APPLICATION OF NUMERICAL DERIVATION FORMULAS TO THE NUMERICAL INTEGRATION OF DIFFERENTIAL EQUATIONS

BY
D. V. IONESCU

The differential equation is considered

y^{\prime}=f(x,y)

(1)

and either $y(x)$ its integral that satisfies the initial condition $y\left(x_{0}\right)=y_{0}$ We assume that the conditions that ensure the existence and uniqueness of the integral are met. $y(x)$ in the interval $\left[x_{0},x_{0}+a\right]$ .

In the interval ( $x_{0},x_{0}+a$ ) we take the nodes $x_{1},x_{2},\ldots,x_{n}$ so that $x_{1}<x_{2}<x_{3}<\cdots<x_{n}$ and we assume that the integral $y(x)$ is known on the nodes $x_{1},x_{2},\ldots,x_{n}$ The problem of numerical integration of the differential equation (1) consists in giving the practical formula for calculating the integral $y(x)$ , at a point in the interval ( $x_{0},x_{0}+a$ ], using the values of $y(x)$ on the nodes $x_{0},x_{1},\ldots,x_{n}$ and to study the rest of the numerical integration formula.

If the function $f(x,y)$ has partial derivatives with respect to $x$ and $y$ until ordered $n$ , continue in the field $D$ in which the conditions that ensure the existence and uniqueness of the integral are met $y(x)$ , then applying Taylor's formula we will have

	$\displaystyle y(x)=y\left(x_{0}\right)+\frac{x-x_{0}}{1!}y^{\prime}\left(x_{0}\right)+\frac{\left(x-x_{0}\right)^{2}}{2!}y^{\prime\prime}\left(x_{0}\right)+\ldots$
	$\displaystyle\quad+\frac{\left(x-x_{0}\right)^{n}}{n!}y^{(n)}\left(x_{0}\right)+\int_{x_{0}}^{x}\frac{(xs)^{n}}{n!}y^{(n+1)}(s)ds$		(2)

The problem of numerical integration of the differential equation (1), as stated above, is reduced by formula (2) to the problem of numerical derivation of a function, which is formulated as follows: given the function $f(x)$ continues
in the interval $\left[x_{0},x_{0}+a\right]$ and having continuous derivatives in this interval of all orders, which will intervene in the calculations, to give practical formulas for calculating the derivatives $f^{\prime}(x),f^{\prime\prime}(x),\ldots,f^{(n)}(x)$ in the node $x_{0}$ , with the help of the values of $f(x)$ on the nodes $x_{0},x_{1},\ldots,x_{n}$ and calculate the rest of the numerical derivation formula.

The paper is divided into two chapters. The first deals with numerical derivation formulas, and the second deals with their application to the numerical integration of differential equations of the form (1).

Chapter I

NUMERICAL DERIVATION FORMULAS

§ 1. Numerical derivation formulas that come from the expression of a divided difference in the form of a definite integral

1.

In a comprehensive work on quadrature formulas, which can be obtained using the generalized integration by parts formula, as done by J. Radon [1], I gave an expression for the difference divided by the order $n$ on repeated nodes in integral form [2], showing that

	$\displaystyle{[\underbrace{a,\ldots,a}_{\alpha\text{ times }},\underbrace{x_{1},\ldots,x_{1}}_{k_{1}\text{ times }},\ldots,x_{k_{p}\text{ times }}^{x_{p},\ldots,x_{p}},\underbrace{b,\ldots,b}_{\beta\text{ times }};f(x)]=}$
	$\displaystyle=\int_{a}^{b}\varphi(x)f^{\left(k_{1}+\ldots+k_{p}+\alpha+\beta-1\right)}(x)dz$		(3)

where the function $\varphi(x)$ is determined by a certain system of differential equations that integrate with certain boundary conditions in $a,x_{1},\ldots x_{p}$ and $b$ . Solving equation (3) in relation $\mathrm{cu}f^{(\alpha-1)}(a)$ a numerical derivation formula is obtained, from which it is deduced $f^{(a-1)}(a)$ using the function values $f(x)$ and its successive derivatives, which enter equation (3), on the nodes $a,x_{1},\ldots,x_{p},b$ .

Prof. T. Popoviciu [3] and Ş. E. Mikeladze [4] proceeded in the same way, but starting from other expressions of the difference divided by the first member of formula (3).

We will summarize the result from [2], for the case of a single multiple node, with some additions.
2. Let $f(x)$ a class function $C^{n+p}$ , defined in the interval $[a,b]^{1}$ ) in which we take the nodes $x_{0},x_{1},\ldots,x_{n}$ so that we have $x_{0}<x_{1}<\ldots<x_{n}$ At intervals $\left[x_{0},x_{1}\right],\left[x_{1},x_{2}\right],\ldots,\left[x_{n-1},x_{n}\right]$ we attach the functions $\varphi_{1}(x),\varphi_{2}(x),\ldots\varphi_{n}(x)$ and the constants $\lambda_{1},\lambda_{2},\ldots,\lambda_{n}$ and we form the differential equations

\varphi_{1}^{(n+p-1)}(x)=\lambda_{1},\quad\varphi_{2}^{(n+p-1)}(x)=\lambda_{2},\ldots,\varphi_{n}^{(n+p-1)}(x)=\lambda_{n}

(4)

⁰⁰footnotetext:¹ ) That is, the function

f(x)

has in the range

[a,b]

successive derivatives up to the order

n+p

, the latter being continuous in this interval.

which integrates with the following boundary conditions:

	$\displaystyle\varphi_{1}\left(x_{0}\right)=0,\quad\varphi_{1}^{\prime}\left(x_{0}\right)=0,\ldots,\varphi_{1}^{(n-2)}\left(x_{0}\right)=0$
	$\displaystyle\varphi_{1}^{(n-1)}\left(x_{0}\right)=(-1)^{p}\frac{\nu_{p}}{p!},\varphi_{1}^{(n)}\left(x_{0} \right)=(-1)^{p-1}\frac{\nu_{p-1}}{(p-1)!},\ldots,\varphi^{(n+p-2)}\left(x_{0}\right)=-\frac{\nu_{1}}{1!},$
	$\displaystyle\varphi_{2}\left(x_{1}\right)=\varphi\left(x_{1}\right),\varphi_{2}^{\prime}\left(x_{1}\right)=\varphi_{ 1}^{\prime}\left(x_{1}\right),\ldots,\varphi_{2}^{(n+p-2)}\left(x_{1}\right)=\varphi_{1}^{(n+p-2)}\left(x_{1}\right),$		(5)
	$\displaystyle\varphi_{n}\left(x_{n-1}\right)=\varphi_{n-1}\left(x_{n-1}\right),\varphi_{n}^{\prime}\left(x_{n-1}\right)=\varphi_{n-1}^{\prime}\left(x_{n-1}\right),\ldots,\varphi_{n}^{(n+p-2)}\left(x_{n-1}\right)=\varphi_{n-1}^{(n+p-2)}\left(x_{n-1}\right),$
	$\displaystyle\varphi_{n}\left(x_{n}\right)=0,\quad\varphi_{n}^{\prime}\left(x_{n}\right)=0,\ldots,\varphi_{n}^{(n+p-2)}\left(x_{n}\right)=0.$

constants $\lambda_{1},\lambda_{2},\ldots,\lambda_{n}$ and $\nu_{1},\nu_{2},\ldots,\nu_{p}$ it is thus determined that the integrals of the differential equations (4) verify the boundary equations (5).

It is obvious that each integral

\int_{x_{0}}^{x_{1}}\varphi_{1}^{(n+p)}(x)f(x)dx,\int_{x_{1}}^{x_{2}}\varphi_{2}^{(n+p)}(x)f(x)dx,\ldots,\int_{x_{n-1}}^{x_{n}}\varphi^{(n+p)}(x)f(x)dx

is zero. Applying to each, the generalized formula for integration by parts and taking their sum, we obtain, taking into account the boundary conditions (5), the numerical derivative formula

	$\displaystyle-\left[\lambda_{1}f\left(x_{0}\right)+\frac{v_{1}}{1!}f^{\prime}\left(x_{0}\right)+\frac{v_{2}}{2!}f^{\prime\prime}\left(x_{0}\right)+\ldots+\frac{v_{p}}{p!}f^{(p)}\left(x_{0}\right)\right]+$
	$\displaystyle+K_{1}f\left(x_{1}\right)+K_{2}f\left(x_{2}\right)+\ldots+K_{n}\left(x_{n}\right)=(-1)^{n+p-1}\int_{x_{0}}^{x_{n}}\varphi(x)f^{(n+p)}(x)dx$		(6)

where the function $\varphi(x)$ coincides in turn with the functions $\varphi_{1}(x),\varphi_{2}(x),\ldots,\varphi_{n}(x)$ in the intervals $\left[x_{0},x_{1}\right],\ldots,\left[x_{n-1},x_{n}\right]$ and where

\lambda_{1}-\lambda_{2}=K_{1},\quad\lambda_{2}-\lambda_{3}=K_{2},\ldots,\lambda_{n-1}-\lambda_{n}=K_{n-1},\lambda_{n}=K_{n}.

(7)

Functions $\varphi_{1}(x),\varphi_{2}(x),\ldots,\varphi_{n}(x)$ are given by the formulas

	$\displaystyle\varphi_{n}(x)=K_{n}\frac{\left(x-x_{n}\right)^{n+p-2}}{(n+p-1)!}$
	$\displaystyle\varphi_{n-1}(x)=K_{n-1}\frac{\left(x-x_{n-1}\right)^{n+p-1}}{(n+p-1)!}+K_{n}\frac{\left(x-x_{n}\right)^{n+p-1}}{(n+p-1)!}$		(8)
	$\displaystyle\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots K_{n}\frac{\left(x-x_{n}\right)^{n+p-1}}{(n+p-1)!}.$

With formulas (8) the differential equations (4) and the boundary conditions (5) at the nodes are verified. $x_{1},x_{2},\ldots,x_{n}$ . Writing that and the boundary conditions at the node $x_{0}$ are satisfied, we have the system of equations

\begin{array}[]{lll}K_{1}\left(x_{1}-x_{0}\right)&+K_{2}\left(x_{2}-x_{0}\right)&+\ldots+K_{n}\left(x_{n}-x_{0}\right)=v_{1}\\ K_{1}\left(x_{1}-x_{0}\right)^{2}&+K_{2}\left(x_{2}-x_{0}\right)^{2}&+\ldots+K_{n}\left(x_{n}-x_{0}\right)^{2}=v_{2}\\ \cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot&+K_{2}\left(x_{2}-x_{0}\right)^{p}&+\ldots+K_{p}\left(x_{n}-x_{0}\right)^{p}=v_{p}\\ K_{1}\left(x_{1}-x_{0}\right)^{p}&+K_{0}&\left(x_{0}\right)\\ K_{1}\left(x_{1}-x_{0}\right)^{p+1}&+K_{2}\left(x_{2}-x_{0}\right)^{p+1}&+\ldots+K_{n}\left(x_{n}-x_{0}\right)^{p+1}=0\\ \cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\\ K_{1}\left(x_{1}-x_{0}\right)^{n+p-1}+K_{2}\left(x_{2}-x_{0}\right)^{n+p-1}+\ldots+K_{n}\left(x_{n}-x_{0}\right)^{n+p-1}=0,\end{array}

which determines $K_{1},K_{2},\ldots,K_{n}$ and $\nu_{1},\nu_{1},\ldots,\nu_{p}$ .

Taking into account equations (7), equations (9) can also be written in the form

	$\displaystyle-\lambda_{1}+K_{1}+\ldots+K_{n}=0$
	$\displaystyle-\lambda_{1}x_{0}-C_{1}^{1}\quad v_{1}\quad+K_{1}x_{1}\quad+\ldots+K_{n}x_{n}=0$		(10)
	$\displaystyle-\lambda_{1}x_{0}^{p}\quad-C_{p}^{1}\quad v_{1}x_{0}^{p-1}\quad-\ldots-C_{p}^{p}\quad v_{p}\quad+K_{1}x_{1}^{p}\quad+\ldots+K_{n}x_{n}^{p}=0$
	$\displaystyle-\lambda_{1}x_{0}^{n+p-1}-C_{n+p-1}^{1}\nu_{1}x_{0}^{n+p-2}-\ldots-C_{n+p-1}^{p}\nu_{p}x_{0}^{n-1}+K_{1}x_{1}^{n+p-1}+\ldots+K_{n}x_{n}^{n+p-1}=0\text{. }$

Equations (9) express that the second term of the numerical derivation formula (6) is zero when $f(x)$ is replaced with

x-x_{0},\left(x-x_{0}\right)^{2},\ldots,\left(x-x_{0}\right)^{n+p-1}

and equations (10) express that the second member of the numerical derivation formula (6) is zero when $f(x)$ is replaced with

1,x,x^{2},\ldots,x^{n+p-1}

Let us consider the matrix formed with the coefficients of system (10), i.e.

\left\|\begin{array}[]{|llllllll||}1&0&&\ldots&0&1&\ldots&1\\ x_{0}&C_{1}^{1}&&\ldots&0&x_{1}&\ldots&x_{n}\\ \vdots&\vdots&&&\vdots&\vdots&&\\ x_{0}^{p}&C_{p}^{1}&x_{0}^{p-1}&\ldots&C_{p}^{p}&x_{1}^{p}&\ldots&x_{n}^{p}\\ \vdots&\vdots&&&\vdots&\vdots&&\vdots\\ x_{0}^{n+p-1}&C_{n+p-1}^{1}x_{0}^{n+p-2}&\ldots&C_{n+p-1}^{p}x_{0}^{n-1}&x_{1}^{n+p-1}&\ldots&x_{n}^{n+p-1}\end{array}\right\|

and let's note with $A_{i}$ the determinant that has the same rows and columns except for column a $i$ -a. All these determinants are different from zero and solving the system (10) with respect to $-\lambda_{1},-v_{1},\ldots,-v_{p},K_{1},K_{2},\ldots,K_{n}$ is obtained

	$\displaystyle\frac{-\lambda_{1}}{A_{1}}=\frac{-v_{1}}{-A_{2}}=\ldots=\frac{-v_{p}}{(-1)^{p}A_{p+1}}=$
	$\displaystyle=\frac{K_{1}}{(-1)^{p+1}A_{p+2}}=\ldots=\frac{K_{n}}{(-1)^{p+n}A_{p+n+1}}.$		(11)

Taking into account that the first member of the numerical derivation formula (6) is a linear combination of the numerators in formulas (11), we deduce that formulas (11) can also be written in the form

	$\displaystyle\frac{-\lambda_{1}}{A_{1}}=\frac{-\nu_{1}}{-A_{2}}=\ldots=\frac{-\nu_{p}}{(-1)^{p}A_{p+1}}=$
	$\displaystyle=\frac{K_{1}}{(-1)^{p+1}A_{p+2}}=\ldots=\frac{K_{n}}{(-1)^{p+n}A_{p+n+1}}=\frac{-\int_{x_{0}}^{x_{n}}\varphi(x)f^{(n+p)}(x)dx}{\Delta}$		(12)

where

\Delta=\left\lvert\,\begin{array}[]{lllllll}1&0&\ldots&0&1&\ldots&1\\ x_{0}&C_{1}^{1}&\ldots&0&x_{1}&\ldots&x_{n}\\ \vdots&\vdots&&\vdots&\vdots&&\vdots\\ x_{0}^{p}&C_{p}^{1}&x_{0}^{p-1}&\ldots&C_{p}^{p}&x_{1}^{p}&\ldots\\ \vdots&\vdots&&\vdots&\vdots&&x_{n}^{p}\\ x_{0}^{n+p-1}&C_{n+p-1}^{1}x_{0}^{n+p-2}&\ldots&C_{n+p-1}^{p}x_{0}^{n-1}&x_{1}^{n+p-1}&\ldots&x_{n}^{n+p-1}\\ f\left(x_{0}\right)&&\frac{f^{\prime}\left(x_{0}\right)}{1!}&\ldots&\frac{f^{(p)}\left(x_{0}\right)}{p!}&f\left(x_{1}\right)&\ldots\end{array}\right.

determinant $\Delta_{1}$ which is obtained from $\Delta$ replacing $f(x)$ with $x^{n+p}$ , is different from zero, because

\Delta_{1}=\left(x_{1}-x_{0}\right)^{p}\left(x_{2}-x_{0}\right)^{p}\ldots\left(x_{n}-x_{0}\right)^{p}V\left(x_{0},x_{1},\ldots,x_{n}\right),

(14)

where $V\left(x_{0},x_{1},\ldots x_{n}\right)$ is the Vandermonde determinant of the numbers $x_{0},x_{1},\ldots,x_{n}$ .

Equating the ratios in (12) with $-\frac{1}{\Delta_{1}}$ , we deduce that the formula for
the eric derivation can also be written as the numerical derivation can also be written as the form

\frac{\Delta}{\Delta_{1}}=\int_{x_{0}}^{x_{n}}\varphi(x)f^{(n+p)}(x)dx

(15)

or, in the form of

[\underbrace{x_{0},x_{0},\ldots,x_{0}}_{p+1\text{ ori }},\quad x_{1},\ldots,x_{n};\quad f(x)]=\int_{x_{0}}^{x_{n}}\varphi(x)f^{(n+p)}(x)dx

(16)

because the divided difference of the function $f(x)$ from the first member of formula (16) is equal to $\frac{\Delta}{\Delta_{1}}$ 3.
Calculation of coefficients $\lambda_{1},\nu_{1},\ldots,\nu_{p},K_{1},\ldots,K_{n}$ from the numerical derivation formula (6). From equation (12) we deduce that we have

		$\displaystyle K_{1}=(-1)^{p}\frac{A_{p+2}}{\Delta_{1}}$
		$\displaystyle K_{2}=(-1)^{p+1}\frac{A_{p+3}}{\Delta_{1}}$
		$\displaystyle\ldots\ldots\ldots$
		$\displaystyle K_{n}=(-1)^{p+n-1}\frac{A_{n+p+1}}{\Delta_{1}}$

Without weight it is shown that in general

\begin{gathered}A_{p+i+1}=\left(x_{1}-x_{0}\right)^{p}\ldots\left(x_{i-1}-x_{0}\right)^{p}\left(x_{i+1}-x_{0}\right)^{p}\ldots\left(x_{n}-x_{0}\right)^{p}\times\\ \times V\left(x_{0},x_{1},\ldots,x_{i-1},x_{i+1},\ldots,x_{n}\right)\end{gathered}

from which it follows that we have

	$\displaystyle K_{1}=(-1)^{p}\frac{1}{\left(x_{1}-x_{0}\right)^{p}}\cdot\frac{V\left(x_{0},x_{2},\ldots,n_{n}\right)}{V\left(x_{0},x_{1},\ldots,x_{n}\right)}$
	$\displaystyle K_{2}=(-1)^{p+1}\frac{1}{\left(x_{2}-x_{0}\right)^{p}}\cdot\frac{V\left(x_{0},x_{1},x_{3},\ldots,x_{n}\right)}{V\left(x_{0},x_{1},\ldots,x_{n}\right)}$		(17)
	$\displaystyle\cdot\cdots\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot$
	$\displaystyle K_{n}=(-1)^{p+n-1}\frac{1}{\left(x_{n}-x\right)_{p}}\cdot\frac{V\left(x_{0},x_{1},\ldots x_{n-1}\right)}{V\left(x_{0},x_{1},\ldots,x_{n}\right)}$

If we take into account that in general we have

	$\displaystyle V\left(x_{0},x_{1},\ldots,x_{n}\right)=$	$\displaystyle\left(x_{i}-x_{0}\right)\left(x_{i}-x_{1}\right)\ldots\left(x_{i}-x_{i-1}\right)\left(x_{i+1}-x_{i}\right)\ldots\left(x_{n}-x_{i}\right)\times$
		$\displaystyle\times V\left(x_{0},x_{1},\ldots,x_{i-1},x_{i+1},\ldots,x_{n}\right)$

formulas (17) reduce to 1a

K_{1}=\frac{(-1)^{p}}{\left(x_{1}-x_{0}\right)^{p+1}\left(x_{2}-x_{1}\right)\left(x_{3}-x_{1}\right)\ldots\left(x_{n}-x_{1}\right)}

	$\displaystyle K_{2}=\frac{(-1)^{p+1}}{\left(x_{2}-x_{0}\right)^{p+1}\left(x_{2}-x_{1}\right)\left(x_{3}-x_{2}\right)\ldots\left(x_{n}-x_{2}\right)}$		(18)
	$\displaystyle K_{n}=\frac{(-1)^{p+n-1}}{\left(x_{n}-x_{0}\right)^{n+1}\left(x_{n}-x_{1}\right)\left(x_{n}-x_{2}\right)\ldots\left(x_{n}-x_{n-1}\right)}$

From these expressions it can be seen that all the denominators being positive, the coefficients $K_{1},K_{2},\ldots,K_{n}$ have alternating signs, $K_{1}$ having the sign of $(-1)^{p}$ .

Coefficients $\lambda_{1},v_{1},\ldots,v_{p}$ are given by the first equation (10) and the first $p$ equations (9). T, Taking into account formulas (17), we will generally have

	$\displaystyle v_{j}=\frac{(-1)^{p}}{V\left(x_{0},x_{1},\ldots,x_{n}\right)}[$	$\displaystyle\frac{1}{\left(x_{1}-x_{0}\right)^{p-j}}V\left(x_{0},x_{2},\ldots,x_{n}\right)-$
	$\displaystyle-$	$\displaystyle\frac{1}{\left(x_{2}-x_{0}\right)^{p-j}}V\left(x_{0},x_{1},x_{3},\ldots,x_{n}\right)+\ldots+$
		$\displaystyle\left.+(-1)^{n-1}\frac{1}{\left(x_{n}-x_{0}\right)^{p-j}}V\left(x_{0},x_{1},\ldots,x_{n-1}\right)\right]$

v_{j}=\frac{(-1)^{p+n-1}}{V\left(x_{0},x_{1},\ldots,x_{n}\right)}\left|\begin{array}[]{cccc}1&1&\cdots&1\\ x_{0}&x_{1}&\cdots&x_{n}\\ \vdots&\vdots&&\vdots\\ x_{0}{}^{n-1}&x_{1}{}^{n-1}&\cdots&x_{n}^{n-1}\\ 0&\frac{1}{\left(x_{1}-x_{0}\right)^{p-j}}&\cdots&\frac{1}{\left(x_{n}-x_{0}\right)^{p-j}}\end{array}\right|

or yet
$v_{j}=(-1)^{p+n-1}\frac{\left(x_{1}-x_{0}\right)\left(x_{2}-x_{0}\right)\ldots\left(x_{n}-x_{0}\right)}{V\left(x_{0},x_{1},\ldots,x_{n}\right)}\left|\begin{array}[]{ccc}1&\cdots&1\\ x_{1}-x_{0}&\cdots&x_{n}-x_{0}\\ \vdots&&\vdots\\ \frac{\left(x_{1}-x_{0}\right)^{n-2}}{\left(x_{1}-x_{0}\right)^{p-j+1}}\cdots&\cdots&\left(x_{n}-x_{0}\right)^{n-2}\\ \left(x_{n}-x_{0}\right)^{p-j+1}\end{array}\right|$ , where $j=0,1,\ldots,p\cdot\left(\lambda_{1}=v_{0}\right)$ .

It is known that
$\left|\begin{array}[]{ccc}1&\ldots&1\\ y_{1}&\ldots&y_{n}\\ \vdots&&\vdots\\ y_{1}^{n-2}&&y_{n}^{n-2}\\ \frac{1}{y_{1}^{l}}&\ldots&\frac{1}{y_{n}^{l}}\end{array}\right|=(-1)^{n-1}\frac{V\left(y_{1}y_{2}\cdots y_{n}\right)}{\left(y_{1},y_{2},\cdots y_{n}\right)}Q_{-1}\left(\frac{1}{y_{1}},\frac{1}{y_{2}},\ldots,\frac{1}{y_{n}}\right)$ ,
where $Q_{l-1}\left(\frac{1}{y_{1}},\frac{1}{y_{2}},\ldots,\frac{1}{y_{n}}\right)$ is a homogeneous polynomial of degree $l-1$ in $\frac{1}{y_{1}},\frac{1}{y_{2}},\ldots,\frac{1}{y_{n}}$ , with coefficients equal to 1 .

Taking this into account, the final expression of the coefficients $v_{j}$ is $v_{j}=\frac{(-1)^{p}}{\left(x_{1}-x_{0}\right)\left(x_{2}-x_{0}\right)\ldots\left(x_{n}-x_{0}\right)}Q_{p-j}\left(\frac{1}{x_{1}-x_{0}},\frac{1}{x_{2}-x_{0}},\cdots,\frac{1}{x_{n}-x_{0}}\right)$ , where $j=0,1,\ldots,n.\left(\lambda_{1}=v_{0}\right)$ .

It is found that all the coefficients $\nu_{j}$ they have his sign $(-1)^{p}$ .
4. Another form of the numerical derivation formula (6). Let us consider the rational function

\frac{1}{\left(x-x_{0}\right)^{p+1}\left(x-x_{1}\right)\ldots\left(x-x_{n}\right)}

which we decompose into simple rational functions highlighting the residues $A_{1},A_{2},\ldots,A_{n}$ relative to the nodes $x_{1},x_{2},\ldots,x_{n}$ We will have
$\frac{1}{\left(x-x_{0}\right)^{p+1}\left(x-x_{1}\right)\ldots\left(x-x_{n}\right)}=\frac{A_{1}}{x-x_{1}}+\frac{A_{2}}{x-x_{2}}+\ldots+\frac{A_{n}}{x-x_{n}}+\ldots$ ,
where

		$\displaystyle A_{1}=\frac{(-1)^{n-1}}{\left(x_{1}-x_{0}\right)^{p+1}\left(x_{2}-x_{1}\right)\ldots\left(x_{n}-x_{1}\right)}$
		$\displaystyle A_{2}=\frac{(-1)^{n-2}}{\left.\left(x_{2}-x_{0}\right)^{p+1}\left(x_{2}-x_{1}\right)\left(x_{3}-x_{2}\right)\ldots x_{n}-x_{2}\right)}$
		$\displaystyle A_{n}=\frac{1}{\left(x_{n}-x_{0}\right)^{p+1}\left(x_{n}-x_{1}\right)\ldots\left(x-x_{n-1}\right)}$

Taking these into account, formulas (18) show that we have

	$\displaystyle K_{1}=(-1)^{n+p-1}A_{1}$
	$\displaystyle K_{2}=(-1)^{n+p-1}A_{2}$		(21)
	$\displaystyle K_{n}=(-1)^{n+p-1}A_{n}$

From the first equation (10) and from the first $p$ equations (9) it follows that

\begin{gathered}\lambda_{1}f\left(x_{0}\right)+\frac{v_{1}}{1!}f^{\prime}\left(x_{0}\right)+\ldots+\frac{v_{p}}{p!}f^{(p)}\left(x_{0}\right)=\\ =\sum_{i=1}^{n}K_{i}\left[f\left(x_{0}\right)+\frac{x_{i}-x_{0}}{1!}f^{\prime}\left(x_{0}\right)+\ldots+\frac{\left(x_{i}-x_{0}\right)^{p}}{p!}f\left(x_{0}\right)\right]\end{gathered}

The numerical derivation formula (6) can therefore be written in the form

	$\displaystyle\sum_{i=1}^{n}K_{i}\left\{f\left(x_{i}\right)-\left[f\left(x_{0}\right)+\frac{x_{i}-x_{0}}{1!}f^{\prime}\left(x_{0}\right)+\ldots+\frac{\left(x_{i}-x_{0}\right)^{p}}{p!}f^{(p)}\left(x_{0}\right)\right]\right\}=$
	$\displaystyle=(-1)^{n+p-1}\int_{x_{0}}^{x_{n}}\varphi(x)f^{(n+p)}(x)dx$		(22)

which only contains the coefficients $K_{i}$ , given by formulas (18), both in the first and second members, due to formulas (8).

But formulas (21) show that the coefficients $K_{1},K_{2},\ldots,K_{n}$ are proportional to $A_{1},A_{2},\ldots,A_{n}$ , which means that the numerical derivation formulas (6) or (22) can also be written in the form

	$\displaystyle\sum_{i=1}^{i}A_{i}\left\{f\left(x_{i}\right)-\left[f\left(x_{0}\right)+\frac{x_{i}-x_{0}}{1!}f^{\prime}\left(x_{0}\right)+\ldots+\frac{\left(x_{i}-x_{0}\right)^{p}}{p!}f^{(p)}\left(x_{0}\right)\right]\right\}=$
	$\displaystyle=(-1)^{n+p-1}\int_{x_{0}}^{x_{n}}\varphi(x)f^{(n+p)}(x)dx$		( $\prime$ )

In this formula the function $\varphi(x)$ coincides in turn with the functions $\varphi_{1}(x),\varphi_{2}(x),\ldots,\varphi_{n}(x)$ in the intervals $\left[x_{0},x_{1}\right],\left[x_{1},x_{2}\right],\ldots,\left[x_{n-1},x_{n}\right]$ where, according to formulas (8), we have
$\varphi_{1}(x)=A_{1}\frac{\left(x-x_{1}\right)^{n+p-1}}{(n+p-1)!}+A_{2}\frac{\left(x-x_{2}\right)^{n+p-1}}{(n+p-1)!}+\ldots+A_{n}\frac{\left(x-x_{n}\right)^{n+p-1}}{(n+p-1)!}$

\varphi_{2}(x)=\quad A_{2}\frac{\left(x-x_{2}\right)^{n+p-1}}{(n+p-1)!}+\ldots+A_{n}\frac{\left(x-x_{n}\right)^{n+p-1}}{(n+p-1)!}

(23)

$\varphi_{n}(x)=$

A_{n}\frac{\left(x-x_{n}\right)^{n+p-1}}{(n+p-1)!}

In formulas (22) and (22') the factor by which -1 multiplies $K_{i}$ is the difference between $f\left(x_{i}\right)$ and the first $p+1$ terms from his development $f(x)$ according to his/her abilities $x_{i}-x_{0}$ using Taylor's formula. This observation provides a practical procedure for writing numerical derivation formulas of the form (6).

Example. Suppose $n=3$ and $x_{1}=x_{0}+h,x_{2}=x_{0}+2h$ , $x_{3}=x_{0}+3h$ and $p=2$ To find the corresponding numerical derivation formula, we start by first decomposing it into simple rational functions.

\frac{1}{\left(x-x_{0}\right)^{3}\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)}=\frac{A_{1}}{x-x_{1}}+\frac{A_{2}}{x-x_{2}}+\frac{A_{3}}{x-x_{3}}+\ldots

where

A_{1}=\frac{1}{2h^{5}},A_{2}=-\frac{1}{8h^{5}},A_{3}=\frac{1}{54h^{5}}.

The numerical derivation formula is therefore

\begin{gathered}\left.\frac{1}{2h^{5}}\left[fx_{1}\right)-f\left(x_{0}\right)-hf^{\prime}\left(x_{0}\right)-\frac{h^{2}}{2!}f^{\prime\prime}\left(x_{0}\right)\right]-\\ -\frac{1}{8h^{5}}\left[f\left(x_{2}\right)-f\left(x_{0}\right)-2hf^{\prime}(x)-\frac{(2h)^{2}}{2!}f^{\prime\prime}\left(x_{0}\right)\right]+\\ +\frac{1}{54h^{5}}\left[f\left(x_{3}\right)-f\left(x_{0}\right)-3hf^{\prime}\left(x_{0}\right)-\frac{(3h)^{2}}{2!}f^{\prime\prime}\left(x_{0}\right)\right]=\int_{x_{0}}^{x_{3}}\varphi(x)f^{(5)}(x)dx,\end{gathered}

MEAN

	$\displaystyle\frac{1}{216h^{5}}\left[108f\left(x_{1}\right)-27f\left(x_{2}\right)\right.$	$\displaystyle\left.+4f\left(x_{3}\right)-85f\left(x_{0}\right)-66hf^{\prime}\left(x_{0}\right)-18h^{2}f^{\prime\prime}(x)\right]=$
		$\displaystyle=\int_{x_{0}}^{x_{3}}\varphi(x)f^{(5)}(x)dx$		(24)

In this formula we have

\begin{array}[]{rlrl}\varphi_{1}(x)=&\frac{1}{2h^{5}}\frac{\left(x-x_{1}\right)^{4}}{4!}-\frac{1}{8h^{5}}\frac{\left(x-x_{2}\right)^{4}}{4!}+\frac{1}{54h^{5}}\frac{\left(x-x_{3}\right)^{4}}{4!}\\ \varphi_{2}(x)=&-\frac{1}{8h^{5}}\frac{\left(x-x_{2}\right)^{4}}{4!}+\frac{1}{54h^{5}}\frac{\left(x-x_{3}\right)^{4}}{4!}\\ \varphi_{3}(x)=&&\frac{1}{54h^{5}}\frac{\left(x-x_{2}\right)^{4}}{4!}\end{array}

5.

constants $\lambda_{1},\lambda_{2},\ldots,\lambda_{n}$ from the differential equations (4) are all different from zero. To prove this, let us observe that from formulas (7) and (21) we have

	$\displaystyle\lambda_{1}=(-1)^{p+n-1}\left(A_{1}+A_{2}+\cdots+A_{n}\right)$
	$\displaystyle\lambda_{2}=(-1)^{p+n-1}\left(A_{2}+A_{3}+\cdots+A_{n}\right)$		(25)
	$\displaystyle\lambda_{n}=(-1)^{p+n-1}A_{n}.$

So to prove that the constants $\lambda_{1},\lambda_{2},\ldots,\lambda_{n}$ are different from zero, it is the same as proving that $A_{1},A_{2},\ldots A_{n}$ being the residues of the rational function (20) relative to the poles $x_{1},x_{2},\ldots,x_{n}$ , the amounts $A_{j}+A_{j+1}+\ldots+A_{n}$ are different from zero.

Let's consider the polynomial

h_{k}(x)=\frac{\left(x-x_{0}\right)^{p+1}\left(x-x_{1}\right)\ldots\left(x-x_{k-1}\right)\left(x-x_{k+1}\right)\ldots\left(x-x_{n}\right)}{\left(x_{k}-x_{0}\right)^{p+1}\left(x_{k}-x_{1}\right)\ldots\left(x_{k}-x_{k-1}\right)\left(x_{k}-x_{k+1}\right)\ldots\left(x_{k}-x_{n}\right)},

which cancels out in the node $x_{0}$ together with the first $p$ derivatives and in the nodes $x_{1},x_{2},\ldots,x_{k-1},x_{k+1},\ldots,x_{n}$ and which takes the value 1 in the node $x_{k}$ The polynomial $h_{k}(x)$ , is a polynomial of degree $p+n$ whose first term is $A_{k}x^{p+n},A_{k}$ being the residue of the pole $x_{k}$ of the rational function (20).

polynomial
$h(x)=h_{j}(x)+h_{j+1}(x)+\ldots+h_{n}(x)=\left(A_{j}+A_{j+1}+\ldots+A_{n}\right)x^{p+n}+\ldots$ has the root $x_{0}$ multiple of the order $p+1$ and we have
again

h\left(x_{j}\right)=1,\quad h\left(x_{j+1}\right)=1,\ldots,h\left(x_{n}\right)=1

h\left(x_{1}\right)=0,\quad h\left(x_{2}\right)=0,\ldots,h\left(x_{j-1}\right)=0.

Applying to the polynomial $h(x)$ Rolle's theorem 1a intervals $\left[x_{0},x_{1}\right],\ldots,\left[x_{j-2},x_{j-1}\right]$ and at intervals $\left[x_{j},x_{j+1}\right],\ldots,\left[x_{n-1},x_{n}\right]$ , we deduce that $h(x)$ has the root $x_{0}$ multiple of the order $p$ and others $n-1$ distinct roots to the right of $x_{0}$ . Repeating this reasoning to the polynomial $h^{\prime}(x)$ , then to $h^{\prime\prime}(x),\ldots$ we deduce that the polynomial $h^{(p)}(x)$ has the root $x_{0}$ and others $n-1$ distinct roots 1st right of $x_{0}$ We have
$h^{(p)}(x)=(n+p)(n+p-1)\ldots(p+1)\left(A_{j}+A_{j+1}+\ldots+A_{n}\right)x^{n}+\ldots$
If $A_{j}+A_{j+1}+\ldots+A_{n}=0$ , then $h(x)$ is a polynomial of degree less than $n$ , which having $n$ distinct roots is identically null. It follows that $h^{(p-1)}(x)=0$ , because $h^{(p-1)}\left(x_{0}\right)=0$ In $\bmod$ analogously it follows that $h^{(p-2)}(x)=0\ldots$ and so on that $h^{\prime}(x)=0$ It would therefore follow that the polynomial $h(x)$ to be a constant, which is impossible because $h\left(x_{0}\right)=0$ , while $h\left(x_{n}\right)=1$ .

So we have $A_{j}+A_{j+1}+\ldots+A_{n}\neq 0$ and therefore all the constants $\lambda_{1},\lambda_{2},\ldots,\lambda_{n}$ are different from zero.

We can specify the sign of the sum $A_{j}+A_{j+1}+\ldots+A_{n}$ For this, we note that in the interval $\left[x_{j-1},x_{j}\right]$ polynomial $h(x)$ is positive. It follows that $h(x)$ has a maximum in the range $\left[x_{j},x_{j+1}\right]$ , a minimum in the range $\left[x_{j+1},x_{j+2}\right],\ldots$ and that in the last interval $\left[x_{n-1},x_{n}\right],h(x)$ has a maximum or a minimum depending on $(-1)^{n-j}$ is -1 or +1 .

If $h(x)$ has a maximum in the range $\left[x_{n-1},x_{n}\right]$ , then $h(x)$ tends towards $-\infty$ when $x$ tends towards $+\infty$ ; therefore $A_{j}+A_{j+1}+\ldots+A_{n}<0$ .

If $h(x)$ has a minimum in the range $\left[x_{n-1},x_{n}\right]$ , then $h(x)$ tends towards $+\infty$ when $x$ tends towards $+\infty$ , so $A_{j}+A_{j+1}+\ldots+A_{n}>0$ .

Therefore the amount $A_{j}+A_{j+1}+\ldots+A_{n}$ has his sign $(-1)^{n-j}$ and we can add according to formulas (25) that $\lambda_{j}$ has his sign $(-1)^{p+j+1}$ .
6. Function $\varphi(x)$ from the numerical derivation formula (6) or (22) or (22') is positive in the interval ( $x_{0},x_{n}$ ). For $n=1$ , this is obvious because formula (6) reduces in this case to Taylor's formula with the remainder written in Lagrange's form.

Let us suppose then $n\geqslant 2$ Function $\varphi(x)$ satisfy in the nodes $x_{0}$ and $x_{n}$ under the following conditions

		$\displaystyle\varphi\left(x_{0}\right)=0,\quad\varphi^{\prime}\left(x_{0}\right)=0,\ldots,\varphi^{(n-2)}\left(x_{0}\right)=0$
		$\displaystyle\varphi\left(x_{n}\right)=0,\quad\varphi^{\prime}\left(x_{n}\right)=0,\ldots,\varphi^{(n+p-2)}\left(x_{n}\right)=0$

and in the interval $\left[x_{n-1},x_{n}\right]$ it coincides with

\varphi_{n}(x)=\lambda_{n}\frac{\left(x-x_{n}\right)^{n+p-1}}{(n+p-1)!}=(-1)^{n+p-1}\lambda_{n}\frac{\left(x_{n}-x\right)^{n+p-1}}{(n+p-1)!}

But it was shown in No. 4 that $\lambda_{n}$ has his sign $(-1)^{n+p+1}$ , from which it follows that $\varphi(x)=\varphi_{n}(x)$ is positive in the interval $\left[x_{n-1}x_{n}\right)$ .

Function $\varphi(x)$ is continuous in the interval $\left[x_{0},x_{n}\right]$ and has continuous derivatives up to the order $n+p-2$ , derivative $\varphi^{(n+p-1)}(x)$ being discontinuous at the nodes $x_{1},x_{2},\ldots,x_{n-1}$ . Function $\varphi(x)$ canceling out at the nodes $x_{0}$ and $x_{n}$ , according to Rolle's theorem, its derivative $\varphi^{\prime}(x)$ has at least one zero in the interval ( $x_{0},x_{n}$ ). Let us show that it can only have one zero.

Indeed, if $\varphi^{\prime}(x)$ would have two zeros in the interval ( $x_{0},x_{n}$ ), successively applying Rolle's theorem would result in $\varphi^{\prime\prime}(x),\ldots,\varphi^{(n-2)}(x)$ to have $3,\ldots,n-1$ zeros in the range ( $x_{0},x_{n}$ ). Continuing the same reasoning, it would follow that $\varphi^{(n-1)}(x),\varphi^{(n)}(x),\ldots,\varphi^{(n+p-2)}(x)$ to have $n$ zeros in the range ( $x_{0},x_{n}$ ). Let us show that this is an impossibility.

We first observe that no zero of $\varphi^{(n+p-2)}(x)$ not found in range1 $\left[x_{n-1},x_{n}\right)$ We also observe that in an interval. $\left(x_{k-1},x_{k}\right]$ , where $k=1,2,\ldots,n-1,\varphi^{(n+p-2)}(x)$ cannot have two zeros, because applying Rolle's theorem it would follow that $\varphi^{(n+p-1)}(x)$ to vanish at a
point in the interval ( $x_{k-1},x_{k}$ ), which is impossible because in this interval we have $\varphi^{(n+p-1)}(x)=\varphi_{k}^{(n+p-1)}(x)=\lambda_{k}\neq 0$ . So in the interval ( $x_{0},x_{n-1}$ ) derivative $\varphi^{(n+p-2)}(x)$ can only have at most $n-1$ zeros, while above it was shown that in the same interval $\varphi^{(n+p-2)}(x)$ has $n$ zeros.

So the hypothesis that $\varphi^{\prime}(x)$ would have two zeros in the interval ( $x_{0},x_{n}$ ) leading to a contradiction, we deduce that $\varphi^{\prime}(x)$ has a single zero in the interval ( $x_{0},x_{n-1}$ ). The function $\varphi(x)$ being positive in the interval ( $x_{n-1},x_{n}$ ) and its derivative $\varphi^{\prime}(x)$ having a single zero in the interval ( $x_{0},x_{n-1}$ ), is positive in the interval ( $x_{0},x_{n}$ ).

Function graph $\varphi(x)$ is given in Fig. 1.

7.

The remainder in the numerical derivation formula (6), (22) or (22'). The second member of the numerical derivation formula is

R=(-1)^{n+p+1}\int_{x_{0}}^{x_{n}}\varphi(x)f^{(n+p)}(x)dx

(26)

and is called the remainder of the formula.
Since it has been shown that $\varphi(x)$ is a positive function in the interval $\left(x_{0},x_{n}\right)$ , the rest can be written in the form

R=(-1)^{n+p+1}f^{(n+p)}(\xi)\int_{x_{0}}^{x_{n}}\varphi(x)dx

where $\xi\in\left(x_{0},x_{n}\right)$ The integral in the second term is calculated using the numerical derivative formula (6), replacing $f(x)$ with the polynomial

f(x)=\frac{\left(x-x_{0}\right)^{p}\left(x-x_{1}\right)\ldots\left(x-x_{n}\right)}{(n+p)!}

Formula (6) gives us

(-1)^{n+p+1}\int_{x_{0}}^{x_{n}}\varphi(x)dx=-\frac{v_{p}}{p!}f^{(p)}\left(x_{0}\right)

\frac{f^{(p)}\left(x_{0}\right)}{p!}=(-1)^{n}\frac{\left(x_{1}-x_{0}\right)\ldots\left(x_{n}-x_{0}\right)}{(n+p)!}

and formula (19) gives

It follows that

v_{p}=\frac{(-1)^{p}}{\left(x_{1}-x_{0}\right)\cdots\left(x_{n}-x_{0}\right)}

\int_{x_{0}}^{x_{n}}\varphi(x)dx=\frac{1}{(n+p)!}

and therefore the rest $R$ given by formula (26) can also be written in the form

R=(-1)^{n+p+1}\frac{f^{(n+p)}(\xi)}{(n+p)!}

(

\prime

)

where $\xi\in\left(x_{0},x_{n}\right)$ If
the numerical derivation formula (6) is written in the form (16) and we denote by $R_{1}$ the second member of this formula, then we have according to the calculations above

R_{1}=f\frac{(n+p)}{(n+p)!}

and we find here a well-known theorem on the divided difference in the first member of formula (16).

If we note with $M_{n+p}$ an upper edge of $\left|f^{(n+p)}(x)\right|$ in the interval $[a,b]$ , then from the formula $\left(26^{\prime}\right)$ we deduce the following evaluation of $|R|$

|R|\leqslant\frac{M_{n+p}}{(n+p)!}

(27)

§ 2. Generalization of AA Markov's formulas and putting the rest of these formulas in the form of a definite integral

8.

Whether $f(x)$ a class function $C^{n+1}$ in the interval $[a,b]$ and $x_{0},x_{1},\ldots,x_{n}$ nodes in this interval, so that $x_{0}<x_{1}<x_{2}<\ldots<x_{n}$ If the nodes $x_{0},x_{1},\ldots,x_{n}$ are in arithmetic progression with the relationship $h$ , Markov's formulas [4],

h^{p}f^{(p)}\left(x_{0}\right)=\sum_{i=p}^{n}A_{i}\Delta^{i}f\left(x_{0}\right)

(28)

I give the derivative. $f^{(p)}\left(x_{0}\right)$ with the help of his values $f(x)$ on the nodes $x_{0},x_{1},\ldots,x_{n}$ , where $p=1,2,\ldots,n$ .

In this paragraph we will establish formulas of the Markov type, without assuming that the nodes $x_{0},x_{1},\ldots,x_{n}$ are in arithmetic progression and we will study the rest of these formulas which we will put in the form of a definite integral.
9. We therefore assume the nodes $x_{0},x_{1},\ldots,x_{n}$ anyway in the interval $[a,b]$ and ordered so that $x_{0}<x_{1}<\ldots<x_{n}$ At intervals $\left[x_{0},x_{1}\right]$ , $\left[x_{1},x_{2}\right],\ldots,\left[x_{n-1},x_{n}\right]$ we attach the functions $\varphi_{1}(x),\varphi_{2}(x),\ldots,\varphi_{n}(x)$ and the constants $\lambda_{1},\lambda_{2},\ldots,\lambda_{n}$ and then we consider the differential equations

\varphi_{1}^{(n)}(x)=\lambda_{1},\varphi_{2}^{(n)}(x)=\lambda_{2},\ldots,\varphi_{n}^{(n)}(x)=\lambda^{n}

(29)

which we integrate with the boundary conditions

	$\displaystyle\varphi_{1}\left(x_{0}\right)=0,\quad\varphi_{1}^{\prime}\left(x_{0}\right)=0,\ldots,\varphi_{1}^{(n-p-1)}\left(x_{0}\right)=0$
	$\displaystyle\varphi_{1}^{(n-p)}\left(x_{0}\right)=(-1)^{p}\frac{\mu}{p!},\quad\varphi_{1}^{(n-p+1)}\left(x_{0}\right)=0,\ldots,\varphi_{1}^{(n-1)}\left(x_{0}\right)=0$		(30)
	$\displaystyle\varphi_{2}\left(x_{1}\right)=\varphi_{1}\left(x_{1}\right),\quad\varphi_{2}^{\prime}\left(x_{1}\right)=\varphi_{1}^{\prime}\left(x_{1}\right),\ldots,\varphi_{2}^{(n-1)}\left(x_{1}\right)=\varphi_{1}^{(n-1)}\left(x_{1}\right)$
	$\displaystyle\varphi_{n}\left(x_{n-1)}\right)=\varphi_{n-1}\left(x_{n-1}\right),\quad\varphi_{n}^{\prime}\left(x_{n-1}\right)=\varphi_{n-1}^{\prime}\left(x_{n-1}\right),\ldots$
	$\displaystyle\quad\varphi_{n}^{(n-1)}\left(x_{n-1}\right)=\varphi_{n-1}^{(n-1)}\left(x_{n-1}\right)$
	$\displaystyle\varphi_{n}\left(x_{n}\right)=0,\quad\varphi_{n}^{\prime}\left(x_{n}\right)=0,\ldots,\varphi_{n}^{(n-1)}\left(x_{n}\right)=0$

constants $\lambda_{1},\lambda_{2},\ldots,\lambda_{n}$ and $\mu$ from the differential equations (29) and the boundary conditions (30) it is determined that the functions $\varphi_{1}(x),\ldots,\varphi_{n}(x)$ to verify both the differential equations and the boundary conditions.

Each integral

\int_{x_{0}}^{x_{1}}\varphi_{1}^{(n+1)}(x)f(x)dx,\int_{x_{1}}^{x_{2}}\varphi_{2}^{(n+1)}(x)f(x)dx,\ldots,\int_{x_{n-1}}^{x_{n}}\varphi_{n}^{(n+1)}(x)f(x)dx

is zero. Applying to each the generalized integration by parts formula and taking their sum, we obtain the numerical derivative formula

	$\displaystyle-\mu\frac{f^{(p)}\left(x_{0}\right)}{p!}-\lambda_{1}f\left(x_{0}\right)+K_{1}f\left(x_{1}\right)+\ldots+K_{n}f\left(x_{n}\right)=$
	$\displaystyle=(-1)^{n}\int_{x_{0}}^{x_{n}}\varphi(x)f^{(n+1)}(x)dx$		(31)

where

K_{1}=\lambda_{1}-\lambda_{2},\quad K_{2}=\lambda_{2}-\lambda_{3},\ldots,K_{n-1}=\lambda_{n-1}-\lambda_{n},K_{n}=\lambda_{n}.

(32)

In establishing formula (31) it was assumed that $1\leqslant p\leqslant n$ To
show that formula (31) makes sense, we need to determine the functions $\varphi_{1}(x),\ldots,\varphi_{n}(x)$ which verifies the differential equations (29) and the boundary conditions (30).

Differential equations (29) and boundary conditions at the nodes $x_{1},x_{2},\ldots,x_{n}$ they give us

	$\displaystyle\varphi_{n}(x)=K_{n}\frac{\left(x-x_{n}\right)^{n}}{n!}$
	$\displaystyle\varphi_{n-1}(x)=K_{n-1}\frac{\left(x-x_{n-1}\right)^{n}}{n!}+K_{n}\frac{\left(x-x_{n}\right)^{n}}{n!}$		(33)
	$\displaystyle\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots K_{n}\frac{\left(x-x_{n}\right)^{n}}{n!}.$

Writing that and the boundary conditions at point $x_{0}$ are satisfied, we have the system of linear and homogeneous equations

	$\displaystyle K_{1}\left(x_{1}-x_{0}\right)+K_{2}\left(x_{2}-x_{0}\right)+\ldots+K_{n}\left(x_{n}-x_{0}\right)=0$
	$\displaystyle K_{1}\left(x_{1}-x_{0}\right)^{p-1}+K_{2}\left(x_{2}-x_{0}\right)^{p-1}+\cdots+K_{n}\left(x_{n}-x_{0}\right)^{p-1}=0$
	$\displaystyle K_{1}\left(x_{1}-x_{0}\right)^{p}+K_{2}\left(x_{2}-x_{0}\right)^{p}+\ldots+K_{n}\left(x_{n}-x_{0}\right)^{p}=\mu$		(34)
	$\displaystyle K_{1}\left(x_{1}-x_{0}\right)^{p+1}+K_{2}\left(x_{2}-x_{0}\right)^{p+1}+\ldots+K_{n}\left(x_{n}-x_{0}\right)^{p+1}=0$
	$\displaystyle K_{1}\left(x_{1}-x_{0}\right)^{n}+K_{2}\left(x_{2}-x_{0}\right)^{n}+\ldots+K_{n}\left(x_{n}-x_{0}\right)^{n}=0,$

which determines the $K_{1},K_{2},\ldots,K_{n}$ , or the system

	$\displaystyle-\lambda_{1}+K_{1}+\ldots+K_{n}=0$
	$\displaystyle-\lambda_{1}x_{0}^{p}-C_{p}^{p}\mu+K_{1}x_{1}^{p}+\ldots+K_{n}x_{n}^{p}=0$
	$\displaystyle-\lambda_{1}x_{0}^{p+1}-C_{p+1}^{p}\mu x_{0}+K_{1}x_{1}^{p+1}+\cdots+K_{n}x_{n}^{p+1}=0$		(35)
	$\displaystyle-\lambda_{1}x_{0}^{n}-C_{n}^{p}\mu x_{0}^{n-p}+K_{1}x_{1}^{n}+\ldots+K_{n}x_{n}^{n}=0,$

which determines the $K_{1},K_{2},\ldots K_{n},\lambda$ and $\mu$ We consider
the matrix

\begin{array}[]{||ccccc||}1&0&1&\ldots&1\\ x_{0}&0&x_{1}&\ldots&x_{n}\\ \vdots&\vdots&\vdots&&\vdots\\ x_{0}^{p}&C_{p}^{p}&x_{1}^{p}&\ldots&x_{n}^{p}\\ \vdots&\vdots&\vdots&&\vdots\\ x_{0}^{n}&C_{n}^{p}x_{0}^{n-p}&x_{1}^{n}&\ldots&x_{n}^{n}\end{array}\|

and we denote by $A_{1},A_{2},\ldots,A_{n+2}$ determinants that have the same rows and columns except $1-\mathrm{a}$ , of $2-\mathrm{a},\ldots$ , of $(n+2)$ -a. All these determinants are different from zero, and solving system (35) in $-\lambda_{1},-\mu,K_{1},\ldots,K_{n}$ we will have

\frac{-\lambda_{1}}{A_{1}}=\frac{-\mu}{-A_{2}}=\frac{K_{1}}{A_{3}}=\ldots=\frac{(-1)^{n+1}}{A_{n+2}}

(36)

The first member of the numerical derivation formula being a linear combination of the numerators in formulas (36), these formulas can also be written in the form

\frac{-\lambda_{1}}{A_{1}}=\frac{-\mu}{-A_{2}}=\frac{K_{1}}{A_{3}}=\ldots=\frac{K_{n}}{(-1)^{n+1}A_{n+2}}=\frac{-\int_{x_{0}}\varphi(x)f^{(n+1)}(x)dx}{\Delta},

(37)

where

\Delta=\left|\begin{array}[]{ccccc}1&0&1&\ldots&1\\ x_{0}&0&x_{1}&\ldots&x_{n}\\ \vdots&\vdots&\vdots&&\vdots\\ x_{0}^{p}&C_{p}^{p}&x_{1}^{p}&\ldots&x_{n}^{p}\\ \vdots&\vdots&\vdots&&\vdots\\ x_{0}^{n}&C_{n}^{p}x_{0}^{n-p}&x_{1}^{n}&\ldots&x_{n}^{n}\\ f\left(x_{0}\right)&\frac{f^{(p)}\left(x_{0}\right)}{p!}&f\left(x_{1}\right)&\ldots&f\left(x_{n}\right)\end{array}\right|.

Let us prove that the determinant $\Delta_{1}$ which is obtained from $\Delta$ replacing $f(x)\penalty 10000\ cux^{n+1}$ is different from zero. For this, we consider the Vandermonde determinant

\left|\begin{array}[]{lllll}1&1&1&\ldots&1\\ x_{0}&x&x_{1}&\ldots&x_{n}\\ \vdots&\vdots&\vdots&&\vdots\\ x_{0}^{p}&x^{p}&x_{1}^{p}&\ldots&x_{n}^{p}\\ \vdots&\vdots&\vdots&&\vdots\\ x_{0}^{n+1}&x^{n+1}&x_{1}^{n+1}&\ldots&x_{n}^{n+1}\end{array}\right|=(-1)^{n}\left(x-x_{0}\right)\left(x-x_{1}\right)\ldots\left(x-x_{n}\right)\times

where $V\left(x_{0},x_{1},\ldots,x_{n}\right)$ is the Vandermonde determinant of distinct numbers $x_{0},x_{1},\ldots,x_{n}$ . Differentiating both members with respect to $x$ of $p$ times, dividing by $p$ ! and then doing $x=x_{0}$ , we obtain the value of the determinant $\Delta_{1}$ Let's note

h(x)=\left(x-x_{0}\right)\left(x-x_{1}\right)\ldots\left(x-x_{n}\right).

We will have

\frac{h^{(p)}\left(x_{0}\right)}{p!}=\Sigma\left(x_{0}-x_{1}\right)\left(x_{0}-x_{2}\right)\ldots\left(x_{0}-x_{n-p+1}\right)

Noting in general

\mu_{k}\left(x_{1}-x_{0},\ldots,x_{n}-x_{0}\right)=\sum\left(x_{1}-x_{0}\right)\left(x_{2}-x_{0}\right)\ldots\left(x_{k}-x_{0}\right)>0

(39)

we will have

\frac{h^{(p)}\left(x_{0}\right)}{p!}=(-1)^{n-p+1}\mu_{n-p+1}\left(x_{1}-x_{0},\ldots,x_{n}-x_{0}\right)

It follows that

\Delta_{1}=(-1)^{p-1}V\left(x_{0},x_{1},\ldots,x_{n}\right)\mu_{n-p+1}\left(x_{1}-x_{0},\ldots,x_{n}-x_{0}\right)\neq 0

(40)

Equating the ratios (37) with $-\frac{1}{\Delta_{1}}$ , we deduce that the numerical derivation formula (31) can also be written in the form

\frac{\Delta}{\Delta_{1}}=\int_{x_{0}}^{x_{n}}\varphi(x)f^{(n+1)}(x)dx

(41)

where $\Delta$ is the determinant (38), and $\Delta_{1}$ is obtained from $\Delta$ replacing $f(x)$ with $x^{n+1}$ .
10. Calculation of coefficients $\lambda_{1},\mu,K_{1},K_{2},\ldots,K_{n}$ From the formulas

\frac{-\lambda_{1}}{A_{1}}=\frac{-\mu}{-A_{2}}=\frac{K_{1}}{A_{3}}=\ldots=\frac{K_{n}}{(-1)^{n+1}A_{n+2}}=-\frac{1}{\Delta_{1}}

it follows that

\lambda_{1}=\frac{A_{1}}{\Delta_{1}},\mu=-\frac{A_{2}}{\Delta_{1}},K_{1}=-\frac{A_{3}}{\Delta_{1}},\ldots,K_{n}=(-1)^{n}\frac{A_{n+2}}{\Delta_{1}}

But

		$\displaystyle A_{1}=(-1)^{p}V\left(x_{1},x_{2},\ldots,x_{n}\right)\mu_{n-p}\left(x_{1}-x_{0},x_{2}-x_{0},\ldots,x_{n}-x_{0}\right)$
		$\displaystyle A_{2}=V\left(x_{0},x_{1},\ldots,x_{n}\right)$
		$\displaystyle A_{3}=(-1)^{p-1}V\left(x_{0},x_{2}x_{3},\ldots,x_{n}\right)\mu_{n-p}\left(x_{2}-x_{0},x_{3}-x_{0},\ldots,x_{n}-x_{0}\right)$
		$\displaystyle A_{4}=(-1)^{p-1}V\left(x_{0},x_{1},x_{3},\ldots,x_{n}\right)\mu_{n-p}\left(x_{1}-x_{0},x_{3}-x_{0},\ldots,x_{n}-x_{0}\right)$
		$\displaystyle\left.\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot x_{n-1}\right)\mu_{n-p}\left(x_{1}-x_{0},x_{2}-x_{0},\ldots,x_{n-1}\right)$
		$\displaystyle A_{n+2}=(-1)^{p-1}V\left(x_{0},x_{1},x_{2},\ldots,x_{n-1}\right.$

Taking into account formula (40) it is deduced that

\mu=\frac{(-1)^{p}}{\mu_{n-p+1}\left(x_{1}-x_{0},\ldots,x_{n}-x_{0}\right)}

(42)

and

	$\displaystyle\lambda_{1}=-\frac{V\left(x_{1},x_{2},x_{3},\ldots,x_{n}\right)}{V\left(x_{0},x_{1},x_{2},\ldots,x_{n}\right)}\cdot\frac{\mu_{n-p}\left(x_{1}-x_{0},x_{2}-x_{0},\ldots,x_{n}-x_{0}\right)}{\mu_{n-p+1}\left(x_{1}-x_{0},x_{2}-x_{0},\ldots,x_{n}-x_{0}\right)}$
	$\displaystyle K_{1}=-\frac{V\left(x_{0},x_{2},x_{3},\ldots,x_{n}\right)}{V\left(x_{0},x_{1},x_{2},\ldots,x_{n}\right)}\cdot\frac{\mu_{n-p}\left(x_{2}-x_{0},x_{3}-x_{0},\ldots,x_{n}-x_{0}\right)}{\mu_{n-p+1}\left(x_{1}-x_{0},x_{2}-x_{0},\ldots,x_{n}-x_{0}\right)}$		(43)
	$\displaystyle\left.\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot x_{n}\right)$
	$\displaystyle K_{n}=(-1)^{n}\frac{V\left(x_{0},x_{1},x_{2},\ldots,x_{n-1}\right)}{V\left(x_{0},x_{1},x_{2},\ldots,x_{n}\right)}\cdot\frac{\mu_{n-p}\left(x_{1}-x_{0},x_{2}-x_{0},\ldots,x_{n-1}-x_{0}\right)}{\mu_{n-p+1}\left(x_{1}-x_{0},x_{2}-x_{0},\ldots,x_{n}-x_{0}\right)}$

But

		$\displaystyle\frac{V\left(x_{1},x_{2},x_{3},\ldots,x_{n}\right)}{V\left(x_{0},x_{1},x_{2},\ldots,x_{n}\right)}=\frac{1}{\left(x_{1}-x_{0}\right)\left(x_{2}-x_{0}\right)\ldots\left(x_{n}-x_{0}\right)}$
		$\displaystyle\frac{V\left(x_{0},x_{2},x_{3},\ldots,x_{n}\right)}{V\left(x_{0},x_{1},x_{2},\ldots,x_{n}\right)}=\frac{-1}{\left(x_{0}-x_{1}\right)\left(x_{2}-x_{1}\right)\ldots\left(x_{n}-x_{1}\right)}$
		$\displaystyle\frac{V\left(x_{0},x_{1},x_{2},\ldots,x_{n-1}\right)}{V\left(x_{0},x_{1},x_{2},\ldots,x_{n}\right)}=\frac{(-1)^{n}}{\left(x_{0}-x_{n}\right)\left(x_{1}-x_{n}\right)\ldots\left(x_{n-1}-x_{n}\right)}$

and if we consider the rational function

\frac{1}{\left(x-x_{0}\right)\left(x-x_{1}\right)\ldots\left(x-x_{n}\right)}=\frac{B_{0}}{x-x_{0}}+\frac{B_{1}}{x-x_{1}}+\ldots+\frac{B_{n}}{x-x_{n}}

(44)

expanded into simple rational functions, we have

		$\displaystyle B_{0}=\frac{(-1)^{n}}{\left(x_{1}-x_{0}\right)\left(x_{2}-x_{0}\right)\ldots\left(x_{n}-x_{0}\right)}$
		$\displaystyle B_{1}=\frac{(-1)^{n}}{\left(x_{0}-x_{1}\right)\left(x_{2}-x_{1}\right)\ldots\left(x_{n}-x_{1}\right)}$
		$\displaystyle B_{n}=\frac{(-1)^{n}}{\left(x_{0}-x_{n}\right)\left(x_{1}-x_{n}\right)\ldots\left(x_{n-1}-x_{n}\right)}$

Returning to formulas (43), we will have

	$\displaystyle-\lambda_{1}=(-1)^{n}B_{0}\frac{\mu_{n-p}\left(x_{1}-x_{0},x_{2}-x_{0},\ldots,x_{n}-x_{0}\right)}{\mu_{n-p+1}\left(x_{1}-x_{0},x_{2}-x_{0},\ldots,x_{n}-x_{0}\right)}$
	$\displaystyle K_{1}=(-1)^{n}B_{1}\frac{\mu_{n-p}\left(x_{2}-x_{0},x_{3}-x_{0},\ldots,x_{n}-x_{0}\right)}{\mu_{n-p+1}\left(x_{1}-x_{0},x_{2}-x_{0},\ldots,x_{n}-x_{0}\right)}$		(45)
	$\displaystyle\left.\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot x_{n-1}-x_{0}\right)$
	$\displaystyle K_{n}=(-1)^{n}B_{u}\frac{\mu_{n-p}\left(x_{1}-x_{0},x_{2}-x_{0},\ldots,x_{n}\right.}{\mu_{n-p+1}\left(x_{1}-x_{0},x_{2}-x_{0},\ldots,x_{n}-x_{0}\right)}$

Because his sign $B_{i}$ is his sign $(-1)^{n+i}$ , from formulas (45) we deduce that $-\lambda_{1}>0,K_{1}<0,K_{2}>0,\ldots,(-1)^{n}K_{n}>0$ , that is $-\lambda_{1},K_{1},\ldots,K_{n}$ have alternating signs, $-\lambda_{1}$ being positive.

From formulas (45) we deduce that the numerical derivation formula (31) can also be written in the form

$\displaystyle(-1)^{n+p+1}\frac{f^{(p)}\left(x_{0}\right)}{p!}$	$\displaystyle+B_{0}\mu_{n-p}\left(x_{1}-x_{0},x_{2}-x_{0},\ldots,x_{n}-x_{0}\right)f\left(x_{0}\right)$
	$\displaystyle+B_{1}\mu_{n-p}\left(x_{2}-x_{0},x_{3}-x_{0},\ldots,x_{n}-x_{0}\right)f\left(x_{1}\right)$
	$\displaystyle+\cdots\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot$	(46)
	$\displaystyle+B_{n}\mu_{n-p}\left(x_{1}-x_{0},x_{2}-x_{0},\ldots,x_{n-1}-x_{0}\right)f\left(x_{n}\right)$
	$\displaystyle=\mu_{n-p+1}\left(x_{1}-x_{0},x_{2}-x_{0},\ldots,x_{n}-x_{0}\right)\int_{x_{0}}^{x_{n}}\varphi(x)f^{(n+1)}(x)dx$

We therefore have a practical rule for writing the numerical derivation formula (46). We calculate the residues $B_{0},B_{1},\ldots,B_{n}$ from formula (44), then the sums are calculated $\mu_{n-p}$ and $\mu_{n-p+1}$ and finally we write formula (46).

Example. Suppose that $n=4$ and $p=2$ , and the nodes $x_{0},x_{1},x_{2}$ , $x_{3},x_{4}$ are in arithmetic progression with the ratio $h$ We have

		$\displaystyle\frac{1}{\left(x-x_{0}\right)\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x-x_{4}\right)}=$
		$\displaystyle=\frac{B_{0}}{x-x_{0}}+\frac{B_{1}}{x-x_{1}}+\frac{B_{2}}{x-x_{2}}+\frac{B_{3}}{x-x_{3}}+\frac{B_{4}}{x-x_{4}}$

where

B_{0}=\frac{1}{24h^{4}},\quad B_{1}=-\frac{1}{6h^{4}},\quad B_{2}=\frac{1}{4h^{4}},\quad B_{3}=-\frac{1}{6h^{4}},\quad B_{4}=\frac{1}{24h^{4}}

On the other hand, from the string

x_{1}-x_{0}=h,\quad x_{2}-x_{0}=2h,\quad x_{3}-x_{0}=3h,\quad x_{4}-x_{0}=4h

we deduce that

\mu_{3}\left(x_{1}-x_{0},x_{2}-x_{0},x_{3}-x_{0},x_{4}-x_{0}\right)=50h^{3}

and

		$\displaystyle\mu_{2}\left(x_{1}-x_{0},x_{2}-x_{0},x_{3}-x_{0},x_{4}-x_{0}\right)=5h^{2}$
		$\displaystyle\mu_{2}\left(x_{2}-x_{0},x_{3}-x_{0},x_{4}-x_{0}\right)=6h^{2}$
		$\displaystyle\mu_{2}\left(x_{1}-x_{0},x_{3}-x_{0},x_{4}-x_{0}\right)=9h^{2}$
		$\displaystyle\mu_{2}\left(x_{1}-x_{0},x_{2}-x_{0},x_{4}-x_{0}\right)=4h^{2}$
		$\displaystyle\mu_{2}\left(x_{1}-x_{0},x_{2}-x_{0},x_{3}-x_{0}\right)=1h^{2}$

Applying formula (46), we will have the following numerical derivation formula

	$\displaystyle\frac{f^{\prime\prime}\left(x_{0}\right)}{2}=\frac{1}{24h^{2}}\left[35f\left(x_{0}\right)-104f\left(x_{1}\right)+114f\left(x_{2}\right)-56f\left(x_{3}\right)+11f\left(x_{4}\right)\right]$
	$\displaystyle-50h^{3}\int_{x_{0}}^{x_{4}}\varphi(x)f^{(5)}(x)dx$		(47)

11.

A new form of the numerical derivative formula (31) or (46). From the definition of divided differences of the function $f(x)$ on the nodes $x_{0},x_{1},\ldots,x_{n}$ , it follows that we can write
$f\left(x_{1}\right)=f\left(x_{0}\right)+\left(x_{1}-x_{0}\right)\left[x_{0},x_{1};f(x)\right]$
$f\left(x_{2}\right)=f\left(x_{0}\right)+\left(x_{2}-x_{0}\right)\left[x_{0},x_{1};f(x)\right]+\left(x_{2}-x_{0}\right)\left(x_{2}-x_{1}\right)\left[x_{0},x_{1},x_{2};f(x)\right]$

	$\displaystyle f\left(x_{n}\right)=$	$\displaystyle f\left(x_{0}\right)+\left(x_{n}-x_{0}\right)\left[x_{0},x_{1};f(x)\right]+\left(x_{n}-x_{0}\right)\left(x_{n}-x_{1}\right)\left[x_{0},x_{1},x_{2};f(x)\right]+$
		$\displaystyle+\ldots+\left(x_{n}-x_{0}\right)\left(x_{n}-x_{1}\right)\ldots\left(x_{n}-x_{n-1}\right)\left[x_{0},x_{1},\ldots,x_{n};f(x)\right]$

Substituting in formula (46) $f\left(x_{1}\right),f\left(x_{2}\right),\ldots,f\left(x_{n}\right)$ with the help of different, divided, this formula takes the following form

	$\displaystyle(-1)^{n+p+1}\frac{f^{(p)}\left(x_{0}\right)}{p!}+D_{0}f\left(x_{0}\right)+D_{1}\left[x_{0},x_{1};f(x)\right]+\ldots+$
$\displaystyle+$	$\displaystyle D_{p}\left[x_{0},x_{1},\ldots,x_{p};f(x)\right]+\ldots+D_{n}\left[x_{0},x_{1},\ldots,x_{n};f(x)\right]=$
$\displaystyle=$	$\displaystyle\mu_{n-p+1}\left(x_{1}-x_{0},x_{2}-x_{0},\ldots,x_{n}-x_{0}\right)\int_{x_{0}}^{x_{n}}\varphi(x)f^{(n+1)}(x)dx$	(48)

Coefficients $D_{0},D_{1},\ldots,D_{n}$ is determined in the following way: replacing $f(x)$ with $1,x,\ldots x^{p-1}$ it is found that $D_{0},D_{1},\ldots,D_{p-1}$ are null. Then to determine the $D_{p+i}$ , where $i=0,1,\ldots,n-p$ , we replace $f(x)\mathrm{cu}$

f_{p+i}(x)=\left(x-x_{0}\right)\left(x-x_{1}\right)\ldots\left(x-x_{p+i-1}\right)

All divided differences in formula (48) are then zero, except for the divided difference $\left[x_{0},x_{1},\ldots,x_{p+i};f_{p+i}(x)\right]$ which is equal to 1. It follows that

D_{p+i}=(-1)^{n+p}\frac{f_{p+i}^{(p)}\left(x_{0}\right)}{p!}

But

\frac{f_{p+i}^{(p)}\left(x_{0}\right)}{p!}=(-1)^{i}\mu_{i}\left(x_{1}-x_{0},\ldots,x_{p+i-1}-x_{0}\right)

and therefore we will have

D_{p+i}=(-1)^{n+p+i}\mu_{i}\left(x_{1}-x_{0},\ldots,x_{p+i-1}-x_{0}\right)

MEAN

		$\displaystyle D_{p}=(-1)^{n+p}$
		$\displaystyle D_{p+1}=(-1)^{n+p+1}\mu_{1}\left(x_{1}-x_{0},\ldots,x_{p}-x_{0}\right)$
		$\displaystyle D_{p+2}=(-1)^{n+p+2}\mu_{2}\left(x_{1}-x_{0},\ldots,x_{p+1}-x_{0}\right)$
		$\displaystyle D_{n}=(-1)^{n+n}\mu_{n-p}\left(x_{1}-x_{0},\ldots,x_{n-1}-x_{0}\right).$

The numerical derivation formula (46) can therefore be written in the form

	$\displaystyle\frac{f^{(p)}\left(x_{0}\right)}{p!}$	$\displaystyle=\left[x_{0},x_{1},\ldots,x_{p};f(x)\right]-\mu_{1}\left(x_{1}-x_{0},\ldots,x_{p}-x_{0}\right)\left[x_{0},x_{1},\ldots,x_{p+1};f(x)\right]$
		$\displaystyle+\mu_{2}\left(x_{1}-x_{0},\ldots,x_{p+1}-x_{0}\right)\left[x_{0},x_{1},\ldots,x_{p+2};f(x)\right]-\ldots+$
		$\displaystyle+(-1)^{n-p}\mu_{n-p}\left(x_{1}-x_{0},\ldots,x_{n-1}-x_{0}\right)\left[x_{0},x_{1},\ldots,x_{n};f(x)\right]$
		$\displaystyle+(-1)^{n-p+1}\mu_{n-p+1}\left(x_{1}-x_{0},\ldots,x_{n}-x_{0}\right)\int_{x_{0}}^{x_{n}}\varphi(x)f^{(n+1)}(x)dx$

where in the second member the divided differences of the function were highlighted $f(x)$ on the nodes $x_{0},x_{1},\ldots,x_{n}$ .

In particular, when the nodes $x_{0},x_{1},\ldots,x_{n}$ are in arithmetic progression with the ratio $h$ , we have

\left[x_{0},x_{1},\ldots,x_{q};f(x)\right]=\frac{\Delta^{q}f(x)}{q!h^{q}}

and

\mu_{i}\left(x_{1}-x_{0},\ldots,x_{p+i-1}-x_{0}\right)=h^{i}\sum 1.2\ldots i

where the sum in the second term is extended to all products of $i$ factors taken from the numbers $1,2,\ldots,p+i-1$ If we denote this sum by $\theta_{i}(1,2,\ldots,p+i-1)$ , we will be able to write

\mu_{i}\left(x_{1}-x_{0},\ldots,x_{p+i-1}-x_{0}\right)=h^{i}\theta_{i}(1,2,\ldots,p+i-1)

and the numerical derivation formula becomes

	$\displaystyle\frac{h^{p}f^{(p)}\left(x_{0}\right)}{p!}=\frac{\Delta^{(p)}f\left(x_{0}\right)}{p!}-\theta_{1}(1,2,\ldots,p)\frac{\Delta^{p+1}f\left(x_{0}\right)}{(p+1)!}+$
	$\displaystyle+\theta_{2}(1,2,\ldots,p+1)\frac{\Delta^{p+2}f\left(x_{0}\right)}{(p+2)!}-\ldots+(-1)^{n-p}\theta_{n-p}(1,2,\ldots,n-1)\frac{\Delta^{n}f\left(x_{0}\right)}{n!}+$
	$\displaystyle+(-1)^{n-p+1}\theta_{n-p+1}(1,2,\ldots,n)h^{n+1}\int_{x_{0}}^{x_{n}}\varphi(x)f^{(n+1)}(x)dx$		(50)

This formula is known as Markov's formula: it expresses the derivative of the order $p$ of the function $f(x)$ in the node $x_{0}$ , using successive differences of the function $f(x)$ in the node $x_{0}$ , on the nodes $x_{0},x_{1},\ldots,x_{n}$ , starting with the difference of order $p$ and ending with the difference of order $n$ Having established Markov's formula, we also highlighted the rest of it written in the form of a definite integral.

The numerical derivation formula (31) or (46) or (49) thus appears as a generalization of Markov's formula, in the sense that in formula (49) the nodes $x_{0},x_{1},\ldots,x_{n}$ they are no longer in arithmetic progression.

It remains for us to study the rest of the numerical derivation formulas (49) and (50).
12. $O$ important property of coefficients $K_{i}$ Let us prove that the sums $K_{j}+K_{j+1}+\ldots+K_{n}$ , where $j=1,2,\ldots,n$ are different from zero.

For this, let us consider the polynomials

h_{k}(x)=\frac{\left(x-x_{0}\right)\left(x-x_{1}\right)\ldots\left(x-x_{k-1}\right)\left(x-x_{k+1}\right)\ldots\left(x-x_{n}\right)}{\left(x_{k}-x_{0}\right)\left(x_{k}-x_{1}\right)\ldots\left(x_{k}-x_{k-1}\right)\left(x_{k}-x_{k+1}\right)\ldots\left(x_{k}-x_{n}\right)}

for $k=j,j+1,\ldots,n$ , and their sum

h(x)=h_{j}(x)+h_{j+1}(x)+\ldots+h_{n}(x)

6 - Studies and research in mathematics

polynomial $h(x)$ is of the degree $n$ and its coefficient $x^{n}$ has his sign $(-1)^{n-j}$ . (No. 4). We have

\begin{array}[]{ll}h\left(x_{0}\right)=0,&h\left(x_{1}\right)=0,\ldots,h\left(x_{j-1}\right)=0\\ h\left(x_{j}\right)=1,&h\left(x_{j+1}\right)+1,\ldots,h\left(x_{n}\right)=0\end{array}

Substituting in the numerical derivation formula (31) for $f(x)$ with $h(x)$ , we will have

K_{j}+K_{j+1}+\ldots+K_{n}=\frac{h^{(p)}\left(x_{0}\right)}{p!}\mu

It was seen (no. 4) that $h^{\prime}(x)$ has $n-1$ distinct roots in the interval ( $x_{0},x_{n}$ ). Applying Rolle's theorem, it is deduced that $h^{\prime\prime}(x)$ has $n-2$ distinct roots in the interval ( $x_{0},x_{n}$ ), … and that $h^{(p)}(x)$ has $n-p$ distinct roots in the interval $x_{0},x_{n}$ ). However $h^{(p)}(x)$ is a polynomial of degree $n-p$ and therefore $h^{(p)}\left(x_{0}\right)\neq 0$ , which proves that the sum $K_{j}+K_{j+1}+\ldots+K_{n}$ is different from zero. We can also specify the sign of this sum. For this, we note that the sign of $h^{(p)}\left(x_{0}\right)$ is his sign $h^{(p)}(x)$ when $x\rightarrow-\infty$ , that is, the sign of $(-1)^{n-j}\cdot(-1)^{n-p}=(-1)^{p+j}$ On the other hand, from formula (42), the sign of $\mu$ is his sign $(-1)^{p}$ . So the sum sign $K_{j}+K_{j+1}++\ldots+K_{n}$ is his sign $(-1)^{i}$ .

Conclusion. Constants $\lambda_{1},\lambda_{2},\ldots,\lambda_{n}$ from the differential equations (29) are different from zero. Indeed, from formulas (32) it follows that

		$\displaystyle\lambda_{1}=K_{1}+K_{2}+\ldots+K_{n}$
		$\displaystyle\cdot\cdots\cdot\cdots\cdot$
		$\displaystyle\lambda_{j}=K_{j}+K_{j+1}+\ldots+K_{n}$
		$\displaystyle\cdot\cdots\cdot\cdot$
		$\displaystyle\lambda_{n}=K_{n}$

and taking into account the above property of the coefficients $K_{1},K_{2},\ldots,K_{n}$ we deduce that $\lambda_{j}$ is different from zero and has the sign of $(-1)^{j}$ .
13. Function $\varphi(x)$ from the numerical derivation formula (31) is positive in ( $x_{0},x_{n}$ ). Let us first observe that the function $\varphi_{n}(x)$ is positive in the interval $\left[x_{n-1},x_{n}\right)$ Indeed, according to formulas (33) we have

\varphi_{n}(x)=K_{n}\frac{\left(x-x_{n}\right)^{n}}{n!}=(-1)^{n}K_{n}\frac{\left(x_{n}-x\right)^{n}}{n!}

But it was shown at No. 12 that $(-1)^{n}K_{n}>0$ , from which it follows that the function $\varphi_{n}(x)$ is positive in the interval $\left[x_{n-1},x_{n}\right)$ ,

function $\varphi(x)$ is continuous and has continuous derivatives up to order $n-1$ in the interval $\left[x_{0},x_{n}\right]$ In addition $\varphi^{(n)}(x)$ is a continuous function on every partial interval $\left(x_{i-1},x_{i}\right)$ and we have $\varphi_{i}^{(n)}(x)=\lambda_{i}\neq 0$ .

1st boundary conditions at the nodes $x_{0}$ and $x_{n}$ are different according to $p=1$ , $1<p<n-1,p=n-1,p=n$ .
$1^{\circ}.p=1$ The boundary conditions are

	$\displaystyle\varphi\left(x_{0}\right)=0,\quad\varphi^{\prime}\left(x_{0}\right)=0,\ldots,\varphi^{(n-2)}\left(x_{0}\right)=0,\quad\varphi^{(n-1)}\left(x_{0}\right)=-\mu\neq 0$
	$\displaystyle\varphi\left(x_{n}\right)=0,\quad\varphi^{\prime}\left(x_{n}\right)=0,\ldots,\varphi^{(n-2)}\left(x_{n}\right)=0,\quad\varphi^{(n-1)}\left(x_{n}\right)=0.$		(51)

$2^{\circ}.1<p<n-1$ The boundary conditions are

	$\displaystyle\varphi\left(x_{0}\right)=0,\quad\varphi^{\prime}\left(x_{0}\right)=0,\ldots,\varphi^{(n-p-1)}\left(x_{0}\right)=0,\varphi^{(n-p)}\left(x_{0}\right)=(-1)^{p}\frac{\mu}{p!}\neq 0$
	$\displaystyle\varphi^{(n-p+1)}\left(x_{0}\right)=0,\ldots,\varphi^{(n-1)}\left(x_{0}\right)=0$		(52)

$\varphi\left(x_{n}\right)=0,\quad\varphi^{\prime}\left(x_{n}\right)=0,\ldots,\varphi^{(n-1)}\left(x_{n}\right)=0$ .
$3^{\circ}\cdot p=n-1$ The first boundary conditions are

	$\displaystyle\varphi\left(x_{0}\right)=0,\quad\varphi^{\prime}\left(x_{0}\right)=(-1)^{n-1}\frac{\mu}{(n-1)!}\neq 0,\quad\varphi^{\prime\prime}\left(x_{0}\right)=0,\ldots,\varphi^{(n-1)}\left(x_{0}\right)=0$
	$\displaystyle\varphi\left(x_{n}\right)=0,\quad\varphi^{\prime}\left(x_{n}\right)=0,\ldots,\varphi^{(n-1)}\left(x_{n}\right)=0$		(53)

	$\displaystyle\varphi\left(x_{0}\right)=(-1)^{n}\frac{\mu}{n!}\neq 0,\quad\varphi^{\prime}\left(x_{0}\right)=0,\ldots,\varphi^{(n-1)}\left(x_{0}\right)=0$
	$\displaystyle\varphi\left(x_{n}\right)=0,\quad\varphi^{\prime}\left(x_{n}\right)=0,\ldots,\varphi^{(n-1)}\left(x_{n}\right)=0$		(54)

In the first three cases $\varphi(x)$ canceling out in $x_{0}$ and $x_{n}$ , according to Rolle's theorem, $\varphi^{\prime}(x)$ has at least one zero in the interval1 ( $x_{0},x_{n}$ ). Let us prove that $\varphi^{\prime}(x)nu$ can have only one zero in the interval ( $x_{0},x_{n-1}$ ).

Let us assume that the derivative $\varphi^{\prime}(x)$ would have two zeros between $x_{0}$ and $x_{n-1}$ and let's prove that this is impossible.

Applying Rolle's theorem successively, it is shown from the boundary conditions (51), (52) and (53) that $\varphi^{(n-1)}(x)$ has $n$ zeros in the range $\left(x_{0},x_{n-1}\right)$ in case $p=1$ , or $n-1$ zeros in the range ( $x_{0},x_{n-1}$ ) when $1<p\leqslant n-1$ . In a partial interval ( $x_{i-1},x_{i}$ ] cannot find two zeros of $\varphi^{(n-1)}(x)$ . for if this were to happen, applying Rolle's theorem it would follow that $\varphi^{(n)}(x)$ to cancel in the interval ( $x_{i-1},x_{i}$ ) which is impossible because in this interval $\varphi^{(n)}(x)=\varphi_{i}^{(n)}(x)=\lambda_{i}\neq 0$ .

If $p=1$ , those $n$ zeros of $\varphi^{(n-1)}(x)$ from interval1 ( $x_{0},x_{n-1}$ ) can only be distributed one at a time in the intervals $\left(x_{0},x_{1}\right],\left(x_{1},x_{2}\right],\ldots$ , $\left(x_{n-2},x_{n-1}\right)$ and this is impossible because we only have $n-1$ intervals. This leads to a contradiction.

If $1<p\leqslant n-1$ , no zero of $\varphi^{(n-1)}(x)$ is not found in the range ( $x_{0},x_{1}$ ], because if a zero were found, having $\varphi^{(n-1)}\left(x_{0}\right)=0$ , we can apply Rolle's theorem and it would follow that $\varphi^{(n)}(x)$ to cancel in the interval
( $x_{0},x_{1}$ ), which is impossible because $\lambda_{1}\neq 0$ The $n-1$ zeros of $\varphi^{(n-1)}(x)$ are therefore found in the interval ( $x_{1},x_{n-1}$ ). However, it was seen above that in each of the $n-2$ intervals ( $x_{1},x_{2}$ ] , … $\left(\alpha_{n-2},x_{n-1}\right)$ there is only one zero of $\varphi^{(n-1)}(x)$ So again we have reached a contradiction.

It follows that in the case $p<n,\varphi^{\prime}(x)$ has only one zero in the interval ( $x_{0},x_{n-1}$ ) and the function $\varphi(x)$ being positive in the interval [ $x_{n-1},x_{n}$ ), is positive throughout the interval ( $x_{0},x_{n}$ ).

Let's also examine the case $p=n$ We note that $\varphi\left(x_{0}\right)=(-1)^{n}\frac{\mu}{n!}>0$ Function $\varphi^{\prime}(x)$ is cancelled in $x_{0}$ and $x_{n}$ , as shown by formulas (54). Applying Rolle's theorem it follows that $\varphi^{\prime\prime}(x)$ has at least one zero in the range $\left(x_{0},x_{n}\right)$ Let us prove that $\varphi^{\prime\prime}(x)$ has only one zero in the interval $\left(x_{0},x_{n}\right)$ .

If $\varphi^{\prime\prime}(x)$ would have two zeros in the interval ( $x_{0},x_{n}$ ), taking into account the boundary conditions (54) and successively applying Rolle's theorem, it would follow that $\varphi^{(n-1)}(x)$ to have $n-1$ zeros in the range ( $x_{0},x_{n-1}$ ), which is impossible as follows from the reasoning made above.

derivative $\varphi^{\prime\prime}(x)$ having a single zero in the interval ( $x_{0},x_{n-1}$ ), it follows that the derivative $\varphi^{\prime}(x)$ does not cancel in the interval ( $x_{0},x_{n-1}$ ), because if it were to be cancelled, taking into account that $\varphi^{\prime}\left(x_{0}\right)=0,\varphi^{\prime}\left(x_{n}\right)=0$ , it would follow that $\varphi^{\prime\prime}(x)$ to cancel at least at two points in the interval ( $x_{0},x_{n-1}$ ), which is impossible. So the function $\varphi^{\prime}(x)$ does not cancel in the interval ( $x_{0},x_{n}$ ) and being negative in the interval $\left[x_{n-1},x_{n}\right)$ , is negative throughout the interval ( $x_{0},x_{n}$ ). It follows that the function $\varphi(x)$ is decreasing in the interval $\left[x_{0},x_{n}\right]$ ; being positive in $x_{0}$ and null in $x_{n}$ , it is positive in the interval ( $x_{0},x_{n}$ ).

Function graph $\varphi(x)$ in the interval $\left[x_{0},x_{n}\right]$ is shown in Fig. 2.

14.

The remainder in the numerical derivation formula (46). Let us denote by $R$ , the remainder in the numerical derivation formula (46), i.e.

R=\mu_{n-p+1}\left(x_{1}-x_{0},\ldots,x_{n}-x_{0}\right)\int_{x_{0}}^{x_{n}}\varphi(x)f^{(n+1)}(x)dx

Because it has been shown that the function $\varphi(x)$ is positive in the interval ( $x_{0},x_{n}$ ), we can apply the average formula, and we will have
where $\xi\in\left(x_{0},x_{n}\right)$ .

R=\mu_{n-p+1}\left(x_{1}-x_{0},\ldots,x_{n}-x_{0}\right)f^{(n+1)}(\xi)\int_{x_{0}}^{x_{n}}\varphi(x)dx

The integral in the second term is calculated using the numerical derivative formula (46), replacing $f(x)$ with

g(x)=\frac{\left(x-x_{0}\right)\left(x-x_{1}\right)\ldots\left(x-x_{n}\right)}{(n+1)!}.

We will have

(-1)^{n+p+1}\frac{g^{(p)}\left(x_{0}\right)}{p!}=\mu_{n-p+1}\left(x_{1}-x_{0},\ldots,x_{n}-x_{0}\right)\int_{x_{0}}^{x_{n}}\varphi(x)dx

But

\frac{g^{(p)}\left(x_{0}\right)}{p!}=\frac{(-1)^{n-p+1}}{(n+1)!}\mu_{n-p+1}\left(x_{1}-x_{0},\ldots,x_{n}-x_{0}\right),

from which it follows that

\int_{x_{0}}^{x_{n}}\varphi(x)dx=\frac{1}{(n+1)!}

(55)

Therefore the rest $R$ from the numerical derivation formula (46) it can also be written as

R=\mu_{n-p+1}\left(x_{1}-x_{0},\ldots,x_{n}-x_{0}\right)\frac{f^{(n+1)}(\xi)}{(n+1)!},

(56)

where $\xi\in\left(x_{0},x_{n}\right)$ Also
, the rest $R_{1}$ from the numerical derivation formula (49) it can also be written in the form

R_{1}=(-1)^{n-p+1}\mu_{n-p+1}\left(x_{1}-x_{0},\ldots,x_{n}-x_{0}\right)\frac{f^{(n+1)}(\xi)}{(n+1)!},

(57)

where $\xi\in\left(x_{0},x_{n}\right)$ From
formulas (56), (57) an evaluation of the remainder results $R$ or $R_{1}$ We have

|R|=\left|R_{1}\right|\leqslant\mu_{n-p+1}\left(x_{1}-x_{0},\ldots,x_{n}-x_{0}\right)\frac{M_{n+1}}{(n+1)!}.

In the case of Markov's formula (50), the rest

R_{2}=(-1)^{n-p+1}\theta_{n-p+1}(1,2,\ldots,n)h^{n+1}\int_{x_{0}}^{x_{n}}\varphi(x)f^{(n+1)}(x)dx

can also be written in the form

R_{2}=(-1)^{n-p+1}\theta_{n-p+1}(1,2,\ldots,n)h^{n+1}\frac{f^{(n+1)}(\xi)}{(n+1)!}

(58)

where $\xi\in\left(x_{0},x_{n}\right)$ .

Chapter II

NUMERICAL INTEGRATION OF DIFFERENT EQUATIONS, IALE

§ 1. Numerical integration formulas resulting from the application of numerical derivative formulas (6)

15.

Before moving on to establishing numerical integration formulas, let's consider Taylor's formula

	$\displaystyle y(x)=$	$\displaystyle y\left(x_{0}\right)+\frac{x-x_{0}}{1!}y^{\prime}\left(x_{0}\right)+\frac{\left(x-x_{0}\right)^{2}}{2!}y^{\prime\prime}\left(x_{0}\right)+\ldots+$
		$\displaystyle+\frac{\left(x-x_{0}\right)^{p}}{p!}y^{(p)}\left(x_{0}\right)+\int_{x_{0}}^{x}\frac{(x-s)^{p}}{p!}y^{(p+1)}(s)ds$		(59)

and let's see what it becomes when they are replaced $y^{\prime}\left(x_{0}\right),y^{\prime\prime}\left(x_{0}\right),\ldots,y^{(p)}\left(x_{0}\right)$ , from the numerical derivation formulas (6).

In the numerical derivation formula

\begin{gathered}-\left[\lambda_{1}f\left(x_{0}\right)+\frac{\nu_{1}}{1!}f^{\prime}\left(x_{0}\right)+\ldots+\frac{\nu_{p}}{p!}f^{(p)}\left(x_{0}\right)\right]\\ +K_{1}f\left(x_{1}\right)+K_{2}f\left(x_{2}\right)+\ldots+K_{n}f\left(x_{n}\right)=(-1)^{n+p-1}\int_{x_{0}}^{x_{n}}\varphi(x)f^{(n+p)}(x)dx\end{gathered}

in which we recall that $v_{p}\neq 0$ , because we have

v_{p}=\frac{(-1)^{p}}{\left(x_{1}-x_{0}\right)\left(x_{2}-x_{0}\right)\ldots\left(x_{n}-x_{0}\right)}

to replace $f(x)$ with $y(x)$ , and the function $\varphi(x)$ let's write it down with $\Phi_{p}(x)$ , to highlight the index $p$ which is the highest order of the derivatives in formula (6). We also recall that

\int_{x_{0}}^{x_{n}}\Phi_{p}(x)dx=\frac{1}{(n+p)!}

(60)

Substituting into formula (6), $p$ with $1,2,3,\ldots$ and solving these equations in terms of $y^{\prime}\left(x_{0}\right),y^{\prime\prime}\left(x_{0}\right),\ldots,y^{(p)}\left(x_{0}\right)$ , we will obtain formulas of the form

	$\displaystyle y^{\prime}\left(x_{0}\right)=B_{10}y\left(x_{0}\right)+B_{11}y\left(x_{1}\right)+\ldots+B_{1n}y\left(x_{n}\right)+$
	$\displaystyle+C_{11}\int_{x_{0}}^{x_{n}}\Phi_{1}(x)y^{(n+1)}(x)dx$
	$\displaystyle y^{\prime\prime}\left(x_{0}\right)=B_{20}y\left(x_{0}\right)^{x_{n}}+B_{21}y\left(x_{1}\right)+\ldots+B_{2n}y\left(x_{n}\right)+$
	$\displaystyle+C_{21}\int_{x_{0}}^{x_{n}}\Phi_{1}(x)y^{(n+1)}(x)dx+C_{22}\int_{x_{0}}^{x_{n}}\Phi_{2}(x)y^{(n+2)}(x)dx$		(61)
	$\displaystyle y^{(p)}\left(x_{0}\right)=B_{p0}y\left(x_{0}\right)+B_{p1}y\left(x_{1}\right)+\ldots+B_{pn}y\left(x_{n}\right)+$
	$\displaystyle+C_{p1}\int_{x_{0}}^{x_{n}}\Phi_{1}(x)y^{(n+1)}(x)dx+\ldots+C_{pp}\int_{x_{0}}^{x_{n}}\Phi_{p}(x)y^{(n+p)}(x)dx$

where $B_{ik}$ and $C_{jl}$ are well-determined numbers, and

C_{11}\neq 0,\quad C_{22}\neq 0,\ldots,C_{pp}\neq 0

(62)

If we substitute in Taylor's formula (59) $y^{\prime}\left(x_{0}\right),y^{\prime\prime}\left(x_{0}\right),\ldots,y^{(p)}\left(x_{0}\right)$ with the second members of formulas (61), we obtain formulas of the form

	$\displaystyle y\left(x_{0}\right)=B_{0}(x)y\left(x_{0}\right)+B_{1}(x)y\left(x_{1}\right)+\ldots+B_{n}(x)y\left(x_{n}\right)+$		(63)
	$\displaystyle+C_{1}(x)\int_{x_{0}}^{x_{n}}\Phi_{1}(x)y^{(n+1)}(x)dx+C_{2}(x)\int_{x_{0}}^{x_{n}}\Phi_{2}(x)y^{(n+2)}(x)dx+\ldots+$
	$\displaystyle+\ldots+C_{p}(x)\int_{x_{0}}^{x_{n}}\Phi_{p}(x)y^{(n+p)}(x)dx+\int_{x_{0}}^{x}\frac{(x-s)^{p}}{p!}y^{(p+1)}(s)ds$

It remains to determine the polynomials $B_{0}(x),B_{1}(x),\ldots,B_{n}(x)$ and $C_{1}(x)$ , $C_{2}(x),\ldots,C_{p}(x)$ .

Determining polynomials $B_{0}(x),B_{1}(x),\ldots,B_{n}(x)$ We will distinguish two cases, as $p<n$ or $p\geqslant n$ .
$1^{\circ}$ The case $p<n$ To determine the $B_{0}(x)$ , we replace in formula (63) by $y(x)$ with

We will get

y_{0}(x)=\left(x-x_{1}\right)\left(x-x_{2}\right)\ldots\left(x-x_{n}\right).

\left(x_{0}-x_{1}\right)\left(x_{0}-x_{2}\right)\ldots\left(x_{0}-x_{n}\right)B_{0}(x)=y_{0}(x)-\int_{x_{0}}^{x}\frac{(x-s)^{p}}{p!}y_{0}^{(p+1)}(s)ds

The difference in the second term is given by Taylor's formula (59), in which we replace $y(x)$ with $y_{0}(x)$ We have

\frac{y_{0}^{(k)}\left(x_{0}\right)}{k!}=(-1)^{n-k}\mu_{n-k}\left(x_{1}-x_{0},x_{2}-x_{0},\ldots,x_{n}-x_{0}\right),

from which it follows that we will have

	$\displaystyle B_{0}(x)=\frac{1}{\left(x_{1}-x_{0}\right)\left(x_{2}-x_{0}\right)\ldots\left(x_{n}-x_{0}\right)}\left\{\mu_{n}\left(x_{1}-x_{0},x_{2}-x_{0},\ldots,x_{n}-x_{0}\right)-\right.$
	$\displaystyle-\mu_{n-1}\left(x_{1}-x_{0},x_{2}-x_{0},\ldots,x_{n}-x_{0}\right)\left(x-x_{0}\right)+$
	$\displaystyle\left.\left.+\ldots\ldots,\ldots,\ldots,\ldots,x_{n}\right)\left(x-x_{0}\right)^{p}\right\}.$

To determine the polynomial $B_{i}(x)$ , we replace in formula (63) by $y(x)\mathrm{cu}$

y_{i}(x)=\left(x-x_{0}\right)\left(x-x_{1}\right)\ldots\left(x-x_{i-1}\right)\left(x-x_{i+1}\right)\ldots\left(x-x_{n}\right)

and we get

\begin{gathered}\left(x_{i}-x_{0}\right)\left(x_{i}-x_{1}\right)\ldots\left(x_{i}-x_{i-1}\right)\left(x_{i}-x_{i+1}\right)\ldots\left(x_{i}-x_{n}\right)B_{i}(x)=\\ =y_{i}(x)-\int_{x_{0}}^{x}\frac{(x-s)^{p}}{p!}y_{i}^{(p+1)}(s)ds\end{gathered}

The second term is also calculated using Taylor's formula (59), in which we replace $y(x)$ with $y_{i}(x)$ We have $y_{i}\left(x_{0}\right)=0$ , and

\frac{y^{(k)}\left(x_{0}\right)}{k!}=(-1)^{n-k}\mu_{n-k}\left(x_{1}-x_{0},\ldots,x_{i-1}-x_{0},x_{i+1}-x_{0},\ldots,x_{n}-x_{0}\right).

It follows that the polynomial $B_{i}(x)$ is

		$\displaystyle B_{i}(x)=\frac{(-1)^{n-1}}{\left(x_{i}-x_{0}\right)\left(x_{i}-x_{1}\right)\ldots\left(x_{i}-x_{i-1}\right)\left(x_{i}-x_{i+1}\right)\ldots\left(x_{i}-x_{n}\right)}$
		$\displaystyle\left\{\mu_{n-1}\left(x_{1}-x_{0},\ldots,x_{i-1}-x_{0},x_{i+1}-x_{0},\ldots,x_{n}-x_{0}\right)\left(x-x_{0}\right)-\right.$
		$\displaystyle-\mu_{n-2}\left(x_{1}-x_{0},\ldots,x_{i-1}-x_{0},x_{i+1}-x_{0},\ldots,x_{n}-x_{0}\right)\left(x-x_{0}\right)^{2}+(5)$
		$\displaystyle+\ldots.\ldots.\ldots......$
		$\displaystyle\left.+(-1)^{p-1}\mu_{n-p}\left(x_{1}-x_{0},\ldots,x_{i-1}-x_{0},x_{i+1}-x_{0},\ldots,x_{n}-x_{0}\right)\left(x-x_{0}\right)^{p}\right\}.$

If we note with $A_{0},A_{1},\ldots,A_{n}$ the residues of the rational function

\frac{1}{\left(x-x_{0}\right)\left(x-x_{1}\right)\ldots\left(x-x_{n}\right)}=\frac{A_{0}}{x-x_{0}}+\frac{A_{1}}{x-x_{1}}+\ldots+\frac{A_{n}}{x-x_{n}}

relative to the poles $x_{0},x_{1},x_{2},\ldots,x_{n}$ , formulas (64) and (65) can also be written in the form

$\displaystyle B_{0}(x)=(-1)^{n}A_{0}$	$\displaystyle\left\{\left(\mu_{n}\left(x_{1}-x_{0},x_{2}-x_{0},\ldots,x_{n}-x_{0}\right)\right.\right.$
	$\displaystyle-\mu_{n-1}\left(x_{1}-x_{0},x_{2}-x_{0},\ldots,x_{n}-x_{0}\right)\left(x-x_{0}\right)+$
	$\displaystyle+\ldots,\ldots x_{n}$
	$\displaystyle\left.+(-1)^{p}\mu_{n-p}\left(x_{1}-x_{0},x_{2}-x_{0},\ldots,x_{n}-x_{0}\right)\left(x-x_{0}\right)^{p}\right\}$	(66)

	$\displaystyle B_{i}(x)$	$\displaystyle=(-1)^{n-1}A_{i}\left\{\mu_{n-1}\left(x_{1}-x_{0},\ldots,x_{i-1}-x_{0},x_{i+1}-x_{0},\ldots,x_{n}-x_{0}\right)\left(x-x_{0}\right)\right.$
		$\displaystyle-\mu_{n-2}\left(x_{1}-x_{0},\ldots,x_{i-1}-x_{0},x_{i+1}-x_{0},\ldots,x_{n}-x_{0}\right)\left(x-x_{0}\right)^{2}$
		$\displaystyle\left.\left.+\ldots,\ldots,x_{i},\ldots,x_{n},\ldots,x_{i+1}-x_{0},\ldots,x_{n}-x_{0}\right)\left(x-x_{0}\right)^{p}\right\},$

where $i=1,2,\ldots,n$ .
$2^{\circ}$ The case $p\geqslant n$ Proceeding as above, we obtain

$\displaystyle B_{0}(x)$	$\displaystyle=\frac{\left(x-x_{1}\right)\left(x-x_{2}\right)\ldots\left(x-x_{n}\right)}{\left(x_{0}-x_{1}\right)\left(x_{0}-x_{2}\right)\ldots\left(x_{0}-x_{n}\right)}$
$\displaystyle B_{1}(x)$	$\displaystyle=\frac{\left(x-x_{0}\right)\left(x-x_{2}\right)\ldots\left(x-x_{n}\right)}{\left(x_{1}-x_{0}\right)\left(x_{1}-x_{2}\right)\ldots\left(x_{1}-x_{n}\right)}$	(67)
$\displaystyle\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\left(x-x_{n-1}\right)$	$\displaystyle\ldots=\frac{\left(x-x_{0}\right)\left(x-x_{1}\right)\ldots\left(x_{n}-x_{n-1}\right)}{\left(x_{n}-x_{0}\right)\left(x_{n}-x_{0}\right)\ldots(x)\ldots(x)}$

from which it follows that

B_{0}(x)y\left(x_{0}\right)+B_{1}(x)y\left(x_{1}\right)+\ldots+B_{n}(x)y\left(x_{n}\right)=L\left[x_{0},x_{1},\ldots,x_{n};y(x)\right],

(

\prime

)

the second term being the Lagrange polynomial relative to the nodes $x_{0},x_{1},\ldots,x_{n}$ .

Determining polynomials $C_{1}(x),C_{2}(x),\ldots,C_{p}(x)$ As above, we distinguish two cases, as $p<n$ or $p\geqslant n$ .
$1^{\circ}$ The case $p<n$ To determine the polynomial $C_{k}(x)$ we replace in formula (63) by $y(x)$ with the polynomial

y_{k}(x)=\frac{\left(x-x_{0}\right)^{k}\left(x-x_{1}\right)\ldots\left(x-x_{n}\right)}{(n+k)!}

where $k=1,2,3,\ldots,p$ .

function $y_{k}(x)$ canceling out at the nodes $x_{0},x_{1},\ldots,x_{n}$ and having $y_{k}^{\prime}\left(x_{0}\right)=0$ , $y_{k}^{\prime\prime}\left(x_{0}\right)=0,\ldots,y^{(k-1)}\left(x_{0}\right)=0$ , from formulas (61) we deduce that $\int_{x_{0}}^{x_{n}}\Phi_{1}(x)y_{k}^{(n+1)}(x)dx=0,\int_{x_{0}}^{x_{n}}\Phi_{2}(x)y_{k}^{(n+2)}(x)dx=0,\ldots,\int_{x_{0}}^{x_{n}}\Phi_{k-1}(x)y_{k}^{(n+k-1)}(x)dx=0$ , while

\int_{x_{0}}^{x_{n}}\Phi_{k}(x)y_{k}^{(n+k)}(x)dx=\int_{x_{0}}^{x_{n}}\Phi_{k}(x)dx=\frac{1}{(n+k)!}

Then, formula (63) gives us

\frac{C_{k}(x)}{(n+k)!}=y_{k}(x)-\int_{x_{0}}^{x}\frac{(x-s)^{p}}{p!}y_{k}^{(p+1)}(s)ds

The second term is calculated using the lni Taylor formula (59), in which we replace $y(x)$ with $y_{k}(x)$ .

y_{k}\left(x_{0}\right)=0,\quad y_{k}^{\prime}\left(x_{0}\right)=0,\ldots,y_{k}^{(k-1)}\left(x_{0}\right)=0

and

\frac{y^{(k+l)}\left(x_{0}\right)}{(k+l)!}=\frac{(-1)^{n-l}}{(n+k)!}\mu_{n-l}\left(x_{1}-x_{0},\ldots,x_{n}-x_{0}\right)

We will definitely have

$\displaystyle C_{k}(x)=($	$\displaystyle-1)^{n}\left(x-x_{0}\right)^{k}\left\{\mu_{n}\left(x_{1}-x_{0},x_{2}-x_{0},\ldots,x_{n}-x_{0}\right)-\right.$
	$\displaystyle\quad-\mu_{n-1}\left(x_{1}-x_{0},x_{2}-x_{0},\ldots,x_{n}-x_{0}\right)\left(x-x_{0}\right)+$
	$\displaystyle+\cdots\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot$
	$\displaystyle\left.+(-1)^{p-k}\mu_{n-p+k}\left(x_{1}-x_{0},x_{2}-x_{0},\ldots,x_{n}-x_{0}\right)\left(x-x_{0}\right)^{p-k}\right\}.$	(68)

$2^{\circ}$ The case $p\geqslant n$ In this case, having $n>1$ , and $n+1\leqslant p+1<n+p$ , formula (63) is written

	$\displaystyle y(x)=B_{0}(x)y\left(x_{0}\right)+B_{1}(x)y\left(x_{1}\right)+\ldots+B_{n}(x)y\left(x_{n}\right)+$
	$\displaystyle+C_{1}(x)\int_{x_{0}}^{x_{n}}\Phi_{1}(x)y^{(n+1)}(x)dx+\ldots+C_{p-n}(x)\int_{x_{0}}^{x_{n}}\Phi_{p-n}(x)y^{(p)}(x)dx+$
	$\displaystyle+C_{p-n+1}(x)\int_{x_{0}}^{x_{n}}\Phi_{p-n+1}(x)y^{(p+1)}(x)dx+\ldots+C_{p}(x)\int_{x_{0}}^{x}\Phi_{p}(x)y^{(n+p)}(x)dx+$
	$\displaystyle+\int_{x_{0}}^{x}\frac{(x-s)^{p}}{p!}y^{(p+1)}(s)ds$		( $\prime$ )

It is obvious that if $p=n$ , the terms written on the second line of formula (63') are missing. Substituting in formula (63') $y(x)$ with

y_{h}(x)=\frac{\left(x-x_{0}\right)^{h}\left(x-x_{1}\right)\ldots\left(x-x_{n}\right)}{(n+h)!}

where $h=1,2,\ldots,p-n$ , and proceeding as above, it is found that

C_{h}(x)=\left(x-x_{0}\right)^{h}\left(x-x_{1}\right)\ldots\left(x-x_{n}\right)

(69)

To determine the polynomial $C_{p-n+l}(x)$ , where $l=1,2,\ldots,n$ , we substitute in the formula ( $63^{\prime}$ ) on $y(x)\mathrm{cu}$

y_{p-n+l}(x)=\frac{\left(x-x_{0}\right)^{p-n+l}\left(x-x_{1}\right)\ldots\left(x-x_{n}\right)}{(p+l)!}

Proceeding as above, we will have

\frac{C_{p-n+l}(x)}{(p+l)!}=y_{p-n+l}(x)-\int_{x_{0}}^{x_{n}}\frac{(x-s)^{p}}{p!}y_{p-n+l}^{(p+1)}(s)ds

The second term is calculated with Taylor's formula (59), in which we replace $y(x)$ with $y_{p-n+l}(x)$ It is found

	$\displaystyle\frac{C_{p-n+l}(x)}{(p+l)!}=\frac{y_{p-n+l}^{(p-n+l)}\left(x_{0}\right)}{(p-n+l)!}(x$	$\displaystyle\left.-x_{0}\right)^{p-n+l}+\frac{y_{p-n+l}^{(p-n+l+1)}\left(x_{0}\right)}{(p-n+l+1)!}\left(x-x_{0}\right)^{p-n+l+1}+\ldots+$
		$\displaystyle+\frac{y_{p-n+l}^{(p)}}{p!}\left(x-x_{0}\right)^{p}$

It is shown that, in general, we have

\frac{y_{p-n+l}^{(j)}\left(x_{0}\right)}{j!}=\frac{(-1)^{p-j+l}}{(p+l)!}\mu_{p-j+l}\left(x_{1}-x_{0},x_{2}-x_{0},\ldots,x_{n}-x_{0}\right)

We thus deduce that

	$\displaystyle C_{p-n+l}(x)$	$\displaystyle=(-1)^{n}\left(x-x_{0}\right)^{p-n+l}\left\{\mu_{n}\left(x_{1}-x_{0},x_{2}-x_{0},\ldots,x_{n}-x_{0}\right)-\right.$
		$\displaystyle-\mu_{n-1}\left(x_{1}-x_{0},x_{2}-x_{0},\ldots,x_{n}-x_{0}\right)\left(x-x_{0}\right)+$
		$\displaystyle+\cdot\cdot\cdot\cdot\cdot\cdot$
		$\displaystyle\left.+(-1)^{n-l}\mu_{l}\left(x_{1}-x_{0},x_{2}-x_{0},\ldots,x_{n}-x_{0}\right)\left(x-x_{0}\right)^{n-l}\right\}$

or changing the $p-n+l$ in $h$ , we will have

$\displaystyle C_{h}(x)=$	$\displaystyle(-1)^{n}(x-x)^{h}\left\{\mu_{n}\left(x_{1}-x_{0},x_{2}-x_{0},\ldots,x_{n}-x_{0}\right)-\right.$
	$\displaystyle-\mu_{n-1}\left(x_{1}-x_{0},x_{2}-x_{0},\ldots,x_{n}-x_{0}\right)\left(x-x_{0}\right)+$
	$\displaystyle+\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot+$
	$\displaystyle\left.+(-1)^{p-h}\mu_{n-p+h}\left(x_{1}-x_{0},x_{2}-x_{0},\ldots,x_{n}-x_{0}\right)\left(x-x_{0}\right)^{p-h}\right\}$	(70)
$\displaystyle\text{ pentru }h$	$\displaystyle=p-n+1,p-n+2,\ldots,p.$

Examples. $1^{\circ}$ Let's assume $n=4,p=2$ and the nodes $x_{0},x_{1},x_{2}$ , $x_{3},x_{4}$ in arithmetic progression with the ratio $h$ .

If in Taylor's formula

y(x)=y\left(x_{0}\right)+\frac{x-x_{0}}{1!}y^{\prime}\left(x_{0}\right)+\frac{\left(x-x_{0}\right)^{2}}{2!}y^{\prime\prime}\left(x_{0}\right)+\int_{x_{0}}^{x}\frac{(x-s)^{2}}{2!}y^{\prime\prime\prime}(s)ds

derivatives are replaced $y^{\prime}\left(x_{0}\right),y^{\prime\prime}\left(x_{0}\right)$ from the numerical derivation formulas, we obtain

	$\displaystyle y(x)=B_{0}(x)y\left(x_{0}\right)+B_{1}(x)y\left(x_{1}\right)+B_{2}(x)y\left(x_{2}\right)+B_{3}(x)y\left(x_{3}\right)+B_{4}(x)y\left(x_{4}\right)+$
	$\displaystyle+C_{1}(x)\int_{x_{0}}^{x_{4}}\Phi_{1}(x)y^{(5)}(x)dx+C_{2}(x)\int_{x_{0}}^{x_{4}}\Phi_{2}(x)y^{(6)}(x)dx+\int_{x_{0}}^{x}\frac{(x-s)^{2}}{2!}y^{\prime\prime\prime}(s)ds$		(71)

To write polynomials $B_{0}(x),\ldots,B_{4}(x)$ , we will apply formulas (66). We decompose into simple rational functions

\begin{gathered}\frac{\left(x-x_{0}\right)\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x-x_{4}\right)}{1}=\frac{A_{0}}{x-x_{0}}+\frac{A_{1}}{x-x_{1}}+\\ +\frac{A_{2}}{x-x_{2}}+\frac{A_{3}}{x-x_{3}}+\frac{A_{4}}{x-x_{4}}\end{gathered}

and we have

A_{0}=\frac{1}{24h^{4}},\quad A_{1}=\frac{-1}{6h^{4}},\quad A_{2}=\frac{1}{4h^{4}},\quad A_{3}=\frac{-1}{6h^{4}},\quad A_{4}=\frac{1}{24h^{4}}

It then follows that

	$\displaystyle B_{0}(x)=$	$\displaystyle\frac{1}{24h^{4}}\{$
		$\displaystyle\quad-\mu_{4}\left(x_{1}-x_{0},x_{2}-x_{0},x_{3}-x_{0},x_{4}-x_{0}\right)-$
		$\displaystyle\quad+\mu_{2}\left(x_{1}-x_{0},x_{2}-x_{0},x_{4}-x_{0}\right)\left(x-x_{0}\right)+$
	$\displaystyle B_{1}(x)=\frac{1}{6h^{4}}$	$\displaystyle\left\{\mu_{3}\left(x_{2}-x_{0},x_{4}-x_{4}\right)\left(x-x_{0}\right)^{2}\right\}$
		$\displaystyle\quad-\mu_{2}\left(x_{2}-x_{0},x_{3}-x_{0}\right)\left(x-x_{0}\right)-$
		$\displaystyle\left.\left.x_{4}-x_{0}\right)\left(x-x_{0}\right)^{2}\right\}$

MEAN

		$\displaystyle B_{0}(x)=\frac{1}{24h^{2}}\left[24h^{2}-50h\left(x-x_{0}\right)+35\left(x-x_{0}\right)^{2}\right]$
		$\displaystyle B_{1}(x)=\frac{1}{6h^{2}}\left[24h\left(x-x_{0}\right)-26\left(x-x_{0}\right)^{2}\right]$

	$\displaystyle B_{2}(x)=\frac{-1}{4h^{2}}\left[12h\left(x-x_{0}\right)-19\left(x-x_{0}\right)^{2}\right]$		(72)
	$\displaystyle B_{3}(x)=\frac{1}{6h^{2}}\left[8h\left(x-x_{0}\right)-14\left(x-x_{0}\right)^{2}\right]$
	$\displaystyle B_{4}(x)=\frac{-1}{24h^{2}}\left[6h\left(x-x_{0}\right)-11\left(x-x_{0}\right)^{2}\right]$

Applying formulas (68), we will have

		$\displaystyle C_{1}(x)=\left(x-x_{0}\right)\left[\mu_{4}\left(x_{1}-x_{0},x_{2}-x_{0},x_{3}-x_{0},x_{4}-x_{0}\right)-\right.$
		$\displaystyle\left.-\mu_{3}\left(x_{1}-x_{0},x_{2}-x_{0},x_{3}-x_{0},x_{4}-x_{0}\right)\left(x-x_{0}\right)\right]$
		$\displaystyle C_{2}(x)=\left(x-x_{0}\right)^{2}\mu_{4}\left(x_{1}-x_{0},x_{2}-x_{0},x_{3}-x_{0},x_{4}-x_{0}\right),$

MEAN

	$\displaystyle C_{1}(x)=\left(x-x_{0}\right)\left[24h^{4}-50h^{3}\left(x-x_{0}\right)\right]$		(73)
	$\displaystyle C_{2}(x)=24h^{4}\left(x-x_{0}\right)^{2}$

$2^{\circ}$ Let's assume $n=2,p=2$ , and the nodes $x_{0},x_{1},x_{0}$ in arithmetic progression with the ratio $h$ If in Taylor's formula

y(x)=y\left(x_{0}\right)+\frac{x-x_{0}}{1!}y^{\prime}\left(x_{0}\right)+\frac{\left(x-x_{0}\right)^{2}}{2!}y^{\prime\prime}\left(x_{0}\right)+\int_{x_{0}}^{x}\frac{(x-s)^{2}}{2!}y^{\prime\prime\prime}(s)ds

derivatives are replaced $y^{\prime}\left(x_{0}\right),y^{\prime\prime}\left(x_{0}\right)$ from the numerical derivation formulas, we obtain

	$\displaystyle y(x)=B_{0}(x)y\left(x_{0}\right)+B_{1}(x)y\left(x_{1}\right)+B_{2}(x)y\left(x_{2}\right)+$
	$\displaystyle+C_{1}(x)\int_{x_{0}}^{x_{2}}\Phi_{1}(x)y^{\prime\prime\prime}(x)dx+C_{2}(x)\int_{x_{0}}^{x_{2}}\Phi_{2}(x)y^{(4)}(x)dx+\int_{x_{0}}^{x}\frac{(x-s)^{2}}{2!}y^{\prime\prime\prime}(s)ds$		(74)

Applying formulas (67) we have

	$\displaystyle B_{0}(x)=\frac{\left(x-x_{1}\right)\left(x-x_{2}\right)}{2h^{2}}=\frac{2h^{2}-3h\left(x-x_{0}\right)+\left(x-x_{0}\right)^{2}}{2h^{2}}$
	$\displaystyle B_{1}(x)=\frac{\left(x-x_{0}\right)\left(x-x_{2}\right)}{-h^{2}}=\frac{4h\left(x-x_{0}\right)-2\left(x-x_{0}\right)^{2}}{2h^{2}}$		(75)
	$\displaystyle B_{2}(x)=\frac{\left(x-x_{0}\right)\left(x-x_{1}\right)}{2h^{2}}=-\frac{h\left(x-x_{0}\right)-\left(x-x_{0}\right)^{2}}{2h^{2}}$

Also, applying formulas (70) we have

		$\displaystyle C_{1}(x)=\left(x-x_{0}\right)\left[\mu_{2}\left(x_{1}-x_{0},x_{2}-x_{0}\right)-\mu_{1}\left(x_{1}-x_{0},x_{2}-x_{0}\right)\left(x-x_{0}\right)\right]$
		$\displaystyle C_{2}(x)=\left(x-x_{0}\right)^{2}\left(\mu_{2}\left(x_{1}-x_{0},x_{2}-x_{0}\right)\right.$

MEAN

	$\displaystyle C_{1}(x)=\left(x-x_{0}\right)\left[2h^{2}-3h\left(x-x_{0}\right)\right]$
	$\displaystyle C_{2}(x)=2h^{2}\left(x-x_{0}\right)^{2}$		(76)

$3^{\circ}$ Let's assume $n=2$ and $p=3$ , and the nodes $x_{0},x_{1},x_{2},x_{3}$ in arithmetic progression with the ratio $h$ If in Taylor's formula

\begin{gathered}y(x)=y\left(x_{0}\right)+\frac{x-x_{0}}{1!}y^{\prime}\left(x_{0}\right)+\frac{\left(x-x_{0}\right)^{2}}{2!}y^{\prime\prime}\left(x_{0}\right)+\frac{\left(x-x_{0}\right)^{3}}{3!}y^{\prime\prime\prime}\left(x_{0}\right)+\\ +\int_{x_{0}}^{x}\frac{(x-s)^{3}}{3!}y^{(4)}(s)ds\end{gathered}

derivatives are replaced $y^{\prime}\left(x_{0}\right),y^{\prime\prime}\left(x_{0}\right),y^{\prime\prime\prime}\left(x_{0}\right)$ from the numerical derivation formulas, we obtain

	$\displaystyle y(x)=B_{0}(x)y\left(x_{0}\right)+B_{1}(x)y\left(x_{1}\right)+B_{2}(x)y\left(x_{2}\right)+$
	$\displaystyle+C_{1}(x)\int_{x_{0}}^{x_{2}}\Phi_{1}(x)y^{\prime\prime\prime}(x)dx+C_{2}(x)\int_{x_{0}}^{x_{2}}\Phi_{2}(x)y^{(4)}(x)dx+C_{3}(x)\int_{x_{0}}^{x_{2}}\Phi_{3}(x)y^{(5)}(x)dx+$
	$\displaystyle+\int_{x_{0}}^{x}\frac{(x-s)^{3}}{3!}y^{(4)}(s)ds$		(77)

Applying formulas (67), we observe that the polynomials $B_{0}(x),B_{1}(x),B_{2}(x)$ are given by formulas (75). Also, applying formulas (69) and (70) we have

	$\displaystyle C_{1}(x)=\left(x-x_{0}\right)\left(x-x_{1}\right)\left(x-x_{2}\right)$
	$\displaystyle C_{2}(x)=\left(x-x_{0}\right)^{2}\left[2h^{2}-3h\left(x-x_{1}\right)\right]$		(78)
	$\displaystyle C_{3}(x)=2h^{2}\left(x-x_{0}\right)^{3}$

16.

The remainder in Lagrange's interpolation formula. It was seen in the previous issue that in the case $p\geqslant n$ , polynomials $B_{0}(x),B_{1}(x),\ldots B_{n}(x)$ from formula (63) are given by formulas (67) and therefore the sum

B_{0}(x)y\left(x_{0}\right)+B_{1}(x)y\left(x_{1}\right)+\ldots+B_{n}(x)y\left(x_{n}\right)

is equal to Lagrange's polynomial $L\left[x_{0},x_{1},\ldots,x_{n};y(x)\right]$ of the function $y(x)$ on the nodes $x_{0},x_{1},\ldots,x_{n}$ This does not happen when $p<n$ .

Formula (63) is therefore written, when $p\geqslant n$ , in the form of

y=L\left[x_{0},x_{1},\ldots,x_{n};y(x)\right]+R(x).

(79)

We thus obtained Lagrange's interpolation formula, together with its remainder given by the formula

	$\displaystyle R(x)=$	$\displaystyle C_{1}(x)\int_{x_{0}}^{x_{n}}\Phi_{0}(x)y^{(n+1)}(x)dx+C_{2}(x)\int_{x_{0}}^{x_{n}}\Phi_{2}(x)y^{(n+2)}(x)dx+\ldots+$
		$\displaystyle+C_{p}(x)\int_{x_{0}}^{x_{n}}\Phi_{p}(x)y^{(n+p)}(x)dx+\int_{x_{0}}^{x}\frac{(x-s)^{p}}{p!}y^{(p+2)}(s)ds$		(80)

This remainder expression is related to the numerical derivation formulas studied in § 1 , of Chapter I, used to go from Taylor's formula to formula (63).

Recall that another expression of the remainder $R(x)$ in Lagrange's interpolation formula, was given by G. Kowa $1\mathrm{ewski}\mathrm{[5]}$ .

We will see further another expression of the remainder in Lagrange's interpolation formula in which only the derivative appears $y^{(n+1)}(x)$ and which is related to the numerical derivation formulas studied in chap. I, § 2.

If we note the $M_{n+1},M_{n+2},\ldots,M_{n+p}$ upper bounds of the absolute values of derivatives $y^{(n+1)}(x),y^{(n+2)}(x),\ldots,y^{(n+p)}(x)$ in the interval $[a,b]$ in which the knots are taken $x_{0},x_{1},\ldots,x_{n}$ and $x$ , from formula (80), we deduce the following evaluation of $|R(x)|$ ,

	$\displaystyle\|R(x)\|\leqslant\left\|C_{1}(x)\right\|\frac{M_{n+1}}{(n+1)!}$	$\displaystyle+\left\|C_{2}(x)\right\|\frac{M_{n+2}}{(n+2)!}+\ldots+\left\|C_{p}(x)\right\|\frac{M_{n+p}}{(n+p)!}+$
		$\displaystyle+\left(x-x_{0}\right)^{p+1}\frac{M_{p+1}}{(p+1)!}$		(81)

which was arrived at taking into account that the functions $\Phi_{1}(x),\ldots,\Phi_{p}(x)$ are positive in the interval $\left(x_{0},x_{n}\right)$ and using formulas (60).
17. Numerical integration of differential equations. Let us consider the differential equation

y^{\prime}(x)=f(x,y)

(82)

and suppose that the function $f(x,y)$ meets in the rectangle $D$ defined by inequality

\left|x-x_{0}\right|<\alpha,\quad\left|y-y_{0}\right|<\beta

the conditions that ensure the existence and uniqueness of the integral $y(x)$ which satisfies the initial condition $y\left(x_{0}\right)=y_{0}$ , in the interval $\left[x_{0},x_{0}+a\right]$ .

If the function $f(x,y)$ has successive partial derivatives with respect to $x$ and with $y$ , continue until the order $p$ in the rectangle $D$ , then from the differential equation (82) we deduce by successive derivatives

	$\displaystyle y^{\prime\prime}(x)=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}f=f_{1}(x,y)$
	$\displaystyle y^{\prime\prime\prime}(x)=\frac{\partial f_{1}}{\partial x}+\frac{\partial f_{1}}{\partial y}f=f_{2}(x,y)$		(83)
	$\displaystyle\cdots\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot$
	$\displaystyle y^{(p)}(x)=\frac{\partial f_{p-2}}{\partial x}+\frac{\partial f_{p-2}}{\partial y}f=f_{p-1}(x,y)$
	$\displaystyle y^{(p+1)}(x)=\frac{\partial f_{p-1}}{\partial x}+\frac{\partial f_{p-1}}{\partial y}f=f_{p}(x,y).$

Taylor's formula (59) in which we replace $y^{\prime}\left(x_{0}\right),y^{\prime\prime}\left(x_{0}\right),\ldots,y^{(p)}\left(x_{0}\right)$ with $f\left(x_{0},y_{0}\right),f_{1}\left(x_{0},y_{0}\right),\ldots,f_{p-1}\left(x_{0},y_{0}\right)$ and on $y^{(p+1)}(x)$ with $f_{p}[x,y(x)]$ , gives us the numerical integration formula

	$\displaystyle y(x)=y_{0}+\frac{x-x_{0}}{1!}f\left(x_{0},y_{0}\right)+\frac{\left(x-x_{0}\right)^{2}}{2!}f_{1}\left(x_{0},y_{0}\right)+\ldots+$
	$\displaystyle+\frac{\left(x-x_{0}\right)^{p}}{p!}f_{p-1}\left(x_{0},y_{0}\right)+R$		(84)

in which the rest is

R=\int_{x_{0}}^{x}\frac{(x-s)^{p}}{p!}f_{p}[s,y(s)]ds

(85)

Noting with $F_{p}$ an upper bound of the function $\left|f_{p}(x,y)\right|$ in the rectangle $D$ , we have for $|R|$ next assessment

|R|\leqslant F_{p}\frac{\left(x-x_{0}\right)^{p+1}}{(p+1)!}

(86)

But his calculation $y^{\prime\prime}\left(x_{0}\right),\ldots,y^{(p)}\left(x_{0}\right)$ it is considered too complicated in practice and other numerical integration formulas are preferred, for example Adams-type formulas.

We obtain numerical integration formulas if we use the numerical derivation formulas in Chapter I, § 1, which led us to formula (63) which replaces Taylor's formula (59).

To be precise, let us assume that the integral $y(x)$ was calculated on the nodes $x_{1},x_{2},\ldots x_{n}$ and that it is therefore known on these nodes. In order to apply formula (63), we will assume that the function $f(x,y)$ has successive partial derivatives with respect to $x$ and with $y$ , continue until the order
$n+p-1$ in the rectangle $D$ Then we can consider the series of formulas (83) up to $y^{(n+p)}(x)=f_{n+p-1}(x,y)$ , functions $f_{1}(x,y),\ldots,f_{n+p-1}(x,y)$ being all continuous in the rectangle $D$ .

Let us apply formula (63), in which we will replace $y^{(p+1)}(x),y^{(n+1)}(x)$ , $\ldots,y^{(n+p)}(x)$ with $f_{p}(x,y),f_{n}(x,y),\ldots f_{n+p-1}(x,y)$ We obtain the numerical integration formula of the differential equation

y(x)=B_{0}(x)y\left(x_{0}\right)+B_{1}(x)y\left(x_{1}\right)+\ldots+B_{n}(x)y\left(x_{n}\right)+R_{1}(x)

(87)

in which $B_{0}(x),B_{1}(x),\ldots,B_{n}(x)$ are polynomials in $x$ of the degree $p$ well-determined, which depend only on the position of the nodes and which are given by formulas (64) and (65) if $p<n$ , or by formulas (67) if $p\geqslant n$ .

REST $R_{1}(x)$ of the numerical integration formula (87) is

	$\displaystyle R_{1}(x)=C_{1}(x)\int_{x_{0}}^{x_{n}}\Phi_{1}(x)f_{n}[x,y(x)]dx+\ldots+C_{p}(x)\int_{x_{0}}^{x_{n}}\Phi_{p}(x)f_{n+p-1}[x,y(x)]dx+$
	$\displaystyle+\int_{x_{0}}^{x}\frac{(x-s)^{p}}{p!}f_{p}[s,y(s)]ds$		(88)

In this formula $C_{1}(x),C_{2}(x),\ldots,C_{p}(x)$ are polynomials of degree $p$ in $x$ , well-determined, which depend only on the position of the nodes and which are given by formulas (68) if $p<n$ , or by formulas (69) and (70) if $p\geqslant n$ Functions $\Phi_{1}(x),\Phi_{2}(x),\ldots,\Phi_{p}(x)$ are well determined in Chapter I, § 1, from the numerical derivation formulas; they also depend on the position of the nodes and for them we have the formulas (60). In the expression of the remainder, only the functions $f_{n}(x,y),\ldots,f_{n+p-1}(x,y),f_{p}(x,y)$ depend on the differential equation (82).

Taking into account the property of functions $\Phi_{1}(x),\ldots,\Phi_{p}(x)$ to be positive in the interval ( $x_{0},x_{n}$ ), (no. 5), we can give a new expression for the remainder $R_{1}(x)$ .

	$\displaystyle R_{1}(x)=C_{1}(x)\frac{f_{n}\left[\xi_{1},y\left(\xi_{1}\right)\right]}{(n+1)!}+C_{2}(x)\frac{f_{n+1}\left[\xi_{2},y\left(\xi_{2}\right)\right]}{(n+2)!}\cdots+$
	$\displaystyle\cdots+C_{p}(x)\frac{f_{n+p-1}\left[\xi_{p},y\left(\xi_{p}\right)\right]}{(n+p)!}+\frac{\left(x-x_{0}\right)^{p+1}}{(p+1)!}f_{p}\left[\xi_{,}y(\xi)\right]$		(89)

where $\xi_{1},\xi_{2},\ldots,\xi_{p}$ are certain points in the interval ( $x_{0},x_{n}$ ) and $\xi$ from the interval ( $x_{0},x$ ).

From formula (89) we also deduce the following evaluation of $\left|R_{1}(x)\right|$ , at the point $x$

	$\displaystyle\left\|R_{1}(x)\right\|\leqslant$	$\displaystyle\left\|C_{1}(x)\right\|\frac{F_{n}}{(n+1)!}+\left\|C_{2}(x)\right\|\frac{F_{n+1}}{(n+2)!}+\ldots+$
		$\displaystyle+\left\|C_{p}(x)\right\|\frac{F_{n+p-1}}{(n+p)!}+\frac{\left(x-x_{0}\right)^{p+1}}{(p+1)!}F_{p}$		(90)

in which it was generally noted with $F_{k}$ an upper edge of $\left|f_{k}(x,y)\right|$ in the rectangle $D$ .

7 - Mathematics studies and research

We have determined the remainder of the numerical integration formula (87), written in the form (88), highlighting how it depends on the function $f(x,y)$ and its successive partial derivatives with respect to $x$ and $y$ , until the order $n+p-1$ , this mode being related to the numerical derivation formulas in Chapter I, § 1, which we applied to obtain formula (63).

With this we have solved for the case considered by us, the problem of determining the remainder of the numerical integration formula (87), using the function $f(x,y)$ and its successive derivatives with respect to $x$ and $y$ .

We will see in the next paragraph another numerical integration formula of the type (87), in which the remainder has another expression related to the numerical derivation formulas, studied in § 2 of Chapter I.

When $p\geqslant n$ , the numerical integration formula (87) is written in the form

y(x)=L\left[x_{0},x_{1},\ldots,x_{n};y(x)\right]+R_{1}(x),

(91)

where $L\left[x_{0},x_{1},\ldots,x_{n};y(x)\right]$ is the Lagrange interpolation polynomial of the integral $y(x)$ , which is known on the nodes $x_{0},x_{1},\ldots x_{n}$ .

It is obvious that we can expect to obtain a numerical integration formula for the differential equation (82), taking as an approximate value of $y(x)$ at the point $x$ , the value of the polynomial of $\mathrm{L}_{\text{agrange }}L\left[x_{0},x_{1},\ldots,x_{n};y(x)\right]$ . But it is important that we have established the rest $R_{1}(x)$ of this numerical integration formula written in the form (88) or (89) and for which we have given the evaluation (90) of its absolute value. The rest $R_{1}(x)$ depends on the choice of numerical derivation formulas and it corresponds to the formulas treated in Chapter I, § 1.

Examples. $1^{\circ}$ Let's assume $n=4,p=2$ and that the nodes $x_{0},x_{1},x_{2}$ , $x_{3},x_{4}$ are in arithmetic progression with the ratio $h$ Applying formulas (71), we derive the numerical integration formula of the differential equation (82)

	$\displaystyle y(x)=B_{0}(x)y\left(x_{0}\right)$	$\displaystyle+B_{1}(x)y\left(x_{1}\right)+B_{2}(x)y\left(x_{2}\right)+B_{3}(x)y\left(x_{3}\right)+$
		$\displaystyle+B_{4}(x)y\left(x_{4}\right)+R_{1}(x)$		(92)

in which the polynomials $B_{0}(x),B_{1}(x),B_{2}(x),B_{3}(x),B_{4}(x)$ are given by formulas (72), the rest $R_{1}(x)$ is

	$\displaystyle R_{1}(x)=C_{1}(x)\int_{x_{0}}^{x_{4}}\Phi_{1}(x)f_{4}[x,y(x)]dx+C_{2}(x)\int_{x_{0}}^{x_{4}}\Phi_{2}(x)f_{5}[x,y(x)]dx+$
	$\displaystyle+\int_{x_{0}}^{x}\frac{(x-s)^{2}}{2!}f_{2}[s,y(s)]ds$		(93)

in which the polynomials $C_{1}(x),C_{2}(x)$ are given by formulas (73).
We have the following evaluation for $\left|R_{1}(x)\right|$

\left|R_{1}(x)\leqslant\left|C_{1}(x)\right|\frac{F_{4}}{5!}+\left|C_{2}(x)\right|\frac{F_{5}}{6!}+\frac{\left(x-x_{0}\right)^{3}}{3!}F_{2}.\right.

(94)

To see the importance of the coefficients of $F_{2},F_{3},F_{4}$ from the evaluation (94) of $\left|R_{1}(x)\right|$ , we give a table relative to $x-x_{0}=\frac{h}{2},\frac{3h}{2},\frac{5h}{2},\frac{7h}{2},\frac{9h}{2},5h$ .

$x-x_{0}$	His evaluation $\left\|R_{1}(x)\right\|$
$\frac{h}{2}$	$\frac{h^{3}}{48}F_{2}+\frac{h^{5}}{240}F_{4}+\frac{h^{6}}{120}F_{5}$
$\frac{3h}{2}$	$\frac{27h^{3}}{48}F_{2}+\frac{153h^{5}}{240}F_{4}+\frac{9h^{6}}{120}F_{5}$
$\frac{5h}{2}$	$\frac{125h^{3}}{48}F_{2}+\frac{505h^{5}}{240}F_{4}+\frac{25h^{6}}{120}F_{5}$
$\frac{7h}{2}$	$\frac{343h^{3}}{48}F_{2}+\frac{1057h^{5}}{240}F_{4}+\frac{49h^{6}}{120}F_{5}$
$\frac{9h}{2}$	$\frac{729h^{3}}{48}F_{2}+\frac{1809h^{5}}{240}F_{4}+\frac{81h^{6}}{120}F_{5}$
$5h$	$\frac{125h^{3}}{6}+\frac{113h^{5}}{12}F_{4}+\frac{5h^{6}}{6}F_{5}$

For $h=0,1$ the previous painting becomes

$x-x_{0}$	His evaluation $\left\|R_{1}(x)\right\|$
0.05	$0,000021F_{2}+0,000000042F_{4}+0,00000000084F_{5}$
0.15		$0,000563F_{2}+0,0000064$	$F_{4}+0,000000075$	$F_{5}$
0.25	0.00265	$F_{2}+0,000022$	$F_{4}+0,00000021$	$F_{5}$
0.35	0.00715	$F_{2}+0,0000441$	$F_{4}+0,00000041$	$F_{5}$
0.45	0.0152	$F_{2}+0,0000754$	$F_{4}+0,000000675$	$F_{5}$
0.50	0.021	$F_{2}+0,000095$	$F_{4}+0,00000084$	$F_{5}$

$2^{\circ}$ Let's assume $n=2,p=2$ and that the nodes $x_{0},x_{1},x_{2}$ are in arithmetic progression with the ratio $h$ From formula (74) we derive the following formula for numerical integration of the differential equation (82)

y(x)=B_{0}(x)y\left(x_{0}\right)+B_{1}(x)y\left(x_{1}\right)+B_{2}(x)y\left(x_{2}\right)+R_{1}(x),

(95)

in which the polynomials $B_{0}(x),B_{1}(x),B_{2}(x)$ are given by formulas (75), and the rest $R_{1}(x)$ is given by the formula

	$\displaystyle R_{1}(x)=C_{1}(x)\int_{x_{0}}^{x_{2}}\Phi_{1}(x)f_{2}[x,y(x)]dx+C_{2}(x)\int_{x_{0}}^{x_{2}}\Phi_{2}(x)f_{3}[x,y(n)]dx+$
	$\displaystyle+\int_{x_{0}}^{x_{2}}\frac{(x-s)^{2}}{2!}f_{2}[s,y(s)]ds$		(96)

in which the polynomials $C_{1}(x),C_{2}(x)$ are given by formulas (76).
We have the following evaluation of $\left|R_{1}(x)\right|$ :

\left|R_{1}(x)\right|\leqslant\left|C_{1}(x)\right|\frac{F_{2}}{3!}+\left|C_{2}(x)\right|\frac{F_{3}}{4!}+\frac{\left(x-x_{0}\right)^{3}}{3!}F_{2}

(97)

We give his ratings $\left|R_{1}(x)\right|$ for $x-x_{0}=\frac{h}{2},\frac{3h}{2},\frac{5h}{2}$ , 3h in the next tableau

$x-x_{0}$	His evaluation $\left\|R_{1}(x)\right\|$	His evaluation $\left\|R_{1}(x)\right\|$ for $h=0,1$
$\frac{h}{2}$	$\frac{13h^{3}}{48}F_{2}+\frac{h^{4}}{2}F_{3}$	$0,000271F_{2}+0,00005F_{3}$
$\frac{3h}{2}$	$\frac{207h^{3}}{48}F_{2}+\frac{9h^{4}}{2}F_{3}$	$0,00432\quad F_{2}+0,00045\quad F_{3}$
$\frac{5h}{2}$	$\frac{785h^{3}}{48}F_{2}+\frac{25h^{4}}{2}F_{3}$	$0,0164\quad F_{2}+0,00125F_{3}$
$3h$	$\frac{51h^{3}}{2}F_{2}+18h^{4}F_{3}$	$0,0255\quad F_{2}+0,0018\quad F_{3}$

$3^{\circ}$ Let's assume $n=2$ and $p=3$ , and the nodes $x_{0},x_{1},x_{2},x_{3}$ in arithmetic progression with the ratio h. From formal (77) we deduce the following formula for numerical integration of the differential equation (82)

y(x)=B_{0}(x)y\left(x_{0}\right)+B_{1}(x)y\left(x_{1}\right)+B_{2}(x)y\left(x_{2}\right)+R_{1}(x)

(98)

where the polynomials $B_{0}(x),B_{1}(x),B_{2}(x)$ are given by formulas (75), and the rest $R_{1}(x)$ is

	$\displaystyle R_{1}(x)$	$\displaystyle=C_{1}(x)\int_{x_{0}}^{x_{3}}\Phi_{1}(x)f_{2}[x,y(x)]dx+C_{2}(x)\int_{x_{0}}^{x_{2}}\Phi_{2}(x)f_{3}[x,y(x)]dx+$
		$\displaystyle+C_{3}(x)\int_{x_{0}}^{x_{2}}\Phi_{3}(x)f_{4}[x,y(x)]dx+\int_{x_{0}}^{x}\frac{(x-s)^{3}}{3!}f_{3}[s,y(s)]ds$		(99)

in which the polynomials $C_{1}(x),C_{2}(x),C_{3}(x)$ are given by formulas (78).
We have the following evaluation for $\left|R_{1}(x)\right|$ :

\left|R_{1}(x)\right|\leqslant\left|C_{1}(x)\right|\frac{F_{2}}{3!}+\left|C_{2}(x)\right|\frac{F_{3}}{4!}+\left|C_{3}(x)\right|\frac{F_{4}}{5!}+\frac{\left(x-x_{0}\right)^{4}}{4!}F_{3}

(100)

following.

We give his ratings $\left|R_{1}(x)\right|$ for $x-x_{0}=\frac{h}{2},\frac{3h}{2},\frac{5h}{2},3h$ in the painting

$x-x_{0}$	His evaluation $\left\|R_{1}(x)\right\|$	His evaluation $\left\|R_{1}(x)\right\|$ for $h=0,1$
$\frac{h}{2}$	$\frac{h^{3}}{16}F_{2}+\frac{49h^{4}}{384}F_{3}+\frac{h^{5}}{4}F_{4}$	$0,000063F_{2}+0,0000128F_{3}+0,0000025F_{4}$
$\frac{3h}{2}$	$\frac{h^{3}}{16}F_{1}+\frac{2241h^{4}}{384}F_{3}+\frac{27h^{5}}{4}F_{4}$	$0,000063F_{2}+0,000584\quad F_{3}+0,0000675F_{4}$
$\frac{5h}{2}$	$\frac{5h^{3}}{16}F_{2}+\frac{13825h^{4}}{384}F_{3}+\frac{125h^{5}}{4}F_{4}$	$0,000313F_{2}+0,0036003F_{3}+0,0003125F_{4}$
3h	$h^{3}F_{2}+\frac{531h^{4}}{8}F_{3}+54h^{5}F_{4}$	$0,001\quad F_{2}+0,0066375F_{3}+0,00054\quad F_{4}$

§ 2. Numerical integration formulas resulting from the application of numerical derivative formulas (31)

18.

The remainder in Lagrange's interpolation formula. Let us resume the numerical derivation formula (31) which we write in the form

	$\displaystyle\frac{f^{(p)}\left(x_{0}\right)}{p!}=-\frac{\lambda_{1}}{\mu}f\left(x_{0}\right)+\frac{K_{1}}{\mu}f\left(x_{1}\right)+\ldots+\frac{K_{n}}{\mu}f\left(x_{n}\right)+$
	$\displaystyle\quad+(-1)^{n-p+1}\mu_{n-p+1}\int_{x_{0}}^{x_{n}}\psi_{p}(x)f^{(n+1)}(x)dx$		(101)

where

\mu=\frac{(-1)^{p}}{\mu_{n-p+1}\left(x_{1}-x_{0},x_{2}-x_{0},\ldots,x_{n}-x_{0}\right)}

and where was it written shorter

\mu_{n-p+1}=\mu_{n-p+1}\left(x_{1}-x_{0},\ldots,x_{n}-x_{0}\right)

and in the second member it was written $\psi_{p}(x)$ in his place $\varphi(x)$ , to take into account the order $p$ derivation from the first member.

It was shown that the function $\psi_{p}(x)$ is positive in the interval ( $x_{0},x_{n}$ ) and that we have

\int_{x_{0}}^{x_{n}}\psi_{p}(x)dx=\frac{1}{(n+1)!}

(102)

In formula (101) let us replace $f(x)$ with $y(x)$ and let's do $p=1,2,\ldots,n$ We will obtain the following formulas:

	$\displaystyle\frac{y^{\prime}\left(x_{0}\right)}{1!}=K_{10}y\left(x_{0}\right)+K_{11}y\left(x_{1}\right)+\ldots+K_{1n}y\left(x_{n}\right)+(-1)^{n}\mu_{n}\int_{x_{0}}^{x_{n}}\psi_{1}(s)y^{(n+1)}(s)ds$
	$\displaystyle\frac{y^{\prime\prime}\left(x_{0}\right)}{2!}=K_{20}y\left(x_{0}\right)+K_{20}y\left(x_{1}\right)+\ldots+K_{2n}y\left(x_{n}\right)+(-1)^{n-1}\mu_{n-1}\int_{x_{0}}^{x_{n}}\psi_{2}(s)y^{(n+1)}(s)ds$
	$\displaystyle\frac{y^{(n)}\left(x_{0}\right)}{n!}=K_{n0}y\left(x_{0}\right)+K_{n1}y\left(x_{1}\right)+\ldots+K_{nn}y\left(x_{n}\right)+(-1)\mu_{1}\int_{x_{0}}^{x_{n}}\psi_{n}(s)y^{(n+1)}(s)ds$		(103)

These formulas being specified, let us consider Taylor's formula

y(x)=y\left(x_{0}\right)+\frac{x-x_{0}}{1!}y^{\prime}\left(x_{0}\right)+\ldots+\frac{\left(x-x_{0}\right)^{n}}{n!}y^{(n)}\left(x_{0}\right)+\int_{x_{0}}^{x}\frac{(x-s)^{n}}{n!}y^{(n+1)}(s)ds

and let's replace $y^{\prime}\left(x_{0}\right),y^{\prime\prime}\left(x_{0}\right),\ldots,y^{(\prime\prime)}\left(x_{0}\right)$ from the numerical derivation formulas (103). We will obtain a formula of the form

y(x)=B_{0}(x)y\left(x_{0}\right)+B_{1}(x)y\left(x_{1}\right)+\ldots+B_{n}(x)y\left(x_{n}\right)+R(x)

(105)

where

R(x)=-\int_{x_{0}}^{x_{n}}\psi(x,s)y^{(n+1)}(s)ds+\int_{x_{0}}^{x}\frac{(x-s)^{n}}{n!}y^{(n+1)}(s)ds

(106)

in which it was noted

	$\displaystyle\psi(x,s)=\mu_{1}\psi_{n}(s)\left(x-x_{0}\right)^{n}-\mu_{2}\psi_{n-1}(s)\left(x-x_{0}\right)^{n-1}+\ldots$
	$\displaystyle+(-1)^{n-1}\mu_{n}\psi_{1}(s)\left(x-x_{0}\right)$		(107)

It is easily shown that the sum of the first $n+1$ terms in formula (105) is the Lagrange interpolation polynomial, on the nodes $x_{0},x_{1},\ldots,x_{n}$ , which means that formula (105) can also be written in the form

y(x)=L\left[x_{0},x_{1},\ldots,x_{n};y(x)\right]+R(x)

(

\prime

)

which is Lagrange's interpolation formula, with its remainder written in the form (106).

If we note with $M_{n+1}$ an upper edge of $\left|y^{(n+1)}(x)\right|$ in the interval $[a,b]$ in which the knots were tied $x_{0},x_{1},\ldots,x_{n}$ and $x$ , from formulas (106), (107) we deduce the following evaluation of $|R(x)|$

|R(x)|\leqslant\frac{M_{n+1}}{(n+1)!}\left[\left(x-x_{0}\right)^{n+1}+\mu_{1}\left(x-x_{0}\right)^{n}+\ldots+\mu_{n}\left(x-x_{0}\right)\right]

or, taking into account $\mu_{1},\mu_{2},\ldots,\mu_{n}$ , this is also written in the form

|R(x)|\leqslant\left(x-x_{0}\right)\left[\left(x-x_{0}\right)+\left(x_{1}-x_{0}\right)\right]\ldots\left[\left(x-x_{0}\right)+\left(x_{n}-x_{0}\right)\right]\frac{M_{n+1}}{\left(n_{3}+1\right)!}

(108)

If the nodes $x_{0},x_{1},\ldots,x_{n}$ are in arithmetic progression with the ratio $h$ , then $\left|R\left(x_{0}+\lambda h\right)\right|$ it is his order $h^{n+1}$ . This follows from formula (108). We will have

\left|R\left(x_{0}+\lambda h\right)\right|\leqslant\frac{\lambda(\lambda+1)\ldots(\lambda+n)}{(n+1)!}M_{n+1}h^{n+1}

(109)

To see how big his coefficient is $M_{n+1}$ , we give a picture in case1 $n=5$ , for $\lambda=\frac{1}{2},\frac{3}{2},\frac{5}{2},\frac{7}{2},\frac{9}{2},\frac{11}{2},6$ for a $h$ some and for $h=0,1$ .

$\lambda$	His evaluation $\left\|R\left(x_{0}+\lambda h\right)\right\|$	His evaluation $\left\|R\left(x_{0}+\lambda h\right)\right\|$ for $h=0,1$
$\frac{1}{2}$	$\frac{231}{1024}h^{6}M_{6}$	$0,00000023M_{6}$
$\frac{3}{2}$	$\frac{3003}{1024}h^{6}M_{6}$	$0,00000294M_{6}$
$\frac{5}{2}$	$\frac{15015}{1024}h^{6}M_{6}$	$0,00001467M_{6}$
$\frac{7}{2}$	$\frac{51051}{1024}h^{6}M_{6}$	$0,00004986M_{6}$
$\frac{9}{2}$	$\frac{138567}{1024}h^{6}M_{6}$	$0,00013532M_{6}$
$\frac{11}{2}$	$\frac{323323}{1024}h^{6}M_{6}$	$0,00031575M_{6}$
6	$462\quad h^{6}M_{6}$	$0,00046200M_{6}$

19.

Numerical integration formula for differential equations. Let's consider the differential equation

y^{\prime}=f(x,y)

(110)

and suppose that the function $f(x,y)$ is continuous together with its partial derivatives with respect to $x$ and $y$ , until the order $n$ , in the rectangle $D$ In this case, successively deriving $y(x)$ , in relation to $x$ , we will obtain as in no. 16:

	$\displaystyle y^{\prime\prime}(x)=f_{1}(x,y)$
	$\displaystyle y^{\prime\prime\prime}(x)=f_{2}(x,y)$		(111)
	$\displaystyle y^{(n+1)}(x)=f_{n}(x,y)$

functions $f_{1}(x,y),f_{2}(x,y),\ldots,f_{n}(x,y)$ being continuous in the rectangle $D$ .
Either $y(x)$ the integral of the differential equation (110) that satisfies the initial condition $y\left(x_{0}\right)=y_{0}$ and which we assume is known on the nodes $x_{1},x_{2},\ldots,x_{n}$ Then, applying Lagrange's interpolation formula (105') to this integral, we obtain

y(x)=L\left[x_{0},x_{1},\ldots,x_{n};y(x)\right]+R_{2}(x),

(112)

in which the rest $R_{2}(x)$ is obtained from formula (106) by replacing $y^{(n+1)}(x)\mathrm{cu}f_{n}[x,y(x)]$ from formulas (111). For $R_{2}(x)$ we therefore have the following expression

R_{2}(x)=-\int_{x_{0}}^{x_{n}}\psi(x,s)f_{n}[s,y(s)]ds+\int_{x_{0}}^{x}\frac{(x-s)^{n}}{n!}f_{n}[s,y(s)]ds

(113)

in which function $\psi(x,s)$ is given by formula (107).
We can also give the following expression for $R_{2}(x)$ , taking into account the properties of functions $\psi_{1}(s),\ldots,\psi_{n}(s)$

	$\displaystyle R_{2}(x)=\frac{1}{(n+1)!}$	$\displaystyle\left\{f_{n}\left[\xi_{0},y\left(\xi_{0}\right)\right]\left(x-x_{0}\right)^{n+1}-\mu_{1}f_{n}\left[\xi_{1},y\left(\xi_{1}\right)\right]\left(x-x_{0}\right)^{n}+\right.$
		$\displaystyle\left.+\ldots+(-1)^{n}\mu_{n}f_{n}\left[\xi_{n},y\left(\xi_{n}\right)\right]\left(x-x_{0}\right)\right\}$		(113)

where $\xi_{0}\in\left(x_{0},x\right)$ and $\xi_{1},\xi_{2},\ldots,\xi_{n}\in\left(x_{0},x_{n}\right)$ From
formula (113') the following evaluation is deduced for $\left|R_{2}(x)\right|$

\left|R_{2}(x)\right|\leqslant\frac{\left(x-x_{0}\right)\left[\left(x-x_{0}\right)+\left(x_{1}-x_{0}\right)\right]\ldots\left[\left(x-x_{0}\right)+\left(x_{n}-x_{0}\right)\right]}{(n+1)!}F_{n}

(114)

where $F_{n}$ is an upper edge of $\mid f_{n}(x,y)$ in the rectangle $D$ We thus
obtained the numerical integration formula of the differential equation (110) in the form (112), in which $y(x)$ is approximately expressed by the Lagrange interpolation polynomial, relative to the nodes $x_{0},x_{1},\ldots,x_{n}$ ,
which is a result we could have expected, but what is important in formula (112) is that we also determined its remainder in the form (113) or (113') and for which its absolute value has the evaluation (114).

If the nodes $x_{0},x_{1},\ldots,x_{n}$ are in arithmetic progression with the ratio $h$ , then $\left|R_{2}\left(x_{0}+\lambda h\right)\right|$ it is his order $h^{n+1}$ This results from formula (114) and we have

\left|R_{2}\left(x_{0}+\lambda h\right)\right|\leqslant\frac{\lambda(\lambda+1)\ldots(\lambda+n)}{(n+1)!}F_{n}h^{n+1}.

(115)

The table from the previous number gives in the case $n=5$ , the values of the coefficients of $F_{n}$ for $h$ some and for $h=0,1$ .
20. Adams' numerical integration formula and the rest ci. We will now give an Adams-type numerical integration formula in the most general form of the differential equation (110), with the assumptions made in no. 18, but changing the $n$ in $n+1$ and using the results obtained by us relative to the remainder of the numerical integration formula (112), where the integral $y(x)$ is approximately replaced by the interpolation polynomial $L\left[x_{0},x_{1},\ldots,x_{n};y(x)\right.$ relative to the nodes $x_{0},x_{1},\ldots,x_{n}$ We will denote this polynomial more briefly by $L_{n}(x)$ and we will write

y(x)=L_{n}(x)+R_{2}(x)

(116)

where $R_{2}(x)$ is the remainder given by formula (113).
For this, we will apply the simple procedure for obtaining the actual Adams numerical integration formula (see [6]).

Whether $x_{n+1}$ a new node to his right $x_{n}$ and $y(x)$ the integral of equation (110) which is known on the nodes $x_{0},x_{1},\ldots x_{n}$ . Substituting in the differential equation (110) for $y$ with $L_{n}(x)+R_{2}(x)$ , we will have

y^{\prime}=g(x)+h(x)

(117)

where was it noted

g(x)=f\left[x,L_{n}(x)\right],h(x)=f\left[x,L_{n}(x)+R_{2}(x)\right]-f\left[x,L_{n}(x)\right].

(118)

We will assume for a moment that the curve $y=L_{n}(x)$ is located in the rectangle $D$ and we will then return to the case when the curve $y=L_{n}(x)$ comes out of the rectangle $D$ .

Integrating both members of formula (117) between $x_{n}$ and $x_{n+1}$ , we will deduce that

y\left(x_{n+1}\right)=y\left(x_{n}\right)+\int_{x_{n}}^{x_{n+1}}g(x)dx+\int_{x_{n}}^{x_{n+1}}h(x)dx

(119)

function $g(x)$ being known on the nodes $x_{0},x_{1},\ldots,x_{n}$ , to the first integral in formula (119) we will apply a quadrature formula with exterior nodes $x_{0},x_{1},\ldots,x_{n}$ and the knot $x_{n}$ [7].

It turns out that we have

	$\displaystyle\int_{x_{n}}^{x_{n+1}}g(x)dx=B_{0}\left[x_{n};g\right]$	$\displaystyle+B_{1}\left[x_{n-1},x_{n};g\right]+\ldots+B_{n}\left[x_{0},x_{1},\ldots,x_{n};g\right]$
		$\displaystyle+(-1)^{n+1}B_{n+1}\frac{g^{(n+1)}(\xi)}{(n+1)!}$		(120)

where $\xi\in\left(x_{0},x_{n+1}\right)$ , and

B_{0}=\int_{x_{n}}^{x_{n+1}}dx,\quad B_{k}=\int_{x_{n}}^{x_{n+1}}\left(x-x_{n}\right)\left(x-x_{n-1}\right)\ldots\left(x-x_{n-k+1}\right)dx

(121)

for $k=1,2,\ldots,n+1$ If
the nodes $x_{0},x_{1},\ldots,x_{n}$ are in arithmetic progression with the ratio $h$ , then formula (120) becomes

	$\displaystyle\int_{x_{n}}^{x_{n+1}}g(x)dx=h\left[g\left(x_{n}\right)+J_{1}\Delta g\left(x_{n-1}\right)+J_{2}\Delta^{2}g\left(x_{n-2}\right)+\ldots\right.$
	$\displaystyle\left.+J_{n}\Delta^{n}g\left(x_{0}\right)\right]+(-1)^{n+1}J_{n+1}h^{n+2}g^{(n+1)}(\xi),$		(122)

where

J_{k}=\frac{1}{k!}\int_{0}^{1}u(u+1)\cdots(u+k-1)du

(123)

Formula (119) becomes the generalized Adams formula with the nodes anyway $x_{0},x_{1},\ldots,x_{n}$

	$\displaystyle y\left(x_{n+1}\right)=$	$\displaystyle y\left(x_{n}\right)+B_{0}\left[x_{n};f\left[x,L_{n}(x)\right]\right]+B_{1}\left[x_{n-1}x_{n};f\left[x,L_{n}(x)\right]\right]$
		$\displaystyle+\ldots+B_{n}\left[x_{0},x_{1},\ldots,x_{n};f\left[x,L_{n}(x)\right]\right]+R_{3}$		(124)

where the rest $R_{3}$ is given by the formula

R_{3}=(-1)^{n+1}B_{n+1}\frac{g^{(n+1)}(\xi)}{(n+1)!}+\int_{x_{n}}^{x_{n+1}}h(x)dx

(125)

in which $g(x)$ and $h(x)$ are given by the formulas (118).
We can give an evaluation of $\left|R_{3}\right|$ From formula (125) we deduce that

\left|R_{3}\right|\leqslant B_{n+1}\frac{\left|g^{(n+1)}(\xi)\right|}{(n+1)!}+\int_{x_{n}}^{x_{n+1}}|h(x)|dx

(126)

But

h(x)=\int_{L_{n}(x)}^{L_{n}(x)+R_{2}(x)}\frac{\partial f}{\partial y}(x,y)dy

and so

|h(x)|\leqslant K\left|R_{2}(x)\right|

where $K$ is an upper edge of $\left|\frac{\partial f}{\partial y}\right|$ in the rectangle $D$ We will therefore have

\int_{x_{n}}^{x_{n+1}}|h(x)|dx\leqslant K\int_{x_{n}}^{x_{n+1}}\left|R_{2}(x)\right|dx

Taking into account formula (114), we deduce that

\begin{gathered}\int_{x_{n}}^{x_{n+1}}|h(x)|dx\leqslant\frac{KF_{n}}{(n+1)!}\int_{x_{n}}^{x_{n+1}}\left(x-x_{0}\right)\left[\left(x-x_{0}\right)+\left(x_{1}-x_{0}\right)\right]\ldots\\ {\left[\left(x-x_{0}\right)+\left(x_{n}-x_{0}\right)\right]dx}\end{gathered}

Noting with $\bar{F}_{n}$ an upper edge of $\left|g^{(n+1)}(x)\right|$ in the interval $\left[x_{0},x_{0}+a\right]$ , we will then deduce from formula (126) that

\left|R_{3}\right|\leqslant\frac{H_{n}}{(n+1)!}

(127)

where was it noted
$H_{n}=\bar{F}_{n}B_{n+1}+KF_{n}\int_{x_{n}}^{x_{n+1}}\left(x-x_{0}\right)\left[\left(x-x_{0}\right)+\left(x_{1}-x_{0}\right)\right]\ldots\left[\left(x-x_{0}\right)+\left(x-x_{0}\right)\right]dx$ Thus
we determined an evaluation of $\left|R_{3}\right|$ in the generalized Adams formula with nodes anyway $x_{0},x_{1},\ldots,x_{n}$ .

The case when the nodes $x_{0},x_{1},\ldots,x_{n}$ are in arithmetic progression with the ratio $h$ In this case, the numerical integration formula (124) is replaced, taking into account formula (122), with

	$\displaystyle y\left(x_{n+1}\right)=1\left(x_{n}\right)+h\left[g\left(x_{n}\right)+J_{1}\Delta g\left(x_{n-1}+J_{2}\Delta^{2}g\left(x_{n-2}\right)+\ldots\right.\right.$
	$\displaystyle\left.+J_{n}\Delta^{n}g\left(x_{0}\right)\right]+R_{4}$		(129)

where

R_{4}=(-1)^{n+1}J_{n+1}h^{n+2}g^{(n+1)}(\xi)+\int_{x_{n}}^{x_{n+1}}h(x)dx

Proceeding as above, we will have for $\left|R_{4}\right|$

\left|R_{4}\right|\leqslant J_{n+1}\bar{F}_{n}h^{n+2}+K\int_{x_{n}}^{x_{n+1}}\left|R_{2}(x)\right|dx

\begin{gathered}\left|R_{4}\right|\leqslant J_{n+1}\bar{F}_{n}h^{n+2}+\\ +\frac{KF_{n}}{(n+1)!}\int_{x_{n}}^{x_{n+1}}\left(x-x_{0}\right)\left[\left(x-x_{0}\right)+\left(x_{1}-x_{0}\right)\right]\ldots\left[\left(x-x_{0}\right)+\left(x_{n}-x_{0}\right)\right]dx\end{gathered}

If the change is made in the integral in the second member

x=x_{0}+nh+hu,

HAVE

\begin{gathered}\int_{x_{n}}^{x_{n+1}}\left(x-x_{0}\right)\left[\left(x-x_{0}\right)+\left(x_{1}-x_{0}\right)\right]\ldots\left[\left(x-x_{0}\right)+\left(x_{n}-x_{0}\right)\right]dx=\\ =h^{n+2}\int_{0}^{1}(u+n)(u+n+1)\ldots(u+2n)du\end{gathered}

so we will finally have

\left|R_{4}\right|\leqslant\bar{H}_{n}h^{n+2},

(130)

where was it noted

\bar{H}_{n}=J_{n+1}\bar{F}_{n}+\frac{KF_{n}}{(n+1)!}\int_{0}^{1}(u+n)(u+n+1)\ldots(u+2n)du

(131)

Thus it was proven that the remainder in Adams' formula with the nodes in arithmetic progression $x_{0},x_{1},\ldots,x_{n},x_{n+1}$ , with the ratio $h$ , it is of the order of $h^{n+2}$

It is very important to compare this result with the rest $R_{2}\left(x_{n+1}\right)$ when the integral $y(x)$ of the differential equation is replaced by the Lagrange polynomial relative to the nodes $x_{0},x_{1},\ldots,x_{n}$ , which according to formula (115) is of the order of $h^{n+2}$ .

From this comparison it follows that if the numerical integration formula (112) is applied, taking as an approximate value of the integral $y(x)$ Lagrange's polynomial relative to the nodes $x_{0},x_{1},\ldots,x_{n}$ , the rest in the node $x_{n+1}$ it is his order $h^{n+1}$ , while if Adams' numerical integration formula (129) is applied, the remainder is of the order of $h^{n+2}$ .

In the case of $n=5$ , which corresponds to the actual numerical integration formula of Adams, the remainder in the node $x_{6}$ it is his order $h^{6}$ in the
numerical integration formula (112), while in Adams' formula it is of the order of $h^{7}$ .

We recall that W. To11mien [8] established that the remainder in Adams' numerical integration formula is of the order of $h^{7}$ What we have shown above constitutes a generalization of this result for any number of nodes in arithmetic progression.

In another paper we will show how to construct numerical integration formulas of the Adams type in general, for which the remainder in the case of nodes $x_{0},x_{1},\ldots,x_{n}$ in arithmetic progression, it is of the order of $h^{n+q}$ , where $q>2$ .
21. The case when the curve $y=L_{n}(x)$ comes out of the rectangle $D$ In this case, we will extend the function $f(x,y)$ above and below the sides of the rectangle $D$ , parallel to the axis $Ox$ , in the following way, as he did $O$ Copper [9].

The function is considered

	$\displaystyle F^{\prime}(x,y)=\varphi_{n}(x)\frac{\left(y-\beta+y_{0}\right)^{n}}{n!}+\varphi_{n-1}(x)\frac{\left(y-\beta+y_{0}\right)^{n-1}}{(n-1)!}+$
	$\displaystyle+\ldots+\varphi_{0}(x)$		(132)

defined for $\left|x-x_{0}\right|\leqslant\alpha$ and $y-y_{0}\geqslant\beta$ . We determine the functions $\varphi_{0}(x)$ , $\varphi_{1}(x),\ldots,\varphi_{n}(x)$ so that

\left.\frac{\partial^{k}F^{\prime}}{\partial y^{k}}\right|_{y=y_{0}+\beta}=\left.\frac{\partial^{k}f}{\partial y^{k}}\right|_{y=y_{0}+\beta}

for $k=0,1,\ldots,n$ It will be enough to take

\varphi_{k}(x)=\left.\frac{\partial^{k}f}{\partial y^{k}}\right|_{y=y_{0}+\beta}\quad(k=0,1,\ldots,n).

From formulas (133) we deduce that we will also have

\left.\frac{\partial^{j}F^{\prime}}{\partial x^{i-k}\partial y^{k}}\right|_{y=y_{0}+s}=\left.\frac{\partial^{j}f}{\partial x^{j-k}\partial y^{k}}\right|_{y=y_{0}+\beta}

(134)

for $j=k+1,k+2,..n$ and $k=0,1,2,\ldots,n-1$ .
The function can also be considered
$F^{\prime\prime}(x,y)=\psi_{n}(x)\frac{\left(y+\beta-y_{0}\right)^{n}}{n!}+\psi_{n-1}(x)\frac{\left(y+\beta-y_{0}\right)^{n-1}}{(n-1)!}+\cdots+\psi_{0}(x)$ , where

\psi_{k}(x)=\left.\frac{\partial^{k}f}{\partial y^{k}}\right|_{y=y_{0}-\beta}\quad(k=0,1,\ldots,n).

It turns out that we have

\left.\frac{\partial^{j}F^{\prime\prime}}{\partial x^{j-k}\partial y^{k}}\right|_{y=y_{0}-\beta}=\left.\frac{\partial^{i}f}{\partial x^{j-k}\partial y^{k}}\right|_{y=y_{0}-\beta}

for $j=k,k+1,\ldots,n$ and $k=0,1,\ldots,n$ .

Whether $y=L_{n}(x)$ Lagrange's polynomial of the integral $y(x)$ of the differential equation $y^{\prime}=f(x,y)$ , which takes the values $y_{0},y_{1},\ldots,y_{n}$ on the nodes $x_{0},x_{1},\ldots,x_{n}$ If the curve $y=L_{n}(x)$ comes out of the rectangle $D$ , we denote by $\beta^{\prime}$ an upper edge of $L_{n}(x)$ and with $\beta^{\prime\prime}$ a lower edge of $L_{n}(x)$ in the interval $\left[x_{0},x_{0}+a\right]$ We denote by $\beta^{*}$ the largest of the numbers $\beta^{\prime}-y_{0},y_{0}-\beta^{\prime\prime}$ In the rectangle $D^{*}$ defined by

\left|x-x_{0}\right|\leqslant\alpha,\quad\left|y-y_{0}\right|\leqslant\beta^{*}

we consider the function

f^{*}(x,y)=\begin{cases}F^{\prime}(x,y)&\text{ dacă }y\geqslant\beta-y_{0}\\ f(x,y)&\text{ dacă }\left|y-y_{0}\right|\leqslant\beta\\ F^{\prime\prime}(x,y)&\text{ dacă }y\leqslant y_{0}-\beta\end{cases}

which is continuous, together with its partial derivatives with respect to $x$ and $y$ until ordered $n$ .

Let us consider the differential equation

Y^{\prime}=f^{*}(x,Y)

(136)

with the initial condition $Y\left(x_{0}\right)=y_{0}$ . Considering the definition of the function $f^{*}(x,y)$ , given by formula (135), and by the differential equation (110) with the same initial condition $y\left(x_{0}\right)=y_{0}$ , it follows that in the interval1 $\left[x_{0},x_{0}+a\right]$ , on which it was proven that there exist integrals $y(x)$ , we have $Y(x)=y(x)$ .

For the differential equation (136), the curve $y=L_{n}(x)$ , which coincides with the curve $y=L_{n}(x)$ relative to equation (110); does not leave the rectangle $D^{*}$ . So everything said in no. 19 remains valid provided that the rectangle is replaced $D$ with $D^{*}$ and the function $f(x,y)$ with the function $f^{*}(x,y)$ In particular, formulas (128), (131) remain valid, but replacing $F_{n}$ and $K$ with the corresponding values relative to the rectangle $D^{*}$ .

\begin{gathered}\text{ Universilatea Babeş - Bolyai, Cluj }\\ \text{ Catedra de ecuații diferentiale }\end{gathered}

BIBLIOGRAPHY

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J. Radon, Restausdrücke bei Interpolations und Quadratur Formeln durch bestimmte Integralen, Monatshefte für Math. und Phys. 42, 389, (1935).
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DV Ionescu, Numerical Quadratures, Chapter III, §5. Technical Publishing House, Bucharest 1957.
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T. Popoviciu, On the remainder in some numerical derivation theorems, Studii si Cerc. Mathem. (Bucharest), III, 1-2, 53-122 (1952).
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G. Kowalewski, Remarque sur l'interpolation Newtonienne, CR de l'Acad. often Sci. Paris, 227, 21-23 (1948).
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DV Ionescu, Quadrature formulas with exterior nodes, Studies and Research in Mathematics (Cluj), IX, 45-135 (1958).
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W. Tollmien, Uber die Fehlerabschätzung bei Adamschen Verfahren zur Integration gewöhnlicher Differential Gleihungen, Zeitschr. for ang. Math. und Mech., 18, 83-90 (1938).
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	$\displaystyle\left\|R_{1}(x)\right\|\leqslant$	$\displaystyle\left\|C_{1}(x)\right\|\frac{F_{n}}{(n+1)!}+\left\|C_{2}(x)\right\|\frac{F_{n+1}}{(n+2)!}+\ldots+$
		$\displaystyle+\left\|C_{p}(x)\right\|\frac{F_{n+p-1}}{(n+p)!}+\frac{\left(x-x_{0}\right)^{p+1}}{(p+1)!}F_{p}$		(90)

Applying numerical derivative formulas to numerical integration of differential equations

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APPLICATION OF NUMERICAL DERIVATION FORMULAS TO THE NUMERICAL INTEGRATION OF DIFFERENTIAL EQUATIONS

Chapter I

NUMERICAL DERIVATION FORMULAS

§ 1. Numerical derivation formulas that come from the expression of a divided difference in the form of a definite integral

§ 2. Generalization of AA Markov's formulas and putting the rest of these formulas in the form of a definite integral

Chapter II

NUMERICAL INTEGRATION OF DIFFERENT EQUATIONS, IALE

§ 1. Numerical integration formulas resulting from the application of numerical derivative formulas (6)

§ 2. Numerical integration formulas resulting from the application of numerical derivative formulas (31)

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