On certain quadrature formulas of maximum degree of accuracy

Tiberiu Popoviciu
Approximation Theory, html, quadrature

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Tiberiu Popoviciu
Institutul de Calcul

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T. Popoviciu, Sur certaines formules de quadrature de degré d’exactitude maximum, Trudy Mat. Inst. Steklov, 134 (1975), pp. 254-259 (in French) Theory of functions and its applications

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1975 a -Popoviciu- Trudy Mat. Inst. Steklov – Sur certaines formules de quadrature de degre d’exactitude maximum

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Trudy Matematicheskogo Instituta imeni V.A. Steklova

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Steklov Mathematical Institute of RAS

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0371-9685 

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3034-1809 

published also as T. Popoviciu, Sur certaines formules de quadrature de degré d’exactitude maximum, Proc. Steklov Inst. Math., 134 (1975), 289-294 (1977) (in French)

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1975 a -Popoviciu- Trudy Mat. Inst. Steklov - On certain quadrature formulas of degree of exactness
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T. POPOVICIU

ON CERTAIN QUADRATURE FORMULAE OF MAXIMUM DEGREE OF ACCURACY

  1. Consider the quadrature formula
(1) ∫ − 1 1 f ( x ) d x = ∑ i = 1 p ∑ I = 0 k i − 1 c i , I f ( I ) ( z i ) + R ( f ) (1) ∫ − 1 1   f ( x ) d x = ∑ i = 1 p   ∑ I = 0 k i − 1   c i , I f ( I ) z i + R ( f ) {:(1)int_(-1)^(1)f(x)dx=sum_(i=1)^(p)sum_(j=0)^(k_(i)-1)c_(i,j)f^((j))(z_(i))+R(f):}\begin{equation*} \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) \tag{1} \end{equation*}(1)∫−11f(x)dx=∑i=1p∑I=0ki−1ci,If(I)(zi)+R(f)
Or z 1 , z 2 , … , z p z 1 , z 2 , … , z p z_(1),z_(2),dots,z_(p)z_{1}, z_{2}, \ldots, z_{p}z1,z2,…,zpare p ( > 0 ) p ( > 0 ) p( > 0)p(>0)p(>0)distinct points of the real axis. These are the nodes of the formula and k 1 , k 2 , … , k p k 1 , k 2 , … , k p k_(1),k_(2),dots,k_(p)k_{1}, k_{2}, \ldots, k_{p}k1,k2,…,kpare p p pppnatural numbers, the orders of multiplicity of the knots z 1 , z 2 , … , z p z 1 , z 2 , … , z p z_(1),z_(2),dots,z_(p)z_{1}, z_{2}, \ldots, z_{p}z1,z2,…,zprespective. We can assume that k i k i k_(i)k_{i}kinodes are merged in the point z i z i z_(i)z_{i}zi, so that z i z i z_(i)z_{i}ziis an order node k i k i k_(i)k_{i}kiof multiplicity. The total number of distinct or non-distinct nodes is equal to k 1 + k 2 + … … + k p = m k 1 + k 2 + … … + k p = m k_(1)+k_(2)+dots dots+k_(p)=mk_{1}+k_{2}+\ldots \ldots+k_{p}=mk1+k2+……+kp=m, Or m m mmmis therefore a natural number such that m ⩾ p ⩾ 1 m ⩾ p ⩾ 1 m >= p >= 1m \geqslant p \geqslant 1m⩾p⩾1.
Regarding the function f f fffwe can assume that it is defined and continuous on an interval containing the points − 1 , 1 − 1 , 1 -1.1-1.1−1,1and the knots z i , i = 1 z i , i = 1 z_(i),i=1z_{i}, i=1zi,i=1, 2 , … , p 2 , … , p 2, dots, p2, \ldots, p2,…,p. Moreover, f f fffadmits a certain number of derivatives, so that the second member of formula (1), where the accents signify successive derivations, has a meaning. c i , I c i , I c_(i,j)c_{i, j}ci,Iare independent coefficients of the function f f fff, the coefficients of formula (1) and R ( f ) R ( f ) R(f)R(f)R(f)is the remainder of the quadrature formula considered. Many authors, including ourselves, have studied [3, 4] the structure of this remainder.
2. Formula (1), or the remainder R ( f ) R ( f ) R(f)R(f)R(f)of this formula, have a degree of accuracy. It is the whole number n ⩾ − 1 n ⩾ − 1 n >= -1n \geqslant-1n⩾−1completely determined by the condition that R ( P ) R ( P ) R(P)R(P)R(P)be zero on any polynomial P P PPPof degree n ∗ n ∗ n^(**)n^{*}n∗and that R ( x n + 1 ) ≠≠ 0 R x n + 1 ≠≠ 0 R(x^(n+1))≠≠0R\left(x^{n+1}\right) \neq \neq 0R(xn+1)≠≠0. We also say that formula (1), or the remainder R ( f ) R ( f ) R(f)R(f)R(f)of this formula, is of degree of accuracy n . R ( f ) n . R ( f ) nR(f)n . R(f)n.R(f)is, moreover, like the other two terms of formula (1), a linear functional (additive and homogeneous) defined on a certain linear set of functions f f fff. In the following we can always assume n ⩾ 0 n ⩾ 0 n >= 0n \geqslant 0n⩾0, the case n = − 1 n = − 1 n=-1n=-1n=−1not intervening. The condition n ⩾ 0 n ⩾ 0 n >= 0n \geqslant 0n⩾0is, moreover, equivalent to equality 2 = ∑ i = 0 p c i , 0 2 = ∑ i = 0 p   c i , 0 2=sum_(i=0)^(p)c_(i,0)2=\sum_{i=0}^{p} c_{i, 0}2=∑i=0pci,0.
3. The knots z i z i z_(i)z_{i}ziand their orders of multiplicity k i , i = 1 , 2 , … , p k i , i = 1 , 2 , … , p k_(i),i=1,2,dots,pk_{i}, i=1,2, \ldots, pki,i=1,2,…,p, given, but any, the condition n ⩾ m − 1 n ⩾ m − 1 n >= m-1n \geqslant m-1n⩾m−1completely determines the quadrature formula (1), so the coefficients c i , I , I = 0 , 1 , … , k i − 1 c i , I , I = 0 , 1 , … , k i − 1 c_(i,j),j=0,1,dots,k_(i)-1c_{i, j}, j=0.1, \ldots, k_{i}-1ci,I,I=0,1,…,ki−1,
i = 1 , 2 , … , p i = 1 , 2 , … , p i=1,2,dots,pi=1,2, \ldots, pi=1,2,…,p, from this formula. We can then easily obtain these coefficients, by applying formula (1) to the Lag-range-Hermite interpolation polynomial of degree m − 1 m − 1 m-1m-1m−1of the function f f fffcorresponding to the nodes z i z i z_(i)z_{i}zi, counted with their respective orders of multiplicity. It is useless to write here this polynomial and the values ​​of the coefficients c i , I q u ′ c i , I q u ′ c_(i,j)qu^(')c_{i, j} qu^{\prime}ci,Iqu′we deduce from this. In some particular cases considered later we will give their explicit values.
In the following we will always assume that it is such a quadrature formula (1).
4. The degree of accuracy of the formula is then equal to n = m + q − 1 n = m + q − 1 n=m+q-1n=m+q-1n=m+q−1, where the non-negative integer q q qqqis characterized by the property that the polynomial
(2) L ( x ) = ∏ i = 1 p ( x − z i ) k i (2) L ( x ) = ∏ i = 1 p   x − z i k i {:(2)l(x)=prod_(i=1)^(p)(x-z_(i))^(k_(i)):}\begin{equation*} l(x)=\prod_{i=1}^{p}\left(x-z_{i}\right)^{k_{i}} \tag{2} \end{equation*}(2)L(x)=∏i=1p(x−zi)ki
is orthogonal to any polynomial of degree q − 1 q − 1 q-1q-1q−1, so that
(3) ∫ − 1 1 Q ( x ) L ( x ) d x = 0 (3) ∫ − 1 1   Q ( x ) L ( x ) d x = 0 {:(3)int_(-1)^(1)Q(x)l(x)dx=0:}\begin{equation*} \int_{-1}^{1} Q(x) l(x) dx=0 \tag{3} \end{equation*}(3)∫−11Q(x)L(x)dx=0
for any polynomial Q Q QQQof degree q − 1 q − 1 q-1q-1q−1and that
∫ − 1 1 x q L ( x ) d x ≠ 0 ∫ − 1 1   x q L ( x ) d x ≠ 0 int_(-1)^(1)x^(q)l(x)dx!=0\int_{-1}^{1} x^{q} l(x) dx \neq 0∫−11xqL(x)dx≠0
For the demonstration see, for example, one of our cited works [4]. Let us retain the property expressed by Lemma
1. If the quadrature formula (1) is of degree of exactness n == m + q − 1 n == m + q − 1 n==m+q-1n= =m+q-1n==m+q−1, Or q ⩾ 0 q ⩾ 0 q >= 0q \geqslant 0q⩾0, the polynomial (2) changes sign, inside the interval [ − 1 , 1 ] [ − 1 , 1 ] [-1,1][-1,1][−1,1], at least q q qqqtimes.
This is an immediate consequence of orthogonality.
5. We can seek to impose on the nodes z i z i z_(i)z_{i}ziadditional conditions to obtain the highest possible order of accuracy. The non-negative integer q q qqqis at most equal to p p ppp[4] and the maximum p p pppof q q qqqcan be achieved. We will then say (so if q = p q = p q=pq=pq=p) that the quadrature formula (1) is of maximum degree of accuracy or that it is of the Gauss type. Such formulas have been studied by several authors among whom we point out L. Tchakaloff [5], P. Turán [6], A. Ossicini [1, 2], we have considered them in our cited works [3, 4]. We note, in passing, that we have also considered analogous problems for linear functionals more general than the first member of formula (1).
Let us retain, in particular, the property expressed by
Theorem 1. For there to exist at least one Gauss-type formula, of the form indicated above, it is necessary and sufficient that the orders of multiplicity k 1 , k 2 , … , k p k 1 , k 2 , … , k p k_(1),k_(2),dots,k_(p)k_{1}, k_{2}, \ldots, k_{p}k1,k2,…,kp, assumed to be given, nodes z 1 , z 2 , … , z p z 1 , z 2 , … , z p z_(1),z_(2),dots,z_(p)z_{1}, z_{2}, \ldots, z_{p}z1,z2,…,zpare all odd.
We can summarize the proof as follows:
The condition is necessary according to Lemma 1.
The condition is also sufficient. Indeed, if the numbers k i k i k_(i)k_{i}kiare all odd, the polynomial of p p pppvariables z 1 , z 2 , … , z p z 1 , z 2 , … , z p z_(1),z_(2),dots,z_(p)z_{1}, z_{2}, \ldots, z_{p}z1,z2,…,zp,
Φ ( z 1 , z 2 , … , z p ) = ∫ − 1 1 L ( x ) ∏ i = 1 p ( x − z i ) d x Φ z 1 , z 2 , … , z p = ∫ − 1 1   L ( x ) ∏ i = 1 p   x − z i d x Phi(z_(1),z_(2),dots,z_(p))=int_(-1)^(1)l(x)prod_(i=1)^(p)(x-z_(i))dx\Phi\left(z_{1}, z_{2}, \ldots, z_{p}\right)=\int_{-1}^{1} l(x) \prod_{i=1}^{p}\left(x-z_{i}\right) dxΦ(z1,z2,…,zp)=∫−11L(x)∏i=1p(x−zi)dx
is positive over the entire space at p p pppdimensions. It therefore has a positive minimum which is reached at a relative minimum point. We can easily see that for an (absolute) minimum point ( z 1 , z 2 , … , z p z 1 , z 2 , … , z p z_(1),z_(2),dots,z_(p)z_{1}, z_{2}, \ldots, z_{p}z1,z2,…,zp) the coordinates z 1 , z 2 , … , z p z 1 , z 2 , … , z p z_(1),z_(2),dots,z_(p)z_{1}, z_{2}, \ldots, z_{p}z1,z2,…,zpare distinct.
On such a point we have
(4) ∂ Φ ∂ z i = 0 , i = 1 , 2 , … , p . (4) ∂ Φ ∂ z i = 0 , i = 1 , 2 , … , p . {:(4)(del Phi)/(delz_(i))=0","quad i=1","2","dots","p.:}\begin{equation*} \frac{\partial \Phi}{\partial z_{i}}=0, \quad i=1,2, \ldots, p . \tag{4} \end{equation*}(4)∂Φ∂zi=0,i=1,2,…,p.
It is then sufficient to note that the system (4) is equivalent to the orthogonality of the polynomial (2) with any polynomial of degree p − 1 p − 1 p-1p-1p−1.
6. It follows from the above that the numbers k i , i = 1 , 2 , … , p k i , i = 1 , 2 , … , p k_(i),i=1,2,dots,pk_{i}, i=1,2, \ldots, pki,i=1,2,…,pbeing assumed always odd, in any formula (1) of the Gauss type, the nodes z 1 , z 2 , … , z p z 1 , z 2 , … , z p z_(1),z_(2),dots,z_(p)z_{1}, z_{2}, \ldots, z_{p}z1,z2,…,zpare distinct and within the interval [ − 1 , 1 ] [ − 1 , 1 ] [-1,1][-1,1][−1,1].
But, to my knowledge, it is not yet known what the exact number of distinct Gauss-type formulas is, for a given system of (odd) multiplicity orders. It seems that this number is equal to the number of distinct permutations of the numbers k 1 , k 2 , … , k p k 1 , k 2 , … , k p k_(1),k_(2),dots,k_(p)k_{1}, k_{2}, \ldots, k_{p}k1,k2,…,kp, to any permutation corresponding to a well-determined Gauss-type formula. This is true if k 1 = k 2 == … = k p k 1 = k 2 == … = k p k_(1)=k_(2)==dots=k_(p)k_{1}=k_{2}= =\ldots=k_{p}k1=k2==…=kp. There is then only one Gauss-type formula.
Two formulas are, moreover, distinct if and only if the corresponding polynomials (2) are different.
The aim of this little work is to solve this problem completely in some special cases, different from the one where the orders of multiplicity are all equal. We thus complete some of our previous results [3].
7. Consider the case p = 2 , k 1 = k , k 2 = 1 p = 2 , k 1 = k , k 2 = 1 p=2,k_(1)=k,k_(2)=1p=2, k_{1}=k, k_{2}=1p=2,k1=k,k2=1. The case k = 3 k = 3 k=3k=3k=3was considered previously [3]. Now suppose that k k kkkbe any odd. The nodes z 1 , z 2 z 1 , z 2 z_(1),z_(2)z_{1}, z_{2}z1,z2are then given by the system
∫ − 1 1 ( x − z 1 ) k + i ( x − z 2 ) d x = 0 , i = 0 , 1 ∫ − 1 1   x − z 1 k + i x − z 2 d x = 0 , i = 0 , 1 int_(-1)^(1)(x-z_(1))^(k+i)(x-z_(2))dx=0,quad i=0,1\int_{-1}^{1}\left(x-z_{1}\right)^{k+i}\left(x-z_{2}\right) d x=0, \quad i=0,1∫−11(x−z1)k+i(x−z2)dx=0,i=0,1
If we pose
(5) 1 − z 1 1 + z 1 = y (5) 1 − z 1 1 + z 1 = y {:(5)(1-z_(1))/(1+z_(1))=y:}\begin{equation*} \frac{1-z_{1}}{1+z_{1}}=y \tag{5} \end{equation*}(5)1−z11+z1=y
the system can be written
y k + 2 + 1 k + 2 + z 1 − z 2 1 + z 1 ⋅ y k + 1 − 1 k + 1 = 0 , (6) y k + 3 − 1 k + 3 + z 1 − z 2 1 + z 1 ⋅ y k + 2 + 1 k + 2 = 0 . y k + 2 + 1 k + 2 + z 1 − z 2 1 + z 1 ⋅ y k + 1 − 1 k + 1 = 0 , (6) y k + 3 − 1 k + 3 + z 1 − z 2 1 + z 1 ⋅ y k + 2 + 1 k + 2 = 0 . {:[(y^(k+2)+1)/(k+2)+(z_(1)-z_(2))/(1+z_(1))*(y^(k+1)-1)/(k+1)=0","],[(6)(y^(k+3)-1)/(k+3)+(z_(1)-z_(2))/(1+z_(1))*(y^(k+2)+1)/(k+2)=0.]:}\begin{align*} & \frac{y^{k+2}+1}{k+2}+\frac{z_{1}-z_{2}}{1+z_{1}} \cdot \frac{y^{k+1}-1}{k+1}=0, \\ & \frac{y^{k+3}-1}{k+3}+\frac{z_{1}-z_{2}}{1+z_{1}} \cdot \frac{y^{k+2}+1}{k+2}=0 . \tag{6} \end{align*}yk+2+1k+2+z1−z21+z1⋅yk+1−1k+1=0,(6)yk+3−1k+3+z1−z21+z1⋅yk+2+1k+2=0.
By eliminating z 2 z 2 z_(2)z_{2}z2, the knot z 1 z 1 z_(1)z_{1}z1is given by (5) and by the reciprocal equation in y y yyy
( k + 2 ) 2 ( y k + 1 − 1 ) ( y k + 3 − 1 ) − ( k + 1 ) ( k + 3 ) ( y k + 2 + 1 ) 2 = (7) = y 2 k + 4 − ( k + 2 ) 2 y k + 3 − 2 ( k + 1 ) ( k + 3 ) y k + 2 − ( k + 2 ) 2 y k + 1 + 1 = 0 ( k + 2 ) 2 y k + 1 − 1 y k + 3 − 1 − ( k + 1 ) ( k + 3 ) y k + 2 + 1 2 = (7) = y 2 k + 4 − ( k + 2 ) 2 y k + 3 − 2 ( k + 1 ) ( k + 3 ) y k + 2 − ( k + 2 ) 2 y k + 1 + 1 = 0 {:[(k+2)^(2)(y^(k+1)-1)(y^(k+3)-1)-(k+1)(k+3)(y^(k+2)+1)^(2)=],[(7)quad=y^(2k+4)-(k+2)^(2)y^(k+3)-2(k+1)(k+3)y^(k+2)-(k+2)^(2)y^(k+1)+1=0]:}\begin{align*} & (k+2)^{2}\left(y^{k+1}-1\right)\left(y^{k+3}-1\right)-(k+1)(k+3)\left(y^{k+2}+1\right)^{2}= \\ & \quad=y^{2 k+4}-(k+2)^{2} y^{k+3}-2(k+1)(k+3) y^{k+2}-(k+2)^{2} y^{k+1}+1=0 \tag{7} \end{align*}(k+2)2(yk+1−1)(yk+3−1)−(k+1)(k+3)(yk+2+1)2=(7)=y2k+4−(k+2)2yk+3−2(k+1)(k+3)yk+2−(k+2)2yk+1+1=0
which has exactly 2 positive and distinct real roots (separated by 1).
It follows that in the case studied we have exactly two Gauss-type formulas which are confused for k = 1 k = 1 k=1k=1k=1in the classical Gauss formula, but are quite distinct for k > 1 k > 1 k > 1k>1k>1. Indeed, the knot z 2 z 2 z_(2)z_{2}z2is obtained from one of the equations (6) and the preceding theory shows us that it will be completely determined, different from z 1 z 1 z_(1)z_{1}z1and strictly between -1 and 1.
For k = 3 k = 3 k=3k=3k=3we have already given the explicit value of z 1 z 1 z_(1)z_{1}z1(so also of z 2 z 2 z_(2)z_{2}z2) and the corresponding formula (1). z 1 z 1 z_(1)z_{1}z1And z 2 z 2 z_(2)z_{2}z2are then quadratic irrationalities. For k > 3 k > 3 k > 3k>3k>3the arithmetic nature of z 1 z 1 z_(1)z_{1}z1And z 2 z 2 z_(2)z_{2}z2seems to be more complicated.
8. To calculate z 1 z 1 z_(1)z_{1}z1we must first solve the reciprocal equation (7). To do this we set
(8) y + 1 y = t . (8) y + 1 y = t . {:(8)y+(1)/(y)=t.:}\begin{equation*} y+\frac{1}{y}=t . \tag{8} \end{equation*}(8)y+1y=t.
SO y s + 1 / y s y s + 1 / y s y^(s)+1//y^(s)y^{s}+1 / y^{s}ys+1/ysis equal to a polynomial P s ( t ) P s ( t ) P_(s)(t)P_{s}(t)Ps(t)in t t ttt, of degree s s sss, given by the recurrence relation
P s + 1 = t P s − P s − 1 , s = 1 , 2 , … ( P 0 = 2 , P 1 = t ) P s + 1 = t P s − P s − 1 , s = 1 , 2 , … P 0 = 2 , P 1 = t P_(s+1)=tP_(s)-P_(s-1),s=1,2,dotsquad(P_(0)=2,P_(1)=t)P_{s+1}=t P_{s}-P_{s-1}, s=1,2, \ldots \quad\left(P_{0}=2, P_{1}=t\right)Ps+1=tPs−Ps−1,s=1,2,…(P0=2,P1=t)
The equation in t t tttis therefore
(9) P k + 2 ( t ) − ( k + 2 ) 2 t − 2 ( k + 1 ) ( k + 3 ) = 0 (9) P k + 2 ( t ) − ( k + 2 ) 2 t − 2 ( k + 1 ) ( k + 3 ) = 0 {:(9)P_(k+2)(t)-(k+2)^(2)t-2(k+1)(k+3)=0:}\begin{equation*} P_{k+2}(t)-(k+2)^{2} t-2(k+1)(k+3)=0 \tag{9} \end{equation*}(9)Pk+2(t)−(k+2)2t−2(k+1)(k+3)=0
From the above it follows that this equation in t t ttthas a real root > 2 > 2 > 2>2>2. We easily check that it also has the double root -2.
For k = 1 k = 1 k=1k=1k=1And k = 3 k = 3 k=3k=3k=3the first member of (9) is
t 3 − 12 t − 16 = ( t + 2 ) 2 ( t − 4 ) t 5 − 5 t 3 + 5 t 2 − 25 t − 48 = ( t + 2 ) 2 ( t − 3 ) ( t 2 − t + 4 ) t 3 − 12 t − 16 = ( t + 2 ) 2 ( t − 4 ) t 5 − 5 t 3 + 5 t 2 − 25 t − 48 = ( t + 2 ) 2 ( t − 3 ) t 2 − t + 4 {:[t^(3)-12 t-16=(t+2)^(2)(t-4)],[t^(5)-5t^(3)+5t^(2)-25 t-48=(t+2)^(2)(t-3)(t^(2)-t+4)]:}\begin{aligned} t^{3}-12 t-16 & =(t+2)^{2}(t-4) \\ t^{5}-5 t^{3}+5 t^{2}-25 t-48 & =(t+2)^{2}(t-3)\left(t^{2}-t+4\right) \end{aligned}t3−12t−16=(t+2)2(t−4)t5−5t3+5t2−25t−48=(t+2)2(t−3)(t2−t+4)
respectively, which gives the well-known reported results.
For k = 5 k = 5 k=5k=5k=5equation (9) becomes
t 7 − 7 t 5 + 14 t 3 − 56 t − 96 = ( t + 2 ) 2 ( t 5 − 4 t 4 + 5 t 3 − 4 t 2 + 10 t − 24 ) = 0 t 7 − 7 t 5 + 14 t 3 − 56 t − 96 = ( t + 2 ) 2 t 5 − 4 t 4 + 5 t 3 − 4 t 2 + 10 t − 24 = 0 t^(7)-7t^(5)+14t^(3)-56 t-96=(t+2)^(2)(t^(5)-4t^(4)+5t^(3)-4t^(2)+10 t-24)=0t^{7}-7 t^{5}+14 t^{3}-56 t-96=(t+2)^{2}\left(t^{5}-4 t^{4}+5 t^{3}-4 t^{2}+10 t-24\right)=0t7−7t5+14t3−56t−96=(t+2)2(t5−4t4+5t3−4t2+10t−24)=0
and we check that this equation no longer has any positive rational roots.
9. Let us again consider the case p = 3 , k 1 = k , k 2 = k 3 = 1 p = 3 , k 1 = k , k 2 = k 3 = 1 p=3,k_(1)=k,k_(2)=k_(3)=1p=3, k_{1}=k, k_{2}=k_{3}=1p=3,k1=k,k2=k3=1. The case k = 3 k = 3 k=3k=3k=3has already been considered previously [3]. Suppose that k k kkkbe any odd. . The nodes z 1 , z 2 , z 3 z 1 , z 2 , z 3 z_(1),z_(2),z_(3)z_{1}, z_{2}, z_{3}z1,z2,z3are given by the system
(10) ∫ − 1 1 ( x − z 1 ) k + i ( x − z 2 ) ( x − z 3 ) d x = 0 , i = 0 , 1 , 2 (10) ∫ − 1 1   x − z 1 k + i x − z 2 x − z 3 d x = 0 , i = 0 , 1 , 2 {:(10)int_(-1)^(1)(x-z_(1))^(k+i)(x-z_(2))(x-z_(3))dx=0","quad i=0","1","2:}\begin{equation*} \int_{-1}^{1}\left(x-z_{1}\right)^{k+i}\left(x-z_{2}\right)\left(x-z_{3}\right) d x=0, \quad i=0,1,2 \tag{10} \end{equation*}(10)∫−11(x−z1)k+i(x−z2)(x−z3)dx=0,i=0,1,2
By writing (5) and eliminating z 2 z 2 z_(2)z_{2}z2And z 3 z 3 z_(3)z_{3}z3, we find that y y yyyis given by the equation
| y k + 3 − 1 k + 3 y k + 2 + 1 k + 2 y k + 1 − 1 k + 1 y k + 4 + 1 k + 4 y k + 3 − 1 k + 3 y k + 2 + 1 k + 2 y k + 5 − 1 k + 5 y k + 4 + 1 k + 4 y k + 3 − 1 k + 3 | = 0 y k + 3 − 1 k + 3 y k + 2 + 1 k + 2 y k + 1 − 1 k + 1 y k + 4 + 1 k + 4 y k + 3 − 1 k + 3 y k + 2 + 1 k + 2 y k + 5 − 1 k + 5 y k + 4 + 1 k + 4 y k + 3 − 1 k + 3 = 0 |[(y^(k+3)-1)/(k+3),(y^(k+2)+1)/(k+2),(y^(k+1)-1)/(k+1)],[(y^(k+4)+1)/(k+4),(y^(k+3)-1)/(k+3),(y^(k+2)+1)/(k+2)],[(y^(k+5)-1)/(k+5),(y^(k+4)+1)/(k+4),(y^(k+3)-1)/(k+3)]|=0\left|\begin{array}{ccc} \frac{y^{k+3}-1}{k+3} & \frac{y^{k+2}+1}{k+2} & \frac{y^{k+1}-1}{k+1} \\ \frac{y^{k+4}+1}{k+4} & \frac{y^{k+3}-1}{k+3} & \frac{y^{k+2}+1}{k+2} \\ \frac{y^{k+5}-1}{k+5} & \frac{y^{k+4}+1}{k+4} & \frac{y^{k+3}-1}{k+3} \end{array}\right|=0|yk+3−1k+3yk+2+1k+2yk+1−1k+1yk+4+1k+4yk+3−1k+3yk+2+1k+2yk+5−1k+5yk+4+1k+4yk+3−1k+3|=0
or, by doing the calculations,
4 y 3 k + 9 − A y 2 k + 8 − B y 2 k + 7 − C y 2 k + 6 − D y 2 k + 5 − E y 2 k + 4 + (11) + E y k + 5 + D y k + 4 + C y k + 3 + B y k + 2 + A y k + 1 − 4 = 0 4 y 3 k + 9 − A y 2 k + 8 − B y 2 k + 7 − C y 2 k + 6 − D y 2 k + 5 − E y 2 k + 4 + (11) + E y k + 5 + D y k + 4 + C y k + 3 + B y k + 2 + A y k + 1 − 4 = 0 {:[4y^(3k+9)-Ay^(2k+8)-By^(2k+7)-Cy^(2k+6)-Dy^(2k+5)-Ey^(2k+4)+],[(11)quad+Ey^(k+5)+Dy^(k+4)+Cy^(k+3)+By^(k+2)+Ay^(k+1)-4=0]:}\begin{align*} & 4 y^{3 k+9}-A y^{2 k+8}-B y^{2 k+7}-C y^{2 k+6}-D y^{2 k+5}-E y^{2 k+4}+ \\ & \quad+E y^{k+5}+D y^{k+4}+C y^{k+3}+B y^{k+2}+A y^{k+1}-4=0 \tag{11} \end{align*}4y3k+9−HASy2k+8−By2k+7−Cy2k+6−Dy2k+5−Ey2k+4+(11)+Eyk+5+Dyk+4+Cyk+3+Byk+2+HASyk+1−4=0
with
A = ( k + 2 ) 2 ( k + 3 ) 2 , B = 4 ( k + 1 ) ( k + 3 ) 2 ( k + 4 ) C = 6 ( k + 2 ) ( k + 4 ) ( k 2 + 6 k + 7 ) , D = 4 ( k + 2 ) ( k + 3 ) 2 ( k + 5 ) E = ( k + 3 ) 2 ( k + 4 ) 2 A = ( k + 2 ) 2 ( k + 3 ) 2 ,      B = 4 ( k + 1 ) ( k + 3 ) 2 ( k + 4 ) C = 6 ( k + 2 ) ( k + 4 ) k 2 + 6 k + 7 ,      D = 4 ( k + 2 ) ( k + 3 ) 2 ( k + 5 ) E = ( k + 3 ) 2 ( k + 4 ) 2      {:[A=(k+2)^(2)(k+3)^(2)",",B=4(k+1)(k+3)^(2)(k+4)],[C=6(k+2)(k+4)(k^(2)+6k+7)",",D=4(k+2)(k+3)^(2)(k+5)],[E=(k+3)^(2)(k+4)^(2),]:}\begin{array}{ll} A=(k+2)^{2}(k+3)^{2}, & B=4(k+1)(k+3)^{2}(k+4) \\ C=6(k+2)(k+4)\left(k^{2}+6 k+7\right), & D=4(k+2)(k+3)^{2}(k+5) \\ E=(k+3)^{2}(k+4)^{2} & \end{array}HAS=(k+2)2(k+3)2,B=4(k+1)(k+3)2(k+4)C=6(k+2)(k+4)(k2+6k+7),D=4(k+2)(k+3)2(k+5)E=(k+3)2(k+4)2
Note that A , B , C , D , E A , B , C , D , E A,B,C,D,EA, B, C, D, EHAS,B,C,D,Eare positive.
If we designate by φ ( y ) φ ( y ) varphi(y)\varphi(y)φ(y)the first member of the equation, we have φ ( 0 ) < 0 φ ( 0 ) < 0 varphi(0) < 0\varphi(0)<0φ(0)<0And φ ′ ( 1 ) = 12 ( k + 3 ) − ( k + 7 ) A − ( k + 5 ) B − ( k + 3 ) C − ( k + 1 ) D − φ ′ ( 1 ) = 12 ( k + 3 ) − ( k + 7 ) A − ( k + 5 ) B − ( k + 3 ) C − ( k + 1 ) D − varphi^(')(1)=12(k+3)-(k+7)A-(k+5)B-(k+3)C-(k+1)D-\varphi^{\prime}(1)=12(k+3)-(k+7) A-(k+5) B-(k+3) C-(k+1) D-φ′(1)=12(k+3)−(k+7)HAS−(k+5)B−(k+3)C−(k+1)D−- ( k − 1 ) E < 0 ( k − 1 ) E < 0 (k-1)E < 0(k-1) E<0(k−1)E<0. Descartes' rule of signs then shows us that the reciprocal equation (11) has exactly 3 real and positive roots which are distinct, one of which is equal to 1.
From orthogonality and Lemma 1 it follows that for a z 1 z 1 z_(1)z_{1}z1determined the polynomial ( x − z 2 ) ( x − z 3 ) x − z 2 x − z 3 (x-z_(2))(x-z_(3))\left(x-z_{2}\right)\left(x-z_{3}\right)(x−z2)(x−z3)is determined completely, z 2 , z 3 z 2 , z 3 z_(2),z_(3)z_{2}, z_{3}z2,z3being distinct, strictly between -1 and 1 and being different from z 1 z 1 z_(1)z_{1}z1.
In this case we therefore have exactly 3 Gauss-type formulas. If k = 1 k = 1 k=1k=1k=1These three formulas are confused with the classical Gauss formula with 3 knots. But if k > 1 k > 1 k > 1k>1k>1The three formulas are distinct. One corresponds to the value 0 of z 1 z 1 z_(1)z_{1}z1and the other two with two values ​​of z 1 z 1 z_(1)z_{1}z1symmetrical about the origin.
Note that in this case the number of distinct permutations of the numbers k 1 , k 2 , k 3 k 1 , k 2 , k 3 k_(1),k_(2),k_(3)k_{1}, k_{2}, k_{3}k1,k2,k3is indeed equal to 3.
10. Still in the particular case considered, the Gauss type formula corresponding to the node z 1 = 0 z 1 = 0 z_(1)=0z_{1}=0z1=0can be obtained explicitly in a fairly simple form.
By posing z 1 = 0 z 1 = 0 z_(1)=0z_{1}=0z1=0in system (10) we obtain ( x − z 2 ) ( x − z 3 ) = x 2 − λ 2 x − z 2 x − z 3 = x 2 − λ 2 (x-z_(2))(x-z_(3))=x^(2)-lambda^(2)\left(x-z_{2}\right)\left(x-z_{3}\right)=x^{2}-\lambda^{2}(x−z2)(x−z3)=x2−λ2, Or λ = ( k + 2 ) / ( k + 4 ) λ = ( k + 2 ) / ( k + 4 ) lambda=sqrt((k+2)//(k+4))\lambda=\sqrt{(k+2) /(k+4)}λ=(k+2)/(k+4). So we have z 2 = ( k + 2 ) / ( k + 4 ) , z 3 == − ( k + 2 ) / ( k + 4 ) z 2 = ( k + 2 ) / ( k + 4 ) , z 3 == − ( k + 2 ) / ( k + 4 ) quadz_(2)=sqrt((k+2)//(k+4)),quadz_(3)==-sqrt((k+2)//(k+4))\quad z_{2}=\sqrt{(k+2) /(k+4)}, \quad z_{3}= =-\sqrt{(k+2) /(k+4)}z2=(k+2)/(k+4),z3==−(k+2)/(k+4).
Let us write the quadrature formula in the form
∫ − 1 1 f ( x ) d x = ∑ j = 0 k − 1 c j f ( j ) ( 0 ) + c f ( λ ) + d f ( − λ ) + R ( f ) . ∫ − 1 1   f ( x ) d x = ∑ j = 0 k − 1   c j f ( j ) ( 0 ) + c f ( λ ) + d f ( − λ ) + R ( f ) . int_(-1)^(1)f(x)dx=sum_(j=0)^(k-1)c_(j)f^((j))(0)+cf(lambda)+df(-lambda)+R(f).\int_{-1}^{1} f(x) d x=\sum_{j=0}^{k-1} c_{j} f^{(j)}(0)+c f(\lambda)+d f(-\lambda)+R(f) .∫−11f(x)dx=∑I=0k−1cIf(I)(0)+cf(λ)+df(−λ)+R(f).
The formula is of degree k + 4 k + 4 k+4k+4k+4of accuracy.
To find the coefficients c c ccc, d d dddlet us take successively f = x k , f = x k + 1 f = x k , f = x k + 1 f=x^(k),f=x^(k+1)f=x^{k}, f=x^{k+1}f=xk,f=xk+1. Considering that k ¯ k ¯ bar(k)\bar{k}k¯is odd, we deduce
(12) c = d = 1 ( k + 2 ) λ k + 1 . (12) c = d = 1 ( k + 2 ) λ k + 1 . {:(12)c=d=(1)/((k+2)lambda^(k+1)).:}\begin{equation*} c=d=\frac{1}{(k+2) \lambda^{k+1}} . \tag{12} \end{equation*}(12)c=d=1(k+2)λk+1.
By taking f = x i f = x i f=x^(i)f=x^{i}f=xiOr i i iiiis odd we find ( k > 1 k > 1 k > 1k>1k>1)
(13) c 1 = c 3 = … = c k − 2 = 0 . (13) c 1 = c 3 = … = c k − 2 = 0 . {:(13)c_(1)=c_(3)=dots=c_(k-2)=0.:}\begin{equation*} c_{1}=c_{3}=\ldots=c_{k-2}=0 . \tag{13} \end{equation*}(13)c1=c3=…=ck−2=0.
Finally if we take f = x 2 j ( x k − 2 j + 1 − λ k − 2 j + 1 ) , j = 0 , 1 , … , k − 1 2 f = x 2 j x k − 2 j + 1 − λ k − 2 j + 1 , j = 0 , 1 , … , k − 1 2 f=x^(2j)(x^(k-2j+1)-lambda^(k-2j+1)),j=0,1,dots,(k-1)/(2)f=x^{2 j}\left(x^{k-2 j+1}-\lambda^{k-2 j+1}\right), j=0,1, \ldots, \frac{k-1}{2}f=x2I(xk−2I+1−λk−2I+1),I=0,1,…,k−12, we get
(14) c 2 j = 2 ( 2 j + 1 ) ! − 2 ( 2 j ) ! ( k + 2 ) λ k − 2 j + 1 , j = 0 , 1 , … , k − 1 2 . (14) c 2 j = 2 ( 2 j + 1 ) ! − 2 ( 2 j ) ! ( k + 2 ) λ k − 2 j + 1 , j = 0 , 1 , … , k − 1 2 . {:(14)c_(2j)=(2)/((2j+1)!)-(2)/((2j)!(k+2)lambda^(k-2j+1))","quad j=0","1","dots","(k-1)/(2).:}\begin{equation*} c_{2 j}=\frac{2}{(2 j+1)!}-\frac{2}{(2 j)!(k+2) \lambda^{k-2 j+1}}, \quad j=0,1, \ldots, \frac{k-1}{2} . \tag{14} \end{equation*}(14)c2I=2(2I+1)!−2(2I)!(k+2)λk−2I+1,I=0,1,…,k−12.
The desired quadrature formula is therefore written as
(15) ∫ − 1 1 f ( x ) d x = ∑ j = 0 k − 1 2 ( 2 ( 2 j + 1 ) ! − 2 ( 2 j ) ! ( k + 2 ) λ k − 2 j + 1 ) f ( 2 j ) ( 0 ) + + 1 ( k + 2 ) λ k + 1 ( f ( λ ) + f ( − λ ) ) + R ( f ) (15) ∫ − 1 1   f ( x ) d x = ∑ j = 0 k − 1 2   2 ( 2 j + 1 ) ! − 2 ( 2 j ) ! ( k + 2 ) λ k − 2 j + 1 f ( 2 j ) ( 0 ) + + 1 ( k + 2 ) λ k + 1 ( f ( λ ) + f ( − λ ) ) + R ( f ) {:[(15)int_(-1)^(1)f(x)dx=sum_(j=0)^((k-1)/(2))((2)/((2j+1)!)-(2)/((2j)!(k+2)lambda^(k-2j+1)))f^((2j))(0)+],[quad+(1)/((k+2)lambda^(k+1))(f(lambda)+f(-lambda))+R(f)]:}\begin{align*} & \int_{-1}^{1} f(x) d x=\sum_{j=0}^{\frac{k-1}{2}}\left(\frac{2}{(2 j+1)!}-\frac{2}{(2 j)!(k+2) \lambda^{k-2 j+1}}\right) f^{(2 j)}(0)+ \tag{15}\\ & \quad+\frac{1}{(k+2) \lambda^{k+1}}(f(\lambda)+f(-\lambda))+R(f) \end{align*}(15)∫−11f(x)dx=∑I=0k−12(2(2I+1)!−2(2I)!(k+2)λk−2I+1)f(2I)(0)++1(k+2)λk+1(f(λ)+f(−λ))+R(f)
The rest R ( f ) R ( f ) R(f)R(f)R(f)is given by the formula (see [3]),
R ( f ) = R ( x k + 5 ) [ ξ 1 , ξ 2 , … , ξ k + f ; f ] R ( f ) = R x k + 5 ξ 1 , ξ 2 , … , ξ k + f ; f R(f)=R(x^(k+5))[xi_(1),xi_(2),dots,xi_(k+f);f]R(f)=R\left(x^{k+5}\right)\left[\xi_{1}, \xi_{2}, \ldots, \xi_{k+\mathrm{f}} ; f\right]R(f)=R(xk+5)[ξ1,ξ2,…,ξk+f;f]
Or ξ 1 , ξ 2 , … , ξ k + 6 ξ 1 , ξ 2 , … , ξ k + 6 xi_(1),xi_(2),dots,xi_(k+6)\xi_{1}, \xi_{2}, \ldots, \xi_{k+6}ξ1,ξ2,…,ξk+6are k + 6 k + 6 k+6k+6k+6distinct points inside the interval [ − 1 , 1 ] [ − 1 , 1 ] [-1,1][-1,1][−1,1]and which depend, in general, on the (continuous) function f ( x ) . [ ξ 1 , ξ 2 , … … , ξ k + 6 ; f ] f ( x ) . ξ 1 , ξ 2 , … … , ξ k + 6 ; f f(x).[xi_(1),xi_(2),dots:}{: dots,xi_(k+6);f]f(x) .\left[\xi_{1}, \xi_{2}, \ldots\right. \left.\ldots, \xi_{k+6} ; f\right]f(x).[ξ1,ξ2,……,ξk+6;f]is the divided difference of the function f f fffon the knots ξ i , i == 1 , 2 , … , k + 6 ξ i , i == 1 , 2 , … , k + 6 xi_(i),i==1,2,dots,k+6\xi_{i}, i= =1,2, \ldots, k+6ξi,i==1,2,…,k+6.
We can obtain the coefficient R ( x k + 5 ) R x k + 5 R(x^(k+5))R\left(x^{k+5}\right)R(xk+5)taking in formula (15) f = x k + 3 ( x 2 − λ 2 ) f = x k + 3 x 2 − λ 2 f=x^(k+3)(x^(2)-lambda^(2))f=x^{k+3}\left(x^{2}-\lambda^{2}\right)f=xk+3(x2−λ2). We thus obtain
R ( x k + 5 ) = R ( x k + 3 ( x 2 − λ 2 ) ) = 2 ( 1 k + 6 − λ 2 k + 4 ) = 8 ( k + 4 ) 2 ( k + 6 ) . R x k + 5 = R x k + 3 x 2 − λ 2 = 2 1 k + 6 − λ 2 k + 4 = 8 ( k + 4 ) 2 ( k + 6 ) . R(x^(k+5))=R(x^(k+3)(x^(2)-lambda^(2)))=2((1)/(k+6)-(lambda^(2))/(k+4))=(8)/((k+4)^(2)(k+6)).R\left(x^{k+5}\right)=R\left(x^{k+3}\left(x^{2}-\lambda^{2}\right)\right)=2\left(\frac{1}{k+6}-\frac{\lambda^{2}}{k+4}\right)=\frac{8}{(k+4)^{2}(k+6)} .R(xk+5)=R(xk+3(x2−λ2))=2(1k+6−λ2k+4)=8(k+4)2(k+6).

BIBLIOGRAPHY

  1. A. Ossicini. Construction of the Gaussian quadratura formula. - Ann. Mast. (IV), 1966, 72, 213-238.
  2. A. Ossicini. The function of influenza in the Gauss problem is based on the quadratura formula. Le Mathematiche, 1968, 23, 7-30.
  3. T. Popoviciu. Asupra uni generalizări a formulai de integrare numerică to him Gauss. Studii și Cercetări ştiinţ. Iasi. Anul 6 (1955), 29-57.
  4. T. Popoviciu. The simplicity of the remainder in certain quadrature formulas. - Mathematica, 1964, 6 (29), 157-184.
  5. L. Tchakaloff. Über eine allgemeine Quadraturformel. - Account returns. Acad. Bulgarian sci, 1948, t. 1, 9-12.
  6. P. Turàn. On the theory of the mechanical quadrature. - Acta Sci. Math., 1950, 12, 29-37.
    • A polynomial of degree g g gggis a polynomial of effective degree ⩽ g ⩽ g <= g\leqslant g⩽g. The unique polynomial of degree - 1 is the zero polynomial.
1975

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