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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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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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T. POPOVICIU
ON CERTAIN QUADRATURE FORMULAE OF MAXIMUM DEGREE OF ACCURACY
- Consider the quadrature formula
{:(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*}
Orz_(1),z_(2),dots,z_(p)z_{1}, z_{2}, \ldots, z_{p}arep( > 0)p(>0)distinct points of the real axis. These are the nodes of the formula andk_(1),k_(2),dots,k_(p)k_{1}, k_{2}, \ldots, k_{p}areppnatural numbers, the orders of multiplicity of the knotsz_(1),z_(2),dots,z_(p)z_{1}, z_{2}, \ldots, z_{p}respective. We can assume thatk_(i)k_{i}nodes are merged in the pointz_(i)z_{i}, so thatz_(i)z_{i}is an order nodek_(i)k_{i}of multiplicity. The total number of distinct or non-distinct nodes is equal tok_(1)+k_(2)+dots dots+k_(p)=mk_{1}+k_{2}+\ldots \ldots+k_{p}=m, Ormmis therefore a natural number such thatm >= p >= 1m \geqslant p \geqslant 1.
Regarding the functionffwe can assume that it is defined and continuous on an interval containing the points-1.1-1.1and the knotsz_(i),i=1z_{i}, i=1,2, dots, p2, \ldots, p. Moreover,ffadmits a certain number of derivatives, so that the second member of formula (1), where the accents signify successive derivations, has a meaning.c_(i,j)c_{i, j}are independent coefficients of the functionff, the coefficients of formula (1) andR(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 remainderR(f)R(f)of this formula, have a degree of accuracy. It is the whole numbern >= -1n \geqslant-1completely determined by the condition thatR(P)R(P)be zero on any polynomialPPof degreen^(**)n^{*}and thatR(x^(n+1))≠≠0R\left(x^{n+1}\right) \neq \neq 0. We also say that formula (1), or the remainderR(f)R(f)of this formula, is of degree of accuracynR(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 functionsff. In the following we can always assumen >= 0n \geqslant 0, the casen=-1n=-1not intervening. The conditionn >= 0n \geqslant 0is, moreover, equivalent to equality2=sum_(i=0)^(p)c_(i,0)2=\sum_{i=0}^{p} c_{i, 0}.
3. The knotsz_(i)z_{i}and their orders of multiplicityk_(i),i=1,2,dots,pk_{i}, i=1,2, \ldots, p, given, but any, the conditionn >= m-1n \geqslant m-1completely determines the quadrature formula (1), so the coefficientsc_(i,j),j=0,1,dots,k_(i)-1c_{i, j}, j=0.1, \ldots, k_{i}-1,
2. Formula (1), or the remainderR(f)R(f)of this formula, have a degree of accuracy. It is the whole numbern >= -1n \geqslant-1completely determined by the condition thatR(P)R(P)be zero on any polynomialPPof degreen^(**)n^{*}and thatR(x^(n+1))≠≠0R\left(x^{n+1}\right) \neq \neq 0. We also say that formula (1), or the remainderR(f)R(f)of this formula, is of degree of accuracynR(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 functionsff. In the following we can always assumen >= 0n \geqslant 0, the casen=-1n=-1not intervening. The conditionn >= 0n \geqslant 0is, moreover, equivalent to equality2=sum_(i=0)^(p)c_(i,0)2=\sum_{i=0}^{p} c_{i, 0}.
3. The knotsz_(i)z_{i}and their orders of multiplicityk_(i),i=1,2,dots,pk_{i}, i=1,2, \ldots, p, given, but any, the conditionn >= m-1n \geqslant m-1completely determines the quadrature formula (1), so the coefficientsc_(i,j),j=0,1,dots,k_(i)-1c_{i, j}, j=0.1, \ldots, k_{i}-1,
i=1,2,dots,pi=1,2, \ldots, p, from this formula. We can then easily obtain these coefficients, by applying formula (1) to the Lag-range-Hermite interpolation polynomial of degreem-1m-1of the functionffcorresponding to the nodesz_(i)z_{i}, counted with their respective orders of multiplicity. It is useless to write here this polynomial and the values ​​of the coefficientsc_(i,j)qu^(')c_{i, j} qu^{\prime}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 ton=m+q-1n=m+q-1, where the non-negative integerqqis characterized by the property that the polynomial
4. The degree of accuracy of the formula is then equal ton=m+q-1n=m+q-1, where the non-negative integerqqis characterized by the property that the polynomial
{:(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*}
is orthogonal to any polynomial of degreeq-1q-1, so that
{:(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*}
for any polynomialQQof degreeq-1q-1and that
int_(-1)^(1)x^(q)l(x)dx!=0\int_{-1}^{1} x^{q} l(x) dx \neq 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 exactnessn==m+q-1n= =m+q-1, Orq >= 0q \geqslant 0, the polynomial (2) changes sign, inside the interval[-1,1][-1,1], at leastqqtimes.
1. If the quadrature formula (1) is of degree of exactnessn==m+q-1n= =m+q-1, Orq >= 0q \geqslant 0, the polynomial (2) changes sign, inside the interval[-1,1][-1,1], at leastqqtimes.
This is an immediate consequence of orthogonality.
5. We can seek to impose on the nodesz_(i)z_{i}additional conditions to obtain the highest possible order of accuracy. The non-negative integerqqis at most equal topp[4] and the maximumppofqqcan be achieved. We will then say (so ifq=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).
5. We can seek to impose on the nodesz_(i)z_{i}additional conditions to obtain the highest possible order of accuracy. The non-negative integerqqis at most equal topp[4] and the maximumppofqqcan be achieved. We will then say (so ifq=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 multiplicityk_(1),k_(2),dots,k_(p)k_{1}, k_{2}, \ldots, k_{p}, assumed to be given, nodesz_(1),z_(2),dots,z_(p)z_{1}, z_{2}, \ldots, z_{p}are all odd.
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 multiplicityk_(1),k_(2),dots,k_(p)k_{1}, k_{2}, \ldots, k_{p}, assumed to be given, nodesz_(1),z_(2),dots,z_(p)z_{1}, z_{2}, \ldots, z_{p}are 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 numbersk_(i)k_{i}are all odd, the polynomial ofppvariablesz_(1),z_(2),dots,z_(p)z_{1}, z_{2}, \ldots, z_{p},
The condition is necessary according to Lemma 1.
The condition is also sufficient. Indeed, if the numbersk_(i)k_{i}are all odd, the polynomial ofppvariablesz_(1),z_(2),dots,z_(p)z_{1}, z_{2}, \ldots, z_{p},
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) dx
is positive over the entire space atppdimensions. 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),dots,z_(p)z_{1}, z_{2}, \ldots, z_{p}) the coordinatesz_(1),z_(2),dots,z_(p)z_{1}, z_{2}, \ldots, z_{p}are distinct.
On such a point we have
{:(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*}
It is then sufficient to note that the system (4) is equivalent to the orthogonality of the polynomial (2) with any polynomial of degreep-1p-1.
6. It follows from the above that the numbersk_(i),i=1,2,dots,pk_{i}, i=1,2, \ldots, pbeing assumed always odd, in any formula (1) of the Gauss type, the nodesz_(1),z_(2),dots,z_(p)z_{1}, z_{2}, \ldots, z_{p}are distinct and within the interval[-1,1][-1,1].
6. It follows from the above that the numbersk_(i),i=1,2,dots,pk_{i}, i=1,2, \ldots, pbeing assumed always odd, in any formula (1) of the Gauss type, the nodesz_(1),z_(2),dots,z_(p)z_{1}, z_{2}, \ldots, z_{p}are distinct and within the interval[-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 numbersk_(1),k_(2),dots,k_(p)k_{1}, k_{2}, \ldots, k_{p}, to any permutation corresponding to a well-determined Gauss-type formula. This is true ifk_(1)=k_(2)==dots=k_(p)k_{1}=k_{2}= =\ldots=k_{p}. 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 casep=2,k_(1)=k,k_(2)=1p=2, k_{1}=k, k_{2}=1. The casek=3k=3was considered previously [3]. Now suppose thatkkbe any odd. The nodesz_(1),z_(2)z_{1}, z_{2}are then given by the system
7. Consider the casep=2,k_(1)=k,k_(2)=1p=2, k_{1}=k, k_{2}=1. The casek=3k=3was considered previously [3]. Now suppose thatkkbe any odd. The nodesz_(1),z_(2)z_{1}, z_{2}are then given by the system
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
If we pose
{:(5)(1-z_(1))/(1+z_(1))=y:}\begin{equation*}
\frac{1-z_{1}}{1+z_{1}}=y \tag{5}
\end{equation*}
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.]:}\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*}
By eliminatingz_(2)z_{2}, the knotz_(1)z_{1}is given by (5) and by the reciprocal equation inyy
{:[(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*}
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 fork=1k=1in the classical Gauss formula, but are quite distinct fork > 1k>1. Indeed, the knotz_(2)z_{2}is obtained from one of the equations (6) and the preceding theory shows us that it will be completely determined, different fromz_(1)z_{1}and strictly between -1 and 1.
It follows that in the case studied we have exactly two Gauss-type formulas which are confused fork=1k=1in the classical Gauss formula, but are quite distinct fork > 1k>1. Indeed, the knotz_(2)z_{2}is obtained from one of the equations (6) and the preceding theory shows us that it will be completely determined, different fromz_(1)z_{1}and strictly between -1 and 1.
Fork=3k=3we have already given the explicit value ofz_(1)z_{1}(so also ofz_(2)z_{2}) and the corresponding formula (1).z_(1)z_{1}Andz_(2)z_{2}are then quadratic irrationalities. Fork > 3k>3the arithmetic nature ofz_(1)z_{1}Andz_(2)z_{2}seems to be more complicated.
8. To calculatez_(1)z_{1}we must first solve the reciprocal equation (7). To do this we set
8. To calculatez_(1)z_{1}we must first solve the reciprocal equation (7). To do this we set
{:(8)y+(1)/(y)=t.:}\begin{equation*}
y+\frac{1}{y}=t . \tag{8}
\end{equation*}
SOy^(s)+1//y^(s)y^{s}+1 / y^{s}is equal to a polynomialP_(s)(t)P_{s}(t)intt, of degreess, given by the recurrence relation
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)
The equation inttis therefore
{:(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*}
From the above it follows that this equation intthas a real root> 2>2. We easily check that it also has the double root -2.
Fork=1k=1Andk=3k=3the first member of (9) is
{:[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}
respectively, which gives the well-known reported results.
Fork=5k=5equation (9) becomes
Fork=5k=5equation (9) becomes
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)=0
and we check that this equation no longer has any positive rational roots.
9. Let us again consider the casep=3,k_(1)=k,k_(2)=k_(3)=1p=3, k_{1}=k, k_{2}=k_{3}=1. The casek=3k=3has already been considered previously [3]. Suppose thatkkbe any odd. . The nodesz_(1),z_(2),z_(3)z_{1}, z_{2}, z_{3}are given by the system
9. Let us again consider the casep=3,k_(1)=k,k_(2)=k_(3)=1p=3, k_{1}=k, k_{2}=k_{3}=1. The casek=3k=3has already been considered previously [3]. Suppose thatkkbe any odd. . The nodesz_(1),z_(2),z_(3)z_{1}, z_{2}, z_{3}are given by the system
{:(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*}
By writing (5) and eliminatingz_(2)z_{2}Andz_(3)z_{3}, we find thatyyis 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\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
or, by doing the calculations,
{:[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*}
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)+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}
Note thatA,B,C,D,EA, B, C, D, Eare positive.
If we designate byvarphi(y)\varphi(y)the first member of the equation, we havevarphi(0) < 0\varphi(0)<0Andvarphi^(')(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--(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 az_(1)z_{1}determined the polynomial(x-z_(2))(x-z_(3))\left(x-z_{2}\right)\left(x-z_{3}\right)is determined completely,z_(2),z_(3)z_{2}, z_{3}being distinct, strictly between -1 and 1 and being different fromz_(1)z_{1}.
In this case we therefore have exactly 3 Gauss-type formulas. Ifk=1k=1These three formulas are confused with the classical Gauss formula with 3 knots. But ifk > 1k>1The three formulas are distinct. One corresponds to the value 0 ofz_(1)z_{1}and the other two with two values ​​ofz_(1)z_{1}symmetrical about the origin.
Note that in this case the number of distinct permutations of the numbersk_(1),k_(2),k_(3)k_{1}, k_{2}, k_{3}is indeed equal to 3.
10. Still in the particular case considered, the Gauss type formula corresponding to the nodez_(1)=0z_{1}=0can be obtained explicitly in a fairly simple form.
10. Still in the particular case considered, the Gauss type formula corresponding to the nodez_(1)=0z_{1}=0can be obtained explicitly in a fairly simple form.
By posingz_(1)=0z_{1}=0in system (10) we obtain(x-z_(2))(x-z_(3))=x^(2)-lambda^(2)\left(x-z_{2}\right)\left(x-z_{3}\right)=x^{2}-\lambda^{2}, Orlambda=sqrt((k+2)//(k+4))\lambda=\sqrt{(k+2) /(k+4)}. So we havequadz_(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)}.
Let us write the quadrature formula in the form
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) .
The formula is of degreek+4k+4of accuracy.
To find the coefficientscc,ddlet us take successivelyf=x^(k),f=x^(k+1)f=x^{k}, f=x^{k+1}. Considering thatbar(k)\bar{k}is odd, we deduce
To find the coefficientscc,ddlet us take successivelyf=x^(k),f=x^(k+1)f=x^{k}, f=x^{k+1}. Considering thatbar(k)\bar{k}is odd, we deduce
{:(12)c=d=(1)/((k+2)lambda^(k+1)).:}\begin{equation*}
c=d=\frac{1}{(k+2) \lambda^{k+1}} . \tag{12}
\end{equation*}
By takingf=x^(i)f=x^{i}Oriiis odd we find (k > 1k>1)
{:(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*}
Finally if we takef=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}, we get
{:(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*}
The desired quadrature formula is therefore written as
{:[(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*}
The restR(f)R(f)is given by the formula (see [3]),
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]
Orxi_(1),xi_(2),dots,xi_(k+6)\xi_{1}, \xi_{2}, \ldots, \xi_{k+6}arek+6k+6distinct points inside the interval[-1,1][-1,1]and which depend, in general, on the (continuous) functionf(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]is the divided difference of the functionffon the knotsxi_(i),i==1,2,dots,k+6\xi_{i}, i= =1,2, \ldots, k+6.
We can obtain the coefficientR(x^(k+5))R\left(x^{k+5}\right)taking in formula (15)f=x^(k+3)(x^(2)-lambda^(2))f=x^{k+3}\left(x^{2}-\lambda^{2}\right). We thus obtain
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)} .
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- A polynomial of degreeggis a polynomial of effective degree<= g\leqslant g. The unique polynomial of degree - 1 is the zero polynomial.