The simplicity of the remainder in certain quadrature formulas

THE SIMPLICITY OF THE REMAINDER IN CERTAIN QUADRATURE FORMULAS

by
TIBERIU POPOVICIU
in Cluj
§ 1.

1.

Let us consider the quadrature formula

\int_{-1}^{1}f(x)dx=\sum_{i=1}^{p}\sum_{j=0}^{k_{i}-1}c_{i,j}f^{(j)}\left(z_{i}\right)+R[f]

(1)

Or $z_{1},z_{2},\ldots,z_{p}$ are $p(\geqq 1)$ distinct points on the real axis; these are the nodes of the formula. $k_{1},k_{2},\ldots,k_{p}$ are $p$ natural numbers and $c_{i,j}$ independent coefficients of the function $f$ .

The function $f$ is defined, continuous and has a continuous derivative of order equal to $\max\left(k_{1}-1,k_{2}-1,\ldots,k_{p}-1\right)$ over an interval $E$ containing the points $-1.1$ and the knots $z_{i},i=1,2,\ldots,p$ The derivative of order 0 is the function itself.

We will refer to $\mathscr{F}$ all of these functions. $\mathscr{F}$ is a linear set and contains, in particular, all polynomials.

The accents in the second part of (1) denote successive derivations.

In what follows, unless expressly stated otherwise, we will assume that $E$ reduces to the smallest closed interval containing the nodes and points $-1.1$ .

The integral of the left-hand side of (1) could be taken between any two finite limits, but we do not restrict the generality by taking these limits to be equal to -1 and 1 respectively. The passage through the interval $[-1,1]$ at any finite integration interval $[A,B]$ is done using the
linear transformation formula $x = 2y - AB / BA$ Such a transformation preserves the continuity, differentiability of all orders, and also any convexity of the functions.
2. The second term $R[f]$ The second term of formula (1) is the remainder of that formula. It is a linear (additive and homogeneous) functional defined on the set $\mathscr{F}$ .

Formula (1) and the remainder $R[f]$ corresponding values have a degree of accuracy. It is the integer $n\geqq-1$ completely determined by the condition that $R[f]$ let zero on any polynomial of degree $n$ and that $R[x^{n+1}] ≠ 0$ It is also said that formula (1) or the remainder $R[f]$ the degree of accuracy of this formula $n$ In the following, we can always assume $n\geqq 0$ the case $n==-1$ , not intervening. The condition $n\geqq 0$ is, moreover, equivalent to $2=\sum_{i=1}^{p}c_{i,0}$ .

The rest $R[f]$ is said to be of simple form if there exists an integer $n(\geqq 0)$ independent of function $f$ , such as one might have

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

(2)

Or $M$ East $\neq 0$ and independent of the function $f\in\mathscr{F}$ , the points $\xi_{1},\xi_{2},\ldots,\xi_{n+2}$ being distinct, within the interval $E$ and dependent, in general, on the function $f$ In this case, the number $n$ is completely determined (is unique) and is precisely the degree of accuracy of $R[f]$ .

The rating $\left[\xi_{1},\xi_{2},\ldots,\xi_{n+2};f\right]$ denotes the difference divided (of order $n+1)$ of the function $f$ on the points, or the nodes, $\xi_{1},\xi_{2},\ldots,\xi_{n+2}$ We assume that the definition and main properties of differences divided over distinct or non-distinct nodes are known.

It is easy to see that, under the previous assumptions, the number $M$ is equal to $R\left[x^{n+1}\right]=R\left[x^{n+1}+P\right]$ , Or $P$ is an arbitrary polynomial of degree $n$ .

If we agree to designate by $D_{k}[f]$ a difference divided by order $k$ of the function $f$ on $k+1$ distinct, unspecified nodes within the interval $E$ , formula (2) can be written

R[f]=R\left[x^{n+1}\right]D_{n+1}[f].

(3)

We introduced the notion of simplicity of a linear functional, of the nature of remainder $R[f]$ (initially under another name) in other works [5, 6]. We have supplemented and clarified this research in a more detailed paper [9]. We ask the reader to refer to this paper for our earlier results, which will often be implicitly used in the following.

This work is devoted to the study of the simplicity of the remainder in certain quadrature formulas of the form (1) (§§ 1-3). These are, in short, applications of our previous results. In the last § (§ 4) we will make some remarks on the simplicity of the remainder in certain quadrature formulas relating to integrals extended over an infinite interval.
3. The numbers $k_{1},k_{2},\ldots,k_{p}$ are the multiplicity orders of the nodes $z_{1},z_{2},\ldots,z_{p}$ respective. We ask $k_{1}+k_{2}+\ldots+k_{p}=m$ and then we have $m\geqq p\geqq 1$ We can assume that $k_{i}$ nodes are confused at the point $z_{i}$ Therefore, that $z_{i}$ is a node of order $k_{i}$ of multiplicity (simple if $k_{i}=1$ , double if $k_{i}=2$ etc). The total number of nodes, distinct or not, is therefore equal to $m$ and we can designate these nodes by $x_{1},x_{2},\ldots,x_{m}$ , by choosing the notations, for example, in such a way that we have $x_{k_{1}+k_{2}+\ldots k_{i-1}+j}=z_{i},j=1,2,\ldots,k_{i},i=1,2,\ldots,p$ (the sum $k_{1}+k_{2}+,\ldots+k_{i-1}$ being replaced by 0 for $i=1$ 4.
In what follows we will always assume that the degree of accuracy of formula (1) is at least equal to $m-1$ With the previous data, this condition completely determines formula (1), the right-hand side of which is then obtained by approximating the function $f$ by its Lagrange-Hermite polynomial on the nodes $x_{i},i=1,2,\ldots,m$ [7]. If, therefore, we designate by $n$ the degree of accuracy of formula (1) and if $n\geqq m-1$ the coefficients $c_{i,j}$ are completely determined. Whatever the given nodes $x_{1},x_{2},\ldots,x_{m}$ Therefore, there always exists one and only one formula (1) with a degree of accuracy $\geqq m-1$ .

Moreover, for given nodes with given multiplicity orders, the calculation of the coefficients $c_{i,j}$ is, in general, quite complicated. Formula (11), which will be given later (no. 7), gives us

c_{i,k_{i}-1}=\frac{1}{\left(k_{i}-1\right)!\prod_{\begin{subarray}{c}j=1\\ j\neq i\end{subarray}}^{p}\left(z_{i}-z_{j}\right)^{k_{j}}}\int_{-1}\frac{l(x)}{x-z_{i}}dx,i=1,2,\ldots,p

(4)

Or
$l(x)$ is the polynomial (6) furthest away.
For $p=1$ this formula becomes

c_{1,k_{1}-1}=\frac{1}{\left(k_{1}-1\right)!}\int_{-1}^{1}\left(x-z_{1}\right)^{k_{1}-1 }dx=\frac{\left(1-z_{1}\right)^{k_{1}}-(-1)^{k_{1}}\left(1+z_{1}\right)^{k_{1}}}{k_{1}!}

(5)

Calculating the other coefficients $c_{i,j}$ is more complicated. We will only give it here in certain very specific cases.
5. The degree of accuracy can easily be delimited at a higher level. $n$ of formula (1) (degree of accuracy $n\geqq m-1$ Let's assume $n=m++q-1$ , then the hypothesis $n\geqq m-1$ comes down to $q\geqq 0$ Consider the polynomial
(6)

l(x)=\prod_{i=1}^{m}\left(x-x_{i}\right)=\prod_{i=1}^{p}\left(x-z_{i}\right)^{k_{i}}

The polynomial $Q(x)=l(x)\prod_{i=1}^{p}\left(x-z_{i}\right)^{\frac{1-(-1)^{k_{i}}}{2}}$ is of degree $m+p$ , is non-negative (on $E)$ and we have $R[Q]=\int_{-1}^{1}Q(x)dx>0$ Therefore, we necessarily have $q\leqq p$ .

This delimitation can be refined by taking into account, to some extent, the distribution of the nodes. If we denote by $p_{1}$ the number $(\leqq p)$ knots $z_{i}$ belonging to the open interval $(-1.1)$ we have $q\leqq p_{1}$ This property can be demonstrated in the same way by taking, instead of $Q(x)$ the polynomial $l(x)\prod_{z_{i}\in(-1,1)}\left(x-z_{i}\right)^{\frac{1-(-1)^{k_{i}}}{2}}$ where the product is extended only to those nodes.

6. We have 1st

THEOREM 1. For the formula (1) (of degree of accuracy $\geqq m-1$ ) or of a degree of accuracy equal to $n=m+q-1$ it is necessary and sufficient that the polynomial $l(x)$ either orthogonal to any polynomial of degree $q-1$ on the interval $[-1,1]$ and that one has

R\left[x^{n+1}\right]=\int_{-1}^{1}x^{q}l(x)dx=\int_{-1}^{1}Q(x)l(x)dx\neq 0

(7)

Or $Q(x) = x^{q} + P(x), P(x)$ being any polynomial of degree $q-1$ The
condition is necessary. This property results from the formula

R[Ql]=\int_{-1}^{1}Q(x)l(x)dx

if $Q(x)$ is any polynomial.

The condition is sufficient. This property follows from the fact that if $Q,S$ are respectively the quotient and remainder of the division of any polynomial $P$ by polynomial (6), we have $R[P]=R\left[Q_{l}\right]+R[S]==R[Ql]$ .

We can put the property expressed by Theorem 1 in another form. Let

P_{i}(x)=\int_{-1}^{x}P_{i-1}(x)dx,i=1,2,\ldots,P_{0}(x)=l(x)

(8)

We then have

P_{i}(x)=\frac{1}{(i-1)!}\int_{-1}^{x}(x-t)^{i-1}l(t)dt,i=1,2,\ldots

And

P_{i}(-1)=0,P_{i}^{(i)}(x)=l(x),i=1,2,\ldots

(9)

And it is easy to see that the property expressed by Theorem 1 can be stated in the following equivalent form:

THEOREM 2. For the formula (1) (of degree of accuracy $\geqq m-1$ ) or of a degree of accuracy equal to $n=m+q-1$ It is necessary and sufficient that:

1.

$P_{1}(1)\neq 0$ For $q=0$ .
2.

$P_{1}(1)=P_{2}(1)=\ldots=P_{q}(1)=0,\quad P_{q+1}(1)\neq 0$ , For $q>0$ .

The number $R\left[x^{n+1}\right]$ is then also given by the formula

R\left[x^{n+1}\right]=(-1)^{q}q!P_{q+1}(1)=(-1)^{q}q!\int_{-1}^{1}P_{q}(x)dx

(10)

7.

The remainder of formula (1) (degree of accuracy $\geqq m-1$ ) is given by

R[f]=\int_{-1}^{1}l(x)\left[x_{1},x_{2},\ldots,x_{m},x;f\right]dx

(11)

If we ask

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

(12)

and if we use the integration by parts formula, we deduce ( $q>0$ )

R[f]=\sum_{i=1}^{q}(-1)^{i-1}P_{i}(1)\omega^{(i-1)}(1)+(-1)^{q}\int_{-1}^{1}P_{q}(x)\omega^{(q)}(x)dx

(13)

We can then deduce
THEOREM 3. If there exists a non-negative integer $q$ such as :

1.

$P_{0}(x)=l(x)$ does not change sign on $[-1,1]$ , For $q=0$
2.

$P_{i}(1)=0,i=1,2,\ldots,q$ And $P_{q}(x)$ does not change sign on $[-1,1]$ , pow $q>0$ ,
then the rest $R[f]$ of formula (1) (assumed to have a degree of accuracy $\geqq m-1)$ is of a degree of accuracy equal to $n=m+q-1$ and is of the simple form.

Taking into account the well-known formula

\omega^{(q)}(x)=q![x_{1},x_{2},\ldots,x_{m},\underbrace{x,x,\ldots,x}_{q+1};f],

(14)

the theorem results from the formula

R[f]=(-1)^{q}q!\int_{-1}^{1}P_{q}(x)[x_{1},x_{2},\ldots,x_{m},\underbrace{x,x,\ldots,x}_{q+1};f]dx

(15)

which is verified under the accepted assumptions, and because of the fact that $R[f]$ is of the simple form if (and only if) $R[f]\neq 0$ for any function $f\in\mathscr{F}$ . convex of order $n$ on $E$ .

A function is said to be convex of order $n$ on $E$ if all its differences divided (of order $n+1$ ) on $n+2$ points, not all of them combined, of $E$ are positive. The property expressed by Theorem 3 is what we can call Steffensen's simplicity criterion, by virtue of the important theorem of J. I. Steffensen [12] relating to the remainder of Cotes' quadrature formula. J. I. Steffensen assumes that the function $f$ has a derivative of order $n+1$ continues, but we will see that the remainder in Cotes' formula is of the simple form under the set hypothesis of the continuity of the function $f$ 8.
The conditions under which the Steffensen criterion is applicable must be specified. We will demonstrate in this work that the Steffensen criterion applies only under the assumption that $f\in\mathbb{F}$ This property is expressed by Theorem 5 further on.

Let us note, in passing, that the number $q$ of theorem 3 may not exist. For example, the remainder of the quadrature formula

	$\displaystyle\int_{-1}^{1}f(x)dx=$	$\displaystyle\frac{2}{3}\left[f\left(-\frac{1}{\sqrt{2}}\right)+f(0)+f\left(\frac{1}{\sqrt{2}}\right)\right]+0\cdot f^{\prime}(0)+R[f]=$		(16)
		$\displaystyle=\frac{2}{3}\left[f\left(-\frac{1}{\sqrt{2}}\right)+f(0)+f\left(\frac{1}{\sqrt{2}}\right)\right]+R[f]$

can exclaim

R[f]=\int_{-1}^{1}x^{2}\left(x^{2}-\frac{1}{2}\right)\left[0,0,-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},x;f\right]dx

(17)

In this case $R\left[x^{4}\right]=\frac{1}{15}\neq 0$ The number $q$ Theorem 3 does not exist, since otherwise it should be equal to 0. But the polynomial $l(x)=P_{0}(x)=x^{2}\left(x^{2}-\frac{1}{2}\right)$ changes sign over the interval $[-1,1]$ We will see later (no. 15) that the remainder of formula (16) is of simple form. This result is obtained by interpreting formula (16) with the value 3 of $m$ instead of the value 4.9.
If we refer to how we obtained formula (15), we see that for Steffensen's criterion to be applicable, it suffices that the function (14) be continuous on the interval $[-1,1]$ . If $q=0$ This additional assumption is even unnecessary since the product $l(x)$ ' $\left[x_{1},x_{2},\ldots,x_{m},x;f\right]$ is equal to the difference between the function $f$ and its Lagrange-Hermite polynomial on the knots $x_{1},x_{2},\ldots,x_{m}$ The result is

THEOREM 4. If we have $l(x)\geqq 0$ on the interval $[-1,1]$ , the remainder of formula (1) (assumed to have a degree of accuracy $\geqq m-1$ ) is the degree of accuracy $m-1$ and is of the simple form.

This is particularly true if the nodes $z_{1},z_{2},\ldots,z_{p}$ are all outside the open interval ( $-1,1$ ), or if all the nodes that belong to $(-1,1)$ , are of even order of multiplicity.
10. Let

k_{i}^{\prime}=\left\{\begin{array}[]{ll}k_{i},&\text{ si }z_{i}\in[-1.1]\\ 0&\text{ si }z_{i}\notin[-1.1]\end{array},i=1,2,\ldots,p\right.

For 1a continuity on $[-1,1]$ of the function (14) i1 it suffices that the function $f(\epsilon\mathscr{F})$ has a derivative, continuous of order $q+\max\left(k_{1}^{\prime},k_{2}^{\prime},\ldots,k_{p}^{\prime}\right)$ on $E$ This obviously happens if

q+\max\left(k_{1}^{\prime},k_{2}^{\prime},\ldots,k_{p}^{\prime}\right)\leqq\max\left(k_{1}-1,k_{2}-1,\ldots,k_{p}-1\right)

according to the definition of the set $\mathscr{F}$ In this case, Steffensen's criterion is therefore applicable. But we will demonstrate that this criterion is always applicable, regardless of condition (19).

Let us consider the functions $\varphi_{n+1,\lambda}(x)=\left(\frac{x-\lambda+|x-\lambda|}{2}\right)^{n}(n\geqq 1)$ which, for any real value of the parameter $\lambda$ admit a continuous derivative of order $n-1$ on the real axis.

By virtue of Theorem 15 of our cited work [9], the Steffensen criterion is certainly applicable if

m+q-1\geqq 1,\quad m-2\geqq\max\left(k_{1}^{\prime},k_{2}^{\prime},\ldots,k_{p}^{\prime}\right)

(20)

The first condition is always met for $p\geqq 2$ and, since $m-p+1\geqq\max\left(k_{1}^{\prime},k_{2}^{\prime},\ldots,k_{p}^{\prime}\right)$ , the second condition is also always verified for $p\geqq 3$ . If $p=2$ , this second condition (20) is satisfied, unless one of the nodes $z_{1},z_{2}$ is simple and the other (simple or not) belongs to the closed interval $[-1,1]$ .

We can now state
THEOREM 5. Steffensen's simplicity criterion (Theorem 3) is still applicable to formula (1) (assumed to have a degree of accuracy $\geqq m-1$ ).

From the foregoing, the theorem is proven, except in the noted exceptions, that is, unless inequalities (20) are not simultaneously satisfied and if, by virtue of Theorem 4, we also have $q>0$ 11.
We could eliminate the noted exceptions by using, instead of Theorem 15 of our work cited [9], a more powerful simplicity criterion, but we will not address this issue here. We will continue with the analysis of the noted exceptions. In this way, we will have the opportunity to establish the existence and uniqueness of certain formulas of the type (1). The proof of Theorem 5 in these exceptions will be indirect and will consist of showing that if the hypotheses of Theorem 3 are verified, the rest is indeed of the simple form.

Taking into account the symmetry of the problem with respect to the interval $[-1,1]$ It suffices to examine the following two exceptional cases:

Case 1. $p=1,-1\leqq z_{1}<1$
Case 2. $p=2,k_{2}=1,-1\leqq z_{1}\leqq 1,z_{1}<z_{2}$

It is also sufficient to eliminate from this the direct examination of cases where $q=0$ 12.
Case $1(p=1)$ To simplify the notation a bit, let's

z_{1}=\alpha,\quad k_{1}=k

(21)

and we can assume $-1\leqq\alpha<1$ .

We have

l(x)=P_{0}(x)=(x-\alpha)^{k},P_{1}(x)=\frac{(x-\alpha)^{k+1}+(-1)^{k}(1+\alpha)^{k+1}}{k+1}

The polynomial $P_{0}(x)$ does not change sign on $[-1,1]$ yes, or $\alpha=-1$ or $k$ is even. If $k$ is odd and $-1<\alpha<1$ the polynomial $P_{0}(x)$ changes sign on $[-1,1]$ and the condition $P_{1}(1)=0$ gives us $\alpha=0$ We then have $P_{1}(x)=\frac{x^{k+1}-1}{k+1}\leqq 0$ on $[-1,1]$ .

It is therefore sufficient to demonstrate the simplicity of the rest of the formula of type (1):

\left(F_{1}\right)\quad p=1,\quad k=\text{ impair, }\alpha=0,\quad q=1,\quad n=k

In the following paragraph, we will give the explicit form of this formula (formula (40)) and, in general, of formulas (1) for which $p=1$ 13.
Case $2\left(p=2,k_{2}=1\right)$ To simplify the notation, we set in this case

z_{1}=\alpha,\quad z_{2}=\beta,\quad 1-\alpha=u,\quad 1+\alpha=v,k_{1}=k

(22)

We have

l(x)=P_{0}(x)=(x-\alpha)^{k}(x-\beta)

P_{1}(x)=\frac{(x-\alpha)^{k+2}-(-1)^{k+2}v^{k+2}}{k+2}-(\beta-\alpha)\frac{(x-\alpha)^{k+1}+(-1)^{k}v^{k+1}}{k+1}

and the polynomial $P_{0}(x)$ does not change sign on $[-1,1]$ in the cases $\alpha=-1,\beta\geqq 1;k=$ peer, $-1<\alpha<1\leqq\beta$ And $\alpha=1<\beta$ .

It remains to demonstrate the simplicity of the remainder in the following formulas of type (1):
( $\mathrm{F}_{2}$ ) $\quad p=2,-1=\alpha<\beta<1,P_{1}(1)=0,q=1,n=k+1$
( $\mathrm{F}_{3}$ ) $\quad p=2,k=$ peer, $-1<\alpha<\beta<1,P_{1}(1)=0,q=1,n=k+1$
( $\mathrm{F}_{4}$ ) $\quad p=2,k=$ odd, $-1<\alpha<\beta=1,P_{1}(1)=0,q=1,n=k+1$
( $\mathrm{F}_{5}$ ) $\quad p=2,k=$ odd, $-1<\alpha<1<\beta,P_{1}(1)=0,q=1,n=k+1$
( $\mathrm{F}_{6}$ ) $\quad p=2,k=$ odd, $-1<\alpha<\beta<1,P_{1}(1)=P_{2}(1)=0,q=2,n==k+2$

In the formulas $\left(\mathrm{F}_{2}\right)-\left(\mathrm{F}_{5}\right)$ the function $P_{1}(x)$ changes the direction of monotony on $[-1,1]$ only once (to the point $\beta$ For $\left(\mathrm{F}_{2}\right)$ , $\left(\mathrm{F}_{3}\right)$ and to the point $\alpha$ For $\left(\mathrm{F}_{4}\right),\left(\mathrm{F}_{5}\right)$ ). Of $P_{1}(1)=0$ It therefore follows that $P_{1}(x)$ does not change sign on $[-1,1]$ .

Legality $P_{1}(1)=0$ , in the case of the formula $\left(\mathrm{F}_{2}\right)$ gives us $\beta=\frac{k}{k+2}$ which is well understood in $(-1,1)$ This demonstrates the existence of a unique such formula.

For the formula $\left(\mathrm{F}_{3}\right)$ 1'equation $P_{1}(1)=0$ becomes
(23)

\beta-\alpha=\frac{k+1}{k+2}\cdot\frac{u^{k+2}-v^{k+2}}{u^{k+1}+v^{k+1}}

from which we deduce
(24)

1-\beta=\frac{u^{k+2}+(2k+3-\alpha)v^{k+1}}{(k+2)\left(u^{k+1}+v^{k+1}\right)}

From (23) it follows, taking into account $-1<\alpha<1$ that we have $\beta-\alpha>0$ respectively $\beta-\alpha<0$ depending on $\alpha<0$ respectively $\alpha>0$ From (23) it follows that $1-\beta>0$ The result is that there is a unique formula. $\left(F_{3}\right)$ for everything $\alpha\in(-1,0)$ .

In the case of formulas $\left(\mathrm{F}_{4}\right),\left(\mathrm{F}_{5}\right)$ equality $P_{1}(1)=0$ becomes
(25)

\beta-\alpha=\frac{k+1}{k+2}\cdot\frac{u^{k+2}+v^{k+2}}{u^{k+1}-v^{k+1}}

who determines $\beta$ depending on $\alpha$ Formula (25) shows us that in order to have $\alpha<\beta(-1<\alpha<1)$ it is necessary and sufficient that $\alpha$ is negative. The derivative of $\beta$ compared to $\alpha$ ,

\frac{d\beta}{d\alpha}=\frac{u^{2k+2}+v^{2k+2}+2(uv)^{k}\left(2^{k^{2}}+4k+1+\alpha^{2}\right)}{(k+2)\left(u^{k+1}-v^{k+1}\right)^{2}}

shows us that $\beta$ is an increasing function of $\alpha$ on $[-1,0)$ It is easy to see that if $\alpha$ varies from -1 to $0(-1\leqq\alpha<0),\beta$ grows from $\frac{h}{h+2}$ has $+\infty$ Therefore, there exists a formula and a set of the form ( $\mathrm{F}_{4}$ ) and for this formula $\alpha$ is equal to a number $a$ between -1 and $0(-1<a<0)$ , very determined. For everything $\alpha\in(a,0)$ There is a formula and a single form ( $\mathrm{F}_{5}$ ).

It remains to be demonstrated that there exists a formula of the form ( $\mathrm{F}_{6}$ Assuming $P_{1}(1)=0$ the polynomial $P_{1}(x)$ changes sign over the interval $[-1,1]$ In this case, only once. So if we have $P_{2}(1)=0$ the polynomial $P_{2}(x)$ does not change sign on $[-1,1]$ Taking into account (25), $P_{2}(1)$ is given by the formula

P_{2}(1)=\frac{-u^{2k+4}-v^{2k+4}+2(uv)^{k+1}\left(2k^{2}+8k+7+\alpha^{2}\right)}{(k+2)^{2}(k+3)\left(u^{k+1}-v^{k+1}\right)}

and is a function of $\alpha$ The derivative of this function is given by the formula

\begin{gathered}(k+2)^{2}(k+3)\left(u^{k+1}-v^{k+1}\right)^{2}\frac{dP_{2}(1)}{d\alpha}=\\ =(k+3)\left(u^{3k+4}+v^{3k+4}\right)-\\ -4\alpha\left[2k^{3}+10k^{2}+3k+3+(k+3)\alpha^{2}\right]\left(u^{k+1}-v^{k+1}\right)+\\ +\left(4k^{3}+20k^{2}+27k+9+(k+3)\alpha^{2}\right]uv\left(u^{k}+v^{k}\right)\end{gathered}

The function $P_{2}(1)$ of $\alpha$ is therefore increasing on the interval $[-1,0)$ It has a negative value. $-\frac{2^{k+3}}{(k+2)^{2}(k+3)}$ For $\alpha=-1$ and is certainly positive for $a\leq\alpha<0$ (since then $P_{1}(x)$ does not change sign on $[-1,1]$ ). It follows that there is one and only one point $\alpha=a^{*}$ in the meantime $(-1,a)$ for which $P_{2}(1)=0$ . If $b^{*}$ is the value of $\beta$ taken from (25) for $\alpha=a^{*}$ , the only formula of the form ( $F_{0}$ ) is obtained for $\alpha=a^{*},\beta=b^{*}$ .

We have therefore demonstrated the existence of the formulas $\left(\mathrm{F}_{2}\right)-\left(\mathrm{F}_{6}\right)$ and even the uniqueness of the formulas $\left(F_{2}\right),\left(F_{4}\right),\left(F_{6}\right).I^{\prime}$ 'existence of the formula $\left(F_{6}\right)$ also results, in a different way, from another of our works [8].

In the following paragraph, we will give the explicit form of some of these formulas (formula (45)) and, in general, formulas for which $p==2,k_{2}=1$ (formula (40)).

For the formulas $\left(F_{1}\right)-\left(F_{6}\right)$ (nos. 12, 13) we have also indicated the corresponding value of $q$ and their degree of accuracy $n$ .

Demonstrating the simplicity of the remainders of the formulas $\left(F_{1}\right)-\left(F_{6}\right)$ will be given in the following paragraph. In this way, Theorem 5 will be proven.

To conclude this section, we will look at some applications.
14. First application. Cotes' formulas are quadrature formulas of the form (1) with all nodes simple and equidistant (and of degree of accuracy). $\geqq m-1$ ).

More generally, let us consider, with JE STEFFENSEN, the formula of the type (1)

\int_{a}^{b}f(x)dx=\sum_{i=1}^{m}c_{i}f\left(z_{i}\right)+R[f]

(26)

Or $m=s-2r+1$ , s being a natural number, $r$ an integer $\leqq\frac{1}{2}s$ and the (simple) knots being given by the formulas

z_{i}=a+(r+i-1)\frac{b-a}{s},\quad i=1,2,\ldots,m

(27)

The coefficients $c_{i},i=1,2,\ldots,m$ are completely determined by the condition that formula (26) has its degree of accuracy $\geqq m-1$ .

We then have the

THEOREM 6. The remainder of Cotes' formula (26) is of degree of exactness $2\left[\frac{m-1}{2}\right]+1$ and is of the simple form.

To demonstrate this theorem, it suffices to modify very slightly the proof given by Steffensen [12], assuming that the function $f$ has a continuous derivative of order $2\left[\frac{m-1}{2}\right]+1$

For demonstration 1a, we distinguish two cases depending on the parity of the number $m$ (or of $s$ )

1.

If $m$ is odd, therefore $s$ is even, we have $\int_{a}^{b}l(x)dx=0$ And

(-1)^{\min(r,1)}\int_{a}^{x}l(x)dx\geqq 0,\text{ pour }x\in[a,b]

(28)

It is precisely this inequality that was demonstrated in a very elegant way by JE STEFFENSEN.

The Steffensen criterion is applicable (we consider the interval $[a,b]$ instead of $[-1,1]$ ). The rest $R[f]$ of formula (26) is of degree of accuracy $m$ and is of the simple form.

Note, moreover, that in this case we have
(29)

(-1)^{\min(r-1,0)}R[f]>0

for any (continuous) function $f$ convex of order $m$ 2.
If $m$ is even, therefore $s$ is odd, still according to JE STEFFENSEN we decompose the remainder

R[f]=\int_{a}^{b}l(x)\left[z_{1},z_{2},\ldots,z_{m};x;f\right]dx

(30)

in short

R[f]=R_{1}[f]+R_{2}[f]

(31)

corresponding to the decomposition $\left[a,b-\frac{b-a}{s}\right]\cup\left[b-\frac{b-a}{s},b\mid\right.$ of the integration interval $[a,b]$ of the second member of (30). We have [12],

R_{1}[f]=\int_{a}^{b-\frac{b-a}{s}}\frac{l(x)}{x-z_{m}}\left[z_{1},z_{2},\ldots,z_{m-1},x;f\right]dx

which is the remainder of a formula of the form (26) corresponding to $m-1$ , therefore to an odd number of nodes, but with the same value of $r$ As a result, we have

(-1)^{\min(r-1,0)}R_{1}[f]>0

(32)

for any (continuous) function $f$ convex of order $m-1$ We
also have

R_{2}[f]=\int_{b-\frac{b-a}{s}}^{b}l(x)\left[z_{1},z_{2},\ldots,z_{m},x;f\right]dx

Here we have $(-1)^{\min(r-1,0)}l(x)\geqq 0$ on the interval $\left[b-\frac{b-a}{s},b\right]$ and, by virtue of the theorem $3,R_{2}[f]$ is the degree of accuracy $m-1$ and of the simple form. Furthermore, we have

(-1)^{\min(r-1,0)}R_{2}[f]>0

(33)

for any (continuous) function $f$ , convex of order $m-1$ From
(31) - (33) it follows that inequality (29) is still verified for any (continuous) function $f$ , convex of order $m-1$ .

We can see that this result is valid even if $s=1$ We then have $R[f]=R_{2}[f]$ .

Theorem 6 is therefore proven. We also see that in formula (2) or (3), in this case, we have $\mathrm{sg}M=\mathrm{sg}R\left[x^{n+1}\right]=\mathrm{sg}(-1)^{\mathrm{min}(r-1,0)}$ . $L_{e}$ sign of the numerical coefficient $M$ depends solely on the number $r$ 15.
Second application. A formula of type (1), with simple nodes and coefficients $c_{i,0},i=1,2,\ldots,p(=m)$ all equal and of the same degree of accuracy $\geq m$ This is called a Chebyshev quadrature formula. Such a formula is therefore of the form

\int_{-1}^{1}f(x)dx=\frac{2}{m}\sum_{i=1}^{m}f\left(z_{i}\right)+R[f]

(34)

where the nodes $z_{1},z_{2},\ldots,z_{n}$ are determined (apart from a permutation) by the condition that the formula in question has the degree of accuracy $\geqq m$ .

We know that such a formula exists only for the values 1, 2, $3,4,5,6,7,9$ of $m$ The nodes are then inside the interval $[-1,1]$ and we can take $E=[-1,1]$ .

We have 1st
theorem 7. The remainder of Chebyshev's formula (34) (for the possible values of $m$ ) is of a degree of accuracy equal to $2\left[\frac{m}{2}\right]+1$ and is of the simple form.

The property results from applying Theorem 3.
The polynomial $l(x)=P_{0}(x)$ is found to be calculated, for example, in the book by VI Krilov [2]. This polynomial is given by the table

		$\displaystyle m=1,x;m=2,x^{2}-\frac{1}{3};m=3,x^{3}-\frac{x}{2};m=4,x^{4}-\frac{2}{3}x^{2}+\frac{1}{15}$
		$\displaystyle m=5,x^{5}-\frac{5}{6}x^{3}+\frac{7x}{72};m=6,x^{6}-x^{4}+\frac{1}{5}x^{2}-\frac{1}{105};$
		$\displaystyle m=7,x^{7}-\frac{7}{6}x^{5}+\frac{119}{360}x^{3}-\frac{149x}{6480};$
		$\displaystyle m=9,x^{9}-\frac{3}{2}x^{7}+\frac{27}{40}x^{5}-\frac{57}{560}x^{3}+\frac{53x}{22400}.$

From this, we can deduce the polynomials. $P_{1}(x)$ correspondents,

		$\displaystyle m=1,\frac{1}{2}\left(x^{2}-1\right);\quad m=2,\frac{x\left(x^{2}-1\right)}{3};m=3,\frac{x^{2}\left(x^{2}-1\right)}{4};$
		$\displaystyle m=4,\frac{1}{45}x\left(x^{2}-1\right)\left(9x^{2}-1\right);\quad m=5,\frac{x^{2}-1}{144}\left[24\left(x^{2}-\frac{1}{8}\right)^{2}+\frac{5}{8}\right];$
		$\displaystyle m=6,\frac{x\left(x^{2}-1\right)\left(15x^{4}-6x^{2}+1\right)}{105};$
		$\displaystyle m=7,\frac{x^{2}-1}{12960}\left[1620x^{2}\left(x^{2}-\frac{5}{18}\right)^{2}+46x^{2}+22\right];$
		$\displaystyle m=9,\frac{x^{2}-1}{44800}\left[4480\left(x^{4}-\frac{7}{16}x^{2}+\frac{15}{512}\right)^{2}+\frac{3035}{32}x^{2}+\frac{59709}{2048}\right].$

We can see that for $m$ odd $P_{1}(x)$ does not change sign on $[-1,1]$ Theorem 7 therefore follows for these values of $m$ . For $m$ even the polynomial $P_{1}(x)$ changes sign on $[-1,1]$ , we deduce, for these values of $m$ , polynomials $P_{2}(x)$ correspondents,

		$\displaystyle m=2,\frac{\left(x^{2}-1\right)^{2}}{12};m=4,\frac{1}{90}\left(x^{2}-1\right)^{2}\left(3x^{2}+1\right)$
		$\displaystyle m=6,\frac{1}{840}\left(x^{2}-1\right)^{2}\left(15x^{4}+2x^{2}+3\right)$

which do not change sign on $[-1,1]$ Theorem 7 therefore also follows for these values of $m$ .

We also give the value of the degree in the following table. $n$ accuracy and coefficient $R\left[x^{n+1}\right]$ Chebyshev's formulas,

$m$	1	2	3	4	5	6	7	9
$n$	1	3	3	5	5	7	7	9
$R\left[x^{n+1}\right]$	$\frac{2}{3}$	$\frac{8}{45}$	$\frac{1}{15}$	$\frac{32}{945}$	$\frac{13}{756}$	$\frac{16}{1575}$	$\frac{281}{48600}$	$\frac{163}{73920}$

For the calculation of $R\left[x^{n+1}\right]$ We can use formula (7) and we can check the results obtained using the formula

R\left[x^{n+1}\right]=\int_{-1}^{1}x^{n+1}dx-\frac{2}{m}\sum_{i=1}^{m}z_{i}^{n+1}=\frac{2}{n+2}-\frac{2}{m}\sum_{i=1}^{m}z_{i}^{n+1},n=2\left[\frac{m}{2}\right]+1

and taking into account

\frac{1-(-1)^{v+1}}{v+1}=\frac{2}{m}\sum_{i=1}^{m}z_{i}^{v},v=1,2,\ldots,2\left[\frac{m}{2}\right]+1

For $m=3$ T'chebycheff's formula (34) is none other than formula (16), the remainder of which is therefore given by the formula $R[f]=\frac{1}{15}D_{4}[f]$ , whatever the function $f$ continues over the interval $[-1,1]$ , the nodes of the divided difference $D_{4}[f]$ which appears in this formula being within this interval (and generally depending on the function $f$ ).
16. Third application. Theorem 3 also allows us to construct formulas (1) having a remainder of a given degree of accuracy $m+q-1(q\geqq 0)$ and of the simple form. It suffices to take as nodes the roots (assumed all to be real) of a polynomial of the form $l(x)=\left[\left(x^{2}-1\right)^{q}Q(x)\right]^{(q)}$ , Or $Q$ is a polynomial of degree $m-q$ whose sign does not change on $[-1,1]$ This polynomial can always be chosen such that $l(x)$ has all its real roots. If, in particular, we take $m=q$ , done if $Q$ is a constant ( $\neq 0$ ), the polynomial $l$ is given by the formula

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

THEOREM 8. The remainder in Gauss's quadrature formula with $m$ nodes, is of degree of accuracy $2m-1$ and is of the simple form, regardless of the function $f$ , continues over the integration interval.

The coefficient $R\left[x^{2m}\right]$ results, by applying formula (15),

\begin{gathered}R\left[x^{2m}\right]=\frac{m!}{(2m)!}\int_{-1}^{1}x^{m}\left[\left(x^{2}-1\right)^{m}\right]^{(m)}dx=-\frac{m!m!}{(2m)!}\int_{-1}^{1}\left(1-x^{2}\right)^{m}dx=\\ =\frac{2^{2m+1}(m!)^{4}}{(2m)!(2m-1)!}\end{gathered}

§ 2

17.

Let's return to formula (1). If we set $R^{*}[f]=R\left[f^{\prime}\right]$ , the linear functional $R^{*}[f]$ is, defined on the differentiable functions $f$ whose derivative belongs to $\mathscr{F}$ .

According to a previous result (see Theorem 13 of our cited work [9]), so that the rest $R[f]$ either degree of accuracy $n$ and in simple form, it is necessary and sufficient that $R^{*}[f]$ either degree of accuracy $n+1$ and of the simple form.

If we assume that $R^{*}[f]$ has the degree of accuracy $m+q$ Or $q\geqq 0$ , We have

R^{*}[f]=\sum_{i=1}^{p^{\prime}-q-1}\mu_{i}\left[y_{i},y_{i+1},\ldots,y_{i+m+q+1};f\right]

(35)

Or :

1.

$p^{\prime}=p+2,p+1$ respectively $p$ depending on the points $-1,1$ are both distinct from the nodes $z_{i}$ , one and only one of the points $-1,1$ coincides with one of the nodes $z_{i}$ , respectively each of the points $-1,1$ coincides with one of the $p$ knots $z_{i}$ .
2.

Inequality $p^{\prime}-q-1\geqq 1$ is verified. It results from the inequality $q\leq p^{\prime}-2$ which is a consequence of the boundaries given to $q$ in the preceding paragraph (no. 5).
3.

$y_{1}\leqq y_{2}\leqq\ldots\leqq y_{m+p^{\prime}}$ These points include $k_{i}+1$ times 1st knot $z_{i}$ , For $i=1,2,\ldots,p$ and each of the points $-1,1$ distinct nodes $z_{i}$ .
4.

The coefficients $\mu_{i},i=1,2,\ldots,p^{\prime}-q-1$ are independent of function 1a $f$ .

The coefficient $R^{*}\left[x^{m+q+1}\right]$ is given by the formula

R^{*}\left[x^{m+q+1}\right]=\sum_{i=1}^{p^{\prime}-q-1}\mu_{i}

(36)

Moreover, formulas (35) and (36) are valid only under the assumption that $R^{*}[f]$ vanishes on any polynomial of degree $m+q$ .

We deduce from this
theorem 9. Under the hypotheses and with the previous notation, if all the coefficients $\mu_{i},i=1,2,\ldots,p^{\prime}-q-1$ are of the same sign (all $\geqq 0$ or all $\leqq 0$ ), the rest $R[f]$ of formula (1) is of degree of accuracy $n=m++q-1$ and is of the simple form.

It follows from the problem data that the $\mu_{i},i=1,2,\ldots,p^{\prime}-q-1$ are never all zero. Formula (36) also shows us that we have

R\left[x^{n+1}\right]=\frac{1}{m+q+1}\sum_{i=1}^{p^{\prime}-q-1}\mu_{i}

(37)

18.

First application. In the case $p^{\prime}-q-1=1$ , Or $q=p^{\prime}-2$ , the second member of (35) contains only one term and the rest $R[f]$ is necessarily of the simple form. The coefficient $\mu_{1}$ (who is $\neq 0$ ) can be calculated by identifying the two sides of equation (35). We thus find

\mu_{1}=-k_{i}!c_{i,k_{i}}\Pi_{j}^{\prime}\left(z_{i}-z_{j}\right)^{k_{j}+1}

(38)

where it was placed $z_{0}=-1,z_{p+1}=1,k_{0}=k_{p+1}=0,c_{0,0}=1,c_{p+1,0}=-1$ , in the product $\Pi^{\prime}$ the value $i$ of $j$ is an exception and where the indices $i,j$ scan the values:

1.

$0,1,\ldots,p+1$ if the points $-1,1$ are both distinct from the nodes ( $p^{\prime}=p+2$ ).
2.

$1,2,\cdots,p+1,z_{1}=-1$ and point 1 is different from the nodes ( $p^{\prime}=p+1$ )
3.

$0,1,\ldots,p,z_{p}=1$ and point -1 is different from the nodes ( $p^{\prime}==p+1$ ).
4.

$1,2,\ldots,p$ And $z_{1}=-1,z_{p}=1$ .
(37) it is easy to compare the value of $R\left[x^{n+1}\right]$ deduced from formulas (37), (38) with that which is obtained from (10).
5.

An important special case is obtained for

l(x)=\frac{(\rho+\sigma+m)!}{(\rho+\sigma+2m)!}\left[(x+1)^{\rho+m}(x-1)^{\sigma+m}\right]^{(m)},

Or $m,\rho,\sigma$ are non-negative integers, not all of which are zero. In this case, we have $p^{\prime}=m+2,q=m$ And $m=0$ The corresponding formula (1) is Obrechkoff's formula [3]. Gauss's formula is also a special case ( $\rho=\sigma=0,m>0$ ).

We therefore have the following generalization of Theorem 8,
Theorem 10. The remainder in the quadrature formula considered with $\rho+\sigma+m$ nodes (including $\rho$ coincide with -1 and $\sigma$ with 1) is the degree of accuracy $n=\rho+\sigma+2m-1$ and is of the simple form, regardless of the function $f$ having on $[-1,1]$ a continuous derivative of maximum order $(\rho,\sigma)$ .

The coefficient $R\left[x^{n+1}\right]$ is calculated as in the specific case of Gauss's formula and we obtain

R\left[x^{\rho+\sigma+2m}\right]=\frac{(-1)^{\sigma}2^{\rho+\sigma+2m+1}m!(\rho+m)!(\sigma+m)!(\rho+\sigma+m)!}{(\rho+\sigma+2m)!(\rho+\sigma+2m+1)}.

20.

The formulas $\left.\left(\mathrm{F}_{1}\right),\left(\mathrm{F}_{2}\right),\mathrm{F}_{4}\right),\left(\mathrm{F}_{6}\right)$ are of the previous form ( $p^{\prime}-q-1=1$ ) and therefore have remainders of the simple form.

When $p=1$ Using notation (21), formula (1) becomes

\int_{-1}^{1}f(x)dx=\sum_{i=0}^{k-1}\frac{u^{k-i}-(-v)^{k-i}}{(k-i)!}f^{(k-1-i)}(\alpha)+R[f]

(39)

The degree of accuracy is equal to $k-1$ , unless $k$ is odd and $\alpha=0$ , when it is equal to $k$ In the latter case, we find the formula ( $F_{1}$ ) which can therefore be written
(40)

\int_{-1}^{1}f(x)dx=2\sum_{i=0}^{\frac{k-1}{2}}\frac{1}{(k-2i)!}f^{(k-1-2i)}(0)+\frac{2}{k+2}D_{k+1}[f](k=\text{ impair })

When, or $k$ is even or $k$ is arbitrary and $\alpha\in(-1,1)$ The remainder of the formula, by virtue of Theorem 3, is of the simple form and is given by the formula

R[f]=\frac{u^{k+1}-(-v)^{k+1}}{k+1}D_{k}[f].

When $k$ is odd and $\alpha\in(-1,0)\cup(0,1)$ The rest is a matter of degree of accuracy. $k-1$ but is not of the simple form. This property results from the study of formula (62) of our work cited [9], a formula which we presented as a generalization of Obrechkoff's formula [3]. We will return to this formula later (no. 23).
21. When $p=2$ And $k_{2}=1$ Using notation (22), formula (1) becomes

	$\displaystyle\int_{-1}^{1}f(x)dx=$		(41)
	$\displaystyle=\sum_{i=0}^{k-1}\frac{1}{(k-1-i)!}\left[\frac{u^{k-i}-(-v)^{k-i}}{k-i}-\frac{1}{(\beta-\alpha)^{i+1}}\cdot\frac{u^{k+1}-v^{k+1}}{k+1}\right]f^{(k-1-i)}(\alpha)+$
	$\displaystyle+\frac{u^{k+1}-(-v)^{k+1}}{k+1}\cdot\frac{f(\beta)}{(\beta-\alpha)^{k}}+R[f].$

The degree of accuracy of this formula is equal to $k,k+1$ Or $k+2$ ( $q$ is respectively equal to 0, 1 or 2). We have

R\left[x^{k+1}\right]=\frac{u^{k+2}-(-v)^{k+2}}{k+2}+(\alpha-\beta)\frac{u^{k+1}-(-v)^{k+1}}{k+1}

(42)

If the degree of accuracy is $>k$ the difference $\alpha-\beta$ is given by the formula

\alpha-\beta=-\frac{k+1}{k+2}\cdot\frac{u^{k+2}-(-v)^{k+2}}{u^{k+1}-(-v)^{k+1}}

which amounts to (23) respectively to (25) depending on that $k$ is even respectively odd and in these cases we have

	$\displaystyle R\left[x^{h+2}\right]=\frac{u^{2k+4}+v^{2k+4}-2(-uv)^{k+1}\left(2k^{2}+8k+7+\alpha^{2}\right)}{(k+2)^{2}(k+3)\left[u^{k+1}-(-v)^{k+1}\right]}$		(43)
	$\displaystyle R\left[x^{k+3}\right]=2\frac{u^{2k+5}-v^{2k+5}+2\alpha(-uv)^{k+1}\left(2k^{2}+10k+11+\alpha^{2}\right)}{(k+2)(k+3)(k+4)\left[u^{k+1}-(-v)^{k+1}\right]}$

For $\alpha=-1,\beta=\frac{k}{k+2}$ taking into account (41), (43), we find the formula $\left(\mathrm{F}_{2}\right)$ which is therefore written

	$\displaystyle\int_{-1}^{1}f(x)dx=\sum_{i=0}^{k-1}\frac{2^{k-i}}{(k-1-i)!}\left[\frac{1}{k-1}-\frac{(k+2)^{i+1}}{(k+1)^{i+2}}\right]f^{(k-1-i)}(\alpha)+$		(45)
	$\displaystyle+\frac{2(k+2)^{k}}{(k+1)^{k+1}}f\left(\frac{k}{k+2}\right)+\frac{2^{k+3}}{(k+2)^{2}(k+3)}D_{k+2}[f]$

The formula $\left(\mathrm{F}_{4}\right)$ is obtained from (41), assuming $k$ odd, $\alpha$ equal to the number $a$ defined in no. 13 and $\beta=1$ The remainder of this formula is equal to

R\left[x^{k+2}\right]D_{k+2}[f]

(46)

where the coefficient $R\left[x^{k+2}\right]$ is given by formula (43), with $k$ odd and $\alpha=a$ .

Similarly, the formula $\left(\mathrm{F}_{6}\right)$ is obtained from (41) assuming $k$ odd and $\alpha=a^{*},\beta=b^{*}$ , Or $a^{*},b^{*}$ are the numbers defined in no. 13. The remainder of this formula is equal to $R\left[x^{k+3}\right]D_{k+3}[f]$ where the coefficient $R\left[x^{k+3}\right]$ is given by formula (44) for $k$ odd and $\alpha=a^{*}$ 22.
Second application. Let us now suppose that $p^{\prime}-q-1=2$ The second member of (35) then contains two terms and the remainder $R[f]$ is of the simple form if and only if $\mu_{1}\mu_{2}\geqq 0$ (therefore, if and only if the two coefficients $\mu_{1},\mu_{2}$ are of the same sign). In accordance with the definition of the number $q$ we must also have $\mu_{1}+\mu_{2}=R^{*}\left[x^{m+q+1}\right]\neq 0$ So $\mu_{1}\mu_{2}<0$ , the rest $R[f]$ is the degree of accuracy $m+q-1$ and is not of simple form. Formula (35) is valid, of course, only under the assumption that $R[f]$ vanishes for any polynomial of degree $m++q-1$ , but when $\mu_{1}\mu_{2}=0$ Or $\mu_{1}+\mu_{2}=0$ We return, by a simple modification of the notations, to the first application (when there is only one term in the second member of (35)).

We have

R^{*}[f]=f(1)-f(-1)-\sum_{i=1}^{p}\sum_{j=0}^{k_{i}-1}c_{i,j}f^{(j+1)}\left(z_{i}\right)

(47)

So if $q=p-1$ , therefore if the degree of accuracy of $R[f]$ East $m++p-2$ and if the nodes $z_{i},i=1,2,\ldots,p$ are all within the open interval $(-1,1)$ , formula (35) can be written ( $y_{1}=-1,y_{m+p+2}=1$ )
(48) $\quad R^{*}[f]=\mu_{1}\left[y_{1},y_{2},\ldots,y_{m+p+1};f\right]+\mu_{2}\left[y_{2},y_{3},\ldots,y_{m+p+2};f\right]=$

	$\displaystyle=\mu_{1}$	$\displaystyle{[-1,\underbrace{z_{1},z_{1},\ldots,z_{1}}_{k_{1}+1},\underbrace{z_{2},z_{2},\ldots,z_{2}}_{k_{2}+1}\cdots,\underbrace{z_{p},z_{p},\ldots,z_{p}}_{k_{p+1}};f]+}$
		$\displaystyle+\mu_{2}[\underbrace{z_{1},z_{1},\ldots,z_{1}}_{k_{1}+1},\underbrace{z_{2},z_{2},\ldots,z_{2}}_{k_{2}+1},\ldots,\underbrace{z_{p},z_{p},\ldots,z_{p}}_{k_{p}+1}1;f]$

where the coefficients $\mu_{1},\mu_{2}$ are given by the formulas

\mu_{1}=(-1)^{m+p-1}\prod_{i=1}^{p}\left(1+z_{i}\right)^{k_{i}+1},\quad\mu_{2}=\prod_{i=1}^{p}\left(1-z_{i}\right)^{k_{i}+1}

(49)

and we can state
THEOREM 11. The remainder of formula (1) where the nodes $z_{i},i=1,2,\ldots p$ are all within the interval $(-1,1)$ and whose degree of accuracy is equal to $m+p-2$ , is of the simple form respectively $n$ 'is not of the simple form depending on that $m+p$ is an odd number or an even number.
23. Let's take formula (39) again for $k$ odd and $\alpha\in(-1,0)\cup(0,1)$ The hypotheses of Theorem 11 are verified and we have $m=k,p==1$ SO $m+p$ is even. It follows that the formula has a degree of accuracy $k-1$ , but the rest of it is not of simple form.

The hypotheses of Theorem 11 are also verified for the formula ( $\mathrm{F}_{3}$ ). In this case $m=k+1,p=2$ and the sum $m+p$ is odd. It follows that the remainder of this formula, of degree of accuracy $k+1$ , is of simple form. This remainder is given by (46) where $R\left[x^{k+2}\right]$ is given by formula (43) where $k$ is even.

An example of a formula of the type $\left(\mathrm{F}_{3}\right)$ is given by $(k=2)$

\int_{-1}^{1}f(x)dx=\frac{2}{25}\left[9f\left(-\frac{1}{3}\right)-5f^{\prime}\left(-\frac{1}{3}\right)+16f\left(\frac{1}{2}\right]+\frac{34}{135}D_{4}[f]\right.

24.

To complete the proof of Theorem 5, we still need to study the formula ( $\mathrm{F}_{5}$ ). In this case we can write

R^{*}[f]=\mu_{1}[-1,\underbrace{\alpha,\alpha,\ldots,\alpha}_{k+1},1,\beta;f]+\mu_{2}[\underbrace{\alpha,\alpha,\ldots,\alpha}_{k+1},1,\beta,\beta;f].

(50)

To calculate the coefficients $\mu_{1},\mu_{2}$ , we identify the coefficients of $f(-1)$ and $f^{\prime}(\beta)$ in (47) and (50). Taking into account (4), we deduce,

\begin{gathered}\mu_{1}=-2(1+\beta)(1+\alpha)^{k+1}<0\\ \mu_{2}=-(\beta-\alpha)^{k+1}(\beta-1)c_{2,0}=-(\beta-\alpha)(\beta-1)\int_{-1}^{1}(x-\alpha)^{k}dx=\\ =-(\beta-\alpha)(\beta-1)\frac{u^{k+1}-v^{k+1}}{k+1}<0\end{gathered}

It follows that the rest of the formula $\left(\mathrm{F}_{5}\right)$ is of the simple form. This remainder is again given by formula (43) where, this time, $k$ is odd.

Theorem 5 is therefore completely proven.
The formula

\int_{-1}^{1}f(x)dx=\frac{2}{13}\left[12f\left(-\frac{1}{6}\right)+f(2)\right]-\frac{11}{9}D_{3}[f]

is an example of a formula of the type ( $\mathrm{F}_{5}$ 25.
When $p=3,z_{1}=-1,z_{2}=\alpha,z_{3}=1$ , We have $p^{\prime}=3$ and we are still in the same situation $p^{\prime}-q-1=1$ or in the case $p^{\prime}--q-1=2$ . For $\alpha\in(-1,1)$ We studied the simplicity of the rest of this formula in a previous work [9]. In this case, when the formula has a degree of accuracy $m-1$ The rest is either in simple form or not in simple form depending on whether $(-1)^{k_{2}+k_{3}+1}\mathrm{sg}\left(c_{1,k_{1}-1}c_{3,k_{3}-1}\right)$ is equal to 1 or -1.

We find the simplicity of the rest of Simpson's formula by taking $k_{1}=k_{3}=1,k_{2}=2,\alpha=0$ . In this case $q=0$ And $c_{1,0}=c_{3,0}=\frac{1}{3}$ are indeed positive. Note that the simplicity of the rest of Simpson's formula is also obtained from Theorem 10 by taking $\rho=\sigma=m=1$ Indeed, this formula can also be obtained by taking $k_{1}=k_{2}=k_{3}=1$ , $\alpha=0$ but then you have to take $q=1$ In the first interpretation, we are in the case $p^{\prime}-q-1=2$ , but in the second interpretation in the case $p^{\prime}-q-1=1$ .

For Simpson's formula, we also have,

\begin{gathered}R^{*}[f]=-\frac{2}{3}[-1,-1,0,0,0,1;f]-\frac{2}{3}[-1,0,0,0,1,1;f]=\\ =-\frac{4}{3}[-1,-1,0,0,1,1;f]\end{gathered}

As another example, consider the quadrature formula of DG SANIKIDZE [11],

	$\displaystyle\int_{-1}^{1}f(x)dx=\frac{1}{210}\left[-51f(-1)-42f^{\prime}(-1)-6f^{\prime\prime}(-1)+\right.$		(51)
	$\displaystyle\left.\quad+432f(0)-48f^{\prime}(0)+40f^{\prime\prime}(0)+39f(1)\right]+R[f]$

In this case $k_{1}=3,k_{2}=3,k_{3}=1,\alpha=0,c_{1,2}=-\frac{1}{35}<0,c_{3,0}==\frac{13}{70}>0$ And $(-1)^{k_{\mathrm{a}}+k_{\mathrm{a}}+1}\mathrm{sg}\left(c_{1,k_{1}-1}\cdot c_{3,k_{3}-1}\right)=1$ .

The remainder, equal to $-\frac{8}{35}D_{7}[f]$ , of this formula is therefore of degree of accuracy 6 and is of the simple form.

Regarding the formula given by DG SANIKIDZE (formula (11)) in his work [11], the simplicity of the remainder cannot be sufficiently recognized*).
26. The application of Theorem 9 generally requires the calculation of the coefficients $\mu_{i}$ of formula (35).

Assuming that formula (1) has a degree of accuracy equal to $m+q-1(q\geq 0)$ we have for everything $j$ verifying inequalities

k=\max\left(k_{1}+1,k_{2}+1,\ldots,k_{p}+1\right)\leqq j\leqq m+q+1

(52)

an equality of the form

R^{*}[f]=\sum_{i=1}^{m+p^{\prime}-j}\mu_{i}^{(j)}\left[y_{1},y_{2},\ldots,y_{i+j};f\right]

(53)

THE $\mu_{i}^{(j)}$ being independent of the function $f$ We
then have

\mu_{i}=\mu_{i}^{(m+q+1)},i=1,2,\ldots,p^{\prime}-q-1.

(54)

Starting with the coefficients $\mu_{i}^{(k)},i=1,2,\ldots,m+p^{\prime}-k$ we can successively calculate the coefficients $\mu_{i}^{(j)}$ For $i=1,2,\ldots,m+p^{\prime}--k,j=k,k+1,\ldots,m+q+1$ using recurrence formulas

	$\displaystyle\mu_{i}^{(j+1)}=-\left(y_{j+i+1}-y_{i}\right)\sum_{v=1}^{i}\mu_{v}^{(j)}$		(55)
	$\displaystyle i=1,2,\ldots,m+p^{\prime}-j-1,\quad j=k,\quad k+1,\ldots,m+q$

Applying this method gives us, in the case of formula (51), $k=3$ ,

		$\displaystyle 0R^{*}[f]=2[-1,-1,-1,-1,;f]-8[-1,-1,-0;f]-$
		$\displaystyle-1[-1,-00;f]+6[-000;f]-$
		$\displaystyle-0[000;f]+4[001;f]-3[011;f]$

⁰⁰footnotetext: *). It follows from the above that in this formula one can always take

\xi=\eta

and, using formulas (55) we find

\begin{gathered}35R^{*}[f]=-12[-1,-1,-1,-1,0,0,0,0,1;f]-\\ -52[-1,-1,-1,0,0,0,0,1,1;f]\end{gathered}

This formula clearly highlights the simplicity of the remainder.
27. The preceding method can be applied, in general, to linear combinations. $A[f]$ of a finite number of values of the function $f$ and some of its derivatives, respecting, of course, the assumptions that were highlighted in our previous work [9]. For example, the linear functional

A[f]=f(1)-f(0)-f^{\prime\prime\prime}(0)

(56)

is of degree of accuracy 0 but, obviously, cannot be put in the form (35). The linear functional (56) is defined for any function $f$ having a continuous derivative of order 3 on an interval $E$ containing $[0,1]$ but $n$ It's not a simple form. Otherwise, for everything $f$ there would be at least one point $\xi$ belonging to the interior of $E$ such as one has

A[f]=f^{\prime}(\xi).

(57)

If we take $f(x)=x^{3}+x^{2}$ equality (57) becomes $3\xi^{2}+2\xi+4=0$ which is not verified by any real value of $\xi$ .

§ 3

28.

If the coefficients $\mu_{i}$ Since the terms in formula (35) are all of the same sign (and not all zero), we can deduce the simplicity of the remainder. $R[f]$ without making the assumption $3\mathrm{du}\mathrm{no}$ .17 on the monotony of the sequence $\left(y_{i}\right)_{i=1}^{m+p^{\prime}}$ .

Thus, we can state results analogous to that expressed by Theorem 11 when the nodes $z_{i}$ , without coinciding with points -1 and 1, are not necessarily all in ( $-1,1$ ). Taking into account (18) and (49), we deduce that ( $z_{i}^{2}\neq 1,i=1,2,\ldots,p$ ) sg $\left(\mu_{1}\mu_{2}\right)==\operatorname{sg}(-1)^{\sum_{i=1}^{p}\left(k_{i}^{\prime}+\mathrm{sg}^{\prime}k_{i}^{\prime}\right)+1}$
Therefore, if the degree of accuracy is $m+p-2$ , the simple p for is not of the form sime that the sum $\sum_{i=1}\left(k_{i}^{\prime}+\operatorname{sg}k_{i}^{\prime}\right)$ is odd or even.
29. Consider the formula of type (1)

\int_{-1}^{1}f(x)dx=\sum_{i=1}^{p}c_{i}f\left(z_{i}\right)+R[f]

(58)

the knots $z_{i}$ being simple, distinct and different from the points $\mathbf{-1,1.\penalty 10000\ Pour\penalty 10000\ }$ Let's clarify the ideas, suppose that $-1<z_{1}<z_{2}<\ldots<z_{p}<1$ So if we assume that the remainder of formula (58) vanishes for every polynomial of degree $p$ , we deduce the formula

R^{*}[f]=-\sum_{i=1}^{p}\left(z_{i}^{2}-1\right)l^{\prime}\left(z_{i}\right)c_{i}\left[-1,z_{1},z_{2},\ldots,z_{p},1,z_{i};f\right]

(59)

Or $l(x)=\prod_{i=1}^{p}\left(x-z_{i}\right)$
The proof of formula (59) presents no difficulties. The difference between the two sides of this inequality is a linear combination of the values at the points $-1,1,z_{1},z_{2},\ldots,z_{p}$ of the function $f$ and which vanishes over any polynomial of degree $p+1$ Therefore, it is identically null.

From formula (59) we deduce
THEOREM 12. If in the quadrature formula (58) the coefficients $c_{i}$ are alternately positive and negative ( $c_{i}c_{i+1}<0,i=1,2,\ldots,p-1$ ) and if the degree of accuracy of this formula is equal to $p$ , its remainder is of simple form.

It is always assumed that $-1<z_{1}<z_{2}<\ldots<z_{p}<1$
In particular, we have

\begin{gathered}\int_{-1}^{1}f(x)dx=2\left[f\left(-\frac{1}{\sqrt{6}}\right)-f(0)+f\left(\frac{1}{\sqrt{6}}\right)\right]+\frac{13}{45}D_{4}[f]\\ \int_{-1}^{1}f(x)dx=2\left[f\left(-\sqrt{\frac{33}{60}}\right)-f\left(-\sqrt{\frac{13}{60}}\right)+f(0)-f\left(\sqrt{\frac{13}{60}}\right)+\right.\\ \left.+f\left(\sqrt{\frac{33}{60}}\right)\right]+\frac{3821}{37800}D_{6}[f]\end{gathered}

which are valid for any function $f$ continues over the interval $[-1,1]$ S. E. MIKELADZE gave these formulas [4], assuming that $f$ admits a derivative $4^{\text{ième }}$ respectively a derivative $6^{\text{ième }}$ continue on $[-1,1]$ .

The existence of formulas of the form considered in Theorem 12 was studied by S.N. Bernstein [1].
30. The preceding results can easily be generalized. We will simply state the following result with respect to formula (1): If $p>1,-1\leqq z_{1}<z_{2}<\ldots<z_{p}\leqq 1$ , if the formula has a degree of
accuracy $m+p^{\prime}-p-2$ and if $(-1)^{k_{i}}c_{i,k_{i-1}}c_{i+1,k_{i+1^{-1}}}>0$ For $i=1,2$ , $\cdots,p-1$ The rest is of simple form.

For example, the remainder of the quadrature formula

\begin{gathered}\int_{-1}^{1}f(x)dx=\frac{1}{8}\left[\left[3f(-1)+3f(1)+16f(0)-3f\left(-\frac{1}{3}\right)-3f\left(\frac{1}{3}\right)\right]+\right.\\ -\frac{46}{135}D_{4}[f]\end{gathered}

is simple.
In the case of the formula

\int_{-1}^{1}f(x)dx=\frac{1}{2}[f(-1)+2f(0)+f(1)]-\frac{1}{3}D_{2}[f]

We have $R^{*}[f]=-[-1,-1,0,1;f]+[-1,0,0,1;f]-[-1,0,1,1;f]==-\frac{1}{2}\{[-1,-1,0,0;f]+[0,0,1,1;f]\}$ and the simplicity of the rest does not result from the previous remark, but rather from theorem 9.

84

31.

In this section we will make some remarks on certain generalizations of Theorem 7. HE SALZER studied [10], as an extension of Chebyshev's formulas (34), the formulas

	$\displaystyle\int_{0}^{\infty}e^{-x}f(x)dx=\frac{1}{m}\sum_{i=1}^{m}f\left(z_{i}\right)+R[f]$		(60)
	$\displaystyle\int_{-\infty}^{\infty}e^{-x^{2}}\cdot f(x)dx=\frac{\sqrt{\pi}}{m}\sum_{i=1}^{m}f\left(z_{i}\right)+R[f]$		(61)

degree of accuracy $\geqq m$ These are Chebyshev's formulas for the semi-infinite and infinite intervals.

We are taking now as a set $\mathscr{F}$ the set of continuous functions $f$ on the interval $E=[0,+\infty)$ respectively on the interval $E=(-\infty,+\infty)$ and for which the integral of the first member 7 exists. $\mathcal{F}$ still contains all polynomials and $R[f]$ is a linear functional defined on $\mathscr{F}$ 32.
Formula (60) does not exist for $m=3$ and formula (61) does not exist for $m=4$ (and for larger values of $m$ )

Chebyshev's formulas (60) for $m=1,2$ and Chebyshev's formulas (61) for $m=1,2,3$ have residues of the simple form and are written respectively

	$\displaystyle\int_{0}^{\infty}e^{-x}f(x)dx=f(1)+D_{2}[f]$		(62)
	$\displaystyle\int_{-\infty}^{\infty}e^{-x}f(x)dx=\frac{1}{2}[f(0)+f(2)]+2D_{3}[f]$		(63)
	$\displaystyle\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-x^{2}}f(x)dx=f(0)+\frac{1}{2}D_{2}[f]$		(64)
	$\displaystyle\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-x^{2}}f(x)dx=\frac{1}{2}\left[f\left(-\frac{1}{\sqrt{2}}\right)+f\left(\frac{1}{\sqrt{2}}\right)\right]+\frac{1}{2}D_{4}[f]$		(65)
	$\displaystyle\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-x^{2}}f(x)dx=\frac{1}{3}\left[f\left(-\frac{\sqrt{3}}{2}\right)+f(0)+f\left(\frac{\sqrt{3}}{2}\right)\right]+\frac{3}{8}D_{4}[f]$		(66)

33.

The proof of formulas (62)-(65) presents no difficulties since these formulas do not change if the terms are added to the right-hand side. $0.f^{\prime}(1),0.f^{\prime}(2),0.f^{\prime}(0),0.f^{\prime}\left(\frac{1}{\sqrt{2}}\right)+0.f^{\prime}\left(-\frac{1}{\sqrt{2}}\right)$ respectively and the rest of these formulas can therefore be written

\begin{gathered}R[f]=\int_{0}^{\infty}(x-1)^{2}e^{-x}[1,1,x;f]dx\\ R[f]=\int_{0}^{\infty}x(x-2)^{2}e^{-x}[0,2,2,x;f]dx\\ \sqrt{\pi}R[f]=\int_{-\infty}^{\infty}x^{2}e^{-x^{2}}[0,0,x;f]dx\\ \sqrt{\pi}R[f]=\int_{-\infty}^{\infty}\left(x^{2}-\frac{1}{2}\right)^{2}e^{-x^{2}}\left[-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},x;f\right]dx\end{gathered}

respectively.

It follows that formulas (62) - (65) are valid for any function $f\in\mathbb{F}$ differentiable.
34. On formula (66) we will establish only a less general result. The rest of this formula can be written

\sqrt{\pi}R[f]=\int_{-\infty}^{\infty}x\left(x^{2}-\frac{3}{4}\right)e^{-x^{2}}\left[0,-\frac{\sqrt{3}}{2},\frac{\sqrt{3}}{2},x;f\right]dx

and the parts integration formula

		$\displaystyle\sqrt{\pi}R[f]=-\left.\left(\frac{x^{2}}{2}+\frac{1}{8}\right)e^{-x^{2}}\left[0,-\frac{\sqrt{3}}{2},\frac{\sqrt{3}}{2},x;f\right]\right\|_{-\infty}^{\infty}+$
		$\displaystyle\quad+\int_{-\infty}^{\infty}\left(\frac{x^{2}}{2}+\frac{1}{8}\right)\left[0,-\frac{\sqrt{3}}{2},\frac{\sqrt{3}}{2},x,x;f\right]e^{-x^{2}}dx$

It follows that formula (66) is true for any function $f$ for which the integral of the second member exists (in particular for any polynomial).

BIBLIOGRAPHY

[1] Bernstein S., Some applications of the parametric method to the study of quadrature formulas. Communic. Kharkow (4) 15, 1, 3-29 (1938). . 1959.

⁵ For higher order convex functions (IX).Bulletin
. Soc Roum. of Sci. 43, 4.4ucrările Asupra formi restului in unele formuch 183-185 (1950).
6] Its. Gen. Ștințitice ale Acad. R. derivare numerică. Studii și Cerc. Matem. 3,
[7] - 53-122 (1952). Supra numerică to him Gauss. Studii
[8] - Asupra unei g
[8] - și Cerc. St. Iași, 6, 29-57 (1955). linear approximation of the analysis. [9] - On the remainder in certain formulas
Mathematica 1 (24), 95-142 (1959). quadrature formulas over semi-infinite [10] Salzer Herbert E., Equally weighted quadvand Phys., XXXIV, 54-63 (1955). and infinite intervals'. Journan of mazdelenimi raznostiami. Sobsc. Acad. [11] Sanikidze DG, Interpolirovanie (1960). [12] Steffensen JF, Interpolation, 1950.

The simplicity of the remainder in certain quadrature formulas

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