ON THE REMAINDER IN SOME NUMERICAL DERIVATION FORMULAS

by TIBERIU POPOVICIU

§ 1. Formulas of maximum accuracy

1.

A numerical derivative formula is a formula that allows the approximate evaluation of the value of the derivative of a given order of a function, at a given point, by a given linear combination of the values of the function and a finite number of its successive derivatives, taken at a finite number of given points.

A numerical derivation formula is therefore a formula of the form

f^{(m+r)}\left(x_{0}\right)=\sum_{j=0}^{r-1}a_{j}f^{(j)}\left(x_{0}\right)+\sum_{i=1}^{s}\sum_{j=0}^{r_{i}-1}a_{i,j}f^{(i)}\left(x_{i}\right)+R

(1)

which allows us to attribute to $f^{(m+r)}\left(x_{0}\right)$ , as an approximate value, sum of the second term (without $R$ ). In this formula the coefficients $a_{j},j=0,1,\ldots$ , $r-1,a_{i,j},j=0,1,\ldots,r_{i}-1,i=1,2,\ldots,s$ are some real constants given, $\operatorname{iar}f^{(i)}\left(x_{0}\right),j=0,1,\ldots,r-1,f^{(i)}\left(x_{i}\right),j=0,1,\ldots,r_{i}-1,i=1,2,\ldots,s$ are the values of the function $f(x)$ and its successive derivatives $f^{\prime}(x),f^{\prime\prime}(x),\ldots$ at the given points $x_{0},x_{1},x_{2},\ldots,x_{s}$ .

number $R$ from the second member of formula (1), is the remainder of this formula. An expression of the remainder, resulting from the nature of the function $f(x)$ , allows an evaluation of the error that is committed by the indicated approximation of the first member.
2. We must specify from the beginning the hypotheses that will be maintained throughout the work and which are related to the data of formula (1). At the same time, we will also fix some names in connection with these data.
$1^{\circ}$ . The points

x_{1},x_{2},\ldots,x_{s}

(2)

on the real axis, assumed distinct and in number of $s\geqq 1$ , we will call them the nodes of the numerical derivation formula.
$2^{\circ}$ . Each node $x_{i}$ an order of multiplicity is attached to it $r_{i}$ . The numbers $r_{1},r_{2},\ldots,r_{s}$ are positive integers.
$3^{\circ}$ The point $x_{0}$ is different from the nodes (2) and we will call it the drift point of formula (1).
$\mathbf{4}^{\circ}$ . The point of derivation $x_{0}$ an order of multiplicity is attached to it $R$ , assumed to be a positive integer or zero.
$5^{\circ}$ Number $m$ , integer position or zero, we will call it the derivative index of formula (1).

Finally, relative to the function $f(x)$ we will make the following hypothesis, noting with $[a,b]$ the smallest closed interval containing the nodes and the derivation point.
$6^{\circ}$ . The real function $f(x)$ , of the real variable $x$ , is defined and continuous in the interval $[a,b]$ . Besides this, it is assumed that it admits the derivatives that actually occur in formula (1) and at least at the points where the values of these derivatives actually appear in formula (1) ¹ ).

The continuity assumption is made for the purpose of studying the rest of formula (1).
Any other assumption made on the problem data will be expressly stated.

There is an infinite multitude $\mathfrak{T}\mathfrak{C}$ of numerical derivation formulas (1), which are obtained by giving the coefficients $a_{i},a_{i,j}$ all possible real values and having all of the following common characteristics:
$1^{\circ}$ . The nodes, $2^{\circ}$ The respective multiplicity orders of the nodes, $3^{\circ}$ . The derivation point, $4^{\circ}$ . The multiplicity order of the derivative point, $5^{\circ}$ . Derivation index.

It is useful to note with $p+1$ the sum of the multiplicity orders of the nodes, so $r_{1}+r_{2}+\ldots+r_{s}=p+1$ . Then $p\geqq 0$ and $p+1$ represents the total number of nodes, each counted with its order of multiplicity. For all formulas in $\mathfrak{N}$ , the numbers $p,r,m$ are the same.
Each formula (1) in $\mathfrak{N}$ We will attach two important characteristic numbers: the order of the formula and the degree of accuracy of the formula.
(3) The order of formula (1) serves to specify the sum of the orders of derivation, which effectively intervene in the second member. For this, we will define the numbers as follows $r_{1}^{\prime},r_{2}^{\prime},\ldots,r_{s}^{\prime},r^{\prime}$ attached to the nodes and the derivation point:
4. $r_{i}^{\prime}-1$ is the highest order of derivation that actually occurs in the node $x_{i}$ , therefore,
$r_{i}^{\prime}=0$ , if all the coefficients $a_{i,j},j=0,1,\ldots,r_{i}-1$ are null;
$a_{i,r:-1}\neq 0,a_{i,j}=0,j=r_{i}^{\prime},r_{i}^{\prime}+1,\ldots,r_{i}-1$ , if the coefficients $a_{i,j}$ they are not all null, $i=1,2,\ldots,s$ .
$r^{\prime}-1$ is the highest order of derivation that actually occurs at the point $x_{0}$ in the second member, therefore,
$r^{\prime}=0$ , if $r=0$ or if all the coefficients $a_{j},j=0,1,\ldots,r-1$ are null,
$a_{r^{\prime}-1}\neq 0,a_{j}=0,j=r^{\prime},r^{\prime}+1,\ldots,r-1$ if the coefficients $a_{j}$ they are not all null.

⁰⁰footnotetext: ¹⁾ This means that the coefficient

a_{j}

a_{i,j}

correspondingly is different from zero.

We then have:

0\leqq r^{\prime}\leqq r,0\leqq r_{i}^{\prime}\leqq r_{i};i=1,2,\ldots,s

equal $r^{\prime}=0$ , respectively $r_{i}^{\prime}=0$ , means that the derivation point, respectively the node $x_{i}$ , does not actually appear in the second member of formula (1). The equality $r^{\prime}=r(>0)$ , respectively $r_{i}^{\prime}=r_{i}$ , means that the coefficient $a_{r-1}$ , respectively $a_{r_{i}-1}$ , is different from zero.

By definition, the sum $r_{1}^{\prime}+r_{2}^{\prime}+\ldots+r_{s}^{\prime}+r^{\prime}$ will be called the order of the numerical derivation formula (1).

The order of formula (1) is at most equal to $p+r+1$
In the crowd $\mathfrak{N}$ there is a formula that has order equal to 0. This is the trivial formula in $\mathscr{O}$ which has all the coefficients $\overline{a_{j}},a_{i,j}$ null. Any other formula in $\mathfrak{K}$ has positive order.

The trivial formula is of no interest for the problem of numerical derivation.
4. The degree of accuracy of formula (1) shows us how good the approximation studied is in the particular case, when the function $f(x)$ reduces to a polynomial of sufficiently small degree.

REST $R$ of formula (1) is the value, given by this formula, of an additive and homogeneous functional. To highlight the function $f(x)$ , we will denote this functional by $R[f]$ .

function $R[f]$ is defined in the field of functions $f(x)$ , which verifies the above hypotheses. This field is a vector field, which contains all functions differentiable a sufficient number of times in the interval $[a,b]$ and therefore, in particular, all polynomials.

By definition, the degree of accuracy of the function $R[f]$ , or of formula (1), is an integer $n\geq-1$ , so that:
$1^{\circ}.R[P]=0$ , for any polynomial $\boldsymbol{P}(x)$ of the degree $n^{1}$ ).
$2^{\circ}.R[P]\neq 0$ , for at least one polynomial $P(x)$ of the degree $n+1$ The
definition obviously applies to any functional that is defined for all polynomials.

Due to the additivity and homogeneity of the functional, the degree of accuracy $n$ can also be defined by the following properties:
$1^{\circ}$ If $R[1]\neq 0$ , we have $n=-1$ .
$2^{\circ}$ If $R[1]=0,n$ is the smallest integer, such that

R\left[x^{i}\right]=0,i=0,1,\ldots.n,R\left[x^{n+1}\right]\neq 0.

The following observation is useful: because the degree of accuracy of the functional $R[f]$ to be $>n$ , it is necessary and sufficient for us to have $R[P]=0$ for any polynomial $P(x)$ of the degree $n$ and $R\left(P_{1}\right)=0$ for at least one polynoma $P_{1}(x)$ of effective degree $n+1$ Indeed, any polynomial $Q(x)$ of the degree $n+1$ is of the form $Q(x)=CP_{1}(x)+P(x)$ , where $C$ is a constant and $P(x)$ a polynomial of degree $n$ We then have $R[Q]=CR\left[P_{1}\right]+R[P]=0$ .

Let's introduce the polynomial

l(x)=\left(x-x_{1}\right)^{\gamma_{1}}\left(x-x_{2}\right)^{\gamma_{2}}\ldots\left(x-x_{s}\right)^{\gamma_{s}}

(3)

⁰⁰footnotetext: ¹⁾ A polynomial of degree -1 coincides with the identical zero polynomial.

An immediate calculation then gives us (enf. (1)).

R\left[\left(x-x_{0}\right)^{m^{\prime}+r}l(x)\right]=\left[\left(x-x_{0}\right)^{m^{\prime}+r}l(x)\right]_{x=x_{0}}^{(m+r)}=\frac{(m+r)!}{\left(m-m^{\prime}\right)!}l^{\left(m-m^{\prime}\right)}\left(x_{0}\right)

(4)

if $m^{\prime}$ is an integer $\geq 0$ For $m^{\prime}>m$ this expression is equal to 0 .

For $m^{\prime}=m$ , the number (4) is different from 0 based on the assumptions made. It follows from this that the degree of accuracy of the functional $R[f]$ It also follows that this degree of accuracy is $\leqq p+r+m$ .

It is clear that the degree of accuracy is uniquely determined.
The degree of accuracy of the trivial formula in $\mathfrak{N}$ it's the same as $m+r-1$ , which can be easily verified.
5. Let us consider the interpolation polynomial of degree $p+r$

L(\underbrace{x_{0},x_{0},\ldots,x_{0}}_{r},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f\mid x)

(5)

relative to the function $f(x)$ and to the system $p+r+1$ nodes, formed from the point $x_{0}$ repeated by $r$ times and from the points

\quad x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime}

(6)

which represent the nodes (2), each repeated as many times as its multiplicity order indicates. System (6) therefore represents a renumbering of the nodes (2), taking into account their multiplicities. To fix the notations it can be assumed that:

x_{r_{1}+r_{2}+\ldots+r_{i-1}+j}^{\prime}=x_{i},j=1,2,\ldots,r_{i},i=1,2,\ldots,s.

(7)

If $r=0$ , the point $\mathfrak{x}_{0}$ is not a node of the polynomial and does not appear in the notation (5).

The polynomial (5) is the lowest degree polynomial that coincides with the function $f(x)$ in the past $p+r+1$ given nodes. This means that if a node is repeated by $k$ times, polyne mul (5), together with its first $k-1$ derivatives, coincides with the function $f(x)$ and with his first $k-1$ respective derivatives in this node. The polynomial (5) is therefore in general the Lagrange-Hermite polynomial [2], [3], which corresponds to the function and the nodes highlighted by the notation (5), that is, the (unique) polynomial of the lowest degree that verifies the conditions:

\begin{gathered}L^{(i)}(\underbrace{x_{0},x_{0},\ldots,x_{0}}_{r},x_{1},x_{2},\ldots,x_{p+1};f\mid x_{i})=f^{(i)}\left(x_{v}\right)\\ i=0,1,\ldots,r_{i}-1,i=0,1,\ldots,s\quad\left(r_{0}=r\right)\end{gathered}

The explicit form of the pclinc m (5) $\in$ is well known [3], [11]. We will recall some of these results in the form used here.

\displaystyle\qquad\begin{aligned} L\left(\frac{d}{f_{0}},x_{0},\ldots,x_{0}\right.&\left.,x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f\mid x\right)=\sum_{i=0}^{r-1}C_{i}(x)f^{(j)}\left(x_{0}\right)+\\ &+\sum_{i=1}^{s}\sum_{i=0}^{r_{i}-1}C_{i,j}(x)f^{(j)}\left(x_{i}\right)\end{aligned}

where the first sum in the second term vanishes if $r=0$ . Here $C_{j}(x)=C_{0,j}(x)$ , $C_{i,i}(x)$ are polynomials of degree $p+r$ , completely determined by the conditions
$1^{\circ}.C_{i,j}^{(j)}{}_{j}\left(x_{i}\right)=1$ ,
$2^{\circ}.C_{i,j}^{(u)}\left(x_{v}\right)=0$ , for $|u-j|+|v-i|\neq 0$ .

j=0,1,\ldots,r_{i}-1,\quad i=0,1,\ldots,s\quad\left(r_{0}=r\right).

If we put

\left.\begin{array}[]{rl}c_{i}&=C_{j}^{(m+r)}\left(x_{0}\right),j=0,1,\ldots,r-1,\\ c_{i,j}&=C_{i,j}^{(m+r)}\left(x_{0}\right),j=0,1,\ldots,r_{i}-1,i=1,2,\ldots,s\end{array}\right\}

HAVE

L^{(m+r)}\underbrace{\left(x_{0},x_{0},\ldots,x_{0}\right.}_{r},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f\mid x_{0})=\sum_{j=0}^{r-1}c_{j}f^{(j)}\left(x_{0}\right)+

(10)

+\sum_{i=1}^{s}\sum_{i=0}^{r_{i}-1}c_{i,i}f^{(i)}\left(x_{i}\right)

and it turns out that

f^{(m+r)}\left(x_{0}\right)=L^{(m+r)}\underbrace{\left(x_{0},x_{0},\ldots,x_{0}\right.}_{r}x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f\mid x_{0})+R

(E)

is a numerical derivation formula belonging to the set $\mathfrak{N}$
Based on formulas (9) it is clear that for $m>p$ , formula (E) reduces to the trivial formula. =
6. We propose to determine the degree of accuracy of formula (E). For this we will need the following lemma:

Let me 1. If two derivatives of consecutive orders $\mathrm{i},\mathrm{i}+1$ of a polynomial with all real roots have a common root, this root is a root of the polynomial of order multiplicity at least equal to $1+2$ .

Indeed, if the common root of the derivatives is $a$ , the polynomial can be written in the form
$\alpha_{0}+\alpha_{1}(x-a)+\alpha_{2}(x-a)^{2}+\ldots+\alpha_{i-1}(x-a)^{i-1}+\alpha_{i+2}(x-a)^{i+2}+\ldots$
and if the coefficients $\alpha_{0},\alpha_{1},\ldots,\alpha_{i-1}$ were not all zero, the polynomial would present $o$ gap of at least 2 terms, which is in contradiction with the reality of all roots. We assume, of course, that the effective order of the polynomial is at least $i$ .

We now observe that, if $f(x)$ is a polynomial of degree $p+r$ , we have

L\underbrace{\left(x_{0},x_{0},\ldots,x_{0}\right.}_{r},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f\mid x)=f(x)

(11)

and for formula (E) we then have $R[f]=0$ This shows us that the degree of accuracy of formula (E) is $\geq p+r$ It is immediately seen that the conclusion holds even if (E) is reduced to the trivial formula.

If $m\leq p$ , formula (4) gives us, in particular,

	$\displaystyle R\left[\left(x-x_{0}\right)^{r}l(x)\right]$	$\displaystyle=\frac{(m+r)!}{m!}l^{(m)}\left(x_{0}\right),R\left[\left(x-x_{0}\right)^{r+1}l(x)\right]=$
		$\displaystyle=\frac{(m+r!)}{(m-1)!}l^{(m-1)}\left(x_{0}\right)$		(12)

and polynomials $\left(x-x_{0}\right)^{r}l(x),\left(x-x_{0}\right)^{r+1}l(x)$ are respectively of the effective degrees $p+r+1,p+r+2$ .

If $m=0$ , we have $l\left(x_{0}\right)\neq 0$ , based on the assumptions made, and if $m>0$ , taking into account lemma 1, it is seen that we cannot have at the same time

l^{(m)}\left(x_{0}\right)=0,\quad l^{(m-1)}\left(x_{0}\right)=0.

We therefore deduce the following theorem:
Theorem 1. The degree of accuracy of the formula $(E)$ it is always $\geq\mathrm{p}+\mathrm{r}$ .

If $\mathrm{m}>\mathrm{p}$ this degree of accuracy is equal to $\mathrm{m}+\mathrm{r}-1$ If
$\mathrm{m}\leqq\mathrm{p}$ , the degree of accuracy is equal to $\mathrm{p}+\mathrm{r}$ if $1^{(\mathrm{m})}\left(\mathrm{x}_{0}\right)\neq 0$ and is equal to p + r + 1 if $1^{(\mathrm{m})}\left(\mathrm{x}_{0}\right)=0$ .

It also follows from this that (E) reduces to the trivial formula if and only if $m>p$ .
7. Let (1) be a formula from $\mathfrak{M}$ , with the degree of accuracy $\geqq p+r$ . We will show that this formula reduces to formula (E). For this, let us assume the opposite. Then, at least one of the coefficients $a_{j}=a_{0,i},a_{i,j}$ is different from the coefficient $c_{j}=c_{0,j},c_{i,j}$ respectively. For the moment, let us note with $R[f]$ the rest of the formula (1) and egg $R_{1}[f]$ the rest of the formula (E).

To establish the ideas, let's assume that

a_{\beta,j}-c_{\beta,i}\left\{\begin{array}[]{l}\neq 0,j=\alpha\\ =0,j=\alpha+1,\alpha+2,\ldots,r_{i}-1.\quad\left(r_{0}=r\right)\end{array}\right.

Subtracting formula (E) from formula (1) member by member and solving with respect to $f^{\left(\alpha_{1}\right)}\left(x_{\beta}\right)$ , we find a formula of the form

f^{(\alpha)}\left(x_{\beta}\right)=\sum_{j=0}^{\alpha-1}a_{j}^{\prime}f^{(j)}\left(x_{\beta}\right)+\sum_{i=1}^{s}\sum_{i=0}^{r_{i}-1}a_{i_{i},j}^{\prime}f^{(j)}\left(x_{i}\right)+R_{2}[f]

(14)

where, if $\beta\neq 0$ in the second sum from the second member instead of $x_{\beta},r_{\beta}$ get along $x_{0},r$ respectively. The rest $R_{2}[f]$ of formula (14) is

R_{2}[f]=\frac{R_{1}[f]-R[f]}{a_{\beta,\alpha}-c_{\beta,\alpha}}

(15)

We have, of course, $\alpha\leqq r_{\beta}-1\left(r_{0}=r\right)$ For formula
(14), the numbers $p,r,m$ are $p+r-r_{\beta},\max\{\alpha-1,0\},0$ respectively, so, based on an observation from point 4, the degree of accuracy
of this formula is $\leq p+r-r_{\beta}+\max\{\alpha-1,0\}\leq p+r-r_{\beta}+\alpha\leq<p+r-1$ . But from (15) it is seen that the degree of accuracy is $\geq p+r$ This contradiction shows us that hypothesis (13) is inadmissible.

We can therefore state the following theorem:
Theorem 2. In the set $\mathfrak{I}\mathfrak{C}$ of the numerical derivation formulas (1) there is one formula, and only one, of maximum accuracy, and this formula is the formula $(E)$ .

For this reason, formula (E) will be called a maximum accuracy numerical derivation formula.
8. We can now specify the order of the formula ( $\mathbb{E}$ ). This order is $\leq p+r+1$ The order of the formula can also be lower than $p+r+1$ , because certain coefficients $c_{j},c_{i,j}$ can be null.

If $m>p$ , the order is 0.
If $m\leq p$ , the order is a positive number. Suppose this order were $p+r+1-\alpha$ , where $\alpha\geq 0$ In this case, formula (E) is at the same time a formula for numerical derivation from a set $\mathfrak{N}_{1}$ , which is deduced from $\mathscr{N}$ , modifying the characteristics by reducing a number $\alpha$ of nodes (6) or and the point $x_{0}$ a certain number of times, but maintaining the amount $m+r$ Number $p+r$ , by passing into $\mathfrak{N}_{1}$ BECOMES $p+r-\alpha$ But the formula having the degree of accuracy $\geq p+r\geq p+r-\alpha$ is a formula of maximum accuracy in $\mathfrak{N}_{1}$ Consequently, or the formula reduces to the trivial formula in $\mathscr{N}_{1}$ , or has an accuracy level at most equal to $p+r+1-\alpha$ The first hypothesis is inadmissible because it would result in the formula being reduced to the trivial formula in $\mathfrak{I}\mathfrak{C}$ , which is a saying with the hypothesis $m\leq p$ From the second hypothesis it follows that $\alpha\leq 1$ . sall $\alpha=1$ .

case $\alpha=1$ occurs if one of the coefficients $c_{r-1},\,c_{i,r_{i}-1},\,i=1,2,\ldots,s$ is zero. Moreover, if one of these coefficients is zero, the others are necessarily different from zero.

In this case, the formula passes from the set $\mathfrak{O}$ in the crowd $\mathfrak{O}_{1}$ , by suppressing once the derivation point or once a node $x_{i}$ Through this passage, the numbers $m,r$ are $m+1,r-1$ respectively, or remain unchanged.

The first case can only occur if $m<p$ , because otherwise we would end up contradicting the fact that the formula in $\mathfrak{O}_{t}$ it does not reduce to the trivial formula.

Also, by moving into $\mathscr{N}_{1}$ , the polynomial $l(x)$ remains unchanged or becomes $\frac{l(x)}{x-x_{i}}$ .

The formula considered has $\mathscr{N}_{1}$ degree of accuracy $\leq p+r$ and it turns out that this degree of accuracy is even $p+r$ , that is, in $\mathfrak{K}_{1}$ the formula is found when its degree of accuracy increases by one unit.

From the foregoing it follows that we will have $c_{r-1}=0$ , if and only if $l^{(m+1)}\left(x_{0}\right)=0$ This case is only possible if $r>0$ and then all the coefficients $c_{i,r_{i}-1},i=1,2,\ldots,s$ are different from zero, and if $r>1$ we also have $c_{r-2}\neq 0$ We also have $c_{i,r_{i}-1}=0$ , if and only if $\left[\frac{l(x)}{x-x_{i}}\right]_{x=x_{0}}^{(m)}=0$ and then all the coefficients $c_{\alpha,r_{\alpha}-1}\neq 0,\alpha=1,2,\ldots,i\leftarrow 1,i+1,\ldots,s$ We also have $c_{r-1}\neq 0$ if $r>0$ and $c_{i,r_{i}-2}\neq 0$ if $r_{i}>1$ .

From the above discussion the following theorem results:

Theorem-3. If $\mathrm{m}\leq\mathrm{p}$ , the degree of accuracy of the formula ( $E$ ) is at most equal to its order. The degree of accuracy and the order are equal if and only if $\mathrm{x}_{0}$ is a root, different from the nodes, of one of the polynomials

l^{(m)}(x),l^{(m+1)}(x),\left\lfloor\frac{l(x)}{x-x_{i}}\right]^{(m)},\quad i=1,2,\ldots,s

(16)

In general, the degree of accuracy is $p+r$ and the order is $p+r+1$ . The degree of accuracy and the order are both equal to $p+r+1$ , if $x_{0}$ is a root of the first polynomial (16). The degree of accuracy and the order are both equal to $p+r$ , if $x_{0}$ is a root of one of the last $s+1$ polynomials (16).
9. From the previous analysis it follows that any two of the polynomials (16) cannot have a common root other than a node. This can also be proven directly. If $m=0$ , the polynomials, except possibly the second one, do not have roots different from the nodes. Also, the property is immediate if $m=p$ , because then

l^{(p+1)}\left(x_{0}\right)=(p+1)!\neq 0,\left[\frac{l(x)}{x-x_{i}}\right]^{(p)}=p!\neq 0,i=1,2,\ldots,s

Let us examine the general case.
From Lemma 1 it follows that any common root of the first two polynomials (16) is a knot of order multiplicity at least equal to $m+2$ .

Highlighting one of the latest $s$ polynomials (16), we can write

l^{(m)}(x)=\left[\frac{l(x)}{x-x_{i}}\right]^{(m)}\left(x-x_{i}\right)+m\left[\frac{l(x)}{x-x_{i}}\right]^{(m-1)}

It can be seen from this that any common root of the polynomials $l^{(m)}(x),\left[\frac{l(x)}{x-x_{i}}\right]^{(m)}$ is a common root of the polynomials $\left[\frac{l(x)}{x-x_{i}}\right]^{(m)},\left[\frac{l(x)}{x-x_{i}}\right]^{(m-1)}$ and, based on Lemma 1 ; this can only be a node of order multiplicity at least equal to $m+1$ .

We also have

		$\displaystyle{\left[\frac{l(x)}{x-x_{\alpha}}\right]^{(m)}=\left[\frac{l(x)}{\left(x-x_{\alpha}\right)\left(x-x_{\beta}\right)}\right]^{(m)}\left(x-x_{\beta}\right)+m\left[\frac{l(x)}{\left(x-x_{\alpha}\right)\left(x-x_{\beta}\right)}\right]^{(m-1)}}$
		$\displaystyle{\left[\frac{l(x)}{x-x_{\beta}}\right]^{(m)}=\left[\frac{l(x)}{\left(x-x_{\alpha}\right)\left(x-x_{\beta}\right)}\right]^{(m)}\left(x-x_{\alpha}\right)+m\left[\frac{l(x)}{\left(x-x_{\alpha}\right)\left(x-x_{\beta}\right)}\right]^{(m-1)}}$

from which it is seen that, if $\alpha\neq\beta$ , any common root of these polynomials is a common root of the polynomials

\left[\frac{l(x)}{\left(x-x_{\alpha}\right)\left(x-x_{\beta}\right)}\right]^{(m)},\left[\frac{l(x)}{\left(x-x_{\alpha}\right)\left(x-x_{\beta}\right)}\right]^{(m-1)}

and so, based on Lemma 1, it is a node of order multiplicity at least equal to $m+1$ .

Finally we have

l^{(m+1)}(x)=\left[\frac{l(x)}{x-x_{i}}\right]^{(m+1)}\left(x-x_{i}\right)+(m+1)\left[\frac{l(x)}{x-x_{i}}\right]^{(m)}

(17)

which shows us that any common root of polynomials

l^{(m+1)}(x),\left[\frac{l(x)}{x-x_{i}}\right]^{(m)}

(18)

is a common root of the polynomials

\left[\frac{l(x)}{x-x_{i}}\right]^{(m+1)}\left(x-x_{i}\right),\quad\left[\frac{l(x)}{x-x_{i}}\right]^{(n)}

It follows that any root of the polynomials (18) is either the node $x_{i}$ , or, based on Lemma 1, a node of multiplicity order at least equal to $m+2$ .

From the previous analysis it also follows that the property of the coefficients $c_{r-1}$ , $c_{r-2}$ (if $r>1$ ) of not being both zero is a consequence of the fact that the polynomials $l^{(m+1)}(x),l^{(m+2)}(x)$ cannot have a common root other than a knot. This latter fact follows from Lemma 1, which states that any common root of these polynomials is a knot of order multiplicity at least c gal with $m+3$ .

Also, the fact that $c_{i,r_{i}-1},c_{i,r_{i}-2}$ (if $r_{i}>1$ ) cannot both be zero results from the fact that the polynomials

\left[\frac{l(x)}{x-x_{i}}\right]^{(m)},\left[\frac{l(x)}{\left(x-x_{i}\right)^{2}}\right]^{(m)}

(19)

cannot have a common root other than a node. However, we have

\left[\frac{l(x)}{x-x_{i}}\right]^{(m)}=\left[\frac{l(x)}{\left(x-x_{i}\right)^{2}}\right]^{(m)}\left(x-x_{i}\right)+m\left[\frac{l(x)}{\left(x-x_{i}\right)^{2}}\right]^{(m-1)}

from which it is seen that any common root of the polynomials (19) is a common root of the polynomials

\left[\frac{l(x)}{\left(x-x_{i}\right)^{2}}\right]^{(m)},\left[\frac{l(x)}{\left(x-x_{i}\right)^{2}}\right]^{(m-1)}

so, based on Lemma 1, a node of multiplicity order equal to at least $m+1$ .

Theorem 3 and the previous results can also be established directly, by calculating the coefficients $c_{r-1},c_{r-2},c_{i,r_{i}-1},c_{i,r_{i}-2}i=1,2,\ldots,s$ of the polynomial (10).
10. Let us return to formula (E). We will say that this formula is exceptional if its order is equal to its degree of accuracy.

Formula (E) becomes exceptional in three different aspects: $1^{\circ}$ The degree of accuracy increases by one unit, $2^{\circ}$ Number $r^{\prime}$ , attached to the derivation point, decreases by one unit from $r\left(r^{\prime}=r-1\right),3^{\circ}$ Number $r_{i}^{\prime}$ , attached to a node $x_{i}$ , decreases by one unit from $r_{i}^{\prime}\left(r_{i}^{\prime}=r_{i}-1\right)$ .

However, from the results of point 8 it follows that the three aspects return to each other by passing the formula from the set $\mathfrak{N}\mathfrak{C}$ in an analogous crowd $\mathfrak{N}_{\mathbf{1}}$ , through the following changes in its characteristics $\mathfrak{N}$ .

If formula (E) looks like $1^{\circ}$ in $\mathfrak{M}$ , then it presents:
The aspect $2^{\circ}$ in the crowd $\mathfrak{N}_{1}$ , obtained by increasing the multiplicity order of the derivative point by one unit and decreasing the derivative index by one unit.

Appearance $3^{\circ}$ in the crowd $\mathscr{N}\mathfrak{c}_{1}$ , obtained by increasing the multiplicity order of the respective node by one unit.

If the formula looks like $2^{\circ}$ in $\mathscr{IK}$ , it presents:
The appearance $1^{\circ}$ in the crowd $\mathfrak{N}_{1}$ , obtained by decreasing the multiplicity order of the derivative point by one unit and increasing the derivative index by one unit.

Appearance $3^{\circ}$ in the crowd $\mathfrak{N}_{1}$ , obtained by decreasing the multiplicity order of the derivation point by one unit, increasing the derivation index by one unit, and increasing the multiplicity order of the respective node by one unit.

If the formula looks like $3^{\circ}$ in $\mathfrak{Ir}$ , it presents:
The appearance $1^{\circ}$ in the crowd $\mathfrak{N}\mathfrak{C}_{1}$ , obtained by decreasing the multiplicity order of the respective node by one unit.

Appearance $2^{\circ}$ in the crowd $\mathfrak{N}_{1}$ , obtained by increasing the multiplicity order of the derivation point by one unit, decreasing the derivation index by one unit, and decreasing the multiplicity order of the respective node by one unit.
11. The equivalence of the three aspects of exceptionality can be expressed by formulas. To establish these formulas, let us consider the general formulas:

	$\displaystyle L\left(\alpha_{1},\alpha_{2},\ldots,\alpha_{k+1};f\mid x\right)=L\left(\alpha_{1},\alpha_{2},\ldots,\alpha_{k};f\mid x\right)+$		(20)
	$\displaystyle+\prod_{i=1}^{k}\left(x-x_{i}\right)\cdot\left[\alpha_{1},\alpha_{2},\ldots,\alpha_{k+1};f\right]$
	$\displaystyle L\left(\alpha_{1},\alpha_{2},\ldots,\alpha_{k+1};f\mid x\right)=L\left(\alpha_{1},\alpha_{2},\ldots,\alpha_{k},\beta;f\mid x\right)+$
	$\displaystyle+\left(\alpha_{k+1}-\beta\right)\prod_{i=1}^{k}\left(x-\alpha_{i}\right)\cdot\left[\alpha_{1},\alpha_{2},\ldots,\alpha_{k+1},\beta;f\right]$		(21)

In these formulas, $\left[\alpha_{1},\alpha_{2},\ldots,\alpha_{k+1};f\right]$ is the difference divided by the order $k$ of the function $f(x)$ on those $k+1$ knots $\alpha_{1},\alpha_{2},\ldots,\alpha_{k+1}$ , distinct or not. We assume that the definition and main properties of divided differences are known, as well as of interpolation polynomials on distinct or not nodes [6].

Formula (20) gives us, in particular

		$\displaystyle L(\underbrace{x_{0},x_{0},\ldots,x_{0}}_{r},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f\mid x)=$
		$\displaystyle=L(\underbrace{\left(x_{0},x_{0},\ldots,x_{0}\right.}_{r+1},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f\mid x)-$
		$\displaystyle\quad-\left(x-x_{0}\right)^{r}l(x)[\underbrace{x_{0},x_{0},\ldots,x_{0}}_{r+1},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f]=$
		$\displaystyle=L(\underbrace{x_{0},x_{0},\ldots,x_{0}}_{r},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime},x_{i};f\mid x)-$
		$\displaystyle\quad-\left(x-x_{0}\right)^{r}l(x)[\underbrace{x_{0},x_{0},\ldots,x_{0}}_{r},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime},x_{i};f].$

Deriving from $m+r$ times and taking into account $l^{(m)}\left(x_{0}\right)=0$ , we deduce

	$\displaystyle L^{(m+r)}$	$\displaystyle\underbrace{\left(x_{0},x_{0},\ldots,x_{0}\right.}_{r},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f\mid x_{0})=$
		$\displaystyle=L^{(m+r)}\underbrace{\left(x_{0},x_{0},\ldots,x_{0}\right.}_{r+1},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f\mid x_{0})=$
		$\displaystyle=L^{(m+r)}\underbrace{\left(x_{0},x_{0},\ldots,x_{0}\right.}_{r},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime},x_{i};f\mid x_{0}),\left[l^{(m)}\left(x_{0}\right)=0\right]$

which indicates the transition of the formula from the aspect $1^{\circ}$ to the aspects $2^{\circ}$ and $3^{\circ}$ respectively. Also, formulas (20), (21) give us

		$\displaystyle L\underbrace{\left(x_{0},x_{0},\ldots,x_{0}\right.}_{r},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f\mid x)=$
		$\displaystyle\quad=L\underbrace{\left(x_{0},x_{0},\ldots,x_{0}\right.}_{r-1},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f\mid x)+$
		$\displaystyle\quad+\left(x-x_{0}\right)^{r-1}l(x)[\underbrace{x_{0},x_{0},\ldots,x_{0}}_{r},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f]=$
		$\displaystyle\quad=L\left(x_{0},x_{0},\ldots,x_{0},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime},x_{i};f\mid x\right)+$
		$\displaystyle+\left(x_{0}-x_{i}\right)\left(x-x_{0}\right)^{r-1}l(x)\underbrace{x_{0},x_{0},\ldots,x_{0}}_{r},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime},x_{i};f]$

Deriving from $m+r$ times and taking into account $l^{(m+1)}\left(x_{0}\right)=0$ , we deduce

		$\displaystyle L^{(m+r)}\underbrace{\left(x_{0},x_{0},\ldots,x_{0}\right.}_{r},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f\mid x_{0})=$
		$\displaystyle\quad=L^{(m+r)}\underbrace{\left(x_{0},x_{0},\ldots,x_{0}\right.}_{r-1},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f\mid x_{0})=$
		$\displaystyle\quad=L^{(m+r)}\underbrace{\left(x_{0},x_{0},\ldots,x_{0}\right.}_{r-1},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime},x_{i};f\mid x_{0}),\quad\left[l^{(m+1)}\left(x_{0}\right)=0\right]$

which indicates the transition of the formula from the aspect $2^{\circ}$ to the aspects $1^{\circ}$ and $3^{\circ}$ respectively.
Formulas (20), (21) also give us

		$\displaystyle L(\underbrace{}_{0},x_{0},\ldots,x_{0},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f\mid x)=$
		$\displaystyle\quad=L\underbrace{\left(x_{0},x_{0},\ldots,x_{0}\right.}_{r},x_{1}^{\prime\prime},x_{2}^{\prime\prime},\ldots,x_{p}^{\prime\prime};f\mid x)+$
		$\displaystyle\quad+\left(x-x_{0}\right)^{r}\frac{l(x)}{x-x_{i}}[\underbrace{x_{0},x_{0},\ldots,x_{0}}_{r},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f\mid x)=-$
		$\displaystyle\quad=L(\underbrace{x_{0},x_{0},\ldots x_{0}}_{r},x_{1}^{\prime\prime},x_{2}^{\prime\prime},\ldots,x_{p}^{\prime\prime};f\mid x)+$
		$\displaystyle\quad+\left(x_{i}-x_{0}\right)\left(x-x_{0}\right)^{r}\frac{l(x)}{x-x_{i}}[\underbrace{}_{\frac{x_{0},x_{0},\ldots,x_{0}}{r+1}},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f]$

where $x_{1}^{\prime\prime},x_{2}^{\prime\prime},\ldots,x_{p}^{\prime\prime}$ are the nodes (6), of which the node was once deleted $x_{i}$ . Deriving from $m+r$ times and taking into account $\left[\frac{l(x)}{x-x_{i}}\right]_{x=x_{0}}^{(m)}=0$ , it is deduced

	$\displaystyle L^{(m+r)}$	$\displaystyle\underbrace{\left(x_{0},x_{0},\ldots,x_{0}\right.}_{r},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f\mid x_{0})=$
		$\displaystyle=L^{(m+r)}\underbrace{\left(x_{0},x_{0},\ldots,x_{0}\right.}_{r},x_{1}^{\prime\prime},x_{2}^{\prime\prime},\ldots,x_{p}^{\prime\prime};f\mid x_{0})=$
		$\displaystyle=L^{(m+r)}(\underbrace{x_{0},x_{0},\ldots,x_{0}}_{r0},x_{1}^{\prime\prime},x_{2}^{\prime\prime},\ldots,x_{p}^{\prime\prime};f\mid x_{0}),\left[\left(\frac{l(x)}{x-x_{i}}\right)_{x=x_{0}}^{(m)}=0\right]$

which indicates the transition from the aspect $3^{\circ}$ to the aspects $1^{\circ}$ and $2^{\circ}$ respectively.
The notion of a formula of exceptional maximum accuracy is thus completely clarified.
12. Let $\mathfrak{T}^{0}$ gathering of crowds $\mathfrak{N}$ , obtained by varying the derivation point $x_{0}$ and leaving his other characteristics unchanged $\mathfrak{N}$ There are then a finite number of exceptional formulas in $\mathfrak{K}^{0}$ This number is equal to the total number of distinct node roots of the polynomials (16) if $r>0$ and is equal to the total number of distinct roots of nodes of the first and last $s$ polynomials (16) if $r=0$ .

Roots of the polynomial $l^{(m)}(x)$ are of two kinds. Some, let us call them improper, come from the multiplicity of knots. These are all knots, if $m=0$ , and are those nodes which are also roots of the polynomial $l^{(m-1)}(x)$ , if $m>0$ . Any improper root is counted with its multiplicity order and coincides, based on Lemma 1, with a node of multiplicity order equal to $m+1$ , at least. More precisely, the number of improper roots that coincide with the node $x_{i}$ is equal to $\max\left\{0,r_{i}-m\right\}$ Thus, the total number of improper roots of the polynomial $l^{(m)}(x)$ is equal to

\sum_{i=1}^{s}\max\left\{0,r_{i}-m\right\}.

(22)

His other roots $l^{(m)}(x)$ , let's call them proper, are simple roots. Such roots can only exist if $m>0$ . A proper root may coincide with a node, but in any case not with the smallest or largest of these nodes. This latter property results from the fact that the roots of the derivative of a polynomial with all real roots always belong to the smallest closed interval containing all the roots of the polynomial. Based on this property, if $l^{(m)}(x)$ it cancels out in an extreme node, $l^{(n-1)}(x)$ also cancels out at this node. This node therefore has a multiplicity order at least equal to $m+1$ and so it is an improper root. If $x_{i}$ is its own root, we must have $r_{i}\leqq m-1$ For $r_{i}=m$ HAVE $l^{(m)}\left(x_{i}\right)\neq 0$ .

The number of its own roots $l^{(m)}(x)$ is equal to

N_{m}=p+1-m-\sum_{i=1}^{s}\max\left\{0,r_{i}-m\right\}

If $m=0$ , we have $N_{m}=0$ If $m=1$ , we have $N_{m}=s-1$ and all proper roots are distinct from nodes. If $m>1$ , at most max $\{0,s-2\}$ and of course at most $p+1-m$ , eigenroots coincide with a node.

The number of improper roots of the polynomial $\left[\frac{l(x)}{x-x_{i}}\right]^{(m)}$ will be equal to

\sum_{i=1}^{s}\max\left\{0,r_{i}-m\right\}-\max\left\{0,r_{i}-m\right\}+\max\left\{0,r_{i}-m-1\right\}.

The eigenroots of this polynomial are simple and, for the same reason as above, different from the smallest and largest nodes. It follows that the total number of eigenroots of the last $s$ polynomials (16) is equal to $(s-1)\left(N_{m}-1\right)+N_{m+1}$ . The results from point 9 show us that any two of the polynomials (16) cannot have a common root other than a node. Also, formula (17) shows us that, if $x_{i}$ is a proper root of one of the polynomials (18), then it is a proper root of both polynomials (18), but it is not a proper root of any other polynomial (16). If
a node is a proper root of the polynomial $l^{(m)}(x)$ , then this node is not a root of any other polynomial (16). Finally, we observe that the maximum - max $\{0,s-2\}$ nodes can be eigenroots of polynomials (16).

From the foregoing it follows that we can state the following theorem:
Theorem 4. The number N of formulas (E) in $\mathfrak{N}^{0}$ , whose degree of accuracy is equal to $\mathrm{p}+\mathrm{r}+1$ , is equal to 0 for $\mathrm{m}=0$ , equal to s -1 for $\mathrm{m}=1$ and check the inequalities

N_{m}-\max\{0,s-2\}\leq N^{\prime}\leqq N_{m}

for $\mathrm{m}>1$ Number
$\mathrm{N}^{\prime\prime}$ of formulas $(E)$ FROM $\mathfrak{N}^{0}$ , whose order is equal to $\mathrm{p}+\cdots$ is equal to 0 if $\mathrm{m}=0,\mathrm{r}=0$ , equal to $\mathrm{s}-1$ if $\mathrm{m}=0,\mathrm{r}>0$ and check the inequalities.

\begin{gathered}\quad(s-1)\left(N_{m}-1\right)+N_{m+1}-\max\{0,s-2\}\leq\\ \leq N^{\prime\prime}\leq\left\{\begin{array}[]{l}(s-1)\left(N_{m}-1\right)+N_{m+1},\text{ dacă }r=0\\ (s-1)\left(N_{m}-1\right)+2N_{m+1},\text{ dacă }r>0\end{array}\right.\end{gathered}

for $\mathrm{m}>0$ .
Number N of exceptional (E) formulas in $\mathfrak{N}^{0}$ is equal to 0 if $\mathrm{m}=\mathrm{r}=0$ , equal to s -1 if $\mathrm{m}=0,\mathrm{r}>0$ and check the inequalities

\begin{gathered}(s-1)\left(N_{m}-1\right)+N_{m}+N_{m+1}-\max\{0,s-2\}\leq\\ \leq N\leq\left\{\begin{array}[]{l}(s-1)\left(N_{m}-1\right)+N_{m}+N_{m+1},\text{ dacă }r=0\\ (s-1)\left(N_{m}-1\right)+N_{m}+2N_{m+1},\text{ dacă }r>0\end{array}\right.\end{gathered}

for $\mathrm{m}>0$ .
Max number $\{0,s-2\}$ from the infcric delimitations can be changed depending on $p,m,s$ , but we do not deal with this issue.
13. We can delimit the number (22) quite conveniently in terms of only $p,m,s$ For this we will use the following lemma:

Lemma 2. If the numbers $\alpha_{1},\alpha_{2},\ldots,\alpha_{k},\beta$ are nonnegative, then the inequality holds...

\max\left\{0,\sum_{i=1}^{k}\alpha_{i}-k\beta\right\}\leq\sum_{i=1}^{k}\max\left\{0,\alpha_{i}-\beta\right\}\leq\max\left\{0,\sum_{i=1}^{k}\alpha_{i}-\beta\right\}

(25)

It is enough to prove the property for $k=2$ , because for $k$ some will result by complete induction.

For $k=2$ , the inequalities return to ¹ )

\left|\alpha_{1}+\alpha_{2}-2\beta\right|\leq\left|\alpha_{1}-\beta\right|+\left|\alpha_{2}-\beta\right|\leq\left|\alpha_{1}+\alpha_{2}-\beta\right|+\beta

which is immediately verified ² ).

⁰⁰footnotetext: 1. We have

\max\{0,\alpha\}=\tfrac{1}{2}(\alpha+|\alpha|)

.
2. Each member is linear in

\beta

, in the intervals determined by the points

0,\alpha_{1},\alpha_{2},\tfrac{\alpha_{1}+\alpha_{2}}{2},\alpha_{1}+\alpha_{2},+\infty

It is therefore sufficient to verify the inequalities for the finite extrema of these intervals and for

\beta>\alpha_{1}+\alpha_{2}

The first inequality (23) goes back to a classical inequality and is true whatever the numbers are. $\alpha_{1},\alpha_{2},\ldots,\alpha_{k},\beta$ .

Returning to the delimitation of the sum (22), which for $m=0$ is equal to $p+1$ , we have, based on Lemma 2 ,
$\max\{0,p+1-sm\}\leq\sum_{i=1}^{s}\max\left\{0,r_{i}-m\right\}\leq\max\{0,p+2-s-m\}(m>0)$ , because $r_{i}-1\geqq 0,m-1\geqq 0$ .

We now observe that $\left.{}^{1}\right)\alpha-\max\{\beta,\gamma\}=\min\{\alpha-\beta,\alpha-\gamma\}$ and we deduce

	$\displaystyle\min\{p+1-m,s-1\}$	$\displaystyle\leqq N_{m}\leqq\min\{p+1-m,(s-1)m\},$		(24)
	$\displaystyle m$	$\displaystyle>0\quad\left(N_{0}=0\right).$

The numbers $(s-1)\left(N_{m}-1\right),N_{m},N_{m+1}$ are null for $s=1$ If
$m=p$ , the only point $x_{0}$ for which the formula may be exceptional is

\frac{1}{p+1}\sum_{i=1}^{s}r_{i}x_{i}

(25)

because this is the root of $i^{(p)}(x)$ If
$s>1,\quad 0<m<p$ , we have
$(s-1)[\min\{p+1-m,s-1\}-1]-\max\{0,s-2\}\geq 0,\min\{p-ms-1\}\geq 1$ which shows us, taking into account (24), that we have

(s-1)\left(N_{m}-1\right)+N_{m+1}-\max\{0,s-2\}\geq 1.

Therefore, taking into account theorem 4, we deduce the following theorem:
Theorem 5. In the respective set there are not always formulas $(E)$ exceptional, except for the following three cases:

$1^{\circ}s=1.2^{\circ}m=p$ and point (25) coincides with a node. $3^{\circ}m=0$ and $r=0$ .
14. If in formula (1) we have $m+r>0$ and if the coefficients $a_{0}$ (if $r>0)a_{i,0},i=1,2\ldots,s$ are all zeros, the formula is at the same time a numerical derivation formula for the derivative $f^{\prime}(x)$ of the function $f(x)$ . In this case, we will say that formula (1) is reducible. Otherwise, we will say that the formula is irreducible. Formula (1) is therefore reducible, respectively irreducible, as in this formula none of the values of the function at the nodes and at the point of derivation appear, respectively at least one actually appears.

It is interesting to examine which maximum accuracy formulas are reducible. To do this, let us examine the reducibility of formula (E). Let us assume that formula (E) is non-trivial and is reducible. We must then have $r=0$ , or $r>1$ and $m+r>0$ . Let's net with ( $\mathrm{E}^{*}$ ) this form-
¹ ) We have

\max\{\alpha,\beta\}=\frac{\alpha+\beta+|\alpha-\beta|}{2},\min\{\alpha,\beta\}=\frac{\alpha+\beta-|\alpha-\beta|}{2}.

mulă, viewed as a numerical derivation formula for the function $I^{\prime}(x)=g(x)$ . So using the notations (10), we have

g^{(m+r-1)}\left(x_{0}\right)=\sum_{j=1}^{r-1}c_{j}g^{(j-1)}\left(x_{0}\right)+\sum_{i=1}^{s}\sum_{j=1}^{r_{i}-1}c_{i,j}g^{(j-1)}\left(x_{0}\right)+R

(*)

Based on what has been established above, we can assume that the order of formula (E) is $p+r+1$ Let us denote by $p^{*},r^{*},s^{*}$ the numbers $p,r,s$ , corresponding to the formula (E*). All these facts imply $r_{i}>1,i=1,2,\ldots,s$ and then we have $p^{*}=p-s,s^{*}=s,r^{*}=0$ or $r^{*}=r-1$ as $r^{\prime}=0$ or $r>1$ .

The degree of accuracy of the formula ( $\mathrm{E}^{*}$ ) is $p+r-1$ or $p+r$ , as formula (E) is not or is exceptional, because putting $g(x)=f^{r}(x)$ in $\left(\mathrm{E}^{*}\right)$ we find the formula $(\mathrm{E})$ , and $\left(x^{i}\right)^{\prime}=ix^{i-1}$ .

It is now immediately verified that $p+r-1\geqq p^{*}+r^{*}$ , which shows us that (E*) is a formula of maximum accuracy. Since this formula does not reduce to the trivial formula, its degree of accuracy is $p^{*}+r^{*}$ or $p^{*}+r^{*}+1$ , as it is not or is exceptional. We have evaluated the degree of accuracy in two different ways and the results obtained must coincide. If $s=1$ , the formulas (E), (E*) cannot be exceptional and then we must have $p+r-1=p^{*}+r^{*}$ and it is immediately verified that this equality holds if and only if $r=0$ If $s>1$ , we have $p+r-1>p^{*}+r^{*}$ and equality between degrees of accuracy can only occur if $p+r-1=p^{*}+r^{*}+1$ This equality only holds if $r=0$ and $s=2$ .

In short, formula (E) cannot be reduced except in the following
two cases: in this case, none of the formulas (E), (E*) is exceptional. Note that an exceptional formula (E) is all In the case

$2^{\circ}$ , the exceptional conditions are written.

l^{(m)}\left(x_{0}\right)\neq 0,\quad\left[\frac{l(x)}{\left(x-x_{1}\right)\left(x-x_{2}\right)}\right]_{x=x_{2}}^{(m-2)}=0

(26)

15.

It remains to prove that in the two cases above formula (E) is effectively reducible.

In case $1^{\circ}$ HAVE

L(\underbrace{x_{1},x_{1},\ldots,x_{1}}_{r_{1}};f\mid x)=\sum_{j=0}^{r_{1}-1}\frac{\left(x-x_{1}\right)^{j}}{j!}f^{(j)}\left(x_{1}\right)

so $\mathrm{C}_{1,0}(x)=1$ , from where $G_{1,0}^{(m)}\left(x_{0}\right)=c_{1,0}=0(m>0)$
In the case of $2^{\circ}$ , polynomials $C_{1,0}^{\prime}(x),C_{2,0}^{\prime}(x)$ , of the degree $p-1$ , ⁹ based on formulas (8) are divided by the polynomial

\frac{l(x)}{\left(x-x_{1}\right)\left(x-x_{2}\right)}

(27)

so they differ only by some constant factors from the polynomial (27). These constant factors are also different from zero, also based on formulas (8) ¹ ). The conditions $c_{1,0}=c_{2,0}=0$ are therefore equivalent to the second relation (26). The first relation (26) is a consequence of the previous analysis.

The simultaneity of conditions (26) can also be demonstrated directly. We have

l^{\prime}(x)=\left[r_{1}\left(x-x_{2}\right)+r_{2}\left(x-x_{1}\right)\right]\frac{l(x)}{\left(x-x_{1}\right)\left(x-x_{2}\right)}

from where do we deduce

	$\displaystyle l^{(m)}(x)=$	$\displaystyle{\left[r_{1}\left(x-x_{2}\right)+r_{2}\left(x-x_{1}\right)\right]\left[\frac{l(x)}{\left(x-x_{1}\right)\left(x-x_{2}\right)}\right]^{(m-1)}+}$
		$\displaystyle+(p+1)(m-1)\left[\frac{l(x)}{\left(x-x_{1}\right)\left(x-x_{2}\right)}\right]^{(m-2)}$

From this it is seen that any common root of the polynomials $l^{(m)}(x)$ ,

		$\displaystyle{\left[\frac{l(x)}{\left(x-x_{1}\right)\left(x-x_{2}\right)}\right]^{(m-2)}\text{ este o rădăcină comună a polinoamelor }}$
		$\displaystyle{\left[\frac{l(x)}{\left(x-x_{1}\right)\left(x-x_{2}\right)}\right]^{(m-1)},\left[\frac{l(x)}{\left(x-x_{1}\right)\left(x-x_{2}\right)}\right]^{(m-2)}}$

and such a root, by Lemma 1, necessarily coincides with one of the nodes $x_{1},x_{2}$ .

Finally, we can state the following theorem:
Theorem 6. If $\mathrm{m}\leq\mathrm{p}$ , formula (E) is reducible if and only if:
$1^{\circ}\cdot\mathrm{r}=0,\mathrm{\penalty 10000\ s}=1,\mathrm{\penalty 10000\ s}au$
$2^{\circ}.\mathrm{r}=0,\mathrm{\penalty 10000\ s}=2$ and $\mathrm{x}_{0}$ is a root, different from the nodes $\mathrm{x}_{1},\mathrm{x}_{2}$ of the polynomial

\left[\frac{l(x)}{\left(x-x_{1}\right)\left(x-x_{2}\right)}\right]^{(m-1)}

The properties are expressed by the formulas

\begin{gathered}L^{(m)}(\underbrace{x_{1},x_{1},\ldots,x_{1}}_{r_{1}};f\mid x)=L^{(m-1)}(\underbrace{x_{1};x_{1},\ldots,x_{1}}_{r_{1}-1};f^{\prime}\mid x)\\ \begin{array}[]{c}L^{(m)}(\underbrace{x_{1},x_{1},\ldots,x_{1}}_{r_{1}},\underbrace{x_{2},x_{2},\ldots,x_{2}}_{r_{2}};f\mid x_{0})=\\ \left.=L^{(m-1)}\right)\end{array}\\ \underbrace{x_{1},x_{1},\ldots,x_{1}}_{r_{1}-1},\underbrace{x_{2},x_{2},\ldots,x_{2}}_{r_{2}-1};f^{\prime}\mid x_{0})\end{gathered}

where $x_{0}$ -check the second condition (26).

⁰⁰footnotetext: 1. From

C_{1,0}(x)+C_{2,0}(x)=1

it follows that these factors are of opposite signs and equal in absolute value.

§ 2. Some considerations on the rest

16.

In certain important particular cases, the rest $R$ of formula (1) is put in a simple and convenient form. In the case of the Lagrange-Hermite interpolation formula

f\left(x_{0}\right)=L\left(x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};fx_{0}\right)+R

(28)

HAVE

R=R[f]=l\left(x_{0}\right)\left[x_{0},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f\right]

(29)

where $l(x)$ is the polynomial (3) already defined and which in the notation (7) can also be written

l(x)=\left(x-x_{1}^{\prime}\right)\left(x-x_{2}^{\prime}\right)\ldots\left(x-x_{p+1}^{\prime}\right).

The rest of formula (28), in the form (29), is expressed as a product of a non-zero number and independent of the function $f(x)$ and a difference divided by the order $p+1$ of the function $f(x)$ .

Based on formulas

\left[\alpha_{1},\alpha_{2},\ldots,\alpha_{k+1};x^{i}\right]=\left\{\begin{array}[]{l}0,\text{ pentru }i=0,1,\ldots,k-1\\ 1,\text{ pentru }i=k\end{array}\right.

it is visible that formula (28) has the degree of accuracy $p$ Putting $f(x)=l(x)=x^{n+1}+\ldots$ in (28) and taking into account formula (11), we see that the factor of the divided difference in (29) is equal to

R[l]=R\left[x^{p+1}\right]=l\left(x_{0}\right)

17.

We will say that the rest of formula (1) or that the functional $R[f]$ is of simple form if we have

R[f]=M\cdot\left[\xi_{1},\xi_{2},\ldots,\xi_{n+2};f\right]

(31)

whatever the function $f(x)$ , where $n$ is the degree of accuracy of the formula, $M$ a non-zero constant independent of the function $f(x)$ , and $\xi_{1},\xi_{2},\ldots,\xi_{n-1},n+2$ distinct points in the open interval $(a,b)$ .

In this case, based on formulas (30), the coefficient $M$ is given by the equality $M=R\left[x^{n+1}\right]=R[P]$ , where $P(x)$ is any polynomial of degree $n+1$ , with the first coefficient 1 .

points $\xi_{i}$ generally depend on the function $f(x)$ , and the interval $[a,b]$ This is the one specified in § 1.

For example, as we will see below, the rest of formula (28) is of simple form.

To simplify the notations, we will denote by $D_{k}[f]0$ difference divided by the order $k$ of the function $f(x)$ on $k+1$ arbitrary distinct nodes in the interval ( $a,b$ ). So if formula (1) has the degree of accuracy $n$ and if restyl is of simple form, we have $R[f]=R\left[x^{n+1}\right]\cdot D_{n+1}[f].8$

The previous definition remains for $n\geq 0$ . We will also maintain 0 for
the extremes of the interval $[a,b]$ . This convention simplifies the exposition. Moreover, in the almost trivial case when $n=-1$ , it is generally easy to recognize whether the point can be chosen or not $\xi_{1}$ , within the interval

In the case when the remainder is of simple form, formula (31) gives us sufficient indications on the structure of this remainder, based on the properties of divided differences, properties that we expose elsewhere. These properties allow us to express the remainder in an analogous way, with the divided differences of the derivative of a given order of the function $f(x)$ , when this derivative exists in the interval ( $a,b$ ).
18. Based on a previous result [9], for recognizing the simple form of the remainder we can apply the following criterion:
C. For the additive and homogeneous functional R [f], with the degree of accuracy n, to be of simple form it is necessary and sufficient that R [f] is non-zero for any function $\mathrm{f}(\mathrm{x})$ convex of order n in $[\mathrm{a},\mathrm{b}]^{1}$ ).

The definition of the simple form and the C criterion apply not only to the rest of formula (1) but also to an additive and homogeneous functional $R[f]$ , defined in a vector field $F$ of continuous functions in the interval $[a,b]$ and containing all polynomials.

A function $f(x)$ it is said to be unconquered by the order $n$ in the interval $[a,b]$ , if all its differences divided by $n+2$ distinct points in $[a,b]$ are nonnegative. The function is said, in particular, to be convex of order $n$ in this interval, if all these divided differences are positive. We will point out in their place the properties used here of non-concave and convex functions.

A non-concave function of order $n\geq 1$ , so in particular a convex function of order $n\geqslant 1$ , is continuous in the open interval ( $a,b$ ). If $n>1$ , it admits continuous derivatives of the orders $1,2,\ldots,n-1$ in the interval ( $a,b$ ) In general, such a function does not have a derivative of the order $n$ at all points of the interval ( $a,b$ ).

In general, the field $F$ of the functional $R[f]$ does not contain all convex functions of order $n$ in ( $a,b$ ) because, on the one hand, by hypothesis the field $F$ is made up of continuous functions in $[a,b]$ , on the other hand the functional may depend on values of the derivatives of order $>n-1$ of the function $f(x)$ Criterion C is, however, always applicable because from the demonstration of this criterion [7], it follows that it is sufficient to consider only functions of order $n$ which belong to $F$ .
19. In the statement of the criterion $C$ condition $R[f]\neq 0$ can be replaced with the condition $R\left[f_{1}\right]R\left[f_{2}\right]>0$ for any two functions $f_{1}(x)$ and $f_{2}(x)\in F$ , convex of the order $n$ This property results from the following lemma, very useful in applying the C criterion.

Lemma 3. If n is the degree of accuracy of the additive and homogeneous functional R[f] and if we can find two functions $\mathrm{f}_{1}(\mathrm{x}),\mathrm{f}_{2}(\mathrm{x})$ unconquered by order $\mathrm{n}^{\prime}\leqq\mathrm{n}$ in the interval $[\mathrm{a},\mathrm{b}]$ so that

R\left[f_{1}\right]R\left[f_{2}\right]<0

(32)

then there is a function $\mathrm{f}(\mathrm{x})$ , convex of order $\mathrm{n}^{\prime}$ in $[\mathrm{a},\mathrm{b}]$ , so that

R[f]=0

(33)

⁰⁰footnotetext: ¹⁾ The criterion in this form is sufficient here. In general, from the fact that

R[f]\neq 0

, for any convex function of order

n

, it follows that the degree of accuracy is

n

It is known that a linear combination, with positive coefficients, of several (a finite number) non-concave functions of order $n^{\prime}$ , of which at least one is convex of order $n^{\prime}$ , is a convex function of order $n^{\prime}$ .

Let us now observe that the function $(x-a)^{n+1}$ is convex of order $n^{\prime}$ in the interval $[a,b]$ .

On the other hand, $R\!\left[(x-a)^{n+1}\right]\neq 0$ , because of the definition of the degree of accuracy.

It follows that the functions …

		$\displaystyle f_{1}^{*}(x)=f_{1}(x)+\frac{1}{2}\left\|\frac{R\left[f_{1}\right]}{R\left[(x-a)^{n+1}\right]}\right\|(x-a)^{n+1}$
		$\displaystyle f_{2}^{*}(x)=f_{2}(x)+\frac{1}{2}\left\|\frac{R\left[f_{2}\right]}{R\left[(x-a)^{n+1}\right]}\right\|(x-a)^{n+1}$

are convex of the order $n^{\prime}$
Taking into account (32), a simple calculation shows us that

R\left[f_{1}^{*}\right]R\left[f_{2}^{*}\right]=\frac{3}{4}R\left[f_{1}\right]R\left[f_{2}\right]<0.

It follows that the function

f(x)=f_{1}^{*}(x)-\frac{R\left[f_{1}^{*}\right]}{R\left[f_{2}^{*}\right]}f_{2}^{*}(x)

is convex of order $n^{\prime}$ and, it is immediately verified, satisfies equality (33).
Lemma 3 is therefore proven.
For $n^{\prime}=n$ , applying the criterion C , the following theorem is deduced:
Theorem 7. If n is the degree of accuracy of the functional R[f], adjectival : and homogeneous and if we can find two functions $\mathrm{f}_{1}(\mathrm{x}),\mathrm{f}_{2}(\mathrm{x})\in\mathrm{F}$ , non-concave of order n in $[\mathrm{a},\mathrm{b}]$ , so that we have (32), then the functional is not of simple form.
20. The remainder $R$ of formula (1) can be studied in a more symmetrical form

R[f]=\sum_{i=0}^{s}\sum_{j=0}^{T_{m}}b_{i,j}f^{(j)}\left(x_{i}\right)

(34)

where for the moment we can assume

a=x_{0}\leqslant x_{1}<\ldots<x_{s}\leqslant b

Of course, the field $F$ of the functional (34) consists of the functions $f(x)$ continue in $[a,b]$ and admitting the derivatives at the points where they actually intervene in the expression of $R[f]$ .

We will seek to establish some properties regarding the simple form of the functional (34).

Let's assume that the degree of accuracy is $n$ We will show that the functional cannot effectively depend on the derivatives of order $>n+1$ in case its form is simple. For this it will be enough, based on Theorem 7, to construct two functions in the opposite case $f_{1}(x),f_{2}(x)\in F$ non-concave of the order $n$ , so that we have inequality (32).

⁰⁰footnotetext: ¹ ) It can be shown, for example, that the derivative of the order

n^{\prime}+1

of the function is positive in the interval

[a,b]

Let us therefore suppose that $b_{i,k}\neq 0,b_{i,j}=0,j=k+1,k+2,\ldots$ , where $k\geq n+2$ .

Whether $\varepsilon$ a fairly small positive number and which will be further specified in the course of the demonstration.

Let's define the continuous function $\varphi_{\varepsilon}(x)$ in the following way:
$1^{\circ}.\varphi_{\varepsilon}\left(x_{i}\right)=1$ .
$2^{\circ}.\varphi_{\varepsilon}(x)=0$ for $x\in\left[a,x_{i}-\varepsilon\right]\cup\left[x_{i}+\varepsilon,b\right],x\in[a+2\varepsilon,b]$ , respectively $x\in[a,b-2\varepsilon]$ , as $0<i<s,i=0$ , respectively $i=s$ .
$3^{\circ}\varphi_{\varepsilon}(x)$ is linear in each of the remaining intervals of $[a,b]$ so in $\left[x_{i}-\varepsilon,x_{i}\right]$ and $\left[x_{i},x_{i}+\varepsilon\right],[a,a+2\varepsilon]$ , respectively $[b-2\varepsilon,b]$ , as $0<i<s,i=0$ , respectively $i=s$ .

We then have $\varphi_{\varepsilon}(x)\geqq 0$ for $x\in[a,b]$ Let us
now consider the function

f_{\varepsilon}(x)=\int_{a}^{x}\frac{(x-t)^{k-1}}{(k-1)!}\varphi_{\varepsilon}(t)dt

Then, if $\varepsilon<\max\left\{x_{1}-x_{0},x_{s}-x_{s-1},\frac{x_{s}-x_{0}}{2}\right\}$ HAVE

f_{\varepsilon}^{(li)}(x)=\varphi_{\varepsilon}(x),\text{ pentru }x\in[a,b]

$0=f_{\varepsilon}^{(l-1)}(a)\leqq f_{\varepsilon}^{(l-1)}(x)\leqq f_{\varepsilon}^{(l-1)}(b)=\int_{a}^{b}\varphi_{\varepsilon}(t)dt=\varepsilon,\quad$ for $\quad x\in[a,b]$

f_{\varepsilon}^{(j)}(x)=\int_{a}^{x}\frac{(x-t)^{k-2-j}}{(k-2-j)!}f_{\varepsilon}^{(k-1)}(t)dt,\quad\text{ pentru }\quad j=0,1,\ldots,k-2

It then immediately follows that
$0\leqq f_{\varepsilon}^{(i)}(x)\leqq\varepsilon\frac{(b-a)^{k-1-j}}{(k-1-j)!}$ , for $x\in[a,b]$ and $j=0,1,\ldots,k-1$ .

Be it now $A$ some non-zero constant and consider the function

f(x)=Af_{\varepsilon}(x)+|A|\frac{\varepsilon(b-a)^{k-n-2}}{(n+1)!(k-n-2)!}(x-a)^{n+1}

(36)

We then have

f^{(n+1)}(x)=Af_{\varepsilon}^{(n+1)}(x)+|A|\frac{\varepsilon(b-a)^{k-n-2}}{(k-n-2)!}

and therefore, based on (35),

f^{(n+1)}(x)\geq 0,\quad x\in[a,b].

This inequality shows us that the function $f(x)$ is non-concave of the order $n$ in $[a,b]$ .

Taking into account (36), we deduce
$f^{(j)}(x)=\left\{\begin{array}[]{c}Af_{\varepsilon}^{(j)}(x)+|A|\frac{\varepsilon(b-a)^{k-n-2}}{(n+1-j)!(k-n-2)!}(x-a)^{n+1-j},\\ \quad j=0,1,\ldots,n+1\\ Af_{\varepsilon}^{(j)}(x),j=n+2,n+3,\ldots,k-1\end{array}\right.$
for $x\in[a,b]$
Taking into account (35), we have

\left|f^{(j)}(x)\right|\leq 2|A|(b-a)^{k-1-i}\zeta,\quad j=0,1,\ldots,k-1,x\in[a,b].

(37)

If we now assume that

\frac{1}{2}\min_{i=0,1,\ldots,s}\left\{x_{i+1}-x_{i}\right\}.

(38)

we also have

f^{(j)}\left(x_{\alpha}\right)=0,\text{ pentru }j\geq k,\alpha=0,1,\ldots,i-1,i+1,\ldots,s.

(39)

Taking into account $f^{(k)}\left(x_{i}\right)=A$ and from (34), (37), (39), we deduce

\left|R[t]-b_{i,k}A\right|<\left(\sum_{i=0}^{s}\sum_{j=1}^{k-1}\left|b_{i,j}\right|\right)2|A|\cdot\max\left\{1,(b-a)^{k-1}\right\}\cdot\varepsilon

Choosing $\varepsilon$ small enough and in any case such that inequality (38) is verified, we deduce

\left|R[t]-b_{i,k}A\right|<\frac{\left|b_{i,k}A\right|}{2}

which shows us that then $R[f]$ has his sign $b_{i,k}A$ But $A$ was chosen randomly. Taking $A=1$ and noting with $f_{1}(x)$ function (36) thus obtained, and taking the aroi $A=-1$ and noting with $/_{2}(x)$ function (36) thus obtained, inequality (32) is verified.

We can, however, state the following theorem:
Theorem 8. If the degree of accuracy of the functional (34) is n, this junctional cannot be of simple form unless it contains derivative effects of order $>\mathrm{n}+1$ .

In other words, the functional cannot be of simple form unless $b_{i,j}=0$ for $j>n+1,i=0,1,\ldots,s$ .

We will examine in particular the functional (32) with the degree of accuracy $-1,0$ or 1.
21. Suppose that the degree of accuracy is -1. Theorem 8, which also applies here, shows us that the functional (32) is necessarily of the form

R[f]=\sum_{i=0}^{s}b_{i}f\left(x_{i}\right)

(40)

if it is of simple form.

For the functional (40) to be of simple form, it is necessary that the coefficients $b_{0},b_{1},\ldots,b_{s}$ be of the same sign ¹ ). Indeed, let us assume the opposite, and let $b_{\alpha}b_{\beta}<0$ Then, inequality (32) is verified only for $j_{1}(x)$ we take a continuous non-negative function, such that $j_{1}\left(x_{\alpha}\right)=1,j_{1}\left(x_{i}\right)=0$ , $j=0,1,\ldots,\alpha-1,\alpha+1,\ldots,s$ and for $t_{2}(x)$ we take a continuous non-negative function, so it $f_{2}\left(x_{\beta}\right)=1,f_{2}\left(x_{j}\right)=0,j=0,1,\ldots,\beta-1,\beta+1,\ldots,s$ .

The condition is also sufficient, because if it is fulfilled, then

R[1]=\sum_{i=0}^{s}b_{i}\neq 0

it results that $R[y]\neq 0$ for any positive function $f(x)$ .
We therefore deduce the following theorem:
Theorem 9. The necessary and sufficient condition for the functional (40) to be of degree of accuracy -1 and to be of simple form is that all the coefficients $\mathrm{b}_{i},\mathrm{i}=0,1,\ldots,\mathrm{\penalty 10000\ s}$ be of the same sign, without all being zero.

If the condition is met, then we have

\sum_{i=0}^{s}b_{i}f\left(x_{i}\right)=\left(\sum_{i=0}^{s}b_{i}\right)D_{0}[f]

which also expresses a well-known property of continuous functions. Here $D_{0}[f]=f(\xi)$ The point $\xi$ generally belongs to the closed interval $[a,b]$ If, for example, $s>1$ and at least one of the coefficients $b_{2},b_{3},\ldots,b_{s-1}$ is different from zero, the point $\xi$ can be chosen in the range $(a,b)$ .

As an application, note that the necessary and sufficient condition for formula (1) to be of degree of accuracy -1 and with the remainder in simple form is that $m=r=0$ and that $a_{i,0}\leqq 0,a_{i,j}=0$ , for $j=1,2,\ldots,r_{i}-1,i=1,2,\ldots,s$ .
22. If the functional (34) is of degree of accuracy 0 and of simple form, then, according to Theorem 8, it is necessarily of the form

R[f]=\sum_{i=0}^{s}\left[b_{i}f\left(x_{0}\right)+b_{i}^{\prime}f^{\prime}\left(x_{i}\right)\right].

(41)

The conditions for the degree of accuracy to be 0 are

\sum_{i=0}^{s}b_{i}=0,\quad\sum_{i=0}^{s}\left(b_{i}x_{0}+b_{i}^{\prime}\right)\neq 0.

(42)

Let's now look for necessary conditions for the functional to also be of simple form.

The first relation (42) is necessary because 1 and -1 are both non-decreasing functions and we have

R[1]\cdot R[-1]=-\left(\sum_{i=0}^{s}b_{i}\right)^{2}

and if the condition were not fulfilled, we would satisfy inequality (32).
Also, the second condition (42) is necessary, because otherwise we would have $R[x]=0$ , and $x$ is an increasing function.

⁰⁰footnotetext: ¹⁾ This means that the numbers are all non-negative or all non-positive. A single number is considered to always have the same sign.

Be it now $f(x)$ a continuous, non-decreasing function equal to 0 for $x\in\left[a,x_{i-1}+\varepsilon\right]$ and equal to 1 for $x\in\left[x_{i}-\varepsilon,b\right],\in$ being a positive number, often small $\left[\varepsilon<\frac{1}{2}\left(x_{i}-x_{i-1}\right)\right]$ We then have

R[f]=b_{i}+b_{i+1}+\ldots+b_{s}

Based on Theorem 7, it follows that the numbers

b_{i}+b_{i+1}+\ldots+b_{s},\quad i=1,2,,\ldots s

(43)

be of the same sign.
Then let the function $f(x)$ equal to 0 for $x\leq x_{i}-\varepsilon$ , equal to $2\varepsilon$ for $x\geqq x_{i}+\varepsilon$ and linear in the interval $\left[x_{i}-\varepsilon,x_{i}+\varepsilon\right]$ , where $\varepsilon$ is a fairly small positive number, $\varepsilon<\min_{i=1,2,3}\left\{x_{i}-x_{i-1}\right\}$ This function is non-decreasing and we have

R[f]=\varepsilon\left[\left(b_{i}+b_{i+1}+\ldots+b_{s}\right)+b_{i+1}+b_{i+2}+\ldots+b_{s}\right]+b_{i}^{\prime}

Since e can be taken as small as desired, it is seen, also based on Theorem 7, that the numbers

b_{0}^{\prime},b_{1}^{\prime},\ldots,b_{s}^{\prime}

(44)

together with the numbers (43) must be of the same sign.
Finally, we will highlight another necessary condition. Let $f(x)$ function defined as follows:

\begin{gathered}f(x)=x-\frac{\left(x-x_{i-1}\right)\left(x-x_{i}\right)\left(2x-x_{i-1}-x_{i}\right)}{\left(x_{i}-x_{i-1}\right)^{2}},\text{ pentru }x\in\left[x_{i-1},x_{i}\right]\\ i=1,2,\ldots,s.\end{gathered}

We then have
$f^{\prime}(x)=-\frac{6}{\left(x_{i}-x_{i-1}\right)^{2}}\left(x-x_{i-1}\right)\left(x-x_{i}\right)$ , for $x\in\left[x_{i-1},x_{i}\right],i=1,2,\ldots,s$ and it is seen that the function is continuous and differentiable in the interval $[a,b]$ But $f\left(x_{i}\right)=x_{i},f^{\prime}\left(x_{i}\right)=0,i=0,1,\ldots,s$ and $f^{\prime}(x)>0$ , for $x\in\left(x_{i-1},x_{i}\right)i=1,2,\ldots,s$ so the function $f(x)$ is increasing and we have

R[f]=\sum_{i=0}^{s}b_{i}x_{i}

We therefore deduce the necessary condition

b_{i}x_{i}\neq 0

(45)

Let us now show that the established conditions are also sufficient.
Based on the first relation (42) we can write

R[f]=\sum_{i=1}^{s}\left(x_{i}-x_{i-1}\right)\left(b_{i}+b_{i+1}+\ldots+b_{s}\right)\left[x_{i-1},x_{i};f\right]+\sum_{i=0}^{s}b_{i}^{\prime}f^{\prime}\left(x_{i}\right)

(46)

which is obtained by applying Abel's transformation formula.

We now observe that

\sum_{i=1}^{s}\left(x_{i}-x_{i-1}\right)\left(b_{i}+b_{i+1}+\ldots+b_{s}\right)=\sum_{i=0}^{s}b_{i}x_{i}

what is achieved by doing $f(x)=x$ in (41) and (46). Condition (15) therefore implies that the numbers (43) are not all zero.

If now $f(x)$ is an increasing function, we have

\left[x_{i-1},\dot{x}_{i};f\right]>0,i=1,2,\ldots,s,f^{\prime}\left(x_{i}\right)\geqq 0,i=0,1,\ldots.s

so the first relation (42), the condition (45) and the fact that the numbers (43), (44) are of the same sign, imply, based on formula (46), that $R[f]\neq 0$ .

Note that the first relation (42) and condition (45) entail the second condition (42), if the numbers (43), (44) are of the same sign.

We therefore have the following theorem:
Theorem 10. The necessary and sufficient conditions for the functional (41) to be of degree of accuracy 0 and of simple form are that we have

\sum_{i=0}^{s}b_{i}=0,\quad\sum_{i=0}^{s}b_{i}x_{i}\neq 0

and like the numbers

b_{s},b_{s}+b_{s-1},\ldots,b_{1}+b_{2}+\ldots+b_{s},b_{0}^{\prime},b_{1}^{\prime},\ldots,b_{s}^{\prime}

have the same sign.
23. In order for formula (1) to have the degree of accuracy 0 and to be of simple form, it is necessary to have $m+r\leq 1$ and $a_{i,j}=0,j=2,3,\ldots,r_{i}-1$ , $i=1,2,\ldots,s$ To write the other conditions, the mutual position of the nodes and the point of derivation must be taken into account.

To fix the ideas, we can assume that the nodes are in ascending order, so

x_{1}<x_{2}<\ldots<x_{s}

(47)

We must distinguish three cases here:
$1^{\circ}.m=r=0$ Assuming $x_{i-1}<x_{0}<x_{i}$ , the necessary and sufficient conditions are then

\sum_{i=1}^{s}a_{i,0}=1,\quad\sum_{i=1}^{s}a_{i,0}x_{i}\neq x_{0}

and like the numbers

\left.\begin{array}[]{c}-\left(a_{1,0}+a_{2,0}+\ldots+a_{j,0}\right),j=1,2,\ldots,i-1\\ a_{j,0}+a_{i+1,0}+\ldots+a_{s,0},j=i,i+1,\ldots,s\\ a_{i,1},j=1,2,\ldots,s\end{array}\right\}

be of the same sign.
If $i=1$ or $i=s+1$ the first and second series of numbers (48) disappear. In these cases we have $x_{0}<x_{1}$ , respectively $x_{0}>x_{s}$ .
$2^{\circ}.m=0,r=1$ Assuming again $x_{i-1}<x_{0}<x_{i}$ the necessary and sufficient conditions are

a_{0}+\sum_{i=0}^{s}a_{i,0}=0,\quad a_{0}x_{0}+\sum_{i=1}^{s}a_{i,0}x_{i}\neq 0

and that all numbers (48) are non-positive, with the above restriction, if $i=1$ or $i=s+1$ .
3. $m=1,r=0$ The necessary and sufficient conditions are

\sum_{i=1}^{s}a_{i,0}=0,\quad\sum_{i=1}^{s}a_{i,0}x_{i}\neq 0

and like the numbers

a_{i,0}+a_{i+1,0}+\ldots+a_{s,0},\quad i=2,3,\ldots,s;\quad a_{i,1},\quad i=1,2,\ldots,s

be all non-positive.
There is only one formula (E) of degree of accuracy 0, which is of simple form and which we will indicate in § 5.
24. If the functional (34) is of degree of accuracy 1 and of simple form, then, according to Theorem 8, it is necessarily of the form

R[f]=\sum_{i=0}^{s}\left[b_{i}f\left(x_{i}\right)+b_{i}^{\prime}f^{\prime}\left(x_{i}\right)+b_{i}^{\prime\prime}f^{\prime\prime}\left(x_{i}\right)\right]

(49)

The conditions for the degree of accuracy to be 1 are

\sum_{i=0}^{s}b_{i}=0,\quad\sum_{i=0}^{s}\left(b_{i}x_{i}+b_{i}^{\prime}\right)=0,\quad\sum_{i=0}^{s}\left(b_{i}x_{i}^{2}+2b_{i}^{\prime}x_{i}+2b_{i}^{\prime\prime}\right)\neq 0

(50)

Let us again look for the necessary conditions for functional (49) to be of simple form.

The first two conditions (50) are necessary because the functions $1,-1,x,-x$ are non-concave of order 1 and

R[1]R[-1]=-\left(\sum_{i=0}^{s}b_{i}\right)^{2},\quad R[x]R[-x]=-\left[\sum_{i=0}^{s}\left(b_{i}x_{i}+b_{i}^{\prime}\right)\right]^{2}

Let's consider the function

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

where $\varepsilon$ is a positive number $<x_{i}-x_{i-1}$ This function is non-concave of order 1 in $[a,b]$ and we have

\begin{array}[]{r}R[f]=\lambda_{i+1}+\lambda_{i+2}+\ldots+\lambda_{s}+b_{i}^{\prime}+b_{i+1}^{\prime}+\ldots+b_{s}^{\prime}+\\ +\varepsilon\left(b_{i}+b_{i+1}+\ldots+b_{s}\right)\end{array}

where did I put

\lambda_{i}=\left(x_{i}-x_{i-1}\right)\left(b_{i}+b_{i+1}+\ldots+b_{s}\right),\quad i=1,2,\ldots,s

Also, the function

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

is non-concave of order 1 in $[a,b]$ and we have ( $\varepsilon<x_{i}-x_{i-1}$ )

\begin{array}[]{r}R[f]=\lambda_{i}+\lambda_{i+1}+\ldots+\lambda_{s}+b_{i}^{\prime}+b_{i+1}^{\prime}+\ldots+b_{s}^{\prime}+\\ +\varepsilon\left(b_{i}+b_{i+1}+\ldots+b_{s}\right)\end{array}

Formulas (51) ; (52) hold no matter how small $\varepsilon$ positive. It follows immediately, based on Theorem 8, that the numbers ¹ )

\left\{\begin{array}[]{l}\lambda_{i}+\lambda_{i+1}+\ldots+\lambda_{s}+b_{i}^{\prime}+b_{i+1}^{\prime}+\ldots+b_{s}^{\prime}\\ \lambda_{i+1}+\lambda_{i+2}+\ldots+\lambda_{s}+b_{i}^{\prime}+b_{i+1}^{\prime}+\ldots+b_{s}^{\prime}\end{array}\quad i=1,2,\ldots,s\right.

must be of the same sign.
Be it now $f(x)$ null function for $x\leqq x_{i}-\varepsilon$ , equal to $\left(x-x_{i}+\varepsilon\right)^{2}$ in the interval $\left[x_{i}-\varepsilon,x_{i}+\varepsilon\right]$ and equal to $4\varepsilon\left(x-x_{i}\right)$ for $x\geq x_{i}+\varepsilon$ This is a continuous non-concave function of order 1 in $\left.[a,b]^{2}\right)$ and for $\varepsilon$ quite small $\left(\varepsilon<x_{i}-x_{i-1},x_{i+1}-x_{i}\right)$ HAVE

	$\displaystyle R[f]=\varepsilon^{2}b_{i}+2\varepsilon b_{i}^{\prime}+4\varepsilon\left(\lambda_{i+1}+\lambda_{i+2}\right.$	$\displaystyle+\ldots+\lambda_{s}+b_{i+1}^{\prime}+$
		$\displaystyle\left.+b_{i+2}^{\prime}+\ldots+b_{s}^{\prime}\right)+2b_{i}^{\prime\prime}$

number $\varepsilon$ can be however small positive, also based on theorem 8, we see that the numbers

b_{0}^{\prime\prime},b_{1}^{\prime\prime},\ldots,b_{s}^{\prime\prime}

(54)

are of the same sign as the numbers (53).
Finally, as in the case of $n=0$ , we will highlight another necessary condition. Let $f(x)$ function defined as follows:

f(x)=x^{2}-\frac{\left(x-x_{i-1}\right)^{2}\left(x-x_{i}\right)^{2}}{\left(x_{i}-x_{i-1}\right)^{2}},\text{ pentru }x\in\left[x_{i-1},x_{i}\right],\quad i=1,2,\ldots,s

We then have
$f^{\prime\prime}(x)=-\frac{12}{\left(x_{i}-x_{i-1}\right)^{2}}\left(x-x_{i-1}\right)\left(x-x_{i}\right)$ , for $x\in\left[x_{i-1},x_{i}\right],i=1,2,\ldots,s$ It is seen
that the function is continuous and has a second derivative, continuous in the interval $[a,b]$ But $f\left(x_{i}\right)=x_{i}^{2},f^{\prime}\left(x_{i}\right)=2x_{i},f^{\prime\prime}\left(x_{i}\right)=0,i=0,1,\ldots,s$ and $f^{\prime\prime}(x)>0$ for $x\in\left(x_{i-1},x_{i}\right),i=1,2,\ldots,s$ , so the function $f(x)$ is convex of order 1 and we have

R[f]=\sum_{i=0}^{s}\left(b_{i}x_{i}^{2}+2b_{i}^{\prime}x_{i}\right)

¹ ) For $i=s$ , the second number reduces to $b_{s}^{\prime}$ .
² ) It is useful to make a graphical representation of this function, as well as of the other auxiliary functions used here.

We also have the necessary condition

\sum_{i=0}^{s}\left(b_{i}x_{i}^{2}+2b_{i}^{\prime}x_{i}\right)\neq 0

(55)

Let us now show that the conditions found are also sufficient. This follows from the formula

	$\displaystyle R[f]=\sum_{i=1}^{s}\left(x_{i}\right.$	$\displaystyle\left.-x_{i-1}\right)\left(\lambda_{i}+\lambda_{i+1}+\ldots+\lambda_{s}+b_{i}^{\prime}+b_{i+1}^{\prime}+\right.$
		$\displaystyle\left.+\ldots+b_{s}^{\prime}\right)\cdot\left[x_{i-1},x_{i-1},x_{i};f\right]+\sum_{i=1}^{s}\left(x_{i}-x_{i-1}\right)\left(\lambda_{i+1}+\lambda_{i+2}+\right.$
		$\displaystyle\left.+\ldots+\lambda_{s}+b_{i}^{\prime}+b_{i+1}^{\prime}+\ldots+b_{s}^{\prime}\right)\cdot\left[x_{i-1},x_{i},x_{i};f\right]+\sum_{i=0}^{s}b^{\prime\prime}f^{\prime\prime}\left(x_{i}\right)$

which is valid when the first two conditions (50) are satisfied. This formula is a particular case of the general formula for the transformation of divided differences and is obtained by applying Abel's transformation formula twice [6]. If we put $f(x)=x^{2}$ , we deduce

\begin{gathered}\sum_{i=1}^{s}\left(x_{i}-x_{i-1}\right)\left(\lambda_{i}+\lambda_{i+1}+\ldots+\lambda_{s}+b_{i}^{\prime}+b_{i+1}^{\prime}+\ldots+b_{s}^{\prime}\right)+\\ +\sum_{i=1}^{s}\left(x_{i}-x_{i-1}\right)\left(\lambda_{i+1}+\lambda_{i+2}+\ldots+\lambda_{s}+b_{i}^{\prime}+b_{i+1}^{\prime}+\ldots+b_{s}^{\prime}\right)=\\ =\sum_{i=0}^{s}\left(b_{i}x_{i}^{2}+2b_{i}^{\prime}x_{i}\right)\end{gathered}

and then, based on hypothesis (55), at least one of the numbers (53) is different from zero.

If $f(x)$ is a convex function of order 1, we have

\begin{gathered}{\left[x_{i-1},x_{i-1},x_{i};f\right]>0,\quad\left[x_{i-1},x_{i},x_{i};f\right]>0,\quad i=1,2\ldots,s}\\ f^{\prime\prime}\left(x_{i}\right)\geq 0,i=0,1\ldots,s\end{gathered}

so the first two equalities (50), condition (55) and the fact that the numbers (53), (54) are of the same sign imply $R[f]\neq 0$ .

Note that under the same conditions the third relation (50) results.
Theorem 11. The necessary and sufficient conditions for the functional (49) to be of degree of accuracy 1 and of simple form are that the first two relations (50) are verified, that the inequality (55) is verified and that the numbers (53), (54) have the same sign.
25. For the formula (1) to be of degree of accuracy 1 and of simple form, it is necessary that $m+r\leq 2$ and as $a_{i,j}=0,j=3,4\ldots,r_{i}-1,i=1,2\ldots s$ The expression of the other conditions depends on the mutual positions of the nodes and the derivation point.

Let us assume that the nodes are in increasing order, so that we have (47): We will then be able to express the necessary and sufficient conditions in terms of the values of
$m,r$ and the position of the point $x_{0}$ with respect to the nodes. These conditions are quite complicated. It will therefore suffice to show in each case how we bring the remainder to the form (49), studied above, and for which the points $x_{0},x_{1}\ldots x_{s}$ are in ascending order. The conditions will then be expressed using the coefficients $b_{i},b_{i}^{\prime},b_{i}^{\prime\prime}$ .

We must distinguish 6 cases. We will generally assume that $x_{i-1}<x_{0}<x1^{\circ}.m=r=0$ A series of necessary and sufficient conditions become

\sum_{i=1}^{s}a_{i,0}=1,\sum_{i=1}^{s}\left(a_{i,0}x_{i}+a_{i,1}\right)=x_{0},\sum_{i=2}^{s}\left(a_{i,0}x_{i}^{2}+2a_{i,1}x_{i}\right)\neq x_{0}^{2}

and the other conditions are deduced by reducing the remainder to the form (49) ¹ ) which comes down to replacing in (49) the sequence of points $x_{0},x_{1},\ldots,x_{s}$ egg $x_{1},x_{2},\ldots,x_{i-1}$ , $x_{0},x_{i},\ldots x_{s}$ and then putting

\left.\begin{array}[]{ll}b_{i}=a_{i+1,0},&b_{j}^{\prime}=a_{j+1,1},\quad b_{j}^{\prime\prime}=a_{j+1,2},\quad j=0,1,\ldots i-2\\ b_{i-1}=-1,&b_{i-1}^{\prime}=b_{i-1}^{\prime\prime}=0\\ b_{j}=a_{j,0},&b_{i}^{\prime}=a_{j,1},\quad b_{j}^{\prime\prime}=a_{j,2},\quad j=i,i+1,\ldots,s\end{array}\right\}

$2^{\circ}.m=0,r=1$ . Analogous reduction, except that the second series of formulas (56) becomes

b_{i-1}=a_{0},\quad b_{i-1}^{\prime}=-1,\quad b_{i-1}^{\prime\prime}=0

3.

$m=1,r=0$ . Analogous reduction, the second series of formulas (56) becomes

b_{i-1}=0,\quad b_{i-1}^{\prime}=-1,\quad b_{i-1}^{\prime\prime}=0

4.

$m=0,r=2$ Analogous reduction with

b_{i-1}=a_{0},\quad b_{i-1}^{\prime}=a_{1},\quad b_{i-1}^{\prime\prime}=-1

5.

$m=r=1$ Analogous reduction with

b_{i-1}=a_{0},\quad b_{i-1}^{\prime}=0,\quad b_{i-1}^{\prime\prime}=-1

6.

$m=2,r=0$ Analogous reduction with

b_{i-1}=b_{i-1}^{\prime}=0,\quad b_{i-1}^{\prime\prime}=-1

In cases $4^{\circ},5^{\circ},6^{\circ}$ the sign of all the corresponding numbers (53), (54) is non-positive.

If $i=1$ so $x_{0}<x_{1}$ , the first series of formulas (56) is suppressed, and if $i=s+1$ so $x_{s}<x_{0}$ , the third series of formulas (56) is deleted.

There are 5 formulas (E) of accuracy level 1 that we will report in § 5.

⁰⁰footnotetext: ¹ ) More precisely (for the sake of simplicity of exposition) the remainder changed by sign is brought to this form, a fact which does not restrict the generality of the problem.

§ 3. A criterion for recognizing the simple form of the remainder of a formula of maximum accuracy

26.

Let us resume the numerical derivation formula (E). We will first assume $r=0$ , so we will consider the formula

f^{(m)}\left(x_{0}\right)=L^{(m)}\left(x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f\mid x_{0}\right)+R

(57)

To study the rest of this formula, we can start from the case $m=0$ , first considering formula (28).

Because at some point it will be convenient to consider the point $x_{0}$ as a variable, we will then, for greater clarity, denote it by $x$ Formula (29) shows us that if we put

R(x)=l(x)\left[x,x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f\right]

(58)

the rest of formula (28) is $R=R\left(x_{0}\right)$ , and the rest of formula (57) will be given by the equality

R=R[f]=R^{(m)}\left(x_{0}\right)

(59)

Derivative of the order $m,R^{(m)}(x)$ can be calculated from formula (58). We have

R^{(m)}(x)=\sum_{i=0}^{m}\binom{m}{i}l^{(m-i)}(x)\left[x,x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f\right]^{(i)}

(

\prime

)

But [5]

\left[x,\alpha_{1},\alpha_{2},\ldots,\alpha_{li};j\right]^{(i)}=i!\underbrace{[x,x,\ldots,x}_{i+1},\alpha_{1},\alpha_{2},\ldots,\alpha_{k};f]

(60)

from which the more general formula is deduced

	$\displaystyle{[\underbrace{x,x,\ldots,x}_{j},\alpha_{1},\alpha_{2},\ldots,\alpha_{k};f]^{(i)}=}$		(61)
	$\displaystyle\qquad=\frac{(i+j-1)!}{(j-1)!}[\underbrace{x,x,\ldots,x}_{i+j},\alpha_{1},\alpha_{2},\ldots,\alpha_{1};\rho]$

where $\alpha_{1},\alpha_{2},\ldots,\alpha_{k}$ are fixed nodes and $x$ is 0 variable.
Taking into account (60), formula (59') becomes

R^{(m)}(x)=m!\sum_{i=0}^{m}\frac{l^{(m-i)}(x)}{(m-i)!}\underbrace{[x,x,\ldots,x}_{i+1},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f]

(62)

27.

We will now transform formula (62) into another formula that will prove useful.

We can write

R^{(m)}(x)=\sum_{i=0}^{m}A_{m,i}(x)[\underbrace{x,x,\ldots,x}_{i+1},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1-i}^{\prime};n]

(63)

where the coefficients $A_{m,i}(x),i=0,1,\ldots,m$ are polynomials in $x$ , completely determined and independent of the function $f(x)$ .

This formula results from a general transformation formula [6] applied to the additive and homogeneous functional (62), defined for the functions $f(x)$ data on points

\underbrace{x,x,\ldots,x}_{m+1},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime}

(64)

taken in this order.
Formula (63) is deduced from formula (62), conveniently applying to the divided differences in the second member the recurrence formula of divided differences until all these divided differences are expressed linearly and homogeneously in terms of divided differences of the order $p+1$ taken as many as $p+2$ consecutive points in the string (64).

Polynomials $A_{m,i}(x),i=0,1,\ldots,m$ can be calculated explicitly by putting them in a convenient form. Deriving formula (63) with respect to $x$ and taking into account formulas (59) and (61), we deduce

	$\displaystyle\sum_{i=0}^{m+1}A_{m+1,i}(x)[\underbrace{x,x,\ldots,x}_{i+1},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1-i}^{\prime};f]=$
$\displaystyle=$	$\displaystyle\sum_{i=0}^{m}A_{m_{i}^{\prime}}^{\prime}(x)[\underbrace{x,x,\ldots,x}_{i+1},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1-i}^{\prime};f]+$	(65)
$\displaystyle+$	$\displaystyle\sum_{i=0}^{m}(i+1)A_{m,i}(x)[\underbrace{x,x,\ldots,x}_{i+2},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1-i}^{\prime};f]$

Taking into account the recurrence formula

	$\displaystyle\left.\frac{[x,x,\ldots,x}{i+2},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1-i}^{\prime};f\right]=$
	$\displaystyle\quad=\frac{1}{x-x_{p+1-i}^{\prime}}\left\{\frac{\left[x,x,\ldots,x,x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p-i}^{\prime};f\right]}{i+2}\right.$		(66)
	$\displaystyle\quad-\left[\frac{\left.x,x,\ldots,x,x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1-i}^{\prime};f\right]}{i+1}\right.$

and identifying the coefficients in (65) we deduce

	$\displaystyle A_{m+1,i}(x)=A_{m,i}^{\prime}(x)-(i+1)\frac{A_{m,i}(x)}{x-x_{p+1-i}^{\prime}}+i\frac{A_{m,i-1}(x)}{x-x_{p+2-i}^{\prime}}$		(67)
	$\displaystyle i=0,1,\ldots,m+1,\quad\left[A_{m,1}(x)=A_{m,m+1}(x)=0\right]$

28.

To find the explicit form of polynomials $A_{m,i}(x)$ we will introduce some more useful notations. We put

l_{0}(x)=1,\quad l_{i}(x)=\left(x-x_{1}^{\prime}\right)\left(x-x_{2}^{\prime}\right),\ldots,\left(x-x_{i}^{\prime}\right),\quad i=1,2,\ldots,p+1

We then have $l_{p+1}(x)=l(x)$ and

l_{i+1}(x)=\left(x-x_{i+1}^{\prime}\right)l_{i}(x),\quad i=0,1,\ldots,p

(68)

If, for the sake of simplicity, we agree to put

l_{i,\alpha}(x)=\frac{1}{(i-\alpha)!}l_{i}^{(i-\alpha)}(x),\quad\alpha=0,1,\ldots,i

(69)

from (68) we immediately deduce the relations

	$\displaystyle l_{i+1,\alpha+1}(x)$	$\displaystyle=l_{i,\alpha+1}(x)+\left(x-x_{i+1}^{\prime}\right)l_{i,\alpha}(x)$		(70)
	$\displaystyle l_{i,\alpha}^{(\beta)}(x)$	$\displaystyle=\frac{(i-\alpha+\beta)!}{(i-\alpha)!}l_{i,\alpha-\beta}(x)$		(74)

It is always taken into account in calculations $l_{i,\alpha}(x)=0$ , if $\alpha<0$ or $\alpha>i$
We can write the nodes (6) in the form

x_{i}^{\prime}=x_{0}+h_{i},\quad i=1,2,\ldots,p+1

(72)

then $h_{1},h_{2},\ldots,h_{p+1}$ are the deviations of the respective nodes from the derivation point $x_{0}$ .

We will note with $H_{i},i=0,1,\ldots,p+1\left(H_{0}=1\right)$ fundamental symmetric polynomials of deviations $h_{i},i=1,2,\ldots,p+1$ and we will always put $H_{i}=0$ , if $i<0$ or $i>p+1$ We will also denote by $H_{i,\alpha}$ , $\alpha=0,1,\ldots,i$ the fundamental symmetric polynomials of the first $i$ DEVIATIONS $h_{1},h_{2},\ldots,h_{i}\left(H_{i,0}=1,(i\geqslant 0)\right.$ and $H_{i,\alpha}=0$ for $\alpha<0$ and $\left.\alpha>i\right)$ We then have the formulas $H_{p+1,\alpha}=H_{\alpha},\alpha=0,1,\ldots,p+1$ and

l_{i,\alpha}\left(x_{0}\right)=(-1)^{\alpha}H_{i,\alpha},\alpha=0,1,\ldots,i,i=0,1,\ldots,p+1

(73)

Considering the above, we will prove that we have

A_{m,i}(x)=m!\left(x-x_{p+1-i}^{\prime}\right)l_{p-i,p-m}(x),i=0,1,\ldots,m

(74)

For this we note that $A_{0,0}(x)=l(x)=\left(x-x_{p+1}^{\prime}\right)l_{p,p}(x)$ and therefore for $m=0$ , formulas (74) are true. Let us assume that these formulas are true for $m$ and let's demonstrate them for $m+1$ . Taking into account (67) and (74), we deduce

	$\displaystyle A_{m+1,i}(x)$	$\displaystyle=m!\left[l_{p-i,p-m}(x)+\left(x-x_{p+1-i}^{\prime}\right)l_{p-i,p-m}^{\prime}(x)-\right.$
		$\displaystyle\left.-(i+1)l_{p-i,p-m}(x)+il_{p-i+1,p-m}(x)\right]$		(75)

But from (70), (71) it follows that

\begin{gathered}l_{p-i,p-m}^{\prime}(x)=(m-i+1)l_{p-i,p-m-1}(x)\\ l_{p-i+1,p-m}(x)=l_{p-i,p-m}(x)+\left(x-x_{p+1-i}^{\prime}\right)l_{p-i,p-m}(x)\end{gathered}

and substituting in (75), we find

A_{m+1,i}(x)=(m+1)!\left(x-x_{p+1-i}^{\prime}\right)l_{p-i,p-m-1}(x),i=0,1,\ldots,m+1

which shows that formulas (74) are general.
29. Applying Abel's transformation formula to formula (63), we have

	$\displaystyle R^{(m)}(x)=\sum_{i=0}^{m-1}\left[\sum_{\alpha=1}^{m-i}A_{m,i+\alpha}(x)\right]\left\{\frac{\left[x,x,\ldots,x,x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p-i}^{\prime};f\right]}{i+2}\right.$		(76)
	$\displaystyle\left.-\left[\frac{x,x,\ldots,x}{i+1},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1-i}^{\prime};f\right]\right\}+\left[\sum_{i=0}^{m}A_{m,i}(x)\right]\left[x,x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1};f\right].$

If, however, in formula (63) we put $f(x)=x^{p+1}$ and taking into account (30) we deduce

\sum_{i=0}^{m}A_{m,i}(x)=l^{(m)}(x)

(77)

We observe that from (70), (74) it follows

\begin{gathered}\sum_{\alpha=1}^{m-i}A_{m,i+\alpha}(x)=m!\sum_{\alpha=1}^{m-i}\left[l_{p+1-i-\alpha,p-m+1}(x)-l_{p+i-\alpha,p-m+1}(x)\right]=\\ =m!l_{p-i,p-m+1}(x)\end{gathered}

Taking into account the recurrence formula (66) and (77), formula (76) becomes

	$\displaystyle R^{(m)}(x)=m\sum_{i=0}^{m-1}A_{m-1,i}(x)[\underbrace{x,x,\ldots,x}_{i+2},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1-i}^{\prime};f]+$		(78)
	$\displaystyle+l^{(m)}(x)\left[x,x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f\right].$

Formulas (63), (78) remain, of course, whatever $x$ (and if $x$ coincides with one of the nodes $x_{i}^{\prime}$ ).
30. Numbers $A_{m,i}(x),i=0,1,\ldots,m$ cannot all be zero. Otherwise, we would have $R[f]=0$ , whatever the function $f(x)$ , which is in contradiction with the fact that formula (57) has, based on the assumptions made in point 2, a certain degree of accuracy. At least one of the numbers $A_{m,i}(x),i=0,1,\ldots,m$ is different from zero. This result, as well as the following ones, holds if we slightly broaden the assumptions from point 2. We can assume, more generally, that $x_{0}$ may also coincide with a node. But we must then assume that

m\geq r_{i}\text{ dacă }x_{0}=x_{i},\quad i=1,2,\ldots,s

(79)

If $x_{0}=x_{i}$ and $m<r_{i}$ the rest of formula (57) is null whatever the function $f(x)$ .
Suppose the numbers $A_{m,i}\left(x_{0}\right)$ are all of the same sign ¹ ). Then, from the above observation and from (77), it follows that $l^{(m)}\left(x_{0}\right)\neq 0$ Therefore, the degree of accuracy of formula (57) is $p$ To fix the ideas, let's assume that

A_{m,j}\left(x_{0}\right)\neq 0

(80)

From formula (74) it then follows that

x_{0}\neq x_{p+1-i}^{\prime}

(81)

⁰⁰footnotetext: ¹ ) See reference ¹ ), on page 75.

If now $f(x)$ is a convex function of order $p$ , we have

\underbrace{\left[x_{0},x_{0},\ldots,x_{0},\right.}_{i+1}x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1-i}^{\prime};f]\geq 0,i=0,1,\ldots,m.

(82)

But, a convex function of order $p(\geqq 0)$ enjoys the property that differences divided by the order $p+1$ its, on nodes not all confused, are positive. From (81) it follows that

\left[\frac{x_{0},x_{0},\ldots,x_{0}}{j+1},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1-i}^{\prime};f\right]>0.

(83)

From formulas (74), (80), (82), (83) it then follows that we have $R[f]\neq 0$ Based on criterion C
we can state the following theorem:
Theorem 12. If the numbers $\mathrm{A}_{\mathrm{m},1}\left(\mathrm{x}_{0}\right),\mathrm{i}=0,1,\ldots,\mathrm{\penalty 10000\ m}$ are all of the same sign, the degree of accuracy of formula (57) is p and the rest is of simple form.

Under the conditions of the theorem, the remainder is therefore of the form

R=l^{(m)}\left(x_{0}\right)D_{p+1}[f].

If we assume that $l^{(m)}\left(x_{0}\right)=0$ , an absolutely analogous demonstration based on formula (78), allows us to state

Theorem 13. If ${}^{(\mathrm{m})}\left(\mathrm{x}_{0}\right)=0$ and if the numbers $\mathrm{A}_{\mathrm{m}-1,\mathrm{i}}\left(\mathrm{x}_{0}\right)$ , $\mathrm{i}=0,1,\ldots,\mathrm{\penalty 10000\ m}-1$ are of the same sign, the degree of accuracy of formula (57) is equal to p + 1 and the remainder is of simple form.

Under the conditions of the theorem, the remainder is of the form

R=ml^{m-11}\left(x_{0}\right)D_{p+2}[f]

because from (62) we deduce

R\left[x^{p+2}\right]=ml^{(m-1)}\left(x_{0}\right).

Theorems 12, 13 are valid under hypothesis (79). Indeed, if $r_{i}\leqq m$ and $x_{0}=x_{i}$ , equalities $l^{(m)}\left(x_{0}\right)=0$ and $l^{(m-1)}\left(x_{0}\right)=0$ are incompatible by Lemma 1.
31. Let us now consider the general case of formula (E). We will show how this case can be reduced to the case $r=0$ .

If we consider the polynomial of degree $r-1$

L(x)=L\underbrace{\left(x_{0},x_{0},\ldots,x_{0}\right.}_{r};f\mid x)

(84)

and if we take into account formulas (28), (29), we deduce

\frac{f(x)-L(x)}{\left(x-x_{0}\right)^{r}}=[\underbrace{x,x_{0},x_{0},\ldots,x_{0}}_{r};f].

(85)

Now applying the formula

L(\underbrace{x_{0},x_{0},\ldots,x_{0}}_{r},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f\mid x)=

(86)

$=l(x)L(\underbrace{x_{0},x_{0},\ldots,x_{0}}_{r};\left.\frac{f(x)}{l(x)}\right\rvert\,x)+\left(x-x_{0}\right)^{r}L\left(x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};\left.\frac{f(x)}{\left(x-x_{0}\right)^{r}}\right\rvert\,x\right)$
Service $f(x)-L(x)$ and taking into account the relationships

\begin{gathered}L(\underbrace{x_{0},x_{0}}_{r},\ldots,x_{0};\left.\frac{f(x)-L(x)}{\left(x-x_{0}\right)^{r}}\right\rvert\,x)=0\\ L(\underbrace{x_{0},x_{0},\ldots,x_{0}}_{r},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f(x)-L(x)\mid x)=\\ =L(\underbrace{x_{0},x_{0},\ldots,x_{0}}_{r},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f\mid x)-L(x)\end{gathered}

deduce

\begin{gathered}L(\underbrace{x_{0},x_{0},\ldots,x_{0}}_{r},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f\mid x)=\\ =L(x)+\left(x-x_{0}\right)^{r}L(x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};[x,\underbrace{x_{0},x_{0},\ldots,x_{0}}_{r};f]\mid x)\end{gathered}

from where

	$\displaystyle L^{(m+r)}(\underbrace{x_{0},x_{0},\ldots,x_{0}}_{r},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f\mid x_{0})=$
	$\displaystyle\Rightarrow\frac{(m+r)!}{m!}L^{(m)}(x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};[x,\underbrace{x_{0},x_{0},\ldots,x_{0}}_{r};f]\mid x_{0})$		(87)

But, from (85) it also follows

f^{(m+r)}\left(x_{0}\right)=\frac{(m+r)!}{m!}[x,\underbrace{x_{0},x_{0},\ldots,x_{0}}_{r}]_{x=x_{0}}^{(m)}

(88)

Let's note with $R[f]$ the rest of the formula (E) and with $R_{1}[f]$ the rest of formula (57). We then have, taking into account (87) and (88),

R[f]=\frac{(m+r)!}{m!}R_{1}[[x,\underbrace{x_{0},x_{0},\ldots,x_{0}}_{\boldsymbol{0}};f]]

(89)

and formula (E) can be written

	$\displaystyle f^{(m+r)}\left(x_{0}\right)=\frac{(m+r)!}{m!}L^{(m)}(x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};[x,\underbrace{x_{0},x_{0},\ldots,x_{0}}_{r};f]\mid x_{0})+$
	$\displaystyle+\frac{(m+r)!}{m!}R_{1}[[x,\underbrace{x_{0},x_{0},\ldots,x_{0}}_{r};f]]$		(90)

32.

To proceed further, we will first demonstrate

Le ma 4. If the additive and homogeneous functional $\mathrm{R}_{1}$ [f] is of degree of accuracy $\mathrm{n}(\geqq 0)$ and of simple form, then the additive and homogeneous functional

R[f]=K\cdot R_{1}\left[\left[x,\alpha_{1},\alpha_{2},\ldots,\alpha_{k};f\right]\right]

(91)

where $K\neq 0$ is a constant (independent of the function $f(x)$ ) and $\alpha_{1},\alpha_{2},\ldots,\alpha_{k}$ there are k fixed points, it is of degree of accuracy n + k and it is of simple form.

The proof is easy. We have $\left[x,\alpha_{1},\alpha_{2},\ldots,\alpha_{k};x^{i}\right]=0$ , for $i=0,1,\ldots,k-1$ , and for $i\geq k$ this divided difference is a polynomial of degree $i-k$ , with the first coefficient 1 , so

\left[x,\alpha_{1},\alpha_{2},\ldots,\alpha_{k};x^{i}\right]=x^{i-k}+\ldots

From here follows the property relative to the degree of accuracy.
Let us now suppose that $f(x)$ is a convex function of order $n+k$ We then say that the function $x\left[x,\alpha_{1},\alpha_{2},\ldots,\alpha_{k};f\right]$ is convex of order n. Indeed, we have
$\left[\beta_{1},\beta_{2},\ldots,\beta_{n+2};\left[x,\alpha_{1},\alpha_{2},\ldots,\alpha_{k};f\right]\right]=\left[\beta_{1},\beta_{2},\ldots,\beta_{n+2},\alpha_{1},\alpha_{2},\ldots\alpha_{k};f\right]>0$
if the nodes are not all confused. Equality (91) shows us that $R[f]\neq 0$ , for any function $f(x)$ convex of the order $n+k$ , because $R_{\mathbf{1}}[\varphi]\neq 0$ for any function $\varphi(x)$ convex of the order $n$ , so in particular, $R_{1}\left[\left[x,\alpha_{1},\alpha_{2},\ldots,\alpha_{k};f\right]\right]\neq 0$ .

Lemma 4 is completely proven.
Taking into account formula (89), from theorems 12, 13 we deduce, based on the previous lemma

Theorem 14. If $\mathrm{r}\geqslant 0$ is integer and if the numbers $\mathrm{A}_{\mathrm{m},\mathrm{i}}\left(\mathrm{x}_{0}\right)$ , $\mathrm{i}=0,1,\ldots,\mathrm{\penalty 10000\ m}$ are of the same sign, the numerical derivation formula $(E)$ has the degree of accuracy $\mathrm{p}+\mathrm{r}$ and the rest of simple form.

Based on formula (12) the remainder is then written

R=\frac{(m+r)!}{m!}l^{(m)}\left(x_{0}\right)D_{p+r+1}[f]

(92)

Theorem 15. If $\mathrm{r}\geq 0$ is whole, if $1^{(\mathrm{m})}\left(\mathrm{x}_{0}\right)=0$ and if the numbers $\mathrm{A}_{\mathrm{m}-1,\mathrm{i}}\left(\mathrm{x}_{0}\right),\mathrm{i}=0,1,\ldots,\mathrm{\penalty 10000\ m}-1$ are of the same sign, the numerical derivation formula (E) has the degree of accuracy $\mathrm{p}+\mathrm{r}+1$ and the rest of simple form.

Based on formulas (12) the remainder is then written

R=\frac{(m+r)!}{(m-1)!}l^{(m-1)}\left(x_{0}\right)D_{p+r+2}[f]

Theorems 14, 15 are valid under the hypothesis expressed by relation (79).
33. From what has been said it follows that formula (E) certainly has the remainder of simple form if $m=0$ . Formula (E) for $m=0$ has the degree of accuracy $p+r$ and the rest

R=l\left(x_{0}\right)D_{p+r+1}[f]

in the assumptions of point 2. Under the same conditions, formula (E) has the degree of accuracy $p+r+1$ and the rest

R=l\left(x_{0}\right)D_{p+r+2}[f]

if $m=1$ and $x_{0}$ is a root, different from nodes, of $l^{\prime}(x)$ .
The condition imposed on numbers $A_{m,i}\left(x_{0}\right),i=0,1,\ldots,m$ , respectively the numbers $A_{m-1},i\left(x_{0}\right),i=0,1,\ldots,m-1$ , is sufficient for formula (E) to have the remainder in simple form. Number formation $A_{m,i}\left(x_{0}\right)$ depends on the order in which the nodes are taken (6). More precisely, a number system $i=0,1,\ldots,m$ is characterized by the order in which the clouds are taken. In order for formula (E) to be of a certain degree of accuracy $p+r$ and to be of simple form, it is sufficient that one of these systems be added to the remaining numbers of the same sign. We do not examine here the necessity of this general form. However, it follows from § 4 that in the case of $m=1$ the condition is conditions in

§ 4. On some applications of the preceding results

We will apply the previous formulas to the following 3 examples: 34. Example 1. The point of derivation $x_{0}$ is outside the smallest open interval containing the nodes (2).

If we assume that, more generally, the hypothesis expressed by condition (79) is fulfilled, we have

l^{(m)}\left(x_{0}\right)\neq 0

(93)

so formula (E) is of degree of accuracy $p+\mathrm{r}$ and, in particular, is never exceptional. Indeed, the roots of $l^{(m)}(x)$ belong to the smallest closed interval containing the nodes (2). If $x_{0}$ does not belong to this interval, (93) is proven. Suppose that $x_{0}$ coincides with an extreme node $x_{i}$ . In this case $l^{(\alpha)}\left(x_{i}\right)\neq 0$ for $\alpha\geq r_{i}$ But $m\geq r_{i}$ , based on hypothesis (79), so (93) results this time as well.

Deviations ¹ ) $h_{i},i=1,2,\ldots,p+1$ of the nodes of the derivation point are in this case of the same sign, namely non-positive, respectively non-negative, as

x_{0}\geqq x_{i}^{\prime},i=1,2,\ldots,p+1,\text{ respectiv }x_{0}\leqq x_{i}^{\prime},i=1,2,\ldots,p+1

Formula (73) then shows us that

l_{\alpha,\gamma}\left(x_{0}\right)=(-1)_{\Sigma}^{\gamma}H_{\alpha,\gamma}\geq 0,\operatorname{respectiv}(-1)^{\dot{\gamma}}l_{\alpha,\gamma}\left(x_{0}\right)=H_{\alpha,\gamma}\geq 0,

whatever $\alpha$ and $\gamma$ .

⁰⁰footnotetext: ¹⁾ See point 28.

Taking into account formulas (74), we then have, respectively,

\begin{gathered}A_{m,i}\left(x_{0}\right)=(-1)^{p-m+1}m!h_{p+1-i}H_{p-i,p-m}\geqq 0,i=0,1,\ldots,m,\\ (-1)^{p-m+1}A_{m,i}\left(x_{0}\right)=m!h_{p+1-i}H_{p-i,p-m}\geqq 0,i=0,1,\ldots,m.\end{gathered}

All the hypotheses in which Theorem 14 was established are therefore fulfilled and we can state

Theorem 16. If $\mathrm{r}\geq 0$ is integer, m satisfies the condition expressed by (79) and $\mathrm{x}_{0}$ is outside the smallest open interval containing the nodes (2), we have the numerical derivation formula ( $E$ ), with the degree of accuracy $\mathrm{p}+\mathrm{r}$ and with the remainder of the form (92).

The assumptions under which the theorem is true require that in the case when $s=1$ , the point $x_{0}$ to be different from the nodes.
35. Example 2. The nodes (2) are symmetrically distributed with respect to the derivation point $x_{0}$ To simplify the language we will say, in this case, that formula (E) is a symmetric formula.

In the symmetric case, we can assume that the point of derivation $x_{0}$ is different from the nodes, so that the deviations $h^{i}$ are different from zero. The number of nodes is then even and equal to $2q=p+1$ , where $q>0$ For greater clarity we will denote by $\pm t_{i},i=1,2,\ldots,q$ deviations, where $t_{1},t_{2},\ldots,t_{q}$ are $q$ positive numbers (distinct or not). We will denote by $T_{i},i=0,1,\ldots,q$ fundamental symmetric polynomials of numbers $t_{1}^{2},t_{2}^{2},\ldots,t_{\eta}^{2}\left(T_{0}=1,T_{i}=0\right.$ , for $i<0$ and for $i>q$ ). We will also denote by $T_{i,\alpha},\alpha=0,1,\ldots,i$ fundamental symmetric polynomials of numbers $t_{1}^{2},t_{2}^{2},\ldots,t_{i}^{2}\left(T_{i,0}=1\right.$ , $T_{i,\alpha}=0$ , for $\alpha<0$ and for $\alpha>i$ ). We then have $T_{q,i}=T_{i},i=0,1,\ldots,q$ .

l^{(m)}\left(x_{0}\right)=\begin{cases}0,&\text{ dacă }m\text{ este impar }\\ (-1)^{\frac{m}{2}}m!T_{q-\frac{m}{2}}&\text{ dacă }m\text{ este par. }\end{cases}

It follows that $l^{(m)}\left(x_{0}\right)\neq 0$ or $=0$ , as $m$ is even or odd. Formula (E) is in this case exceptional if $r>0$ , and the results from point 11 show us that it is enough to consider only the case when $m$ it is even, the case $m$ odd reducing to this by changing the number $r$ .

In case the derivation index $m$ is even, we will denote it by $2\mu$ , so $m=2\mu$ , where $0\leq\mu<q$ .

Let's now take the nodes in the following order:

x_{0}-t_{1},x_{0}+t_{1},x_{0}-t_{2},x_{0}+t_{2},\ldots,x_{0}-t_{q},x_{0}+t_{q}

We have, so but ¹ )

h_{\alpha}=x_{\alpha}^{\prime}-x_{0}=(-1)^{\alpha}t_{\left[\frac{\alpha+1}{2}\right]},\quad\alpha=1,2,\ldots,2q.

(94)

To calculate numbers $A_{2\mu,i}\left(x_{0}\right)$ we will first calculate the corresponding numbers (73), where $H_{i,\alpha}$ correspond to deviations (94).

⁰⁰footnotetext: ¹ ) [z] means the largest integer contained in z.

If $\alpha$ it is hair, $\mathrm{H}_{\alpha,\gamma},\gamma=0,1,\ldots,\alpha$ are the fundamental symmetric polynomials of the numbers $\pm t_{i},i=1,2,\ldots,\frac{\alpha}{2}$ An elementary calculation shows us that

H_{\alpha,\gamma}=\left\{\begin{array}[]{ll}0,&\text{ dacă }\gamma\text{ este impar }\\ (-1)^{\frac{\gamma}{2}}T_{\frac{\alpha}{2}},\frac{\gamma}{2},&\text{ dacă }\gamma\text{ este par. }\end{array}\quad(\alpha\text{ par })\right.

If $\alpha$ it is odd $H_{\alpha,\gamma},\gamma=0,1,\ldots,\alpha$ are the fundamental symmetric polynomials of the numbers $\pm t_{i},i=1,2,\ldots,\frac{\alpha-1}{2}$ and $\frac{t_{\alpha+1}}{2}$ . Formulas (70), (73) together with (94), (95) give us

H_{\alpha,\gamma}=\begin{cases}(-1)^{\frac{\gamma-1}{2}}t_{\frac{\alpha+1}{2}}T_{\frac{\alpha-1}{2},\frac{\gamma-1}{2},}&\text{ dacă }\gamma\text{ este impar }\\ (-1)^{\frac{\gamma}{2}}T_{\frac{\alpha-1}{2},\frac{\gamma}{2}},&\text{ dacă }\gamma\text{ este par. }\end{cases}

Formulas (74) show us that we have

A_{2\mu,i}\left(x_{0}\right)=\left\{\begin{array}[]{cc}0,&\text{ dacă }i\text{ este impar }\\ (-1)^{q-\mu}(2\mu)!t_{q-\frac{i}{2}}^{2}T_{q-1-\frac{i}{2},q-\mu-1},&\text{ dacă }i\text{ este par. }\end{array}\right.

It is clear, however, that we have

(-1)^{q-\mu}A_{2\mu,i}\left(x_{0}\right)\geqq 0,\quad i=0,1,\ldots,2q

and therefore that all the conditions of Theorem 14 are fulfilled. We can then
state Theorem 17. If $\mathrm{r}\geqq 0$ is an integer, if $\mu$ is an integer, so $0\leq\mu<q$ , and $\mathrm{t}_{1},\mathrm{t}_{2},\ldots,\mathrm{t}_{q},q\geqq 1$ positive numbers, we have the numerical derivative formula ¹ )

\begin{gathered}f^{(2\mu+r)}\left(x_{0}\right)=L^{(2\mu+r)}(x_{0},\underbrace{x_{0},\ldots,x_{0}}_{r},x_{0}\pm t_{1},x_{0}\pm t_{2},\ldots,x_{0}\pm t_{q};f\mid x_{0})+\\ +(-1)^{q-\mu}(2\mu+r)!T_{q-\mu}D_{2q+r}[f]\end{gathered}

the degree of accuracy $2q+r-1$ .
36. Example 3. Derivation index $m$ is equal to 1.

To examine this case, we will assume that nodes (2), and therefore (6), are in non-decreasing order, so

x_{1}^{\prime}\leqq x_{2}^{\prime}\leqq\ldots\leqq x_{p+1}^{\prime}

as well as

x_{1}<x_{2}<\ldots<x_{s}.

(96)

⁰⁰footnotetext: 1. To simplify the notations in the interpolation polynomial and in the divided difference, we denote by

x_{0}\pm t_{i}

the two symmetrical nodes

x_{0}-t_{i},x_{0}+t_{i}

provided $m=1$ GET INVOLVED $p\geqq 1$ We will assume $s>1$ , so that the nodes are not all confused. Otherwise, we return to example 1 studied above, which then exhausts the problem.

For simplicity, let us denote by $\varphi(x)$ polynomial $l_{p}(x)$ So we have

\varphi(x)=\left(x-x_{p+1}^{\prime}\right)\varphi(x)

(97)

and

A_{1,0}(x)=\left(x-x_{p+1}^{\prime}\right)\varphi^{\prime}(x),\quad A_{1,1}(x)=\varphi(x)

from where

A_{1,0}(x)A_{1,1}(x)=l(x)\varphi^{\prime}(x)

if the polynomials $A_{1,0}(x),A_{1,1}(x)$ are constructed by taking the nodes in order (6).

The condition that these polynomials be of the same sign therefore becomes

l(x)\varphi^{\prime}(x)\geqq 0.

(98)

Analogously, if we put

l(x)=\left(x-x_{1}^{\prime}\right)\psi(x)

\psi(x)=\left(x-x_{2}^{\prime}\right)\left(x-x_{3}^{\prime}\right)\ldots\left(x-x_{p+1}^{\prime}\right)

and if we form the polynomials $A_{1,0}(x),A_{1,1}(x)$ taking the knots in reverse order $x_{p+1}^{\prime},x_{p}^{\prime},\ldots,x_{1}^{\prime}$ , the condition for these polynomials to be of the same sign is that

l(x)\Psi^{\prime}(x)\geqq 0.

(99)

The hypothesis expressed by condition (79) here comes back to the fact that $x_{0}$ does not coincide with a node which is not simple.

Both inequalities (98), (99) are verified for $x=x_{0}$ , if $x_{0}$ coincides with a node or if $x_{0}$ is outside the smallest interval containing the nodes. We are then in the case of example 1 above.

If we have

\varphi^{\prime}(x)\psi^{\prime}(x)\leqq 0

(100)

one of the inequalities (98), (99) is verified for $x=x_{0}$ . So if, in this case, $x_{0}$ does not coincide with a node that is not simple, our formula has the degree of accuracy $p+r$ and the rest of simple form. It will therefore be sufficient to examine the point $x_{0}$ from the open interval ( $x_{1},x_{s}$ ), for which inequality (100) is not verified. If in addition, in this case, the degree of accuracy is $p+r$ , we will see that the remainder is not simple.
37. To show this, we will rely on criterion C and first prove

Lemma 5. If $\mathrm{x}_{0}\in\left(\mathrm{x}_{1},\mathrm{x}_{s}\right)$ and if

\varphi^{\prime}\left(x_{0}\right)\Psi^{\prime}\left(x_{0}\right)>0

(101)

we can find a convex function $\mathbf{f}(\mathbf{x})$ of the order $\mathrm{p}+\mathrm{r}$ in the interval $\left[\mathbf{x}_{1},\mathbf{x}_{s}\right]$ , so that we have $\mathrm{R}[\mathrm{f}]=0$ .

Let us assume that condition (79) is fulfilled.
Based on Lemma 3, it is enough to construct two functions $f_{1}(x),f_{2}(x)$ non-concave of the order $p+r$ , so that inequality (32) is verified. For this, let us consider the functions

\varphi_{\lambda}(x)=\left\{\begin{array}[]{cl}0,&\text{ pentru }x\in\left[x_{1},\lambda\right]\\ (x-\lambda)^{p+r},&\text{ pentru }x\in\left[\lambda,x_{s}\right]\end{array}\right.

defined for $\lambda\in\left[x_{1},x_{s}\right]$ .
Function $\varphi_{\lambda}(x)$ is non-concave of the order $p+r$ in [ $x_{1},x_{s}$ ]. If we look at $\lambda$ as a variable, $R\left[\varphi_{\lambda}\right]$ is a polynomial of degree $p+r$ in relation to $x-\lambda$ , in any interval that does not contain $x_{0}$ and the nodes.

	We have the formula
	$\displaystyle\qquad\begin{array}[]{l}R\left[\varphi_{\lambda}\right]=\left(x_{0}-x_{s}\right)\varphi^{\prime}\left(x_{0}\right)[\underbrace{x_{0},x_{0},\ldots,x_{0}}_{0},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p}^{\prime};\varphi_{\lambda}]+\\ \qquad+\varphi\left(x_{0}\right)[\underbrace{x_{0},x_{0},\ldots,x_{0}}_{r},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};\varphi_{\lambda}]\end{array}$

If we assume $x_{s-1}<\lambda<x_{s}$ and $x_{s}<\lambda$ , then it is seen that $R\left[\varphi_{\lambda}\right]$ is a polynomial in $x_{s}-\lambda$ , which divides by $\left(x_{s}-\lambda\right)^{p+r-r_{s}+1}$ and the coefficients of this power of $x_{s}-\lambda$ come only from the first term of the second member of formula (103). But then we have

		$\displaystyle{[\underbrace{x_{0},x_{0},\ldots,x_{0}}_{r},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};\varphi_{\lambda}]=}$
		$\displaystyle=[\underbrace{x_{s},x_{s},\ldots,x_{s}}_{r_{s}};\frac{(x-\lambda)^{p+r}}{\left(x-x_{0}\right)^{r+1}k_{s}(x)}]=$
		$\displaystyle=\frac{1}{\left(r_{s}-1\right)!}\left[\frac{(x-\lambda)^{p+r}}{\left(x-x_{0}\right)^{r+1}k_{s}(x)}\right]_{x=x_{0}}^{\left(r_{s}-1\right)}$

polynomial $k_{s}(x)$ being given by $l(x)=\left(x-x_{0}\right)^{r_{s}}k_{s}(x)$ .
Doing the calculations we find

R\left[\varphi_{\lambda}\right]=\left(x_{s}-\lambda\right)^{p+r-r_{s}+1}\left[\binom{p+r}{r_{s}-1}\frac{\left(x_{0}-x_{s}\right)\varphi^{\prime}\left(x_{0}\right)}{\left(x-x_{0}\right)^{r+1}k_{s}\left(x_{s}\right)}+\cdots\right]

the unwritten terms being divisible by $x_{s}-\lambda$ From this
it is seen that, if $\lambda$ it is close enough to $x_{s}$ HAVE $R\left[\varphi_{\lambda}\right]\neq 0$ , and more precisely ¹ )

\operatorname{sg}R\left[\varphi_{\lambda}\right]=-\operatorname{sg}\varphi^{\prime}\left(x_{0}\right).

(104)

⁰⁰footnotetext: 1. We put sg

z=-1,0,1

, as

z<0,z=0,z>0

We have the fundamental relationship

\operatorname{sg}z_{1}z_{2}=\operatorname{sg}z_{1}\cdot\operatorname{sg}z_{2}

function $\psi_{\lambda}(x)=\varphi_{\lambda}(x)-(x-\lambda)^{p+r}$ is also non-concave of the order $p+r$ If we assume $\lambda<x_{0},\lambda\in\left(x_{1},x_{2}\right)$ and we use the formula

	$\displaystyle R\left[\psi_{\lambda}\right]=\left(x_{0}-x_{1}\right)\varphi^{\prime}\left(x_{0}\right)$	$\displaystyle{[\underbrace{x_{0},x_{0},\ldots,x_{0}}_{r+1},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};\psi_{\lambda}]+}$
		$\displaystyle+\psi\left(x_{0}\right)\left[\frac{x_{0},x_{0},\ldots,x_{0}}{r+2},x_{2}^{\prime},x_{3}^{\prime},\ldots,x_{p+1}^{\prime};\psi_{\lambda}\right]$

we find, as above, that the polynomial $R\left[\psi_{\lambda}\right]$ in $x_{1}-\lambda$ is of the form

R\left[\psi_{\lambda}\right]=-\left(x_{1}-\lambda\right)^{p+r-r_{1}+1}\left[\binom{p+r}{r_{1}-1}\frac{\left(x_{0}-x_{1}\right)\psi^{\prime}\left(x_{0}\right)}{\left(x_{1}-x_{0}\right)^{r+1}k_{1}\left(x_{1}\right)}+\cdots\right]

where the polynomial $k_{1}(x)$ is given by $l(x)=\left(x-x_{1}\right)^{r_{1}}k_{1}(x)$ , and the unordered terms are divided by $x_{1}-\lambda$ From this it is seen that, if $\lambda$ it is close enough to $x_{1}$ , we have $R\left[\psi_{\lambda}\right]\neq 0$ , and more precisely

\operatorname{sg}R\left[\psi_{\lambda}\right]=\operatorname{sg}\psi^{\prime}\left(x_{0}\right)

(105)

Formulas (104), (105) were deduced in addition to the hypotheses $\varphi^{\prime}\left(x_{0}\right)\neq 0$ , $\psi^{\prime}\left(x_{0}\right)\neq 0$ .

Let us now consider the functions

f_{1}(x)=\varphi_{\lambda_{s}}(x),\quad f_{2}(x)=\psi_{\lambda_{1}}(x)

Then, if $\lambda_{1}$ it is close enough to $x_{1}$ and $\lambda_{s}$ close enough to $x_{s}$ , from (101), (104), (105), we deduce

\operatorname{sg}R\left[f_{1}\right]R\left[f_{2}\right]=-\operatorname{sg}\varphi^{\prime}\left(x_{0}\right)\psi^{\prime}\left(x_{0}\right)=-1

and inequality (32) is verified.
Lemma 5 is completely proven.
From the above it follows that, if $m=1$ and if $x_{0}$ does not coincide with a node that is not simple, the numerical derivation formulas (E) present one of the following 3 aspects:
$1^{\circ}$ If $l^{\prime}\left(x_{0}\right)=0$ , the degree of accuracy is $p+r+1$ , with the rest of the form simple.
$2^{\circ}$ If $x_{0}\in\left(x_{1},x_{s}\right),l^{\prime}\left(x_{0}\right)\neq 0$ and if inequality (101) is verified, the degree of accuracy is $p+r$ , but the rest is not of simple form.
$3^{\circ}$ For all other values of $x_{0}$ , the degree of accuracy is $p+r$ and the rest of the simple form.
38. We can determine the position of the point more precisely $x_{0}$ after the three reported cases.

Whether $y_{1},y_{2},\ldots,y_{s-1}$ the distinct roots of the nodes, of the derivative $l^{\prime}(x)$ These roots are separated from the nodes (96) and we can assume

x_{i}<y_{i}<x_{i+1},\quad j=1,2,\ldots,s-1

polynomial $\varphi^{\prime}(x)$ has $s-2$ , respectively $s-1$ , different roots of nodes as $r_{s}=1$ , respectively $r_{s}>1$ Let us denote these roots, in their ascending order, by $z_{1}^{\prime},z_{2}^{\prime},\ldots,z_{s-1}^{\prime}$ , where $z_{s-1}^{\prime}$ does not exist if $r_{s}=1$ We then have

x_{j}<z_{j}^{\prime}<y_{i+1},\quad j=1,2,\ldots,s-2

and if $r_{s}>1$ , these inequalities also hold for $j=s-1$ From the well -
known property of the variation of the roots of the derivative of a polynomial with all real roots [8], it follows that

y_{j}<z_{j}^{\prime},\quad j=1,2,\ldots,s-2

(106)

inequality that also occurs for $j=s-1$ , if $r_{s}>1$
Let us now, by definition , $z_{s-1}^{\prime}=x_{s}$ , if $r_{s}=1$ . Then, inequality (106) is always true for $j=s-1$ .

In the intervals between the nodes, $\varphi^{\prime}(x)$ , so and $l(x)\varphi^{\prime}(x)$ , changes sign, passing through the points $z_{1}^{\prime},z_{2}^{\prime},\ldots$ But, from (97) it follows that

l\left(y_{i}\right)\varphi^{\prime}\left(y_{j}\right)=-\varphi^{2}\left(y_{j}\right)<0

It follows, therefore, that inequality (98) is verified only in the intervals

\left(-\infty,x_{1}\right],\left[z_{1}^{\prime},x_{2}\right],\left[z_{2}^{\prime},x_{3}\right],\ldots,\left[z_{s-2}^{\prime},x_{s-1}\right],\left[z_{s-1}^{\prime},\infty\right)

Let us also note with $z_{1}^{\prime\prime},z_{2}^{\prime\prime},\ldots,z_{s-1}^{\prime\prime}$ the roots, different from the nodes, of $\Psi^{\prime}(x)$ , where $z_{1}^{\prime\prime}$ does not exist if $r_{1}=1$ As above, it is seen that

x_{j}<z_{j}^{\prime\prime}<y_{j}<x_{j+1},\quad j=1,2,\ldots,s-1

inequalities that are always true, agreeing to put, by definition, $z_{1}^{\prime\prime}=x_{1}$ if $r_{1}=1$ .

It then follows, as above, that inequality (105) is verified only in the intervals

\left(-\infty,z_{1}^{\prime\prime}\right],\left[x_{2},z_{2}^{\prime\prime}\right],\left[x_{3},z_{3}^{\prime\prime}\right],\ldots,\left[x_{s-1},z_{s-1}^{\prime\prime}\right],\left[x_{s},\infty\right)

From the previous analysis it finally results
that Te o rema 18. If $\mathbf{r}\geq 0$ is an integer, if $\mathbf{x}_{0}$ does not coincide with a node that is not simple and if $\mathrm{m}=1$ , the numerical derivation formula (E) is:
$1^{\circ}$ . With the degree of accuracy $\mathrm{p}+\mathrm{r}+1$ and like the rest of the simple form, if $\mathrm{x}_{0}$ coincide $cu$ one of the points $\mathrm{y}_{1},\mathrm{y}_{2},\ldots,\mathrm{y}_{5-1}$ .
$2^{\circ}$ . With the degree of accuracy $\mathrm{p}+\mathrm{r}$ and with the rest of the simple form, if $\mathrm{x}_{0}$ belongs to one of the intervals

\left.\left(-\infty,z_{1}^{\prime\prime},\right]z_{s-1}^{\prime},\infty\right),\left[z_{j}^{\prime},z_{j+1}^{\prime\prime}\right],\quad j=1,2,\ldots,s-2

$3^{\circ}.Cu$ degree of accuracy $\mathrm{p}+\mathrm{r}$ , but with a remainder different from the simple form, if $\dot{\mathrm{x}}_{0}$ belongs to one of the intervals

\left(z_{i}^{\prime\prime},y_{i}\right),\left(y_{i},z_{i}^{\prime}\right),\quad j=1,2,\ldots,s-1

The rest, in cases $1^{0}$ and $2^{0}$ respectively is

R=(r+1)!l\left(x_{0}\right)D_{p+r+2}[f],R=(r+1)!l^{\prime}\left(x_{0}\right)D_{p+r+1}[f]

These results were found, in a different way and in a less general form, by GD Birkhoff [1].
39. From the previous results a conclusion can be drawn about formula (E) in the case when $m=2$ In this case, if $l^{\prime\prime}(x)=0$ and the numbers $A_{1,0}\left(x_{0}\right)$ , $A_{1,1}\left(x_{0}\right)$ are of the same sign, the rest of the formula is of simple form. To satisfy the sign condition it will be sufficient to prove the following lemma:

Lemma 6. The roots of the second derivative $1^{\prime\prime}(\mathrm{x})$ of the polynomial $1(\mathrm{x})$ are all contained in the open intervals

\left(-\infty,z_{1}^{\prime\prime}\right),\left(z_{s-1}^{\prime},\infty\right),\left(z_{j}^{\prime},z_{j+1}^{\prime\prime}\right),\quad j=1,2,\ldots,s-2

The proof is simple, if we rely on the way the roots of the derivative vary when the roots of the polynomial vary [8]. Let us prove, for example, that in the interval $\left[y_{i},z_{i}^{\prime}\right]$ polynomial $l^{\prime\prime}(x)$ it is not canceled.

This is obvious if $j=s-1$ and $r_{s}=1$ , because then $y_{s-1}$ is the largest root of $l^{\prime}(x)$ and this root is simple.

In the other cases we observe that $l^{\prime\prime}(y_{j})\neq 0$ , because $y_{i}$ is a simple root of $l^{\prime}(x)$ , and from (112) we deduce …

l^{\prime\prime}\left(z_{j}^{\prime}\right)=\varphi^{\prime\prime}\left(z_{j}^{\prime}\right)\left(z_{j}^{\prime}-x_{s}\right)\neq 0

because then $z_{j}^{\prime}\neq x_{s}$ and $\varphi^{\prime\prime}\left(z_{j}^{\prime}\right)\neq 0$ because $z_{j}^{\prime}$ is a simple root of $\varphi^{\prime}(x)$ .

In the open range $\left(y_{j},z_{j}^{\prime}\right)$ , the polynomial $l^{\prime\prime}(x)$ can have at most $o$ root because, otherwise, $l^{\prime}(x)$ should have a root in this interval, which is impossible. Suppose there were a root of $l^{\prime\prime}(x)$ in the interval ( $y_{j},z_{j}^{\prime}$ ). This root would remain in the interval ( $y_{j},z_{j}^{\prime}$ ), while the first $r_{1}+r_{2}+\ldots+r_{j}-1$ knots and the last $r_{j+1}+r_{j+2}++\ldots+r_{s}-2$ nodes would vary. Making the first ones tend towards $-\infty$ , and on the last ones to $+\infty$ , we see that the property should be true in the particular case

p=2,x_{1}<x_{2}\leq x_{3},\quad j=1

(107)

because, during this variation, $y_{j},z_{j}^{\prime}$ they remain a finite distance apart on the real axis.

In case (107) we have however $y_{1}<z_{1}^{\prime}=\frac{x_{1}+x_{2}}{2}$ and the root $\frac{x_{1}+x_{2}+x_{3}}{3}$ his/her $l^{\prime\prime}(x)$ is greater than $z_{1}^{\prime}$ The property is demonstrated.

polynomial $l^{\prime\prime}(x)$ therefore it does not cancel in the intervals $\left[y_{j},z_{j}^{\prime}\right]j=1,2,\ldots,s-1$ It is also proven that this polynomial does not vanish in the intervals

\left[z_{j}^{\prime\prime},y_{i}\right],\quad j=1,2,\ldots,s-1

Lemma 6 is completely proven.
We therefore deduce
Theorem 19. If $\mathrm{r}\geq 0$ is an integer, if $\mathrm{x}_{0}$ does not coincide with a node that is not simple and if $\mathrm{m}=2,\mathrm{l}^{\prime\prime}\left(\mathrm{x}_{0}\right)=0$ , numerical derivation formula $(E)$ has the degree of accuracy $\mathrm{p}+\mathrm{r}+1$ and the rest of simple form.

The rest of the formula is

R=(r+2)!\cdot l^{\prime}\left(x_{0}\right)D_{p+r+2}[f]

§5. On some explicit formulas for numerical derivation

The practical use of numerical derivation formulas (E) depends largely on the explicit form under which the respective derivative of the interpolation polynomial is put. The speed and accuracy of the actual numerical calculation depend on this explicit form. Also, the eventual use of numerical tables and calculating machines requires a thorough study of these explicit forms. This problem has a very extensive literature. We limit ourselves here to citing the research of SE Micheladze [4] and JF Steffensen [10].

We will examine two kinds of such explicit forms:
$1^{\circ}$ Numerical derivative formulas without differences.
$2^{\circ}$ . Numerical differentiation formulas with differences.
In this paper we are mainly interested in giving a completion of the results from the previous §§. In a subsequent paper we will resume other important examples.

Formulas without differences

40.

Maximum accuracy numerical derivation formulas can be classified according to the values of the numbers $p,m,r,r_{1},r_{2},\ldots,r_{s}$ which fall into the characteristics of this formula, as well as according to their particular nature, such as: degree of accuracy, reducibility, exceptionality, symmetry. In particular, symmetric formulas are of particular interest and their study has been resumed recently by SE Micheladze [4].

We will always assume $m\leq p$ .
Ignoring the values of the nodes and their mutual order of magnitude, a system of values of $p,m,r$ there are as many types of formulas corresponding to it as there are ways we can choose the orders of multiplicity $r_{1},r_{2},\ldots,r^{s}$ with an amount equal to $p+1$ This number is equal to the number $\Omega_{p+1}$ of solutions in non-negative integers $\alpha_{1},\alpha_{2},\ldots,\alpha_{p+1}$ of the Diophantine equation

\alpha_{1}+2\alpha_{2}+3\alpha_{3}+\ldots+p\alpha_{p}+(p+1)\alpha_{p+1}=p+1

Because $m$ take the values $0,1,\ldots,p$ , the number of formulas for $p$ and $r$ give it to you too $(p+1)\Omega_{p+1}$ .

It follows that the number of formulas (E) with degree of exartity $e(\geq 0)$ and non-exceptional (general) is equal to

\Omega_{1}+2\Omega_{2}+3\Omega_{3}+\ldots+(e+1)\Omega_{e+1}

To enumerate the exceptional formulas (E), it is sufficient, based on the results from point 10, to consider only those exceptional formulas that result from increasing the degree of accuracy by one unit ( ¹ ). The number of these formulas for $p,r$ given is $p\left(\Omega_{p+1}-1\right)$ , because such formulas cannot exist for $s=1$ and for $m=0$ Instead, for $s>1$ and $m>0$ , based on inequalities (24), there are such formulas. For $p,r,m$ give (and $s>1$ ) these
¹⁾ Formulas presenting the aspect $1^{\circ}$ (point 10).
formulas vanish only for particular mutual positions of the nodes. In enumerating exceptional formulas we do not distinguish between the different roots, distinct from the nodes, of the polynomial $l^{(m)}(x)$ It then follows that the number of exceptional formulas with the degree of accuracy $e(\geq 2)$ is

\Omega_{2}+2\Omega_{3}+3\Omega_{4}+\ldots+(e-1)\Omega_{s}-\frac{e(e-1)}{2}

To enumerate the symmetric formulas, we take $p=2q-1,q>0$ We have seen that we can assume $m=2\mu$ For $q,r,\mu$ yes there will be $\Omega_{q}$ types of such formulas. For $q,r$ given we will have then $q\Omega_{q}$ such formulas. Finally, the degree of accuracy of the formula being $2q+r-1$ , the number formula 1 . symmetrical s, with the degree of accuracy $e(\geq 1)$ is

\Omega_{1}+2\Omega_{2}+3\Omega_{3}+\ldots+\left[\frac{e+1}{2}\right]\Omega_{\left[\frac{e+1}{2}\right]}

If $e$ is even, all these formulas are exceptional. However, if $e$ is odd for $r=0$ , the formulas are not exceptional. So in this case, $\frac{e+1}{2}\Omega_{\frac{e+1}{2}}$ of these formulas are unexceptional.

To enumerate the reducible formulas, we distinguish two cases. Some with all the nodes confused ( $r=0,s=1$ ). For a $p$ given exists $p$ such formulas, corresponding to the values $1,2,\ldots,p$ his/hers $m$ Their degree of accuracy is $p$ ; therefore, there is $e$ such formulas of degree of accuracy $e(\geq 1)$ The other reducible formulas have two distinct nodes ( $r=0,s=2$ ), which may present $\left[\frac{p+1}{2}\right]$ different types. Of these, however, only at $\left[\frac{p+1}{2}\right]-1$ reducible formulas can correspond, because if $r_{1}=1$ , $r_{2}=p$ , the polynomial (27) has all its roots mixed up. Also, $m$ can only take the values $2,3,\ldots,p-1$ For a $p(\geq 3)$ given, so we have $(p-2)\left(\left[\frac{p+1}{2}\right]-1\right)$ such formulas. These formulas being non-exceptional, there are $(e-2)\left(\left[\frac{e+1}{2}\right]-1\right)$ such formulas with the degree of accuracy $e(\geq 3)$ .

The following table summarizes the above discussion, relative to the number of maximum accuracy formulas, by their specified nature, from the degree of accuracy $0$ to the degree of accuracy $5$ inclusive.

We take into account the following values of the numbers,

\Omega_{1}=1,\quad\Omega_{2}=2,\quad\Omega_{3}=3,\quad\Omega_{4}=5,\quad\Omega_{5}=7,\quad\Omega_{6}=11

41.

To obtain the formulas of maximum accuracy, we can start from the case $r=0,s=p+1$ . This last condition is equivalent to the fact that all the nodes are simple, that is, the points (6) are distinct. We then have

L\left(x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f\mid x\right)=\sum_{i=1}^{p+1}\frac{l(x)}{\left(x-x_{i}^{\prime}\right)l^{\prime}\left(x_{i}^{\prime}\right)}f\left(x_{i}^{\prime}\right)

(108)

To express the coefficients of formula (E) using deviations $h_{i}$ of the nodes of the derivation point, we introduce, in addition to the notations of point 28, the numbers $H_{\alpha}^{(i)},\alpha=0,1,\ldots,p$ what are the fundamental symmetric polynomials of $h_{1},h_{2},\ldots,h_{i-1},h_{i+1},\ldots,h_{p+1}\left(H_{0}^{(i)}=1,H_{\alpha}^{(i)}=0\right.$ for $\alpha<0$ and for $\alpha>p$ ). Either

\lambda(x)=\left(x-h_{1}\right)\left(x-h_{2}\right)\ldots\left(x-h_{p+1}\right).

We then have $l(x)=\lambda\left(x-x_{0}\right)$ , so $l^{\prime}\left(x_{i}^{\prime}\right)=\lambda^{\prime}\left(h_{i}\right)$ and formula (73) gives us

l^{(m)}\left(x_{0}\right)=(-1)^{p+1-m}m!H_{p+1-m}

as well as

\left.\left[\frac{l(x)}{x-x_{i}^{\prime}}\right]_{x=x_{0}}^{(m)}=(-1)^{p-m}m\right\rvert\,H_{p-m}^{(i)},\quad i=1,2,\ldots,p+1

Taking into account formula (108), we deduce

\frac{(-1)^{p-m}}{m!}f^{(m)}\left(x_{0}\right)=\sum_{i=1}^{p+1}\frac{H_{p-m}^{(i)}}{\lambda^{\prime}\left(\bar{h}_{i}\right)}f\left(x_{0}+h_{i}\right)+R

(109)

Passing the case $r=0$ in case $r>0$ is done using formula (90). Formula (84) gives us

L(x)=L\underbrace{\left(x_{0},x_{0},\ldots,x_{0}\right.}_{r};f\mid x)=\sum_{i=0}^{r-1}\frac{\left(x-x_{0}\right)^{j}}{j\mid}f^{(j)}\left(x_{0}\right)

(110)

and from (85) we then deduce

[x_{0}+h_{i},\underbrace{x_{0},x_{0},\ldots,x_{0}}_{r};f]=\frac{1}{h_{i}^{r}}\left[f\left(x_{0}+h_{i}\right)-\sum_{i=0}^{r-1}\frac{h_{i}^{i}}{j!}f^{(j)}\left(x_{0}\right)\right]

(111)

so taking into account formulas (90) and (109) we deduce

	$\displaystyle\frac{(-1)^{p-m}}{(m+r)!}f^{(m+r)}\left(x_{0}\right)=-\sum_{j=0}^{r-1}\left[\sum_{i=1}^{p+1}\frac{H_{p-m}^{(i)}}{h_{i}^{r-j}\lambda^{\prime}\left(h_{i}\right)}\right]\cdot\frac{f^{(j)}\left(x_{0}\right)}{j!}+$		(112)
	$\displaystyle+\sum_{i=1}^{p+1}\frac{H_{p-m}^{(i)}}{h_{i}^{r}\lambda^{\prime}\left(h_{i}\right)}f\left(x_{0}+h_{i}\right)+R$

If $H_{p+1-m}\neq 0$ , this formula has the degree of accuracy $p+r$ , and if the remainder is of simple form, we have

R=-H_{p+1-m}D_{p+r+1}[f].

If $H_{p+1-m}=0$ , the formula becomes exceptional and has the degree of accuracy $p+r+1$ and if the remainder is of simple form, we have

R=H_{p+2-m}D_{p+r_{+}+2}[f]

Starting from formula (111), established in the case when the deviations $h$ are distinct (and different from zero), we obtain the other types of numerical derivation formulas (E) by making, in convenient groups, the deviations tend towards each other.
42. To evaluate the coefficients of $f^{(j)}\left(x_{0}\right),j=0,1,\ldots,r-1$ in formula (112), we can use formula (86). Referring to the notations in formula (10), we deduce from (86)

\sum_{j=0}^{r-1}c_{j}f^{(j)}\left(x_{0}\right)=[l(x)L(\underbrace{x_{0},x_{0},\ldots,x_{0}}_{r};\left.\frac{f(x)}{l(x)}\right\rvert\,x)]_{x=x_{0}}^{(m+r)}

(113)

Taking into account (110) we have

	$\displaystyle{[l(x)L(\underbrace{x_{0},x_{0},\ldots,x_{0}}_{r};\left.\frac{f(x)}{l(x)}\right\rvert\,x)]_{x=x_{0}}^{(m+r)}=}$		(114)
	$\displaystyle=\sum_{j=0}^{r-1}\frac{1}{j!}\left[l(x)\left(x-x_{0}\right)\right]_{x=x_{0}}^{(m+r)}\cdot\left[\frac{f(x)}{l(x)}\right]_{x+x_{0}}^{(j)}$

However, based on formulas (69), (73),

\left[l(x)\left(x-x_{0}\right)^{j}\right]_{x=x_{\mathrm{f}}}^{(m+r)}=(-1)^{p+1-m-r+j}(m+r)!H_{p+1-m-r+j}

From the comparison of formulas (113), (114) we deduce therefore

c_{r-1-j}=\frac{(-1)^{p-m-j}(m+r)!}{(r-1-j)!}\sum_{\alpha=0}^{j}H_{p-m-j+\alpha}\frac{(-1)^{\alpha}}{\alpha!}\left[\frac{1}{l(x)}\right]_{x=x_{0}}^{(\alpha)}.

(115)

But, from the equality

l(x)\frac{1}{l(x)}=1

deduce

\left.\begin{array}[]{l}(-1)^{p+1}H_{p+1}\left[\frac{1}{l(x)}\right]_{x=x_{0}}=1\\ \sum_{\alpha=0}^{i}H_{p+1-i+\alpha}\frac{(-1)^{\alpha}}{\alpha!}\left[\left.\frac{1}{l(x)}\right|_{a=x_{0}}^{(\alpha)}=0,\quad i=1,2,\ldots\right.\end{array}\right\}

Eliminating the

\frac{(-1)^{\alpha}}{\alpha!}\left[\frac{1}{l(x)}\right]_{x=x_{\theta}}^{(\alpha)},\quad\alpha=0,1,\ldots,j

from equation (115) and from the first $j+1$ equations (116), we deduce

		$\displaystyle c_{r-1-j}=\frac{(-1)^{m+1}(m+r)!}{(r-1-j)!H_{p+1}^{i+1}}\left\|\begin{array}[]{llllll}H_{p}&H_{p+1}&0&0\ldots\ldots&0\\ H_{p-1}&H_{p}&H_{p+1}&0\ldots\ldots&0\\ \ldots\ldots\ldots\ldots\ldots\ldots&\ldots\ldots&\ldots&\ldots&\ldots\\ H_{p+1-j}&H_{p+2-j}&\ldots\ldots&\ldots&H_{p+1}\\ H_{p-m-j}&H_{p-m-j+1}&\ldots\ldots&\ldots&H_{p-m}\end{array}\right\|$
		$\displaystyle\quad j=1,\ldots,r-1.$

Comparing this formula with (112) we finally deduce

	$\displaystyle\sum_{i=1}^{p+1}\frac{H_{p-m}^{(i)}}{h_{i}^{j+1}\lambda^{\prime}\left(h_{i}\right)}=\frac{(-1)^{p}}{H_{p+1}^{j+1}}$	$\displaystyle\left\lvert\,\begin{array}[]{ccccc}H_{p}&H_{p+1}&0&0&\ldots\\ H_{p-1}&H_{p}&H_{p+1}&0&\ldots\\ \cdots&\ldots&\ldots&0\\ H_{p+1-j}&H_{p+2-j}&\ldots&\ldots&\ldots\\ H_{p-m-j}&H_{p-m-j+1}&\ldots&\ldots&\ldots\end{array}H_{p+1}\right.$		(117)
		$\displaystyle\quad j=0,1,\ldots,r-1.$

43.

Let's consider some particular cases
$1^{\circ}$ For $r=0,s=1$ , we have

	$\displaystyle f^{(m)}\left(x_{0}\right)=\sum_{j=0}^{p-m}(-1)^{i}\frac{h^{i}}{j!}f^{(m+j)}\left(x_{0}+h\right)+$	$\displaystyle(118)$		(118)
		$\displaystyle+(-1)^{p+1-m}\frac{(p+1)!}{(p+1-m)!}h^{p+1-m}D_{p+1}[f]$

This is Taylor's formula, with a new expression for the remainder. For $m>0$ , we obtain the first series of reducible formulas.
$2^{\circ}$ For $p=m=0$ , we obtain the formula

\frac{1}{r!}f^{(r)}\left(x_{0}\right)=-\sum_{i=0}^{r-1}\frac{1}{j!h^{r-j}}f^{(j)}\left(x_{0}\right)+\frac{1}{r!}f\left(x_{0}+h\right)-hD_{r+1}[f]

(119)

This formula also results from Taylor's formula (118) (for $m=0$ and with a change in notation). For $p\Rightarrow m=r=0$ , the two formulas (118), (119) coincide.
$3^{\circ}$ Let's assume that $p=1,m=0$ A simple calculation shows us that the determinant in the second member of formula (117) is equal to

\frac{h_{1}^{j+2}-h_{2}^{j+2}}{h_{1}-h_{2}}

and we derive the formula

	$\displaystyle\frac{1}{r!}f^{(r)}\left(x_{0}\right)=-\sum_{j=0}^{r-1}\frac{1}{j!\left(h_{1}h_{2}\right)^{r-j}}\cdot\frac{h_{1}^{r-j+1}-h_{2}^{r-j+1}}{h_{1}-h_{2}}f^{(j)}\left(x_{0}\right)-$
	$\displaystyle\quad-\frac{1}{\left(h_{1}h_{2}\right)^{r}\left(h_{1}-h_{2}\right)}\left[h_{2}^{r+1}f\left(x_{0}+h_{1}\right)-h_{1}^{r+1}f\left(x_{0}+h_{2}\right)\right]+h_{1}h_{2}D_{r+2}[f].$		(120)

From this, the limit formula ( $\left.h_{2}\rightarrow h_{1}=h\right)$

	$\displaystyle\frac{1}{r!}f^{(r)}\left(x_{0}\right)=$	$\displaystyle-\sum_{j=0}^{r-1}\frac{r-j+1}{j!h^{r-j}}f^{(j)}\left(x_{0}\right)+\frac{r+1}{h^{r}}f\left(x_{0}+h\right)-$
		$\displaystyle-\frac{1}{h^{r-1}}f^{\prime}\left(x_{0}+h\right)+h^{2}D_{r+2}[f]$		(121)

$4^{\circ}$ For $p=m=1$ , the determinant in the second member of formula (116) is equal to

\frac{h_{1}^{i+1}-h_{2}^{i+1}}{h_{1}-h_{2}}

and we deduce the formula

	$\displaystyle\frac{1}{(r+1)!}f^{(r+1)}\left(x_{0}\right)=\sum_{i=0}^{r-1}\frac{1}{j!\left(h_{1}h_{2}\right)^{r-j}}\cdot\frac{h_{1}^{r-j}-h_{2}^{r-j}}{h_{1}-h_{2}}f^{(j)}\left(x_{0}\right)+$
	$\displaystyle\quad+\frac{1}{\left(h_{1}h_{2}\right)^{r}\left(h_{1}-h_{2}\right)}\left[h_{2}^{r}f\left(x_{0}+h_{1}\right)-h_{1}^{r}f\left(x_{0}+h_{2}\right)\right]+R$		(122)

This formula is of the degree of accuracy $r+1$ , if $h_{1}+h_{2}\neq 0$ , but the rest is generally not of simple form. From this formula we also deduce the limit formula ( $h_{2}\rightarrow h_{1}=h$ )

	$\displaystyle\frac{1}{(r+1)!}f^{(r+1)}\left(x_{0}\right)=\sum_{j=0}^{r-1}\frac{r-i}{j!h^{r-i+1}}f^{(j)}\left(x_{0}\right)-\frac{r}{h^{r+1}}f\left(x_{0}+h\right)+$
	$\displaystyle+\frac{1}{h^{r}}f^{\prime}\left(x_{0}+h\right)-2hD_{r+2}[f]$		(123)

If $h_{1}+h_{2}=0$ , formula (122) becomes exceptional and can be written in the form

\begin{array}[]{r}\frac{1}{(r+1)!}f^{(r+1)}\left(x_{0}\right)=-\sum_{j=0}^{\left[\frac{r-1}{2}\right]}\frac{1}{(r-1-2j)!}\cdot\frac{1}{h^{2i+2}}f^{(r-1-2j)}\left(x_{0}\right)+\\ +\frac{1}{2h^{r+1}}\left[f\left(x_{0}+h\right)+(-1)^{r+1}f\left(x_{0}-h\right)\right]-h^{2}D_{r+3}[f]\end{array}

44.

Among the formulas of the degree of accuracy $\leqq 5$ we will only write explicitly those for which $s\Rightarrow 1$ , so for which the nodes are all confused, except of course those that have been signaled. The rest of these formulas are of simple form. For the degrees of accuracy $0,1,2,3,4,5$ , there is respectively $1,3,6,10,15,21$ such formulas. Of these formulas 1, $3,4,5$ respectively 6 are deduced from formula (118) and one of formulas (119), (121), (123). We group the $0,0,0,3,7$ respectively the remaining 12 formulas according to their degree of accuracy. In parentheses, after the formula, we indicate in turn the corresponding values of the numbers $p,r,m$ For the sake of simplicity, we write the particulars, assuming $x_{0}=0$ . Moving on to his case $x_{0}$ any, through a simple linear transformation.

Accuracy level 3:
(F1) $\quad f^{\prime}(0)=\frac{3}{h}[f(h)-f(0)]-2f^{\prime}(h)+\frac{h}{2}f^{\prime\prime}(h)-h^{3}D_{4}[f]\quad(2,1,0)$
(F2) $f^{\prime\prime}(0)=\frac{6}{h^{2}}[f(0)-f(h)]+\frac{6}{h}f^{\prime}(h)-2f^{\prime\prime}(h)+6h^{2}D_{4}[f]\quad(2,1,1)$
(F3) $f^{\prime\prime\prime}(0)=\frac{6}{h^{3}}[f(h)-f(0)]-\frac{6}{h^{2}}f^{\prime}(h)+\frac{3}{h}f^{\prime\prime}(h)-18hD_{4}[f]\quad(2,1,2)$
Accuracy level 4:
(F4) $f^{\prime}(0)=\frac{4}{h}[f(h)-f(0)]-3f^{\prime}(h)+hf^{\prime\prime}(h)-\frac{h^{2}}{6}f^{\prime\prime\prime}(h)+h^{4}D_{5}[f](3,1,0)$
(F5) $f^{\prime\prime}(0)=\frac{12}{h^{2}}[f(0)-f(h)]+\frac{12}{h}f^{\prime}(h)-5f^{\prime\prime}(h)+hf^{\prime\prime\prime}(h)-8h^{3}D_{5}[f](3,1,1)$
(F6) $f^{\prime\prime\prime}(0)=\frac{24}{h^{3}}[f(h)-f(0)]-\frac{24}{h^{2}}f^{\prime}(h)+\frac{12}{h}f^{\prime\prime}(h)-3f^{\prime\prime\prime}(h)+$

+36h^{2}D_{5}[f]

(3,1,2)

\begin{array}[]{r}f^{IV}(x)=\frac{24}{h^{4}}[f(0)-f(h)]+\frac{24}{h^{3}}f^{\prime}(h)-\frac{12}{h^{2}}f^{\prime\prime}(h)+\\ \frac{4}{h}f^{\prime\prime\prime}(h)-96hD_{5}[f]\end{array}

(F8) $f^{\prime\prime}(0)=\frac{12}{h^{2}}[f(h)-f(0)]-\frac{6}{h}\left[f^{\prime}(0)+f^{\prime}(h)\right]+f^{\prime\prime}(h)-2h^{3}D_{5}[f](2,2,0)$ .
(F9) $f^{\prime\prime\prime}(0)=\frac{48}{h^{3}}[f(0)-f(h)]+\frac{18}{h^{2}}f^{\prime}(0)+\frac{80}{h^{2}}f^{\prime}(h)-$

-\frac{6}{h}f^{\prime\prime}(h)+18h^{2}D_{5}[\gamma]

(F10) $f^{IV}(0)=\frac{72}{h^{4}}[f(h)-f(0)]-\frac{24}{h^{3}}f^{\prime}(0)-\frac{48}{h^{3}}f^{\prime}(h)+$

\frac{12}{h^{2}}f^{\prime\prime}(h)-72hD_{5}[f]

Accuracy level 5:
(F11) $f^{\prime}(0)=\frac{5}{h}[f(h)-f(0)]-4f^{\prime}(h)+\frac{3}{2}hf^{\prime\prime}(h)-$

\frac{1}{3}h^{2}f^{\prime\prime\prime}(h)+\frac{h^{3}}{24}f^{IV}(h)-h^{5}D_{6}[f]

(4,1,0)

(F12) $f^{\prime\prime}(0)=\frac{20}{h^{2}}[f(0)-(fh)]+\frac{20}{h}f^{\prime}(h)-9f^{\prime\prime}(h)+$

+\frac{7}{3}hf^{\prime\prime\prime}(h)-\frac{h^{2}}{3}f^{IV}(h)+10h^{4}D_{6}[f]

(F13) $f^{\prime\prime\prime}(0)=\frac{60}{h^{3}}[f(h)-f(0)]-\frac{60}{h^{2}}f^{\prime}(h)+\frac{30}{h}f^{\prime\prime}(h)-$

-9f^{\prime\prime\prime}(h)+\frac{3}{2}hf^{lV}(h)-60h^{3}D_{6}[f]

(4,1,2)

(F14) $f^{\prime V}(0)=\frac{120}{h^{4}}[f(0)-f(h)]+\frac{120}{h^{3}}f^{\prime}(h)-\frac{60}{h^{2}}f^{\prime\prime}(h)+$

+\frac{20}{h}f^{\prime\prime\prime}(h)-4f^{lV}(h)+240h^{2}D_{6}[f]

(4,1,3)

(F15) $f^{V}(0)=\frac{120}{h^{5}}[f(h)-f(0)]-\frac{120}{h^{4}}f^{\prime}(h)+\frac{60}{h^{3}}f^{\prime\prime}(h)-$
$(4,1,4)$ .

-\frac{20}{h^{2}}f^{\prime\prime\prime}(h)+\frac{5}{h}f^{IV}(h)-600hD_{6}[f]

(F16) $f^{\prime\prime}(0)=\frac{20}{h^{2}}[f(h)-f(0)]-\frac{8}{h}f^{\prime}(0)-\frac{12}{h}f^{\prime}(h)+$
(F17) $f^{\prime\prime\prime}(0)=\frac{120}{h^{3}}[f(0)-f(h)]+\frac{36}{h^{2}}f^{\prime}(0)+\frac{84}{h^{2}}f^{\prime}(h)-$

-\frac{24}{h}f^{\prime\prime}(h)+3f^{\prime\prime\prime}(h)-24h^{3}D_{6}[f]

(3,2,1)

(F18) $f^{IV}(0)=\frac{360}{h^{4}}[f(h)-f(0)]-\frac{96}{h^{3}}f^{\prime}(0)-\frac{264}{h^{3}}f^{\prime}(h)+$

\frac{84}{h^{2}}f^{\prime\prime}(h)-\frac{12}{h}f^{\prime\prime\prime}(h)+144h^{2}D_{6}[f]

(3,2,2)

(F19) $f^{V}(0)=\frac{480}{h^{5}}[f(0)-f(h)]+\frac{120}{h^{4}}f^{\prime}(0)+\frac{360}{h^{4}}f^{\prime}(h)-$

-\frac{120}{h^{3}}j^{\prime\prime}(h)+\frac{20}{h^{2}}f^{\prime\prime\prime}(h)-480hD_{6}[\not]

(3,2,3)

f^{\prime\prime\prime}(0)=\frac{60}{h^{3}}[f(h)-f(0)]-\frac{36}{h^{2}}f^{\prime}(0)-\frac{9}{h}f^{\prime\prime}(0)-

(F20)

-\frac{24}{h^{2}}f^{\prime}(h)+\frac{3}{h}f^{\prime\prime}(h)-6h^{3}D_{6}[f]

(2,3,0)

(F21) $\quad f^{\prime V}(0)=\frac{360}{h^{4}}[f(0)-f(h)]+\frac{192}{h^{3}}f^{\prime}(0)+\frac{36}{h^{2}}f^{\prime\prime}(0)+$

+\frac{168}{h^{3}}f^{\prime}(h)-\frac{24}{h^{2}}f^{\prime\prime}(h)+72h^{2}D_{6}[f]

(2,3,1)

(F22) $f^{V}(0)=\frac{720}{h^{5}}[f(h)-f(0)]-\frac{360}{h^{4}}f^{\prime}(0)-\frac{60}{h^{3}}f^{\prime\prime}(0)-$

-\frac{360}{h^{\mathbf{4}}}f^{\prime}(h)+\frac{60}{h^{3}}f^{\prime\prime}(h)-360hD_{6}[f]

(2,3,2)

45.

Exceptional formulas are obtained by imposing the restriction on deviations $H_{p+1-m}=0$ We will write the exceptional degree of accuracy formulas $\leq 5$ , but only those corresponding to $s=2$ , without those resulting from formula (124). In this case, the two distinct deviations $h_{1},h_{2}$ are linked by a relationship that allows us to express them in the form $\alpha h,\beta h(\alpha,\beta$ being two fixed numbers). The ratio of the numbers $\alpha,\beta$ is not always rational. The rest of these formulas are of simple form when $m=1,2$ , or when the formula is symmetric. For the degrees of accuracy $3,4,5$ we have respectively $2,8,16$ such formulas, of which 6 the simple form of the remainder does not result from the preceding ones. In parentheses, we indicate the values of $p,r,m$ , as well as the restriction to which it is subject $h_{1}$ and $h_{2}$ , the order of multiplicity of the node $x_{2}$ being at least equal to his $x_{1}$ .

+3f^{\prime\prime}(h)-\frac{h}{3}f^{\prime\prime\prime}(h)+2h^{4}D_{6}[f]

Accuracy level 3:

f^{\prime}(0)=\frac{4}{9h}[f(h)-f(-2h)]-\frac{1}{3}f^{\prime}(-2h)-4h^{3}D_{4}[f]

(F23)

\left(2,0,1;2h_{1}+h_{2}=0\right)

	$\displaystyle f^{\prime\prime}(0)=\frac{2}{9h^{2}}[f(-2h)-f(h)]+\frac{2}{3h}f^{\prime}(h)-6h^{2}D_{4}[f]$		(F24)
	$\displaystyle\left(2,0,2;h_{1}+2h_{3}=0\right)$

Accuracy level 4:

(F25) $\quad f^{\prime}(0)=\frac{27}{64h}[f(h)-f(-3h)]-\frac{11}{16}f^{\prime}(-3h)-$

-\frac{3h}{8}f^{\prime\prime}(-3h)-27h^{4}D_{5}[f]\quad\left(3,0,1;3h_{1}+h_{2}=0\right)

f^{\prime}(0)=\frac{3}{4h}[f(h)-f(-h)]-\frac{1}{4}\left[f^{\prime}(h)+f^{\prime}(-h)\right]+h^{4}D_{5}[f]

(F26)

\left(3,0,1;h_{1}+h_{2}=0\right)

(F27)

f^{\prime\prime}(0)=\frac{3}{4h^{2}}[f(h)-f(-h)]-\frac{3}{2h}f^{\prime}(-h)-\frac{1}{2}f^{\prime\prime}(-h)-4h^{3}D_{5}[f]

\left(3,0,2;h_{1}+h_{2}=0\right)

(F28)

		$\displaystyle f^{\prime\prime}(0)=\frac{2\sqrt{3}+3}{3h^{2}}\left\{f(h)-f(-[2-\sqrt{3}]h)-(2-\sqrt{3})hf^{\prime}(h)-\right.$
		$\displaystyle\left.-hf^{\prime}(-[2-\sqrt{3}]h)\right\}+4(3\sqrt{3}-5)h^{3}D_{5}[f]$
		$\displaystyle\left(3,0,2;h_{1}^{2}+4h_{1}h_{2}+h_{2}^{2}=0\right)$

(F29)

\begin{array}[]{r}f^{\prime\prime\prime}(0)=\frac{3}{32h^{3}}[f(h)-f(-3h)]-\frac{3}{8h^{2}}f^{\prime}(h)+\frac{3}{4h}f^{\prime\prime}(h)+R\\ \left(3,0,3;h_{1}+3h_{2}=0\right)\end{array}

(F30)

f^{\prime\prime\prime}(0)=\frac{3}{2h^{3}}[f(-h)-f(h)]+\frac{3}{2h^{2}}\left[f^{\prime}(h)+f^{\prime}(-h)\right]-12h^{2}D_{5}[f]

\left(3,0,3;h_{1}+h_{2}=0\right)

(F31)

\begin{array}[]{r}f^{\prime\prime}(0)=\frac{1}{18h^{2}}[16f(h)+11f(-2h)-27f(0)]+\frac{1}{3h}f^{\prime}(-2h)-\\ -8h^{3}D_{5}[f]\quad\left(2,1,1;2h_{1}+h_{2}=0\right)\end{array}

(F32)

\begin{array}[]{r}f^{\prime\prime\prime}(0)=\frac{1}{3h^{3}}[9f(0)-8f(h)-f(-2h)]+\frac{2}{h^{2}}f^{\prime}(h)-18h^{2}D_{5}[7]\\ \left.(2,1,2);h_{1}+2h_{2}=0\right)\end{array}

Accuracy level 5:

	$\displaystyle f^{\prime}(0)$	$\displaystyle=\frac{256}{625h}[f(h)-f(-4h)]-\frac{131}{125}f^{\prime}(-4h)-\frac{28}{25}hf^{\prime\prime}(-4h)-$		(F33)
		$\displaystyle-\frac{8}{15}h^{2}f^{\prime\prime\prime}(-4h)-256h^{5}D_{6}[f]\quad\left(4,0,1;4h_{1}+h_{2}=0\right)$

(F34)

	$\displaystyle f^{\prime}(0)=$	$\displaystyle\frac{216}{625h}[f(2h)-f(-3h)]-\frac{1}{125}\left[27f^{\prime}(2h)+64f^{\prime}(-3h)\right]-$
		$\displaystyle-\frac{6h}{25}f^{\prime\prime}(-3h)+08h^{5}D_{6}[f]\quad\left(4,0,1;3h_{1}+2h_{2}=0\right)$

(F35) $f^{\prime\prime}(0)=\frac{108}{625h^{2}}[f(2h)-f(-3h)]-\frac{108}{125h}f^{\prime}(-3h)-\frac{29}{25}f^{\prime\prime}(-3h)-$

-\frac{3h}{5}f^{\prime\prime\prime}(-3h)-270h^{4}D_{6}[f]\quad\left(4,0,2;3h_{1}+2h_{2}=0\right)

(F36) $\quad f^{\prime\prime}(0)=\frac{9(117+62\sqrt{6})}{625h^{2}}\left[f(h)-f\left(-\frac{3-\sqrt{6}}{3}h\right)\right]++\frac{9(54+19\sqrt{6})}{250h}\left[(\sqrt{6}-2)f^{\prime}(h)+\frac{\sqrt{6}}{3}f^{\prime}\left(-\frac{3-\sqrt{6}}{3}h\right)\right]+\frac{3\sqrt{6}-2}{25}f^{\prime\prime}$

-\frac{2}{3}(3+4\sqrt{6})h^{4}D_{6}[f]\quad\left(4,0,2;3h_{1}^{2}+6h_{1}h_{2}+h_{2}^{2}=0\right)

(h)

(F37) $f^{\prime\prime\prime}(0)=\frac{48}{625h^{3}}[f(3h)-f(-2h)]-\frac{48}{125h^{2}}f^{\prime}(-2h)-\frac{24}{25h}f^{\prime\prime}(-2h)-$

-\frac{3}{5}f^{\prime\prime\prime}(-2h)+R\quad\left(4,0,3;2h_{1}+3h_{2}=0\right)

(F38) $\quad f^{\prime\prime\prime}(0)=\frac{6(311+129\sqrt{6})}{625h^{3}}[f(-[3-\sqrt{6}]h)-f(h)]+$

		$\displaystyle+\frac{3(81+34\sqrt{6})}{125h^{2}}f^{\prime}(-[3-\sqrt{6}]h)+\frac{3(107+48\sqrt{6})}{125h^{2}}f^{\prime}(h)-$
		$\displaystyle\quad-\frac{3(2+3\sqrt{6})}{25}f^{\prime\prime}(h)+R\quad\left(4,0,3;h_{1}^{2}+6h_{1}h_{2}+3h_{2}^{2}=0\right)$

(F39) $\quad f^{IV}(0)=\frac{24}{625h^{4}}[f(-4h)-f(h)]+\frac{24}{125h^{3}}f^{\prime}(h)-\frac{12}{25h^{2}}f^{\prime\prime}(h)+$

+\frac{4}{5h}f^{\prime\prime\prime}(h)+R\quad\left(4,0,4;h_{1}+4h_{2}=0\right)

(F40) $\quad f^{IV}(0)+\frac{72}{625h^{4}}[f(-2h)-f(3h)]+\frac{24}{125h^{3}}\left[f^{\prime}(3h)+2f^{\prime}(-2h)\right]+$

+\frac{12}{25h^{2}}t^{\prime\prime}(-2h)+R\quad\left(4,0,4;2h_{1}+3h_{2}=0\right)

(F41) $\quad f^{\prime\prime}(0)=\frac{1}{96h^{2}}[81f(h)+47f(-3h)-128f(0)]+\frac{5}{8h}f^{\prime}(-3h)+$

+\frac{1}{4}f^{\prime\prime}(-3h)-54h^{4}D_{6}[f]\quad\left(3,1,1;3h_{1}+h_{2}=0\right)

(F42) $\quad f^{\prime\prime}(0)=\frac{2}{h^{2}}[f(-h)-2f(0)+f(h)]+\frac{1}{2h}\left[f^{\prime}(-h)-f^{\prime}(h)\right]+2h^{4}D_{6}[f]$

\left(3,1,1;h_{1}+h_{2}=0\right)

(F43) $\quad f^{\prime\prime\prime}(0)=\frac{1}{4h^{3}}[48f(0)-9f(-h)-39f(H)]+\frac{15}{2h}f^{\prime}(h)-\frac{3}{2}f^{\prime\prime}(h)+$

12h^{4}D_{6}[f]\quad\left(3,1,2;h_{1}+h_{2}=0\right)

(F44) $\quad f^{\prime\prime\prime}(0)=\frac{-12(5+3\sqrt{3})}{h^{3}}f(0)+\frac{3(1+\sqrt{3})}{h^{3}}f(h)++\frac{3(19+11\sqrt{3})}{h^{3}}f([\sqrt{3}-2]h)-\frac{\sqrt{3}}{h^{2}}f^{\prime}(h)+\frac{12+7\sqrt{3}}{h^{2}}f^{\prime}([\sqrt{3}-2]h)++12(3\sqrt{3}-5)h^{3}D_{6}[f]\quad\left(3,1,2;h_{1}^{2}+4h_{1}h_{2}+h_{2}^{2}=0\right)$
(F45)

	$\displaystyle f^{\prime V}(0)=$	$\displaystyle\frac{1}{8h^{4}}[f(-3h)+3f(h)-4f(0)]-$
		$\displaystyle-\frac{15}{2h^{3}}f^{\prime}(h)+\frac{3}{h^{2}}f^{\prime\prime}(h)+R\quad\left(3,1,3;h_{1}+3h_{2}=0\right)$

(F46) $\quad f^{\prime V}(0)=\frac{12}{h^{4}}[2f(0)-f(h)-f(-h)]+\frac{6}{h^{3}}\left[f^{\prime}(h)-f^{\prime}(-h)\right]-$

-48h^{2}D_{6}[f]\quad\left(3,1,3,h_{1}+h_{2}=0\right)

(F47) $\quad f^{\prime\prime\prime}(0)=\frac{1}{6h^{3}}[16f(h)-9f(0)-7f(-2h)]-$

-\frac{1}{2h^{2}}\left[9f^{\prime}(0)+f^{\prime}(-2h)\right]-24h^{3}D_{6}[f]\quad\left(2,2,1;2h_{1}+h_{2}=0\right)

(F48)

\begin{array}[]{r}f^{IV}(0)=\frac{2}{3h^{4}}[27f(0)+f(-2h)-28f(h)]+\frac{4}{\dot{h}^{3}}\left[3f^{\prime}(0)+2f^{\prime}(h)\right]-\\ -72h^{2}D_{6}[f]\quad\left(2,2,2;h_{1}+2h_{2}=0\right)\end{array}

46.

The reducible formulas in the second series (with $s=2$ ) are easily obtained from exceptional formulas because such a formula is exceptional as a formula for numerical derivation of the function $f^{\prime}(x)$ A reducible formula with the characteristics $p,(r=)0,m$ is obtained from the exceptional formula corresponding to the respective characteristics $p-2,0,m-1$ The 9 reducible formulas of degree of accuracy $\leq 5$ , from this series, are obtained from formula (124) for $r=0$ and from formulas (F23) - (F30). These formulas are

Accuracy level 3:

(F49)

f^{\prime\prime}(0)=\frac{1}{2h}\left[f^{\prime}(h)-f^{\prime}(-h)\right]-4h^{2}D_{4}[f]

(3,0,2)

Accuracy level 4:
(F50)

\begin{array}[]{r}f^{\prime\prime}(0)=\frac{4}{9h}\left[f^{\prime}(h)-f^{\prime}(-2h)\right]-\frac{1}{3}f^{\prime\prime}(-2h)-20h^{3}D_{5}[f;\\ \left(4,0,2;2h_{1}+h_{2}=0\right)\end{array}

(F51)

f^{\prime\prime\prime}(0)=\frac{2}{9h^{2}}\left[f^{\prime}(-2h)-f^{\prime}(h)\right]+\frac{2}{3h}f^{\prime\prime}(h)-30h^{2}D_{5}[f]

$\left(4,0,3;h_{1}+2h_{2}=0\right)$
Accuracy level 5:

	$\displaystyle f^{\prime\prime}(0)=\frac{27}{64h}\left[f^{\prime}(h)-f^{\prime}(-3h)\right]-\frac{11}{16}f^{\prime\prime}(-2h)-\frac{3h}{8}f^{\prime\prime\prime}(-3h)-$		(F52)
	$\displaystyle-162h^{4}D_{6}[f]\quad\left(5,0,2;3h_{1}+h_{2}=0\right)$

(F53)

\begin{array}[]{r}f^{\prime\prime}(0)=\frac{3}{4h}\left[f^{\prime}(h)-f^{\prime}(-h)\right]-\frac{1}{4}\left[f^{\prime\prime}(h)+f^{\prime\prime}(-h)\right]+6h^{4}D_{6}[f]\\ \left(5,0,2;h_{1}+h_{2}=0\right)\end{array}

(F54) $\quad f^{\prime\prime\prime}(0)=\frac{3}{4h^{2}}\left[f^{\prime}(h)-f^{\prime}(-h)\right]-\frac{3}{2h}f^{\prime\prime}(-h)-$

-\frac{1}{2}f^{\prime\prime\prime}(-h)-24h^{3}D_{6}[f]\quad\left(5,0,3;h_{1}+h_{2}=0\right)

(F55)

\begin{array}[]{r}f^{\prime\prime\prime}(0)=\frac{2\sqrt{3}+3}{3h^{2}}\left\{f^{\prime}(h)-f^{\prime}(-[2-\sqrt{3}]h)-(2-\sqrt{3})hf^{\prime\prime}(h)-\right.\\ \left.-hf^{\prime\prime}(-[2-\sqrt{3}]h)\right\}+24(3\sqrt{3}-5)h^{3}D_{6}[f]\\ \left(5,0,3;h_{1}^{2}+4h_{1}h_{2}+h_{2}^{2}=0\right)\end{array}

(F56) $\quad f^{\prime}V(0)=\frac{3}{32h^{3}}\left[f^{\prime}(h)-f^{\prime}(-3h)\right]-\frac{3}{8h^{2}}f^{\prime\prime}(h)+\frac{3}{4h}f^{\prime\prime\prime}(h)+\boldsymbol{R}$

\left(5,0,4;h_{1}+3h_{2}=0\right)

(F57)

\begin{array}[]{r}f^{IV}(0)=\frac{3}{2h^{3}}\left[f^{\prime}(-h)-f^{\prime}(h)\right]+\frac{3}{2h^{2}}\left[f^{\prime\prime}(h)+f^{\prime\prime}(-h)\right]-72h^{2}D_{6}[f]\\ \left(5,0,4;h_{1}+h_{2}=0\right)\end{array}

The simplicity of the remainder, in the case of formulas (F49), (F53), (F57), results from the symmetry of these formulas.

The remainders of formulas (F50), (F51), (F52), (F54) and (F55) can be written respectively as

\begin{gathered}-4h^{3}D_{4}\left[f^{\prime}\right],\quad-6h^{2}D_{4}\left[f^{\prime}\right],\quad-27h^{4}D_{5}\left[f^{\prime}\right],\quad-4h^{3}D_{5}\left[f^{\prime}\right],\\ 4(3\sqrt{3}-5)h^{3}D_{5}\left[f^{\prime}\right].\end{gathered}

But the functional ( $n\geqq 2$ )

R[f]=\left[\alpha_{1},\alpha_{2},\ldots,\alpha_{n};f^{\prime}\right]

where $\alpha_{1},\alpha_{2},\ldots,\hat{\alpha}_{n}$ are $n$ fixed points, not all confused, is of a degree of accuracy $n-1$ and is of simple form. The property relative to the degree of accuracy results from formula (30). The simplicity results from the fact that, if $f(x)$ is a convex function of order $n=1,f^{\prime}(x)$ is a convex function of order $n-2$ We have in this case $R\left[x^{n}\right]=n$ , so

\left[\alpha_{1},\alpha_{2},\ldots,\alpha_{n};f^{\prime}\right]=nD_{n}[f]

RANGE $[a,b]$ being the smallest closed interval containing the points $\alpha_{1},\alpha_{2},\ldots,\alpha_{n}$ . From here the formula results

D_{n-1}\left[f^{\prime}\right]=nD_{n}[f]

whose meaning is clear and from which we have deduced the remainders of formulas (F50), (F51), (F52), (F54) and (F55).
47. The symmetrical formulas (E) can be written using the notations of point 35, writing the deviations in the form $\pm t_{i},i=1,2,\ldots,q$ If we denote by $T_{\alpha}^{(i)},\alpha=0,1,\ldots,q-1$ the fundamental symmetric polynomials of $t_{1}^{2},t_{2}^{2},\ldots,t_{i-1}^{2},t_{i+1}^{2},\ldots,t_{l}^{2}\left(T_{0}^{(i)}=1,T_{\alpha}^{(i)}=0\right.$ , for $\alpha<0$ and $\left.\alpha>q-1\right)$ and if we take into account (94), we deduce

\begin{gathered}H_{2q-2\mu-1}^{(2i)}=(-1)^{q-\mu}t_{\iota}T_{q-\mu-1}^{(i)},H_{2q-2\mu-1}^{(2i-1)}=(-1)^{q-\mu-1}t_{i}T_{q-\mu-1}^{(i)}\\ i=1,2,\ldots,q\end{gathered}

We also have

H_{\alpha}=\left\{\begin{array}[]{cc}0,&\text{ pentru }\alpha\text{ impar }\\ (-1)^{\frac{\alpha}{2}}T_{\frac{\alpha}{2}}&\text{ pentru }\alpha\text{ par. }\end{array}\right.

Let's put

\rho(x)=\left(x-t_{1}^{2}\right)\left(x-t_{2}^{2}\right)\ldots\left(x-t_{q}^{2}\right)

then $\lambda(x)=\rho\left(x^{2}\right)$ , whence, taking into account (94),

\lambda^{\prime}\left(h_{2i}\right)=2t_{i}\rho^{\prime}\left(t_{i}^{2}\right),\quad\lambda^{\prime}\left(h_{2i-1}\right)=-2t_{i}\rho^{\prime}\left(t_{i}^{2}\right).

Doing the calculations, formula (109) becomes

\frac{(-1)^{q-\mu-1}}{(2\mu)!}f^{2\mu}\left(x_{0}\right)=\sum_{i=1}^{q}\frac{T_{\eta-\mu-1}^{(i)}}{2\rho^{\prime}\left(t_{i}^{2}\right)}\left[f\left(x_{0}+t_{i}\right)+f\left(x_{0}^{\prime}-t_{i}\right)\right]-T_{q-\mu}D_{2q}[f]

To move on to the case $r>0$ , we take into account formulas (85) and $(90)$ We note that

\begin{gathered}{[x_{0}+t_{i},\underbrace{x_{0},x_{0},\ldots,x_{0}}_{r};f]+[x_{0}-t_{i},\underbrace{x_{0},x_{0},\ldots,x_{0}}_{r};f]=}\\ =2[x_{0}\pm t_{i},\underbrace{x_{0},x_{0},\ldots,x_{0}}_{r-1};f]\end{gathered}

the second member reducing by definition to $f\left(x_{0}+t_{i}\right)+f\left(x_{0}-t_{i}\right)$ , when $r=0$ . We therefore deduce ¹ )

	$\displaystyle\frac{(-1)^{q-\mu-1}}{(2\mu+r)!}f^{(2\mu+r)}\left(x_{0}\right)=$	$\displaystyle\sum_{i=1}^{q}\frac{T_{q-\mu-1}^{(i)}}{\rho^{\prime}\left(t_{i}^{2}\right)}[x_{0}\pm t_{i},\underbrace{x_{0},x_{0},\ldots,x_{0}}_{r-1};f]-$
		$\displaystyle-T_{q-\mu}D_{2q+r}[f].$		(126)

Taking into account (111), we also deduce

\begin{gathered}{[x_{0}\pm t_{i},\underbrace{x_{0},x_{0},\ldots,x_{0}}_{r-1};f]=\frac{1}{2t_{i}^{r}}\left[f\left(x_{0}+t_{i}\right)+(-1)^{r}f\left(x_{0}-t_{i}\right)-\right.}\\ \left.-2\sum_{i=0}^{\left[\frac{r}{2}\right]}\frac{t_{i}^{r-2i}}{(r-2j)!}f^{(r-2j)}\left(x_{0}\right)\right]\end{gathered}

Substituting into (126), we finally obtain the formula

	$\displaystyle\frac{(-1)^{q-\mu-1}}{(2\mu+r)!}f^{(2\mu+r)}\left(x_{0}\right)=-\sum_{j=0}^{\left[\frac{r}{2}\right]}\left[\sum_{i=1}^{\eta}\frac{T_{q-\mu-1}^{(i)}}{t_{i}^{2j}\rho^{\prime}\left(t_{i}^{2}\right)}\right]\frac{f^{(r-2j)}\left(x_{0}\right)}{(r-2j)!}+$
	$\displaystyle+\sum_{i=1}^{q}\frac{T_{q-\mu-1}^{(i)}}{2t_{i}^{r}\rho^{\prime}\left(t_{i}^{2}\right)}\left[f\left(x_{0}+t_{i}\right)+(-1)^{r}f\left(x_{0}-t_{i}\right)\right]-T_{q-\mu}D_{2q+r}[f].$		(127)

⁰⁰footnotetext: ¹ ) See the abbreviated notation at ^-1 ), on page 39.

His coefficients $f^{(r-2j)}\left(x_{0}\right),j=0,1,\ldots,\left[\frac{r}{2}\right]$ in this formula can be calculated as above, in the case of formula (112). By making the calculations and following the above procedure, we deduce

\begin{gathered}\sum_{i=1}^{q}\frac{T_{q-\mu-1}^{(i)}}{t_{i}^{2j}p^{\prime}\left(t_{\imath}^{2}\right)}=\frac{(-1)^{q-1}}{T_{q}^{j}}\left\lvert\,\begin{array}[]{ccccc}T_{q-1}&T_{q}&0&0&\ldots\\ T_{q-2}&T_{q-1}&T_{q}&0&\ldots\\ \cdots&\cdots&\cdots&0\\ T_{q-j+1}&T_{q-i+2}&\cdots&\cdots&\cdots\\ T_{q-\mu-i}&T_{q-\mu-j+1}&\cdots&\cdots&T_{q}\\ &=0,1,\ldots,\left[\frac{r}{2}\right]\end{array}\right.\end{gathered}

This result can also be deduced from (117), taking into account (125). Formula (127) is valid as long as the deviations $\pm t_{i}$ are distinct. The formulas of the other types are obtained by making the deviations $t_{i},i=1,2,\ldots,q$ . tend, in groups, towards each other.
48. From (120) we deduce the symmetric formula ( $h_{1}+h_{2}=0$ )

\frac{1}{r!}f^{(r)}\left(x_{0}\right)=-\sum_{j=1}^{\left[\frac{r}{2}\right]}\frac{f^{(r-2i)}\left(x_{0}\right)}{(r-2j)!h^{2j}}+\frac{1}{2h^{r}}\left[f\left(x_{0}+h\right)+(-1)^{r}f\left(x_{0}-h\right)\right]-h^{2}D_{r+2}[f].

This, for $r=0,1,2,3,4$ , gives us the symmetric formulas of degree of accuracy $1,2,3,4,5$ ,

\begin{array}[]{ll}\text{ (F58) }f(0)=\frac{1}{2}[f(h)+f](-h)-h^{2}D_{2}[f]&(1,0,0)\\ \left(\text{ F59) }f^{\prime}(0)=\frac{1}{2h}[f(h)-f(-h)]-h^{2}D_{3}[f]\right.&(1,1,0)\\ \text{ F60) }\frac{1}{2}f^{\prime\prime}(0)=-\frac{f(0)}{h^{2}}+\frac{1}{2h^{2}}[f(h)+f(-h)]-h^{2}D_{4}[f]&(1,2,0)\\ \text{ F61) }\frac{1}{6}f^{\prime\prime\prime}(0)=-\frac{f^{\prime}(0)}{h^{2}}+\frac{1}{2h^{3}}[f(h)-f(-h)]-h^{2}D_{5}[f]&(1,3,0)\\ \text{ F62) }\frac{1}{24}f^{IV}(0)=-\frac{f(0)}{h^{4}}-\frac{f^{\prime\prime}(0)}{2h^{2}}+\frac{1}{2h^{4}}[f(h)+f(-h)]-h^{2}D_{6}[f]&(1,4,0)\end{array}

Apart from this, we also have the following symmetric formulas of degree of accuracy $\leq 5$ :

Accuracy level 3:

\text{ (F63) }f(0)=\frac{u^{2}[f(t)+f(-t)]-t^{2}[f(u)+f(-u)]}{2\left(u^{2}-t^{2}\right)}+t^{2}u^{2}D_{4}[f]\quad(-3,0,0)

(F64) $f(0)=\frac{f(h)+f(-h)}{2}-\frac{h}{4}\left[f^{\prime}(h)-f^{\prime}(-h)\right]+h^{4}D_{4}[f]$
(F65) $f^{\prime\prime}(0)=\frac{f(t)+f(-t)-f(u)-f(-u)}{t^{2}-u^{2}}-2\left(t^{2}+u^{2}\right)D_{4}[f]\quad(3,0,2)$ .
To these is added the formula (F49).
Degree of accuracy 4 :
(F66) $f^{\prime}(0)=\frac{u^{3}[f(t)-f(-t)]-t^{3}[f(u)-f(-u)]}{2ut\left(u^{2}-t^{2}\right)}+t^{2}u^{2}D_{5}[f]\quad(3,1,0)$ .
(F67) $f^{\prime\prime\prime}(0)=3\frac{u[f(t)-f(-t)]-t[f(u)-f(-u)]}{ut\left(t^{2}-u^{2}\right)}-6\left(t^{2}+u^{2}\right)D_{5}[f].(3,1,2)$
To these are added the formulas (F26), (F30).
Accuracy level 5:
(F68) $f(0)=\frac{\sum u^{2}\nu^{2}\left(\nu^{2}-u^{2}\right)[f(t)+f(-t)]}{\left(\nu^{2}-u^{2}\right)\left(\nu^{2}-t^{2}\right)\left(u^{2}-t^{2}\right)}-t^{2}u^{2}\nu^{2}D_{6}[f]$
(F69) $f(0)=\frac{u^{2}\left(u^{2}-2t^{2}\right)[f(t)+f(-t)]+t^{4}[f(u)+f(-u)]}{2\left(u^{2}-t^{2}\right)^{2}}-$

-\frac{u^{2}t}{4\left(u^{2}-t^{2}\right)}\left[f^{\prime}(t)-f^{\prime}(-t)\right]-t^{4}u^{2}D_{6}[f]

(5,0,0)

(F70) $f(0)=\frac{f(h)+f(-h)}{2}-\frac{5h}{16}\left[f^{\prime}(h)-f^{\prime}(-h)\right]+$

+\frac{h^{2}}{16}\left[f^{\prime\prime}(h)+f^{\prime\prime}(-h)\right]-h^{6}D_{6}[f]

(5,0,0)

(F71) $f^{\prime\prime}(0)=\frac{\Sigma\left(u^{4}-p^{4}\right)[f(t)+f(-t)]}{\left(v^{2}-u^{2}\right)\left(v^{2}-t^{2}\right)\left(u^{2}-t^{2}\right)}+2\left(\Sigma t^{2}u^{2}\right)D_{6}[f]$
(F72) $f^{\prime\prime}(0)=\frac{2t^{2}}{\left(u^{2}-t^{2}\right)^{2}}[f(t)+f(-t)-f(u)-f(-u)]+$

\frac{u^{2}+t^{2}}{2t\left(ut^{2}-t^{2}\right)}\left[f^{\prime}(t)-f^{\prime}(-t)\right]+2t^{2}\left(t^{2}+2u^{2}\right)D_{6}[/]

(5,0,2)

(F73) $f^{\prime\prime}(0)=\frac{12\Sigma\left(v^{2}-u^{2}\right)[f(t)+f(-t)]}{\left(v^{2}-u^{2}\right)\left(v^{2}-t^{2}\right)\left(u^{2}-t^{2}\right)}-24\left(\Sigma t^{2}\right)D_{6}[f]$
(F74) $f^{IV}(0)=\frac{12}{\left(u^{2}-t^{2}\right)^{2}}[f(u)+f(-u)-f(t)-f(-t)]-$

\frac{6}{t\left(u^{2}-t^{2}\right)}\left[f^{\prime}(t)-f^{\prime}(-t)\right]-24\left(2t^{2}+u^{2}\right)D_{6}[/]

(5,0,4)

(F75) $f^{\prime\prime}(0)=-2\frac{u^{2}+t^{2}}{u^{2}t^{2}}f(0)+\frac{u^{4}[/(t)+f(-t)]-t^{4}[f(u)-f(-u)]}{t^{2}u^{2}\left(u^{2}-t^{2}\right)}+$

+2t^{2}u^{2}D_{6}[f]

(3,2,0)

(F76) $f^{IV}(0)=\frac{24}{u^{2}t^{2}}f(0)-12\frac{u^{2}[f(t)+f(-t)]-t^{2}[f(u)+f(-u)]}{t^{2}u^{2}\left(u^{2}-t^{2}\right)}$

-24\left(t^{2}+u^{2}\right)D_{6}[f]

(3,2,2)

To these are added formulas (F42), (F46), (F53) and (F57). Deviations $h,t,u,v,\ldots$ which in a formula are assumed to be different from each other and different from zero. The summation in formulas (F68), (F71) and (F73) refers to the circular permutations of the letters $t,u,0$ . For the sake of simplicity, I have also assumed here $x_{0}=0$ .

Formulas with differences

49.

We will briefly indicate how these formulas are obtained, without now worrying about the form of the remainder. If we use the previous notations, Newton's formula

L\left(x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f\mid x\right)=\sum_{i=0}^{p}\left(x-x_{1}^{\prime}\right)\left(x-x_{2}^{\prime}\right)\ldots\left(x-x_{i}^{\prime}\right)\left[x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{i+1}^{\prime};f\right]

gives us

		$\displaystyle L^{(m)}\left(x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f\mid x_{0}\right)=$
		$\displaystyle\quad=m!\sum_{i=m}^{n}(-1)^{i-m}H_{i,i-m}\left[h_{1},h_{2},\ldots,h_{i+1};f\left(x_{0}+x\right)\right]$

because, taking into account (72), we have ¹ )

\left[x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{i+1}^{\prime};f(x)\right]=\left[h_{1},h_{2},\ldots,h_{i+1};f\left(x_{0}+x\right)\right].

We derive the following numerical derivation formula

f^{(m)}\left(x_{0}\right)=m!\sum_{i=m}^{n}(-1)^{i-m}H_{i,i-m}\left[h_{1},h_{2},\ldots,h_{i+1};f\left(x_{0}+x\right)\right]+R

(128)

If we use formula (21), we deduce
$L\left(x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p}^{\prime};f\mid x\right)=L\left(x_{0},x_{1},x_{2}^{\prime},\ldots,x_{p}^{\prime};f\mid x\right)+$

+\left(x_{p+1}^{\prime}-x_{0}\right)\left(x-x_{1}^{\prime}\right)\left(x-x_{2}^{\prime}\right)\ldots\left(x-x_{p}^{\prime}\right)\left[x_{0},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f\right]

from where we obtain the numerical derivation formula

$\displaystyle f^{(m)}\left(x_{0}\right)=$	$\displaystyle m!$	(129)
	$\displaystyle\sum_{j=m-1}^{p-1}(-1)^{i-m+1}H_{i,i-m+1}\left[0,h_{1},h_{2},\ldots,h_{i+1};f\left(x_{0}+x\right)\right]+$
	$\displaystyle+m!h_{p+1}H_{p,p-m}\left[0,h_{1},h_{2},\ldots,h_{p+1};f\left(x_{0}+x\right)\right]+R$

1.

For greater clarity, we also highlight the variable $x$ in the notation of the divided difference (and of the interpolation polynomial).

50.

Let's see how we get to the case. $r>0$ From the general formula ¹ )

	$\displaystyle L\left(x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};fg\mid x\right)=$		(130)
	$\displaystyle\quad=\sum_{i=0}^{p}L\left(x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};l_{i}(x)g(x)\mid x\right)\left[x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{i+1}^{\prime};f\right]$

if $x_{0}$ is different from the nodes, we deduce

		$\displaystyle L\left(x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};\left.\frac{f(x)}{\left(x-x_{0}\right)^{r}}\right\rvert\,x\right)=$
		$\displaystyle\quad=\sum_{i=0}^{p}L\left(x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};\left.\frac{l_{i}(x)}{\left(x-x_{0}\right)^{r}}\right\rvert\,x\right)\left[x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f\right]$

Taking into account formulas (86) and (113), we obtain

	$\displaystyle L^{(m+r)}(\underbrace{x_{0},x_{0},\ldots,x_{0},}$	$\displaystyle\left.x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f\mid x_{0}\right)=\sum_{j=0}^{r-1}c_{i}f^{(j)}\left(x_{0}\right)+$
		$\displaystyle+\frac{(m+r)!}{m!}\sum_{i=0}^{p}I_{i}^{(r)}\left[h_{1},h_{2},\ldots,h_{p+1};f\left(x_{0}+x\right)\right]$

where the coefficients $I_{i}^{(r)}$ are given by the formula

I_{i}^{(r)}=L^{(m)}\left(x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};\left.\frac{l_{i}(x)}{\left(x-x_{0}\right)^{r}}\right\rvert\,x_{0}\right),\quad i=0,1,\ldots,p

From formula (130) it can also be deduced that
$L\left(x_{0},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};fg\mid x\right)=L\left(x_{0},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};g\mid x\right)f\left(x_{0}\right)+$

+\left(x-x_{0}\right)\sum_{i=0}^{p}L\left(x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};l_{i}(x)g(x)\mid x\right)\left[x_{0},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};\lceil].\right.

Replacing here the function $g(x)$ with $\left(x-x_{0}\right)g(x)$ , we deduce $\left.{}^{2}\right)$
$L\left(x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};fg\mid x\right)=L\left(x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};g\mid x\right)f\left(x_{0}\right)+$

+\sum_{i=0}^{p}L\left(x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};l_{i}(x)\left(x-x_{0}\right)g(x)\mid x\right)\left[x_{0},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{i+}^{\prime}\right.

We therefore also have the formula

		$\displaystyle L\left(x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};\left.\frac{f(x)}{\left(x-x_{0}\right)^{r}}\right\rvert\,x\right)=L\left(x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};\left.\frac{1}{\left(x-x_{0}\right)^{r}}\right\rvert\,x\right)f\left(x_{0}\right)+$
		$\displaystyle\quad+\sum_{i=0}^{p}L\left(x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};\left.\frac{l_{i}(x)}{\left(x-x_{0}\right)^{r-1}}\right\rvert\,x\right)\left[x_{0},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{i+1}^{\prime};f\right]$

¹ ) The equality results from the fact that the two polynomials of degree $p$ , from the first and second members, coincide in the nodes $x_{i}^{\prime}$ .
² ) We can proceed as in formula (130).

We deduce from this that, if $r>0$ and if $x_{0}$ is different from nodes,

\begin{gathered}L^{(m+r)}(\underbrace{x_{0},x_{0},\ldots,x_{0}}_{r},x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{p+1}^{\prime};f\mid x_{0})=\sum_{j=0}^{r-1}c_{j}f^{(j)}\left(x_{0}\right)+\\ +\frac{(m+r)!}{m!}\left\{I_{0}^{(i)}/\left(x_{0}\right)+\sum_{i=0}^{p}I_{i}^{(r-1)}\left[0,h_{1},h_{2},\ldots,h_{i+1};f\left(x_{0}+x\right)\right]\right\}\end{gathered}

Finally, we have the following numerical derivation formulas

	$\displaystyle f^{(m+r)}\left(x_{0}\right)=\sum_{j=0}^{r-1}c_{j}f^{(j)}\left(x_{0}\right)+$
	$\displaystyle+\frac{(m+r)!}{m!}\sum_{i=0}^{r}I_{i}^{r)}\left[h_{1},h_{2},\ldots,h_{i+1};f\left(x_{0}+x\right)\right]+R$		(131)
	$\displaystyle f^{(m+r)}\left(x_{0}\right)=\sum_{j=0}^{r-1}c_{i}f^{(j)}\left(x_{0}\right)+$		(132)
	$\displaystyle+\frac{(m+r)!}{m!}\left\{I_{0}^{(r)}f\left(x_{0}\right)+\sum_{i=0}^{n}I_{i}^{(r-1)}\left[0,h_{1},h_{2},\ldots,h_{i+1};f\left(x_{0}+x\right)\right]+R.\right.$

What are the efficiencies? $I_{i}^{(r)}$ can be calculated using the recurrence formula

I_{i}^{(r)}=I_{i+1}^{(r+1)}+h_{i+1}I_{i}^{(r+1)}\left(I_{p+1}^{(r)}=0\right)

giving him $r$ , successively, the values $-1,0,1,2,\ldots$ and noting that

\begin{gathered}I_{0}^{(0)}=I_{1}^{(0)}=\ldots=I_{m-1}^{(0)}=0,I_{i}^{(0)}=m!(-1)^{i-m}H_{i,i-m}\\ i=m,m+1,\ldots,p.\end{gathered}

Coefficients $c_{j},j=0,1,\ldots,r-1$ were calculated above.
51. Formula (131) is convenient, in particular, if the nodes are equidistant, and formula (132), if the nodes and the derivation point form an equidistant system.

To illustrate what has been said, we will consider a particular case.
Let us suppose that $r=0,m=1$ , that the nodes are equidistant and symmetrical with respect to $x$ Using the notations (94), putting

t_{i}=(2i-1)h,\quad i=1,2,\ldots,q

(133)

and noting that

H_{i,i-1}=h_{1}h_{2}\ldots h_{i}\left(\frac{1}{h_{1}}+\frac{1}{h_{2}}+\ldots+\frac{1}{h_{i}}\right)

deduce

H_{i,i-1}=\left\{\begin{array}[]{cc}0,&\text{ pentru }i\text{ par }\\ (-1)^{\frac{i-1}{2}1^{2}\cdot 3^{2}}&\ldots(i-2)^{2}h^{i-1},\text{ pentru }i\text{ impar }\\ &\left(H_{1,0}=1\right)\end{array}\right.

Formula (128) becomes

f^{\prime}\left(x_{0}\right)=\sum_{i=1}^{q}H_{2i-1,2i-2}\left[h_{1},h_{2},\ldots,h_{2i};f\left(x_{0}+x\right)\right]+R

Introducing the usual notation of differences and taking into account (94), (133), we deduce

\left[h_{1},h_{2},\ldots,h_{2i};f\left(x_{0}+x\right)\right]=\frac{\Delta_{2h}^{2i-1}f\left(x_{0}-[2i-1]h\right)}{(2i-1)!(2h)^{2i-1}},i=1,2,\ldots,q.

If we also use relations (134), we finally have the numerical derivative formula

\begin{gathered}f^{\prime}\left(x_{0}\right)=\sum_{i=1}^{q}\frac{(-1)^{i-1}(2i-2)!}{(2i-1)[(i-1)!]^{2}2^{4i-3}h}\Delta_{2h}^{2i-1}/\left(x_{0}-[2i-1]h\right)+\\ +1^{2}\cdot 3^{2}\cdots(2q-1)^{2}h^{2q}D_{2q+1}[f]\end{gathered}

Let us assume that the nodes are symmetric with respect to $x_{0}$ and that together with this point it forms a system of equidistant points. Then we can use formula (129).

Instead of (133) we take

\iota_{i}=ih,i=1,2,\ldots q

We then have

H_{i,i}=\left\{\begin{array}[]{l}(-1)^{\frac{i+1}{2}}\left[\left(\frac{i-1}{2}\right)!\right]^{2}\frac{i+1}{2}h^{i},\text{ pentru }i\text{ impar }\\ (-1)^{\frac{i}{2}}\left[\left(\frac{i}{2}\right)!\right]^{2}h^{i},\text{ pentru }i\text{ par }\end{array}\right.

and

\left[0,h_{1},h_{2},\ldots,h_{i+1};n\right]=\left\{\begin{array}[]{l}\frac{\Delta_{h}^{i+1}/\left(x_{0}-\frac{i+1}{2}h\right)}{(i+1)!h^{i+1}},\text{ pentru }i\text{ impar }\\ \frac{\Delta_{h}^{i+1}/\left(x_{0}-\frac{i+2}{2}h\right)}{(i+1)!h^{i+1}},\text{ pentru }i\text{ par }\end{array}\right.

and we derive the numerical derivation formula

		$\displaystyle f^{\prime}\left(x_{0}\right)=\sum_{i=0}^{q-1}\frac{(-1)^{i}(i!)^{2}}{(2i+1)!2h}\left[2\Delta_{h}^{2i+1}f\left(x_{0}-[i+1]h\right)+\right.$
		$\displaystyle\left.\quad+\Delta_{h}^{2i+2}f\left(x_{0}-[i+1]h\right)\right]+(q!)^{2}D_{2q+1}[!]$

Mathematical Sciences Department of the RPR Academy Branch, Cluj

On the remainder in some numerical derivation formulas

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ON THE REMAINDER IN SOME NUMERICAL DERIVATION FORMULAS

§ 1. Formulas of maximum accuracy

§ 2. Some considerations on the rest

§ 3. A criterion for recognizing the simple form of the remainder of a formula of maximum accuracy

§ 4. On some applications of the preceding results

§5. On some explicit formulas for numerical derivation

Formulas without differences

Accuracy level 4:

Accuracy level 3:

Formulas with differences

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