TIBERIU POPOVICIU

1. 1.

   Let's assume that the rest$R[f]$of a linear approximation formula or a linear functional defined on the vector space$S$, formed by functions$f=f(x)$, defined and continuous on an interval$I$. The functions$f$and the linear functional$R[f]$are real and$S$contains all polynomials.

   Report issue for preceding element

We say that$R[f]$is of the simple form if there exists an integer$n\geqq-1$. such that we have

Or$K=R\left[x^{n+1}\right]$East$\neq 0$, independent of the function$f$And$\xi_{i},i=1,2,\ldots\ldots,n+2$are$n+2$distinct points of the interval$I$(which can, in general, depend on the function$f$and located even inside$I$if$n\geqq 0$). The notation$\left[\xi_{1},\xi_{2},\ldots,\xi_{n+2};f\right]$denotes the divided difference of the function$f$on the knots$\xi_{1},\xi_{2},\ldots,\xi_{n+2}$. For these notions and the few properties which will follow we ask the reader to refer to our previous work, in particular, to our work in the previous volume of this journal [3].

In this case$n$is the degree of accuracy of the remainder and enjoys the property (characteristic) that$R[f]$is zero on any polynomial of degree$n$, but$R\left[x^{n+1}\right]\neq 0$.

Let us recall that the necessary and sufficient condition for$R[f]$, assumed from the degree of accuracy$n$, either of the simple form is that we have$R\left[f^{\prime}\right]\neq 0$for everything$f\in S$, which is convex of order$n^{\prime}$(on I). In this case it is, moreover, necessary that$R[f]$keep his sign for$f$convex of order$n$. Noting that the function$x^{n+1}$is indeed convex of order$n$, the previous condition can also be written
(2)

Condition (2), for all$f\in S$convex of order$n$, is therefore necessary and sufficient for$R[f]$either of the simple form (1). Note that for this it is also necessary (but not sufficient) that we have$R\left[x^{n+1}\right]\neq 0$and
(3)

for any function$/\in S$, non-concave of order$n$.
mule.
2. If$R[/]$is of the simple form (1), we can delimit it by the for-

Besides, if$f$has a derivative of order$n\not-1$(bounded) on$I$, the number
(5) is given by the equality

But, delimitation (4) is valid in a more general case. In particular, we will demonstrate that:

The delimitation (4) is valid if$R[t]$is the degree of accuracy$n$and if inequality (3) is verified for any function$f\in S$, non-concave of order$n$.

We have$R\left[x^{n+1}\right]\neq 0$and for the demonstration we can assume that$R\left[x^{n+1}\right]>0$. Let us then consider the linear functional (also defined on$S$)
(6)

Or$x_{1},x_{2},\ldots,x_{n+2}$are$n+2$fixed distinct points (independent of the function$f$) of the interval$I$And$\varepsilon$is any positive number. We will show that$R_{1}[f]$is of the simple form (1). Indeed, if we take into account the fact that the difference divided on$n+2$nodes (not all combined) of a convex function of order$n$is, by positive definition, we deduce that$R_{1}[f]>0$, For$f\in S$convex of order$n$. The property is demonstrated. Taking into account (5) and (6), also writing the corresponding delimitation (4) for$R_{1}[t]$, we get

This inequality being true whatever the positive number$\varepsilon$, it follows that we have (4) and the sought property is demonstrated. If we have$R\left[x^{n+1}\right]<0$, the demonstration is completely analogous. We then take in (6) for$\varepsilon$any negative number.
3. To apply the previous property it is sufficient to know criteria allowing us to affirm that (under the hypothesis$R\left[x^{n+1}\right]\not z^{\prime}0$) inequality
(3) is verified for any function$f\in S$non-concave order$n$. We will present here such a criterion which results from the remarkable property of SN Bernstein approximation polynomials of preserving the convexity characteristics of functions [2].

Let's suppose that$I=[0,1]$and that the elements of$S$have a derivative of order$j(\geq 0)$continue on$[0,1]$. Consider the linear functional$R[f]$, of the degree of accuracy$n$and which is limited in relation to the norine

Let's ask

Under the previous assumptions, we have the following property:
For inequality (3) to hold for any function$f\in S$non-concave order$n$, it is (necessary and it is) sufficient that one has$R\left[x^{n+1}\right]R\left[\pi_{k,l}\right]\geqq 0$, whatever the non-negative integers$k$And$l$.

Note that$\pi_{k,l}^{(n+1)}=x^{k}(1-x)^{l}$. If

is the SN Bernstein polynomial of degree$m$, for its order derivative$n+1$We have ($m\geqslant n+1$),
$B_{m}^{(n+1)}=\frac{(m-1)!(n+1)!}{m^{n}(m-n-1)!}\sum_{i=0}^{m-1}\binom{m-n-1}{i}\left[\frac{i}{m},\frac{i+1}{m},\ldots,\frac{i+n+1}{m};f\right]\pi_{i,m-n-1-i}^{(n+1)}$from where

Or$\beta_{m}$is a polynomial of degree$n$.
According to sn BERNSTEIN [1] and S. WIGERT [4] if the derivative$f^{(l)}$of order$i(\geqq 0)$of the function$f$exists and is continuous on$[0,1]$, the sequel$\left(B_{m}^{(i)}\right)$tends, for$m\rightarrow\infty$, uniformly on$[0,1]$, towards$f^{(i)}$. It follows that$R\left[B_{m}\right]\rightarrow\rightarrow R[f]$For$m\rightarrow\infty$, therefore
(8)

If we notice that

and that the differences divided on$n+2$nodes of a non-concave function of order$n$are non-negative, it follows that

for a function$f$non-concave order$n$. Taking into account (8), the desired property results.

7. To give an application either$R[f]$the remainder in the digital quadrature formula

Or$f$has a continuous third-order derivative on$[0,1]$.

In this case$R[f]$is the degree of accuracy$n=5$and is bounded with respect to the norm (7) for$j=3$. In our case

We deduce

and a simple calculation gives us

The delimitation (4) is therefore applicable in this case and we have

If the derivative of order$6$,$f^{(6)}$, exists on$[0,1]$, We have