On the remainder of certain quadrature formulas

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T. Popoviciu
Institutul de Calcul

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T. Popoviciu, Sur le reste de certaines formules de quadrature, Aequationes Math., 2 (1969) nos. 2-3, pp. 265-268 (in French) http://doi.org/10.1007/BF01817710

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Note: paper dedicated to M.A. Ostrowski on the occasion of his 75th anniversary.

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On the remainder of certain quadrature formulas

Tiberiu Popoviciu in Cluj (Romania)

Dedicated to MA Ostrowski on the occasion of his 75th birthday

  1. 1.

    The remainder can be expressed in several ways.R[f]R[f]of the quadrature formula

hasbf(x)𝑑x=α=1nHASαf(xα)+R[f]\int_{a}^{b}f(x)dx=\sum_{\alpha=1}^{n}A_{\alpha}f\left(x_{\alpha}\right)+R[f] (1)

Orxα(α=1.2,,n)x_{\alpha}(\alpha=1,2,\ldots,n)are distinct points of the real axis andHASα,α=1.2,,nA_{\alpha},\alpha=1,2,\ldots,ngiven real constants, independent of the functionff. We will assume thathashasAndbb, Orhas<ba<b, are finite, that all the functions considered are real, and we will denote byIIan interval containing the pointshas,b,xα,(α=1.2,,n)a,b,x_{\alpha},(\alpha=1,2,\ldots,n).

Eithermmthe degree of accuracy of the remainderR[f]R[f](or the quadrature formula (1)), therefore the number, well determined by the conditions:R[1]=R[x]==R[xm]=0R[1]=R[x]=\cdots=R\left[x^{m}\right]=0,R[xm+1]0R\left[x^{m+1}\right]\neq 0. IfR[1]0R[1]\neq 0we takem=1m=-1AndifR[1]=0\operatorname{si}R[1]=0We have0m2n10\leqq m\leqq 2n-1.

Ifffis a continuous function on the intervalIIwe have [1]

R[f]=HAS[ξ1,ξ2,,ξm+2;f]+B[ξ1,ξ2,,ξm+2;f]R[f]=A\left[\xi_{1},\xi_{2},\ldots,\xi_{m+2};f\right]+B\left[\xi_{1}^{\prime},\xi_{2}^{\prime},\ldots,\xi_{m+2}^{\prime};f\right] (2)

The pointsξα\xi_{\alpha}, on the one hand, and the pointsξα\xi_{\alpha}^{\prime}, on the other hand, are distinct but generally depend on the functionff. The constantsHAS,BA,Bare independent of the functionff. We haveHAS+B=R[xm+1]A+B=R\left[x^{m+1}\right]And[y1,y2,,yr;f]\left[y_{1},y_{2},\ldots,y_{r};f\right]denotes the divided difference, of orderr1r-1, of the functionffon the points (or nodes)y1,y2,,yry_{1},y_{2},\ldots,y_{r}. Ifm0m\geqq 0the pointsξα,ξα\xi_{\alpha},\xi_{\alpha}^{\prime}, can be chosen within the intervalIIand if in addition the functionffhas a (m+1m+1)-th derivativef(m+1)f^{(m+1)}insideII, we have

R[f]=HASf(m+1)(ξ)(m+1)!+Bf(m+1)(ξ)(m+1)!R[f]=A\underset{(m+1)!}{f^{(m+1)}(\xi)}+B\frac{f^{(m+1)}\left(\xi^{\prime}\right)}{(m+1)!} (3)

ξ,ξ\xi,\xi^{\prime}, being two points inside the intervalIIAndHAS,BA,Bbeing, moreover, the same constants as in (2).

If we can takeB=0B=0in (2) or in (3), we say that the remainderR[f]R[f]is of the simple form. We have previously given necessary and sufficient conditions for this to be so [1], but the rest is not always of the simple form.

We shall show that even if the remainder is not of the simple form, in the preceding sense, we can, in certain cases, introduce
modified divided differences so that the remainder becomes of the simple form with respect to these new divided differences. The definition of the divided difference which will occur here will result from the following.
2. Let alwaysmmthe degree of accuracy ofR[f]R[f]and designate byk(0kn)k(0\leqq k\leqq n)the number of pointsxαx_{\alpha}included in the open interval (has,ba,b). Ifk>0k>0we can assume that the pointsx1,x2,,xkx_{1},x_{2},\ldots,x_{k}, are in (has,ba,b) and ifk<nk<nthat the pointsxk+1x_{k+1},xk+2,,xnx_{k+2},\ldots,x_{n}, are outside of (has,ba,b). Let us designate byc,d(chas<bd)c,d(c\leqq a<b\leqq d), the ends of the intervalIIand consider the polynomials

Pα=(xc)αR[(xc)α]R[(xc)m+1](xc)m+1(α=0.1,,n+k)P_{\alpha}=(x-c)^{\alpha}-\frac{R\left[(x-c)^{\alpha}\right]}{R\left[(x-c)^{m+1}\right]}(x-c)^{m+1}\quad(\alpha=0,1,\ldots,n+k)

which are well defined sinceR[(xc)m+1]=R[xm+1]0R\left[(x-c)^{m+1}\right]=R\left[x^{m+1}\right]\neq 0and which verify the equalities

R[Pα]=0,(α=0.1,,n+k).R\left[P_{\alpha}\right]=0,(\alpha=0,1,\ldots,n+k). (4)

It is easy to see that we must havemn+k1m\leqq n+k-1. Indeed if we take the polynomial

L=(xx1)2(xx2)2(xxk)2(xxk+1)(xxk+2)(xxn)l=\left(x-x_{1}\right)^{2}\left(x-x_{2}\right)^{2}\ldots\left(x-x_{k}\right)^{2}\left(x-x_{k+1}\right)\left(x-x_{k+2}\right)\ldots\left(x-x_{n}\right)

(whose form is easy to write ifk=0k=0Ork=nk=n), We have

R[L]=hasbL(x)𝑑x0R[l]=\int_{a}^{b}l(x)dx\neq 0 (5)

Let us now designate byD(y1,y2,,yn+k+1;f)D\left(y_{1},y_{2},\ldots,y_{n+k+1};f\right)the determinant of function valuesP0,P1,,Pm,Pm+2,Pm+3,,Pn+k,fP_{0},P_{1},\ldots,P_{m},P_{m+2},P_{m+3},\ldots,P_{n+k},f(ifm=1m=-1functionsP1,P2,P_{1},P_{2},\ldots,Pn+k,f)\left.P_{n+k},f\right)on the pointsy1,y2,,yn+k+1y_{1},y_{2},\ldots,y_{n+k+1}, and byD(y1,y2,,yn+k)D^{*}\left(y_{1},y_{2},\ldots,y_{n+k}\right)the minor corresponding to the elementf(yn+k+1)f\left(y_{n+k+1}\right)of this determinant, therefore the determinant of the values ​​of the functionsP0,P1,,Pm,Pm+2,Pm+3,,Pn+kP_{0},P_{1},\ldots,P_{m},P_{m+2},P_{m+3},\ldots,P_{n+k}on the pointsy1,y2,,yn+ky_{1},y_{2},\ldots,y_{n+k}. We haveD(y1,y2,,yn+k+1;xm+1)=(1)n+km1V(y1,y2,,yn+k+1)D\left(y_{1},y_{2},\ldots,y_{n+k+1};x^{m+1}\right)=(-1)^{n+k-m-1}V\left(y_{1},y_{2},\ldots,y_{n+k+1}\right)OrV(y1,y2,V\left(y_{1},y_{2},\ldots\right.,yn+k+1y_{n+k+1}) is the Vandermonde determinant of numbersy1,y2,,yn+k+1y_{1},y_{2},\ldots,y_{n+k+1}.

The new divided difference we introduce is defined by the formula

[y1,y2,,yn+k+1;f]=D(y1,y2,,yn+k+1;f)D(y1,y2,,yn+k+1;xm+1).\left[y_{1},y_{2},\ldots,y_{n+k+1};f\right]^{*}=\frac{D\left(y_{1},y_{2},\ldots,y_{n+k+1};f\right)}{D\left(y_{1},y_{2},\ldots,y_{n+k+1};x^{m+1}\right)}. (6)

In this definition we have assumed that the pointsyαy_{\alpha}are distinct. But we can extend the definition also to the case of pointsyαy_{\alpha}not necessarily distinct by a passage to the limit in formula (6). This amounts to maintaining the definition formula (6) by suitably modifying the determinantDD(and also the determinants
D,V)\left.D^{*},V\right)as we explained previously [1]. This requires the existence of a certain number of derivatives of the functionff. In this way the divided difference (6) is defined whatever the pointsyαy_{\alpha}distinct or not.
3. With the previous notations consider the continuous functionφ(x)\varphi(x)which outside the pointsxαx_{\alpha}is equal to the linear combinationL(x)[x1,x1,x2,x2,l(x)\left[x_{1},x_{1},x_{2},x_{2},\ldots\right.,xk,xk,xk+1,xk+2,,xn,x;f]\left.x_{k},x_{k},x_{k+1},x_{k+2},\ldots,x_{n},x;f\right]^{*}functionsP0,P1,Pm,Pm+1,Pm+2,,Pn+k,fP_{0},P_{1}\ldots,P_{m},P_{m+1},P_{m+2},\ldots,P_{n+k},f. We then haveφ(xα)=0(α=1.2,,n)\varphi\left(x_{\alpha}\right)=0(\alpha=1,2,\ldots,n)so also

R[φ]=hasbφ(x)𝑑xR[\varphi]=\int_{a}^{b}\varphi(x)dx

and taking into account (4), (5), we have

R[f]=R[xm+1]R[L]hasbφ(x)𝑑xR[f]=\frac{R\left[x^{m+1}\right]}{R[l]}\int_{a}^{b}\varphi(x)dx

Noting that the functionLldoes not change sign on(has,b)(a,b), it follows that

R[f]=R[xm+1][ξ1,ξ2,,ξn+k+1;f]R[f]=R\left[x^{m+1}\right]\left[\xi_{1},\xi_{2},\ldots,\xi_{n+k+1};f\right]^{*} (7)

where theξα\xi_{\alpha}are distinct points of the interior ofII(generally depending on the functionff), provided that the mean theorems can be applied to divided differences (6). I refer the reader for these theorems to a previous work [1]. This condition is surely verified if:
(H). The determinantD(y1,y2,,yn+k)D^{*}\left(y_{1},y_{2},\ldots,y_{n+k}\right)East0\neq 0, whatever the pointsy1y_{1},y2,,yn+ky_{2},\ldots,y_{n+k}distinct or not.

In the previous demonstration of formula (7) we assumed that the function continuesffhas a derivative (at least on the pointsx1,x2,,xkx_{1},x_{2},\ldots,x_{k}). We can easily see that the result obtained is valid under the sole hypothesis of continuity on the closed interval[has,b][a,b]of the functionff.

If we designate byW(g1,g2,,gs)W\left(g_{1},g_{2},\ldots,g_{s}\right)the Wronskian of functionsg1,g2,,gsg_{1},g_{2},\ldots,g_{s}, if we further assume that the functionffhave a(n+k)(n+k)-th derivative on the interior ofIIand if the condition(H)(H)is verified, we have (the Wronskian has its rows and columns in the usual order)

R[f]=(1)n+km11!2!(n+k)!R[xm+1]{W(P0,P1,,Pm,Pm+2,Pm+3,,Pn+k,f)}x=ξR[f]=\frac{(-1)^{n+k-m-1}}{1!2!\ldots(n+k)!}R\left[x^{m+1}\right]\left\{W\left(P_{0},P_{1},\ldots,P_{m},P_{m+2},P_{m+3},\ldots,P_{n+k},f\right)\right\}_{x=\xi}

ξ\xibeing an interior point ofII. Note that the second member of this formula
is of the form

α=0n+km1Qα(ξ)f(m+1+α)(ξ)\sum_{\alpha=0}^{n+k-m-1}Q_{\alpha}(\xi)f^{(m+1+\alpha)}(\xi)

OrQαQ_{\alpha}is a polynomial of degreeα\alphaindependent of functionf(α=0.1,f(\alpha=0,1,\ldots,n+km1n+k-m-1).
4. Ifp0=1p_{0}=1Andpα(α=0.1,,n+k)p_{\alpha}(\alpha=0,1,\ldots,n+k)are the fundamental symmetric functions of numbersyαc,(α=1.2,,n+k)y_{\alpha}-c,(\alpha=1,2,\ldots,n+k), We have

D(y1,y2,,yn+k)=\displaystyle D^{*}\left(y_{1},y_{2},\ldots,y_{n+k}\right)=
=V(y1,y2,,yn+k)α=1n+km1(1)αR[(xc)m+1+α]R[(xc)m+1]Pn+km1α\displaystyle\quad=V\left(y_{1},y_{2},\ldots,y_{n+k}\right)\sum_{\alpha=1}^{n+k-m-1}(-1)^{\alpha}\frac{R\left[(x-c)^{m+1+\alpha}\right]}{R\left[(x-c)^{m+1}\right]}P_{n+k-m-1-\alpha}

It follows that the condition (HH) is checked if the numbers

α=0n+km1(1)αR[(xc)m+1+α]R[(xc)m+1](sn+km1α)(dc)n+km1α(s=0.1,,n+k)\begin{array}[]{r}\sum_{\alpha=0}^{n+k-m-1}(-1)^{\alpha}\frac{R\left[(x-c)^{m+1+\alpha}\right]}{R\left[(x-c)^{m+1}\right]}\binom{s}{n+k-m-1-\alpha}(d-c)^{n+k-m-1-\alpha}\\ (s=0,1,\ldots,n+k)\end{array}

are all positive or all negative.
Example. Consider the quadrature formula

02f(x)𝑑x=2f(0)+6f(1)+2f(2)5+R[f]\int_{0}^{2}f(x)dx=\frac{2f(0)+6f(1)+2f(2)}{5}+R[f]

In this case we havek=m=1k=m=1and we can takec=0,d=2c=0,d=2. The rest is not of the usual simple form (we haveR[f]=7375f"(ξ)32375f"(ξ)R[f]=\frac{7}{375}f^{\prime\prime}(\xi)-\frac{32}{375}f^{\prime\prime}\left(\xi^{\prime}\right)if the second derivative exists). In this case the numbers (8) are all of the same sign and different from 0, so the condition(H)(H)is verified. If the functionffis continuous on[0.2][0,2]and has a fourth derivative on(0.2)(0,2), We have

R[f]=160[4f"(ξ)4(ξ1)f′′′(ξ)+(2ξ24ξ+3)fIV(ξ)],0<ξ<2R[f]=-\frac{1}{60}\left[4f^{\prime\prime}(\xi)-4(\xi-1)f^{\prime\prime\prime}(\xi)+\left(2\xi^{2}-4\xi+3\right)f^{\mathrm{IV}}(\xi)\right],\quad 0<\xi<2
  1. 5.

    The above considerations can easily be extended to quadrature formulas where the second member also linearly contains some of the values ​​on the pointsxαx_{\alpha}successive derivatives of the functionffand also to linear approximation formulas corresponding to more general linear functionals.

BIBLIOGRAPHY

[1] Popoviciu, T., On the remainder in some linear approximation formulas of analysis, Mathematica, ClujII(24), 95-142 (1959).

Institute of Computing

1969

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