Posts by Tiberiu Popoviciu

Abstract

Authors

Tiberiu Popoviciu
Institutul de Calcul

Keywords

?

Paper coordinates

T. Popoviciu, Remarques sur le reste de certaines formules d’approximation d’une différence divisée par des dérivées, Bul. Inst. Politehn. Iaşi (N.S.), 13(17) (1967) fasc. 3-4, pp. 103-109 (in French)

PDF

About this paper

Journal

Buletinul Institutului Politehnic din Iaşi

Publisher Name

published by the “Gheorghe Asachi” Technical University of Iaşi

DOI

 

Print ISSN

 

Online ISSN

 

[/vc_column]

??

Paper (preprint) in HTML form

1967 b -Popoviciu- Bul. Inst. Politehn. Iasi - Remarks on the rest of certain d_approx formulas
Original text
Rate this translation
Your feedback will be used to help improve Google Translate

REMARKS ON THE REMAINDER OF CERTAIN FORMULAS FOR APPROXIMATING A DIFFERENCE DIVIDED BY DERIVATIVES

BYTIBERIU POPOVICIUmember of the Academy of the Socialist Republic of Romania

  1. Let us consider a function f f fffdefined and having a derivative n n nnn-th continuous on the finite and closed interval [ has , b ] , ( has < b ) [ has , b ] , ( has < b ) [a,b],(a < b)[a, b],(a<b)[has,b],(has<b).
Let has = x 1 x 2 x n x n 1 = b , n + 1 has = x 1 x 2 x n x n 1 = b , n + 1 a=x_(1) <= x_(2) <= dots <= x_(n) <= x_(n-1)=b,n+1a=x_{1} \leqq x_{2} \leqq \ldots \leqq x_{n} \leqq x_{n-1}=b, n+1has=x1x2xnxn1=b,n+1distinct or non-distinct points belonging to the interval [ has , b ] [ has , b ] [a,b][a, b][has,b]such as the extremities has has hashashas, b b bbbare among these points. It follows that n ( > 0 ) n ( > 0 ) n( > 0)n(>0)n(>0)is a natural number. Let us denote by has = x 1 = x 1 < x 2 < < x p = x n + 1 = b has = x 1 = x 1 < x 2 < < x p = x n + 1 = b a=x_(1)=x_(1)^(') < x_(2)^(') < dots < x_(p)^(')=x_(n+1)=ba=x_{1}=x_{1}^{\prime}<x_{2}^{\prime}<\ldots<x_{p}^{\prime}=x_{n+1}=bhas=x1=x1<x2<<xp=xn+1=bthe distinct points among the points x α x α x_(alpha)x_{\alpha}xαand be k 1 , k 2 , , k p k 1 , k 2 , , k p k_(1),k_(2),dots,k_(p)k_{1}, k_{2}, \ldots, k_{p}k1,k2,,kptheir respective multiplicity orders. Therefore, we have 2 p n + 1 , k 1 , k 2 , , k p 2 p n + 1 , k 1 , k 2 , , k p 2 <= p <= n+1,k_(1),k_(2),dots,k_(p)2 \leqq p \leqq n+1, k_{1}, k_{2}, \ldots, k_{p}2pn+1,k1,k2,,kpare natural numbers, k 1 + k 2 + + + k p = n + 1 k 1 + k 2 + + + k p = n + 1 k_(1)+k_(2)+dots++k_(p)=n+1k_{1}+k_{2}+\ldots+ +k_{p}=n+1k1+k2+++kp=n+1And max ( k 1 , k 2 , , k p ) n max k 1 , k 2 , , k p n max(k_(1),k_(2),dots,k_(p)) <= n\max \left(k_{1}, k_{2}, \ldots, k_{p}\right) \leqq nmax(k1,k2,,kp)n.
In accordance with the point system x α , x α x α , x α x_(alpha),x_(alpha)^(')x_{\alpha}, x_{\alpha}^{\prime}xα,xα, We have
x k 1 + k 2 + + k α 1 + β = x α , ( β = 1 , 2 , , k α ; α = 1 , 2 , , p ) , x k 1 + k 2 + + k α 1 + β = x α , β = 1 , 2 , , k α ; α = 1 , 2 , , p , x_(k_(1)+k_(2)+dots+k_(alpha-1)+beta)=x_(alpha)^('),(beta=1,2,dots,k_(alpha);alpha=1,2,dots,p),x_{k_{1}+k_{2}+\ldots+k_{\alpha-1}+\beta}=x_{\alpha}^{\prime},\left(\beta=1,2, \ldots, k_{\alpha} ; \alpha=1,2, \ldots, p\right),xk1+k2++kα1+β=xα,(β=1,2,,kα;α=1,2,,p),
( k 1 + k 2 + + k α 1 k 1 + k 2 + + k α 1 k_(1)+k_(2)+cdots+k_(alpha-1)k_{1}+k_{2}+\cdots+k_{\alpha-1}k1+k2++kα1is replaced by 0 if α = 1 α = 1 alpha=1\alpha=1α=1
There exists a continuous function G ( x ) G ( x ) G(x)G(x)G(x), completely determined, such that one has
(1) [ x 1 , x 2 , , x n + 1 ; f ] = has b G ( x ) f ( n ) ( x ) d x , (1) x 1 , x 2 , , x n + 1 ; f = has b G ( x ) f ( n ) ( x ) d x , {:(1)[x_(1),x_(2),dots,x_(n+1);f]=int_(a)^(b)G(x)f^((n))(x)dx",":}\begin{equation*} \left[x_{1}, x_{2}, \ldots, x_{n+1}; f\right]=\int_{a}^{b} G(x) f^{(n)}(x) \mathrm{d} x, \tag{1} \end{equation*}(1)[x1,x2,,xn+1;f]=hasbG(x)f(n)(x)dx,
where we have designated by [ x 1 , x 2 , , x n + 1 ; f ] x 1 , x 2 , , x n + 1 ; f [x_(1),x_(2),dots,x_(n+1);f]\left[x_{1}, x_{2}, \ldots, x_{n+1}; f\right][x1,x2,,xn+1;f]the divided difference of the function f f fffon the nodes x α , ( α = 1 , 2 , , n + 1 ) x α , ( α = 1 , 2 , , n + 1 ) x_(alpha),(alpha=1,2,dots,n+1)x_{\alpha},(\alpha=1,2, \ldots, n+1)xα,(α=1,2,,n+1).
Formula (1) is a special case of a formula from R. v. Mises [4]. When n = max ( k 1 , k 2 , , k p ) n = max k 1 , k 2 , , k p n=max(k_(1),k_(2),dots,k_(p))n=\max \left(k_{1}, k_{2}, \ldots, k_{p}\right)n=max(k1,k2,,kp)which happens if and only if p = 2 p = 2 p=2p=2p=2And k 1 = 1 k 1 = 1 k_(1)=1k_{1}=1k1=1Or k 2 = 1 k 2 = 1 k_(2)=1k_{2}=1k2=1, the property results from the formula ( 1 k n 1 k n 1 <= k <= n1 \leqq k \leqq n1kn)
[ has , has , , has , b , b , , b ; f ] = [ has , has , , has , b , b , , b ; f ] = [a,a,dots,a,b,b,dots,b;f]=[a, a, \ldots, a, b, b, \ldots, b ; f]=[has,has,,has,b,b,,b;f]=
(2) = 1 ( k 1 ) ! ( n k ) ! ( b a ) n a b ( x a ) n k ( b x ) k 1 f ( n ) ( x ) d x (2) = 1 ( k 1 ) ! ( n k ) ! ( b a ) n a b ( x a ) n k ( b x ) k 1 f ( n ) ( x ) d x {:(2)=(1)/((k-1)!(n-k)!(b-a)^(n))int_(a)^(b)(x-a)^(n-k)(b-x)^(k-1)f^((n))(x)dx:}\begin{equation*} =\frac{1}{(k-1)!(n-k)!(b-a)^{n}} \int_{a}^{b}(x-a)^{n-k}(b-x)^{k-1} f^{(n)}(x) \mathrm{d} x \tag{2} \end{equation*}(2)=1(k1)!(nk)!(bhas)nhasb(xhas)nk(bx)k1f(n)(x)dx
which is easy to obtain, by calculating the integral of the right-hand side by repeated integrations by parts.
2. Formula (1) can also be deduced from another by G. Kowalewski [3] relating to the remainder of the Lagrange interpolation formula (the Lagranze-Hernite formula in general).
To simplify, let's assume that the nodes x α , ( α = 1 , 2 , , n + 1 ) x α , ( α = 1 , 2 , , n + 1 ) x_(alpha),(alpha=1,2,dots,n+1)x_{\alpha},(\alpha=1,2, \ldots, n+1)xα,(α=1,2,,n+1), are distinct, therefore that a = x 1 < x 2 < < x n < x n + 1 = b a = x 1 < x 2 < < x n < x n + 1 = b a=x_(1) < x_(2) < dots < x_(n) < x_(n+1)=ba=x_{1}<x_{2}<\ldots<x_{n}<x_{n+1}=bhas=x1<x2<<xn<xn+1=bLet's ask. l ( x ) == α = 1 [ ( x x α ) l ( x ) == α = 1 [ x x α l(x)==prod_(alpha=1)^([)(x-x_(alpha))l(x)= =\prod_{\alpha=1}^{[ }\left(x-x_{\alpha}\right)L(x)==α=1[(xxα)and consider the fundamental interpolation polynomials l α ( x ) = l ( x ) ( x x α ) l ( x α ) , ( α = 1 , 2 , , n ) l α ( x ) = l ( x ) x x α l x α , ( α = 1 , 2 , , n ) l_(alpha)(x)=(l(x))/((x-x_(alpha))l^(')(x_(alpha))),(alpha=1,2,dots,n)l_{\alpha}(x)=\frac{l(x)}{\left(x-x_{\alpha}\right) l^{\prime}\left(x_{\alpha}\right)},(\alpha=1,2, \ldots, n)Lα(x)=L(x)(xxα)L(xα),(α=1,2,,n), related to the nodes x α , ( α = 1 , 2 , , n ) x α , ( α = 1 , 2 , , n ) x_(alpha),(alpha=1,2,dots,n)x_{\alpha},(\alpha=1,2, \ldots, n)xα,(α=1,2,,n)Finally, let us designate by L ( x 1 , x 2 , , x n ; f x ) L x 1 , x 2 , , x n ; f x L(x_(1),x_(2),dots,x_(n);f∣x)L\left(x_{1}, x_{2}, \ldots, x_{n} ; f \mid x\right)L(x1,x2,,xn;f|x)the Lagrange interpolation polynomial of the function f f fffon these nodes. G. Kowalewski obtains [3] the remainder f ( x ) L ( x 1 , x 2 , , x n ; f x ) = l ( x ) [ x 1 , x 2 , , x n , x ; f ] f ( x ) L x 1 , x 2 , , x n ; f x = l ( x ) x 1 , x 2 , , x n , x ; f f(x)-L(x_(1),x_(2),dots,x_(n);f∣x)=l(x)[x_(1),x_(2),dots,x_(n),x;f]f(x)-L\left(x_{1}, x_{2}, \ldots, x_{n} ; f \mid x\right)=l(x)\left[x_{1}, x_{2}, \ldots, x_{n}, x ; f\right]f(x)L(x1,x2,,xn;f|x)=L(x)[x1,x2,,xn,x;f]of the Lagrange interpolation formula in the following form
(3) l ( x ) [ x 1 , x 2 , , x n , x ; f ] = α = 1 n l α ( x ) x α x ( x α u ) n 1 ( n 1 ) ! f ( n ) ( u ) d u (3) l ( x ) x 1 , x 2 , , x n , x ; f = α = 1 n l α ( x ) x α x x α u n 1 ( n 1 ) ! f ( n ) ( u ) d u {:(3)l(x)[x_(1),x_(2),dots,x_(n),x;f]=sum_(alpha=1)^(n)l_(alpha)(x)int_(x_(alpha))^(x)((x_(alpha)-u)^(n-1))/((n-1)!)f^((n))(u)du:}\begin{equation*} l(x)\left[x_{1}, x_{2}, \ldots, x_{n}, x ; f\right]=\sum_{\alpha=1}^{n} l_{\alpha}(x) \int_{x_{\alpha}}^{x} \frac{\left(x_{\alpha}-u\right)^{n-1}}{(n-1)!} f^{(n)}(u) \mathrm{d} u \tag{3} \end{equation*}(3)L(x)[x1,x2,,xn,x;f]=α=1nLα(x)xαx(xαu)n1(n1)!f(n)(u)du
If we now ask L ( x ) = l ( x ) ( x x n + 1 ) L ( x ) = l ( x ) x x n + 1 L(x)=l(x)(x-x_(n+1))L(x)=l(x)\left(x-x_{n+1}\right)L(x)=L(x)(xxn+1)we obtain
l ( x n + 1 ) = L ( x n + 1 ) , l α ( x n + 1 ) = L ( x n + 1 ) L ( x α ) , ( 0 = 1 , 2 , , n ) . l x n + 1 = L x n + 1 , l α x n + 1 = L x n + 1 L x α , ( 0 = 1 , 2 , , n ) . l(x_(n+1))=L^(')(x_(n+1)),quadl_(alpha)(x_(n+1))=-(L^(')(x_(n+1)))/(L^(')(x_(alpha))),quad(0=1,2,dots,n).l\left(x_{n+1}\right)=L^{\prime}\left(x_{n+1}\right), \quad l_{\alpha}\left(x_{n+1}\right)=-\frac{L^{\prime}\left(x_{n+1}\right)}{L^{\prime}\left(x_{\alpha}\right)}, \quad(0=1,2, \ldots, n) .L(xn+1)=L(xn+1),Lα(xn+1)=L(xn+1)L(xα),(0=1,2,,n).
By asking x = x n 1 1 = b x = x n 1 1 = b x=x_(n-1-1)=bx=x_{n-1-1}=bx=xn11=bIn (3), we deduce
(4) [ x 1 , x 2 , , x n + 1 ; f ] = α = 1 n 1 L ( x α ) x α b ( x α u ) n 1 ( n 1 ) ! f ( n ) ( u ) d u . (4) x 1 , x 2 , , x n + 1 ; f = α = 1 n 1 L x α x α b x α u n 1 ( n 1 ) ! f ( n ) ( u ) d u . {:(4)[x_(1),x_(2),dots,x_(n+1);f]=-sum_(alpha=1)^(n)(1)/(L^(')(x_(alpha)))int_(x_(alpha))^(b)((x_(alpha)-u)^(n-1))/((n-1)!)f^((n))(u)du.:}\begin{equation*} \left[x_{1}, x_{2}, \ldots, x_{n+1} ; f\right]=-\sum_{\alpha=1}^{n} \frac{1}{L^{\prime}\left(x_{\alpha}\right)} \int_{x_{\alpha}}^{b} \frac{\left(x_{\alpha}-u\right)^{n-1}}{(n-1)!} f^{(n)}(u) \mathrm{d} u . \tag{4} \end{equation*}(4)[x1,x2,,xn+1;f]=α=1n1L(xα)xαb(xαu)n1(n1)!f(n)(u)du.
  1. Formula (4) shows us that if the nodes are distinct the function G ( x ) G ( x ) G(x)G(x)G(x)of formula (1) is continuous and even, if n > 2 n > 2 n > 2n>2n>2, has a continuous derivative of order n 2 n 2 n-2n-2n2on [ a , b ] [ a , b ] [a,b][a, b][has,b]and reduces to a polynomial of degree n 1 n 1 n-1n-1n1on each of the partial intervals [ x α , x α + 1 ] , ( α = 1 , 2 , , n ) x α , x α + 1 , ( α = 1 , 2 , , n ) [x_(alpha),x_(alpha+1)],(alpha=1,2,dots,n)\left[x_{\alpha}, x_{\alpha+1}\right],(\alpha=1,2, \ldots, n)[xα,xα+1],(α=1,2,,n)We formerly called such a function an elementary function
    of order n 1 n 1 n-1n-1n1and we have shown its importance in the theory of higher-order convex functions [6]. Today they are also called "spline" functions of degree n 1 n 1 n-1n-1n1IJ Schoenberg used them [9], [10] in interesting research on approximate quadrature.
When the nodes x α , ( α = 1 , 2 , , n + 1 ) x α , ( α = 1 , 2 , , n + 1 ) x_(alpha),(alpha=1,2,dots,n+1)x_{\alpha},(\alpha=1,2, \ldots, n+1)xα,(α=1,2,,n+1)They are not distinct but are grouped into distinct nodes. x α x α x_(alpha)^(')x_{\alpha}^{\prime}xαorder k α k α k_(alpha)k_{\alpha}kαof respective multiplicity, the previous properties are only partially verified. Let's extend the function G ( x ) G ( x ) G(x)G(x)G(x)by the value 0 outside the interval [ a , b ] [ a , b ] [a,b][a, b][has,b]The function G ( x ) G ( x ) G(x)G(x)G(x)thus extended is defined on ( , + , + -oo,+oo-\infty,+\infty,+), is continuous on the open interval ( a , b a , b a,ba, bhas,band reduces to a polynomial of degree n 1 n 1 n-1n-1n1on each of the intervals ( , x 1 ) , ( x p , + ) , ( x α , x α + 1 ) , ( α = 1 , x 1 , x p , + , x α , x α + 1 , ( α = 1 (-oo,x_(1)^(')),(x_(p)^('),+oo),(x_(alpha)^('),x_(alpha+1)^(')),(alpha=1\left(-\infty, x_{1}^{\prime}\right),\left(x_{p}^{\prime},+\infty\right),\left(x_{\alpha}^{\prime}, x_{\alpha+1}^{\prime}\right),(\alpha=1(,x1),(xp,+),(xα,xα+1),(α=1, 2 , , p 1 ) 2 , , p 1 ) 2,dots,p-1)2, \ldots, p-1)2,,p1)On the node x α , ( α = 1 , 2 , , p ) x α , ( α = 1 , 2 , , p ) x_(alpha)^('),(alpha=1,2,dots,p)x_{\alpha}^{\prime},(\alpha=1,2, \ldots, p)xα,(α=1,2,,p)the extended function G ˙ ( x ) G ˙ ( x ) G^(˙)(x)\dot{G}(x)G˙(x)is continuous if n 1 + k α n 1 + k α n >= 1+k_(alpha)n \geqq 1+k_{\alpha}n1+kαand has a continuous derivative of order n 1 k k n 1 k k n-1-k_(k)n-1-k_{k}n1kkif n > 1 + k α n > 1 + k α n > 1+k_(alpha)n>1+k_{\alpha}n>1+kα4.
The function G ( x ) G ( x ) G(x)G(x)G(x)of formula (1) is non-negative and has a positive integral over [ a , b ] [ a , b ] [a,b][a, b][has,b]Indeed, if we ask f = x n f = x n f=x^(n)f=x^{n}f=xnwe deduce
a b G ( x ) d x = 1 n ! a b G ( x ) d x = 1 n ! int_(a)^(b)G(x)dx=(1)/(n!)\int_{a}^{b} G(x) \mathrm{d} x=\frac{1}{n!}hasbG(x)dx=1n!
The non-negativity of G ( x ) G ( x ) G(x)G(x)G(x)on [ a , b ] [ a , b ] [a,b][a, b][has,b]results from the Cauchy mean formula
(5) [ x 1 , x 2 , , x n + 1 ; f ] = 1 n ! f ( n ) ( ξ ) , ξ ( a , b ) . (5) x 1 , x 2 , , x n + 1 ; f = 1 n ! f ( n ) ( ξ ) , ξ ( a , b ) . {:(5)[x_(1),x_(2),dots,x_(n+1);f]=-(1)/(n!)f^((n))(xi)","quad xi in(a","b).:}\begin{equation*} \left[x_{1}, x_{2}, \ldots, x_{n+1} ; f\right]=-\frac{1}{n!} f^{(n)}(\xi), \quad \xi \in(a, b) . \tag{5} \end{equation*}(5)[x1,x2,,xn+1;f]=1n!f(n)(ξ),ξ(has,b).
Indeed, if the function continues G ( x ) G ( x ) G(x)G(x)G(x)is not non-negative; there exists a subinterval [ α , β ] [ α , β ] [alpha,beta][\alpha, \beta][α,β]of non-zero length [ a , b ] [ a , b ] [a,b][a, b][has,b]on which G ( x ) < 0 G ( x ) < 0 G(x) < 0G(x)<0G(x)<0So then f ( x ) f ( x ) f(x)f(x)f(x)a (continuous) function, defined on [ a , b ] [ a , b ] [a,b][a, b][has,b]whose derivative n n nnn-th is
(6) f ( n ) ( x ) = { 0 pour x [ a , α ] [ β , b ] , ( x α ) ( β x ) 0 pour x ( α , β ) . (6) f ( n ) ( x ) = 0  pour  x [ a , α ] [ β , b ] , ( x α ) ( β x ) 0  pour  x ( α , β ) . {:(6)f^((n))(x)={[0," pour "x in[a","alpha]uu[beta","b]","],[(x-alpha)(beta-x)!=0," pour "x in(alpha","beta).]:}:}f^{(n)}(x)=\left\{\begin{array}{cl} 0 & \text { pour } x \in[a, \alpha] \cup[\beta, b], \tag{6}\\ (x-\alpha)(\beta-x) \neq 0 & \text { pour } x \in(\alpha, \beta) . \end{array}\right.(6)f(n)(x)={0 For x[has,α][β,b],(xα)(βx)0 For x(α,β).
On the one hand, from (5) it follows that [ x 1 , x 2 , , x n + 1 ; f ] 0 x 1 , x 2 , , x n + 1 ; f 0 [x_(1),x_(2),dots,x_(n+1);f] >= 0\left[x_{1}, x_{2}, \ldots, x_{n+1} ; f\right] \geqslant 0[x1,x2,,xn+1;f]0(the function f f fffis non-concave of order n 1 n 1 n-1n-1n1). On the other hand, from (6) it follows that
a b G ( x ) f ( n ) ( x ) d x = a β G ( x ) f ( n ) ( x ) d x < 0 a b G ( x ) f ( n ) ( x ) d x = a β G ( x ) f ( n ) ( x ) d x < 0 int_(a)^(b)G(x)f^((n))(x)dx=int_(a)^(beta)G(x)f^((n))(x)dx < 0\int_{a}^{b} G(x) f^{(n)}(x) \mathrm{d} x=\int_{a}^{\beta} G(x) f^{(n)}(x) \mathrm{d} x<0hasbG(x)f(n)(x)dx=hasβG(x)f(n)(x)dx<0
The properties of the function G ( x ) G ( x ) G(x)G(x)G(x)of formula (1) have also been studied by DV Ionescu [2].
5. Cauchy's formula (59) suggests the approximation formula
(7) [ x 1 , x 2 , , x n + 1 ; f ] 1 n ! f ( n ) ( x 0 ) , (7) x 1 , x 2 , , x n + 1 ; f 1 n ! f ( n ) x 0 , {:(7)[x_(1),x_(2),dots,x_(n+1);f]~~(1)/(n!)f^((n))(x_(0))",":}\begin{equation*} \left[x_{1}, x_{2}, \ldots, x_{n+1} ; f\right] \approx \frac{1}{n!} f^{(n)}\left(x_{0}\right), \tag{7} \end{equation*}(7)[x1,x2,,xn+1;f]1n!f(n)(x0),
Or x 0 x 0 x_(0)x_{0}x0is a given point. The degree of accuracy of this formula is n n >= n\geqq nnand this degree of accuracy is > n > n > n>n>nif and only if in (7) we take x 0 = 1 n + 1 α = 1 n + 1 x α x 0 = 1 n + 1 α = 1 n + 1 x α x_(0)=(1)/(n+1)sum_(alpha=1)^(n+1)x_(alpha)x_{0}=\frac{1}{n+1} \sum_{\alpha=1}^{n+1} x_{\alpha}x0=1n+1α=1n+1xαSo the degree of accuracy is n + 1 n + 1 n+1n+1n+1and we have the approximation formula
(8) [ x 1 , x 2 , , x n + 1 ; f ] = 1 n ! f ( n ) ( x 1 + x 2 + + x n + 1 n 1 ) + R (8) x 1 , x 2 , , x n + 1 ; f = 1 n ! f ( n ) x 1 + x 2 + + x n + 1 n 1 + R {:(8)[x_(1),x_(2),dots,x_(n+1);f]=(1)/(n!)f^((n))((x_(1)+x_(2)+cdots+x_(n+1))/(n-1))+R:}\begin{equation*} \left[x_{1}, x_{2}, \ldots, x_{n+1} ; f\right]=\frac{1}{n!} f^{(n)}\left(\frac{x_{1}+x_{2}+\cdots+x_{n+1}}{n-1}\right)+R \tag{8} \end{equation*}(8)[x1,x2,,xn+1;f]=1n!f(n)(x1+x2++xn+1n1)+R
This formula is a Gaussian formula [7] and therefore has a remainder of simple form [8]. A simple calculation gives us the remainder R R RRRin the form
R = 1 ( n + 1 ) ( n + 2 ) ! [ ( n + 1 ) α = 1 n + 1 x α 2 ( α = 1 n + 1 x α ) 2 ] [ ξ 1 , ξ 2 , ξ 3 ; f ( n ) ] R = 1 ( n + 1 ) ( n + 2 ) ! ( n + 1 ) α = 1 n + 1 x α 2 α = 1 n + 1 x α 2 ξ 1 , ξ 2 , ξ 3 ; f ( n ) R=(1)/((n+1)*(n+2)!)[(n+1)sum_(alpha=1)^(n+1)x_(alpha)^(2)-(sum_(alpha=1)^(n+1)x_(alpha))^(2)]*[xi_(1),xi_(2),xi_(3);f^((n))]R=\frac{1}{(n+1) \cdot(n+2)!}\left[(n+1) \sum_{\alpha=1}^{n+1} x_{\alpha}^{2}-\left(\sum_{\alpha=1}^{n+1} x_{\alpha}\right)^{2}\right] \cdot\left[\xi_{1}, \xi_{2}, \xi_{3} ; f^{(n)}\right]R=1(n+1)(n+2)![(n+1)α=1n+1xα2(α=1n+1xα)2][ξ1,ξ2,ξ3;f(n)]
Or ξ 1 , ξ 2 , ξ 3 ξ 1 , ξ 2 , ξ 3 xi_(1),xi_(2),xi_(3)\xi_{1}, \xi_{2}, \xi_{3}ξ1,ξ2,ξ3are three distinct points of the interval ( a , b a , b a,ba, bhas,b(generally dependent on the function) f f fff).
If the function f f fffhas a derivative of order n + 2 n + 2 n+2n+2n+2on ( a , b ) ( a , b ) (a,b)(a, b)(has,b), we also have
(9) R = 1 2 ( n + 1 ) ( n + 2 ) ! [ ( n + 1 ) α = 1 n + 1 x α 2 ( α = 1 n + 1 x α ) 2 ] f ( n + 2 ) ( ξ ) , ξ ξ ( a , b ) R = 1 2 ( n + 1 ) ( n + 2 ) ! ( n + 1 ) α = 1 n + 1 x α 2 α = 1 n + 1 x α 2 f ( n + 2 ) ( ξ ) , ξ ξ ( a , b ) quad R=(1)/(2(n+1)*(n+2)!)[(n+1)sum_(alpha=1)^(n+1)x_(alpha)^(2)-(sum_(alpha=1)^(n+1)x_(alpha))^(2)]f^((n+2))(xi),quad xi xi(a,b)\quad R=\frac{1}{2(n+1) \cdot(n+2)!}\left[(n+1) \sum_{\alpha=1}^{n+1} x_{\alpha}^{2}-\left(\sum_{\alpha=1}^{n+1} x_{\alpha}\right)^{2}\right] f^{(n+2)}(\xi), \quad \xi \xi(a, b)R=12(n+1)(n+2)![(n+1)α=1n+1xα2(α=1n+1xα)2]f(n+2)(ξ),ξξ(has,b).
Formula (8) was examined in the particular case p = 2 p = 2 p=2p=2p=2And k 1 = 1 k 1 = 1 k_(1)=1k_{1}=1k1=1Or k 2 = 1 k 2 = 1 k_(2)=1k_{2}=1k2=1, by Laura Gotusso [1] who obtained, in this case, the remainder with some imprecision. Laura Gotusso's correct formula is obtained by setting x 1 = x 2 = = x n = x , x n + 1 = x + h x 1 = x 2 = = x n = x , x n + 1 = x + h x_(1)=x_(2)=cdots=x_(n)=x,x_(n+1)=x+hx_{1}=x_{2}=\cdots=x_{n}=x, x_{n+1}=x+hx1=x2==xn=x,xn+1=x+hOr x 1 = x + h , x 2 = x 3 = == x n + 1 = x x 1 = x + h , x 2 = x 3 = == x n + 1 = x x_(1)=x+h,x_(2)=x_(3)=cdots==x_(n+1)=xx_{1}=x+h, x_{2}=x_{3}=\cdots= =x_{n+1}=xx1=x+h,x2=x3===xn+1=xin (8) and (9). This gives us the approximation formula (with remainder)
f ( x + h ) = α = 0 n 1 h α α ! f ( α ) ( x ) + h n n ! f ( n ) ( x + h n + 1 ) + n h n + 2 2 ( n + 1 ) ( n + 2 ) ! f ( n + 2 ) ( ξ ) f ( x + h ) = α = 0 n 1 h α α ! f ( α ) ( x ) + h n n ! f ( n ) x + h n + 1 + n h n + 2 2 ( n + 1 ) ( n + 2 ) ! f ( n + 2 ) ( ξ ) f(x+h)=sum_(alpha=0)^(n-1)(h^(alpha))/(alpha!)f^((alpha))(x)+(h^(n))/(n!)f^((n))(x+(h)/(n+1))+(nh^(n+2))/(2(n+1)*(n+2)!)f^((n+2))(xi)f(x+h)=\sum_{\alpha=0}^{n-1} \frac{h^{\alpha}}{\alpha!} f^{(\alpha)}(x)+\frac{h^{n}}{n!} f^{(n)}\left(x+\frac{h}{n+1}\right)+\frac{n h^{n+2}}{2(n+1) \cdot(n+2)!} f^{(n+2)}(\xi)f(x+h)=α=0n1hαα!f(α)(x)+hnn!f(n)(x+hn+1)+nhn+22(n+1)(n+2)!f(n+2)(ξ), Or ξ ξ xi\xiξis inside the smallest interval containing the points x , x + h x , x + h x,x+hx, x+hx,x+h6.
The previous results can be obtained without using formula (1). Indeed, we have shown [5] that if the function f f fffis convex of order n 1 n 1 n-1n-1n1we have inequality
[ x 1 , x 2 , , x n + 1 ; f ] > 1 n ! f ( n ) ( x 1 + x 2 + + x n + 1 n + 1 ) x 1 , x 2 , , x n + 1 ; f > 1 n ! f ( n ) x 1 + x 2 + + x n + 1 n + 1 [x_(1),x_(2),dots,x_(n+1);f] > (1)/(n!)f^((n))((x_(1)+x_(2)+cdots+x_(n+1))/(n+1))\left[x_{1}, x_{2}, \ldots, x_{n+1} ; f\right]>\frac{1}{n!} f^{(n)}\left(\frac{x_{1}+x_{2}+\cdots+x_{n+1}}{n+1}\right)[x1,x2,,xn+1;f]>1n!f(n)(x1+x2++xn+1n+1)
(Or x α x α x_(alpha)x_{\alpha}xαare not all confused). The simplicity of the rest of formula (8) then follows [8].
7. Other approximation formulas for the divided difference (1) can be obtained by applying any quadrature formula to the integral of the right-hand side. I will limit myself to examining one more particular case.
Let's start with some preliminary calculations. The moments
c n = a b G ( x ) x n d x , ( n = 0 , 1 , ) c n = a b G ( x ) x n d x , ( n = 0 , 1 , ) c_(n)=int_(a)^(b)G(x)x^(n)dx,quad(n=0,1,dots)c_{n}=\int_{a}^{b} G(x) x^{n} \mathrm{~d} x, \quad(n=0,1, \ldots)cn=hasbG(x)xn dx,(n=0,1,)
can be calculated using well-known symmetric functions w r = α 1 + α 2 + + α n + 1 = r x 1 α 1 x 2 α 2 x n + 1 α n + 1 , ( r = 0 , 1 , ; w 0 = 1 ) w r = α 1 + α 2 + + α n + 1 = r x 1 α 1 x 2 α 2 x n + 1 α n + 1 , r = 0 , 1 , ; w 0 = 1 w_(r)=sum_(alpha_(1)+alpha_(2)+dots+alpha_(n+1)=r)x_(1)^(alpha_(1))x_(2)^(alpha_(2))dotsx_(n+1)^(alpha_(n+1)),(r=0,1,dots;w_(0)=1)w_{r}=\sum_{\alpha_{1}+\alpha_{2}+\ldots+\alpha_{n+1}=r} x_{1}^{\alpha_{1}} x_{2}^{\alpha_{2}} \ldots x_{n+1}^{\alpha_{n+1}},\left(r=0,1, \ldots ; w_{0}=1\right)wr=α1+α2++αn+1=rx1α1x2α2xn+1αn+1,(r=0,1,;w0=1), the summation being extended to the non-negative integer solutions of the Diophantine equation a 1 + α 2 + + α n + 1 = r a 1 + α 2 + + α n + 1 = r a_(1)+alpha_(2)+cdots+alpha_(n+1)=ra_{1}+\alpha_{2}+\cdots+\alpha_{n+1}=rhas1+α2++αn+1=rWe can calculate the w r w r w_(r)w_{r}wrusing the recurrence formula
(10) w r p 1 w r 1 + p 2 w r 2 + ( 1 ) n + 1 p n + 1 w r n 1 = 0 , ( r = 1 , 2 , ) , (10) w r p 1 w r 1 + p 2 w r 2 + ( 1 ) n + 1 p n + 1 w r n 1 = 0 , ( r = 1 , 2 , ) , {:(10)w_(r)-p_(1)w_(r-1)+p_(2)w_(r-2)-cdots+(-1)^(n+1)p_(n+1)w_(r-n-1)=0","(r=1","2","dots)",":}\begin{equation*} w_{r}-p_{1} w_{r-1}+p_{2} w_{r-2}-\cdots+(-1)^{n+1} p_{n+1} w_{r-n-1}=0,(r=1,2, \ldots), \tag{10} \end{equation*}(10)wrp1wr1+p2wr2+(1)n+1pn+1wrn1=0,(r=1,2,),
by asking w 0 = 1 , w 1 = w 2 = = w n = 0 , p 1 , p 2 , , p n + 1 w 0 = 1 , w 1 = w 2 = = w n = 0 , p 1 , p 2 , , p n + 1 w_(0)=1,w_(-1)=w_(-2)=cdots=w_(-n)=0,p_(1),p_(2),dots,p_(n+1)w_{0}=1, w_{-1}=w_{-2}=\cdots=w_{-n}=0, p_{1}, p_{2}, \ldots, p_{n+1}w0=1,w1=w2==wn=0,p1,p2,,pn+1being the fundamental symmetric functions of x 1 , x 2 , , x n + 1 x 1 , x 2 , , x n + 1 x_(1),x_(2),dots,x_(n+1)x_{1}, x_{2}, \ldots, x_{n+1}x1,x2,,xn+1.
If we ask f = x n + r f = x n + r f=x^(n+r)f=x^{n+r}f=xn+rIn (1), we obtain,
hence
w r = [ x 1 , x 2 , , x n + r ; x n r ] = ( n + r ) ! r ! a b G ( x ) x r d x c r = r ! ( n + r ) ! w r , ( r = 0 , 1 , ; 0 ! = 1 ) w r = x 1 , x 2 , , x n + r ; x n r = ( n + r ) ! r ! a b G ( x ) x r d x c r = r ! ( n + r ) ! w r , ( r = 0 , 1 , ; 0 ! = 1 ) {:[w_(r)=[x_(1),x_(2),dots,x_(n+r);x^(n-r)]=((n+r)!)/(r!)int_(a)^(b)G(x)x^(r)dx],[c_(r)=(r!)/((n+r)!)w_(r)","(r=0","1","dots;0!=1)]:}\begin{gathered} w_{r}=\left[x_{1}, x_{2}, \ldots, x_{n+r} ; x^{n-r}\right]=\frac{(n+r)!}{r!} \int_{a}^{b} G(x) x^{r} \mathrm{~d} x \\ c_{r}=\frac{r!}{(n+r)!} w_{r},(r=0,1, \ldots ; 0!=1) \end{gathered}wr=[x1,x2,,xn+r;xnr]=(n+r)!r!hasbG(x)xr dxcr=r!(n+r)!wr,(r=0,1,;0!=1)
  1. Suppose the nodes x α , ( α = 1 , 2 , , n + 1 ) x α , ( α = 1 , 2 , , n + 1 ) x_(alpha),(alpha=1,2,dots,n+1)x_{\alpha},(\alpha=1,2, \ldots, n+1)xα,(α=1,2,,n+1), are symmetrically distributed with respect to the origin, therefore that a = b , x α + x n + 2 α = 0 a = b , x α + x n + 2 α = 0 a=-b,x_(alpha)+x_(n+2-alpha)=0a=-b, x_{\alpha}+x_{n+2-\alpha}=0has=b,xα+xn+2α=0, ( α = 1 , 2 , , n + 1 ) ( α = 1 , 2 , , n + 1 ) (alpha=1,2,dots,n+1)(\alpha=1,2, \ldots, n+1)(α=1,2,,n+1)In this case we have p α = 0 p α = 0 p_(alpha)=0p_{\alpha}=0pα=0For α α alpha\alphaαodd and from (10) it follows that w r = 0 w r = 0 w_(r)=0w_{r}=0wr=0, therefore also c r = 0 c r = 0 c_(r)=0c_{r}=0cr=0, For r r rrrany odd number.
We have the approximation formula ( m 1 m 1 m >= 1m \geqq 1m1)
(11) b b G ( x ) f ( x ) d x b b G ( x ) [ α = 0 2 m 1 x α α ! f ( α ) ( 0 ) ] d x = α = 0 m 1 c 2 α ( 2 α ) ! f ( 2 α ) ( 0 ) (11) b b G ( x ) f ( x ) d x b b G ( x ) α = 0 2 m 1 x α α ! f ( α ) ( 0 ) d x = α = 0 m 1 c 2 α ( 2 α ) ! f ( 2 α ) ( 0 ) {:(11)int_(-b)^(b)G(x)f(x)dx~~int_(-b)^(b)G(x)[sum_(alpha=0)^(2m-1)(x^(alpha))/(alpha!)f^((alpha))(0)]dx=sum_(alpha=0)^(m-1)(c_(2alpha))/((2alpha)!)f^((2alpha))(0):}\begin{equation*} \int_{-b}^{b} G(x) f(x) \mathrm{d} x \approx \int_{-b}^{b} G(x)\left[\sum_{\alpha=0}^{2 m-1} \frac{x^{\alpha}}{\alpha!} f^{(\alpha)}(0)\right] \mathrm{d} x=\sum_{\alpha=0}^{m-1} \frac{c_{2 \alpha}}{(2 \alpha)!} f^{(2 \alpha)}(0) \tag{11} \end{equation*}(11)bbG(x)f(x)dxbbG(x)[α=02m1xαα!f(α)(0)]dx=α=0m1c2α(2α)!f(2α)(0)
of which the rest
(12) b b G ( x ) x 2 m [ x , 0 , 0 , , 0 2 m ; f ] d x (12) b b G ( x ) x 2 m [ x , 0 , 0 , , 0 2 m ; f ] d x {:(12)int_(-b)^(b)G(x)x^(2m)[x","ubrace(0,0,dots,0ubrace)_(2m);f]dx:}\begin{equation*} \int_{-b}^{b} G(x) x^{2 m}[x, \underbrace{0,0, \ldots, 0}_{2 m} ; f] \mathrm{d} x \tag{12} \end{equation*}(12)bbG(x)x2m[x,0,0,,02m;f]dx
is the degree of accuracy 2 m 1 2 m 1 2m-12 m-12m1and is indeed of the simple form, therefore of the form
(13) c 2 m [ ξ 1 , ξ 2 , , ξ 2 m + 1 ; f ] , (13) c 2 m ξ 1 , ξ 2 , , ξ 2 m + 1 ; f , {:(13)c_(2m)[xi_(1),xi_(2),dots,xi_(2m+1);f]",":}\begin{equation*} c_{2 m}\left[\xi_{1}, \xi_{2}, \ldots, \xi_{2 m+1} ; f\right], \tag{13} \end{equation*}(13)c2m[ξ1,ξ2,,ξ2m+1;f],
THE ξ α ξ α xi_(alpha)\xi_{\alpha}ξαbeing 2 m + 1 2 m + 1 2m+12 m+12m+1distinct points of the interval ( b , b b , b -b,b-b, bb,b). In (12) we assume that the function f f fffhas a continuous derivative of order 2 m 1 2 m 1 2m-12 m-12m1, but the rest is of the form (13) even if f f fffhas a continuous derivative of order 2 m 2 2 m 2 2m-22 m-22m2only. The rest can therefore be in the form
2 c 2 m ( 2 m ) ! [ ξ 1 , ξ 2 , ξ 3 ; f ( 2 m 2 ) ] 2 c 2 m ( 2 m ) ! ξ 1 , ξ 2 , ξ 3 ; f ( 2 m 2 ) (2c_(2m))/((2m)!)[xi_(1),xi_(2),xi_(3);f^((2m-2))]\frac{2 c_{2 m}}{(2 m)!}\left[\xi_{1}, \xi_{2}, \xi_{3} ; f^{(2 m-2)}\right]2c2m(2m)![ξ1,ξ2,ξ3;f(2m2)]
ξ 1 , ξ 2 , ξ 3 ξ 1 , ξ 2 , ξ 3 xi_(1),xi_(2),xi_(3)\xi_{1}, \xi_{2}, \xi_{3}ξ1,ξ2,ξ3being three distinct points of the interval ( b , b b , b -b,b-b, bb,b).
We deduce that if the nodes x α , ( α = 1 , 2 , , n + 1 ) x α , ( α = 1 , 2 , , n + 1 ) x_(alpha),(alpha=1,2,dots,n+1)x_{\alpha},(\alpha=1,2, \ldots, n+1)xα,(α=1,2,,n+1), are symmetrically distributed with respect to the origin and if x n + 1 = x 1 = b ( > 0 ) x n + 1 = x 1 = b ( > 0 ) x_(n+1)=-x_(1)=b( > 0)x_{n+1}=-x_{1}=b(>0)xn+1=x1=b(>0)we have the approximation formula
[ x 1 , x 2 , , x n + 1 ; f ] = α = 0 m 1 w 2 α ( n + 2 α ) ! f ( n + 2 α ) ( 0 ) + + 2 w 2 m ( n + 2 m ) ! [ ξ 1 , ξ 2 , ξ 3 ; f ( n + 2 m 2 ) ] x 1 , x 2 , , x n + 1 ; f = α = 0 m 1 w 2 α ( n + 2 α ) ! f ( n + 2 α ) ( 0 ) + + 2 w 2 m ( n + 2 m ) ! ξ 1 , ξ 2 , ξ 3 ; f ( n + 2 m 2 ) {:[[x_(1),x_(2),dots,x_(n+1);f]=sum_(alpha=0)^(m-1)(w_(2alpha))/((n+2alpha)!)f^((n+2alpha))(0)+],[quad+(2w_(2m))/((n+2m)!)[xi_(1),xi_(2),xi_(3);f^((n+2m-2))]]:}\begin{aligned} & {\left[x_{1}, x_{2}, \ldots, x_{n+1} ; f\right]=\sum_{\alpha=0}^{m-1} \frac{w_{2 \alpha}}{(n+2 \alpha)!} f^{(n+2 \alpha)}(0)+} \\ & \quad+\frac{2 w_{2 m}}{(n+2 m)!}\left[\xi_{1}, \xi_{2}, \xi_{3} ; f^{(n+2 m-2)}\right] \end{aligned}[x1,x2,,xn+1;f]=α=0m1w2α(n+2α)!f(n+2α)(0)++2w2m(n+2m)![ξ1,ξ2,ξ3;f(n+2m2)]

Or f f fffhas a continuous derivative of order n + 2 m 2 sur ( b , b ) n + 2 m 2 sur ( b , b ) n+2m-2sur(-b,b)n+2 m-2 \operatorname{sur}(-b, b)n+2m2on(b,b)And ξ 1 , ξ 2 , ξ 3 ξ 1 , ξ 2 , ξ 3 xi_(1),xi_(2),xi_(3)\xi_{1}, \xi_{2}, \xi_{3}ξ1,ξ2,ξ3are three distinct points in this interval.
Formula (11) is still of the Gaussian type, according to the definition of formulas of this type [7].
Regue on 13 II 1967
Babeş-Bolyai University

BIBLIOGRAPHY

  1. Gotusso L., Una valutasione approssimata del fin complementare della formula di Taylor. Atti del Seminario Mat. e Fizico dell’Univ. di Modena, 1964, XIII, pp. 221-229. 2. Ionescu DV, Cuadraturi Numerice. Ed. Tehn. Buc., 1957.
  2. Kowalewski G., Interpolation und genäherte Quadratur. Leipzig u. Berlin, 1932.
  3. v. Mises s R., Über allgemeine Quadraturformeln. J.f. queen u. angew. Math., 1936, 174, S. 56-67.

Related Posts

Remarks on certain mean value formulas

Abstract ? AuthorsTiberiu Popoviciu (Institutul de Calcul) Original title (in French) Remarques sur certaines formules de la moyenne Keywords? Cite…

Proposed problems

Abstract AuthorsR. Askey, R. DeVore, G. Freud, J. Musielak, J. Peetre Tiberiu Popoviciu Tiberiu Popoviciu Keywords? Paper coordinatesProposed problems, Contributed…

On the calculation of a limit

Abstract AuthorsT. Popoviciu Institutul de Calcul Keywords? Paper coordinatesT. Popoviciu, Asupra calculului unei limite, Gazeta Matem. Ser. A, 71 (1971) no. 1, pp.…