Access the full text.
Sign up today, get DeepDyve free for 14 days.
H. Begehr, Hanxing Lin, Hua Liu, B. Shupeyeva (2020)
An iterative real sequence based on and providing a plane parqueting and harmonic Green functionsComplex Variables and Elliptic Equations, 66
1IntroductionWhile sequences are known as mappings from the set of natural numbers, double (two-sided) sequences are based on the entire numbers.Although a double sequence can be rearranged in a sequence, such a rearrangement does not necessarily result in a convergent sequence even if the two ends of the double sequence are convergent (when the indices tend to +∞{+\infty}or -∞{-\infty}).Of course, other combinations of the limit behavior are possible.Double sequences appear in a natural way in the cases of iteratively given sequences if the iteration recipe allows to determine besides the successors from the predecessors also the predecessors from their followers.Repeatedly, iteratively given sequences appear when applying the parqueting-reflection principle to certain circular domains of the (complex) plane; see e.g. [1].A pair of double sequences arises when repeatedly reflecting a certain circular rectangle at its boundary parts [2].This pair of double sequences consists of sequences, the tails of which are, on one hand, natural numbers sequences (see [3]) and, on the other hand, made from rational numbers.Sequences can be extended to ones with complex numbers bearing similar properties as their origins.2Recurrence relationsWith a0=b0=1{a_{0}=b_{0}=1}for k∈ℕ{k\in\mathbb{N}}, the relationsa2k=3a2k-1+b2k-1,\displaystyle a_{2k}=3a_{2k-1}+b_{2k-1},a2k+1=a2k+b2k,\displaystyle\hskip 10.0pta_{2k+1}=a_{2k}+b_{2k},b2k=a2k-1+b2k-1,\displaystyle b_{2k}=a_{2k-1}+b_{2k-1},b2k+1=a2k+3b2k\displaystyle\hskip 10.0ptb_{2k+1}=a_{2k}+3b_{2k}determine two real sequences.Obviously, both sets of equations may be solved for the lower indexed numbers as2a2k-1=a2k-b2k,\displaystyle 2a_{2k-1}=a_{2k}-b_{2k},2a2k=3a2k+1-b2k+1,\displaystyle\hskip 10.0pt2a_{2k}=3a_{2k+1}-b_{2k+1},2b2k-1=3b2k-a2k,\displaystyle 2b_{2k-1}=3b_{2k}-a_{2k},2b2k=b2k+1-a2k+1.\displaystyle\hskip 10.0pt2b_{2k}=b_{2k+1}-a_{2k+1}.Hence, (ak,bk){(a_{k},b_{k})}is a pair of double sequences (ak){(a_{k})}, (bk){(b_{k})}, their first numbers being…,a-4=-54,\displaystyle\dots,a_{-4}=\frac{-5}{4},a-3=-12,\displaystyle\hskip 10.0pta_{-3}=\frac{-1}{2},a-2 =-12,\displaystyle a_{-2}\hskip 10.0pt=\frac{-1}{2},a-1=0,\displaystyle a_{-1}=0,a0=1,\displaystyle\hskip 10.0pta_{0}=1,a1 =2,\displaystyle a_{1}\hskip 10.0pt=2,a2=10,\displaystyle a_{2}=10,a3=16,\displaystyle\hskip 10.0pta_{3}=16,a4 =76,…,\displaystyle a_{4}\hskip 10.0pt=76,\dots,…,b-4=14,\displaystyle\dots,b_{-4}=\frac{1}{4},b-3=1,\displaystyle\hskip 10.0ptb_{-3}=1,b-2 =12,\displaystyle b_{-2}\hskip 10.0pt=\frac{1}{2},b-1=1,\displaystyle b_{-1}=1,b0=1,\displaystyle\hskip 10.0ptb_{0}=1,b1 =4,\displaystyle b_{1}\hskip 10.0pt=4,b2=6,\displaystyle b_{2}=6,b3=28,\displaystyle\hskip 10.0ptb_{3}=28,b4 =44,….\displaystyle b_{4}\hskip 10.0pt=44,\dots.3Properties of the double sequencesIn order to determine the particular explicit numbers in the sequences, some properties are investigated.Lemma 3.1.For any k∈Z{k\in\mathbb{Z}}the relations 3b2k2-a2k2=2k+1{3b_{2k}^{2}-a_{2k}^{2}=2^{k+1}}and b2k+12-3a2k+1=22k+2{b_{2k+1}^{2}-3a_{2k+1}=2^{2k+2}}hold.Proof.For k=0{k=0}and k=1{k=1}, the relations 3b02-a02=2{3b_{0}^{2}-a_{0}^{2}=2}and b12-3a12=22{b_{1}^{2}-3a_{1}^{2}=2^{2}}are true.By the recurrence relations, for 0<k{0<k},3b2k2-a2k2=3(a2k-1+b2k-1)2-(3a2k-1+b2k-1)2=2(b2k-12-3a2k-12)=2[(a2k-2+3b2k-2)2-3(a2k-2+b2k-2)2]=4(3b2k-22-a2k-22),\displaystyle\begin{split}\displaystyle 3b_{2k}^{2}-a_{2k}^{2}&\displaystyle=3%(a_{2k-1}+b_{2k-1})^{2}-(3a_{2k-1}+b_{2k-1})^{2}=2(b_{2k-1}^{2}-3a_{2k-1}^{2})%\\&\displaystyle=2[(a_{2k-2}+3b_{2k-2})^{2}-3(a_{2k-2}+b_{2k-2})^{2}]=4(3b_{2k-2%}^{2}-a_{2k-2}^{2}),\end{split}b2k+12-3a2k+1=(a2k+3b2k)2-3(a2k+b2k)2=2(3b2k2-a2k2)=2[3(a2k-1+b2k-1)2-(3a2k-1+b2k-1)2]=4(b2k-12-3a2k-12)\displaystyle\begin{split}\displaystyle b_{2k+1}^{2}-3a_{2k+1}&\displaystyle=(%a_{2k}+3b_{2k})^{2}-3(a_{2k}+b_{2k})^{2}=2(3b_{2k}^{2}-a_{2k}^{2})\\&\displaystyle=2[3(a_{2k-1}+b_{2k-1})^{2}-(3a_{2k-1}+b_{2k-1})^{2}]=4(b_{2k-1}%^{2}-3a_{2k-1}^{2})\end{split}hold, and for k<0{k<0},3b2k2-a2k2=34(b2k+1-a2k+1)2-14(3a2k+1-b2k+1)2=12(b2k+12-3a2k+12)=18[(3b2k+2-a2k+2)2-3(a2k+2-b2k+2)2]=14(3b2k+22-a2k+22),\displaystyle\begin{split}\displaystyle 3b_{2k}^{2}-a_{2k}^{2}&\displaystyle=%\frac{3}{4}(b_{2k+1}-a_{2k+1})^{2}-\frac{1}{4}(3a_{2k+1}-b_{2k+1})^{2}=\frac{1%}{2}(b_{2k+1}^{2}-3a_{2k+1}^{2})\\&\displaystyle=\frac{1}{8}[(3b_{2k+2}-a_{2k+2})^{2}-3(a_{2k+2}-b_{2k+2})^{2}]=%\frac{1}{4}(3b_{2k+2}^{2}-a_{2k+2}^{2}),\end{split}b2k-12-3a2k-12=14(3b2k-a2k)2-34(a2k-b2k)2=12(3b2k2-a2k2)=18[3(b2k+1-a2k+1)2-(3a2k+1-b2k+1)2]=14(b2k+12-3a2k+12).∎\displaystyle\begin{split}\displaystyle b_{2k-1}^{2}-3a_{2k-1}^{2}&%\displaystyle=\frac{1}{4}(3b_{2k}-a_{2k})^{2}-\frac{3}{4}(a_{2k-b_{2k}})^{2}=%\frac{1}{2}(3b_{2k}^{2}-a_{2k}^{2})\\&\displaystyle=\frac{1}{8}[3(b_{2k+1}-a_{2k+1})^{2}-(3a_{2k+1}-b_{2k+1})^{2}]=%\frac{1}{4}(b_{2k+1}^{2}-3a_{2k+1}^{2}).\qed\end{split}Remark 3.2.Both formulas from the last lemma are unified as31+[k2]-[k+12]bk2-3[k+12]-[k2]ak2=2k+1.3^{1+[\frac{k}{2}]-[\frac{k+1}{2}]}b_{k}^{2}-3^{[\frac{k+1}{2}]-[\frac{k}{2}]}%a_{k}^{2}=2^{k+1}.For convenience, the notation m1=3{m_{1}=\sqrt{3}}will be further used.Lemma 3.3.For k∈Z{k\in\mathbb{Z}}, we have m1b2k±a2k=(m1±1)2k+1{m_{1}b_{2k}\pm a_{2k}=(m_{1}\pm 1)^{2k+1}}and b2k+1±m1a2k+1=(m1±1)2k+2{b_{2k+1}\pm m_{1}a_{2k+1}=(m_{1}\pm 1)^{2k+2}}.Proof.For k=0{k=0}and k=1{k=1}, the relations m1b0±a0=m1±1{m_{1}b_{0}\pm a_{0}=m_{1}\pm 1}and b1±m1a1=4±2m1=(m1±1)2{b_{1}\pm m_{1}a_{1}=4\pm 2m_{1}=(m_{1}\pm 1)^{2}}hold.For 0<k{0<k},m1b2k±a2k=m1(a2n-1+b2k-1)±(3a2n-1+b2n-1)=(m1±1)(b2k-1±m1a2n-1)=(m1±1)[(a2k-2+3b2k-2)±m1(a2k-2+b2k-2)]=(m1±1)2(m1b2n-2±a2k-2),\displaystyle\begin{split}\displaystyle m_{1}b_{2k}\pm a_{2k}&\displaystyle=m_%{1}(a_{2n-1}+b_{2k-1})\pm(3a_{2n-1}+b_{2n-1})=(m_{1}\pm 1)(b_{2k-1}\pm m_{1}a_%{2n-1})\\&\displaystyle=(m_{1}\pm 1)[(a_{2k-2}+3b_{2k-2})\pm m_{1}(a_{2k-2}+b_{2k-2})]=%(m_{1}\pm 1)^{2}(m_{1}b_{2n-2}\pm a_{2k-2}),\end{split}b2k+1±m1a2k+1=a2k+3b2k±m1(a2k+b2k)=(m1±1)(m1b2k±a2k)=(m1±1)[m1(a2k-1+b2k-1)±(3a2k-1+b2k-1)]=(m1±1)2(b2k-1±m1a2k-1),\displaystyle\begin{split}\displaystyle b_{2k+1}\pm m_{1}a_{2k+1}&%\displaystyle=a_{2k}+3b_{2k}\pm m_{1}(a_{2k}+b_{2k})=(m_{1}\pm 1)(m_{1}b_{2k}%\pm a_{2k})\\&\displaystyle=(m_{1}\pm 1)[m_{1}(a_{2k-1}+b_{2k-1})\pm(3a_{2k-1}+b_{2k-1})]=(%m_{1}\pm 1)^{2}(b_{2k-1}\pm m_{1}a_{2k-1}),\end{split}and for k<0{k<0}, taking into account (m1±1)(m1∓1)=2{(m_{1}\pm 1)(m_{1}\mp 1)=2},m1b2k±a2k=m12(b2k+1-a2k+1)±12(3a2k+1-b2k+1)=12[(m1∓1)b2k+1-m1(1∓m1)a2k+1]=m1∓12(b2k+1±m1a2k+1)=m1∓12[3b2k+2-a2k+2±m1(a2k+2-b2k+2)]=m1∓14[(3∓m1)b2k+2-(1∓m1)a2k+2]=(m1∓1)24(m1b2k+2±a2k+2)=((m1∓1)24)-k(m1b0±a0)=((m1∓1)24)-k(m1±1)=((m1∓1)24)-k-1m1∓12=((m1∓1)2)-2k-1=(m1±1)2k+1,\displaystyle\begin{split}\displaystyle m_{1}b_{2k}\pm a_{2k}&\displaystyle=%\frac{m_{1}}{2(b_{2k+1}-a_{2k+1})}\pm\frac{1}{2}(3a_{2k+1}-b_{2k+1})\\&\displaystyle=\frac{1}{2}[(m_{1}\mp 1)b_{2k+1}-m_{1}(1\mp m_{1})a_{2k+1}]=%\frac{m_{1}\mp 1}{2}(b_{2k+1}\pm m_{1}a_{2k+1})\\&\displaystyle=\frac{m_{1}\mp 1}{2}[3b_{2k+2}-a_{2k+2}\pm m_{1}(a_{2k+2}-b_{2k%+2})]\\&\displaystyle=\frac{m_{1}\mp 1}{4}[(3\mp m_{1})b_{2k+2}-(1\mp m_{1})a_{2k+2}]%=\frac{(m_{1}\mp 1)^{2}}{4}(m_{1}b_{2k+2}\pm a_{2k+2})\\&\displaystyle=\biggl{(}\frac{(m_{1}\mp 1)^{2}}{4}\biggr{)}^{-k}(m_{1}b_{0}\pma%_{0})=\biggl{(}\frac{(m_{1}\mp 1)^{2}}{4}\biggr{)}^{-k}(m_{1}\pm 1)\\&\displaystyle=\biggl{(}\frac{(m_{1}\mp 1)^{2}}{4}\biggr{)}^{-k-1}\frac{m_{1}%\mp 1}{2}=\biggl{(}\frac{(m_{1}\mp 1)}{2}\biggr{)}^{-2k-1}=(m_{1}\pm 1)^{2k+1}%,\end{split}b2k-1±m1a2k-1=12[3b2k-a2k±m1(a2k-b2k)]=m1∓12(m1b2k±a2k)=m1∓14[m1(b2k+1-a2k+1)±(3a2k+1-b2k+1)]=(m1∓1)24(b2k+1±m1a2k+1)=((m1∓1)24)-k(b-1±a-1)=((m1∓1)24)-k=(4∓2m14)-k=(m1∓12)-2k=(m1±1)2k.∎\displaystyle\begin{split}\displaystyle b_{2k-1}\pm m_{1}a_{2k-1}&%\displaystyle=\frac{1}{2}[3b_{2k}-a_{2k}\pm m_{1}(a_{2k}-b_{2k})]=\frac{m_{1}%\mp 1}{2}(m_{1}b_{2k}\pm a_{2k})\\&\displaystyle=\frac{m_{1}\mp 1}{4}[m_{1}(b_{2k+1}-a_{2k+1})\pm(3a_{2k+1}-b_{2%k+1})]=\frac{(m_{1}\mp 1)^{2}}{4}(b_{2k+1}\pm m_{1}a_{2k+1})\\&\displaystyle=\biggl{(}\frac{(m_{1}\mp 1)^{2}}{4}\biggr{)}^{-k}(b_{-1}\pm a_{%-1})=\biggl{(}\frac{(m_{1}\mp 1)^{2}}{4}\biggr{)}^{-k}\\&\displaystyle=\biggl{(}\frac{4\mp 2m_{1}}{4}\biggr{)}^{-k}=\biggl{(}\frac{m_{%1}\mp 1}{2}\biggr{)}^{-2k}=(m_{1}\pm 1)^{2k}.\qed\end{split}With the formulas from Lemma 3.3, the terms of the sequences are determined.Theorem 3.4.For k∈Z{k\in\mathbb{Z}}, the double sequences (ak){(a_{k})}and (bk){(b_{k})}are given via2a2k=(m1+1)2k+1-(m1-1)2k+1,\displaystyle 2a_{2k}=(m_{1}+1)^{2k+1}-(m_{1}-1)^{2k+1},2m1a2k+1=(m1+1)2k+2-(m1-1)2k+2,\displaystyle\hskip 10.0pt2m_{1}a_{2k+1}=(m_{1}+1)^{2k+2}-(m_{1}-1)^{2k+2},2m1b2k=(m1+1)2k+1+(m1-1)2k+1,\displaystyle 2m_{1}b_{2k}=(m_{1}+1)^{2k+1}+(m_{1}-1)^{2k+1},2b2k+1=(m1+1)2k+2+(m1-1)2k+2.\displaystyle\hskip 10.0pt2b_{2k+1}=(m_{1}+1)^{2k+2}+(m_{1}-1)^{2k+2}.4Complex versionThe double sequence pair (ak,bk){(a_{k},b_{k})}of rational numbers handled in ℚ(3){\mathbb{Q}(\sqrt{3})}has the counterpart in ℚ(3,i){\mathbb{Q}(\sqrt{3},i)}.For k∈ℤ{k\in\mathbb{Z}}, define complex numbers asc2k=(-1)ka2k,c2k+1=(-1)k+1im1a2k+1,d2k=(-1)k+1im1b2k,d2k+1=(-1)k+1b2k+1.c_{2k}=(-1)^{k}a_{2k},\quad c_{2k+1}=(-1)^{k+1}im_{1}a_{2k+1},\quad d_{2k}=(-1%)^{k+1}im_{1}b_{2k},\quad d_{2k+1}=(-1)^{k+1}b_{2k+1}.Then the recursion relations for ak{a_{k}}and bk{b_{k}}are reflected into ones for ck{c_{k}}and dk{d_{k}}asck=-im1ck-1+dk-1,\displaystyle c_{k}=-im_{1}c_{k-1}+d_{k-1},dk=-ck-1-im1dk-1\displaystyle\hskip 10.0ptd_{k}=-c_{k-1}-im_{1}d_{k-1}withc0=1,d0=-im1,\hskip 10.0pt\text{with}\ c_{0}=1,\,d_{0}=-im_{1},2ck-1=im1ck+dk,\displaystyle 2c_{k-1}=im_{1}c_{k}+d_{k},2dk-1=im1dk-ck\displaystyle\hskip 10.0pt2d_{k-1}=im_{1}d_{k}-c_{k}withc-1=0,d-1=1.\hskip 10.0pt\text{with}\ c_{-1}=0,\,d_{-1}=1.The first terms are…,c-4=-54,\displaystyle\dots,c_{-4}=-\frac{5}{4},c-3=12im1,\displaystyle\hskip 10.0ptc_{-3}=\frac{1}{2}im_{1},c-2 =12,\displaystyle c_{-2}\hskip 10.0pt=\frac{1}{2},c-1=0,\displaystyle c_{-1}=0,c0=1,\displaystyle\hskip 10.0ptc_{0}=1,c1 =-2im1,\displaystyle c_{1}\hskip 10.0pt=-2im_{1},c2=10,\displaystyle c_{2}=10,c3=-16im1,\displaystyle\hskip 10.0ptc_{3}=-16im_{1},c4 =76,…,\displaystyle c_{4}\hskip 10.0pt=76,\dots,…,d-4=-34im1,\displaystyle\dots,d_{-4}=-\frac{3}{4}im_{1},d-3=-1,\displaystyle\hskip 10.0ptd_{-3}=-1,d-2 =12im1,\displaystyle d_{-2}\hskip 10.0pt=\frac{1}{2}im_{1},d-1=1,\displaystyle d_{-1}=1,d0=-im1,\displaystyle\hskip 10.0ptd_{0}=-im_{1},d1 =-4,\displaystyle d_{1}\hskip 10.0pt=-4,d2=6im1,\displaystyle d_{2}=6im_{1},d3=28,\displaystyle\hskip 10.0ptd_{3}=28,d4 =-44im1,….\displaystyle d_{4}\hskip 10.0pt=-44im_{1},\dots.Their properties are|dk|2-|ck|2=2k+1,|ck|+|dk|=(m1+1)k+1,|dk|-|ck|=(m1-1)k+1,\displaystyle\lvert d_{k}\rvert^{2}-\lvert c_{k}\rvert^{2}=2^{k+1},\quad\lvertc%_{k}\rvert+\lvert d_{k}\rvert=(m_{1}+1)^{k+1},\quad\lvert d_{k}\rvert-\lvert c%_{k}\rvert=(m_{1}-1)^{k+1},ck2+dk2=(-2)k+1,ckdk¯+ck¯dk=0,-ickdk¯=ick¯dk=|ckdk|\displaystyle c_{k}^{2}+d_{k}^{2}=(-2)^{k+1},\quad c_{k}\overline{d_{k}}+%\overline{c_{k}}d_{k}=0,\quad-ic_{k}\overline{d_{k}}=i\overline{c_{k}}d_{k}=%\lvert c_{k}d_{k}\rvertfor k∈ℤ{k\in\mathbb{Z}}.
Georgian Mathematical Journal – de Gruyter
Published: Apr 1, 2022
Keywords: Two-sided sequences; natural numbers sequences; complex sequences; 40B05; 40A05
You can share this free article with as many people as you like with the url below! We hope you enjoy this feature!
Read and print from thousands of top scholarly journals.
Already have an account? Log in
Bookmark this article. You can see your Bookmarks on your DeepDyve Library.
To save an article, log in first, or sign up for a DeepDyve account if you don’t already have one.
Copy and paste the desired citation format or use the link below to download a file formatted for EndNote
Access the full text.
Sign up today, get DeepDyve free for 14 days.
All DeepDyve websites use cookies to improve your online experience. They were placed on your computer when you launched this website. You can change your cookie settings through your browser.