DIFFERENCES, DIVIDED AND DERIVED

Tiberiu Popoviciu paste

1.
- •
  
  Either $A[f]=A[f(x)]$ a linear functional, therefore additive and bonogenous, defined on a vector space $S$ functions $f=f(x)$ , real, of the real variable $x$ defined and continuous on an interval $I$ We will designate the end gate by a and by $b$ the right endpoint of interval I. In what follows we will always assume that the elements of $S$ satisfy all the differentiability properties necessary for the linear functionals under consideration to be meaningful. We will always assume that $S$ contains all polynomials. We always assume that $a<b$ .

Any linear functional $A[f]$ to a degree $d$ exactiode well determined. This degree of accuracy is the integer $n\geq-1$ , or the improper number $n=\infty$ characterized by the property:
$1^{\circ}.n=-1$ . if $A[1]\neq 0$
2 . $A\left[x^{i}\right]=0,i=0,1,\ldots,n,A\left[x^{n+1}\right]\neq 0$ if $A[1]=0$ and if at least one of the numbers $A\left[x^{i}\right],t=0,1,\ldots$ is different from zero.
$3^{\circ}.n=\infty$ if $A\left[x^{i}\right]=0,i=0,1,\ldots$
In the cases $1^{\circ}$ And $2^{\circ}$ The degree of accuracy is finite. This case occurs if and only if $A[f]$ is not zero on at least one polynomial. In the case $3^{\circ}$ , the degree of accuracy is infinite and then $A[f]$ is zero for any polynomial.

For a linear functional $A[f]$ either zero on any polynomial of degree $n$ , it is necessary and sufficient that its degree of accuracy be equal to $n$ at least (we always assume that $n<\infty$ A polynomial of degree $n$ is of the form $\alpha_{0}x^{n}+\alpha_{1}x^{n-1}-\mid\ldots+\alpha_{n}$ the coefficients $\alpha_{0},\alpha_{1},\ldots,\alpha_{n}$ being: any real numbers. If the highest coefficient $\alpha_{0}$ East $\neq 0$ The polynomial is said to have effective degree $n$ 9.
- We will focus, in particular, on linear functionals $A[f]$ which are equal to a linear combination of the values, over a finite number of points, of the function $f$ and a finite number of its derivatives of various orders. Such a linear functional is of the form
(1)

A[f]=\sum_{i=1}^{p}\sum_{i=0}^{k_{i-1}}a_{i,l}f^{(l)}\left(z_{i}\right)

Or, $p,k_{1},k_{2},\ldots,k_{p}$ are given natural numbers, $z_{i},i=1,2,\ldots,p$ , $p$ distinct points of the interval $I$ And $a_{i,j},j=0,1,\ldots,k_{i}-1,i=1,2,\ldots$ , $p$ numbers independent of the function $f$ The points $z_{t}$ are the nodes and the numbers $a_{i,j}$ are the coefficients of the linear functional (1).

In expression (1) and relative to the node $z_{i}$ include the values of the function and its $k_{i}-1$ first derivatives, therefore of its first $k_{i}$ derivatives if we agree that the function itself is its own derivative of order 0, on this point. For this reason we agree that in $z_{i}$ be confused $k_{i}$ knots. So $k_{i}$ is the multiplicity order of the knot $z_{i}$ (it's a simple knot if $k_{i}=1$ , double if $k_{i}=2$ etc). We can also say that $z_{j}$ is a node of order $k_{i}$ of multiplicity. In this way, the total number of distinct or non-distinct nodes (i.e., each node counted with its multiplicity order) is equal to $m = k_1 + k_2 + ... + k$ The number $m$ East $\geqq p$ and is equal to $p$ if and only if all the nodes are simple.

THE $m$ knots, simple or not, can be designated by $x_{1},x_{2},\ldots,x_{m}$ Among these points, exactly $k_{i}$ coincide with $z_{i}$ For $i=1,2,\ldots,p$ In this way, we have numbered a certain permutation of the nodes. In principle, the permutation, and therefore the numbering of the nodes, is arbitrary. However, there are certain preferred numbering systems, which we will call normal numbering systems. In a normal numbering system, for all $i$ , THE $k_{i}$ knots $x_{j}$ which coincide with $z_{i}$ are numbered with $k_{i}$ consecutive indices. A normal numbering system is, for example, $x_{k_{1}+k_{2}+\ldots+k_{i-1}+v}==z_{i},v=1,2,\ldots,k_{j},i=1,2,\ldots,p\left(k_{0}=0\right)$ In particular, if the following $x_{1},x_{2},\ldots,x_{n}$ is monotonic (non-decreasing or non-increasing), the numbering is normal.
3. - The (identically) zero functional on $S$ is of the form (1), where all the coefficients $a_{i,j}$ are equal to 0. This linear functional has a degree of accuracy equal to $\infty$

A linear functional of the form (1) does not completely determine the system of nodes. $z_{1}$ with their respective multiplicity orders. Indeed, we can add any finite number of nodes without the linear functional under consideration being modified. It suffices to demonstrate this property for a single node. $x_{0}$ added to the previous ones. So we can add to $A[f]$ without changing its values, the term $0.f(x_{0})$ if $x_{0}$ does not coincide with the author of the nodes $z_{i}$ and the term 0. $f^{\left(k_{i}\right)}\left(z_{i}\right)$ if $x_{0}=z_{i}$ .

Let us then consider a linear functional (1) which does not have all its coefficients zero. We can assume, without restricting the generality, that
(2)

a_{i},k_{i-1}\neq 0,i=1,2,\ldots,p.

In this case, the nodes are reduced to their smallest number since, on the one hand, if conditions (2) are met, we can remove a certain number of nodes without modifying the functional $A[f]$ Furthermore ,
such node deletions are not possible if conditions (2) are not all met. It is easy to see how the minimum number of nodes can be obtained.

Consider the polynomial of degree $m$
(3)

l(x)=\prod_{v=1}^{m}\left(x-x_{v}\right).

SO*)

\begin{gathered}A\left[\frac{l(x)}{x-x_{i}}\right]=a_{i,k_{i}-1}\left[\frac{l(x)}{x-x_{i}}\right]_{x-x_{i}} ^{\left(k_{i}-1\right)}=a_{i,k_{i}-1}\left(k_{i}-1\right)!\prod_{v=1}^{p}\left(z_{i}-z_{v}\right)^{k_{v}}\\ i=1,2,\ldots,p\end{gathered}

which, according to hypothesis (2), are all different from zero. We therefore have Lemma
1. - The linear functional (1), where the coefficients $a_{i,j}$ are not all zero, has a degree of accuracy (finite and) at most equal to $m-2$ .

It follows that if the linear functional (1) has a greater degree of accuracy than $m-2$ it is identically zero.
4. - If the linear functional (1) has a degree of accuracy equal to $m-2$ , it reduces, apart from a non-zero factor independent of the function $f$ , to the difference divided by order $m-1$ on the $m$ knots $x_{1},x_{2},\ldots,x_{m}$ of the function $f$ This divided difference will be denoted by
(4)

\left[x_{1},x_{2},\ldots,x_{m};f\right]

out by

The divided difference is a linear functional of the form (1) determined completely by the conditions of vanishing on any polynomial of degree $m-2$ and to reduce to 1 over the polynomial $x^{n-1}$ .

Divided differences possess various properties and satisfy well-known formulas. We will recall the main formulas that will be used later.

The divided difference is symmetric with respect to the nodes on which it is defined. It follows that in notation (4) the numbering of the nodes is irrelevant.

We have the recurrence relation
(5)

\left[t_{1},t_{2},\ldots,t_{Q+1};f\right]=\frac{\left[t_{2},t_{1}+\ldots,t_{Q+1};f\right]-\left[t_{1},t_{2},\ldots,t_{Q};f\right]}{t_{Q+1}-t_{1}}

⁰⁰footnotetext: *)

\sum_{v=1}^{p}i,\frac{p}{|i|}

meant that in the sum resp. the product la val i of the index is exclae.

which is the relationship between divided differences of order 0 and divided differences of order $a-1$ Formula (5) is valid only under the condition that the nodes $t_{1},t_{0+1}$ , are distinct, assuming, of course, that the divided differences shown therein have meaning.

If all the nodes of a divided difference of order a coincide with the same point $L$ , this difference divided is equal to $\frac{1}{\rho^{!}}f^{(\phi)}(\phi)$ We therefore have formula
(b)

[\underbrace{t}_{\ell+1}t_{1}\ldots,t;f]=\frac{1}{e!}f^{(p)}(t).

We also have the formula for decomposition
(7) $\left[t_{1},t_{2},\ldots,t_{Q},t_{1},t_{2},\ldots,t_{Q};f\right]=\left[t_{1},t_{2},\ldots,t _{Q};\frac{f(x)}{\left(x-t_{1}\right)\left(x-t_{2}\right)\ldots\left(x-t_{Q}\right)}\right]+$

-\left[i_{1},i_{2},\ldots,i_{0^{\prime}};\frac{(x-1)(x-1)}{(x-1)\ldots\left(x-i_{0}\right)}\right]

which is valid provided that none of the nodes $t_{1},t_{2},\ldots,t_{6}$ do not coincide with one of the nodes $i_{1},i_{2},\ldots,i_{Q^{\prime}}$ .

We also have the translation formula
(8)

\begin{gathered}{\left[t_{1},t_{2},\ldots,t_{Q},t_{1},t_{2},\ldots,t_{Q};f(x)\left(x-t_{1}\right)\left(x-t_{2}\right)\ldots\left(x-t_{Q}\right)\right]=}\\ =\left[t_{1},t_{2},\ldots,t_{0};f\right]\end{gathered}

The previous formulas allow us to find the coefficients. $c_{i,j}$ of the difference divided (1),

\left[x_{1},x_{2},\ldots,x_{m};f\right]=\sum_{i=1}^{p}\sum_{j=0}^{i_{t}-1}c_{i,j}f^{(h)}\left(z_{i}\right)

(9)

If we ask

h_{i}(x)=\frac{u(x)}{\left(x-z_{i}\right)^{k_{i}}}=\frac{l_{i}}{\mid v=1}\left(x-z_{v}\right)^{k_{v}},i=1,2,\ldots,p

Or $l(x)$ is the polynomial (3), by appropriately applying, and several times if necessary, formulas (6), (7), 11011, we deduce

\begin{gathered}{\left[x_{1},x_{2},\ldots,x_{m};f\right]=\sum_{i=1}^{p}\left[\underline{z}_{i,}z_{i},\ldots,z_{i};\frac{l}{h_{i}}\right]=}\\ =\sum_{i=1}^{p}\frac{1}{(h-1)!}-\left[\frac{f(x)}{h_{i}(x)}\right]_{x-z_{i}}^{(h,-1)}\end{gathered}

We have done

\begin{gathered}c_{i,j}=\frac{1}{\left(h_{i}-1\right)!}\cdot\left(k_{i}-1\right)\left[\frac{1}{l_{i}(n)}\right]_{z=s_{l}}^{\left(k_{i}-1-j\right)}\\ j=0,1,\ldots,h_{i}-1,i=1,2,\ldots,p.\end{gathered}

We have, in particular.

c_{t_{t},k-1}=\frac{1}{\left(h_{t}-1\right)1}\cdot\frac{1}{\frac{p}{|i|}\left(z-z_{v}\right)^{h_{v}}},\quad i=1,2,\ldots,p

We see that, in the case of the divided difference (4), conditions (2) are satisfied. It follows that, in the case of the divided difference, the notation (4) precisely highlights the system of nodes with the minimum number.
5. - Let us denote by $L\left(x_{1},x_{2},\ldots,x_{n};f\mid x\right)$ , the Lagrange-Termite polynomial relating to the function $f$ and on the nodes $x_{1},x_{2},\ldots,x_{m}$ This is the (unique) polynomial. $L(x)$ degree $m-1$ who verifies the equalities

L^{(j)}\left(z_{i}\right)=f^{(j)}\left(z_{i}\right),\quad j=0,1,\ldots k_{i}\ldots 1,i=1,2,\ldots,p.

(10)

We have*)
(11)

=\sum_{v=0}^{m-1}\left(x-x_{1}\right)\left(x-x_{2}\right)\ldots\left(x-x_{v}\right)\cdot\left[x_{1},x_{2},\ldots,x_{v+1};f\right].

From (10) it follows that
(12)

A[f]=A\left[L\left(x_{1},x_{2},\ldots,x_{m};f\mid x\right)\right]

ct, account tenaut of formula (11),

A[f]=\sum_{v=0}^{m-1}a_{v}\left[x_{1},x_{2},\ldots,x_{v-1-1};f\right]

(13)

\text{ (14) }\quad a_{v}=A\left[\left(x-x_{1}\right)\left(x-x_{2}\right),\ldots\left(x-x_{v}\right)\right],\quad v=0,1,\ldots,m-1.

If the linear functional under consideration has a degree of accuracy less than or equal to $n$ ( $0\leq n\leq m-2$ ) we have $a_{v}=0,v=0,1,\ldots,n$ and vice versa. If it has a degree of accuracy equal to $n(-1\leqq n\leqq m-2)$ Furthermore, we have $a_{n-1}=A\left[x^{\prime\prime+1}\right]\neq 0$ and vice versa. This property can be stated in the form of

The same 2. - For the linear functional (1) to have a degree of accuracy at least equal to $n$ , it is necessary and sufficient that in its expression in the form (13) Zon has $a_{0}=a_{1}=\ldots=a_{n}=0$ . So that the degree of exact-

⁰⁰footnotetext: 4. If

v=0

the product

\left(x-x_{1}\right)\left(x-x_{2}\right)\ldots(x-1v)

is replaced by 1. These same conventions apply even more to analogous formulas

titude is exactly equal to $n$ It is necessary and sufficient that, in addition, one has $a_{n+1}\neq 0$ 6.
- The previous result is true for any numbering of the nodes.

Now suppose that the following $x_{1},x_{2},\ldots,x_{m}$ (therefore the respective numbering) of the nodes depends on the property that if $i,j$ are any indices $1,2,\ldots,m$ ,

j-i\geqq n+2\Rightarrow x_{i}\neq x_{j}.

(15)

Applying formula (6) to divided differences $\left[x_{1},x_{2},\ldots,x_{4};f\right]$ , $v=n+2,n+3,\ldots,m-1$ (if $m\geq n-3$ ), when necessary, even several times (if $m>n-3$ ), we deduce the formula ( $m\geq n+3$ ),

	$\displaystyle A[f]=\sum_{v=0}^{m}a_{v}\left[x_{1},x_{2},\ldots,x_{v+1};f\right]+$		(16)
	$\displaystyle\quad+\sum_{i=1}^{m-1}\mu_{i}\left[x_{i},x_{i+1},\ldots,x_{i+n+1};f\right]$

where the coefficients $a_{y}$ , given by (14), and the coefficients $\mu_{i},i=1,2,\ldots m-n-1$ are independent of the function $f$ .

We can then state
Lemma 3. - For the linear functional (1) to have a degree of exactness at least equal to $n\geq\max\left(k_{1},k_{2},\ldots,k_{p}\right)-2$ (done so that it is zero on your degree polytome $n\geqq\max\left(k_{1},k_{2},\ldots,k_{p}\right)-2$ ) $il$ false and it suffices which soil of the shape

A[f]=\sum_{i=1}^{m-n-1}\mu_{i}\left[x_{i},x_{i+1},\ldots,x_{i+n+1};f\right]

(17)

where the $\mu_{i},i=1,2,\ldots,m-n-1$ sound independent coefficients of the function $f$ .

So that, under the same conditions, the degree of exachilude is equal to $n$ Furthermore, it is necessary and sufficient that one has $\sum_{i=1}^{m-n-1}\mu_{i}\neq 0$ .

The condition is necessary. Indeed, on the one hand, under the assumptions of the lemma, we can find a numbering of the nodes such that conditions (15) are satisfied. Such a numbering is, for example, any normal numbering*). On the other hand, then from formula (16) it follows formula (17).

⁰⁰footnotetext: *). There may be numbering systems, different from normal numbering, for which conditions (15) are met. If, for example, we have

p=3,k_{1}=3,h_{2}=h_{3}=2

(SO

m=7

n=3

, the permutations (

P

z_{1},z_{1},z_{2},z_{2},z_{1},z_{3},z_{3},\left(P^{\prime}\right):z_{1},z_{2},z_{1},z_{1},z_{2},z_{3},z_{3}

give numberings that verify the conditions (15). These two numberings differ in that the formula (17) corresponding to

(P)

leads to the same divided differences as the permato

(P*):z_{1},z_{1},z_{1},z_{2},z_{2},z_{3},z_{3},qui

corresponds to a nominal numbering, whereas formula (17) corresponds to (

P

) leads to divided differences that are not the same (datts lent ettsentile). It follows that the numbering corresponding to the permutation (i) in a certain way, vedacibbe to a normal numbering, while it is not clear.

$I_{1}a$ This condition is sufficient. Indeed, any difference divided by order $n+1$ is the degree of accuracy $n$ , therefore it vanishes on any polynomial of degree $n$ . The same is true for any linear combination of such divided differences.

The sufficiency of the last condition of the lemma results from the formula $A\left[x^{n-1}\right]=\sum_{i=1}^{m-n-1}\mu_{i}$ .
$\mathrm{I}_{\mathrm{r}}\mathrm{a}$ condition $n\geqq\max\left(k_{1},k_{2},\ldots,k_{p}\right)-2$ The lemma is essential. If this condition is not met, there may not exist a relation of the form (17). This follows easily from the fact that if the nodes of a linear functional of the form (17) are reduced to their minimum number, among these nodes there is none that has a multiplicity order. $>n+2$ Moreover, if $n<\max\left(k_{1},k_{2},\ldots,k_{p}\right)-2$ , there is no numbering that verifies property (15).

II.

7.
- •
  
  We will recall the notion of a linear functional of simple form. Ira linear functional $A[f]$ , defined on space $S$ is said to be of simple form if there exists an integer $n\geqq-1$ such as, for your $f_{\epsilon}S$ , one has

A[f]=K.\left[\xi_{1},\xi_{2},\ldots,\xi_{n+2};f\right]

(18)

Or $K$ is a coefficient other than 0 and independent of the function $f$ and the $\xi_{1},\xi_{2},\ldots,\xi_{n+2}$ are $n+2$ distinct points of the interval $I$ , if $n\geqq 0$ even from within the interval $I$ and which, in general, can depend on the function $f$ The fact that, for $n\geq 0$ , the points $\xi_{i}$ can be chosen within the interval $I$ results from the average properties of divided differences [6]. In this case, the degree of accuracy of $A[f]$ is necessarily equal to $n$ It follows that if a linear functional is of the simple form, it is of this form for only one value of $n$ There is an important property that characterizes functionals of simple form [4] and that can be stated in the form of:

Lemma 4. - For the linear functional $A[f]$ either in the simple form, it is necessary and sufficient that there exists an integer $n\geqq-1$ such as one has $A[f]\neq 0$ for everything $f_{\in}S$ , convex of order $n$ .

The property of being of simple form is therefore very closely linked to the notion of a higher-order convex function.

A function defined on $I$ is said to be convex of order $n$ if all its differences divided by order $n+1$ , on $n+2$ distinct nodes (belonging to $I$ ) are positive. The function is said to be non-concave of order $n$ (on $I$ ) if all its differences divided in order $n+1$ at distinct points (or not) are non-negative. A convex function of order $n$ is a non-concave function of order $n$ particular.

The number $n$ of lemma 4 is that which appears in the corresponding formula (18). The coefficient $K$ of this formula is equal to $A\left[x^{13+1}\right]$ or
to $A[f]$ Or $f$ is an arbitrary polynomial of degree $n+1$ with the highest coefficient equal to 1.

If the linear functional $A[f]$ is of the simple form and if the function $f$ has a derivative of order $n+1$ (For $n\geqq 0$ ) on the interior of the interval $I$ , we have

A[f]=\frac{K}{(n+1)!}f^{(n+1)}(\xi),

(19)

Or $K$ is a coefficient independent of the function $f$ (which is also equal to that shown in formula (18)) and $\xi$ a point of $I$ , if $n\geq 0$ , even from within $I$ and which generally depends on the function $f$ .

We find ourselves in a well-known classic case if $A[f]$ is the remainder in Taylor's formula. Formula (19) is then the classical form of the remainder given by Lagrange.
B. - We will recall some properties of higher-order convex functions. Any convex function of order $n>0$ on $I$ is continuous on the inside of $I$ and if $n>1$ it has a continuous derivative of order $n-1$ on the inside of $I$ If the derivative $f^{(n+1)}(x)$ , of order $n+1$ , exists, the condition $f^{(n+1)}(x)\geq 0$ on $I$ is necessary and sufficient for the function to be non-concave of order $n$ on $I$ This condition is only necessary and the condition $f^{(n+1)}(x)>0$ on $I$ is only sufficient for the function $f$ either convex of order $n$ on $I$ . If $f^{(n+1)}(x)\geq 0$ on $I$ and if there is no non-zero interval sound of $I$ on which $f^{(n+1)}(x)$ either no, $f$ is convex of order $n$ on I. In particular we have the

Lemma 5. - For a polynomial $f$ of effective degree $>n$ convex soil of order $n$ on $I$ , it is necessary and sufficient that one $f^{(n+1)}(x)\geq 0$ on $I$ .

The condition is sufficient since the derivative of order $n+1$ of a polynomial of effective degree $>n$ is not identically zero, therefore can only be zero over a finite number ( $\geq 0$ ) of points. This derivative therefore cannot be identically zero on any subinterval of positive length. A polynomial of effective degree $n$ is a polynomial of degree $n$ which does not reduce (over an interval of positive length) to a polynomial of degree $n-1$ .

Convexity of order -1 is equivalent to positivity, and non-concavity of order -1 to non-negativity of the function. Convexity of order 0 is equivalent to increasing, and non-concavity of order 0 to non-decreasing of the function.
9. - A convex function of order $n$ enjoys the property that any difference divided by order $n-1$ of this function on $n-1-2$ nodes which are not all coincident, is positive, provided, of course, that this divided difference exists*)

Consider a linear functional of the form (17). By Lemma 4, it follows that if all the coefficients $\mu_{i},i=1,2,\ldots$ , $m-n-1$ are $\geq 0$ or are they all $\leq 0$ and if there is at least one

⁰⁰footnotetext: *)

\mathrm{J}_{4}

existence an sem of nr ₁ , therefore in the sense that the adnette function effectively less derivatives which occur in the expression (1) of the divided difference considered.

coefficient $\mu_{i}$ different from zero for which the nodes $x_{i},x_{i+1},\ldots,x_{i+n+1}$ of the corresponding divided difference are not all confused, then the linear functional (17) is of the degree of accuracy $n$ and is of the simple form.

The condition that the coefficients $\mu_{i}$ being of the same sign is not, in general, necessary for the linear functional (17) to have the degree of accuracy $n$ and be of the simple form.

Let us now suppose that $x_{1}\leqq x_{2}\leqq\ldots\leqq x_{m}$ and that the multiplicity orders of distinct nodes are $\leqq n$ -t 2. According to some results already obtained [5], it follows that if $n=-1,0$ or 1, the condition that all the coefficients $\mu_{i}$ that the linear functionals have the same sign and that there exists at least one $i$ for which $\mu_{i}\neq 0$ and the knots $x_{i},x_{i+1},\ldots,x_{i+n+1}$ that they are not all confused, is necessary and sufficient for the linear functional in question to have the degree of accuracy $n$ and in simple form. Of course, for $n=-1$ , the last condition, therefore, that the $x_{i},x_{i+1},\ldots,x_{i+n+1}$ that they are not confused, does not arise. We will repeat here the demonstration that we have, moreover, given, with certain non-essential modifications, in our work cited [ $\tilde{b}$ ].

From the above, it suffices to show that if the linear functional (17) is of the degree of exactness $n$ (For $n=-1,0$ or 1) and is of the simple form, none of the coefficients $\mu_{i}$ cannot be different from zero and of opposite sign to the number $A\left[x^{n+1}\right]=\sum_{i=1}^{m-n-1}\mu_{i}$ (which is necessarily $\neq 0$ Assuming $\sum_{i=1}^{m-i}\mu_{i}\neq 0$ Therefore, property is equivalent to the fact that inequalities $\left(\sum_{v=1}^{m-1}\mu_{v}\right)\mu_{i}\geq 0,i=1,2,\ldots,n-n-1$ are verified. For the demonstration we take into account that if $f$ is a non-concave function of order $n$ , it is necessary that $A[f]$ does not change sign (that it is constantly $\geq 0$ or constantly $\leqq 0$ More precisely, that, under the hypothesis $\sum_{i=1}^{m-1}\mu_{i}\neq 0$ , 1'on ait $\left(\sum_{\nu=1}^{m-1}\mu_{\nu}\right)A[f]\geq 0$ , for any function $f$ non-concave of order $n$ 10.
– For the demonstration we will distinguish three cases, depending on the values $-1,0,1$ of $n$ .

Case 1. $n=-1$ We can assume $x_{1}<x_{2}<\ldots<x_{m}$ and the linear functional (17) reduces to $A[f]=\sum_{i=1}^{nt}\mu_{t}f\left(x_{i}\right)$ . If $0<\varepsilon<<\min_{1=1,2,\ldots,m-1}\left(x_{i+1}-x_{i}\right)$ , the continuous function

f_{i}(x)=\frac{1}{2\varepsilon}\left(\left|x-x_{i}+\varepsilon\right|+\left|x-x_{i}-\varepsilon\right|-2\left|x-x_{i}\right|\right)

is mon-negative, reduces to 1 out of $x_{1}$ and 0 on the other nodes. We have therefore $A\left[f_{i}\right]=\mu_{i},i=1,2,\ldots,m$ It follows that $\left(\sum_{v=1}^{m}\mu_{v}\right)\mu_{i}\geq 0,i==1,2,\ldots,m$ , which demonstrates ownership.

Case 2. $n=0$ We can stipulate, without restricting the generality, that all the moods are double. Let us therefore suppose that $m$ either even and $x_{2i-1}=x_{2t},i=1,2,\ldots,\frac{1}{2},x_{1}<x_{9}<\ldots<x_{ij-1}$ The linear functional (17) reduces to $A[f]=\sum_{f=1}^{n-1}\mu_{f}\left[x_{i},x_{i+1};f\right]$ The first case, where there are a few wins and all the nodes are simple, is included in the previous case as a special case. If, for example, there is a double knot $x_{2i-1}=x_{2i}$ , we have a simple node that coincides with this point, we just need to take $\mu_{2-1}=0$ in the previous formula. Then the derivative of the function $f$ on this point disappears in the expression of $A[f]$ .

We must now distinguish between two cases, depending on the parity of the index $i$ the coefficient $\mu_{1}$ .
$1^{\circ}$ . Either $i$ even. So the nodes $x_{i},x_{i+1}$ are distinct $\left(x_{i}<x_{i+1}\right)$ and the continuous function

f_{i}(x)=\frac{1}{2}\left(x_{i+1}-x_{i}+\left|3x-2x_{i}-x_{i+1}\right|-\left|3x-x_{i}-2x_{i+1}\right|\right)

i=2,4,\ldots,m-2

is non-decreasing and gives us $A\left[f_{l}\right]=\mu_{l}$ We have done

	$\displaystyle\left(\sum_{v=1}^{m-1}\mu_{v}\right)\mu_{i}\geqq 0$		(20)
	$\displaystyle\text{ pour }i=2,4,\ldots,m-2.$

$2^{\circ}$ . Either $i$ odd. So $x_{i-1}<x_{i}=x_{i+1}<x_{i+2}$ If $0<\varepsilon<<\min\left(x_{i}-x_{i-1},x_{i+2}-x_{i+1}\right)$ , the continuous function

\begin{gathered}f_{i}(x)=\frac{1}{2}\left(2\varepsilon+\left|x-x_{i}+\varepsilon\right|-\left|x-x_{i}-\varepsilon\right|\right)\\ i=1,3,\ldots,m-1\end{gathered}

is non-decreasing and we have $A\left[f_{i}\right]=\varepsilon\left(\frac{\mu_{i-1}}{x_{i}-x_{i-1}}+\frac{\mu_{i+1}}{x_{i+2}-x_{i+1}}\right)+\mu_{i}$ . If $\mu_{i}\neq 0$ , For $\varepsilon$ small enough, $A\left[f_{i}\right]$ is also $\neq 0$ and of the same sign with $\mu_{i}$ We deduce that inequality (20) is also true for $i=1$ . $3,\ldots,m-1$ .

Inequality (20) is therefore true for $i=1,2,\ldots,m-1$ and ownership is demonstrated.

Case 3. $n=1$ We can assume, without restricting the generality, that all knots are triple. Therefore $m$ a multiple of 3 and are $x_{3i-2}=x_{3i-1}=x_{3i},i=1,2,\ldots,\frac{m}{3},x_{1}<x_{4}<x_{7}<\ldots<x_{m-2}$ The linear functional (17) reduces to $A[f]=\sum_{i=1}^{m-2}\mu_{i}\left[x_{i},x_{i+1},x_{i+2};f\right]$ The case where some or all of the knots are double or single is included in the preceding case as a special case. If, for example, instead of the triple knot $x_{3i-2}=x_{3i-1}=x_{3i}$ we have a double knot coinciding with this point, just take $\mu_{3i-2}=0$ in the previous formula. Then the
second derivative of the function $f$ on this point disappears in the expression of $A\lceil f$ If instead of a double knot, we have a simple knot at this point, we simply take $\mu_{3i-2}=0$ And $\mu_{3i-1},\mu_{3i-3}$ so that 1 we have $\left(x_{3i+1}-x_{3}\right)\mu_{3i-3}=\left(x_{3i-2}-x_{3i-3}\right)\mu_{3i-1}$ , so that in the expression of $A[f]$ disappearance anssi the first derivative of $f$ on this point.

Here again, we will distinguish between two cases, depending on the values of the index. $i$ of $\mu_{i}$ compared to the divisor 3.
$1^{\circ}$ Let us consider the pair of coefficients $\mu_{i},\mu_{i+1}$ Or $i+1$ is a multiple of 3. We have $x_{i+1}<x_{i+2}$ and the continuum function

\begin{gathered}f_{i}(x)=\left(x_{i+2}-x_{i+1}\right)\frac{x-\lambda+|x-\lambda|}{2}\\ i=2,5,8,\ldots,m-4\end{gathered}

Or $x_{i+1}<\lambda<x_{i+2}$ is non-concave of order 1 and we have

A\left[f_{i}\right]=\frac{\mu_{i}\left(x_{i+2}-\lambda\right)+\mu_{i+1}\left(\lambda-x_{i+1}\right)}{x_{i+2}-x_{i+1}}

We can see that if $\mu_{i}\neq 0$ And $\lambda$ is sulistically close $x_{i+1},A\left[f_{i}\right]$ East $\neq 0$ and has the same sign as $\mu_{1}$ and if $\mu_{i+1}\neq 0$ And $\lambda$ is close enough to $x_{i+2},A\left[f_{i}\right]$ East $\neq 0$ and has the same sign as $\mu_{i+1}$ It follows that

\left(\sum_{v=1}^{m-2}\mu_{v}\right)\mu_{v}\geqq 0

(21)

For $i=2,3,5,6,8,9,\ldots,m-4,m-3$ .
$2^{\circ}$ Now suppose that $i$ either congruent to 1 modulo 3. Then $x_{i}==x_{i+1}=x_{i+2}$ . If $0<\varepsilon<\min\left(x_{i}-x_{i-1},x_{i+3}-x_{i+2}\right)$ , the continuous function

\begin{gathered}f_{i}(x)=\frac{1}{2}\left[4\varepsilon\left(x-x_{i}\right)+\left(x-x_{i}+\varepsilon\right)\left|x-x_{i}+\varepsilon\right|-\right.\\ \left.-\left(x-x_{i}-\varepsilon\right)\cdot\left|x-x_{i}-\varepsilon\right|\right]\\ i=1,1,7,\ldots,m-2\end{gathered}

is non-concave of order 1 and we have

A\left[f_{i}\right]=\varepsilon^{2}\left[\frac{\mu_{i-2}-\mu_{i-1}}{\left(x_{i}-x_{i-1}\right)^{2}}+\frac{\mu_{i+2}-\mu_{i+1}}{\left(x_{i+3}-x_{i+2}\right)^{2}}\right]+2\varepsilon\left[\frac{\mu_{i-1}}{x_{i}-x_{i-1}}+\frac{\mu_{i+1}}{x_{i+1}-x_{i+2}}\right]+\mu_{i}.

If $\mu_{i}\neq 0$ , for small enough, $A\left[f_{i}\right]$ East $\neq 0$ and is of the same sign as $\mu_{i}$ It follows that inequality (21.) is also true for $i=1,4,7$ , $\ldots,m-2$ Inequality (21) is therefore true for $i=1,2,\ldots,m-2$ and the property is demonstrated.
11. - The property demonstrated for $n=-1,0$ and 1 is no longer true for $n>1$ To unravel this property, it suffices to show that if $n>1,x_{1}<x_{2}<\ldots<x_{n+4}$ and if $\mu^{\prime},\mu^{\prime\prime}$ are two sufficiently large positive numbers ( $\mu^{\prime}+\mu^{\prime\prime}>1$ ), the linear functional
(22) $\mu^{\prime}\left[x_{1},x_{2},\ldots,x_{n+2};f\right]-\left[x_{2},x_{3},\ldots,x_{n+3};f\right]+\mu^{\prime\prime}\left[x_{3},x_{1},\ldots,x_{n+4};f\right]$
is (degree of accuracy) $n$ et) de la forme simple. En effet, introduisons entre les noends $x_{i}$ encore $n+3$ noends, en formant ainsi la suite de noeuds $x_{1}<x_{2}<\ldots<x_{2n+7},x_{2i-1}=x_{i},i=1,2,\ldots,n+4$ . Des formules de moyenne des différences divisées [3] il résulte que

		$\displaystyle{\left[x_{1},x_{2},\ldots,x_{n+2};f\right]=\sum_{i=1}^{n+2}\alpha_{i}\left[x_{i},x_{i+1}^{\prime},\ldots,x_{i+n+1};f\right]}$
		$\displaystyle{\left[x_{2},x_{2},\ldots,x_{n+3};f\right]=\sum_{i=3}^{n+4}\beta_{i}\left[x_{i}^{\prime},x_{i+1}^{\prime},\ldots,x_{i+n+1}^{\prime};f\right]}$

\left[x_{3},x_{4},\ldots,x_{n+4};f\right]=\sum_{i=5}^{n+6}\gamma_{i}\left[x_{i}^{\prime},x_{i+1}^{\prime},\ldots,x_{i+n+1}^{\prime};f\right]

où $\alpha_{i},\beta_{i},\gamma_{i}$ sont des coefficients positifs, indépendants de la fonction $f$ (et $\sum_{i=1}^{n+2}\alpha_{i}=\sum_{i=3}^{n+4}\beta_{i}=\sum_{i=5}^{n+6}\gamma_{i}=1$ ). La fonctionnelle linéaire (22) peut donc s’écrire sous la forme

\sum_{i=1}^{n+6}\left(\mu^{\prime}\alpha_{i}+\mu^{\prime\prime}\gamma_{i}-\beta_{i}\right)\left[x_{i}^{\prime},x_{i+1}^{\prime},\ldots,x_{i+n+1}^{\prime};f\right]

ou $\alpha_{n+3}=\alpha_{n+4}=\alpha_{n+5}=\alpha_{n+6}=\beta_{1}=\beta_{2}=\beta_{n+5}=\beta_{n+6}=\gamma_{1}=\gamma_{2}=\gamma_{3}==\gamma_{4}=0$ . La propriété résulte du fait qu’il n’existe aucun indice $i$ pour lequel les coefficients $\alpha_{i},\gamma_{i}$ soient nuls à la fois (on voit facilement que ceci n’est plus vrai pour $n=0$ où $=1$ ).

Enfin rappelons que pour qu’une fonctionnelle linéaire de la forme (1) ait un degré d’exactitude $n$ et pour qu’elle soit de la forme simple, il faut que les ordres de multiplicité $k_{1},k_{2},\ldots,k_{0}$ des noeuds, supposés réduits à leur nombre minimum, soient tous $\leqq n+2[5]$ .

III.

12.
- •
  
  Nous allons nous occuper du reste de certaines formules drapproximation pour la fonctionnelle linéaire $A[f]$ . Ces formules peuvent être considérées conme des généralisations de la formule d’interpolation de Lagrange, qui a comme cas particulier la formule de Taylor.

Soit $A[f]$ une fonctionnelle linéaire définie sur l’espace $S$ (voir ur. 1). Nous considérons une suite finie ou infinie de points
(23)

y_{0},y_{1},\ldots

distincts on non. Nous considérons une section
(24)

y_{0},y_{1},\ldots,y_{s}

de cette suite et le polynome de Lagrange-Hermite $L\left(y_{0},y_{1},\ldots,y_{s};f\mid x\right)$ sur ces points et relativement à la fonction $f$ . Pour tout $x\in I$ ce polynome est une fonctionnelle linéaire de la forme (1). Plus exactement, ce polynome pent être mis sous la forme (1), où les $a_{i,j}$ sont des polynomes indépendants de la fonction $f$ , le nombre total des noeuds, distincts ou non, étant égal à $s+1^{*}$ ).

Nous avons alors la formule d’approximation

A[f]=A\left[L\left(y_{0},y_{1},\ldots,y_{s};f\mid x\right)\right]+R_{s}[f]

(25)

où $R_{s}[f]$ est le reste de cette formule.
La formule (11) nous donne
*) Il existe des valeurs de $x$ (en nombre fini), pour lesquels le nowbre minimum des noeuds est plus petit que $s+1$ .
où

	$\displaystyle A[f]$	$\displaystyle=\sum_{v=0}^{s}c_{v}\left[y_{0},y_{1},\ldots,y_{v};f\right]+R_{s}[f]$		(26)
	$\displaystyle c_{v}$	$\displaystyle=A\left[\prod_{i=0}^{v-1}\left(x-y_{i}\right)\right],v=0,1,\ldots,s$		(27)

La formule (25) est completement caractérisée par le fait qu’elle est de la forme (26), avec un reste $R_{s}[f]$ fonctionnelle linéaire de degré d’exactitude au moins égal à s. En effet, pour tout polynome $f$ de degré $s$ , le polynome $L\left(y_{0},y_{1},\ldots,y_{s};f\mid x\right)$ se réduit à $f$ , donc $R_{s}[f]$ est nul. Alors les coefficients $c_{v}$ , donnés par la formule (26) sont bien déterminés et, pour $v$ donné, $c_{v}$ est indépendant de $s$ .

Nous supposons, bien entendu, que les conditions d’existence, données au ur. 1, soient vérifiées pour la fonctionnelle linéaire $R_{s}[f]$ . Ainsi, les points (24), ou bien les points (23) sils interviennent tous, appartiennent à l’intervalle $I$ . Les différences divisées $\left[y_{0},y_{1},\ldots,y_{v};f\right],v=0$ , $1,\ldots,s$ , existent au sens expliqué au nr. 4, etc.

Il est clair que si le reste $R_{s}[f]$ est définie, tous les restes précédents $R_{0}[f],R_{1}[f],\ldots,R_{s-1}[f]$ sont également des fonctionnelles linéaires définies sur $S$ .

Dans ce qui suit nous allons étudier quelques cas où le reste $R_{s}[f]$ de la formule d’approximation (25) est de la forme simple.
13. - Considérons une fonctionnelle linéaire $A[f]$ of the form (1). Unless otherwise stated, we will deal exclusively with linear functionals of this form. The rest $R_{s}[f]$ The formula (25) then has the same form. The multiplicity orders of the nodes can all be taken $\leqq\max\left(k_{1},k_{2},\ldots,k_{p},s+1\right)$ , so if $s+2\Longrightarrow\max\left(k_{1},k_{2},\ldots,k_{p}\right)$ We can apply Lemma 3 and the linear functional $R_{s}[f]$ is a linear combination of divided differences of order $s+1$ To put $R_{s}[f]$ indeed in form (17), it is sufficient first to carry out a suitable numbering of the nodes so that the corresponding condition (15) is verified.

To go further, we will distinguish between the cases where the sequence (24) and the sequence of knots $z_{1},z_{2},\ldots,z_{p}$ of the linear functional $A[f]$ have on common terms. In the following we will examine only the cases where the coincidence took place with only one of the terms $z_{i}$ . Either $z_{1}$ this node, whose multiplicity order is $k_{1}$ and suppose that $x_{1},x_{2},\ldots,x_{m}$ either a normal numbering of the nodes of $A[f]$ , Or $x_{1}=x_{2}=\ldots=x_{h_{1}}=z_{1}$ So (if $p>1$ ) no term of the sequence (24) does not coincide with one of the points $x_{k_{1}+1},x_{k_{1}+2},\ldots,x_{in}$ .

Either $k\left(0\leqq k\leqq k_{1}\right)$ the smallest of the number $k_{1}$ and the number of terms in sequence (24) equal to $z_{1}$ Equality $k=0$ means that none of the terms in sequence (24) coincide with a node $z_{i}$ . If $k>0$ , among the points (24) there are at least $k$ which coincide with $z_{1}$ Let us designate by $s^{r}$ the smallest index such as the sequence $y_{0},y_{1},\ldots,y_{s^{\prime}-1}$ (having $s^{\prime}$ (terms) contains at least $k$ terms equal to $z_{1}$ . We have $s^{\prime}\equiv k$ and if $k=0$ we can take $s^{\prime}=0$ .

The nodes of the linear function $R_{s}[f]$ can be written in 1a sequence*) $y_{s},y_{s-1},\ldots,y_{0},x_{k+1},x_{k+2},\ldots,x_{m}$ Their number is equal to $s+1+m-k$ and the numbering corresponding to this sequence satisfies condition (15) (with $n=s$ ) if $s\geq s^{\prime}+m-k-1(\geq m-1)$ It follows that if $m\leqq k$ the linear functional $R_{s}[f]$ is zero identically**). But the inequality $m\leq k$ takes place if and only if all the nodes $x_{i}$ are confused with the same point $z_{1}$ and the sutte (24) contains at least $m$ terms equal to $z_{1}$ .

Otherwise, doic si out bien $p>1$ or $p=1,k<k_{1}$ (in both cases we have $m>k$ ), we have the formula

R_{s}[f]=\sum_{i=k+1}^{m}\mu_{i}^{(s)}\left[y_{0},y_{1},\ldots,y_{s+1+k-i},x_{k+1},x_{k+2},\ldots,x_{i};f\right]

(28)

where the coefficients $\mu_{i}^{(i)}$ are independent of the function $f$ We
can calculate the coefficient $\mu_{\mu}^{(s)}$ in the following way. That is $x_{m}=z_{\mu}$ Therefore, taking into account formula (1), the coefficient of $f^{\left(k_{\mu}-1\right)}\left(z_{\mu}\right)$ in the first member of (28) is equal to $a_{\mu,k_{\mu}-1}$ and the same coefficient on the second member is equal to $\mu_{n}^{(s)}$ multiplied by

\begin{gathered}\frac{1}{\left(k_{\mu}-1\right)!}\cdot\frac{1}{P^{\prime}\left(z_{\mu}\right)\prod_{v=k+1}^{m-k_{\mu}}\left(z_{\mu}-x_{v}\right)}\text{, si }p>1(\text{ alors }\mu\neq 1)\\ \frac{1}{\left(k_{1}-1\right)!}\cdot\frac{k!}{P^{(k)\left(z_{1}\right)}},\text{ si }p=1,k<k_{1},\end{gathered}

out $P(x)=\prod_{v=0}^{3+1+k-m}\left(x-y_{v}\right)$
It follows that

\mu_{n}^{(s)}=\left\{\begin{array}[]{l}\left(k_{\mu}-1\right)!P\left(z_{\mu}\right)_{v=k+1}^{m-k}\prod_{\mu}\left(z_{\mu}-x_{\nu}\right)a_{\mu,k_{\mu}-1}\text{ si }p>1.\\ \frac{\left(h_{1}-1\right)!}{k!}P^{(k)}\left(z_{1}\right)a_{1,k_{1}-1}\text{ si }p=1,k<k_{1}.\end{array}\right.

Assuming, therefore, that condition (2) is satisfied, we have $\mu_{n}^{(s)}\neq 0$
It follows that if the previous assumptions are verified and if the coefficients $\mu_{i}^{(s)},i=k+1,k+2,\ldots,m$ are all of the same sign, the rest $R_{s}[f]$ is of degree of accuracy s and is of the simple form. The coefficient $K$ of the corresponding formula (18) is equal to $R_{s}\left[x^{s+1}\right]=A\left[\prod_{v=0}^{s}\left(x-y_{v}\right)\right]$ .

⁰⁰footnotetext: *) These are not necessarily the nodes reduced to their minimum quantity.
*) The property will not subsist if

s<s^{\prime}+m-k-1

In what follows we will always assume, unless otherwise stated, that for the linear functional $A[f]$ condition (2) is satisfied.
14. - We obtain an interesting special case by taking for $A[f]$ the difference divided (4). To state the respective property, we will set $a^{\prime}=\min\left(z_{1},z_{2},\ldots,z_{p}\right),b^{\prime}=\max\left(z_{1},z_{2},\ldots,z_{p}\right)$ . SO $\left[a^{\prime},b^{\prime}\right]\subseteq I$ is the smallest closed interval that contains the nodes of the functional $A[f]$ We then have the

THEORHMORAL 1. - If we accept the preceding hypotheses and notations: $1^{\circ}$ the points (24) sound on well lows $\leq a^{\prime}$ , or lows $\geq b^{\prime},2^{\circ}$ Now we have $p>1$ or bten $p=1$ of $k<k_{1},3^{\circ}$ . s $=k,s\geq m-1$ , the rest $R_{s}[f]$ of the approximation formula
(29)

\begin{gathered}{\left[x_{1},x_{2},\ldots,x_{m};f\right]=}\\ =\sum_{y=1}^{s}\left[x_{1},x_{2},\ldots,x_{m};\prod_{i=1}^{v-1}\left(x-y_{i}\right)\right]\cdot\left[y_{0},y_{1},\ldots,y_{v};f\right]+R_{s}[f]\end{gathered}

a the degree of accuracy s ob ost of the simple form.
For the demonstration it will suffice to verify that the coefficients $\left[t_{i}^{(*)}\right.$ The coefficients of the corresponding formula (28) all have the same sign. We will calculate these coefficients.

Let's calculate, in general, the coefficients $\mu_{4}^{(s)}$ of formula (28) for $A[f]$ of the form (1), assuming that the conditions $2^{\circ}$ And $3^{\circ}$ of theorem 1 are verified. To perform the calculation, note that we have*)

A\left[/(x)\left(x-z_{1}\right)^{k}\right]=\sum_{i=k+1}^{m}\mu_{i}^{(5)}\left[x_{k+1},x_{k+2},\ldots,x_{i};\frac{f(x)}{s+1+k-i}\left(x-y_{v}\right)\right]

(30)

If $k<k_{1}$ This formula results, by applying formulas (7), (8), through the identification of the parts of the expressions of $R_{5}[f]$ taken from (26) and (28) and which contain only the terms corresponding to the nodes ${z_{i}}$ . If $h=k_{j}$ The formula results in the same way, by identifying the terms that originate from the nodes. $z_{z},z_{a},\ldots,z_{p}$ .

Frenons naintenant comme fonction $f$ the polynomial
$\left(x_{i}-y_{s+1+k-i}\right)\prod_{v=k}^{s+k-i}\left(x-y_{v}\right)\prod_{v=k+1}^{i-1}\left(x-x_{v}\right)$ , Or $\prod_{v=k+1}^{i-1}\left(x-x_{v}\right)$ For $i=\ldots+k$ And $\prod_{v=k}^{s+k-m}\left(x-y_{r}\right)$ for $=m-1$ are replaced by 1. Then the right-hand side of (30) reduces to $\mu_{i}^{[s]}$ and we obtain

	$\displaystyle k^{(s)}=\left(x_{i}-y_{s+1+k-i}\right)A\left[\prod_{v=1}^{i-1}\left(x-x_{v}\right)\prod_{v=k}^{s+k-i}\left(x-y_{v}\right)\right]$		(31)
	$\displaystyle i=k+1,k+2,\ldots,m.$

*) If $h=0,A\left[f(x)\left(x-z_{1}\right)^{h}\right]$ boils down to $d[f]$ .

Returning to Theorem 1, we have in this case $A[f]==\left[x_{1},x_{2},\ldots,x_{m};f\right]$ and taking into account formula (8),

	$\displaystyle\mu_{i}^{(s)}=\left(x_{i}-y_{s+1+k-i}\right)\cdot\left[x_{i},x_{i+1},\ldots,x_{m};\prod_{v=k}^{s+k-i}\left(x-y_{v}\right)\right],$		(32)
	$\displaystyle i=k+1,k+2,\ldots,m.$

But the polynomial $\prod_{v=k}^{s+k-i}\left(x-y_{v}\right)$ has a derivative of order $m--i$ negative on the interval $\left(a^{\prime},b^{\prime}\right)$ if the points (24) are to the right of $b^{\prime}$ And $s-n+1$ is odd. In other cases, compatible with the hypotheses of Theorem 1, this derivative is positive on ( $a^{\prime},b^{\prime}$ ). It follows that the coefficients $\mu^{(s)}$ are positive if the points (24) are either to the left of $a^{\prime}$ or to the right of $b^{\prime}$ And $s-m+1$ is odd and they are negative if the points (24) are to the right of $b^{\prime}$ And $s-m+1$ is even. We assumed $a^{\prime}<b^{\prime}$ . When $a^{\prime}=b^{\prime}$ we are in the case $p=1,k<k_{1}$ and it is easy to see that the property is still true.

Theorem 1 is therefore proven.
In the case of Theorem 1, in formulas (27) we have $c_{0}=c_{1}==\ldots=c_{m-2}=0$ , done if $s<m-1$ we have $R_{s}[f]=\left[x_{1},x_{2},\ldots,x_{m};f\right]$ .

Theorem 1 generalizes certain properties of H. D. Kloostermann [1]. These properties are obtained for

		$\displaystyle x_{i}=x+(i-1)h,i=1,2,\ldots,m(h\neq 0),y_{i}=x,i=0,1,\ldots,s$
		$\displaystyle x_{i}=x,i=1,2,\ldots,m,y_{i}=x+ih,i=0,1,\ldots,s(h\neq 0)$

respectively, and if, moreover, we assume that the function $f$ admits a continuous derivative of order $s+1$ within the smallest interval containing the points $x_{i},y_{i}$ 15.
- Let us return to formula (28) and consider the conditions under which this formula was established. We can then find a simple relationship between the coefficients $\mu_{i}^{(s)},\mu_{i}^{(s+1)}$ . We have

\begin{gathered}R_{s}[f]-c_{s+1}\left[y_{0},y_{1},\ldots,y_{s+1};f\right]=R_{s+1}[f]=\\ =\prod_{i=k+1}^{m}\frac{\mu_{i}^{(s+1)}}{x_{i}-y_{s+2+k-i}}\left\{\left[y_{0},y_{1},\ldots,y_{s+1+k-i},x_{k+1},x_{k+2},\ldots,x_{i};f\right]-\right.\\ \left.-\left[y_{0},y_{1},\ldots,y_{s+2+k-i},x_{k+1},x_{k+2},\ldots,x_{i-1};f\right]\right\}\end{gathered}

where the second difference divided reduces to $\left[y_{0},y_{1},\ldots,y_{s+t};f\right]$ For $i=k+1$ The formula makes sense, since under the stated assumptions, $x_{i}\neq y_{s+2+k-i},i=k+1,k+2,\ldots,m$ .

By comparing with formula (28), we deduce,

	$\displaystyle\mu_{i}^{(s+1)}=\left(x_{i}-y_{s+2+k-i}\right)\left(\mu_{i}^{(s)}+\mu_{i+1}^{(s)}+\cdots+\mu_{m}^{(s)}\right)$		(33)
	$\displaystyle i=k+1,k+2,\ldots,m.$

These formulas allow us to state the

THEOREM 2. - Under the assumptions under which formula (28) was established, and if: $1^{\circ}$ Let's look at the points (24) $\leqq a^{\prime}$ oh well, they are all $\geq b^{\prime}2^{\circ}$ there is a valley $s_{0}$ of s pow in which all the coefficients $\mu_{-}^{(s)}$ are from the same sign.
The rest $R_{s}[f]$ of the approximation formula (25) is the degree of accuracy $s$ and is of the simple form for $s\geq s_{0}$ .

Indeed, under the conditions of the theorem, we see that if the coefficients $\mu_{i}^{(s)}$ all have the same sign, the coefficients $\mu_{i}^{(s+1)}$ are also all of the same sign.
16. - One might wonder if they always exist, for a linear functional $A[f]$ , of the form (1) e.g., values of $s$ for which the rest $R_{s}[f]$ is it of simple form, or is this remainder of simple form for a sufficiently large s? We will give an example to show that the answer is negative.

Either $A[f]=f(0)+f^{\prime}(0)$ and let's take the points $y_{v}=(v+1)(v+2)$ , $v=0,1,\ldots$ In this case we have ( $s\geq 0$ ),

\begin{gathered}R_{s}[f]=(-1)^{s}s!(s+1)!\left\{\left[0,0,y_{0},y_{1},\ldots,y_{s-1};f\right]-\right.\\ \left.-(s+2)\left[0,y_{0},y_{1},\ldots,y_{s};f\right]\right\}\end{gathered}

Based on previous results, $R_{s}[f]$ , which is the degree of accuracy $s$ , is not of the simple form for any value of $s\geq 3$ 17.
- The preceding example demonstrates that the following property is of some interest,

THEOREM 3. - Under the hypotheses under which the formula (28) was established and if all the points (23) coincide at a single point not belonging to the interval ( $a^{\prime},b^{\prime}$ ),
the restc $R_{s}[f]$ is of the degree of accuracy s and is of the simple form for $s$ large enough.

In this case we have $k=0$ (if points (24) are outside of $\left[a^{\prime},b^{\prime}\right]$ ) Or $h=h_{1}$ (if the points (24) all coincide with $a^{\prime}$ or all with $b^{\prime}$ ). It will suffice to give the demonstration in the case $k=0$ .

So be it $k=0$ Formulas (33) become ( $s\geq m-1$ )

\mu_{i}^{(s+1)}=\left(x_{i}-y_{0}\right)\left(\mu_{i}^{(s)}+\mu_{i+1}^{(s)}+\ldots+\mu_{m}^{(s)}\right),i=1,2,\ldots,n

(31)

We will now choose the notations so that the following $x_{1},x_{2},\ldots,x_{n2}$ either non-decreasing or non-increasing depending on whether $y_{0}<a^{\prime}$ resp. $y_{0}>b^{\prime}$ So, the numbers $x_{i}-y_{0}$ are different from zero, of the same sign, and the sequence $\left|x_{1}-y_{0}\right|,\left|x_{2}-y_{0}\right|\ldots\ldots\left|x_{m}-y_{0}\right|$ , of their absolute values is non-decreasing.

From (34) we deduce

\mu_{i}^{(s)}=\sum_{i=i}^{mt}M_{i,j}^{(\xi)}\mu_{j}^{(m-1)},i=1,2,\ldots,m.

(35)

where the triangular matrix $\left(M_{i,j}^{(s)}\right)$ is the $(s-n-1)^{\text{ème }}$ power of the triangular matrix
u) $\left(\begin{array}[]{ccc}x_{1}-y_{0}&x_{1}-y_{0}&\ldots\\ 0&x_{1}-y_{0}&\ldots\\ 0&\ldots&\ldots\\ 0&0&\ldots\end{array}\right.$

Let us designate by $W_{i}\left(z_{1},z_{2},\ldots,z_{r}\right)$ the symmetric function $\sum z_{1}^{a_{1}}z_{2}^{a_{2}}\ldots z_{t^{\prime}}^{a_{r}}$ , the sum being extended to the solutions in non-negative integers of the equation in $\alpha_{i},\alpha_{1}+\alpha_{2}+\ldots+\alpha_{r}=i\quad\left(W_{0}\left(z_{1},z_{2},\ldots,z_{r}\right)=1\right)$ , We have

	$\displaystyle\Delta I_{i,j}^{(o)}-\left(x_{i}-y_{0}\right)W_{s-m}\left(x_{i}-y_{0},x_{i+1}-y_{0},\ldots,x_{j}-y_{0}\right),$		(36)
	$\displaystyle j=i,i+1,\ldots,mi=1,2,\ldots,m.$

18.
- •
  
  Before going any further, we will establish a lemma that is of interest, regardless of the application we give it here.

Lemma 6. – If your non-negative numbers $z_{1},z_{2},\ldots,z_{r-1}(r>1)$ sond compris dans l'iniervalle $\left[0,z_{r}\right]$ , nons awons l'inequality

\frac{F_{1}\left(\varepsilon_{1},\varepsilon_{2},\ldots,\varepsilon_{r}\right)}{(r-1+i)}\geq\frac{W_{i}\left(\varepsilon_{1},\varepsilon_{2},\ldots,z_{r-1}\right)}{(r-2+i)}

(37)

the gählé dani waie si el senlement si ou bien $i=0$ or $i>0$ down the wombres $z_{1},z_{2},\ldots,z_{r}$ sone éganex.

The property is innádiate for $i=0$ and for $i>0$ And $z_{1}=z_{2}==\ldots=r_{r-1}=0$ .

Suppose that $z_{1},z_{2},\ldots,z_{r-1}$ not all of them are useless. So we necessarily have $z_{r}>0$ We admit

\begin{gathered}W_{i}\left(z_{1},z_{2},\ldots,z_{r}\right)=\left[z_{1},z_{2},\ldots,z_{r};x^{r-1+i}\right]\\ W_{i}\left(z_{1},z_{2},\ldots,z_{r-1}\right)=\left[z_{1},z_{2},\ldots,z_{r-1};x^{r-2+i}\right]=\\ =\left[z_{1},z_{2},\ldots,z_{r};x^{r-2+i}\left(t-z_{r}\right)\right]\end{gathered}

oil
$(r-1+i)\left[\frac{w_{i}\left(r_{1},r_{2},\ldots,r_{r}\right)}{\left(r-\frac{1}{1}+i\right)}-\frac{w_{i}\left(r_{1},r_{2},\ldots,r_{r-1}\right)}{\left(r-\frac{2}{i}+i\right)}\right]=W_{i}\left(r_{1},\hat{w}_{2},\ldots,r_{r}\right)-$
$-\frac{y-1+i}{y-1}W_{i}\left(z_{1},z_{2},\ldots,z_{1-1}\right)-\frac{1}{y-1}\left[z_{1},z_{2},\ldots,z_{r};x^{r-2+i}\left((r-1-1)z_{r}\cdots i,1\right)\right]$ But
the derivative $(r-1)^{\text{bine }}$ of the polynomial $x^{r-2+i}\left[(r-\cdots 1-\mid-i)z_{r}-ix\right]$ is equal $\frac{(v-1-1)!}{(i-1)!}x^{-1}\left(z_{r}-x\right)$ , which is positive on Lindervalle $\left(0,z_{r}\right)$ It follows that the poiynome in question is convex of order $\psi-2$ Inequality (37) follows immediately.

The case of equality is easy to study.
$\mathrm{J}_{\mathrm{e}}$ The lenume a is therefore proven.
The number ( $r-1+i$ ) is precisely the number of terms of the symmetric function $\left.W_{i}\left(z_{1},z_{2},\ldots,z_{r}\right)^{1}\right)$

If $0<z_{1}\leqq z_{2}\leqq\ldots\leqq z_{r}$ This is the case that particularly interests us in the proof of Theorem 3; we can deduce a remarkable inequality. In this case, we have
$(v-1)W_{i}\left(z_{1},z_{2},\ldots,z_{v}\right)-(v-1-1)W_{i}\left(z_{1},z_{2},\ldots,z_{v}-1\right)\geq 0,v-2,3,,,v_{.}$ If we add these inequalities member by member ,
we deduce $(r>1)$ ,

\frac{W_{i}\left(z_{1},z_{2},\ldots,z_{i}\right)}{\sum_{v=1}^{T}W_{i}\left(z_{1},z_{2},\ldots,z_{v}\right)}\geq\frac{i-1}{r-1}

(38)

19.
- •
  
  Let us return to the proof of Theorem 3. Taking into account (36) and (38), we deduce

\frac{M_{i,m}^{(s)}}{\sum_{j=i}^{m-1}M_{i,j}^{(s)}}\geq\frac{s-m-1}{m-i}\geq\frac{s-m+1}{m-1},i=1,2,\ldots,m-1

(39)

and from formula (36) we obtain

\begin{gathered}\mu_{i}^{(s)}=\left\{\begin{array}[]{c}M_{i,m}^{(s)}\\ \sum_{j=1}^{m-1}M_{i,j}^{(s)}-1,-1,-\frac{\sum_{j=i}^{n-1}\mu_{j}^{(m-1)}M_{i,j}^{(s)}}{\sum_{j=1}^{n-1}M_{i,j}^{(s)}}\end{array}\right\}\sum_{j=i}^{m-1}M_{i,j}^{(s)}\\ i=1,2,\ldots,mi-1.\end{gathered}

We now notice that: $1^{\circ}$ the sums $\sum_{j=i}^{m-1}M_{i,j}^{(j)},i=1,2,\ldots$ … $m$ - 1 are different from zero and have the same sign. $2^{\circ}$ the quotient $\sum_{j=1}^{m-1}\mu_{j}^{(m-1)}M_{i,j}^{(s)}/\sum_{j=1}^{m-1}M_{i,j}^{(s)}$ is a weighted arithmetic mean of the numbers $\mu_{i}^{(m-1)},\mu_{i+1}^{(m-1)},\ldots,\mu_{n-1}^{(m-1)}$ These numbers remain between two fixed numbers, independent of $s\left(\right.$ between ${}_{i=1,2,\ldots,m-1}^{\mu_{i}^{(m-1)}}$ And $\left.\max_{i=1,2,\ldots,m-1}^{\mu_{i}^{(m-1)}}\right)$ ,
*) Inequality (07) can also be written in the form of an inequality between two nominal values.

\sqrt[i]{\frac{W_{i}\left(z_{1},z_{2},\ldots,z_{r}\right)}{(r-1+i)}}\geq\sqrt[i]{\frac{F_{i}\left(z_{1},z_{2},\ldots,z_{1}-1\right)}{(r-2+i)}}

$3^{\circ}\mu_{m}^{(m-1)}$ is different from zero, under the hypotheses of Theorem 3 (see note 13). From (39) it therefore follows that, for

s>(m-1)\left(1+\max_{i-1,2\ldots\ldots m-1}\left|\frac{\mu_{i}^{(m-1)}}{\mu_{m}^{(m-1)}}\right|\right),

all numbers $\mu^{(s)},i=1,2,\ldots,m$ are different from zero and have the same sign (their sign is that of $\left[\operatorname{sg}\left(x_{\mathrm{i}}-y_{0}\right)\right]^{s-m}\cdot\operatorname{sg}\mu_{m}^{(m-1)}$ ).

Theorem 3 is thus proven for $k=0$ . For $k=k_{1}$ (in this case $p>1$ ), the demonstration is done in a completely analogous way, based on formulas ( 33 ) for $k=k_{1}$ .

Theorem 3 is therefore proven.
20. - Pont domer nut example takenous the linear functional

		$\displaystyle A[f]=f(0)-f(0)-0,8\left[f^{\prime}(0)-\mid f^{\prime}(0)\right]$
		$\displaystyle\quad-1,2\left[f^{\prime}(1)-\mid-f^{\prime}(5)\right]-2,2\times f^{\prime}(3)$

which is the remainder of Hardy's quadrature formula [2]

\int_{0}^{6}f(x)dx=0,28[f(0)+f(6)]+1,62[f(1)+f(5)]+2,2\times f(3)+R^{*}[f]

applied to the function $f^{\prime}(x)\left(R^{*}\left[f^{\prime}\right]=A[f]\right)$ .
$A[f]$ is of the degree of accuracy 6, but is not of the simple form [7]. If we consider the Taylor development

A[f]=\sum_{v=1}^{s}A\left[x^{v}\right]\frac{f^{(v)}(0)}{v\mid}+R_{s}[f]

by virtue of theorem 3, the remainder $R_{s}[f]$ is clu degree of accuracy s and is of the simple form for sufficiently large s.

In this case, we have $p=5,k_{1}=k_{2}=k_{3}=k_{1}=k_{0}=2,m=10$ , $h=s^{\prime}=2,x_{1}=x_{2}=0,x_{3}=x_{4}=1,x_{5}=x_{6}=3,x_{7}=x_{8}=5,x_{9}=x_{10}==6,y_{0}=y_{1}=\ldots=0$ We can apply formulas (31) and we find

\begin{array}[]{ll}\mu_{3}^{(s)}=A\left[x^{s}\right],&\mu_{7}^{(s)}=5A\left[x^{s-4}(x-1)^{2}(x-3)^{2}\right]\\ \mu_{4}^{(s)}=A\left[x^{s)}(x-1)\right],&\mu_{8}^{(s)}=5A\left[x^{s-5}(x-1)^{2}(x-3)^{2}(x-5)\right]\\ \mu_{5}^{(s)}=3A\left[x^{s-2}(x-1)^{2}\right],&\mu_{9}^{(s)}=6A\left[x^{s-6}(x-1)^{2}(x-3)^{2}(x-5)^{2}\right]\\ \mu_{6}^{(s)}=3A\left[x^{s-3}(x-1)^{2}(x-3)\right],&\mu_{10}^{(s)}=6A\left[x^{s-7}(x-1)^{2}(x-3)^{2}(x-5)^{2}(x-6)\right]\end{array}

Using these formulas, we can calculate the coefficients. $\mu_{i}^{(9)}$ , by doing $s=9$ taking into account that $A[f]$ has a degree of accuracy of 6 and by calculating the numbers

A\left[x^{7}\right]=64,8\quad A\left[x^{8}\right]=1555,2\quad A\left[x^{9}\right]=19828,8

To calculate the coefficients $\mu_{i}^{(s)}$ For $s>9$ , we apply the recurrence formulas (33) which become here

		$\displaystyle\mu_{3}^{(s+1)}=\mu_{3}^{(s)}+\mu_{4}^{(s)}+\mu_{5}^{(s)}+\mu_{6}^{(s)}+\mu_{7}^{(s)}+\mu_{8}^{(s)}+\mu_{9}^{(s)}+\mu_{10}^{(s)}$
		$\displaystyle\mu_{4}^{(s+1)}=\mu_{4}^{(s)}+\mu_{5}^{(s)}+\mu_{6}^{(s)}+\mu_{9}^{(s)}+\mu_{8}^{(s)}+\mu_{9}^{(s)}+\mu_{10}^{(s)}$
		$\displaystyle\mu_{5}^{(s+1)}=3\left(\mu_{5}^{(s)}+\mu_{6}^{(s)}+\mu_{7}^{(s)}+\mu_{8}^{(s)}+\mu_{9}^{(s)}+\mu_{10}^{(s)}\right)$
		$\displaystyle\mu_{6}^{(s)}=3\left(\mu_{6}^{(s)}+\mu_{7}^{(s)}+\mu_{8}^{(s)}+\mu_{9}^{(s)}+\mu_{10}^{(s)}\right)$
		$\displaystyle\mu_{7}^{(s+1)}=5\left(\mu_{7}^{(s)}+\mu_{8}^{(s)}+\mu_{9}^{(s)}+\mu_{10}^{(s)}\right)$
		$\displaystyle\mu_{8}^{(s+1)}=5\left(\mu_{8}^{(s)}+\mu_{9}^{(s)}+\mu_{10}^{(s)}\right)$
		$\displaystyle\mu_{9}^{(s+1)}=6\left(\mu_{9}^{(s)}+\mu_{10}^{(s)}\right),\quad\mu_{10}^{(s+1)}=6\mu_{10}^{(s)}$
		$\displaystyle\quad s=9,0,\ldots$

Simply perform the calculations up to the value of 13. $s$ and we find the values of the coefficients $\mu_{i}^{(i)}$ included in the table

$i\backslash 8$	9	10	11	12	13
3	19828.8	174960	1108792.8	3888777.6	-19594484.2
4	18273.6	155131.2	933832.8	2779984.8	-23483260.8
5	50349.6	4111572.8	2336104.8	5538456	-78789736.8
6	37519.2	259524	1104386.4	-1469858.4	-95405104.8
7	44064	244944	543024	-7971696	-151659216
8	18144	24624	-681696	-10686816	-111800736
9	988.8	-79315.2	-965770.2	-8734003.2	-70439987.2
10	-13608	-81648	-489888	-2939328	-17635988

We can see that the rest is of the simple form for $s\geq 13$ .

BIBIOLOGY:

[1] Kloostermann, HD Derivatives and finite differences. Duke Math, Journat, 17, 109-186 (1950).
[3] Popovieiu, T., Introduction to the theory of divided differences, Bull. Math, from Soc. Roumaine des sc., 42, 65-78 (1940).
[4] - Asupra formed restului in ande formula of aprotimare ale analizer. Lacr. His. Gen, ştiittifice, Acad. RPR, 183-185, 1950.
[5] - Asupra restudui an whele forwale of deriouse numerica. Studio si Cerc. Mat., III, 53-122 (1952).
[6] - Folylonos figguényes hözépérpékléletéröl. Magic. Tud. Akad., II oszt. közlem., IV, 353-356 (1954).
[7] - Sw the remainder in certain linear approximation formals of the analysis. Mathematica, 1 (24), 95-142 (1959).
Received 1st 28. XI. 1959.

Divided differences and derivatives

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DIFFERENCES, DIVIDED AND DERIVED

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