In this paper, we consider explicit representations and convergence of Hermite-&nbsp;Lagrange Interpolation on two disjoint sets of nodes, which are obtained by&nbsp;projecting vertically the zeros of (1- x2) Pn&nbsp;(? ,? )&nbsp;(x) and Pn&nbsp;(&nbsp;? ,? )(x)&nbsp; respectively on the unit&nbsp;circle, where Pn(&nbsp;?, ?)&nbsp; (x) stands for Jacobi polynomials

Bahadur, Swarnima

English

KHALSA PUBLICATIONS

  ISSN 2347-1921                                                           1817 | P a g e                                                           J u n e  2 1 ,  2 0 1 4   Hermite – Lagrange Interpolation on a Unit Circle 1Swarnima Bahadur and 2Manisha Shukla Department of Mathematics and Astronomy University of Lucknow, Lucknow 226007, India 1swarnimabahadur@ymail.com, 2manishashukla2626@gmail.com  ABSTRACT: In this paper , we consider the explicit representation and convergence of Hermite-Lagrange interpolation on two disjoint sets of nodes, which are obtained by projecting vertically the zeros of  1-x2 Pn 𝛼 ,𝛽  𝑥  and Pn 𝛼 ,𝛽 ′ 𝑥  respectively on the unit circle , where Pn 𝛼 ,𝛽  𝑥  stands for Jacobi polynomial.  Mathematics subject classification: 41A05, 30E10. Keywords: Jacobi polynomial; explicit representation; Convergence.                          Council for Innovative Research Peer Review Research Publishing System Journal: Journal of Advances in Mathematics Vol 9, No 1 editor@cirjam.org www.cirjam.com, www.cirworld.com    ISSN 2347-1921                                                           1818 | P a g e                                                           J u n e  2 1 ,  2 0 1 4   1. INTRODUCTION: In a paper, T. N. T. Goodman and A. Sharma [8] considered convergence and divergence behaviour of Hermite interpolation in the circle of radius 𝜌32. In addition, T. N. T .Goodman, K.G. Ivanov and A. Sharma [9] considered the behaviour of the  Hermite  interpolant  in  the roots  of  unity.  In 1998,   S.  Bahadur and K.  K.  Mathur   [2]   proved the convergence of   Quasi - Hermite interpolation   on the   nodes obtained  by  projecting  vertically  the  zeros  of             1 − 𝑥2 𝑃𝑛 𝑥 on the unit circle, where 𝑃𝑛 𝑥  stands for nth Legendre polynomial. Later on, the authors [6] considered the convergence of Hermite interpolation.  Earlier, in 1975 L.G. Pál [13] proved that, when function values are prescribed on one set of n points and derivative values on another set of n-1 points, then there exists no unique polynomial of degree ≤ 2n-2, but by prescribing function value at one more point not belonging to the former set of n points, there exists a unique polynomial of degree ≤ 2𝑛 − 1. Later on, many authors ([3] - [16]) have dealt with the above method of interpolation on the various sets of nodes. In 2006, M. Lenard [11] considered the weighted (0, 2) Pál-type interpolation problems on the zeros of Legendre polynomial 𝑃𝑛 𝑥  and gave the explicit formulae. In another paper, first author considered ([3] [4]) (0,1;0) and (0; 0,1) interpolation problem for the vertically projected zeros of the Legendre polynomial on a unit circle. Also, P. Mathur [12] considered (0,1; 0) interpolation on infinite interval. Many other mathematicians also worked in the same direction. This has motivated us to consider Hermite -Lagrange interpolation on the unit circle. In this paper, we consider two pair wise disjoint sets  𝑧𝑘 𝑘=02𝑛+1 and  𝑡𝑘 𝑘=12𝑛−2 which are the vertically projected zeros of  1-x2 Pn 𝛼 ,𝛽  𝑥  and Pn 𝛼 ,𝛽 ′ 𝑥  respectively on the unit circle. Let 𝒁𝒏 =  𝑧𝑘 ∶ 𝑘 = 0 1 2𝑛 + 1  satisfying:  }11,,sincos                       ,1,1{ Z      (1.1)120nnkzzizzzkknkkkn 𝑻𝒏 =  𝑡𝑘 ∶ 𝑘 = 1 1 2𝑛 − 2  such that   111,,sincos                       ,1,1{T     1.2120nnkttitttkknkkkn In section 2, we give some preliminaries and in section 3, we describe the problem and obtained the regularity of the same. In section 4, we give the explicit formulae of the interpolatory polynomials. In section 5 and 6, estimation of interpolatory polynomials and convergence are given respectively. 2. PRELIMINARIES: In this section, we shall give some well-known results, which we shall use. The differential equation satisfied by   xPn , is  1.2                   0121 ,,,2  xPnnxPxxPx nnn    (2.2)       nnnnkk zzzPKzzzW   21 2,21 (2.3)       12,*22121   nnnnkk zzzPKtzzH  We shall require the fundamental polynomials of Lagrange interpolation based on the nodes as zeroes of R(z) and W(z) are given by: (2.4)              1210,  nkzzzRzRzLkkk where,      zWzzR 12      ISSN 2347-1921                                                           1819 | P a g e                                                           J u n e  2 1 ,  2 0 1 4   (2.5)              2211,  nktztHzHzlkkk  We will also use the following results (2.6)                nkzzxPKzWzW nkkknnkknn11,1211 22,   We will also use the following well-known inequalities (see [8])                                            2,21,1,2,223,212k1,212     ~    ,1     ,     ~      ~x-1      1,1-for   x x-1       13.2 12.211.210.29.28.27.2noxPnkxPcnxPxnoxPnkxPnknoxPnknnnknn 3. THE PROBLEM Let  𝑧𝑘 𝑘=02𝑛+1 and  𝑡𝑘 𝑘=12𝑛−2 be the two disjoint set of nodes obtained by projecting vertically the zeros of the  1-x2 Pn 𝛼 ,𝛽  𝑥  and Pn 𝛼 ,𝛽 ′ 𝑥  on the unit circle respectively, we determine the interpolatory polynomials 𝑄𝑛 𝑧  of degree ≤ 6𝑛 + 1 satisfying the conditions: (3.1)         𝑄𝑛 𝑧𝑘 = 𝛼𝑘 , 𝑘 = 0 1 2𝑛 + 1𝑄′ 𝑛 𝑧𝑘 = 𝛽𝑘 , 𝑘 = 0 1 2𝑛 + 1𝑄𝑛 𝑡𝑘 = 𝛼𝑘∗ , 𝑘 = 1 1 2𝑛 − 2  where 𝛼𝑘  ,  𝛽𝑘  𝑎𝑛𝑑 𝛼𝑘∗  are arbitrary complex numbers. In addition, we are interested in establishing the convergence theorem for the same. Regularity: Theorem 1: Hermite - Lagrange interpolation is regular on  𝑍𝑛  and  𝑇𝑛  . Proof: It is sufficient, if we show the unique solution of (3.1) is 𝑄𝑛 𝑧 ≡ 0 , when all data 𝛼𝑘 = 𝛽𝑘 = 𝛼𝑘∗ = 0. Clearly, in this case, we have 𝑄𝑛 𝑧 = 𝑅 𝑧 𝐻 𝑧 𝑞 𝑧 , where q(z) is a polynomial of degree   2n +1. Obviously, 𝑄𝑛 𝑧𝑘 = 0 and  𝑄𝑛 𝑡𝑘 =0. Then from 𝑄𝑛′  𝑧𝑘 = 0, we get  𝑞 𝑧𝑘 = 0 . Therefore, we have 𝑞 𝑧 =  𝑎𝑧 + 𝑏 𝑊 𝑧 , where  𝑎 and b are arbitrary constants. As, 𝑞 ±1 = 0, we get  −𝑎 + 𝑏 𝑊 −1 = 0  𝑎 + 𝑏 𝑊 1 = 0 as 𝑊 1 = 𝑊 −1 = 𝐾𝑛 . We get  𝑎 = 𝑏 = 0 It indicates 𝑄𝑛 𝑧 ≡ 𝑞 𝑧 ≡ 0. Hence, the theorem follows. 4. EXPLICIT REPRESENTATION OF INTERPOLATORY POLYNOMIALS We shall write 𝑄𝑛 𝑧  satisfying (3.1) as: (4.1)   𝑄𝑛 𝑧 =  𝛼𝑘𝐴𝑘 𝑧 +   𝛽𝑘𝐵𝑘 𝑧 2𝑛+1𝑘=02𝑛+1𝑘=0 +  𝛼𝑘∗𝐶𝑘 𝑧 2𝑛−2𝑘=1   where ,𝐴𝑘 𝑧  , 𝐵𝑘 𝑧  and 𝐶𝑘 𝑧  are unique fundamental polynomial each of degree atmost 6n+1 determined by the following conditions:   ISSN 2347-1921                                                           1820 | P a g e                                                           J u n e  2 1 ,  2 0 1 4   For k= 0 (1) 2n+1 (4.2)             𝐴𝑘 𝑧𝑗  = 𝛿𝑗𝑘  , 𝑗 = 0 1 2𝑛 + 1𝐴𝑘′  𝑧𝑗  = 0 , 𝑗 = 0 1 2𝑛 + 1𝐴𝑘 𝑡𝑗  = 0 , 𝑗 = 1 1 2𝑛 − 2  For k= 0(1)2n+1 (4.3)             𝐵𝑘 𝑧𝑗  = 0 , 𝑗 = 0 1 2𝑛 + 1𝐵𝑘′  𝑧𝑗  = 𝛿𝑗𝑘  , 𝑗 = 0 1 2𝑛 + 1𝐵𝑘 𝑡𝑗  = 0 , 𝑗 = 1 1 2𝑛 − 2  For k= 1(1)2n-2 (4.4)            𝐶𝑘 𝑧𝑗  = 0 , 𝑗 = 0 1 2𝑛 + 1𝐶𝑘′  𝑧𝑗  = 0 , 𝑗 = 0 1 2𝑛 + 1 𝐶 𝑡𝑗  = 𝛿𝑗𝑘  , 𝑗 = 1 1 2𝑛 − 2  Theorem 2: For k=1(1)2n-2, we have, (4.5)    𝐶𝑘 𝑧 =𝑅2 𝑧 𝑙𝑘 𝑧 𝑅2 𝑡𝑘  Theorem 3: For k=0(1)2n+1, we have, (4.6)    𝐵𝑘 𝑧 = 𝑧2−1 𝑅 𝑧 𝐻 𝑧 𝐿𝑘 𝑧  𝑧𝑘2−1 𝑅′  𝑧𝑘 𝐻 𝑧𝑘  Theorem 4: For k=0(1)2n+1, we have, (4.7)    𝐴𝑘 𝑧 =𝐻 𝑧 𝐿𝑘2  𝑧 𝐻 𝑧𝑘 −  𝐻 ′  𝑧𝑘 𝐻 𝑧𝑘 + 2𝐿′𝑘 𝑧𝑘  𝐵𝑘 𝑧  One can prove theorem 2, 3 and 4 owing to (4.4), (4.3) and (4.2) respectively.  5.  ESTIMATION OF FUNDAMENTAL POLYNOMIALS: Lemma 1: [6] Let  zLk be given by (2.4). Then    1201zlogmax           1.5nkk nczL  where c is a constant independent of n and z. Lemma 2: Let  zlk be given by (2.5). Then     221221211zmax          2.5nknkk kzl where c is a constant independent of n and z. Proof: From maximal principal, we know 𝜆𝑛 = max 𝑧 =1𝜆𝑛 𝑧  𝜆𝑛 =  221nkk zl  Let 𝑧 = 𝑥 + 𝑖𝑦 and  𝑧 = 1, then we know, for  0 ≤ arg 𝑧 < 𝜋 and 𝑘 = 1,2, ……… , 𝑛 − 1.  zlk =     kk tztHzH  Using (2.5) and  𝑧 = 1 and after some computation, we get   ISSN 2347-1921                                                           1821 | P a g e                                                           J u n e  2 1 ,  2 0 1 4    zlk =    1−𝑥2 12𝑃𝑛 𝛼 ,𝛽 ′  𝑥  1−𝑥𝑢𝑘+ 1−𝑥2 12 1−𝑢𝑘2 12 12 2 1−𝑢𝑘2 𝑃𝑛 𝛼 ,𝛽 ′  𝑢𝑘  𝑥−𝑢𝑘               ≤ 1 − 𝑥2 12  𝑃𝑛 𝛼,𝛽 ′ 𝑥   1 − 𝑥𝑢𝑘 12 1 − 𝑢𝑘2  𝑃𝑛 𝛼,𝛽 ′ 𝑢𝑘   𝑥 − 𝑢𝑘 = 𝐺𝑘 𝑥  Also,      zl kn 𝐺𝑘 𝑥  Similarly, for 𝜋 ≤ arg 𝑧 < 2𝜋 and k=1, 2, … ,n-1   zlk 𝐺𝑘 𝑥 ,       zl kn 𝐺𝑘 𝑥  Therefore,  for a fixed  𝑧 = 𝑥 + 𝑖𝑦 and  𝑧 = 1 and −1 < 𝑥 < 1 𝜆𝑛 𝑧 ≤ 2  𝐺𝑘 𝑥 𝑛𝑘=1+  𝑙0 𝑧  +  𝑙2𝑛+1 𝑧               = 2  Gk x  uk−x ≥12 1−uk2 12+ 2  Gk x + 2 uk−x <12 1−uk2 12 Using (2.12) and (2.13), we get the required result. Lemma 3: For 𝐶𝑘 𝑧  be given by (4.5) and  𝑧 ≤ 1, we have (5.3)     𝐶𝑘 𝑧  ≤ 𝑐 log 𝑛     ,   𝛼2𝑛−2𝑘=1 , 𝛽 ≤ −12 where ,c is a constant independent of n and z. Proof:   𝐶𝑘 𝑧  =2𝑛−2𝑘=1  𝑅2 𝑧 𝑙𝑘 𝑧 𝑅2 𝑡𝑘  = 𝑧2−1 2𝑊2 𝑧 𝑙𝑘   𝑧  𝑡𝑘2−1 2𝑊2 𝑡𝑘   Using (2.8), (2.10), (2.12) and Lemma 2, we get the required result. Lemma 4: For 𝐵𝑘 𝑧  be given by (4.6) and  𝑧 ≤ 1, we have (5.4)         𝐵𝑘 𝑧  ≤𝑐 log 𝑛𝑛2𝑛+1𝑘=0 ;  𝛼, 𝛽 ≤ −12 where c is a constant independent of n and z. Proof: Proof of this Lemma is similar to Lemma 3. Lemma 5: For 𝐴𝑘 𝑧  be given by (4.7) and  𝑧 ≤ 1, we have (5.5)     𝐴𝑘 𝑧  ≤ 𝑐 log 𝑛     , 𝑤ℎ𝑒𝑟𝑒   𝛼2𝑛+1𝑘=0 , 𝛽 ≤ −12 where, c is a constant independent of n and z. Proof: For  𝑧 ≤ 1, we get (5.6)      𝐻 𝑧 𝐿𝑘2  𝑧 𝐻 𝑧𝑘 ≤  𝑐𝑘−3𝛼+322𝑛+1𝑘=02𝑛+1𝑘=0  and (5.7)         𝐻 ′  𝑧𝑘 𝐻 𝑧𝑘 + 2𝐿′𝑘 𝑧𝑘  2𝑛+1𝑘=0 ≤ 𝑐𝑛 Using (5.6) and (5.7) in (4.7), we get the required result. 6. CONVERGENCE: Let f (z) be continuous in  𝑧 ≤ 1 and analytic in  𝑧 < 1 and 𝜔 𝑓, 𝛿  be the modulus of continuity of 𝑓 𝑒𝑖𝑥 .   ISSN 2347-1921                                                           1822 | P a g e                                                           J u n e  2 1 ,  2 0 1 4   Theorem 5: Let f (z) be continuous in 1z   and analytic in 1z  . Let the arbitrary number𝑠 𝛽𝑘 ‘s be such that: (6.1)        𝛽𝑘  = 𝑜(𝑛𝜔(𝑓, 𝑛−1) )   , 𝑘 = 0 1 2𝑛 + 1 Then {𝑄𝑛 } be defined by:   6.2          𝑄𝑛 𝑧 =  𝑓 𝑧𝑘 2𝑛+1𝑘=0𝐴𝑘 𝑧 +  𝛽𝑘2𝑛+1𝑘=0𝐵𝑘 𝑧 +  𝑓 𝑡𝑘 𝐶𝑘 𝑧 2𝑛−2𝑘=1  satisfies the relation  6.3            𝑄𝑛 𝑧 − 𝑓 𝑧  = 𝑜 𝜔 𝑓,𝑛−1 log𝑛   where, 𝜔 𝑓, 𝑛−1  is the modulus of continuity of f(z). Remark: To prove theorem 5, we shall need the following: Let f (z) be continuous in 1z   and analytic in 1z . Then there exists a polynomial  zFn  of degree 6n+1 satisfying Jackson’s inequality             20ez    ,     ,         6.4 i1  nfczFzf n  And also an inequality due to O. Kiš [3]           Imfor    ,     ,F         5.6 1mn nfcnz m Proof: Since 𝑄𝑛 𝑧 be given by (6.2) is a uniquely determined polynomial of degree ≤ 6𝑛 + 1, the polynomial 𝐹𝑛 𝑧   satisfying (6.4) and (6.5) can be expressed as: 𝐹𝑛 𝑧 =  𝐹𝑛 𝑧𝑘 2𝑛+1𝑘=0𝐴𝑘 𝑧 +  𝐹𝑛 ′ 𝑧𝑘 2𝑛+1𝑘=0𝐵𝑘 𝑧 +  𝐹𝑛 𝑡𝑘 𝐶𝑘 𝑧 2𝑛−2𝑘=1 Then,  𝑄𝑛 𝑧 − 𝑓 𝑧  ≤  𝑄𝑛 𝑧 − 𝐹𝑛 𝑧  +  𝐹𝑛 𝑧 − 𝑓 𝑧   ≤   𝑓 𝑧𝑘 − 𝐹𝑛 𝑧𝑘  2𝑛+1𝑘=0 𝐴𝑘 𝑧  +    𝛽𝑘  +  𝐹𝑛 ′ 𝑧𝑘   2𝑛+1𝑘=0 𝐵𝑘 𝑧  +   𝑓 𝑡𝑘 − 𝐹𝑛   𝑡𝑘  2𝑛−2𝑘=1 𝐶𝑘 𝑧  +  𝐹𝑛 𝑧 − 𝑓 𝑧   Using   𝑧 = 𝑒𝑖𝜃  0 ≤ 𝜃 < 2𝜋  , (6.1), (6.4), (6.5) and Lemma 2, 3 and 5, we get (6.3). REFERENCES: [1] Akhilagi  M. R. and Sharma A., Some Pál-type interpolation Problems , Approx. , Optimization and computing Theory and Applications (A. G. Law and C. L. Wang Ceds ) Elsevier Science Publisher B.V.(North Holland), IMACS,1990,37- 40. [2] Bahadur S. and Mathur K. K.; Quasi-Hermite interpolation on the unit circle; Ganita, 49(1), 1998. 1- 6. [3] Bahadur S., Pál-type (0, 1; 0)-interpolation on unit circle, Adv. 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ISSN 2347-1921                                                           1823 | P a g e                                                           J u n e  2 1 ,  2 0 1 4   [11] Lenard M., On weighted (0, 2)-type interpolation process, Annals Univ. Buda pest, Sect.Comp.16 (1996) 291- 301. [12] Mathur P., (0, 1; 0)-interpolation on infinite interval (−∞, ∞), Analysis in Theory and Application, 22 : 2, 2006, 105- 113. [13] Pál L.G., A new modification to Hermite-Fejér interpolation, Analysis Math, (1975),197- 205. [14] Szegö G.; Orthogonal Polynomials; Amer. Math. Soc. Coll. Publ. (New York, 1959) . [15] Srivastava V. ,Mathur N., Mathur P., A new kind of Pál-type interpolation II ,Int. J. contemp.Math.Sciences,Vol.6, 2011,no. 45,2237- 2246. [16] Xie T.F., On Páls problem, Chinese Quart.J.Math., 7 (1992), 48- 52. 

Hermite Lagrange Interpolation on the Unit Circle

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Hermite Lagrange Interpolation on the Unit Circle

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