ON THE PRESERVATION, BY L. FEJÉR'S INTERPOLATION POLYNOME, OF THE SIGN OR MONOTONY OF THE FUNCTION

BY
TIBERIU POPOVICIU
in Cluj

§ 1. Preservation of the sign of the function

1.

Let

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

(1)

$n(\geqslant 2)$ points on the real axis, $f=f(x)$ a real function of the real variable $x$ , defined on a linear set $E$ containing the points (1) and consider the interpolation polynomial

F[f\mid x]=\sum_{i=1}^{n}f\left(x_{i}\right)h_{t}

(2)

corresponding to the function $f$ on the (interpolation) nodes (1).
The polynomial (2) (of L. Fejér) is the first term of the Lagrange-Hermite polynomial

H[f\mid x]=\sum_{i=1}^{n}f\left(x_{i}\right)h_{i}+\sum_{i=1}^{n}f^{\prime}\left(x_{i}\right)k_{t}

(3)

degree $2n-1$ taking, with the function $f$ and its derivative $f^{\prime}$ the same values on the nodes.

In (3) the $h_{i}$ are the fundamental interpolation polynomials of the first and the $k_{i}$ The fundamental interpolation polynomials of the second kind corresponding to the nodes (1). These polynomials are given by the formulas

Several authors, especially L. Fejér [1], have studied cases, very important for the convergence of interpolation polynomial sequences, where the polynomials (4) are non-negative on a certain interval $I$ .

In this first $\int$ We reproduce, with some additions, the results of L. Fejér [1]. We also revisit the cases where the nodes are the roots of the classical orthogonal polynomials of Jacobi, Laguerre and Hermite considered by L. Fejér [1] and G. Szegö [4].

These results are useful for better highlighting the properties that we will study in $\S 2$ 2.
Either $I$ an arbitrary interval. We say that the interpolation polynomial $F[f\mid x]$ preserves the sign of the function (more precisely: preserves the sign of the function) $f$ on the interval $I$ ) if it is non-negative on $I$ for any function $f$ , non-negative on points (1). This is always the case if $I$ reduces to a node. If the interval $I$ is non-zero ¹ ) or if it reduces to a point other than a node, so that $F[f\mid x]$ preserves the sign of the function; it is necessary and sufficient that linear functions

v_{i}=1-\frac{l^{\prime\prime}\left(x_{i}\right)}{l^{\prime}\left(x_{i}\right)}\left(x-x_{i}\right),\quad i=1,2,\ldots,n

(7)

be non-negative on $L$ Otherwise, we say that the interpolation polynomial $F[f\mid x]$ does not preserve the sign of the function (more precisely: does not preserve the sign of the function $f$ on the interval $I$ ). In this case, there is at least one point of $I$ on which at least one of the polynomials (7) takes a negative value.

The well-known formula

\sum_{i=1}^{n}h_{i}=1

(8)

This shows us that the polynomials (4) cannot all vanish at the same point. It follows that if $F[f\mid x]$ preserves the sign of the function; it is also positive on $I$ for any positive function at the points (1). It is easy to see that, under the same hypothesis, the polynomial $F[f\mid x]$ is non-positive respectively negative on $I$ for any function-

⁰⁰footnotetext: 1. Therefore, it has a non-zero length. Similarly, we say that the interval is zero if it reduces to a single point.

	$\displaystyle h_{i}=\left[\frac{l(x)}{\left(x-x_{i}\right)l^{\prime}\left(x_{i}\right)}\right]^{2}\left[1-\frac{l^{\ prime\prime}\left(x_{i}\right)}{l^{\prime}\left(x_{i}\right)}\left(x-x_{i}\right)\right],i=1,2,\ldots,n$		(4)
	$\displaystyle k_{i}=\left[\frac{l(x)}{\left(x-x_{i}\right)l^{\prime}\left(x_{i}\right)}\right]^{2}\left(x-x_{i}\right),\quad i=1,2,\ldots,n$		(5)
	$\displaystyle l(x)=\left(x-x_{1}\right)\left(x-x_{2}\right)\ldots\left(x-x_{n}\right).$		(Or)

tton $f$ non-positive and negative respectively at points (1). This is, of course, an interval $I$ non-zero. More simply, we could say that if the interpolation polynomial $F[f\mid x]$ If it preserves non-negativity, then it also preserves the non-positivity, positivity, and negativity of the function.
3. The important notion of conjugate points of the nodes, introduced by L. Fejér, allows for a complete discussion of the conservation problem using the polynomial. $F[f\mid x]$ , of the sign of the function.

The conjugate points of the nodes (1) are the points

X_{i}=x_{i}+\frac{l^{\prime}\left(x_{i}\right)}{l^{n}\left(x_{i}\right)},i=1,2,\ldots,n

(6)

$X_{l}$ being the conjugate of the point $x_{l}{}^{1}$
The point $X_{i}$ exists if $l^{\prime\prime}\left(x_{i}\right)\neq 0$ . If $l^{\prime\prime}\left(x_{i}\right)=0$ we can take $X_{i}==+\infty$ and then all the following results remain valid when operating with the improper number $+\infty$ as is usual in mathematical analysis.

The point $X_{i}$ never coincides with $x_{i}$ Inequality
$v_{i}\geqq 0$ on $I$ is equivalent to inequality

\left(X_{i}-x_{i}\right)\left(X_{i}--x\right)\geqq 0\text{ on }I.

We have

\frac{l^{\prime\prime}\left(x_{i}\right)}{l^{\prime}\left(x_{i}\right)}=2\sum_{j=1}^{n}\frac{1}{x_{i}-x_{j}},i=1,2,\ldots,n

(10)

where the accent' at the sign $\sum_{d}^{\prime}$ means that the value $i$ of the index $j$ is excluded.
From inequalities (1) and (10) it follows that

\frac{l^{\prime\prime}\left(x_{1}\right)}{l^{\prime}\left(x_{1}\right)}<0,\frac{l^{\prime\prime}\left(x_{n}\right)}{l^{\prime}\left(x_{n}\right)}>0

so also $X_{1}<x_{1},X_{n}>x_{n}$ The points $X_{1},X_{n}$ are finished and we have $I\subseteq\left[X_{1},X_{n}\right]$ if the polynomial $F[f\mid x]$ preserves the sign of the function over the interval $I$ .

In particular, the interpolation polynomial $F[f\mid x]$ does not preserve the sign of the function over any infinite interval.

This property also results from formula (8) and the fact that none of the polynomials (4) reduces to a constant ( ² ). Indeed, at least one of these polynomials must tend towards $+\infty$ and at least one towards $-\infty$ when $x\rightarrow+\infty$ Or $x\rightarrow-\infty$ .

⁰⁰footnotetext: ¹⁾ Where (1) are the roots of the Chebyshev polynomial

\cos(n\arccos x)

the points

x_{i},X_{i}\left(=\frac{1}{x_{i}}\right)

are harmonically conjugate with respect to the points

-1,1

{}^{\text{2 }}

) Only from (8) and from the fact that at least one of the polynomials (4) does not reduce

4. If $l^{\prime\prime}\left(x_{i}\right)=0$ we have $v_{i}=1>0$ for everything $x$ Let p $i_{1},i_{2},\ldots,i_{k}$ the index values $i$ for which $\frac{l^{\prime\prime}\left(x_{i}\right)}{l^{\prime}(x)}<0$ and by $j_{1},j_{2},\ldots,j_{m}$ the index values $i$ for which $\frac{l^{\prime\prime}\left(x_{i}\right)}{l^{\prime}\left(x_{i}\right)}>0$ . We have $k+m\leqq n$ According to (11) such indices exist (therefore $k\geqq 1,m\geqq 1$ This can also be seen by noting that $\sum_{i=1}^{n}\frac{l^{\prime\prime}\left(x_{i}\right)}{l^{\prime}\left(x_{i}\right)}=0$ and that $l^{\prime\prime}\left(x_{i}\right),i=1,2,\ldots,n$ cannot all be zero ¹ ).

Two cases can then arise:
4.1. We have

X_{i_{r}}\leqq X_{j_{s}},r=1,2,\ldots,k,s=1,2,\ldots,m

So that $F[f\mid x]$ preserves the sign of the function over the interval $I$ It is necessary and sufficient that we have $I\subseteq\left[Z^{-},Z^{+}\right]$ Or

Z^{-}=\max_{r=1,2\ldots,k}X_{i_{r}},\quad Z^{+}=\min_{s=1,2,\ldots,m}X_{j_{s}}

The largest interval (assumed non-zero) on which the polynomial $F[f\mid x]$ preserves the sign of the function is the interval $\left[Z^{-},Z^{+}\right]$ .

In the case of nodes $-1,-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},1(n=4)$ We have $Z^{-}==Z^{+}=0$ , which does not coincide with a node, and in the case of nodes $-1,-\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}},1(n=5)$ we still have $Z^{-}=Z^{+}=0$ , which coincides with a node.
4.2. There is an index $r$ and a clue $s$ such as

X_{t_{r}}>X_{J_{s}}.

(12)

In this case, there is no interval $I$ , not reducing to a knot, on which the polynomial $F[f\mid x]$ preserves the sign of the function,
5. If the interval $I$ is finite, which can always be assumed to be closed, but arbitrary, the results of L. Fejér show us that the preservation of the sign can indeed take place. This is the case if the nodes (1) are normally distributed with respect to the interval $I$ . If $I=[a,b]$ , $a<b$ , we say that the nodes (1) are normally distributed with respect to $I$ , if all the points $x_{i}$ belong to the interval $I$ and all the conjugate points $X_{i}$ are outside the (open) interval ( $a,b$ ). It is easy to see that, in this case, the polynomial $F[f\mid x]$ preserves the sign of the function on $I$ .

⁰⁰footnotetext: ¹ ) Because the polynomial

l^{\prime\prime}

is of degree

n-2

and not identically zero.

We can also obtain a somewhat opposite result.
We have property
I. If $n\geq 4$ , we can construct systems of nodes (1) such that e interpolation polynomial $F[f\mid x]$ does not preserve the sign of the function over any interval $I$ , null or not, and not reducing to a knot.

This property is a consequence of
Lemma 1. If $n\geqq 4$ and if the nodes (1) satisfy the inequalities
(13)

x_{2}-x_{1}\leqq\frac{x_{3}-x_{2}}{n-1},x_{n}-x_{n-1}\leqq\frac{x_{n-1}-x_{n-2}}{n-1}

the interpolation polynomial $F[f\mid x]$ does not preserve the sign of the function over any interval $I$ , null or not, and not reducing to a knot.

By virtue of inequalities (1) and (13), we have

	$\displaystyle\frac{1}{2}\cdot\frac{l^{\prime\prime}\left(x_{2}\right)}{l^{\prime}\left(x_{2}\right)}=$	$\displaystyle\frac{1}{x_{2}-x_{1}}-\sum_{i=3}^{n}\frac{1}{x_{i}-x_{2}}>\frac{1}{x_{2}-x_{1}}-\frac{n-2}{x_{3}-x_{2}}\geqq$
		$\displaystyle\geqq\frac{n-1}{x_{3}-x_{2}}-\frac{n-2}{x_{3}-x_{2}}=\frac{1}{x_{3}-x_{2}}>0$		(14)

SO $\frac{l^{\prime\prime}\left(x_{2}\right)}{l^{\prime}\left(x_{2}\right)}>\frac{2}{x_{3}-x_{2}}>0$ Similarly, we see that

\frac{l^{n}\left(x_{n-1}\right)}{l^{\prime}\left(x_{n-1}\right)}<\frac{-2}{x_{n-1}-x_{n-2}}<0

It follows that it suffices to demonstrate the corresponding inequality (12), therefore inequality
(15)

X_{n-1}>X_{2}.

From (14) we deduce that
therefore

\frac{2}{x_{3}-x_{2}}-\frac{l^{\prime\prime}\left(x_{2}\right)}{l^{\prime}\left(x_{2}\right)}<\frac{2}{x_{3}-x_{2}}-\frac{2}{x_{3}-x_{2}}=0

\frac{x_{3}-x_{2}}{2}-\frac{l^{\prime}\left(x_{2}\right)}{l^{\prime\prime}\left(x_{2}\right)}>0

(16)

and it is demonstrated in the same way that

\frac{x_{n-1}-x_{n-2}}{2}+\frac{l^{\prime}\left(x_{n-1}\right)}{l^{\prime\prime}\left(x_{n-1}\right)}>0.

(17)

Taking into account (1), (16) and (17), we have

\begin{gathered}X_{n-1}-X_{2}=x_{n-1}-x_{2}+\frac{l^{\prime}\left(x_{n-1}\right)}{l^{\prime\prime}\left(x_{n-1}\right)}-\frac{l^{\prime}\left(x_{2}\right)}{l^{\prime\prime}\left(x_{2}\right)}=\frac{x_{n-1}+x_{n-2}-x_{3}-x_{2}}{2}+\\ +\left[\frac{x_{3}-x_{2}}{2}-\frac{l^{\prime}\left(x_{2}\right)}{l^{\prime\prime}\left(x_{2}\right)}\right]+\left[\frac{x_{n-1}-x_{n-2}}{2}+\frac{l^{\prime}\left(x_{n-1}\right)}{l^{\prime\prime}\left(x_{n-1}\right)}\right]>0,\end{gathered}

which demonstrates inequality (15), therefore Lemma 1.6
. The property studied above is not true for $n=2$ and for $n==3$ In these cases, intervals always exist. $I$ , non-zero, on which the interpolation polynomial $F[f\mid x]$ preserves the sign of the function.
Q.1. For $n=2$ , We have

X_{1}=x_{1}-\frac{x_{2}-x_{1}}{2},\quad X_{2}=x_{2}+\frac{x_{2}-x_{1}}{2}

and the diaterpolation polynomial $F[f\mid x]$ preserves the sign of the function over the interval $I$ if and only if $I\subseteq\left[X_{1},X_{2}\right]$ 6.2
. For $n=3$ The discussion is a bit more complicated and depends on the parameter $\varrho=\frac{x_{2}-x_{1}}{x_{3}-x_{1}}$ which remains between 0 and $1(0<\varrho<1)$ . We have

		$\displaystyle X_{1}=x_{1}-\frac{x_{2}-x_{1}}{2(1+\varrho)},X_{3}=x_{3}+\frac{x_{3}-x_{2}}{2(2-\varrho)}$
		$\displaystyle X_{2}=x_{2}+\frac{1-\varrho}{2(1-2\varrho)}\left(x_{2}-x_{1}\right)=x_{2}-\frac{\varrho}{2(2\varrho-1)}\left(x_{3}-x_{2}\right);\left(\varrho\neq\frac{1}{2}\right),$
		$\displaystyle X_{2}=+\infty\left(\varrho=\frac{1}{2}\right),$

and we deduce that the interpolation polynomial $F[f\mid x]$ preserves the sign of the function over the interval $I$ if and only if:
6.2.1. $I\subseteq\left[X_{1},X_{3}\right]$ when $\varrho=\frac{1}{2}$
6.2.2 . $I\subseteq\left[X_{1},\min\left(X_{2},X_{3}\right)\right]$ when $e<\frac{1}{2}$ In this case we have $X_{1}<X_{2},X_{1}<X_{3}$ and we have $X_{2}<$ , =, respectively $>X_{3}$ depending on $\varrho<,=$ , respectively $>\frac{7-2\sqrt{6}}{5}(=0,420\ldots)$ 6.2.3
. I $\subseteq\left[\max\left(X_{1},X_{2}\right),X_{3}\right]$ when $\varrho>\frac{1}{2}$ In this case we have $X_{1}<<X_{3},X_{2}<X_{3}$ and we have $X_{1}<$ , =, respectively $>X_{2}$ depending on $\varrho>,=$ , respectively $<\frac{2\sqrt{6}-2}{5}(=0,579\ldots)$ 6.3
Let us also consider the case $n=4$ , the nodes being symmetrically distributed with respect to a certain point on the real axis, therefore assuming that $x_{2}+x_{3}=x_{1}+x_{4}$ In this case, we can already see the various circumstances that can arise. The discussion depends on the parameter. $\tau=\frac{x_{3}-x_{2}}{x_{2}-x_{1}}=\frac{x_{3}-x_{2}}{x_{4}-x_{3}}$ , which remains positive. If we ask $x_{2}-x_{1}=u$ , We have $x_{3}-x_{2}=\tau u,x_{4}-x_{3}=u$ And

		$\displaystyle X_{1}=x_{1}-\frac{u(\tau+1)(\tau+2)}{2\left(\tau^{2}+5\tau+5\right)},X_{2}=x_{2}+\frac{u\tau(\tau+1)}{2\left(\tau^{2}-\tau-1\right)}$
		$\displaystyle X_{3}=x_{3}-\frac{u\tau(\tau+1)}{2\left(\tau^{2}-\tau-1\right)},\quad X_{4}=x_{4}+\frac{u(\tau+1)(\tau+2)}{2\left(\tau^{2}+5\tau+5\right)}$

We deduce from this

\begin{gathered}X_{2}-X_{1}=u\frac{2\tau^{2}(\tau+2)^{2}-5\tau(\tau+2)-6}{\left(\tau^{2}-\tau-1\right)\left(\tau^{2}+5\tau+5\right)}\\ X_{4}-X_{2}=u(\tau+1)\frac{\tau^{2}(\tau+2)^{2}-7\tau(\tau+2)-6}{\left(\tau^{2}-\tau-1\right)\left(\tau^{2}+5\tau+5\right)}\end{gathered}

X_{3}-X_{2}=u\tau\frac{\tau^{2}-2\tau-2}{\tau^{2}-\tau-1}.

If we ask

\begin{gathered}\tau_{1}=\frac{\sqrt{9+\sqrt{73}}}{2}-1=1,094\ldots\\ \tau_{2}=\frac{\sqrt{9+\sqrt{73}}}{\sqrt{2}}-1=1,961\ldots\\ \tau_{3}=1+\sqrt{3}=2,732\ldots\end{gathered}

numbers $\tau_{1},\tau_{2},\tau_{3}$ are the (unique) values of $\tau$ for which we have $X_{2}=X_{1},X_{2}=X_{4},X_{2}=X_{3}$ respectively. It then follows that the interval $\left[Z^{-},Z^{+}\right]$ reduces to $\left[X_{2},X_{3}\right],\left[X_{1},X_{4}\right]$ respectively $\left[X_{3},X_{2}\right]$ depending on $0\leqq\tau\leqq\tau_{1},\tau_{1}\leqq\tau\leqq\tau_{2}$ respectively $\tau_{2}\leqq\tau\leqq\tau_{3}$ . If $\tau>\tau_{3}$ the interval $\left[Z^{-},Z^{+}\right]$ does not exist.

Other notable values of $\tau$ are

\tau_{4}=1,\tau_{5}=\frac{3+\sqrt{33}}{4}=2,186\ldots,\tau_{6}=2,

for which we have $X_{2}=x_{1}$ (And $X_{3}=x_{4}$ ), $X_{2}=x_{3}$ (And $X_{3}=x_{2}$ ), $X_{2}=x_{4}$ (And $X_{3}=x_{1}$ ) respectively.
7. There are important special cases in which the existence of an interval [ can be affirmed $Z^{-},Z^{+}$ ], non-zero, on which the interpolation polynomial (2) preserves the sign of the function.

Let us designate by $i^{\prime}$ the biggest index $i_{1},i_{2},\ldots,i_{k}$ and by $i^{\prime\prime}$ the smallest of the clues $j_{1},j_{2},\ldots,j_{m}$ defined in no. 4. We have $Z^{-}<<x_{i^{\prime}}$ And $x_{i^{\prime\prime}}<Z^{+}$ Therefore: if $i^{\prime}<i^{\prime\prime}$ the interval $\left[Z^{-},Z^{+}\right]$ exists and is not $nul$ (we have $\left[x_{i^{\prime}},x_{i^{\prime\prime}}\right]\subset\left[Z^{-},Z^{+}\right]$ ).

Inequality $i^{\prime}<i^{\prime\prime}$ is checked if a function exists $\Psi(x)$ defined and non-decreasing on an interval containing the nodes and such that we have

\frac{l^{\prime\prime}\left(x_{i}\right)}{l^{\prime}\left(x_{i}\right)}=\Psi\left(x_{i}\right),\quad i=1,2,\ldots,n.

In this case, moreover, $i_{r}=r,r=1,2\ldots,k,j_{s}=n-m+s,s==1,2,\ldots,m,i^{\prime}=k,i^{n}=n-m+1$ and when $\Psi(x)$ is increasing, we even have $i^{\prime\prime}-i^{\prime}\leqq 2(m+k\geqq n-1)$ .

The function $\overline{\bar{\Psi}}(x)$ exists and is even increasing in the following cases:
7.1. Consider $p(\geqq 2)$ distinct points $a_{1}<a_{2}<\ldots<a_{p}$ of the real axis and are $\sigma_{1},\sigma_{2},\ldots,\sigma_{p},p$ So, for each $q=1,2,\ldots,p-1$ , there exists one and only one polynomial of degree $n$ of the form (6), the nodes (1) being included in the interval ( $a_{q},a_{q+1}$ ) and satisfying the differential equation

l^{n}+\left(\sum_{j=1}^{n}\frac{\sigma_{j}}{x-a_{j}}\right)l^{\prime}+\frac{C(x)}{A(x)}l=0

(18)

Or $A=\left(x-a_{1}\right)\left(x-a_{2}\right)\ldots\left(x-a_{p}\right)$ And $C$ is a polynomial of degree $p-2$ In this case ,
we can take

\Psi(x)=-\sum_{j=1}^{p}\frac{\sigma_{j}}{x-a_{j}}

which is indeed an increasing function on the interval ( $a_{q},a_{q+1}$ 7.2 .
If, in particular, $p=2,a_{1}=-1,a_{2}=1^{1}$ ) and if we ask $\sigma_{1}=\beta+1,\sigma_{2}=\alpha+1$ , the differential equation (18) becomes
(19) $\left(1-x^{2}\right)l^{\prime\prime}-[(\alpha+\beta\div 2)x-\beta+\alpha]l^{\prime}+n(n+\alpha+\beta+1)l=0$ ,

⁰⁰footnotetext: ¹ ) The hypothesis

a_{1}=-1,a_{2}=1\mathrm{n}

'is not an essential restriction. By a linear transformation we can pass to a finite interval [

a_{1},a_{2}

] any.

the actual parameters $\alpha,\beta$ verifying inequalities $\alpha>-1,\beta>-1$ In this case (1) are the roots of the Jacobi polynomials of degree $n$ and parameters $\alpha,\beta$ . We have

x_{k}<\frac{\beta-\alpha}{\alpha+\beta+2}<x_{n-m+1},

so also

Z^{-}<\frac{\beta-\alpha}{\alpha+\beta+2}<Z^{+}.

7.3. If (1) are the roots of the Laguerre polynomial of degree $n$ parameter $a>-1$ , the polynomial (6) satisfies the differential equation

xl^{n}-(x-a-1)l^{\prime}+nl=0.

(20)

In this case, the nodes are positive and we can take $\Psi(x)=1-\frac{a+1}{x}$ which is indeed an increasing function for $x>0$ . We have $Z^{-}<\alpha+1<Z^{+}$ 7.4 .
If (1) are the roots of the Hermite polynomial of degree $n$ , the polynomial (6) satisfies the differential equation

l^{\prime\prime}-2xl^{\prime}+2nl=0.

We can therefore take $\Psi(x)=2x$ which is also an increasing function. In this case $Z^{-}<0<Z^{+}$ 7.5
. We are still in the previous case if the nodes are equidistant. Let $x_{i}=x_{1}+(i-1)h(h>0),i=1,2,\ldots,n$ the knots.

The formula

\frac{l^{\prime\prime}\left(x_{i+1}\right)}{l^{\prime}\left(x_{i+1}\right)}-\frac{l^{n}\left(x_{i}\right)}{l^{\prime}\left(x_{i}\right)}=\frac{2n}{i(n-i)h}>0,i=1,2,\ldots,n-1

shows us that the following $\left(\frac{l^{\prime\prime}\left(x_{i}\right)}{l^{\prime}\left(x_{i}\right)}\right)_{i=1}^{n}$ is increasing. In this case, we have $i^{\prime}=\left[\frac{n}{2}\right],i^{\prime\prime}=\left[\frac{n+1}{2}\right]+1$ And

Z^{-}<x_{1}+\left(\left[\frac{n}{2}\right]-1\right)h<x_{1}+\left[\frac{n+1}{2}\right]h<Z^{+}.

8.

L. Fejér studied [1] the case of Jacobi polynomials in great detail, specifying the extremities $Z^{-},Z^{+}$ of the interval $\left[Z,Z^{+}\right]$ Thus, when inequalities

-1<\alpha\leqq 0,\quad-1<\beta\leqq 0

(21)

are verified, we can obtain a more precise result. Indeed, if we assume that

(\alpha+\beta+2)x_{i}-\beta+\alpha\neq 0,\quad(\alpha+\beta+2)x_{i+1}-\beta+\alpha\neq 0

We have

\frac{X_{i+1}-X_{i}}{x_{i+1}-x_{i}}=\frac{L\left(x_{i},x_{i+1}\right)}{\left[(\alpha+\beta+2)x_{i}-\beta+\alpha\right]\left[(\alpha+\beta+2)x_{i+1}-\beta+\alpha\right]}

(22)

\begin{gathered}L(u,v)=(\alpha+\beta+1)(\alpha+\beta+2)uv-(\beta-\alpha)(\alpha+\beta+1)(u+v)+\\ +(\beta-\alpha)^{2}-(\alpha+\beta+2)\end{gathered}

Note that $L(u,v)$ is a linear function of $u$ and $v$ It follows that if $u,v\in(-1,1),L(u,v)$ remains between the smallest and largest of the numbers $L(1,1),L(-1,-1),L(1,-1)=L(-1,1)$ But, taking into account $(21),L(1,1)=4\alpha(\alpha+1)\leqq 0,L(-1,-1)=4\beta(\beta+1)\leqq\leqq 0,L(1,-1)=L(-1,1)=-4(\alpha+1)(\beta+1)<0$ It is easy to deduce that we have $L(u,v)<0$ For $-1<u,v<1$ .

So we have $L\left(x_{t},x_{t+1}\right)<0,i=1,2,\ldots,n-1$ and formula (22) and the results of no. 4 show us that $Z^{-}=X_{1},Z^{+}=X_{n}$ .

The previous results can also occur for other parameter values. $\alpha,\beta$ , for example for $\alpha=\beta=-1$ . If $-2<\alpha$ , $-2<\beta$ , the differential equation (19) is still satisfied by a polynomial of degree $n$ having all its roots real and distinct. The (1) are then the roots of a generalized Jacobi polynomial of degree $n$ [5]. But, these roots are continuous functions of $\alpha,\beta$ and the previous results remain valid if $\alpha,\beta$ differ sufficiently little from -1. $n\geqq 4$ and if $\alpha\rightarrow-2,\beta\rightarrow-2$ we have $x_{1}\rightarrow-1,x_{2}\rightarrow-1,x_{n-1}\rightarrow 1,x_{n}\rightarrow 1$ and we deduce that if $\alpha,\beta$ are sufficiently close to -2 we are within the conditions of Lemma 1 and the interval $\left[Z^{-},Z^{+}\right]$ does not exist.
9. In the case of the roots of the Laguerre polynomial, the numbers can also be specified. $Z^{-},Z^{+}$ By studying the function $x+\frac{1}{\Psi(x)}=x++\frac{x-\alpha-1}{x}$ It is easy to find that

Z^{-}\leqq\alpha+1-\sqrt{\alpha+1}<\alpha+1+\sqrt{a+1}\leqq Z^{+}

However, these boundaries are rather imprecise. Indeed, by doing $\alpha\rightarrow-1$ From this we only deduce $Z^{-}\leqq Z^{+}$ , although, as we will see later, here the strict inequality holds. The differential equation (20) has as its solution a polynomial of degree $n$ having all its real and distinct roots for $-2<a$ .

For $-2<a\leqq-1$ we are in the case of generalized Laguerre polynomials of parameter $\alpha[5]$ . If $\alpha=-1$ we have $x_{1}=0$ And $x_{1}>$
$>0,i=2,3,\ldots,n$ are the roots of the Laguerre polynomial of degree $n-1$ and parameter 1. If $-2<\alpha<-1$ we have $x_{1}<0$ And $x_{i}>0,i==2,3,\ldots,n$ . When $\alpha\rightarrow-2,x_{1},x_{2}$ both tend towards 0 and the $x_{i},i=3,4,\ldots,n$ to the (positive) roots of the Laguerre polynomial of degree $n-2$ and parameter 2.

For $\alpha=-1$ , We have

\frac{l^{\prime\prime}\left(x_{1}\right)}{l^{\prime}\left(x_{1}\right)}=-(n-1),\frac{l^{\prime\prime}\left(x_{i}\right)}{l^{\prime}\left(x_{i}\right)}=1,\quad i=2,3,\ldots,n

So we have $-\frac{1}{n-1}=Z^{-}<1<Z^{+}=x_{2}+1$ .
For $-2<a<-1$ , We have

\frac{l^{\prime\prime}\left(x_{i}\right)}{l^{\prime}\left(x_{i}\right)}=\frac{x_{i}-a-1}{x_{i}},\quad i=1,2,\ldots,n

which is negative for $i=1$ and positive for $i=2,3,\ldots,n$ . For $i>1$ , we have

\frac{X_{i+1}-X_{i}}{x_{i+1}-x_{i}}=1-\frac{a+1}{\left(x_{i}-a-1\right)\left(x_{i+1}-a-1\right)}>0

and we deduce that

\frac{x_{1}\left(x_{1}-a\right)}{x_{1}-a-1}=X_{1}=Z^{-}<0<Z^{+}=X_{2}=\frac{x_{2}\left(x_{2}-a\right)}{x_{2}-a-1}.

The interval $\left[Z,Z^{+}\right]$ Therefore, it exists and is non-zero for $-2<\alpha$ It can easily be seen that the length of this interval tends towards 0 for $\alpha\rightarrow-2$ .

Let us also say a word about the case where (1) are the roots of the Hermite polynomial of degree $n$ In this case, the function $x+\frac{1}{\psi(x)}=x+\frac{1}{2x}$ has a maximum equal to $-\sqrt{2}$ For $x<0$ and a minimum equal to $\sqrt{2}$ For $x>0$ We can therefore deduce that $Z^{-}\leqq-\sqrt{2}<\sqrt{2}\leqq Z^{+}$ .

§2. Preservation of the monotonicity of the function

10.

We will say that the interpolation polynomial (2) preserves the monotonicity of the function (more precisely: preserves the monotonicity of the function $f$ on the interval $I$ ) if it is non-decreasing on $I$ for any function $f$ non-decreasing on the points (1). For this to be the case, it is necessary and sufficient that we have $F^{\prime}[f,x]\geqq 0$ on $I$ for any function $f$ non-decreasing at points (1). The interval $I$ can be assumed to be non-zero

From formula (8) it follows that

\sum_{i=1}^{n}h_{i}^{\prime}=0

(23)

identically in $x$ The transformation of Abel then gives us
$F^{\prime}[f\mid x]=\sum_{i=1}^{n-1}\left(\sum_{j=i+1}^{n}h_{j}^{\prime}\right)\left(f\left(x_{i+1}\right)-f\left(x_{i}\right)\right)=\sum_{i=1}^{n-1}\left(-\sum_{j=1}^{i}h_{j}^{\prime}\right)\left(f\left(x_{i+1}\right)-f\left(x_{i}\right)\right)$ We deduce
that the interpolation polynomial $F[f\mid x]$ preserves the monotonicity of the function if and only if

\sum_{j=1}^{i}h_{j}^{\prime}\leqq 0,\quad i=1,2,\ldots,n-1

(24)

on $I$ or if and only if

\sum_{\begin{subarray}{c}m\\ i-i+1\end{subarray}}^{n}h_{j}^{\prime}\geqq 0,\quad i=1,2,\ldots,n-1

(

\prime

)

on $I$ If
conditions (24) or (24') are not met, we can say that the interpolation polynomial (2) does not preserve the monotonicity of the function. We can then find a function $f$ non-decreasing on the points (1) such that the polynomial (2) is decreasing on a (non-zero) subinterval of $I$ .

The polynomial $h_{1}$ is of effective degree $2n-1\geqq 3$ , from which it follows that if $F[f\mid x]$ preserves the monotonicity of the function, it is increasing on $I$ for any function $f$ increasing at points (1). It is easily shown that, under the same hypothesis, $F[f\mid x]$ is non-increasing respectively decreasing on $I$ for any function $f$ non-increasing and decreasing respectively at points (1). We can therefore say that if $F[f\mid x]$ preserves the non-decreasing nature of the function; it also preserves the non-increasing, increasing, and decreasing nature of the function.
11. We have the following property:
II. The interpolation polynomial $F[f\mid x]$ does not preserve the monotonicity of the function on any neighborhood ¹ ) of a node.

Indeed, the polynomial $h_{1}$ has all its real roots and more precisely the point $X_{1}\left(<x_{1}\right)$ such as simple roots and nodes $x_{2},x_{3},\ldots,x_{n}$ as double roots. It follows that the derivative $h_{1}^{\prime}$ of the polynomial $h_{1}$ cancels out, changing sign, at the points $x_{2},x_{3},\ldots,x_{n}$ We can see in the same way that the derivative of the polynomial $h_{n}$ cancels out, changing sign, at the points $x_{1},x_{2},\ldots,x_{n-1}$ Property II then results, taking into account the first condition (24) and the last condition ( $24^{\prime}$ ).

⁰⁰footnotetext: ¹ ) By a neighborhood of a point we mean an interval containing that point as an interior point.

We have $h_{1}^{\prime}\left(x_{1}\right)=0$ and the polynomial $h_{1}$ is increasing on the intervals $\left(-\infty,x_{1}\right),\left(x_{n},\infty\right)$ (And $h_{n}$ is decreasing over the same intervals). It follows that the interpolation polynomial $F[f\mid x]$ cannot maintain the monotonicity of the function over an interval $I$ that if $I\subseteq\left[x_{1},x_{n}\right]$ , but this necessary condition is not, in general, sufficient.
12. We also have the following property:
III. There exists a right neighborhood of $x_{1}$ and a left neighborhood ¹ ) of $x_{n}$ on which the interpolation polynomial $F[f\mid x]$ preserves the monotony of the function.

It suffices to demonstrate that in the neighborhoods considered the polynomials $\sum_{j=1}h_{j},i=1,2,\ldots,n-1$ are decreasing.

Let us consider the polynomial $P_{i}=\sum_{j=1}^{i}h_{j}(1\leqq i\leqq n-1)$ This is the Lagrange-Hermite polynomial (3) which takes the value 1 at the nodes. $x_{1},x_{2},\ldots,x_{i}$ the value 0 on the nodes $x_{i+1},x_{i+2},\ldots,x_{n}$ and whose derivative is zero at all nodes. $P_{i}$ is of effective degree $2n-1$ and its derivative has all its roots real and simple. These roots are the nodes and, according to Rolle's theorem, a simple root in each of the intervals

\left(x_{1},x_{2}\right),\left(x_{2},x_{3}\right),\ldots,\left(x_{i-1},x_{i}\right),\left(x_{i+1},x_{i+2}\right),\ldots,\left(x_{n-1},x_{n}\right).

The polynomial $P_{i}$ is alternately increasing and decreasing in the $2n-1$ intervals determined by the roots of $P_{i}^{\prime}$ .

On the interval $\left(x_{i},x_{i+1}\right)P_{i}$ is obviously decreasing, therefore it follows that it is decreasing in a right-hand neighborhood of $x_{1}$ and in a left-hand neighborhood of $x_{n}$ .

Property III is therefore proven. A closer examination of the shape of the polynomials $P_{i}$ allows us to further specify the pro-
) A right (left) neighborhood of $x$ is a (non-zero) interval having the point $x$ as left (right) extremity.
Property. Let us designate by $\xi_{1}^{(i)}<\xi_{2}^{(i)}<\ldots<\xi_{n-2}^{(i)}$ THE $n-2$ different roots of the polynomial nodes $P_{i}^{\prime}$ Let's ask.

		$\displaystyle\eta_{i}^{\prime}=\min\left(\xi_{i}^{(i+1)},\xi_{i}^{(i+2)},\ldots,\xi_{i}^{(n-1)}\right),i=1,2,\ldots,n-2$
		$\displaystyle\eta_{i}^{\prime\prime}=\max\left(\xi_{i-1}^{(1)},\xi_{i-1}^{(2)},\ldots,\xi_{i-1}^{(i-1)}\right),i=2,3,\ldots,n-1.$

We then have $x_{1}<\eta_{1}^{\prime}<x_{2},x_{i}<\eta_{i}^{\prime},\eta_{i}^{\prime\prime}<x_{i+1},i=2,3,\ldots,n-2$ , $x_{n-1}<\eta_{n-1}^{\prime\prime}<x_{n}$ .

Let us designate by $I_{l}$ the interval $\left[\eta_{i}^{\prime\prime},\eta_{i}^{\prime}\right]$ when $\eta_{i}^{\prime\prime}<\eta_{i}^{\prime},i=2,3,..,n-2$ and let's also ask $I_{1}=\left[x_{1},\eta_{1}^{\prime}\right],I_{n-1}=\left\lfloor\eta_{n-1}^{\prime\prime},x_{n}\right\rfloor$ Then the interpolation polynomial $F[f\mid x]$ preserves the monotonicity of the function over any interval $I_{l}$ which exists, in particular therefore on each of the intervals $I_{1},I_{n-1}$ .

Note. The polynomial $P_{t}$ is analogous to the Lagrange-Hermite polynomial $Q$ degree $2n-2$ which takes the value 1 on the nodes $x_{1},x_{2}$ , $\ldots,x_{i}$ the value 0 on the nodes $x_{i+1},x_{i+2},\ldots,x_{n}$ and whose derivative is zero at the points $x_{1},x_{2},\ldots,x_{i-1},x_{i+1},\ldots,x_{n}$ The polynomial $Q$ intervenes in the demonstration of the famous Chebyshev-Markoff-Stieltjes separation theorem (see, for example, G. Szegö [5]).

The polynomial $P_{i}$ is increasing for $x>x_{n}$ , the coefficient of $x^{2n-1}$ Therefore, the value in this polynomial is positive. This property is expressed by the inequalities

\sum_{j=1}^{i}\frac{l^{\prime\prime}\left(x_{j}\right)}{l^{\prime}3\left(x_{j}\right)}<0,i=1,2,\ldots,n-1.

As a result of (8), for $i=n$ The first member of this formula is equal to 0.
13. We will examine the problem of the existence of intervals $I_{i}$ (For $1<i<n-1$ ) in a few specific cases.

We have
Lemma 2. If

\sum_{j=1}^{i}h_{j}^{\prime}\left(x_{0}\right)<0,i=1,2,\ldots,n-1,

there is a neighborhood of the point $x_{0}$ on which the interpolation polynomial (2) preserves the monotonicity of the function.

The property follows immediately from the continuity of the polynomials $\sim\sum_{j=1}^{i}h_{j}^{\prime},i=1,2,\ldots,n-1$ and condition (24).

We also have
Lemma 3. If we have

h_{1}^{\prime}\left(x_{0}\right)<0,h_{n}^{\prime}\left(x_{0}\right)>0

(25)

and if the following
(26)

h_{1}^{\prime}\left(x_{0}\right),h_{2}^{\prime}\left(x_{0}\right),\ldots,h_{n}^{\prime}\left(x_{0}\right)

exhibits a (single) sign variation, we can find a neighborhood of the point $x_{0}$ on which the interpolation polynomial (2) preserves the monotonicity of the function.

Indeed, either $k$ the largest index for which $h_{k}\left(x_{0}\right)<0$ . We have $1\leqq k\leqq n-1$ And

h_{j}^{\prime}\left(x_{0}\right)\left\{\begin{array}[]{l}\leqq 0,\text{ pour }j=1,2,\ldots,k\\ \geqq 0,\text{ pour }j=k+1,k+2,\ldots,n\end{array}\right.

Taking into account (23) we deduce

		$\displaystyle\sum_{=1}^{i}h_{j}^{\prime}\left(x_{0}\right)\leqq h_{1}^{\prime}\left(x_{0}\right)<0\text{ pour }i=1,2,\ldots,k,$
		$\displaystyle\sum_{j=1}^{i}h_{j}^{\prime}\left(x_{0}\right)=-\sum_{j=i+1}^{n}h_{j}^{\prime}\left(x_{0}\right)\leqq-h_{n}^{\prime}\left(x_{0}\right)<0,\text{ pour }i=k+1,k+2,\ldots,n-1$

and Lemma 3 follows from Lemma 2.
14. Using the previous results, let us examine the preservation of monotonicity in a neighborhood of a root of the derivative of the polynomial $l$ .

Let us therefore suppose that $\xi$ either a root of the polynomial $l^{\prime}$ and let us form the sequence (26) for $x_{0}=\xi$ Taking into account $l^{\prime}(\xi)=0$ And $l(\xi)\neq 0$ , We have

	$\displaystyle h_{i}^{\prime}(\xi)=\frac{l^{2}(\xi)}{\left(\xi-x_{i}\right)^{2}l^{\prime 2}\left(x_{i}\right)}\left[\frac{l^{\prime\prime}\left(x_{i}\right)}{l^{\prime}\left(x_{i}\right)}-\frac{2}{\xi-x_{i}}\right],i=1,2,\ldots,n$		(27)
	$\displaystyle\operatorname{sg}h_{i}^{\prime}(\xi)=\operatorname{sg}\left[\frac{l^{\prime\prime}\left(x_{i}\right)}{l^{\prime}\left(x_{i}\right)}-\frac{2}{\xi-x_{i}}\right]-\operatorname{sg}\left(x_{i}-\xi\right)\operatorname{sg}\left[2+\left(x_{i}-\xi\right)\frac{l^{\prime}\left(x_{i}\right)}{l^{\prime\prime}\left(x_{i}\right)}\right]=$		(28)
	$\displaystyle=\operatorname{sg}\left(x_{i}-\xi\right)\operatorname{sg}\left[1+\left(x_{i}-\xi\right)\sum_{i=1}^{i}\frac{1}{x_{i}-x_{j}}\right],i=1,2,\ldots,n$

Taking into account (11) and $x_{1}<\xi<x_{n}$ We deduce that conditions (25) are always satisfied. But, the following

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

presents exactly a change in sign. It follows that
IV. If $\xi$ is a root of the derivative of polynomial (6) and if

2+\left(x_{i}-\xi\right)\frac{l^{\prime\prime}\left(x_{i}\right)}{l^{\prime}\left(x_{i}\right)}\geqq 0,i=1,2,\ldots,n

(29)

there is a neighborhood of the point $\xi$ on which the interpolation polynomial $F[f\mid x]$ preserves the monotonicity of the function.
15. From the preceding analysis it follows that, for $n=2$ and for $n=3$ The following property is true
: V. The interpolation polynomial $F[f\mid x]$ preserves the monotonicity of the function in a suitable neighborhood of each of the roots of the derivative $l^{\prime}$ of the polynomial $l$ .

Moreover, if $n=2$ , We have

h_{1}^{\prime}=-h_{2}^{\prime}=\frac{6\left(x-x_{1}\right)\left(x-x_{2}\right)}{\left(x_{2}-x_{1}\right)^{2}}

and the interpolation polynomial $F[f\mid x]$ preserves the monotonicity of the function on the interval $I$ if and only if $I\subseteq\left[x_{1},x_{2}\right]$ .

For $n=3$ Property V results from the fact that, in this case, the sequence (26) for $x_{0}=\xi$ presents exactly a change in sign, regardless of the value of $h_{2}^{\prime}(\xi)$ Moreover, it can be demonstrated by direct calculation that $h_{1}^{\prime}(\xi)+h_{2}^{\prime}(\xi)<0$ , in this case.

Let's consider the case again $n=4$ , since the nodes are symmetrically distributed, therefore $x_{1}+x_{4}=x_{2}+x_{3}$ Without restricting the generality of the problem, we can take $x_{1}=-1,x_{2}=-\varrho,x_{3}=\varrho,x_{4}=1$ , Or $\varrho$ is a positive number $<1$ Indeed, by a linear transformation, which always preserves the monotonicity and symmetry of the distribution, we reduce the case of arbitrary nodes to the case $x_{1}=-1,x_{4}=1$ .

A calculation, which there is no need to reproduce in detail, gives us
$h_{1}^{\prime}+h_{2}^{\prime}=\frac{\left(x^{2}-\varrho^{2}\right)\left(x^{2}-1\right)}{4\varrho^{3}(1+\varrho)^{3}}\left[7\left(\varrho^{2}+3\varrho+1\right)x^{2}-\left(3\varrho^{4}+9\varrho^{3}+11\varrho^{2}+9\varrho+3\right)\right]$
The roots of the polynomial $l^{\prime}$ are $-\sqrt{\frac{\varrho^{2}+1}{2}},0$ And $\sqrt{\frac{\varrho^{2}+1}{2}}$ .
For $\xi=-\sqrt{\frac{\varrho^{2}+1}{2}}$ and for $\xi=\sqrt{\frac{\varrho^{2}+1}{2}}$ We have

h_{1}^{\prime}(\xi)+h_{2}^{\prime}(\xi)=-\frac{(1-\varrho)^{4}\left(\varrho^{2}+5\varrho+1\right)}{32\varrho^{3}(1+\varrho)}<0

and for $\xi=0$ We have

h_{1}^{\prime}(\xi)+h_{2}^{\prime}(\xi)=-\frac{3\varrho^{4}+9\varrho^{3}+11\varrho^{2}+9\varrho+3}{4\varrho(1+\varrho)^{3}}<0

In all cases, the conditions of Lemma 2 are verified for $x_{0}=\xi$ It follows that property V is also true when $n=4$ and the nodes are symmetrically distributed ⁽¹ ).
16. By (7) we have

2+\left(x_{i}-\xi\right)\frac{l^{\prime\prime}\left(x_{i}\right)}{l^{\prime}\left(x_{i}\right)}=1+v_{i}(\xi),i=1,2,\ldots,n.

But, when the nodes are normally distributed with respect to an interval (no. 5), we have $v_{i}(\xi)>0,i=1,2,\ldots,n$ Therefore, inequalities (29) are, a fortiori, verified. We thus have the following property
VI. If the nodes are normally distributed with respect to an interval, the interpolation polynomial $F[f\mid x]$ preserves (the sign and also) the monotonicity of the function in a suitable neighborhood of each of the roots of the derivative $l^{\prime}$ of the polynomial $l$ .

Let us also examine inequalities (29) in the case where the nodes are the roots of the Jacobi polynomial of degree $n$ . Since $-1<\xi<1$ In this case we have

2+\left(x_{i}-\xi\right)\frac{l^{n}\left(x_{t}\right)}{l^{\prime}\left(x_{i}\right)}=2+\left(x_{t}-\xi\right)\frac{(\alpha+\beta+2)x_{i}-\beta+\alpha}{1-x_{i}^{2}}

which remains between

		$\displaystyle\frac{2-\beta+\alpha+(\alpha+\beta)x_{i}}{1-x_{i}}=\frac{(1-\beta)\left(1-x_{i}\right)+(1+\alpha)\left(1+x_{i}\right)}{1-x_{i}}$
		$\displaystyle\frac{2+\beta-\alpha-(\alpha+\beta)x_{i}}{1+x_{i}}=\frac{(1-\alpha)\left(1+x_{i}\right)+(1+\beta)\left(1-x_{i}\right)}{1+x_{i}}$

We can see that if $-1<\alpha\leqq 1,-1<\beta\leqq 1$ , inequalities (29) are verified. It is easy to see that these inequalities are verified even if $-1\leqq\alpha\leqq 1,-1\leqq\beta\leqq 1$ It follows that we have the following property
VII. If the nodes are the roots of the facobi polynomial of degree n and whose parameters satisfy the inequalities $-1\leqq\alpha\leqq 1,-1\leqq\beta\leqq 1$ the interpolation polynomial $F[f\mid x]$ preserves the monotonicity of the function in a suitable neighborhood of each of the polynomial's roots $l^{\prime}$ .
§3. On some other problems of preserving the shape of the function by interpolation
17. We can seek to study the preservation of the convexity (usual, that is to say of order 1) of the function by the interpolation polynomial (2). $\square$
¹ ) The problem if the property $V$ Whether or not this is generally true remains to be determined.

The definition of conservation by ${}^{F}[f\mid x]$ of non-concavity of order 1 is quite analogous to the definition of conservation of non-negativity and non-decrease [3].

If the interpolation polynomial $F[f\mid x]$ It retains first-order non-concavity; it must also retain first-order non-convexity. But, the function $f=x$ is simultaneously non-concave and non-convex of order 1 on any interval. The function $F[x\mid x]$ must therefore reduce to a polynomial of degree 1. From (3) it then follows that the polynomial $F[x\mid x]=x-\sum_{i=1}^{n}k_{i}$ must be of degree 1. This is impossible since the coefficient of $x^{2n-1}$ in this polynomial is equal to

\sum_{i=1}^{n}\frac{1}{l^{\prime 2}\left(x_{i}\right)}>0.

(30)

We therefore have the following property:
VIII. The interpolation polynomial $F[f\mid x]$ cannot preserve the first-order non-concavity of the function on any (non-zero) interval.
18. When we study the preservation by $F[f\mid x]$ of the sign, monotonicity or convexity of order 1, the fact that we assume the function $f$ Defining only on the points (1) is not an essential restriction. Indeed, such a function can always be extended on any linear set containing the points $x_{i}$ If, therefore, we assume that the set $E$ is arbitrary, we recover the previous results.
19. Finally, we can pose the same problems of preserving the shape of the function for interpolation polynomials other than polynomial (2) of $\mathbf{L}$ . Fejér. In a previous work we dealt, in this sense, with the Lagrange polynomial [3].

Let us consider here the interpolation polynomial of G. Grünwald [2]

G[f\mid x]=\sum_{i=1}^{n}f\left(x_{i}\right)\left[\frac{l(x)}{\left(x-x_{i}\right)l^{\prime}\left(x_{i}\right)}\right]^{2}

which is closely related to Fejér's polynomial (2).
The interpolation polynomial $G[f\mid x]$ (Obviously) preserves the sign of the function over any interval. For it to also preserve monotonicity, it must reduce to a constant for the function. $f=1$ . But, $G[1\mid x]$ is a polynomial of effective degree $2n-2\geqq 2$ since the coefficient of $x^{2n-2}$ is still equal to (30). Therefore, we have the following property:
IX. The interpolation polynomial $G[f\mid x]$ cannot maintain the monotonicity of the function on any (non-zero) interval.

Cluj, November 20, 1961

BIBLIOGRAPHY

1.

Fejér Leopold - Lagrangesche Interpolation and the zugehörigen konjugierten Punkte. Math. Annalen, 106, 1932, 1-55.
2.

Grunwald G. — Qn the theory of interpolation, Acta Math. 75, 1943, 219-245.
3.

Popoviciu T. - On the conservation of the shape of convexity of a function by its interpolation polynomials, Mathematica, 3 (26) (in press).
p. - Uber gewisse Interpolationspolynome, die zu den Gacobischen und Laguer4. Szegö G. -schen Abszissen gehören, Math. Zeitschrift, 35, 1932, 579-602. Orthogonal Polynomials, 1959.

On the conservation by the interpolation polynomial of L. Féjer, of the sign or the monotonicity of the function

Abstract

Authors

Keywords

Paper coordinates

PDF

About this paper

Journal

Publisher Name

DOI

Print ISSN

Online ISSN

References

Paper (preprint) in HTML form

ON THE PRESERVATION, BY L. FEJÉR'S INTERPOLATION POLYNOME, OF THE SIGN OR MONOTONY OF THE FUNCTION

§ 1. Preservation of the sign of the function

§2. Preservation of the monotonicity of the function

BIBLIOGRAPHY

Related Posts