Abstract

A rational divide-and-conquer relation, which is a natural generalization of the classical divide-and-conquer relation, is a recursive equation of the form 𝑓(𝑏𝑛)=𝑅(𝑓(𝑛),𝑓(𝑛),,𝑓(𝑏1)𝑛)+𝑔(𝑛), where 𝑏 is a positive integer 2; 𝑅 a rational function in 𝑏1 variables and 𝑔 a given function. Closed-form solutions of certain rational divide-and-conquer relations which can be used to characterize the trigonometric cotangent-tangent and the hyperbolic cotangent-tangent function solutions are derived and their global behaviors are investigated.

1. Introduction

The classical divide-and-conquer relation is a recursive relation of the form ([13]) 𝑛𝐹(𝑛)=𝑎𝐹𝑏+𝐺(𝑛),(1.1) where 𝑎,𝑏(2) are positive integers and 𝐺(𝑛) is a given function. This class of recurrence relations arises frequently in the analysis of recursive computer algorithms. Such algorithms split a problem of size 𝑛 into 𝑎 subproblems each of size [𝑛/𝑏], with 𝐺(𝑛) extra operations being required when this split of a problem of size 𝑛 into smaller problems is made. Although, there are certain cases, see for example the table on page 273 of [3], where the relation (1.1) can be solved explicitly, it is generally impossible to solve (1.1) for all values of 𝑛. However, when a starting value F(𝑏𝜆) is given, a solution for 𝑛=𝑏𝑘(𝑘>𝜆) can be found by making a change of variables  𝐹(𝑏𝑘)=𝜙(𝑘) which turns (1.1) into a first order difference equation of the form ([1, page 137]) 𝜙𝑏(𝑘)=𝑎𝜙(𝑘1)+𝐺𝑘,(1.2) and this last recursive equation can be easily solved. Another aspect of importance in the study of divide-and-conquer relations deals with the size of 𝐹(𝑛) which is used in analyzing the complexity of corresponding divide-and-conquer algorithms ([2, Section 5.3]).

Generalizing the above notion, by a rational divide-and-conquer (RDAC) relation, we refer to a recursive relation of the form 𝑓(𝑏𝑛)=𝑅(𝑓(𝑛),𝑓(2𝑛),,𝑓(𝑏1)𝑛)+𝑔(𝑛),(1.3) where 𝑏,𝑏2,𝑅(𝑥1,,𝑥𝑏1) a rational function in 𝑥1,,𝑥𝑏1, and 𝑔(𝑛) a given function. Here we aim to find explicit closed form solutions of certain nonlinear divide-and-conquer relations which is closely related to identities of the trigonometric and hyperbolic cotangent identities. Our investigation arises from an observation that the trigonometric cotangent function satisfies, among a number of other identities, the following identity: cot(3𝐴)=cot2𝐴cot𝐴1,cot2𝐴+cot𝐴(1.4) which leads to an RDAC relation of the form 𝑥3𝑛=𝑥2𝑛𝑥𝑛1𝑥2𝑛+𝑥𝑛.(1.5) This relation can be rewritten as 𝑥3𝑛𝑖𝑥3𝑛=𝑥+𝑖𝑛𝑖𝑥𝑛𝑥+𝑖2𝑛𝑖𝑥2𝑛+𝑖𝑖=1,(1.6) which is a simpler looking RDAC relation of the form 𝑈3𝑛=𝑈𝑛𝑈2𝑛𝑈𝑛𝑥=𝑛𝑖𝑥𝑛+𝑖(1.7) that can be immediately solved. Let us mention in passing that similar substitution techniques have been employed earlier in [4, 5].

Our first objective here is to find, in the next section, a closed form solution of 𝑈𝑏𝑛=𝑈𝛼1𝑛𝑈𝛼22𝑛𝑈𝛼𝑏1(𝑏1)𝑛,(1.8) an RDAC relation generalizing (1.7). Experiences from (1.7) with the cotangent function lead us to apply the results from our first objective to use such RDAC relations to characterize the trigonometric and hyperbolic tangent and cotangent functions, and this will be carried out in the following section as applications.

2. Closed Form Solutions

Before stating our main result, it is convenient to introduce a new notation. For 𝑘, let us write 𝛼1𝑉+𝛼2𝑉2++𝛼𝑏1𝑉(𝑏1)𝑘=𝑖1+𝑖2++𝑖𝑏1=𝑘𝑘𝑖1,𝑖2,,𝑖𝑏1𝛼𝑖11𝛼𝑖22𝛼𝑖𝑏1𝑏1𝑉1𝑖12𝑖2(𝑏1)𝑖𝑏1,(2.1) where 𝑘𝑖1,𝑖2,,𝑖𝑏1=𝑘!/𝑖1!𝑖2!𝑖𝑏1! denote the customary multinomial coefficients. Our main result is:

Theorem 2.1. Let 𝑏,𝑏2, and 𝛼1,,𝛼𝑏1. If the sequence {𝑈𝑛}𝑛0 satisfies the RDAC relation 𝑈𝑏𝑛=𝑈𝛼1𝑛𝑈𝛼22𝑛𝑈𝛼𝑏1(𝑏1)𝑛(𝑛1),(2.2) then for 0(mod𝑏), one has 𝑈𝑏𝑘𝑖1+𝑖2++𝑖𝑏1=𝑈𝑘𝑖1,𝑖2,,𝑖𝑏1𝛼𝑖11𝛼𝑖22𝛼𝑖𝑏1𝑏11𝑖12𝑖2(𝑏1)𝑖𝑏1(𝑘).(2.3)

Proof. Taking principal logarithms of (2.2), the relation becomes 𝑉𝑏𝑛=𝛼1𝑉𝑛+𝛼2𝑉2𝑛++𝛼𝑏1𝑉(𝑏1)𝑛(𝑛1),(2.4) where 𝑉𝑖=log𝑈𝑖. For 0(mod𝑏), evaluating (2.4) at 𝑛=𝑏, we get 𝑉𝑏2=𝛼1𝑉𝑏+𝛼2𝑉𝑏2++𝛼𝑏1𝑉𝑏(𝑏1)=𝛼1𝛼1𝑉+𝛼2𝑉2++𝛼𝑏1𝑉(𝑏1)+𝛼2𝛼1𝑉2+𝛼2𝑉22++𝛼𝑏1𝑉2(𝑏1)++𝛼𝑏1𝛼1𝑉(𝑏1)+𝛼2𝑉2(𝑏1)++𝛼𝑏1𝑉(𝑏1)2=𝛼21𝑉+𝛼22𝑉22++𝛼2𝑏1𝑉(𝑏1)2+2𝛼1𝛼2𝑉2++2𝛼𝑖𝛼𝑗𝑉𝑖𝑗++2𝛼𝑏2𝛼𝑏1𝑉(𝑏2)(𝑏1).(2.5) Using the notation introduced above, we see at once that 𝑉𝑏=𝛼1𝑉+𝛼2𝑉2++𝛼𝑏1𝑉(𝑏1)1𝑉𝑏2=𝛼1𝑉+𝛼2𝑉2++𝛼𝑏1𝑉(𝑏1)2.(2.6) To finish the proof, we need only show that for all 𝑘𝑉𝑏𝑘=𝛼1𝑉+𝛼2𝑉2++𝛼𝑏1𝑉(𝑏1)𝑘.(2.7) We proceed by induction. For any 0(mod𝑏), assume that (2.7) holds up to 𝑘. Thus, by (2.4) and the induction hypothesis one has 𝑉𝑏𝑘+1=𝛼1𝑉𝑏𝑘+𝛼2𝑉𝑏𝑘2++𝛼𝑏1𝑉𝑏𝑘(𝑏1)=𝛼1𝛼1𝑉+𝛼2𝑉2++𝛼𝑏1𝑉(𝑏1)𝑘+𝛼2𝛼1𝑉2+𝛼2𝑉22++𝛼𝑏1𝑉2(𝑏1)𝑘++𝛼𝑏1𝛼1𝑉(𝑏1)+𝛼2𝑉2(𝑏1)++𝛼𝑏1𝑉(𝑏1)2𝑘=𝛼1𝑖1+𝑖2++𝑖𝑏1=𝑘𝑘𝑖1,𝑖2,,𝑖𝑏1𝛼𝑖11𝛼𝑖22𝛼𝑖𝑏1𝑏1𝑉1𝑖12𝑖2(𝑏1)𝑖𝑏1+𝛼2𝑖1+𝑖2++𝑖𝑏1=𝑘𝑘𝑖1,𝑖2,,𝑖𝑏1𝛼𝑖11𝛼𝑖22𝛼𝑖𝑏1𝑏1𝑉(1𝑖12)(2𝑖22)((𝑏1)𝑖𝑏12)++𝛼𝑏1𝑖1+𝑖2++𝑖𝑏1=𝑘𝑘𝑖1,𝑖2,,𝑖𝑏1𝛼𝑖11𝛼𝑖22𝛼𝑖𝑏1𝑏1𝑉(1𝑖1(𝑏1))(2𝑖2(𝑏1))(𝑏1)𝑖𝑏1+1=𝑖1+𝑖2++𝑖𝑏1=𝑘+1𝑘𝑖1,𝑖2,,𝑖𝑏1𝛼𝑖11𝛼𝑖22𝛼𝑖𝑏1𝑏1𝑉1𝑖12𝑖2(𝑏1)𝑖𝑏1.(2.8)

The cases 𝑏=2 and 3 are of particular interest and we record them here for future reference.

Corollary 2.2. (I) Let 𝛼. If the sequence {𝑈𝑛}𝑛0 satisfies the RDAC relation 𝑈2𝑛=𝑈𝛼𝑛(𝑛1),(2.9) then for 0(mod2), one has 𝑈2𝑘=𝑈𝛼𝑘(𝑘).(2.10)
(II) Let 𝛼1,𝛼2. If the sequence {𝑈𝑛}𝑛0 satisfies the RDAC relation 𝑈3𝑛=𝑈𝛼1𝑛𝑈𝛼22𝑛(𝑛1),(2.11) then for 0(mod3), one has 𝑈3𝑘=𝑈𝑘0𝛼𝑘1𝛼02𝑈𝑘1𝛼1𝑘1𝛼22𝑈𝑘𝑘𝛼01𝛼𝑘22𝑘(𝑘).(2.12)

3. Applications

We now apply the result of Theorem 2.1 and Corollary 2.2 to several RDAC relations including those that can be used to characterize the trigonometric and hyperbolic tangent and cotangent functions.

Proposition 3.1. (I) Suppose that the sequence {𝑥𝑛}𝑛0 satisfies an RDAC relation of the form 𝑥2𝑛=𝑥2𝑛12𝑥𝑛(𝑛1).(3.1) For 0(mod2) and 𝑘, if the condition 2𝑘𝜃0(mod2𝜋) is fulfilled, then 𝑥2𝑘=cot2𝑘𝜃22=cot𝑘arccot𝑥,(3.2) where 𝜃=2arccot𝑥.(3.3)
(II) Assume that the sequence {𝑥𝑛}𝑛0 satisfies an RDAC relation of the form 𝑥2𝑛=2𝑥𝑛1𝑥2𝑛(𝑛0).(3.4) For 0(mod2) and 𝑘, if 2𝑘𝜃 is not an odd multiple of 𝜋, then 𝑥2𝑘=tan2𝑘𝜃22=tan𝑘arctan𝑥,(3.5) where 𝜃=2arctan𝑥.(3.6)
(III) If the sequence {𝑥𝑛}𝑛0 satisfies 𝑥2𝑛=𝑥2𝑛+12𝑥𝑛(𝑛0),(3.7) then, for 0(mod2), 𝑘, one has 𝑥2𝑘=coth2𝑘𝜃22=coth𝑘arccoth𝑥,(3.8) where 𝜃=2arccoth𝑥.(3.9)
(IV) If the sequence {𝑥𝑛}𝑛0 satisfies 𝑥2𝑛=2𝑥𝑛1+𝑥2𝑛(𝑛0),(3.10) then, for 0(mod2), 𝑘, one has 𝑥2𝑘=tanh2𝑘𝜃22=tanh𝑘arctanh𝑥,(3.11) where 𝜃=2arctanh𝑥.(3.12)

Proof. (I) As seen in Section 1, the RDAC relation (3.1) is equivalent to 𝑈2𝑛=𝑈2𝑛𝑈𝑛=𝑥𝑛𝑖𝑥𝑛+𝑖,(3.13) whose solution is, by virtue of Corollary 2.2, 𝑈2𝑘=𝑈2𝑘. Thus, 𝑥2𝑘𝑖𝑥2𝑘=𝑥+𝑖𝑖𝑥+𝑖2𝑘.(3.14) Setting 𝑒𝑖𝜃=(𝑥𝑖)/(𝑥+𝑖), one has 𝑥2𝑘=𝑖1+𝑒𝑖2𝑘𝜃1𝑒𝑖2𝑘𝜃=cot2𝑘𝜃22=cot𝑘arccot𝑥,(3.15) provided 2𝑘𝜃0(mod2𝜋).
(II) Substituting 𝑥𝑛 by 1/𝑥𝑛 turns (3.4) into (3.1) and so the result follows at once from part (I).
(III) Substituting 𝑥𝑛 by 𝑖𝑥𝑛 in (3.7) turns it into a rational recursive equation of the form (3.1) and so part (I) yields the desired result.
(IV) Replacing 𝑥𝑛 by 𝑖𝑥𝑛 in (3.10), we get a rational recursive equation of the form (3.4) and part (II) yields the result.

Remark 3.2. Although the substitution 𝑥𝑛 by 1/𝑥𝑛 employed in part (II) of Proposition 3.1 allows us to obtain a closed form solution of the RDAC relation (3.4), there remains a difficulty should there exist integer 𝑁 such that 𝑥𝑁=0. To overcome this shortcoming, we may either interpret the infinite value of the two expressions on both sides of the solution as equal or repeat the technique used in the proof of Proposition 3.1 to solve (3.4).

Proposition 3.3. (I) Suppose that the sequence {𝑥𝑛}𝑛0 satisfies an RDAC relation of the form 𝑥3𝑛=𝑥2𝑛𝑥𝑛1𝑥2𝑛+𝑥𝑛(𝑛1).(3.16) For 0(mod3) and 𝑘, if the condition 𝑘0𝜃+𝑘1𝜃2𝑘𝑘𝜃++2𝑘0(mod2𝜋)(3.17) is fulfilled, then 𝑥3𝑘=cot𝑘0𝜃𝑘1𝜃2𝑘𝑘𝜃2𝑘2𝑘0=cotarccot𝑥+𝑘1arccot𝑥2𝑘𝑘++arccot𝑥2𝑘,(3.18) where 𝜃𝑗=2arccot𝑥𝑗𝑗,2,,2𝑘.(3.19)
(II) Assume that the sequence {𝑥𝑛}𝑛0 satisfies an RDAC relation of the form 𝑥3𝑛=𝑥𝑛+𝑥2𝑛1𝑥𝑛𝑥2𝑛(𝑛0).(3.20) For 0(mod3) and 𝑘, if 𝜃+𝑘1𝜃2++𝑘𝑘𝜃2𝑘 is not an odd multiple of 𝜋, then 𝑥3𝑘=tan𝑘0𝜃𝑘1𝜃2𝑘𝑘𝜃2𝑘2𝑘0=tanarctan𝑥+𝑘1arctan𝑥2𝑘𝑘++arctan𝑥2𝑘,(3.21) where 𝜃𝑗=2arctan𝑥𝑗𝑗,2,,2𝑘.(3.22)
(III) If the sequence {𝑥𝑛}𝑛0 satisfies 𝑥3𝑛=𝑥𝑛𝑥2𝑛+1𝑥𝑛+𝑥2𝑛(𝑛0),(3.23) then, for 0(mod3),𝑘, one has 𝑥3𝑘=coth𝑘0𝜃𝑘1𝜃2𝑘𝑘𝜃2𝑘2𝑘0=cotharccoth𝑥+𝑘1arccoth𝑥2𝑘𝑘++arccoth𝑥2𝑘,(3.24) where 𝜃𝑗=2arccoth𝑥𝑗𝑗,2,,2𝑘.(3.25)
(IV) If the sequence {𝑥𝑛}𝑛0 satisfies 𝑥3𝑛=𝑥𝑛+𝑥2𝑛1+𝑥𝑛𝑥2𝑛(𝑛0),(3.26) then, for 0mod3, 𝑘, one has 𝑥3𝑘=tanh𝑘0𝜃𝑘1𝜃2𝑘𝑘𝜃2𝑘2𝑘0=tanharctanh𝑥+𝑘1arctanh𝑥2𝑘𝑘++arctanh𝑥2𝑘,(3.27) where 𝜃𝑗=2arctanh𝑥𝑗𝑗,2,,2𝑘.(3.28)

Proof. (I) As seen in Section 1, the RDAC relation (3.16) is equivalent to 𝑈3𝑛=𝑈𝑛𝑈2𝑛𝑈𝑛=𝑥𝑛𝑖𝑥𝑛+𝑖,(3.29) whose solution is, by virtue of Corollary 2.2, 𝑈3𝑘=𝑈𝑘0𝑈𝑘12𝑈𝑘𝑘2𝑘. Thus, 𝑥3𝑘𝑖𝑥3𝑘=𝑥+𝑖𝑖𝑥+𝑖𝑘0𝑥2𝑖𝑥2+𝑖𝑘1𝑥2𝑘𝑖𝑥2𝑘+𝑖𝑘𝑘.(3.30) Setting 𝑒𝑖𝜃𝑗=(𝑥𝑗𝑖)/(𝑥𝑗+𝑖)(𝑗{,2,,2𝑘}), one has 𝑥3𝑘=𝑖1+𝑒𝑖𝑘0𝜃𝑘𝑘𝜃++2𝑘1𝑒𝑖𝑘0𝜃𝑘𝑘𝜃++2𝑘=cot𝑘0𝜃𝑘𝑘𝜃2𝑘2𝑘0=cotarccot𝑥+𝑘1arccot𝑥2𝑘𝑘++arccot𝑥2𝑘,(3.31) provided 𝑘0𝜃++𝑘𝑘𝜃2𝑘0(mod2𝜋).
(II) Substituting 𝑥𝑛 by 1/𝑥𝑛 turns (3.20) into (3.16) and so the result follows at once from part (I).
(III) Substituting 𝑥𝑛 by 𝑖𝑥𝑛 in (3.23) turns it into a rational recursive equation of the form (3.16), and so part (I) yields the desired result.
(IV) Replacing 𝑥𝑛 by 𝑖𝑥𝑛 in (3.26), we get a rational recursive equation of the form (3.20), and part (II) yields the result.

Remark 3.4. As in the remark following Proposition 3.1, the substitution 𝑥𝑛 by 1/𝑥𝑛 in part (II) causes no harm should there exists integer 𝑁 such that 𝑥𝑁=0 by either interpreting the infinite value of the expressions on both sides of the solution as equal. Alternately, we may repeat the technique used in the proof of Proposition 3.3 to solve (3.20).

As for the case of general 𝑏, an entirely analogous proof as that in Proposition 3.3, which we omit here, leads to Proposition 3.5.

Proposition 3.5. Let 𝑏,𝑏2.
(I) Suppose that the sequence {𝑥𝑛}𝑛0 satisfies an RDAC relation of the form 𝑥𝑏𝑛=𝑥(𝑏1)𝑛𝑥𝑛1𝑥(𝑏1)𝑛+𝑥𝑛(𝑛1).(3.32) For 0(mod𝑏) and 𝑘, if the condition 𝑘0𝜃+𝑘1𝜃(𝑏1)𝑘𝑘𝜃++(𝑏1)𝑘0mod2𝜋(3.33) is fulfilled, then 𝑥𝑏𝑘=cot𝑘0𝜃𝑘1𝜃(𝑏1)𝑘𝑘𝜃(𝑏1)𝑘2𝑘0=cotarccot𝑥+𝑘1arccot𝑥(𝑏1)𝑘𝑘++arccot𝑥(𝑏1)𝑘,(3.34) where 𝜃𝑗=2arccot𝑥𝑗𝑗,(𝑏1),,(𝑏1)𝑘.(3.35)
(II) Assume that the sequence {𝑥𝑛}𝑛0 satisfies an RDAC relation of the form 𝑥𝑏𝑛=𝑥𝑛+𝑥(𝑏1)𝑛1𝑥𝑛𝑥(𝑏1)𝑛(𝑛0).(3.36) For 0(mod𝑏) and 𝑘, if 𝜃+𝑘1𝜃(𝑏1)++𝑘𝑘𝜃(𝑏1)𝑘 is not an odd multiple of 𝜋, then 𝑥𝑏𝑘=tan𝑘0𝜃𝑘1𝜃(𝑏1)𝑘𝑘𝜃(𝑏1)𝑘2𝑘0=tanarctan𝑥+𝑘1arctan𝑥(𝑏1)𝑘𝑘++arctan𝑥(𝑏1)𝑘,(3.37) where 𝜃𝑗=2arctan𝑥𝑗𝑗,(𝑏1),,(𝑏1)𝑘.(3.38)
(III) If the sequence {𝑥𝑛}𝑛0 satisfies 𝑥𝑏𝑛=𝑥𝑛𝑥(𝑏1)𝑛+1𝑥𝑛+𝑥(𝑏1)𝑛(𝑛0),(3.39) then, for 0(mod𝑏), 𝑘, one has 𝑥𝑏𝑘=coth𝑘0𝜃𝑘1𝜃(𝑏1)𝑘𝑘𝜃(𝑏1)𝑘2𝑘0=cotharccoth𝑥+𝑘1arccoth𝑥(𝑏1)𝑘𝑘++arccoth𝑥(𝑏1)𝑘,(3.40) where 𝜃𝑗=2arccoth𝑥𝑗𝑗,(𝑏1),,(𝑏1)𝑘.(3.41)
(IV) If the sequence {𝑥𝑛}𝑛0 satisfies 𝑥𝑏𝑛=𝑥𝑛+𝑥(𝑏1)𝑛1+𝑥𝑛𝑥(𝑏1)𝑛(𝑛0),(3.42) then, for 0(mod𝑏), 𝑘, one has 𝑥𝑏𝑘=tanh𝑘0𝜃𝑘1𝜃(𝑏1)𝑘𝑘𝜃(𝑏1)𝑘2𝑘0=tanharctanh𝑥+𝑘1arctanh𝑥(𝑏1)𝑘𝑘++arctanh𝑥(𝑏1)𝑘,(3.43) where 𝜃𝑗=2arctanh𝑥𝑗𝑗,(𝑏1),,(𝑏1)𝑘.(3.44)

When all the exponents 𝛼𝑗 in (2.2) are equal to 1, RDAC relations, even more general than those in Proposition 3.5 which can be explicitly solved by our device, are given in the next proposition.

Proposition 3.6. Let 𝑏,  𝑏2,  𝑤{0}. If the sequence {𝑥𝑛}𝑛0 satisfies 𝑥𝑏𝑛𝑤=𝑥𝑛𝑥+𝑤2𝑛𝑥+𝑤(𝑏1)𝑛+𝑥+𝑤𝑛𝑥𝑤2𝑛𝑥𝑤(𝑏1)𝑛𝑤𝑥𝑛𝑥+𝑤2𝑛𝑥+𝑤(𝑏1)𝑛𝑥+𝑤𝑛𝑥𝑤2𝑛𝑥𝑤(𝑏1)𝑛𝑤,(3.45) then for 0(mod𝑏) and 𝑘 one has 𝑥𝑏𝑘𝐴=𝑤++𝐴𝐴+𝐴,(3.46) provided the values exist, where 𝐴+=𝑖1+𝑖2++𝑖𝑏1=𝑥1𝑖12𝑖2(𝑏1)𝑖𝑏1+𝑤𝑘𝑖1,𝑖2,,𝑖𝑏1𝐴=𝑖1+𝑖2++𝑖𝑏1=𝑥1𝑖12𝑖2(𝑏1)𝑖𝑏1𝑤𝑘𝑖1,𝑖2,,𝑖𝑏1.(3.47)

Proof. Rewriting (3.45), we get 𝑥𝑏𝑛𝑤𝑥𝑏𝑛=𝑥+𝑤𝑛𝑤𝑥𝑛𝑥+𝑤2𝑛𝑤𝑥2𝑛𝑥+𝑤(𝑏1)𝑛𝑤𝑥(𝑏1)𝑛+𝑤(3.48) or 𝑈𝑏𝑛=𝑈𝑛𝑈2𝑛𝑈(𝑏1)𝑛𝑈𝑗=𝑥𝑗𝑤𝑥𝑗.+𝑤,𝑗{𝑛,2𝑛,,(𝑏1)𝑛}(3.49) Theorem 2.1 thus yields for 0(mod𝑏) and 𝑘𝑈𝑏𝑘=𝑖1+𝑖2++𝑖𝑏1=𝑈𝑘𝑖1,𝑖2,,𝑖𝑏11𝑖12𝑖2(𝑏1)𝑖𝑏1,(3.50) that is, 𝑥𝑏𝑘𝑤𝑥𝑏𝑘=+𝑤𝑖1+𝑖2++𝑖𝑏1=𝑥1𝑖12𝑖2(𝑏1)𝑖𝑏1𝑤𝑥1𝑖12𝑖2(𝑏1)𝑖𝑏1+𝑤𝑘𝑖1,𝑖2,,𝑖𝑏1,(3.51) and the result follows.

4. Global Behaviors

It is often desirable to know about global behaviors of the solutions of recursive equations, such as those in [6]. Using the explicit forms found above, this question is easily solved for RDAC relations in Proposition 3.5 with 𝑏=2.

Proposition 4.1. Let the notation be as in Proposition 3.1.
(I) Suppose that the sequence {𝑥𝑛}𝑛0 satisfies an RDAC relation of the form 𝑥2𝑛=𝑥2𝑛12𝑥n(𝑛1).(4.1) For each fixed 0(mod2) and 𝑘, (a)if 𝜃 is a rational multiple of 𝜋, then either {𝑥2𝑘}𝑘 diverges in finitely many steps or {𝑥2𝑘} is periodic;(b)if 𝜃 is not a rational multiple of 𝜋, then 𝑥2𝑘 exists for all 𝑘 and the sequence {𝑥2𝑘}𝑘 is never periodic.
(II) Suppose that the sequence {𝑥𝑛}𝑛0 satisfies an RDAC relation of the form 𝑥2𝑛=2𝑥𝑛1𝑥2𝑛(𝑛0).(4.2) For each fixed 0(mod2), (a)if 𝜃 is a rational multiple of 𝜋, then either {𝑥2𝑘}𝑘 diverges in finitely many steps or {𝑥2𝑘} is periodic; (b)if 𝜃 is not a rational multiple of 𝜋, then 𝑥2𝑘 exists for all 𝑘 and the sequence {𝑥2𝑘}𝑘 is never periodic.
(III) Suppose that the sequence {𝑥𝑛}n0 satisfies 𝑥2𝑛=𝑥2𝑛+12𝑥𝑛(𝑛0).(4.3) For each fixed 0(mod2), (a)if 𝜃=0, then the sequence {𝑥2𝑘}𝑘1 does not exist; (b)if 𝜃>0, the sequence {𝑥2𝑘} is strictly decreasing in the interval [coth(𝜃),1); (c)if 𝜃<0, then {𝑥2𝑘}𝑘1 is strictly increasing in the interval [coth(𝜃),1).
(IV) Suppose that the sequence {𝑥𝑛}𝑛0 satisfies 𝑥2𝑛=2𝑥𝑛1+𝑥2𝑛(𝑛0).(4.4) For each fixed 0(mod2), (z)if 𝜃=0, then {𝑥2𝑘}𝑘 is the zero sequence; (b)if 𝜃>0, then the sequence {𝑥2𝑘} is strictly increasing in [tanh(𝜃),1); (c)if 𝜃<0, the sequence is strictly decreasing in [tanh(𝜃),1).

Proof. (I) From part (I) of Proposition 3.1, we know that 𝑥2𝑘=cot2𝑘1𝜃(4.5) provided 2𝑘𝜃0(mod2𝜋). Consider the case where 𝜃 is a rational multiple of 𝜋, say, 𝜃=𝑚𝜋𝑡with𝑚,𝑡(>0),gcd(𝑚,𝑡)=1.(4.6) If 𝑡 is a multiple of 2, then it is easily checked that {𝑥2𝑘}𝑘 diverges in finitely many steps. If 𝑡2 is not a multiple of 2, let 𝑡=2𝑣𝑇, where 2𝑣𝑡,𝑇3. Observe that for all large 𝑛, when evaluating the values of cotangent, we need only look at 2𝑛𝜃=2𝑛𝑚𝜋𝑡=2𝑛𝑣𝑚𝜋𝑇(mod2𝜋),(4.7) which is equivalent to looking at 𝐺𝑛=2𝑛𝑣𝑚(mod2𝑇).(4.8) Since each 𝐺𝑛 takes at most 2𝑇 values and the sequence {𝑥2𝑘}𝑘 is infinite, there are positive integers 𝑁1<𝑁2 such that 𝐺𝑁1=𝐺𝑁2, which in turn implies that {𝑥2𝑘}𝑘 is periodic.
Finally, if 𝜃 is not a rational multiple of 𝜋, then 2𝑘1𝜃 is not a multiple of 𝜋 showing that the sequence {𝑥2𝑘}𝑘 is well defined and never periodic.
The proof of part (II) is similar to that of part (I).
(III) If 𝜃=0, then the values 𝑥2𝑘=coth(2𝑘𝜃) become infinite for all 𝑘 and part (a) follows. Since 𝑥2𝑘=coth(2𝑘𝜃) is a strictly decreasing (resp. increasing) function of 𝑘 according as 𝜃>0 (resp. 𝜃<0), the results in (b) and (c) are immediate.
(IV) If 𝜃=0, then 𝑥2𝑘=tanh(𝜃)=0. Arguments for the other two cases 𝜃>0 and 𝜃<0 are similar to those in part (III).

Note from Proposition 4.1 that global behaviors of solutions in the case 𝑏=2 depend solely on the single value 𝜃. The situation when 𝑏3 is more complex since their global behaviors depend heavily on the variable 𝑘 as we see in the following illustration. Keeping the notation of Proposition 3.5, suppose that the sequence {𝑥𝑛}𝑛0 satisfies an RDAC relation of the form x𝑏𝑛=𝑥(𝑏1)𝑛𝑥𝑛1𝑥(𝑏1)𝑛+𝑥𝑛(𝑛1).(4.9) From part (I) of Proposition 3.5, we know that 𝑥𝑏𝑘=cot𝑘0𝜃𝑘1𝜃(𝑏1)𝑘𝑘𝜃(𝑏1)𝑘2.(4.10) This explicit form shows that, for each fixed 0(mod𝑏), the behavior of 𝑥𝑏𝑘 considered as a function of 𝑘 depends on all 𝜃,,𝜃(𝑏1)𝑘, and we can merely infer that the values 𝑥𝑏𝑘 are well defined (i.e., finite) if and only if 𝑘0𝜃+𝑘1𝜃(𝑏1)𝑘𝑘𝜃++(𝑏1)𝑘(mod2𝜋).(4.11)

Acknowledgment

This paper is supported by the Commission on Higher Education, the Thailand Research Fund RTA5180005, and the Centre of Excellence in Mathematics, Thailand.