Remarks on the conservation of sign and monotonicity by certain interpolation polynomials of a function of one variable

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T. Popoviciu
Institutul de calcul

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T. Popoviciu, Remarques sur la conservation du signe et de la monotonie par certains polynomes d’interpolation d’une fonction d’une variable, Ann. Univ. Sci. Budapest. Eötvös Sect. Math., t. 34 (1960/1961), pp. 241-246 (in French)

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REMARKS ON THE CONSERVATION OF SIGN AND MONOTONICITY BY CERTAIN INTERPOLATION POLYNOMIALS OF A FUNCTION OF ONE VARIABLE

By
TIBERIU POPOVICIU
(Cluj, Romanian RP)
(Received April 14, 1960.)
Dedicated to the memory of l. Fejér

We know the particular interest that L. Fejér has shown in problems of interpolation by polynomials. He himself has obtained results of great importance in this field. In the following I propose to make some very simple remarks on some of these problems.

  1. 1.

    Consider the linear operator

Φ[fx]=i=1nφi(x)f(xi),\Phi[f\mid x]=\sum_{i=1}^{n}\varphi_{i}(x)f\left(x_{i}\right), (1)

defined on the space of functionsff, real and of a real variablexx, defined on an intervalIIcontaining the distinct nodes
(2)

x1,x2,,xnx_{1},x_{2},\ldots,x_{n}

and where
(3)

φ1(x),φ2(x),,φn(x)\varphi_{1}(x),\varphi_{2}(x),\ldots,\varphi_{n}(x)

are real functions of the real variablexx, defined on the intervalII^{\prime}. To fix the notations, we will always assume that
(4)

x1<x2<<xn.x_{1}<x_{2}<\ldots<x_{n}.

If, in particular, the (3) are polynomials we can take forII^{\prime}the real axis (,-\infty,\infty).

By giving toxxthe valuex0x_{0}we deduce from (1) the linear functionalΦ[fx0]\Phi\left[f\mid x_{0}\right].

We will say that operator (1) preserves the sign of the functionffif we haveΦ[fx]0\Phi[f\mid x]\geqq 0onII^{\prime}for any functionffnon-negative onII. For this to be so, it is necessary and sufficient that the linear functionalΦ[fx0]\Phi\left[f\mid x_{0}\right]preserves the sign of the functiontt, so that it is non-negative for any functionttnon-negative onIIand for everythingx0Ix_{0}\in I^{\prime}.

It is always possible to construct a functionffnon-negative onIIand taking on the nodes (2) any non-negative values. Such is, for example, any function represented by a suitable polygonal line joining the representative points corresponding to the nodes. It follows that the necessary and sufficient condition for the linear functionalΦ[fx0]\Phi\left[f\mid x_{0}\right]preserves the sign of the functionffis that the sequence (3) is non-negative forx=x0x=x_{0}.

16 Annals

We will say that operator (1) preserves the monotonicity of the functionffif the function ofx,Φ[fx]x,\Phi[f\mid x]is non-decreasing onII^{\prime}for any functionffnon-decreasing onII.

Suppose we have

i=1nφi(x)=1\sum_{i=1}^{n}\varphi_{i}(x)=1 (5)

and that the functions (3) are differentiable onII^{\prime}. So, the derivativeΦ[fx]\Phi^{\prime}[f\mid x]of function (1) can be written

Φ[fx]\displaystyle\Phi^{\prime}[f\mid x] =i=1n1(I=i+1nφI(x))|f(xi+1)f(xi)|=\displaystyle=\sum_{i=1}^{n-1}\left(\sum_{j=i+1}^{n}\varphi_{j}^{\prime}(x)\right)\left|f\left(x_{i+1}\right)-f\left(x_{i}\right)\right|=
=i=1n1(I=1iφI(x))(f(xi+1)f(xi))\displaystyle=\sum_{i=1}^{n-1}\left(-\sum_{j=1}^{i}\varphi_{j}^{\prime}(x)\right)\mid\left(f\left(x_{i+1}\right)-f\left(x_{i}\right)\right)

Note that we can always construct a functionffnon-decreasing onIIand taking on the nodes (2) values ​​forming any non-decreasing sequence. Such is again, for example, any function represented by a suitable polygonal line joining the representative points corresponding to the nodes. It follows that the necessary and sufficient condition for (1), under the preceding hypotheses, to preserve the monotony of the functionffis that the sequence of derivatives of functions (3) has all its non-positive partial sequences, so that we have

I=1iφI(x)0,i=1,2,,n,xI\sum_{j=1}^{i}\varphi_{j}^{\prime}(x)\leqq 0,i=1,2,\ldots,n,\quad x\in I^{\prime} (6)

where fori=ni=nequality is valid identically inxx.
2. Suppose, in particular, that (1) is the Lagrange polynomialL[fx]L[f\mid x]of the functionffon the nodes (2). Then the functions (3) reduce to the fundamental interpolation polynomials

φi(x)=Li(x),i=1,2,,n\varphi_{i}(x)=l_{i}(x),\quad i=1,2,\ldots,n (7)

Or

Li(x)=L(x)(xxi)L(xi),i=1,2,,n,L(x)=i=1n(xxi).l_{i}(x)=\frac{l(x)}{\left(x-x_{i}\right)l^{\prime}\left(x_{i}\right)},\quad i=1,2,\ldots,n,\quad l(x)=\prod_{i=1}^{n}\left(x-x_{i}\right). (8)

We then have the following property:
I. Ifn3n\geqq 3and ifx0x_{0}is different from the nodes (2), the linear functionalL[fx0]L\left[f\mid x_{0}\right]does not preserve the sign of the functionff.

Still assuming that condition (4) is verified, this property results only from the fact that the first 3 polynomials (8),L1(x),L2(x),L3(x)l_{1}(x),l_{2}(x),l_{3}(x),
cannot have the same sign (cannot all be0\geqq 0or all of them0)\leq 0)Forx=x0x=x_{0}. Indeed, we have

L1(x)L2(x)=L2(x)(xx1)(xx2)L(x1)L(x2)<0, For x(,x1)(x2,)\displaystyle l_{1}(x)l_{2}(x)=\frac{l^{2}(x)}{\left(x-x_{1}\right)\left(x-x_{2}\right)l^{\prime}\left(x_{1}\right)l^{\prime}\left(x_{2}\right)}<0,\text{ for }x\in\left(-\infty,x_{1}\right)\cup\left(x_{2},\infty\right)
L1(x)L3(x)=L2(x)(xx1)(xx3)L(x1)L(x3)<0, For x(x1,x3)\displaystyle l_{1}(x)l_{3}(x)=\frac{l^{2}(x)}{\left(x-x_{1}\right)\left(x-x_{3}\right)l^{\prime}\left(x_{1}\right)l^{\prime}\left(x_{3}\right)}<0,\text{ for }x\in\left(x_{1},x_{3}\right)

xxbeing always different from the nodes.
We will also give another demonstration of property I. Letx0x_{0}different from the nodes and is a positive number small enough so that the closed interval[x0has,x0+has]\left[x_{0}-a,x_{0}+a\right]does not contain any nodes. Let us construct (which is always possible) a positive functionffwhich takes the same values ​​as the polynomial of degree2,(xx0)2has22,\left(x-x_{0}\right)^{2}-a^{2}on the nodes. We haveL[fx0]==L[(xx0)2has2x0]=has2<0L\left[f\mid x_{0}\right]==L\left[\left(x-x_{0}\right)^{2}-a^{2}\mid x_{0}\right]=-a^{2}<0and the property is demonstrated.

Forn=1n=1We haveL[fx]==f(x1)L[f\mid x]==f\left(x_{1}\right)and the linear functionalL[fx0]L\left[f\mid x_{0}\right]preserves the sign of the functionfffor everythingx0x_{0}.

Forn=2n=2We have

L[fx]=xx2x1x2f(x1)+xx1x2x1f(x2)L[f\mid x]=\frac{x-x_{2}}{x_{1}-x_{2}}f\left(x_{1}\right)+\frac{x-x_{1}}{x_{2}-x_{1}}f\left(x_{2}\right)

and the linear functionalL[fx0]L\left[f\mid x_{0}\right]preserves the sign of the functionttif and only ifx0x_{0}belongs to the interval[x1,x2]\left[x_{1},x_{2}\right]knots (x1<x2x_{1}<x_{2}).
3. In the case (7) of the Lagrange polynomial, the equality (5) is indeed verified and the functions (3) (which are then polynomials) are everywhere differentiable.

We have the following property:
II. Ifn4n\geqq 4, the Lagrange polynomialL[fx]L[f\mid x]does not maintain the monotony of the functionffon any interval (non-zero)II^{\prime}.

We will give a demonstration analogous to the second demonstration of property I.

It is clear that it is sufficient to demonstrate the property for any intervalII^{\prime}closed on the left. So thenx0x_{0}the left end ofII^{\prime}and, ifx0<xnx_{0}<x_{n}, eitherxrx_{r}. the first term of the sequence (2) located to the right ofx0x_{0}(we have (4)). We can then always construct a functionffnon-decreasing taking the values, forming a non-decreasing sequence, of the polynomial of degree 2 or 3,

P(x)={(xx0)2(xxr) if x0<xn(xx0)2 if x0xnP(x)=\left\{\begin{array}[]{l}\left(x-x_{0}\right)^{2}\left(x-x_{r}\right)\text{ if }x_{0}<x_{n}\\ -\left(x-x_{0}\right)^{2}\text{ if }x_{0}\geqq x_{n}\end{array}\right.

on the knots. ButL[Px]=P(x)L[P\mid x]=P(x)and the polynomialPPis not non-decreasing onII^{\prime}. On the contrary, this polynomial is decreasing

on[x0,x0+2xr3] if x0<xn and on [x0,) if x0xn\operatorname{on}\left[x_{0},\frac{x_{0}+2x_{r}}{3}\right]\text{ if }x_{0}<x_{n}\text{ and on }\left[x_{0},\infty\right)\text{ if }x_{0}\geqq x_{n}

Forn=1n=1and forn=2n=2property II is not true. In these cases we haveL[fx]=0L^{\prime}[f\mid x]=0AndL[fx]=f(x2)f(x1)x2x1L^{\prime}[f\mid x]=\frac{f\left(x_{2}\right)-f\left(x_{1}\right)}{x_{2}-x_{1}}respectively andL[fx]L[f\mid x]preserves the monotony offfover the entire intervalII^{\prime}. Ifn=3n=3, a simple calculation shows us that the conditions (6) for the conservation of monotony become

2xx2x30,2xx1x202x-x_{2}-x_{3}\leq 0,\quad 2x-x_{1}-x_{2}\geqq 0

and we see thatL[fx]L[f\mid x]maintains the monotony of the functionffonII^{\prime}if and only if this interval is a subinterval of

[x1+x22;x2+x32](x1<x2<x3)\left[\frac{x_{1}+x_{2}}{2};\frac{x_{2}+x_{3}}{2}\right]\left(x_{1}<x_{2}<x_{3}\right)
  1. 4.

    Now suppose that (1) is the Fejér polynomialF[fx]F[f\mid x]of the functionffon the nodes (2). Then the functions (3) reduce to the fundamental Lagrange-Hermite interpolation polynomials of the first kind

pi(x)=[1L"(xi)L(xi)(xxi)]Li2(x),i=1,2,,np_{i}(x)=\left[1-\frac{l^{\prime\prime}\left(x_{i}\right)}{l^{\prime}\left(x_{i}\right)}\left(x-x_{i}\right)\right]l_{i}^{2}(x),\quad i=1,2,\ldots,n (9)

We know that thenF[fx]F[f\mid x]can well preserve the sign of the functionffover the entire finite intervalII^{\prime}.

Let's suppose thatIIis reduced to the interval[1,1][-1,1]. Then L. Fejér [1] demonstrated that if the polynomialL(x)l(x)checks the differential equation

(1x2)L"(x)(λ+1)xL(x)+n(n+λ)L(x)=0\left(1-x_{2}\right)l^{\prime\prime}(x)-(\lambda+1)xl^{\prime}(x)+n(n+\lambda)l(x)=0 (10)

ultraspherical polynomials and if0λ10\leq\lambda\leq 1, the operatorF[fx]F[f\mid x]preserves the sign of the functionffon the interval[1,1][-1,1].

WhenL(x)l(x)verifies the differential equation (10) we will say that we are in the ultraspherical case of parameterλ(λ>1)\lambda(\lambda>-1). Besides, ifλ>1\lambda>-1the polynomialL(x)l(x)of degreennwhich verifies the differential equation (10) has all its real roots, distinct and included in the interval (1,1-1.1). In particular, we are in the case of Legendre ifλ=1\lambda=1and in the case of Tchebycheff ifλ=0\lambda=0. Then the nodes are the roots of the Legendre polynomial resp. of the Tchebycheff polynomial of the first kind of degreenn.

In particular thereforeF[fx]F[f\mid x]preserves the sign of the functionffon the interval[1,1][-1,1]in the case of Legendre and in the case of Tchebycheff.
5. Still in the case of the operatorF[fx]F[f\mid x]of Fejér, suppose that the numbern=2mn=2mof the nodes is even and that these nodes are distributed symmetrically with respect to the origin. We then haveL(0)0,L(0)=0,x2=xI,L2m+1I(0)=LI(0)0,L(x2m+1I)=L(xI),L"(x2m+1I)=L"(xI)l(0)\neq 0,l^{\prime}(0)=0,x_{2}=-x_{j},l_{2m+1-j}(0)=l_{j}(0)\neq 0,l^{\prime}\left(x_{2m+1-j}\right)=-l^{\prime}\left(x_{j}\right),l^{\prime\prime}\left(x_{2m+1-j}\right)=l^{\prime\prime}\left(x_{j}\right),I=1,2,,mj=1,2,\ldots,mAnd

{φi(0)=[1+xiL"(xi)L(xi)]Li2(0),i=1,2,,2mφi(0)=[2xi+L"(xi)L(xi)]Li2(0)=1xi[φi(0)+Li2(0)]i=1,2,,2m\left\{\begin{array}[]{c}\varphi_{i}(0)=\left[1+x_{i}\frac{l^{\prime\prime}\left(x_{i}\right)}{l^{\prime}\left(x_{i}\right)}\right]l_{i}^{2}(0),\quad i=1,2,\ldots,2m\\ \varphi_{i}^{\prime}(0)=\left[\frac{2}{x_{i}}+\frac{l^{\prime\prime}\left(x_{i}\right)}{l^{\prime}\left(x_{i}\right)}\right]l_{i}^{2}(0)=\frac{1}{x_{i}}\left[\varphi_{i}(0)+l_{i}^{2}(0)\right]\\ i=1,2,\ldots,2m\end{array}\right.

We deduce that theφi(0),i=1,2,,2m\varphi_{i}(0),i=1,2,\ldots,2mare positive if and only if

1+xiL"(xi)L(xi)>0,i=1,2,,2m1+x_{i}\frac{l^{\prime\prime}\left(x_{i}\right)}{l^{\prime}\left(x_{i}\right)}>0,\quad i=1,2,\ldots,2m (12)

We have the following property:
III. If the numbern=2mn=2mof the nodes is even, if these nodes are symmetrically distributed with respect to the origin and if the inequalities (12) are verified, boperatorF[fx]F[f\mid x]de Fejér preserves the sign and also preserves the monotony of the hundred non-zero interval[has,has],(has>0)[-a,a],(a>0), having its center in the origin.

The conservation of the sign results from the fact that we haveφi(0)>0,i=1\varphi_{i}(0)>0,i=1,2,,2m2,\ldots,2m, as a result of (11) and the fact that theφi(x)\varphi_{i}(x)are continuous functions.

To demonstrate the property relating to the conservation of monotony it is sufficient to demonstrate the inequalities (6) forI=[has,has]I^{\prime}=[-a,a]. To be sure of the existence of such a numberhasait is sufficient, due to the continuity of functionsφi(x)\varphi_{i}(x), to demonstrate that, in the case considered, we have

I=0iφI(0)<0,i=1,2,,2m1\sum_{j=0}^{i}\varphi_{j}^{\prime}(0)<0,\quad i=1,2,\ldots,2m-1 (13)

But, taking into account (11) and the reported symmetry of the knots, we deduce

φI(0)=φ2m+1I(0)<0,I=1,2,,m\varphi_{j}^{\prime}(0)=-\varphi_{2m+1-j}^{\prime}(0)<0,j=1,2,\ldots,m

from which inequalities (13) immediately follow.
Property III is therefore demonstrated.
6. In particular, if we are in the ultraspherical case the nodes are symmetrically distributed with respect to the origin.

Differential equation (10) shows us that in this case, ifλ>1\lambda>-1,

1+xLL"(xi)L(xi)=1+λxi21xi2>0,i=1,2,,n1+x_{l}\frac{l^{\prime\prime}\left(x_{i}\right)}{l^{\prime}\left(x_{i}\right)}=\frac{1+\lambda x_{i}^{2}}{1-x_{i}^{2}}>0,\quad i=1,2,\ldots,n

so conditions (12) are verified.
It follows that we have the following property:
IV. If we are in the ultraspherical case of parameterλ>1\lambda>-1and if the numbern=2mn=2mof nodes is even, the operatorF[fx]F[f\mid x]de Fejér preserves the sign and also preserves the monotony of the functionffon a certain interval won null[has,has](has>0)[-a,a](a>0), having its center in the origin.

The property is true, in particular, in the case of Legendre and in the case of Tschebycheff.

The previous results are to be compared with the important property of conservation of convexities of all orders, which enjoy on the interval[0,1][0,1]SN Bernstein polynomials

B[fx]=i=0n(ni)f(in)xi(1x)niB[f\mid x]=\sum_{i=0}^{n}\binom{n}{i}f\left(\frac{i}{n}\right)x^{i}(1-x)^{n-i}

which are also of the form (l). We formerly obtained these properties [2].

Bibliography

[1] L. Fejér, Über Weierstrassche Approximation besonders durch Hermitesche Interpolation, Mathematische Annalen, 102 (1930), 707-725.
[2] T. Popoviciu, On the approximation of higher order convex functions, Mathematica; 10 (1934), 49-54.

1960

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