthogonal polynomial sequences into orthogonal Laurent polynomial

sequences, thesis (Ph.D.), University of Colorado at Boulder, 1–124 (1997)5. Hagler, B.A.: Formulas for the moments of some strong moment distributions. Orthogonal functions,

moment theory, and continued fractions (Campinas, 1996), LectureNotes inPure andAppliedMathematics,vol. 199, pp. 179–186, Dekker, New York (1998)

123


6. Hagler, B.A., Jones, W.B., Thron, W.J.: Orthogonal Laurent polynomials of Jacobi, Hermite, and Laguerretypes. Orthogonal functions, moment theory, and continued fractions, (Campinas, 1996), Lecture Notes inPure and Applied Mathematics, vol. 199, pp. 187–208, Dekker, New York (1998)

7. Wilson, E.B.:AdvancedCalculus.AText upon Select Parts ofDifferential Calculus,Differential Equations,Integral Calculus, Theory of Functions, with Numerous Exercises. Dover Publications Inc., New York(1959)

123

N-Tupling Transformations and Invariant Deﬁnite · N-Tupling Transformations and Invariant...

Documents

Transcript of N-Tupling Transformations and Invariant Deﬁnite · N-Tupling Transformations and Invariant...