GENERALIZATION OF SOME INTERPOLATION FORMULAS FOR FUNCTIONS OF MULTIPLE VARIABLES AND SOME CONSIDERATIONS ON GAUSS'S NUMERICAL INTEGRATION FORMULA

DD STANCU
Communication presented by T. POPOVIOIU, corresponding member of the RPR Academy,
during the 4th Congress of Romanian Mathematicians from May 27-June 4, 1956

and

1.

The importance of the approximate calculation of a function, defined directly, through its properties, or as a solution of a certain differential equation, is particularly great in technical applications.

The theory of polynomial interpolation plays an important role in this direction. The extensive use of interpolation polynomials is justified by their simple analytical structure, the possibility of drawing up systematic and simple calculation programs, as well as the sufficiently good precision to which they lead.
2. In the case of polynomial interpolation of functions of several variables, however, quite great difficulties are encountered. Even if the interpolation nodes are distinct, it could happen that a certain interpolation polynomial does not exist or is not unique. However, it is easily shown that if the nodes are not located on a hypersurface of an order equal to the degree of the interpolation polynomial, then its existence and uniqueness are assured.

From a practical point of view it is useful to give certain concrete schemes of nodes relative to which the interpolation polynomial is perfectly determined and in addition to obtain for it an effective expression convenient for applications, for calculations.
3. We will seek in what follows to give some more general distributions of nodes than those that have been used so far, distributions that allow a wide variety of node networks to be used.
4. Let us first deal with the case of two variables.

Whether $f(x,y)$ a function defined and bounded in a certain domain $D_{2}$ from the plane. Suppose that the values of the function are known $f(x,y)$ on the following points of this area

M_{ik}\left(x_{i},y_{ik}\right),\quad\left(k=1,m_{i};i=1,\overline{(n)}{}^{\star}\right)

Lagrange's interpolation formula relative to the variable $x$ Service $f(x,y)$ on the string of values

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

is written in the form

f(x,y)=\sum_{i=1}^{n}\frac{u(x)}{\left(x-x_{i}\right)u^{\prime}\left(x_{i}\right)}f\left(x_{i},y\right)+u(x)\left[x,x_{1},\ldots,x_{n};f(x,y)\right]

where

u(x)=\coprod_{i=1}^{n}\left(x-x_{i}\right)

and

\left[x_{1},x_{2},\ldots,x_{n};f\right]=\sum_{i=1}^{n}\frac{f\left(x_{i},y\right)}{u^{\prime}\left(x_{i}\right)}

is the difference divided by the order $n-1$ of the function $f(x,y)$ for his value $x:x_{1},x_{2},\ldots,x_{n}$ .

But the function $\varphi_{i}(y)=f\left(x_{i},y\right)$ can also be developed using the same interpolation formula using the values

y_{i1},y_{i2},\ldots,y_{i,m_{i}}.

It is obtained

\begin{array}[]{r}\varphi_{i}(y)=f\left(x_{i},y\right)=\sum_{k=1}^{m_{i}}\frac{v_{i}(y)}{\left(y-y_{ik }\right)v_{i}^{\prime}\left(y_{ik}\right)}f\left(x_{i},y\right)+v_{i}(y)\left[y,y_{i1},\ldots\right.\\ \left.,y_{i,m_{i}};f\left(x_{i},y\right)\right]\end{array}

where

v_{i}(y)=\prod_{p=1}^{m_{i}}\left(y-y_{ip}\right)

Replacing this expression with $f\left(x_{i},y\right)$ in (2), we find the following interpolation formula

f(x,y)=L_{2}(x,y)+R_{2}(x,y)

where

L_{2}(x,y)=\sum_{i=1}^{n}\sum_{k=1}^{m_{i}}\frac{u(x)}{\left(x-x_{i}\right)u^{\prime}\left(x_{i}\right)}\frac{v_{i}(y)}{\left(y-y_{ik}\right)v_{i}^{\prime}\left(u_{ik}\right)}f\left(x_{i},y_{ik}\right)

⁰⁰footnotetext:

\left.{}^{*}\right)

Prin

N=\overline{\alpha,\beta}

vom ințelege că

N

ia succesiv valorile

\alpha,\alpha+1,\ldots,\beta-1,\beta

is the minimum degree interpolation polynomial that coincides with the function $f(x,y)$ on the nodes (1), and

	$\displaystyle R_{2}(x,y)=\dot{u}(x)\left[x,x_{1},\ldots,x_{n};f(x,y)\right]+\sum_{i=1}^{n}l_{i}(x)v_{i}(y)\left[y,y_{i1},\ldots\right.$
	$\displaystyle\left.\ldots,y_{i,m_{i}};f\left(x_{i},y\right)\right]$		(9)

l_{i}(x)=\frac{u(x)}{\left(x-x_{i}\right)u^{\prime}\left(x_{i}\right)},

(10)

is the remainder of this interpolation formula.
5. We emphasize that the interpolation formula (7) is much more general than the classical Lagrange interpolation formula for two variables, since the ordinates (4) depend on the abscissa $x_{i}$ both as a value and as a number (what we specified through indices).

The well-known Lagrange interpolation formula is obtained if we make the following customizations

y_{ik}=y_{k},m_{i}=m\quad(i=\overline{1,n}).

(11)

6.

Let's give an example. Assuming that
$n=5,m_{1}=3,m_{2}=4,m_{3}=5,m_{4}=$

=4,m_{5}=3

let's write the interpolation polynomial on the nodes

\begin{gathered}M_{1}(-2h,-h),M_{2}(-2h,0),M_{3}(-2h,h),\\ M_{4}\left(-h,-\frac{3h}{2}\right),M_{5}\left(-h,-\frac{h}{2}\right),M_{6}\left(-h,\frac{h}{2}\right),M_{7}\left(-h,\frac{3h}{2}\right),\\ M_{8}(0,-2h),M_{9}(0,-h),M_{10}(0,0),M_{11}(0,h),M_{12}(0,2h),\\ M_{13}\left(h,-\frac{3h}{2}\right),M_{14}\left(h,-\frac{h}{2}\right),M_{15}\left(h,\frac{h}{2}\right),M_{16}\left(h,\frac{3h}{2}\right),\\ M_{17}(2h,-h),M_{18}(2h,0),M_{19}(2h,h),\end{gathered}

which are the nodes of an hexagonal network (see figure).

get

		$\displaystyle L(x,y)=\frac{1}{48h^{6}}\left(x^{2}-h^{2}\right)x(x-2h)y(y-h)f(-2h,-h)-$
		$\displaystyle\quad-\frac{1}{24h^{6}}\left(x^{2}-h^{2}\right)x(x-2h)\left(y^{2}-h^{2}\right)f(-2h,0)+$
		$\displaystyle\quad+\frac{1}{48h^{6}}\left(x^{2}-h^{2}\right)x(x-2h)(y+h)yf(-2h,h)+$
		$\displaystyle+\frac{1}{36h^{7}}\left(x^{2}-4h^{2}\right)x(x-h)\left(y^{2}-\frac{h^{2}}{4}\right)\left(y-\frac{3h}{2}\right)f\left(-h,-\frac{3h}{2}\right)-$
		$\displaystyle-\frac{1}{12h^{7}}\left(x^{2}-4h^{2}\right)x(x-h)\left(y^{2}-\frac{9h^{2}}{4}\right)\left(y-\frac{h}{2}\right)f\left(-h,-\frac{h}{2}\right)+$
		$\displaystyle+\frac{1}{12h^{7}}\left(x^{2}-4h^{2}\right)x(x-h)\left(y^{2}-\frac{9h^{2}}{4}\right)\left(y+\frac{h}{2}\right)f\left(-h,\frac{h}{2}\right)-$
		$\displaystyle-\frac{1}{36h^{7}}\left(x^{2}-4h^{2}\right)x(x-h)\left(y^{2}-\frac{h^{2}}{4}\right)\left(y+\frac{3h}{2}\right)f\left(-h,\frac{3h}{2}\right)+$
		$\displaystyle+\frac{1}{96h^{8}}\left(x^{2}-4h^{2}\right)\left(x^{2}-h^{2}\right)\left(y^{2}-h^{2}\right)y(y-2h)f(0,-2h)-$
		$\displaystyle\quad-\frac{1}{24h^{8}}\left(x^{2}-4h^{2}\right)\left(x^{2}-h^{2}\right)\left(y^{2}-4h^{2}\right)y(y-h)f(0,-h)+$
		$\displaystyle\quad+\frac{1}{16h^{8}}\left(x^{2}-4h^{2}\right)\left(x^{2}-h^{2}\right)\left(y^{2}-4h^{2}\right)\left(y^{2}-h^{2}\right)f(0,0)-$
		$\displaystyle\quad-\frac{1}{24h^{8}}\left(x^{2}-4h^{2}\right)\left(x^{2}-h^{2}\right)\left(y^{2}-4h^{2}\right)(y+h)yf(0,h)+$
		$\displaystyle\quad+\frac{1}{96h^{8}}\left(x^{2}-4h^{2}\right)\left(x^{2}-h^{2}\right)(y+2h)\left(y^{2}-h^{2}\right)yf(0,2h)+$
		$\displaystyle+\frac{1}{36h^{7}}\left(x^{2}-4h^{2}\right)(x+h)x\left(y^{2}-\frac{h^{2}}{4}\right)\left(y-\frac{3h}{2}\right)f\left(h,-\frac{3h}{2}\right)-$
		$\displaystyle-\frac{1}{12h^{7}}\left(x^{2}-4h^{2}\right)(x+h)x\left(y^{2}-\frac{9h^{2}}{4}\right)\left(y-\frac{h}{2}\right)f\left(h,-\frac{h}{2}\right)+$
		$\displaystyle+\frac{1}{12h^{7}}\left(x^{2}-4h^{2}\right)(x+h)x\left(y^{2}-\frac{9h^{2}}{4}\right)\left(y+\frac{h}{2}\right)f\left(h,\frac{h}{2}\right)-$
		$\displaystyle-\frac{1}{36h^{7}}\left(x^{2}-4h^{2}\right)(x+h)x\left(y+\frac{3h}{2}\right)\left(y^{2}-\frac{h^{2}}{4}\right)f\left(h,\frac{3h}{2}\right)+$

		$\displaystyle+\frac{1}{48h^{6}}(x+2h)\left(x^{2}-h^{2}\right)xy(y-h)f(2h,-h)-$
		$\displaystyle-\frac{1}{24h^{6}}(x+2h)\left(x^{2}-h^{2}\right)x\left(y^{2}-h^{2}\right)f(2h,0)+$
		$\displaystyle+\frac{1}{48h^{6}}(x+2h)\left(x^{2}-h^{2}\right)x(y+h)yf(2h,h)$

7.

In the case of three variables, using the nodes

M_{ikj}\left(x_{i},y_{ik},z_{iki}\right)\left(i=\overline{1,n};k=\overline{1,m_{i}};j=\overline{\left.1,p_{ik}\right)}\right.

(12)

the interpolation formula is obtained

f(x,y,z)=L_{3}(x,y,z)+R_{3}(x,y,z)

(13)

where

L_{3}(x,y,z)=\sum_{i=1}^{n}\sum_{k=1}^{m_{i}}\sum_{j=1}^{p_{ik}}\frac{u(x)}{\left(x-x_{i}\right)u^{\prime}\left(x_{i}\right)}\frac{v_{i}(y)}{\left(y-y_{ik}\right)v_{i}^{\prime}\left(y_{ik}\right)}\frac{w_{ik}(z)}{\left(z-z_{ikj}\right)w_{ik}^{\prime}\left(z_{ikj}\right)}f\left(x_{i},y_{ik},z_{ikj}\right)

(14)

is the minimum degree interpolation polynomial that coincides with the function. $f(x,y,z)$ on the nodes (12), and the rest has the expression

	$\displaystyle\quad R_{3}(x,y,z)=u(x)\left[x,x_{1},\ldots,x_{n};f(x,y,z)\right]+$		(15)
	$\displaystyle\quad+\sum_{i=1}^{n}l_{i}(x)v_{i}(y)\left[y,y_{i1},\ldots,y_{im_{i}};f\left(x_{i},y,z\right)\right]+$
	$\displaystyle+\sum_{i=1}^{n}\sum_{k=1}^{m_{i}}l_{i}(x)h_{ik}(y)w_{ik}(z)\left[z,z_{ik1},\ldots,z_{ikp_{ik}};f\left(x_{i},y_{ik},z\right)\right].$

following

Above, along with the notations already explained, I also used the following:

w_{ik}(z)=\prod_{s=1}^{x_{ik}}\left(z-z_{iks}\right),h_{ik}(y)=\frac{v_{i}(y)}{\left(y-y_{ik}\right)v_{i}^{\prime}\left(y_{ik}\right)}

8.

In the general case, considering the function

f(M)=f\left(t^{1},t^{2},\ldots,t^{s}\right)

limited and confined to a domain $D_{s}$ of Euclidean space $s$ - dimensional $E_{s}$ and the nodes

\left.\begin{array}[]{c}M_{i_{1}i_{2}}\cdots i_{s}=M_{i_{1}i_{2}}\cdots i_{s}\left(t_{i_{1}}^{1},t_{i_{1}i_{2}}^{2}\cdots,t_{i_{1}i_{2}}^{s}\cdots i_{s}\right)\\ \left(i_{1}=\overline{1,n};i_{2}=\overline{1,m_{i_{1}}};\ldots;i_{s}=1,m_{i_{1}}\cdots i_{s-1}\right.\end{array}\right),\penalty 10000\ 25

in number of

N=\sum_{i_{1}=1}^{n}\sum_{i_{2}=1}^{m_{i_{1}}}\ldots\sum_{i_{s-1}=1}^{m_{i_{1}}\ldots i_{s-2}}m_{i_{1}i_{2}\ldots i_{s-1}},

the interpolation formula is obtained

f(M)=L_{s}(M)+R_{s}(M)

where

L_{s}(M)=\sum_{i_{1}=1}^{n}\sum_{i_{2}=1}^{m_{i_{1}}}\ldots\sum_{i_{s}=1}^{m_{i_{1}}\ldots i_{s-1}}l_{i_{1}}^{1}\left(t^{1}\right)\ldots l_{i_{1}\ldots i_{s}}^{s}\left(t^{s}\right)f\left(M_{i_{1}\ldots i_{s}}\right),

(1)

\begin{gathered}l_{i_{1}i_{2}\ldots i_{k}}^{k}\left(t^{k}\right)=\frac{u_{i_{1}i_{2}\ldots i_{k-1}}^{k}\left(c^{k}\right)}{\left(l^{k}-l_{i_{1}\ldots i_{k}}^{k}\right)\dot{u}_{i_{1}\ldots i_{k-1}}^{k}\left(k_{i_{1}\ldots i_{k}}^{k}\right)}\\ u_{i_{1}i_{2}\ldots i_{k-1}}^{k}\left(t^{k}\right)=\prod_{i_{k}=1}^{m_{i_{1}\ldots i_{k-1}}}\left(t^{k}-t_{i_{1}\ldots i_{k}}^{k}\right)\end{gathered}

is the minimum degree interpolation polynomial that coincides with the function $f(1$ on the nodes (16)

The rest of the interpolation formula (17) has the expression

R_{s}(M)=\sum_{k=1}^{s}\sum_{i_{1}=1}^{n}\sum_{i_{2}=1}^{m_{i_{1}}}\cdots\sum_{i_{k-1}=1}^{m_{i_{1}}\ldots i_{k-2}}l_{i_{1}}^{1}\left(t^{1}\right)\ldots l_{i_{1}\ldots i_{k-1}}^{k-1}u_{i_{1}\ldots i_{k-1}}^{k_{1}}\left(t^{k}\right)D_{i_{1}\ldots i_{k-1}}^{k}

where

D_{i_{1}\ldots i_{k-1}}^{k}=\left[t^{k},t_{i_{1}\ldots i_{k-1}11}^{k}\ldots,t_{i_{1}\ldots i_{k-1}m_{k}}^{k};\varphi i_{1}\ldots i_{k-1}\right],

and

\varphi_{i_{1}\ldots i_{k-1}}=f\left(t_{i_{1}}^{1},t_{i_{1}i_{2}}^{2},\ldots,t_{i_{1}\ldots i_{k-1}}^{k},t^{k},\ldots,t^{s}\right)

9.

In the particular case *)

\begin{gathered}n=m_{1}+1,m_{i_{1}}=m_{2}+1,\ldots,m_{i_{1}\ldots i_{s-1}}=m_{s}+1\\ \left(i_{k}=\overline{1,m_{k}};k=\overline{1,s}\right)\end{gathered}

the interpolation polynomial (18) has degree ( $m_{1},m_{2},\ldots,m_{s}$ ).
The corresponding interpolation nodes

		$\displaystyle P_{i_{1}i_{2}\ldots i_{s}}\left(t_{i_{1}}^{1},t_{i_{1}i_{2}}^{2},\ldots,t_{i_{1}i_{2}\ldots i_{s}}^{s}\right)$
		$\displaystyle\left(i_{k}^{-}=\overline{1,m_{k}+1};k=\overline{1,s}\right)$

we will say that it determines a pseudo-network of the order $\left(m_{1},m_{2},\ldots,m_{s}\right)$ ,

His coefficient $\left(t^{1}\right)^{m_{1}}\left(t^{2}\right)^{m_{2}}\cdots\left(t^{s}\right)^{m_{s}}$ from the interpolation polynomial, which is obtained in this case, is

\Delta^{m_{1}m_{2}\ldots m_{s}}(f)=\sum_{i_{1}=1}^{m_{1}+1}\cdots\sum_{i_{s}=1}^{m_{s}+1}\frac{f\left(P_{\left.i_{1}i_{2}\ldots is\right)}\right.}{\prod_{k=1}^{s}\dot{v}_{i_{1}i_{2}\ldots i_{k-1}}^{k}\left(l_{i_{1}i_{2}\ldots i_{k}}^{k}\right)},

(21)

(1) where

We will call expression (21) the partial divided difference of the order $\left(m_{1},m_{2},\ldots,m_{s}\right)$ of the function $f(M)$ on points (20).

We mention on this occasion the following formula which allows us to ⁽¹⁾ see the structure of this divided partial difference

	$\displaystyle\Delta^{m_{1}m_{2}\ldots m_{s}(f)=}$
	$\displaystyle=\left[t_{1}^{1}\ldots,t_{m_{1}+1}^{1};\sum_{i_{1}=1}^{m_{1}+1}l_{i_{1}}^{1}\left(t^{1}\right)\left[t_{i_{1}1}^{2},\ldots,t_{i_{1}m_{2}+1}^{2};\ldots\right.\right.$		(22)
	$\displaystyle\left.\left.\sum_{i_{s-1}=1}^{m_{s-1}+1}l_{i_{1}\ldots i_{s-1}^{s-1}}^{s-1}\left(t^{s-1}\right)\left[t_{i_{1}\ldots i_{s-1}1}^{s}\cdots t_{i_{1}\ldots i_{s-1}m_{s+1}}^{s};f(M)\right]\ldots\right]\right].$

10.

If we further particularize the previous pseudo-network so that it reduces to the so-called Merchaus network *) of order ( $m_{1}$ , $m_{2},\ldots,m_{s}$ ), determined by the nodes

Q_{i_{1}i_{2}\ldots i_{s}}\left(t_{i_{1}}^{1},t_{i_{2}}^{2},\ldots,t_{i_{s}}^{s}\right)\quad\left(i_{k}=\overline{1,m_{k}+1};k=\overline{1,s}\right)

(23)

The previous results are greatly simplified.
The interpolation polynomial (18) reduces to the well-known **) Lagrange interpolation polynomial for $s$ variables.

In this case the partial divided difference (21) takes the form

	$\displaystyle D^{m_{1}m_{2}\ldots m_{s}}(f)=$
	$\displaystyle{\left[\begin{array}[]{l}t_{1}^{1},t_{2}^{1},\ldots,t_{m_{1}+1}^{1}\\ t_{1}^{2},t_{2}^{2},\ldots,t_{m_{2}+1}^{2}\\ \ldots\ldots\ldots\ldots.;f\\ t_{1}^{s},t_{2}^{s},\ldots,t_{m_{s}+1}^{s}\end{array}\right]=\sum_{i_{1}=1}^{m_{1}+1}\cdots\sum_{i_{s}=1}^{m_{s}+1}\frac{f\left(t_{i_{1}}^{1},t_{i_{2}}^{2},\ldots,i_{i_{s}}^{s}\right)}{\prod_{k=1}^{s}\omega^{k}\left(t_{i_{k}}^{k}\right)}}$		(24)
	$\displaystyle\omega^{k}\left(t^{k}\right)=\prod_{i_{k}=1}^{m_{k}+1}\left(t^{k}-t_{i_{k}}^{k}\right)$

(and

⁰⁰footnotetext: *) După numele lui A. Marchaud [2].
**) A se vedea de exemplu: D. L. Berman [3].

Formula (22) simplifies greatly:

\begin{gathered}D^{m_{1}m_{2}\ldots m_{s}}(f)=\\ =\left[t_{1}^{1},\ldots,t_{m_{1}+1}^{1};\left[t_{1}^{2},\ldots,t_{m_{2}+1}^{2};\left[t_{1}^{s},\ldots,t_{m_{s}+1}^{s};f\right]\ldots\right]\right].\end{gathered}

This tells us that in the case of a Marchaud network a partial division difference of the order ( $m_{1},m_{2},\ldots,m_{s}$ ) is a superposition of $s$ differences divided by a variable, the order of superposition being arbitrary.
11. Relative to the partial divided difference (24) we will give some average forms useful for establishing the structure of the remainder of many approximation formulas.

A first average formula is

\left[\begin{array}[]{l}t_{1}^{1},t_{2}^{1},\ldots,t_{m_{1}+1}^{1}\\ t_{1}^{2},t_{2}^{2},\ldots,t_{m_{2}+1}^{2}\\ \ldots\ldots\ldots\\ t_{1}^{8},t_{2}^{s},\ldots,t_{ms+1}^{s}\end{array}\right]=\frac{1}{m_{1}!m_{2}!\ldots m_{s}!}\frac{\partial^{m_{1}\ldots+m_{s}}f\left(\xi^{1},\ldots,\xi^{s}\right)}{\partial\left(\xi^{1}\right)^{m_{1}}\ldots\partial\left(\xi^{s}\right)^{m_{s}}}

where $\xi^{i}$ is contained in the smallest interval containing the numbers ${}_{2}^{i},\ldots,t_{m_{i}+1}^{i}$ .

This formula, for the case $s=2$ , can be seen in [4].
12. Let us consider $s$ natural numbers $p_{1},p_{2},\ldots,p_{8}$ so that.

1\leqq p_{i}\leqq m_{i}+2,(i=\overline{1,s}).

Taking into account formula (24), it is easy to verify that we have

\begin{gathered}{\left[\begin{array}[]{c}t_{1}^{1},\ldots,t_{m_{1}+2}^{1}\\ t_{1}^{2},\ldots,t_{m_{2}+2}^{2}\\ \ldots\ldots\ldots\\ t_{1}^{s},\ldots,t_{m_{s}+2}^{s}\end{array};\prod_{i=1}^{s}\left(t^{i}-t_{p_{i}}^{i}\right)f(M)\right]=}\\ =\left[\begin{array}[]{cc}t_{1}^{1},\ldots,t_{p_{1}-1}^{1},t_{p_{1}+1}^{1},\ldots,t_{m_{1}+2}^{1}&\\ t_{1}^{2},\ldots,t_{p_{2}-1}^{2},t_{p_{2}+1}^{2},\ldots,t_{m_{2}+2}^{2}&;f(M)\\ \ldots..&..\\ t_{1}^{s},\ldots,t_{p_{s}-1}^{s},t_{p_{s}+1}^{s},\ldots,t_{m_{s}+2}^{s}&.\ldots\end{array}\right].\end{gathered}

Using identities

t^{i}-t_{p_{i}}^{i}=\frac{\left(t_{p_{i}}^{i}-t_{1}^{i}\right)\left(t^{i}-t_{m_{i}+2}^{i}\right)+\left(t_{m_{i}+2}^{i}-t_{p_{i}}^{i}\right)\left(t^{i}-t_{1}^{i}\right)}{t_{m_{i}+2}^{i}-t_{1}^{i}},

the previous formula will lead us to the average formula

		$\displaystyle{\left[\begin{array}[]{l}t_{1}^{1},\ldots,t_{p_{1}-1}^{1},t_{p_{1}+1}^{1},\ldots,t_{m_{1}+2}^{1}\\ t_{1}^{2},\ldots,t_{p_{2}-1}^{1},t_{p_{2}+1}^{2},\ldots,t_{m_{2}+2}^{2}\\ \cdot\ldots\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\\ t_{1}^{s},\ldots,t_{p_{s^{-}}-1}^{s},t_{p_{s}+1}^{s},\ldots,t_{p_{s}+2}^{s}\end{array}\right]=}$
		$\displaystyle=\frac{1}{\left.\prod_{i=}^{s}\underline{\iota}_{m_{i}+2}^{i}-l_{1}^{i}\right)}\sum_{j_{1}=1}^{2}\ldots\sum_{j_{s}=1}^{2}A_{j_{1}}^{1}\ldots A_{j_{s}}^{s}D_{j_{1}j_{2}\ldots j_{s}}^{m_{1}m_{2}\ldots m_{s}},$

where

\begin{gathered}A_{1}^{k}=t_{p_{k}}^{k}-t_{1}^{k},A_{2}^{k}=t_{m_{k}+2}^{k}-t_{p_{k}}^{k}\\ D_{j_{1}j_{2}\ldots j_{s}}^{m_{1}m_{2}\ldots m_{s}}=\left[\begin{array}[]{c}t_{j_{1}}^{1},t_{j_{1}+1}^{1},\ldots,t_{j_{1}+m_{1}}^{1}\\ t_{j_{2}}^{2},t_{j_{2}+1}^{2},\ldots,t_{j_{2}+m_{2}}^{2};f(M)\\ \cdot\cdot\cdot\cdot\cdot\cdot\cdot\\ t_{j_{s}}^{s},t_{j_{s}+1}^{s},\ldots,t_{j_{s}+m_{s}}^{s}\end{array}\right].\end{gathered}

For example in the case of $s=2$ This average formula is written

\begin{gathered}{\left[\begin{array}[]{l}x_{1},\ldots,x_{p-1},x_{p+1},\ldots,x_{n+2}\\ y_{1},\ldots,y_{q-1},y_{q+1},\ldots,y_{m+2}\end{array};f(x,y)\right]=}\\ =\frac{1}{\left(x_{n+2}-x_{1}\right)\left(y_{m+2}-y_{1}\right)}\left\{\left(x_{p}-x_{1}\right)\left(y_{q}-y_{1}\right)\left[\begin{array}[]{lll}x_{1},&\ldots,&x_{n+1}\\ y_{1},&\ldots,&y_{m+1}\end{array}\right]+\right.\\ +\left(x_{n+2}-x_{p}\right)\left(y_{q}-y_{1}\right)\left[\begin{array}[]{l}x_{2},\ldots,\\ y_{1},\ldots,x_{m+2}\end{array};f\right]+\\ +\left(x_{p}-x_{1}\right)\left(y_{m+2}-y_{1}\right)\left[\begin{array}[]{l}x_{1},\ldots,x_{n+1}\\ y_{2},\ldots,y_{m+2}\end{array};f\right]+\\ \left.+\left(x_{n+2}-x_{p}\right)\left(y_{m+2}-y_{q}\right)\left[\begin{array}[]{l}x_{2},\ldots,x_{n+2}\\ y_{2},\ldots,y_{m+2}\end{array};f\right]\right\}.\end{gathered}

13.

Taking into account formula (25), it can be immediately extended to $s$ variables, an important average theorem given, in the one-dimensional case, by Prof. T. Popoviciu [5], [6].

Considering the following system of $N=m_{1}m_{2}\ldots m_{s}$ puncture

P_{i_{1}i_{2}\ldots i_{s}}\left(t_{i_{1}}^{1},t_{i_{2}}^{2},\ldots,t_{i_{s}}^{s}\right)\left(i_{k}=\overline{1,m_{k}};k=\overline{1,s)}\right.

(27)

assuming that

t_{1}^{r}<t_{2}^{r}<\ldots<t_{m_{r}}^{r},\quad m_{r}\geqslant n_{r}+2(r=\overline{1,s}),

any partial difference divided by the order $\left(n_{1}+1,n_{2}+1,\ldots,n_{s}\right.$ tandem $\xi^{i}$ is contained in the smallest interval containing the numbers $t^{i}$ , on $N=\prod^{s}\left(n_{k}+1\right)$ points, from those in (27), is an arithmetic mean $\left.\right|_{1}^{i},\ldots,t_{m_{i}+1}^{i}$ .

We note that the numbers $\xi^{1},\xi^{2},\ldots,\xi^{s}$ are the same in all (generalized) derivatives of the following kind:

\begin{gathered}\left.\left\lvert\,\begin{array}[]{l}\overline{t_{i_{1},1}},t_{i_{1},2}^{1},\ldots,t_{i_{1},n_{1}+2}^{1}\\ t_{i_{2},1}^{2},t_{i_{2},2}^{2},\ldots,t_{i_{2},n_{2}+2}^{2};f(M)\\ \ldots\ldots\ldots,\ldots,\ldots,t_{i_{s},n_{s}+2}^{s}\\ t_{i_{s,1}}^{s},t_{i_{s,2}}^{s},\ldots,t_{i}^{s},\ldots,t_{i_{1}+n_{1}+1}^{1}\\ \sum_{j_{1}=1}^{m_{s}}\end{array}\right.\right\}=\\ =\sum_{j_{s}=1}^{m_{1}-n_{1}-1}C_{j_{1}j_{2},\ldots j_{s}}\left[\begin{array}[]{c}t_{j_{1}}^{1},t_{j_{1}+1}^{1},\ldots,t_{s}\\ t_{j_{2}}^{2},t_{j_{2}+1}^{2},\ldots,t_{j_{2}+n_{2}+1}^{2};f(M)\\ \cdot\cdot\ldots.\\ t_{j_{s}}^{2},t_{j_{s}+1}^{s},\ldots,t_{j_{s}+n_{s}+1}^{s}\end{array}\right],\end{gathered}

where the coefficients

C_{j_{1}j_{2}\ldots j_{s}}\geqq 0\quad\left(j_{k}=\overline{1,m_{k}-n_{k}-1};k=\overline{1,s}\right)

are independent of the function $f(M)$ and

\sum_{j_{1}=1}^{m_{1}-n_{1}-1}\cdots\sum_{j_{s}=1}^{m_{s}-n_{s}-1}C_{j_{1}j_{2}\ldots j_{s}}=1,

and

1=i_{p,1}<i_{p,2}<\ldots<i_{p,n_{p}+2}=m_{p}\quad(p=1,2,\ldots,s)

14.

In the particular case of nodes (23) the rest of formula (17) can be expressed using the divided partial differences *).

Based on formula (25) and the average theorems we have, it can be shown **) that this remainder can be expressed, of course assuming that the function $f(M)$ is partially differentiable a sufficient number of times, the form

\begin{gathered}R_{s}(M)=\sum_{\left(m_{1}+1\right)!}\frac{u^{1}\left(l^{1}\right)}{\left(m_{1}\right)}\frac{\partial^{m_{1}+1}f\left(\xi^{1},t^{2},\ldots,l^{s}\right)}{\partial\left(\xi^{1}\right)^{m_{1}+1}}-\\ -\sum\frac{u^{1}\left(l^{1}\right)u^{2}\left(l^{2}\right)}{\left(m_{1}+1\right)!\left(m_{2}+1\right)!}\frac{\partial^{m_{1}+m_{2}+2}f\left(\xi^{1},\xi^{2},l^{3},\ldots,l^{s}\right)}{\partial\left(\xi^{1}\right)^{m_{1}+1}\partial\left(\xi^{2}\right)^{m_{2}+1}}+\\ +\sum\frac{u^{1}\left(l^{1}\right)u^{2}\left(l^{2}\right)u^{3}\left(l^{3}\right)}{\left(m_{1}+1\right)!\left(m_{2}+1\right)!\left(m_{3}+1\right)!}\frac{\partial^{m_{1}+m_{2}+m_{3}+3}f\left(\xi^{1},\xi^{2},\xi^{3},l^{4},\ldots,l^{s}\right)}{\partial\left(\xi^{1}\right)^{m_{1}+1}\partial\left(\xi^{2}\right)^{m_{2}+1}\partial\left(\xi^{3}\right)^{m_{3}+1}}-\\ +(-1)^{s}\prod_{i=1}^{s}\frac{u^{i}\left(l^{i}\right)}{\left(m_{i}+1\right)!}\frac{\partial^{m_{1}+\ldots+m_{s}+s}f\left(\xi^{1},\xi^{2},\ldots,\xi^{s}\right)}{\partial\left(\xi^{1}\right)^{m_{1}+1}\ldots\partial\left(\xi^{s}\right)^{m_{s}+1}},\end{gathered}

⁰⁰footnotetext: *) Vezi J. F. Steffensen [4].
**) Demonstrația am dat-o in lucrarea [1].

If we apply to
each divided partial difference, which occurs in the expression of the remainder, only the average formula (26), as some authors did in the case of $s=2$ , then this result, which seems to be important both theoretically and practically, cannot be obtained.

yl

15.

In this part of the paper we will deal with Gaussian-type numerical interpolation formulas, using the interpolation formulas that are refined on the particular distribution of nodes from (23).

First we will make an observation on Gauss' quadrature formula.
16. Let us consider the Lagrange-Hernite interpolation polynomial of degree $2n-1$ :

	$\displaystyle L(x)=L\left(x_{1},x_{1},x_{2},x_{2},\right.$	$\displaystyle\left.\ldots,x_{.},x_{n};f\mid x\right)=$
		$\displaystyle=\sum_{i=1}^{n}h_{i}(x)f\left(x_{i}\right)+\sum_{=1}^{n}k_{i}(x)f^{\prime}\left(x_{i}\right)$		(29)

nFor

h_{i}(x)=\left(1-\frac{\omega^{\prime\prime}\left(x_{i}\right)}{\omega^{\prime}\left(x_{i}\right)}\left(x-x_{i}\right)\right)l_{i}^{2}(x),k_{i}(x)=\left(x-x_{i}\right)l_{i}^{2}(x)

\omega(x)=\prod_{i=1}^{n}\left(x-x_{i}\right),l_{i}(x)=\frac{\omega(x)}{\left(x-x_{i}\right)\omega^{\prime}\left(x_{i}\right)}.

A. Markov [7] observed that if the nodes are chosen $x_{1},x_{2},\ldots,x_{n}$ find the roots of Legendre's polynomial

P_{n}(x)=C_{n}\frac{d^{n}}{dx^{n}}\left(x^{2}-1\right)^{n}

(30)

tunci in the quadrature formula

\int_{-1}^{+1}f(x)dx\approx\int_{-1}^{+1}L(x)dx=\sum_{i=1}^{n}A_{i}f\left(x_{i}\right)+\sum_{i=1}^{n}B_{i}f^{\prime}\left(x_{i}\right)

(31)

I divide the coefficients. $B_{i}$ and we arrive at Gauss' quadrature formula

\int_{-1}^{+1}f(x)dx=\sum_{i=1}^{n}A_{i}f\left(x_{i}\right)

(32)

labile for any polynomial $f(x)$ of degree at most $2n-1$ .
labile for any polynomial $f(x)$ of degree at most $2n-1$ .

This observation allowed AA Markov to establish But the expression of the remainder of this quadrature formula in the case when $f(x)$ is some function differentiable from $2n$ times.
17. We will show that this situation also occurs in somewhat more general cases.

Let's consider the distinct numbers $x_{i},\lambda_{j}(i=1,n;j=\overline{1,k})$ .Forms ${}^{\text{si }}$ of Lagrange interpolation relative to these nodes and a function that $f(x)$ it is written

f(x)=L\left(x_{1},x_{2},\ldots,x_{n},\lambda_{1},\ldots,\lambda_{k};f\mid x\right)+R(x)

where

\begin{gathered}R(x)=Q_{n}(x)S_{k}(x)\left[x,x_{1},\ldots,x_{n},\lambda_{1},\ldots,\lambda_{k};f\right],\\ Q_{n}(x)=\prod_{i=1}^{n}\left(x-x_{i}\right),S_{k}(x)=\prod_{i=1}^{k}\left(x-\lambda_{i}\right)\end{gathered}

Using the interpolation formula (33) to calculate the integral *)

I_{1}=\int_{-1}^{+1}f(x)dx

the evasion formula is reached

\int_{-1}^{+1}f(x)dx=\sum_{i=1}^{n}A_{i}f\left(x_{i}\right)+\sum_{j=1}^{k}B_{j}f\left(\lambda_{j}\right)+\int_{-1}^{+1}R(x)dx

It is worth noting that if one takes $Q_{n}(x)\equiv P_{n}(x)$ , that is, if the node $x_{1},x_{2},\ldots,x_{n}$ the roots of the Legendre polynomial are chosen $(?$ and $k\leqq n$ , then we also have

B_{j}=0\quad(j=\overline{1,k})

and, it is obtained, whatever $k\leqq n$ , Gauss' numerical integration formula.

Indeed, if in formula (37) we replace $f(x)$ with

\varphi(x)=P_{n}(x)\frac{S_{k}(x)}{\left(x-\lambda_{j}\right)S_{k}^{\prime}\left(\lambda_{j}\right)}

is obtained
$\int_{-1}^{+1}\varphi(x)dx=B_{j}P_{n}\left(\lambda_{1}\right)+\int_{-1}^{+1}P_{n}(x)S_{k}(x)\left[x,x_{1},\ldots,x_{n},\lambda_{1},\ldots,\lambda_{k};\varphi(x)\right]$

⁰⁰footnotetext: *) In cele ce urmează vom folosi intervalul de integrare

[-1,+1]

; formulele ca obțin se pot transcrie imediat pentru un interval oarecare

[a,b]

fåcînd schimbarea de vari

x=\frac{b-a}{2}y+\frac{b+a}{2}

	$\displaystyle\int_{-1}^{+1}\varphi(x)dx=\int_{-1}^{+1}P_{n}(x)\frac{S_{k}(x)}{\left(x-\lambda_{j}\right)S_{k}^{\prime}\left(\lambda_{j}\right)}dx=0$
	$\displaystyle{\left[x,x_{1},\ldots,x_{n}\lambda_{1},\lambda_{2},\ldots,\lambda_{k};\varphi(x)\right]\equiv 0}$		(38)

(because $\varphi(x)$ is a polynomial of degree $n+k-1$ and the divided difference (38) has the order $n+k$ .

because $P_{n}\left(\lambda_{j}\right)\neq 0$ , it follows that

B_{j}=0\quad(j=\overline{1,n})

(and the quadrature formula (37) reduces to

\int_{-1}^{+1}f(x)dx=A_{1}f\left(x_{1}\right)+A_{2}f\left(x_{2}\right)+\ldots+A_{n}f\left(x_{n}\right)+\rho

(39)

(and

\rho=\int_{-1}^{+1}R(x)dx

(40)

To determine the coefficients $A_{1},A_{2},\ldots,A_{n}$ let's do in formula (39)

f(x)=\frac{P_{n}(x)}{\left(x-x_{i}\right)P_{n}^{\prime}\left(x_{i}\right)},\quad(i=\overline{1,n})

(41)

get it immediately

A_{i}=\int_{-1}^{+1}\frac{P_{n}(x)dx}{\left(x-x_{i}\right)P_{n}^{\prime}\left(x_{i}\right)},\quad(i=\overline{1,n})

(42)

18.

Given that whatever the polynomial is $S_{k}(x)$ of the degree $\leqq n$ , we have

S_{k}(x)=CP_{n}(x)+\sum_{\alpha=1}^{n}\frac{P_{n}(x)}{\left(x-x_{\alpha}\right)P_{n}^{\prime}\left(x_{\alpha}\right)}S_{k}\left(x_{\alpha}\right)

(43)

potters

\begin{gathered}L(x)=L\left(x_{1},x_{2},\ldots,x_{n},\lambda_{1},\lambda_{2},\ldots,\lambda_{k};f\mid x\right)=\\ =\sum_{i=1}^{n}\frac{P_{n}(x)S_{k}(x)}{\left(x-x_{i}\right)P_{n}^{\prime}\left(x_{i}\right)S_{k}\left(x_{i}\right)}f\left(x_{i}\right)+\sum_{j=1}^{k}\frac{P_{n}(x)S_{k}(x)}{\left(x-\lambda_{j}\right)P_{n}\left(\lambda_{j}\right)S_{k}^{\prime}\left(\lambda_{j}\right)}f\left(\lambda_{j}\right)\end{gathered}

it can be put in the form

\begin{gathered}L(x)=C\sum_{i=1}^{n}P_{n}(x)\frac{P_{n}(x)}{\left(x-x_{i}\right)P_{n}^{\prime}\left(x_{i}\right)S_{k}\left(x_{i}\right)}f\left(x_{i}\right)+\\ +\sum_{i=1}^{n}\sum_{\alpha=1}^{n}\frac{P_{n}(x)}{\left(x-x_{i}\right)P_{n}^{\prime}\left(x_{i}\right)}\frac{P_{n}(x)}{\left(x-x_{\alpha}\right)P_{n}^{\prime}\left(x_{\alpha}\right)}\frac{S_{k}\left(x_{\alpha}\right)}{S_{k}\left(x_{i}\right)}f\left(x_{i}\right)+\\ \quad+\sum_{j=1}^{n}P_{n}(x)\frac{S_{k}(x)f\left(\lambda_{j}\right)}{\left(x-\lambda_{j}\right)P_{n}\left(\lambda_{j}\right)S_{k}^{\prime}\left(\lambda_{j}\right)}.\end{gathered}

Integrating we find that

\int_{-1}^{+1}L(x)dx=\sum_{i=1}^{n}\int_{-1}^{+1}\left(\frac{P_{n}(x)}{\left(x-x_{i}\right)P_{n}^{\prime}\left(x_{i}\right)}\right)^{2}f\left(x_{i}\right)dx

So the coefficients of the quadrature formula (39) can also be expressed by the formula

A_{i}=\int_{-1}^{+1}\left(\frac{P_{n}(x)}{\left(x-x_{i}\right)P_{n}^{\prime}\left(x_{i}\right)}\right)^{2}dx\quad(i=\overline{1,n})

which shows us that the evadration formula (39) has all positive coefficients. This result is due to Stieltjes [8].

Formula (39) is Gauss's numerical integration formula.
For the coefficients of this formula, the following simple expressions can also be given, as shown by Christoffel [9]:

A_{i}=\frac{\|_{2}}{\left(1-x_{i}^{2}\right)\left[P_{n}^{\prime}\left(x_{i}\right)\right]^{2}}(i=\overline{1,n}).

Apart from this, if we take into account that between the roots of the Legendre polynomial there is the relation $x_{i}=-x_{n-i+1}$ , it immediately follows *) that we have such

A_{i}=A_{n-i+1}(i=\overline{1,n}).

Gauss's quadrature formula has, as is well known, a reputation for accuracy. $2n-1$ .

If it is taken $k=n$ , the remainder (40) becomes **)

\rho=\frac{n!}{[(2n)!]^{2}}\int_{-1}^{+1}S_{n}(x)\left[\left(x^{2}-1\right)^{n}\right]^{n}f^{(2n)}(\xi)dx

Here $\xi$ is contained in the smallest interval containing the value: and $x,x_{1},x_{2},\ldots,x_{n},\lambda_{1},\lambda_{2},\ldots,\lambda_{n}$ .
*) See for example [10].
**) The coefficient of $x^{n}$ FROM $P_{n}(x)$ must be equal to 1, as seen from (35), for the reason in (30) we chose $C_{n}=(n!):(2n)!$
where in [11].
19. Above we assumed that the numbers $\lambda_{1},\lambda_{2},\ldots,\lambda_{k}$ are distinct and different from its roots $P_{n}(x)^{*}$ ).

Let us now assume that the numbers $\lambda_{1},\lambda_{2},\ldots,\lambda_{k}$ , are not generally distinct, namely: $\lambda_{j}$ has the order of multiplicity $r_{j}$ , where $j=\overline{1,s}$ and $r_{1}+r_{2}+\ldots+r_{s}=k\leqq n$ .

Let us write the expression of the interpolation polynomial of LagrangeHermite **), on the nodes represented by the roots of the polynomial

\omega(x)=\prod_{i=1}^{n}\prod_{j=1}^{s}\left(x-x_{i}\right)\left(x-\lambda_{j}\right)^{r_{j}}.

(48)

	$\displaystyle L_{H}(x)=\sum_{i=1}^{n}\frac{\omega(x)}{\left(x-x_{i}\right)\omega^{\prime}\left(x_{i}\right)}f\left(x_{i}\right)+$
	$\displaystyle+\sum_{i=1}^{s}\sum_{j=0}^{r_{i}-1}\sum_{k=0}^{r_{i}-j-1}\frac{\omega(x)}{\left(x-\lambda_{i}\right)^{r_{i}}}\left\{\frac{\left(x-\lambda_{i}\right)^{j}}{j!}\left[\frac{\left(x-\lambda_{i}\right)^{k}}{k!}\left(\frac{\left(x-\lambda_{i}\right)^{r_{i}}}{\omega(x)}\right)_{\lambda_{i}}^{(k)}\right]f^{(j)}\left(\lambda_{i}\right)\right\}.$		(49)

In the interpolation formula

f(x)=L_{H}(x)+R(x)

(

\prime

)

the rest has the expression

R(x)=\omega(x)[x,x_{1},\ldots,x_{n},\underbrace{\lambda_{1},\ldots,\lambda_{1}}_{r_{1}},\ldots,\underbrace{\lambda_{s},\ldots,\lambda_{s}}_{r_{s}};f].

(50)

Using formula (49) to calculate integral (36), we arrive at an escapement formula of the form

\int_{-1}^{+1}f(x)dx=\sum_{i=1}^{n}A_{i}f\left(x_{i}\right)+\sum_{i=1}^{s}\sum_{j=0}^{r_{i}-1}B_{ij}f^{(j)}\left(\lambda_{i}\right)+\rho

(51)

\rho=\int_{-1}^{+1}R(x)dx

If $x_{1},x_{2},\ldots,x_{n}$ are the roots of the Legendre polynomial $P_{n}(x)$ , then, taking into account (49), it is observed that

A_{i}=\int_{-1}^{+1}\frac{\omega(x)dx}{\left(x-x_{i}\right)\omega^{\prime}\left(x_{i}\right)},\quad(i=\overline{1,n})

(52)

B_{ij}=0\left(j=\overline{1,r_{i}-1};i=\overline{1,s}\right),

*) If $S_{n}(x)\equiv P_{n}(x)$ one comes across the case of AA Markov [7].
**) The explicit expression of the Lagrange-Hermite polynomial can be seen for example
whatever the roots of the polynomial of degree $k\leqq n$

S_{k}(x)=\prod_{j=1}^{s}\left(x-\lambda_{j}\right)^{r_{j}}.

(53)

Taking into account the introduced notations, the polynomial (48) is written as

\omega(x)=P_{n}(x)S_{k}^{-}(x).

Given that whatever the polynomial is $S_{k}(x)$ , of the degree $k\leqq$ n it can be put in the form (43), we find, as in point 18, expressions (44) for the coefficients $A_{i}$ of the quadrature formula (51).

In the end, we find that even in the case we dealt with in that paragraph, we arrive at Gauss' quadrature formula *)

\int_{-1}^{+1}f(x)dx=\sum_{i=1}^{n}A_{i}f\left(x_{i}\right)+\rho

()

TAKING $k=n$ , the rest of this formula will be

	$\displaystyle\rho=\int_{-1}^{+1}P_{n}(x)S_{n}(x)[x,x_{1},\ldots,x_{n},\underbrace{\lambda_{1},\ldots,\lambda_{1}}_{r_{1}},\ldots,\underbrace{\lambda_{s},\ldots,\lambda_{s}}_{r_{s}};f]dx=$
	$\displaystyle=\frac{1}{(2n)!}\int_{-1}^{+1}P_{n}(x)S_{n}(x)f^{(2n)}(\xi)dx=$
	$\displaystyle=\frac{n!}{[(2n)]^{2}}\int_{-1}^{+1}S_{n}(x)\left[\left(x^{2}-1\right)^{n}\right]^{(n)}f^{(2n)}(\xi)dx.$		(55)

20.

The previous considerations can also be made on multiple integrals. Let us deal, for the sake of ease of exposition, with the case of double integrals.

We propose to establish a cubature formula with a minimum number of terms for the double integral

I_{2}=\iint_{D}f(x,y)dxdy

(56)

where $D$ is the square

-1\leqq x\leqq 1,\quad-1\leqq y\leqq 1

Let's consider the interpolation formula

f(x,y)=L(x,y)+R(x,y),

(5i)

where

L(x,y)=L\left(\begin{array}[]{l}x_{1},\ldots,x_{n},\alpha_{1},\ldots,\alpha_{r}\\ y_{1},\ldots,y_{m},\beta_{1},\ldots,\beta_{s}\end{array};f\binom{x}{y},\right.

⁰⁰footnotetext: *) Intr-o nouă lucrare [12] am dat o importantă generalizare acestei formule clasice lui Gauss.

ember
in which

	$\displaystyle u(x)=\prod_{i=1}^{n}\prod_{j=0}^{r}\left(x-x_{j}\right)\left(x-\alpha_{i}\right)$		(60)
	$\displaystyle v(y)=\prod_{i=1}^{m}\prod_{j=1}^{s}\left(y-y_{i}\right)\left(y-\beta_{i}\right)$

Using the interpolation formula (57) to calculate the integral (56), a cubature formula of the form is obtained

	$\displaystyle\iint_{D}f(x,y)dxdy=\sum_{i=1}^{n}\sum_{k=1}^{m}A_{ik}f\left(x_{i},y_{k}\right)+\sum_{i=1}^{n}\sum_{j=1}^{s}B_{ij}f\left(x_{i},\beta_{j}\right)+$
	$\displaystyle+\quad\sum_{l=1}^{r}\sum_{k=1}^{m}C_{lk}f\left(\alpha_{l},y_{k}\right)+\sum_{l=1}^{r}\sum_{j=1}^{s}D_{lj}f\left(\alpha_{l},\beta_{j}\right)+\rho$		(61)

where

\rho=\iint_{D}R(x,y)dxdy

(62)

If the numbers are chosen

x_{1},x_{2},\ldots,x_{n};y_{1},y_{2},\ldots,y_{\ldots}

so that they are respectively the roots of Legendre's polynomials $P_{n}(x)$ , $P_{n}(y)$ and we note

E_{r}(x)=\prod_{i=1}^{r}\left(x-\alpha_{i}\right),\quad F_{\mathrm{s}}(y)=\coprod_{k=1}^{s}\left(y-\beta_{k}\right),

(63)

it is found that

	$\displaystyle B_{ij}=0,C_{lk}=0,D_{lj}=0$		(64)
	$\displaystyle(i=\overline{1,n};k=\overline{1,m};l=\overline{1,r};j=\overline{1,s})$

The remaining coefficients have the expressions

A_{ik}=\frac{1}{P_{n}^{\prime}\left(x_{i}\right)P_{m}^{\prime}\left(y_{k}\right)E_{r}\left(x_{i}\right)F_{s}\left(y_{k}\right)}\iint_{D}\frac{P_{n}(x)E_{r}(x)P_{m}(y)F_{s}(y)}{\left(x-x_{i}\right)\left(y-y_{k}\right)}dxdy

Since whatever the polynomials (63), with $r\leqq n,s\leqq m$ , we have

\begin{gathered}E_{r}(x)=C_{1}P_{n}(x)+\sum_{\nu=1}^{n}\frac{P_{n}(x)E_{r}\left(x_{\nu}\right)}{\left(x-x_{\nu}\right)P_{n}^{\prime}\left(x_{\nu}\right)}\\ F_{s}(y)=C_{2}P_{m}(y)+\sum_{\mu=1}^{m}\frac{P_{m}(y)F_{s}\left(y_{\mu}\right)}{\left(y-y_{\mu}\right)P_{m}^{\prime}\left(y_{\mu}\right)}\end{gathered}

where $C_{1}\neq 0,C_{2}\neq 0$ if $r=n$ and $s=m$ , and $C_{1}=C_{2}=0$ if $r<n$ and $s<m$ , the expressions for the coefficients (64) are found

A_{ik}=\iint_{D}\left(\frac{P_{n}(x)}{\left(x-x_{i}\right)P_{n}^{\prime}\left(x_{i}\right)}\right)^{2}\left(\frac{P_{m}(y)}{\left(y-y_{k}\right)P_{m}^{\prime}\left(y_{k}\right)}\right)^{2}dxdy

(65)

Taking into account (64), the cubature formula (61) reduces to

\iint_{D}f(x,y)dxdy=\sum_{i=1}^{n}\sum_{k=1}^{m}A_{ik}f\left(x_{i},y_{k}^{*}\right)+\rho

(66)

If we take

f(x,y)=\frac{P_{n}(x)P_{m}(y)}{\left(x-x_{i}\right)\left(y-y_{k}\right)}

the following expressions are also found for the coefficients of this cubature formula

A_{ik}=\iint_{D}\frac{P_{n}(x)}{\left(x-x_{i}\right)P_{n}^{\prime}\left(x_{i}\right)}\frac{P_{m}(y)}{\left(y-y_{k}\right)P_{m}^{\prime}\left(y_{k}\right)}dxdy

((67)

It is observed that

A_{ik}=C_{i}D_{k}

where

C_{i}=\int_{-1}^{+1}\frac{P_{n}(x)}{\left(x-x_{i}\right)P_{n}^{\prime}\left(x_{i}\right)}dx,\quad D_{k}=\int_{-1}^{+1}\frac{P_{m}(y)}{\left(y-y_{k}\right)P_{m}^{\prime}\left(y_{k}\right)}dy

Based on Christoffel's result, reported in point 18, we find e

A_{ik}=\frac{4}{\left(1-x_{i}^{2}\right)\left(1-y_{k}^{2}\right)\left[P_{n}^{\prime}\left(x_{i}\right)P_{m}^{\prime}\left(y_{k}\right)\right]^{2}}

(168)

We also have the relationships

A_{ik}=A_{n-i+1},m-i+1(i=\overline{1,n};k=\overline{1,m})

21.

By a procedure similar to that used in the case of one variable, it is shown that the previous results are preserved even when the roots of the polynomials (63) are not distinct.
22.

Let us now look for the expression of the remainder of the Gauss-type cubature formula (66), which has the degree of accuracy $(2n-1,2m-1)$ .

making $r^{\prime}=n$ , and $s=m$ , formulas (59) and (62) lead us to the following expression of the remainder

\rho=\iint_{D}^{c}R(x,y)dxdy

(69)

where

		$\displaystyle R(x,y)=P_{n}(x)E_{n}(x)\left[x,x_{1},\ldots,x_{n},\alpha_{1},\ldots,\alpha_{n};f\right]+$
		$\displaystyle\quad+P_{m}(y)F_{m}(y)\left[y,y_{1},\ldots,y_{m},\beta_{1},\ldots,\beta_{m};f\right]-$
		$\displaystyle-P_{n}(x)E_{n}(x)P_{m}(y)F_{m}(y)\left[\begin{array}[]{l}x,x_{1},\ldots,x_{n},\alpha_{1},\ldots,\alpha_{n}\\ y,y_{1},\ldots,y_{m},\beta_{1},\ldots,\beta_{m}\end{array};f\right].$

The case treated by AA Markov for one variable corresponds to the case for two variables

E_{n}(x)\equiv P_{n}(x),F_{m}(y)\equiv P_{m}(y)

(

\prime

)

when the previous remainder will become

		$\displaystyle R(x,y)=P_{n}^{2}(x)\left[x,x_{1},x_{1},\ldots,x_{n},x_{n};f\right]+$
		$\displaystyle\quad+P_{m}^{2}(y)\left[y,y_{1},y_{1},\ldots,y_{m},y_{m};f\right]-$
		$\displaystyle-P_{n}^{2}(x)P_{m}^{2}(y)\left[\begin{array}[]{lll}x,x_{1},x_{1},\ldots,x_{n},x_{n}\\ y,y_{1}y_{1},\ldots,y_{m},y_{m}\end{array};f\right]$

We want to find an evaluation of the remainder (69) in this case.
Taking into account formula (25), the additivity property of divided differences and the average formula (26), we can write successively,

\begin{gathered}\iint_{D}\left\{P_{n}^{2}(x)\left[x,x_{1},x_{1},\ldots,x_{n},x_{n};f\right]-P_{n}^{2}(x)P_{m}^{2}(y)\left[\begin{array}[]{l}x,x_{1},x_{1},\ldots,x_{n},x_{n}\\ y,y_{1},y_{1},\ldots,y_{m},y_{m}\end{array}\right]\right]_{n}^{+1}dxdy\\ =\int_{-1}^{+1}\left\{\int_{-1}^{+1}P_{n}^{2}(x)\left[x,x_{1},x_{1},\ldots,x_{n},x_{n};f(x,y)-P_{m}^{2}(y)\left[y,y_{1},y_{1},\ldots,y_{m},y_{m};\right.\right.\right.\\ f(x,y)]]dx\}dy=\\ =\int_{-1}^{+1}\left\{\left[\xi^{\prime},x_{1},x_{1},\ldots,x_{n},x_{n};f\left(\xi^{\prime},y\right)-P_{m}^{2}(y)\left[y,y_{1},y_{1},\ldots,y_{m},y_{m};\right.\right.\right.\\ \left.\left.\left.f\left(\xi^{\prime},y\right)\right]\right]\int_{-1}^{+1}P_{n}^{2}(x)dx\right\}dy=\\ =\int_{-1}^{+1}A_{n}\left\{\frac{1}{(2n)!}\frac{\partial^{2n}f(\xi,y)}{\partial\xi^{2n}}-\frac{P_{m}^{2}(y)}{(2n)!}\left[y,y_{1},y_{1},\ldots,y_{m},y_{m};\frac{\partial^{2n}f(\xi,y)}{\partial\xi^{2n}}\right]\right\}dy=\\ =\frac{2A_{n}}{(2n)!}\frac{\partial^{2n}f\left(\xi,\eta_{1}\right)}{\partial\xi^{2n}}-\frac{A_{n}}{(2n)!}\int_{-1}^{+1}P_{m}^{2}(y)\left[y,y_{1},y_{1},\ldots,y_{n},y_{m};\frac{\partial^{2n}f(\xi,y)}{\partial\xi^{2n}}\right]dy\end{gathered}

where

A_{n}=\int_{-1}^{+1}P_{n}^{2}(x)dx=\frac{2^{2n+1}(n!)^{2}}{(2n)!(2n+1)!}

And $\eta_{1}$ and $\xi$ are respectively included in the interval $[-1,+1]$ and the mth interval containing the values $x,x_{1},\ldots,x_{n}$ .

With these the rest becomes

\begin{gathered}\rho=\frac{2A_{n}}{(2n)!}\frac{\partial^{2n}f\left(\xi,\eta_{1}\right)}{\partial\xi^{2n}}+\int_{-1}^{1}P_{m}^{2}(y)\left[y,y_{1},y_{1},\ldots,y_{m},y_{m};\int_{-1}^{1}f(x,y)dx-\right.\\ \left.-\frac{A_{n}}{(2n)!}\frac{\partial^{2n}f(\xi,y)}{\partial\xi^{2n}}\right]dy=\frac{2A_{n}}{(2n)!}\frac{\partial^{2n}f\left(\xi,\eta_{1}\right)}{\partial\xi^{2n}}+A_{m}\left[\eta^{\prime},y_{1},y_{1},\ldots,y_{m},y_{m};\right.\\ \left.\int_{-1}^{1}f\left(x,\eta^{\prime}\right)dx-\frac{A_{n}}{(2n)!}\frac{\partial^{2n}f\left(\xi,\eta^{\prime}\right)}{\partial\xi^{2n}}\right]=\frac{2A_{n}}{(2n)!}\frac{\partial^{2n}f\left(\xi,\eta_{1}\right)}{\partial\xi^{2n}}+\frac{2A_{m}}{(2m)!}\frac{\partial^{2m}f\left(\xi_{1},\eta\right)}{\partial\eta^{2m}}-\\ -\frac{A_{n}A_{m}}{(2n)!(2m)!}\frac{\partial^{2n+2m}f(\xi,\eta)}{\partial\xi^{2n}\partial\eta^{2m}}.\end{gathered}

In this way we arrive at the following expression for the rest of the cubature formula (66)

\begin{gathered}\rho=\frac{2^{2n+2}}{2n+1}\frac{n!}{[(n+1)(n+2)\ldots 2n]^{3}}\frac{\partial^{2n}f\left(\xi,\eta_{1}\right)}{\partial\xi^{2n}}+\\ +\frac{2^{2m+2}}{2m+1}\frac{m!}{[(m+1)(m+2)\ldots 2m]^{3}}\frac{\partial^{2m}f\left(\xi_{1},\mu\right)}{\partial\eta^{2m}}-\\ -\frac{2^{2n+2m+2}}{(2n+1)(2m+1)}\frac{n!m!}{[(n+1)(n+2)\ldots 2n]^{3}[(m+1)(m+2)\ldots 2m]^{3}}\frac{\partial^{2n+2m}f(\xi)}{\partial\xi^{2n}\partial\eta^{2}}\end{gathered}

Let us now consider some particular cases of the formula (6 $1^{\circ}.n=m=1$ .

\begin{gathered}\iint_{D}f(x,y)dxdy=4f(0,0)+\rho_{1}\\ \rho_{1}=\frac{2}{3}\frac{\partial^{2}f\left(\xi,\eta_{1}\right)}{\partial\xi^{2}}+\frac{2}{3}\frac{\partial^{2}f\left(\xi_{1},\eta\right)}{\partial\eta^{2}}-\frac{1}{9}\frac{\partial^{4}f(\xi,\eta)}{\partial\xi^{2}\partial\eta^{2}}.\end{gathered}

$2^{\circ}.n=m=2$ .
$3^{\circ}.m=n=3$ .

	$\displaystyle\iint_{D}f(x,y)dxdy=\frac{1}{81}\left\{25\left[f\left(-\sqrt{\frac{3}{5}},\sqrt{\frac{3}{5}}\right)+f\left(-\sqrt{\frac{3}{5}},\sqrt{\frac{3}{5}}\right)+\right.\right.$
	$\displaystyle\left.\quad+f\left(\sqrt{\frac{3}{5}},-\sqrt{\frac{3}{5}}\right)+f\left(\sqrt{\frac{3}{5}},\sqrt{\frac{3}{5}}\right)\right]+40\left[f\left(-\sqrt{\frac{3}{5}},0\right)+\right.$
	$\displaystyle\left.\left.+f\left(0,-\sqrt{\frac{3}{5}}\right)+f\left(0,\sqrt{\frac{3}{5}}\right)+f\left(\sqrt{\frac{3}{5}},0\right)\right]+64f(0,0)\right\}+\rho_{3}$		(73)
	$\displaystyle\rho_{3}=\frac{1}{7875}\frac{\partial^{6}f\left(\xi,\eta_{1}\right)}{\partial\xi^{6}}+\frac{1}{7875}\frac{\partial^{6}f\left(\xi_{1},\eta\right)}{\partial\eta^{6}}-\frac{1}{248062500}\frac{\partial^{12}f(\xi,\eta)}{\partial\xi^{6}\partial\eta^{6}}$

23.

If the polynomials $E_{n}(x)$ and $F_{m}(y)$ are not chosen as in determining the remainder. (70), determining the remainder is quite complicated.

Let's give an example.
If one chooses $m=n=2$ we have seen that the formula for cubature (72) is obtained. To evaluate its remainder, let us choose

E_{2}(x)=x^{2}-1,F_{2}(y)=y^{2}-1.

(i) The remainder according to formula (69) and the following will be
where

\rho=\iint_{D}\int_{D}r(x,y)dxdy

\begin{gathered}r(x,y)=\left(x^{2}-1\right)\left(x^{2}-\frac{1}{3}\right)\left[x,-1,-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},1;f\right]+\\ +\left(y^{2}-1\right)\left(y^{2}-\frac{1}{3}\right)\left[y,-1,-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},1;f\right]-\\ -\left(x^{2}-1\right)\left(x^{2}-\frac{1}{3}\right)\left(y^{2}-1\right)\left(y^{2}-\frac{1}{3}\right)\left[\begin{array}[]{l}x,-1,-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},1\\ y,-1,-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},1\end{array}\right].\end{gathered}

*) This formula was also encountered incidentally by Mikeladze

Now some difficulties arise because the polynomials that multiply, the divided differences above do not keep a constant sign in the domain of integration.

By making a convenient decomposition of the integration domain, one can obtain, after some transformations, the following expression for the remainder

$\displaystyle\rho$	$\displaystyle=\frac{14\sqrt{3}}{1215}\frac{\partial^{4}f\left(\xi_{1},\eta\right)}{\partial\xi_{1}^{4}}+\frac{2(9-7\sqrt{3})}{1215}\frac{\partial^{4}f\left(\xi_{2},\eta\right)}{\partial\xi_{2}^{2}}+$
$\displaystyle+$	$\displaystyle\frac{14\sqrt{3}}{1215}\frac{\partial^{4}f\left(\xi,\eta_{1}\right)}{\partial\eta_{1}^{4}}+\frac{2(9-7\sqrt{3})}{1215}\frac{\partial^{4}f\left(\xi,\eta_{2}\right)}{\partial\eta_{2}^{4}}-$
	$\displaystyle-\frac{49}{492075}\frac{\partial^{8}f\left(\xi_{1},\eta_{1}\right)}{\partial\xi_{1}^{4}\partial\eta_{1}^{4}}-\frac{14(3\sqrt{3}-7)}{492075}\frac{\partial^{8}f\left(\xi_{3},\eta_{3}\right)}{\partial\xi_{3}^{4}\partial\eta_{3}^{4}}-$
	$\displaystyle-\frac{2(38-21\sqrt{3})}{492075}\frac{\partial^{8}f\left(\xi_{2},\eta_{2}\right)}{\partial\xi_{2}^{4}\partial\eta_{2}^{4}}$	(7.)

Assuming that in square D we have

\left|\frac{\partial^{4}f(x,y)}{\partial x^{4}}\right|\leqq L,\left|\frac{\partial^{4}f(x,y)}{\partial y^{4}}\right|\leqq M,\left|\frac{\partial^{8}f(x,y)}{\partial x^{4}\partial y^{4}}\right|\leqq N

the following delimitation of this remainder results

|\rho|\leqq\frac{2}{135}(L+M)+\frac{N}{18225}

It is worth noting that the same delimitation is obtained if we start from the expression (72) of the remainder. In fact, the remainder (70) must also be independent of the parameters $\alpha_{i},\beta_{k}$ . In the previous example the correct expression for the remainder is that of (72). The formula (72') differs only in appearance from that of (7.
24. The preceding results can now be extended very easily to the case of several variables.

Let us briefly present some results from the case of three variables.
Let the triple integral be

I_{3}=\iiint_{V}f(x,y,z)dxdydz

where $V$ is the cube

-1\leqq x,y,z\leqq+1

Using the interpolation formula

f(x,y,z)=L(x,y,z)+R(x,y,z)

where

L(x,y,z)=L\left(\left.\begin{array}[]{l}x_{1},\ldots,x_{n},\alpha_{1},\ldots,\alpha_{r}\\ y_{1},\ldots,y_{m},\beta_{1},\ldots,\beta_{s};f\\ z_{1},\ldots,z_{p},\gamma_{1},\ldots,\gamma_{1}\end{array}\right\rvert\,\begin{array}[]{l}x\\ y\\ z\end{array}\right)

is the interpolation polynomial that coincides with the function $f(x,y,z)$ on the hot nodes

\begin{array}[]{cl}M_{ijk}\left(x_{i},y_{j},z_{k}\right)&(i=\overline{1,n};j=\overline{1,m};k=\overline{1,p})\\ P_{\mu\nu\lambda}\left(\alpha_{\mu},\beta_{\nu},\gamma_{\lambda}\right)&(\mu=\overline{1,r};\nu=\overline{1,s};\lambda=\overline{1,t})\end{array}

	$\displaystyle R(x,y,z)$	$\displaystyle=u(x)\left[x,x_{1},\ldots,x_{n},\alpha_{1},\ldots,\alpha_{r};f\right]+$
		$\displaystyle+v(y)\left[y,y_{1},\ldots,y_{m},\beta_{1},\ldots,\beta_{s};f\right]+$
		$\displaystyle+w(z)\left[z,z_{1},\ldots,z_{p},\gamma_{1},\ldots,\gamma_{t};f\right]-$

		$\displaystyle-u(x)v(y)\left[\begin{array}[]{l}x,x_{1},\ldots,x_{n},\alpha_{1},\ldots,\alpha_{r}\\ y,y_{1},\ldots,y_{m},\beta_{1},\ldots,\beta_{s}\end{array};f-\right.$
		$\displaystyle-u(x)w(z)\left[\begin{array}[]{l}x,x_{1},\ldots,x_{n},\alpha_{1},\ldots,\alpha_{r}\\ z,z_{1},\ldots,z_{p},\gamma_{1},\ldots,\gamma_{t}\end{array}\right]-$
		$\displaystyle-v(y)w(z)\left[\begin{array}[]{l}y,y_{1},\ldots,y_{m},\beta_{1},\ldots,\beta_{s}\\ z,z_{1},\ldots,z_{p},\gamma_{1},\ldots,\gamma_{t}\end{array};\right]+$
		$\displaystyle+u(x)v(y)w(z)\left[\begin{array}[]{l}x,x_{1},\ldots,x_{n},\alpha_{1},\ldots,\alpha_{r}\\ y,y_{1},\ldots,y_{m},\beta_{1},\ldots,\beta_{s};f\\ z,z_{1},\ldots,z_{p},\gamma_{1},\ldots,\gamma_{t}\end{array}\right]$

in which

		$\displaystyle u(x)=P_{n}(x)E_{r}(x),E_{r}(x)=\prod_{i=1}^{r}\left(x-\alpha_{i}\right)$
		$\displaystyle v(y)=P_{m}(y)F_{s}(y),F_{s}(y)=\prod_{j=1}^{s}\left(y-\beta_{j}\right)$
		$\displaystyle w(z)=P_{p}(z)G_{l}(z),G_{t}(z)=\prod_{k=1}^{t}\left(z-\gamma_{k}\right)$
		$\displaystyle P_{q}(u)=C_{q}\frac{d^{q}}{du^{q}}\left(u^{2}-1\right)^{q}$

is obtained, regardless of whether $\alpha_{i},\beta_{j},\gamma_{k}(i=1,2,\ldots,r\leqq n;j=1,2,\ldots..,s\leqq m;k=1,2,\ldots,t\leqq p)$ are distinct or not, the cubature formula

\iiint_{V}f(x,y,z)dxdydz=\sum_{i=1}^{n}\sum_{j=1}^{m}\sum_{k=1}^{p}A_{ij:}f\left(x_{i},y_{j},z_{k}\right)+\rho,

(74)

where

\rho=\iiint_{V}R(x,y,z)dxdydz

and

	$\displaystyle A_{ijk}$	$\displaystyle=\iiint_{V}\frac{P_{n}(x)}{\left(x-x_{i}\right)P_{n}^{\prime}\left(x_{i}\right)}\frac{P_{m}(y)}{\left(y-y_{j}\right)P_{m}^{\prime}\left(y_{j}\right)}\frac{P_{p}(z)}{\left(z-z_{k}\right)P_{p}^{\prime}\left(z_{k}\right)}dxdydz=$
		$\displaystyle\left.=\iiint\int_{V}\frac{P_{n}(x)}{\left(x-x_{i}\right)P_{n}^{\prime}\left(x_{i}\right)}\frac{P_{m}(y)}{\left(y-y_{j}\right)P_{m}^{\prime}\left(y_{j}\right)}\frac{P_{p}(z)}{\left(z-z_{k}\right)P_{p}^{\prime}\left(z_{k}\right)}\right)^{2}dxdydz=$
		$\displaystyle=\frac{8}{\left(1-x_{i}^{2}\right)\left(1-y_{j}^{2}\right)\left(1-z_{k}^{2}\right)\left[P_{n}^{\prime}\left(x_{i}\right)P_{m}^{\prime}\left(y_{j}\right)P_{p}^{\prime}\left(z_{k}\right)\right]^{2}}.$

25.

Let us consider two particular cases of the cubature formula (74). $1^{\circ}.n=m=p=1$ .

		$\displaystyle\iiint_{V}f(x,y,z)dxdydz=8f(0,0,0)+\rho$
	$\displaystyle\rho=$	$\displaystyle\frac{4}{3}\left[\frac{\partial^{2}f\left(\xi,\eta^{\prime},\zeta^{\prime}\right)}{\partial\xi^{2}}+\frac{\partial^{2}f\left(\xi^{\prime},\eta,\zeta^{\prime}\right)}{\partial\eta^{2}}+\frac{\partial^{2}f\left(\xi^{\prime},\eta^{\prime},\zeta\right)}{\partial\zeta^{2}}\right]-$
	$\displaystyle-$	$\displaystyle\frac{1}{9}\left[\frac{\partial^{4}f\left(\xi,\eta^{\prime},\zeta^{\prime}\right)}{\partial\xi^{2}\partial\eta^{2}}+\frac{\partial^{4}f\left(\xi,\eta^{\prime},\zeta\right)}{\partial\xi^{2}\partial\zeta^{2}}+\frac{\partial^{4}f\left(\zeta^{\prime},\eta,\zeta\right)}{\partial\eta^{2}\partial\zeta^{2}}\right]+$
	$\displaystyle+$	$\displaystyle\frac{1}{108}\frac{\partial^{6}f(\xi,\eta,\zeta)}{\partial\xi^{2}\partial\eta^{2}\partial\zeta^{2}}$

In these expressions, the Legendre polynomial $P_{q}(u)$ contains numerical factor $C_{q}=\frac{1}{2^{q}\cdot q!}$ .

Between these coefficients there are the relationships

A_{i,j,k}=A_{n-i+1,m-j+1,p-k+1}(i=\overline{1,n};j=\overline{1,m};k=\overline{1,p}).

If it is assumed that

r=n,s=m,t=p

and

E_{n}(x)\equiv P_{n}(x),F_{m}(y)\equiv P_{m}(y),G_{p}(z)\equiv P_{p}(z),

for the rest of the cubature formula (74) the following expression is obtained

\begin{gathered}\rho=\frac{2^{2n+3}}{2n+1}\frac{n!}{[(n+1)(n+2)\ldots 2n]^{3}}\frac{\partial^{2n}f\left(\xi,\eta^{\prime},\zeta^{\prime}\right)}{\partial\xi^{2n}}+\\ +\frac{2^{2m+3}}{2m+1}\frac{m!}{[(m+1)(m+2)\ldots 2m]^{3}}\frac{\partial^{2m}f\left(\xi^{\prime},\eta,\zeta^{\prime}\right)}{\partial\eta^{2m}}+\\ +\frac{2^{2p+3}}{2p+3}\frac{p!}{[(p+1)(p+2)\ldots 2p]^{3}}\frac{\partial^{2p}f\left(\xi^{\prime},\eta^{\prime},\zeta\right)}{\partial\zeta^{2p}}-\\ -\frac{2^{2n+2m+2}}{(2n+1)(2m+1)}\frac{n!m!}{[(n+1)(n+2)\ldots 2n]^{3}[(m+1)(m+2)\ldots 2m]^{3}}\frac{\partial^{2n+2m}f\left(\xi,\eta,\zeta^{\prime}\right)}{\partial\xi^{2n}\partial\eta^{2m}}\\ -\frac{2^{2n+2p+2}}{(2n+1)(2p+1)}\frac{n!p!}{[(n+1)(n+2)\ldots 2n]^{3}[(p+1)(p+2)\ldots 2p]^{3}}\frac{\partial^{2n+2p}f\left(\xi,\eta^{\prime},\right.}{\partial\xi^{2n}\partial\zeta^{2p}}\\ -\frac{2^{2m+2p+2}}{(2m+1)(2p+1)}\frac{m!p!}{[(m+1)(m+2)\ldots 2m]^{3}[(p+1)(p+2)\ldots 2p]^{3}}\frac{\partial^{2m+2p}f\left(\xi^{\prime},\eta,\eta^{\prime}\right.}{\partial\eta^{2m}\partial\zeta^{2p}}\\ \left.+\frac{2^{2n+2m+2p+1}}{(2n+1)(2m+1)(2p+1)}\frac{n!m!p!}{[(n+1)\ldots 2n]^{3}[(m+1)\ldots 2m]^{3}[(p+1)\ldots 2p]^{3}}\frac{\partial^{2n+2m+2p}(\xi,n}{\partial\xi^{2n}\partial\eta^{2m}\partial\zeta^{\prime 2}}\right)\end{gathered}

$2^{\circ}.n=m=p=2$ .

\begin{gathered}\iiint_{V}f(x,y,z)dxdydz=f\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)+f\left(\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)+\\ +f\left(\frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)+f\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}}\right)+f\left(\frac{-1}{\sqrt{3}},\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)+f\left(\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}}\right)+\\ +f\left(\frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}},\frac{-1}{\sqrt{3}}\right)+f\left(\frac{-1}{\sqrt{3}},\frac{-1}{\sqrt{3}},\frac{-1}{\sqrt{3}}\right)+\rho\\ \rho=\frac{4}{135}\left[\frac{\partial^{4}f\left(\xi,\eta^{\prime},\zeta^{\prime}\right)}{\partial\xi^{4}}+\frac{\partial^{4}f\left(\xi^{\prime},\eta,\zeta^{\prime}\right)}{\partial\eta^{4}}+\frac{\partial^{4}f\left(\xi^{\prime},\eta^{\prime},\zeta\right)}{\partial\zeta^{4}}\right]-\\ \frac{1}{18225}\left[\frac{\partial^{8}f\left(\xi,\eta,\zeta^{\prime}\right)}{\partial\xi^{4}\partial\eta^{4}}+\frac{\partial^{8}f\left(\xi,\eta^{\prime},\zeta\right)}{\partial\xi^{4}\partial\zeta^{4}}+\frac{\partial^{8}f\left(\xi^{\prime},\eta,\zeta\right)}{\partial\eta^{4}\partial\zeta^{4}}\right]+\frac{1}{9841500}\frac{\partial^{12}f(\xi,\eta,\zeta)}{\partial\xi^{4}\partial\eta^{4}\partial\zeta^{4}}.\end{gathered}

26.

In a future paper, we will construct, starting from the interpolation formulas we gave in this paper, cubature formulas for double and triple integrals in the case when the integration domain is a regular polygon, a circle, a regular polyhedron or a sphere.

Generalization of some interpolation formulas for functions of several variables and some considerations on the Gauss numerical integration formula

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GENERALIZATION OF SOME INTERPOLATION FORMULAS FOR FUNCTIONS OF MULTIPLE VARIABLES AND SOME CONSIDERATIONS ON GAUSS'S NUMERICAL INTEGRATION FORMULA

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