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Note on constructing a family of solvable sine-type difference equations

El-Sayed, Ahmed; Iričanin, Bratislav; Kosmala, Witold; Stevič, Stevo; Šmarda, Zdeněk

Abstract

We obtain a family of first order sine-type difference equations solvable in closed form in a constructive way, and we present a general solution to each of the equations.

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Ahmed et al. Advances in Difference Equations (2021) 2021:194 https://doi.org/10.1186/s13662-021-03348-2 RESEARCH Open Access Note on constructing a family of solvable sine-type difference equations Ahmed El-Sayed Ahmed1, Bratislav Iriˇ canin2,3, Witold Kosmala4,StevoStevi´ c5,6* and Zdenˇ ek Šmarda7 *Correspondence: [email protected] 5Mathematical Institute of the Serbian Academy of Sciences and Arts, Knez Mihailova 36/III, 11000, Beograd, Serbia 6Department of Medical Research, China Medical University Hospital, China Medical University, Taichung, 40402, Taiwan, Republic of China Full list of author information is available at the end of the article Abstract We obtain a family of first order sine-type difference equations solvable in closed form in a constructive way, and we present a general solution to each of the equations. MSC: Primary 39A20; secondary 39A06; 39A45 Keywords: Solvable difference equation; Closed-form formula; Sine-type difference equation 1 Introduction Throughout the paper we use the standard notations N,N0,Z,R,Cfor the sets of all natural numbers, nonnegative integers, integers, real numbers, and complex numbers, respectively. If l,m∈Z, then instead of using the notation l≤s≤m,s∈Z,wesimply write s=l,mand understand that the variable sbelongs to Z. Iffisaself-mappingofasetX,thenbyf[0](x)wedenotetheidentitymapf(x)=x,x∈X, whereas by f[k](x), where k∈N, we denote the iterated composition f[k](x):=ff···f   ktimes (x)··· for x∈X, consisting of the composition of kfunctions each of which is equal to the given function f. One of the first studied problems connected to recursive relations/difference equations was finding closed-form formulas for their solutions. First results on the problem was obtained at the beginning of the eighteenth century by de Moivre [7,8] and Bernoulli [4] for the case of homogeneous linear difference equations with constant coefficients. The investigation was continued by Euler [9] and to a larger extent by Lagrange (see, e.g.,[12,13]).Someinformationonclassicalsolvabledifferenceequationscanbefoundin manyclassicalbooksoncalculusoffinitedifferencesordifferenceequations,forexample, in [5,10,11,15–17,19], whereas some information on recently studied solvable difference equations and systems of difference equations and their invariants can be found, for example, in [3,20–23,25–35], as well as in the related references quoted therein. Another important figure in the investigation of solvability of difference equations and systems of difference equations of one or several independent variables was Laplace. ©The Author(s) 2021. This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/. Ahmed et al. Advances in Difference Equations (2021) 2021:194 Page 2 of 11 Among other things, he also noticed that some one-dimensional difference equations whichhaveformsofsomewell-knowntrigonometricformulasaresolvableinclosedform. Oneof thedifferenceequationsappearingin [14] isthefollowing nonlinearequationof first order: xn+1 =x2 n–2, n∈N0.(1) Laplacewrotetheinitialvaluex0intheforma+1 a(suchanacanbefoundforanyx0∈C), and based on calculations of the first few members of the sequence concluded that xn=a2n+1 a2n,n∈N0.(2) The closed-form formula in (2) can be regarded as a form of general solution to Eq. (1). By using the linear change of variables xn=2yn,n∈N0,(3) Equation (1) is transformed to the following one: yn+1 =2y2 n–1, n∈N0.(4) Equation(4)canbefoundinmanybooks.Bearinginmindthesimplerelation(3)between solutions to Eqs. (1)and(4), it is highly expected that Eq. (4) was also known to Laplace and some other mathematicians of that period of time. Note that Eq. (4) is similar to the well-known double-angle identity for the cosine, cos2x=2cos2x–1. This observation naturally suggests that the substitution yn=coszn,n∈N0,(5) is used in dealing with Eq. (4). Therefore, the equation is of cosine-type, as well as its relative (1). It seems that difference equations (1)and(4) were among the first nonlinear equations for which some closed-form formulas of their general solutions were found. The difference equation xn+1 =x3 n–3xn,n∈N0,(6) can be also found in some problem books and those dealing with sequences or difference equations. Convergence of solutions to the equation can be studied in several ways. An interestingfact isthatEq. (6)can bealsosolved inclosedforminthesamewayasLaplace did for the case of Eq. (1), that is, by taking x0=a+1 a, calculating the first few members of the sequence, and guessing a formula for general solution to the equation, which can Ahmed et al. Advances in Difference Equations (2021) 2021:194 Page 3 of 11 be verified, say, by the method of induction. But, unlike the case of Eq. (1) the use of the method is less obvious and requires more experience. The above-mentioned facts together with our recent studies of some difference equations and systems of difference equations related to the hyperbolic cotangent function (see, e.g., [29–31] and the related references therein), which are related to some producttype ones (see, e.g., [33–35]), motivated us to construct a sequence of solvable difference equations in closed form extending Eqs. (1)and(6) in a natural way, and to present their solutions in all possible cases. Therefore,in[32]wehaverecentlyconducedadetailedanalysisofconnectionsbetween the quantity a+1 aand polynomials. By using the analysis the following result was proved therein (it should be folklore, but it seems not to have been published in this form in the literature). Theorem1 Consider the difference equation xn+1 =Pk(xn), n∈N0,(7) where k ∈N\{1}and Pkis the polynomial given by Pk(t)=t+√t2–4 2k+t–√t2–4 2k.(8) Then the following statements hold: (a) If x0∈Cis given by x0=a+1 a,(9) then the solution to Eq.(7)with the initial value x0is given by xn=akn+1 akn, for n∈N0. (b) If x0∈Ris such that x0≥2,then the solution to Eq.(7)with the initial value x0is given by xn=x0+x2 0–4 2kn +x0+x2 0–4 2–kn , for n∈N0. (c) If x0∈Ris such that x0≤–2,then the solution to Eq.(7)with the initial value x0is given by xn=Pk(x0)–(Pk(x0))2–4 2kn–1 +Pk(x0)–(Pk(x0))2–4 2–kn–1 , (10) for n∈N. Ahmed et al. Advances in Difference Equations (2021) 2021:194 Page 4 of 11 (d) If |x0|≤2and x0=2cosθfor some θ∈[0,2π),then the solution to Eq.(7)with the initial value x0is given by xn=2cosknθ,n∈N0. Remark1Wewanttosaythatin[32]wasmadeaminoroversight,andusetheopportunity to explain how it should be corrected. Namely, instead of Eq. (10), in [32, Theorem 1] is written the following one: xn=x2 0–2–x0x2 0–4 2kn–1 +x2 0–2–x0x2 0–4 2–kn–1 ,n∈N, which is only true for k=2,thatis,insteadofPk(x0)iswrittenP2(x0)=x2 0–2.Thesame oversightwasmadeinPropositions1–4therein,whereinsteadofP2(x0)shouldhavebeen used Pj(x0), j=3,6, respectively, in the corresponding special cases of Eq. (10). Remark 2 Note also that since the following relation holds: Pk(x0)–(Pk(x0))2–4 2=Pk(x0)+(Pk(x0))2–4 2–1. Equation (10) can also be written it the following form: xn=Pk(x0)+(Pk(x0))2–4 2kn–1 +Pk(x0)+(Pk(x0))2–4 2–kn–1 , for n∈N. One of the main points in [32] is the fact that the polynomial Pksatisfies the following relation: Pka+1 a=ak+1 ak, for every a∈C\{0},andforeachk∈N. This property of the polynomial Pkconsiderably helps in finding closed-form formulas forthesolutionstotheequationsin(7).TheformsofthesolutionsinTheorem1showthat they are related to the cosine and hyperbolic cosine functions. One can construct other typesofdifferenceequationswhosesolutionsarecosineofsomesequencesbyusingother trigonometric relations containing the function (see, e.g., [5, p.169]). Another important feature of the sequence of polynomials Pk,k∈N, is that there is a recursive relation of second order which is satisfied by them. Moreover, the recursive relationissolvable in closedform.Byusing therecursive relation it wasobtained the representation of the polynomials given in (8) (at first sight the representation does not look likeapolynomial,butbyusingthebinomialformulaandsomesimplecalculationsitisnot difficult to see that it really produces some polynomials). Motivatedbyallthefactsabovementioned,andbytheimportanceof thequantitya+1 a intheinvestigationin[32],itisnaturaltoseewhatcanbeobtainedifweusethefollowing relatedquantity:a–1 a,insteadofa+1 a.Thisnoteisdevotedtoinvestigatingoftheproblem. Ahmed et al. Advances in Difference Equations (2021) 2021:194 Page 5 of 11 Ourmainaimhereistoobtainafamily/sequenceoffirstordersine-typedifferenceequations solvable in closed form in a constructive way, and present general solution to each of the equations. 2 Main results This section presents a detailed analysis which leads to a construction of a sequence of polynomials whichwill be effectivelyemployed in the main resultofthisnote.Itis shown thatthesequenceofpolynomialssatisfies arecursiverelationofsecondorderwhich,similartothecaseofthepolynomialsdefinedinTheorem1,canbealsosolvedinclosedform. After the construction of the sequence of polynomials we present and prove the main result in this note, on the existence of a sequence of sine-type difference equations of first order which arealso solvable in closed form. 2.1 Construction of a sequence of polynomials Our considerations in [32] essentially started from the following simple relation: z2+1 z2=z+1 z2–2, z∈C\{0}, (11) which was used by Laplace in [14]. Based on the relation and few other related ones, in [32]weshownthatimportantthingsinthestudythereinweresomerelationsbetweenthe functions f(z)=z+1 z(12) and fk(z)=zk+1 zk,k∈N. Thefunctionin(12),bywhichtheinitialvaluex0inTheorem1isrepresented,isfrequently applied in various fields of mathematics. For example, in solving polynomial equations (see, e.g., [2, p.27]), and in conformal mappings (see, e.g., [1]). Recall that coszand coshz, which are some of the basic analytic functions are defined by using the function. Inamajorityofcasesinthescientificliterature,alongwithcosineandhyperbolic cosine functions, one considers simultaneously their counterparts, that is, the sine and hyperbolic sine functions. Unlike coszand coshz, these two functions are defined by using the function g(z)=z–1 z. (13) However,unlikethecaseoffunctionsf(z)andf(z2),whereaccordingto(11),thefollowing very useful relation exists: fz2=f2(z)–2, there is no such a useful relation between the functions g(z)andg(z2). Note that g2(z)=fz2–2. (14) Ahmed et al. Advances in Difference Equations (2021) 2021:194 Page 6 of 11 On the other hand, there is a useful relation between g(z)andg(z3). Indeed, since z–1 z3=z3–3z+3 z–1 z3, we have gz3=z3–1 z3=z–1 z3+3z–1 z=g3(z)+3g(z). (15) Further, note that z–1 z4=z4–4z2+6– 4 z2+1 z4, from which along with (14) it follows that fz4=z4+1 z4=z–1 z4+4z–1 z2+2=g4(z)+4g2(z)+2, (16) whichisarelationbetween f(z4)andg(z).However,similartoEq.(14),therelationin(16) is also not very useful for our present investigation. Further, we also have z–1 z5=z5–5z3+10z–10 z+5 z3–1 z5, from which along with (15) it follows that gz5=z5–1 z5=z–1 z5+5z–1 z=g5(z)+5g3(z)+5g(z), (17) which is a relation between g(z)andg(z5). From (15)and(17)wehave g(z)=Q1g(z), gz3=Q2g(z), (18) gz5=Q3g(z), where Q1(t)=t, Q2(t)=t3+3t, (19) Q3(t)=t5+5t3+5t. By using the method of induction it can be proved that gz2k–1=Qkg(z),k∈N, (20) Ahmed et al. Advances in Difference Equations (2021) 2021:194 Page 7 of 11 for some polynomials Qk(t) of the following form: Qk(t)=t2k–1 +(2k–1)t2k–3 +k–2  j=1 ajt2j–1, (21) where it seems not quite easy to calculate the coefficients aj,j=1,k–2,inthisway. To present the polynomials Qk(t) in a better form for applications in this note, we use here our idea in [32] on finding a recursive relation which these polynomials satisfy. Now note that z2k–1 –1 z2k–1 z2+1 z2=z2k+1 –1 z2k+1 +z2k–3 –1 z2k–3 , from which along with (14)and(20) it follows that Qk+1g(z)=Q2 1g(z)+2Qkg(z)–Qk–1g(z), for k≥2. Hence,thesequenceofpolynomialsQk(t),satisfiesthefollowingsecondorderrecursive relation: Qk+1(t)=t2+2Qk(t)–Qk–1(t), (22) for k≥2. Byusingrelation(22)andthefirsttwo(initial)conditionsin(19),wecanfindanymemberofthesequenceQk(t).Moreover,Eq.(22)issolvableinclosedformasahomogeneous second order linear difference equation (for a fixed tit is a difference equation with constant coefficients). The polynomial  P2(λ)=λ2–t2+2λ+1 is the characteristic one associated to Eq. (22), and the (characteristic) zeros of the polynomial are λ1=t2+2+t√t2+4 2and λ2=t2+2–t√t2+4 2. (23) Hence, by a well-known theorem general solution to the difference equation (22)hasthe following form: Qk(t)=c1t2+2+t√t2+4 2k+c2t2+2–t√t2+4 2k, (24) for k∈N,wherec1and c2are arbitrary constants. Ahmed et al. Advances in Difference Equations (2021) 2021:194 Page 8 of 11 From (19)and(24)withk=1,2, weobtainthefollowingtwo-dimensionallinearsystem of algebraic equations: c1t2+2+t√t2+4 2+c2t2+2–t√t2+4 2=t, c1t2+2+t√t2+4 22+c2t2+2–t√t2+4 22=t3+3t. (25) The determinant of system (25)is = t2+2+t√t2+4 2t2+2–t√t2+4 2 t2+2+t√t2+4 22t2+2–t√t2+4 22=–t√t2+4. Hence, after some calculations, we have c1=1  tt2+2–t√t2+4 2 t3+3tt2+2–t√t2+4 22=λ2 √t2+4+t 2(26) and c2=1  t2+2+t√t2+4 2t t2+2+t√t2+4 22t3+3t=–λ1 √t2+4–t 2. (27) By using (26)and(27)inEq.(24), as well as Vieta’s formulas, it follows that Qk(t)=√t2+4+t 2t2+2+t√t2+4 2k–1 –√t2+4–t 2t2+2–t√t2+4 2k–1, (28) for k∈N. 2.2 A sequence of solvable sine-type difference equations Now after the construction of polynomials Qk(t), k∈N, which has been done in the previous subsection, we are in a position to state and prove our main result on the existence of a sequence of sine-type difference equations of first order which are solvable in closed form. Theorem2 Consider the difference equation xn+1 =Qk(xn), n∈N0, (29) where k ∈N\{1}and the polynomial Qkis given by (28). Then the following statements hold: Ahmed et al. Advances in Difference Equations (2021) 2021:194 Page 9 of 11 (a) If x0∈C,then the solution to Eq.(29)with the initial value x0is given by xn=x0+x2 0+4 2(2k–1)n +x0–x2 0+4 2–(2k–1)n , (30) for n∈N0. (b) If x0=2isinθ(31) for some θ∈[0,2π),then the solution to Eq.(29)with the initial value x0is given by xn=2isin(2k–1)nθ,n∈N0. (32) Proof (a) First note that for each x0∈Cthere is asuch that the relation x0=a–1 a(33) holds. Indeed,from(33)wehavea2–x0a–1=0, from which it follows that a1=x0+x2 0+4 2and a2=x0–x2 0+4 2. (34) Let a:=a1.Then,fromEq.(29) and by a simple inductive argument, we have xn=Qk(xn–1)=Q[n] k(x0), (35) for n∈N0. By using Eq. (20)in(35)wehave xn=Q[n] kg(a)=Q[n–1] kQkg(a)=Q[n–1] kga2k–1, for n∈N. Byusinga similar argumentandthemethodofmathematical induction it isshownthat xn=ga(2k–1)n, for n∈N, that is, we have xn=a(2k–1)n–1 a(2k–1)n,n∈N0. (36) From (34), (36) and since 1 a=–x0–x2 0+4 2