Interpolation of functional by integral continued C-fractions

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1 Interpolation of functional by integral continue C-fractions Voloymyr L Makarov Institute of Mathematics of the NAS of Ukraine, Kiev, Ukraine makarovimath@gmailcom an Mykhaylo M Pahirya State University of Mukachevo, Ukraine pahirya@gmailcom January 22, 28 arxiv:86852v [mathca] 2 Jan 28 Abstract The functional interpolation problem on a continual set of noes by an integral continue C-fraction is stuie The necessary an sufficient conitions for its solvability are foun As a particular case, the consiere integral continue fraction contains a stanar interpolation continue C-fraction which is use to approximate the functions of one variable Keywors: continuity noes, integral continue C fraction, interpolation of functional AMS subject classification: B7, 4A2, 65D5, 65D5 Introuction Recently, the consierable amount of scientific stuies were evote to the generalization of interpolation theory of functions of real complex variable to the case of functionals an operators in abstract spaces, see the monographs [, 2] for example Continue fractions [] an branche continue fractions introuce by Skorobohat ko VYa [4] were generalize by integral continue fractions which propose by Syvavko MS [5] The problem of interpolation with integral continue fractions was first consiere in the article [6], further expansions an generalizations of this work are containe in the paper [7] Another class of interpolation integral continue fractions has been investigate in the paper [8] This class iffers from the previously stuie integral continue fractions by the fact that n s floor of fraction containing n tuple integral Interpolation integral operator continue fractions in Banach spaces were investigate in the article [9] The natural generalization of the classical Thiele continue fraction to the interpolation integral Thiele type continue fractions were propose in [,, 2] The purpose of this work is the stuy of interpolation of a functional given on the continual set of noes by the integral C type continue fractions Such integral continue fractions contains the interpolation continue C fraction as a particular case, so it is a generalization of one of the types of continue fraction use for interpolation of functions [] 2 Statemet of the problem Assume that xz,x i z C[,],i =,n, are some given functions, x i z x j z Let Fx be a certain functional efine in the space of piecewise continuous functions Q[, ] First we efine continual noes x z = x z, x i z, = x zhz x i z x z, i =,n where, an H is the Heavisie function

2 Let b,a i,b i,i =,n, are numbers, functions, operators, functionals etc We enote a finite continue fraction [] a D n = b, a 2 b b 2 a n as D n = b a b a 2 b 2 Let R R be a compact, R = {x i : x i R,x i x j, i,j =,n} an the function f CR is efine by its values at the points of the set R, y i = fx i, i =,n The function f can be interpolate by the set of values {y i } in ifferent ways, for example, polynomials, splines an continue fractions A problem of function interpolation by interpolation continue C fraction C ICF D n c x = a c ac x x b n a n b n a c 2 x x a c n x x n investigate in monograph [] The coefficients of C ICF are etermine through the interpolation noes R an the set of values of the function {y i } by means of a finite-continue fraction recurrence a c k = x k x k a c = y, a c = y y x x ac k x k x k 2 a c 2 x k x a c 2 x k x,k = 2,n, y k y If all the interpolation noes x i,i =,n, ten to the same value x R, then C ICF 2 transforms into a regular continue C fraction that will correspon to the power series for f aroun the point x Consier the set of integral continue fractions ICF of the form Q n x, = a a z [xz x z ]z a 2 z 2 [xz 2 x z 2,]z 2 a n z n [xz n x n 2 z n,]z n a n z n [xz n x n z n,]z n, 4 where a,a z,a 2 z,,a n z are certain kernels We formulate an interpolation problem as follows: Insie the set of ICF efine by 4 fin a integral continue fraction which satisfies the interpolation conitions at continual noes Fx = Q n x,, Fx i, = Q n x i,,, i =,n, [,] 5 SuchICFwill incluec ICF2asapartialcase SuchanICFiscalleanintegralinterpolation continue C fraction C IICF

3 Interpolation of functional by integral interpolation continue C fraction Define the kernels a,a z,,a n z from the conition that C IICFL 4 satisfies 5 Note that from 4, 5 you can irectly get the expression Q n x k,, = a a z [x k z, x z ]z a 2 z 2 [x k z 2, x z 2,]z 2 a k z k [x k z k, x k z k,]z k, k =,n 6 Theorem Let the functional Fx be n times Gateaux ifferentiable an the following formulas are vali In orer that C IICF 4 to satisfy the continual interpolation conitions 5 it is necessary that its kernels are etermine by the formulas a k = x k x k a k 2 z k 2 [x k z k 2, x k z k 2,]z k 2 a k z k [x k z k, x k 2 z k,]z k a z [x k z, x z ]z, k = 2,n, Fx k, Fx a = Fx, a = x x Fx, a 2 z 2 [x k z 2, x z 2,]z 2 Proof For the case k =, the formulas are obvious When k = m, from 5, 6 we get Fx m, = a a z [x m z, x z ]z a m z m [x m z m, x m 2 z m,]z m By sequentially inverting the continue fraction, we get a 2 z 2 [x m z 2, x z 2,]z 2 7 a m z m [x m z m, x m z m,]z m a m z m [x m z m, x m z m,]z m = a m z m [x m z m, x m 2 z m,]z m a 2 z 2 [x m z 2, x z 2,]z 2 a z [x m z, x z ]z Fx m, Fx

4 The subsequent ifferentiation of both parts of this relation with respect to the variable yiels formula 7 Theorem 2 Let Fx = f xtt 8 an kernels C IICF 4 are etermine by the formulas 7 For C IICF to satisfy the interpolation conitions 5 it is sufficient that the function fs C n, Proof Assume that the kernels are etermine by the formulas 7 From 6 it follows irectly that Q n x, = Fx, Q n x,, = Fx,, [,], for k =, When k = 2 we obtain from 6 the formula Q n x 2,, = a a z [x 2 z, x z ]z a 2 z 2 [x 2 z 2, x z 2,]z 2 9 Next, we substitute value of the kernel a 2 z 2 an fin the fraction s enominator P 2 = a 2 z 2 [x 2 z 2, x z 2,]z 2 = z 2 z 2 a z [x 2 z x z ]z Fx 2, Fx z 2 = where = lim z2 K 2[z 2,x 2 ] K 2 [z 2,x 2 ] = a z [x 2 z x z ]z Fx 2, Fx a z [x 2 z x z ]z z 2 Fx 2,z 2 Fx By applying the L Hospital rule an consiering 7, 8, we get lim K a z 2 [x 2 z 2 x z 2 ] 2[z 2,x 2 ] = lim z 2 z2 Fx 2,z 2 z2 z 2 x 2 z 2 x z 2 z2 x z 2 x z 2 f f x t,z 2 t x z 2 x z 2 x 2 t,z 2 t x z 2 x 2 z 2, a z 2 a z 2 x z 2 x 2 z 2 = =, x 2 z 2 We substitute the calculate value of the limit K 2 [z 2,x 2 ] into the expression for P 2, an then substitute the result into 9 Consequently, we obtain Q n x 2,, = Fx 2,

5 Let k = From 6 we get Q n x,, = Fx a z [x z, x z ]z a 2 z 2 [x z 2, x z 2,]z 2 a z [x z, x 2 z,]z We substitute the value of the kernel a z an evaluate the last enominator to get P = a z [x z, x 2 z,]z = z a 2 z 2 [x z 2,z x z 2,z ]z 2 a z [x z,z x z ]z z Fx = lim,z Fx K z [z,x ] a 2 z 2 [x z 2 x z 2 ]z 2 where a z [x z x z ]z Fx,z Fx a 2 z 2 [x z 2 x z 2 ]z 2 z K [z,x ] = By applying the L Hospital rule, taking into account, we obtain lim K [z,x ] z z a z [x z x z ] Fx,z Fx, a z [x z x z ]z z Fx,z Fx a 2 z [x z x z ] z a z [x z x z ]z Fx,z Fx 2 z Fx,z = x z x z a z [x 2 z x z ] Fx 2,z z Fx,z Fx z x 2 z x z a z [x z x z ] Fx 2,z Fx = Fx z,z x z x z z x 2 z x z Fx,z z Fx z 2,z

6 [x 2 z x z ] Fx,z [x z x z ] Fx 2,z z z [x z x z ] Fx z,z [x z x z ] Fx z,z Using 8 an the mean value theorem, we obtain where f θ lim K x z x z [z,x ] z z x 2 z x z f θ 2 θ 2 = θ = z x s,z sτ 2 a 2 z a 2 z x2 z x z x s,z sτ x s,z sm x s,z s x s,z s =, x z, x 2 s,z s Using equation6 in the general case for k = m we obtain Q n x m,, = Fx a z [x m z, x z ]z x s,z s, x 2 s,z s = x s,z s x s,z s, τ,τ 2, a m z m [x m z m, x m 2 z m,]z m Which permits us to fin the value of the last enominator P m = a 2 z 2 [x m z 2, x z 2,]z 2 a m z m [x m z m, x m z m,]z m a m z m [x m z m, x m z m,]z m To o that we use the values of a m z m from 7 Then P m = z m a m z m [x m z m,z m x m 2 z m,z m ]z m a 2 z 2 [x m z 2,z m x z 2,z m ]z 2 a z [x m z,z m x z ]z Fx m,z m Fx z m =

7 where = lim z m K m[z m,x m ] K m [z m,x m ] = Similarly, one can prove that a 2 z 2 [x m z 2 x z 2 ]z 2 a m z m [x m z m x m 2 z m ]z m a z [x m z x z ]z Fx m, Fx z m a m z m [x m z m x m 2 z m ]z m z m a 2 z 2 [x m z 2 x z 2 ]z 2 lim K m[z m,x m ] z m z m a z [x m z x z ]z z m Fx m,z m Fx a m z m a m z m xm z m x mz m =, x m z m By substituting the obtaine value of P m into we get the sought for result, namely Q n x m,, = Fx m, For brevity we omit the case k >, which can be prove similarly Remark Conitions 8 formulation of Theorem 2 can be change to a more general, when the results of [4, 5] are employe Theorem Assume the conitions of Theorem 2 are vali an =,xz x,x i z x i,i =,n Then C IICF 4 will coincie with the C ICF 2 Proof Taking into account the conitions of the theorem, we obtain the following form of a continue fraction 4 x x a z z Q n x = a From the interpolation conition 5 we get x x a 2 z 2 z 2 x x n, a n z n z n 2 a =fx, ã = a z z = fx fx, ã 2 = x x a 2 z 2 z 2 = x 2 x ãx 2 x fx 2 fx Using the mathematical inuction we inten to show that the formula ã m = a m z m z m = x m x m ãm x m x m 2 ã m 2 x m x m

8 ã 2 x m x ã x m x fx m fx hols true for m =,n For m =,2 formula is obviously vali Let s assume that it is vali for m =,k Then for m = k the continue fraction 2 can be written as fx k = a x k x ã By inverting the continue fraction 4, we get ã k = a k z k z k = x k x k x k x k 2 ã k ãk x k x k 2 ã 2 x k x x k x k ã k z k 4 ã k 2 x k x k ã x k x 5 fx k fx By its form, the right-han sie of 5 coincies with the right-han sie of albeit with ifferent notation, in aition to that, the initial conitions for both formulas are the same Consequently, formula is the same as formula References [] Makarov, VL & Khlobistov, VV 999 Funamentals of the theory of polynomial operator interpolation Kyev: Inst of Mathematics NAS of Ukraine in Russian [2] Makarov, VL, Khlobistov, VV & Yanovich, LA 2 Interpolation of operators Kyev: Naukova umka in Russian [] Jones, W B& Thron W J98 Continue Fractions Analytic Theory an Applications Encyclopeia of Mathematics an its Applications [4] Skorobohat ko, VYa 98 Theory of Branching Continue Fractions an Its Application in Computational Mathematics, Moscow: Nauka in Russian [5] Syvavko, MS 994 Integral continue fractions Kyev: Naukova umka in Ukrainian [6] Mykhal chuk, BR 999 Interpolation of nonlinear functionals by integral continue fractions Ukr Mat ZhVol 5, No pp in Ukrainian [7] Makarov, VL, Khlobistov, VV & Mykhal chuk, BR 2 Interpolation integral continue fractions Ukr MatZh Vol 55, No in Ukrainian [8] Makarov, VL & Demkiv, II28 A new class of interpolation integral continue fractions Dopov Nac aka nauk Ukr, No, pp 7 2 in Ukrainian [9] Makarov, VL, Khlobistov, VV & Demkiv, II 28 Interpolation integral operator fractions in a Banach space Dopov Nac aka nauk Ukr, No, pp 7 2 in Ukrainian

9 [] Makarov, V L & Demkiv, I I 24 Interpolating integral continue fraction of Theile type Mat Metoy Fiz Mekh Polya Vol 57, No 4, pp 44 5 in Ukrainian [] Makarov, VL & Demkiv, II 26 An integral interpolation chain fraction of Thiele type Dopov Nac aka nauk Ukr, No, pp 2 8 in Ukrainian oi: [2] Makarov, V L & Demkiv, I I 26 Abstract interpolation Thiele-type fraction Mat Metoy Fiz Mekh Polya Vol59, No 2, pp 5 57 in Ukrainian [] Pahirya, MM 26 Approximation of functions by continue fractions Uzhhoro: Graza in Ukrainian [4] Averbukh, VI& Smolyanov, OG967 The theory of ifferentiation in linear topological spaces Uspehi Mat Nauk Vol 22, No 6, pp 2-26 in Russian [5] Makarov, VL, Khlobystov, V V, Kashpur, E F & Mikhal chuk, B R 2 Integral Newton-Type Polynomials with Continual Noes Ukr Mat Zh Vol 55, No 6 pp in Ukrainian