1963-IonescuDV-Some-practical-formulas-of-quadrature-in-Communications-Acad.RPRvol.13nr.8-august1963pp
COMMUNICATIONS
OF THE ACADEMY OF THE ROMANIAN PEOPLE'S REPUBLIC Volume XIII, No. 8
August 1963
SOME PRACTICAL QUADRATURE FORMULAS
BY
D. V. IONESCU
Communication presented by academician T. Popoviciu in the journal of May 18, 1963
G. Coulmy [1] considered the following quadrature formula
∫
x
n
x
n
f
(
x )
d
x =
13
h
36
[
f
(
x
0
)
+
f
(
x
n
)
]
+
7
h
6
[ f
(
x
1
)
+ f
(
x
n −
1
)
]
+ h
[ f
(
x
n
)
+ f
(
x
n − 2
)
]
+
(C)
+
35
h
36
[ f
(
x
3
)
+ f
(
x
n − 3
)
]
+ h
[ f
(
x
4
)
+ f
(
x
5
)
+ ⋯
+ f
(
x
n − 4
)
]
∫
x
n
x
n
f ( x )
d
x =
13 h
36
f
x
0
+ f
x
n
+
7 h
6
f
x
1
+ f
x
n − 1
+ h
f
x
n
+ f
x
n − 2
+
(C)
+
35 h
36
f
x
3
+ f
x
n − 3
+ h
f
x
4
+ f
x
5
+ ⋯ + f
x
n − 4
{:[int_(x_(n))^(x_(n))f(x)dx=(13 h)/(36)[f(x_(0))+f(x_(n))]+(7h)/(6)[f(x_(1))+f(x_(n-1))]+h[f(x_(n))+f(x_(n-2))]+],[(C)quad+(35 h)/(36)[f(x_(3))+f(x_(n-3))]+h[f(x_(4))+f(x_(5))+cdots+f(x_(n-4))]]:} \begin{align*}
& \int_{x_{n}}^{x_{n}} f(x) \mathrm{d} x=\frac{13 h}{36}\left[f\left(x_{0}\right)+f\left(x_{n}\right)\right]+\frac{7 h}{6}\left[f\left(x_{1}\right)+f\left(x_{n-1}\right)\right]+h\left[f\left(x_{n}\right)+f\left(x_{n-2}\right)\right]+ \\
& \quad+\frac{35 h}{36}\left[f\left(x_{3}\right)+f\left(x_{n-3}\right)\right]+h\left[f\left(x_{4}\right)+f\left(x_{5}\right)+\cdots+f\left(x_{n-4}\right)\right] \tag{C}
\end{align*} ∫ x n x n f ( x ) d x = 13 h 36 [ f ( x 0 ) + f ( x n ) ] + 7 h 6 [ f ( x 1 ) + f ( x n − 1 ) ] + h [ f ( x n ) + f ( x n − 2 ) ] + (C) + 35 h 36 [ f ( x 3 ) + f ( x n − 3 ) ] + h [ f ( x 4 ) + f ( x 5 ) + ⋯ + f ( x n − 4 ) ] where the nodes
x
0
,
x
1
, …
,
x
n
x
0
,
x
1
, … ,
x
n
x_(0),x_(1),dots,x_(n) x_{0}, x_{1}, \ldots, x_{n} x 0 , x 1 , … , x n are in arithmetic progression with the ratio
h
h h h h , and applied this formula to the problem of tides. It is good to compare formula (C) with the following well-known escapement formulas:
(T)
∫
x
0
x
n
f ( x )
d
x =
h 2
[ f
(
x
a
)
+ f
(
x
n
)
]
+ h
[ f
(
x
1
)
+ f
(
x
2
)
+ … + f
(
x
n − 1
)
]
∫
x
1
x
n
f ( x )
d
x =
5 h
12
f
(
x
0
)
+ f
(
x
n
)
]
+
13 h
12
[ f
(
x
1
)
+ f
(
x
n − 1
)
+ h
[ f
(
x
2
)
+ f
(
x
3
)
+ …
(D)
… + f
(
x
n − 2
)
]
∫
x
0
x
n
f ( x )
d
x =
3 h
8
[ f
(
x
0
)
+ f
(
x
n
)
]
+
7 h
6
[ f
(
x
1
)
+ f
(
x
n − 1
)
]
+
(L)
+
23
h
24
[ f
(
x
2
)
+ f
(
x
n − 2
)
]
+ h
[ f
(
x
3
)
+ f
(
x
4
)
+ … + f
(
x
n − 3
)
]
(T)
∫
x
0
x
n
f ( x )
d
x =
h 2
f
x
a
+ f
x
n
+ h
f
x
1
+ f
x
2
+ … + f
x
n − 1
∫
x
1
x
n
f ( x )
d
x =
5 h
12
f
x
0
+ f
x
n
+
13 h
12
f
x
1
+ f
x
n − 1
+ h
f
x
2
+ f
x
3
+ …
(D)
… + f
x
n − 2
∫
x
0
x
n
f ( x )
d
x =
3 h
8
f
x
0
+ f
x
n
+
7 h
6
f
x
1
+ f
x
n − 1
+
(L)
+
23 h
24
f
x
2
+ f
x
n − 2
+ h
f
x
3
+ f
x
4
+ … + f
x
n − 3
{:[(T)int_(x_(0))^(x_(n))f(x)dx=(h)/(2)[f(x_(a))+f(x_(n))]+h[f(x_(1))+f(x_(2))+dots+f(x_(n-1))]],[{:int_(x_(1))^(x_(n))f(x)dx=(5h)/(12)f(x_(0))+f(x_(n))]+(13 h)/(12)[f(x_(1))+f(x_(n-1))+h[f(x_(2))+f(x_(3))+dots:}],[(D){: dots+f(x_(n-2))]],[int_(x_(0))^(x_(n))f(x)dx=(3h)/(8)[f(x_(0))+f(x_(n))]+(7h)/(6)[f(x_(1))+f(x_(n-1))]+],[(L)quad+(23 h)/(24)[f(x_(2))+f(x_(n-2))]+h[f(x_(3))+f(x_(4))+dots+f(x_(n-3))]]:} \begin{gather*}
\int_{x_{0}}^{x_{n}} f(x) \mathrm{d} x=\frac{h}{2}\left[f\left(x_{a}\right)+f\left(x_{n}\right)\right]+h\left[f\left(x_{1}\right)+f\left(x_{2}\right)+\ldots+f\left(x_{n-1}\right)\right] \tag{T}\\
\left.\int_{x_{1}}^{x_{n}} f(x) \mathrm{d} x=\frac{5 h}{12} f\left(x_{0}\right)+f\left(x_{n}\right)\right]+\frac{13 h}{12}\left[f\left(x_{1}\right)+f\left(x_{n-1}\right)+h\left[f\left(x_{2}\right)+f\left(x_{3}\right)+\ldots\right.\right. \\
\left.\ldots+f\left(x_{n-2}\right)\right] \tag{D}\\
\int_{x_{0}}^{x_{n}} f(x) \mathrm{d} x=\frac{3 h}{8}\left[f\left(x_{0}\right)+f\left(x_{n}\right)\right]+\frac{7 h}{6}\left[f\left(x_{1}\right)+f\left(x_{n-1}\right)\right]+ \\
\quad+\frac{23 h}{24}\left[f\left(x_{2}\right)+f\left(x_{n-2}\right)\right]+h\left[f\left(x_{3}\right)+f\left(x_{4}\right)+\ldots+f\left(x_{n-3}\right)\right] \tag{L}
\end{gather*} (T) ∫ x 0 x n f ( x ) d x = h 2 [ f ( x a ) + f ( x n ) ] + h [ f ( x 1 ) + f ( x 2 ) + … + f ( x n − 1 ) ] ∫ x 1 x n f ( x ) d x = 5 h 12 f ( x 0 ) + f ( x n ) ] + 13 h 12 [ f ( x 1 ) + f ( x n − 1 ) + h [ f ( x 2 ) + f ( x 3 ) + … (D) … + f ( x n − 2 ) ] ∫ x 0 x n f ( x ) d x = 3 h 8 [ f ( x 0 ) + f ( x n ) ] + 7 h 6 [ f ( x 1 ) + f ( x n − 1 ) ] + (L) + 23 h 24 [ f ( x 2 ) + f ( x n − 2 ) ] + h [ f ( x 3 ) + f ( x 4 ) + … + f ( x n − 3 ) ] Formula (T) is derived from the classical trapezoid formula. Formulas (D) and (L) are cited in the book by H. Mineur [2] and are called the Durand and Lacroix formulas.
In this paper we will give the expressions of the ratio in formulas (C), (D) and (L) from which we will draw conclusions regarding the application of one or the other in practice.
I. THE REMAINDER IN FORMULA (C)
To find its remainder, it is assumed that the function
f
(
x
)
f
(
x
)
f(x) f(x) f ( x ) it is classy
C
9
[
x
n
,
x
n
]
C
9
x
n
,
x
n
C^(9)[x_(n),x_(n)] C^{9}\left[x_{n}, x_{n}\right] C 9 [ x n , x n ] and it is a method used consistently in the work [3]. It is found
(1)
R
c
=
∫
e
c
e
∗
φ
(
x
)
f
′
′
(
x
)
d
x
,
(1)
R
c
=
∫
e
c
e
∗
φ
(
x
)
f
′
′
(
x
)
d
x
,
{:(1)R_(c)=int_(e_(c))^(e^(**))varphi(x)f^('')(x)dx",":} \begin{equation*}
R_{c}=\int_{e_{c}}^{e^{*}} \varphi(x) f^{\prime \prime}(x) \mathrm{d} x, \tag{1}
\end{equation*} (1) R c = ∫ and c and ∗ f ( x ) f ′ ′ ( x ) d x , where the function
φ
(
x
)
φ
(
x
)
varphi(x) \varphi(x) f ( x ) coincides on the intervals
[
x
6
,
x
1
]
,
[
x
1
,
x
2
]
,
…
x
6
,
x
1
,
x
1
,
x
2
,
…
[x_(6),x_(1)],[x_(1),x_(2)],dots \left[x_{6}, x_{1}\right],\left[x_{1}, x_{2}\right], \ldots [ x 6 , x 1 ] , [ x 1 , x 2 ] , … with some natural polynomials of the second degree, which we write only for the first half of the interval
[
x
0
,
x
0
]
x
0
,
x
0
[x_(0),x_(0)] \left[x_{0}, x_{0}\right] [ x 0 , x 0 ] :
φ
1
(
x
)
=
(
x
−
x
0
)
2
2
−
13
h
36
(
x
−
x
0
)
,
φ
1
(
x
)
=
(
x
−
x
1
)
Ω
2
−
19
h
36
(
x
−
x
1
)
+
5
h
2
36
φ
1
(
x
)
=
x
−
x
0
2
2
−
13
h
36
x
−
x
0
,
φ
1
(
x
)
=
x
−
x
1
Ω
2
−
19
h
36
x
−
x
1
+
5
h
2
36
varphi_(1)(x)=((x-x_(0))^(2))/(2)-(13 h)/(36)(x-x_(0)),quadvarphi_(1)(x)=((x-x_(1))^(Omega))/(2)-(19 h)/(36)(x-x_(1))+(5h^(2))/(36) \varphi_{1}(x)=\frac{\left(x-x_{0}\right)^{2}}{2}-\frac{13 h}{36}\left(x-x_{0}\right), \quad \varphi_{1}(x)=\frac{\left(x-x_{1}\right)^{\Omega}}{2}-\frac{19 h}{36}\left(x-x_{1}\right)+\frac{5 h^{2}}{36} f 1 ( x ) = ( x − x 0 ) 2 2 − 13 h 36 ( x − x 0 ) , f 1 ( x ) = ( x − x 1 ) Oh 2 − 19 h 36 ( x − x 1 ) + 5 h 2 36 ,
φ
2
(
x
)
=
(
x
−
x
2
)
′
2
−
19
h
36
(
x
−
x
2
)
+
4
h
′
36
,
φ
k
(
x
)
=
(
x
−
x
k
−
1
)
′
2
−
h
2
(
x
−
x
k
−
1
)
+
+
h
2
12
(
k
=
4
,
5
,
…
)
φ
2
(
x
)
=
x
−
x
2
′
2
−
19
h
36
x
−
x
2
+
4
h
′
36
,
φ
k
(
x
)
=
x
−
x
k
−
1
′
2
−
h
2
x
−
x
k
−
1
+
+
h
2
12
(
k
=
4
,
5
,
…
)
varphi_(2)(x)=((x-x_(2))^('))/(2)-(19 h)/(36)(x-x_(2))+(4h^('))/(36),quadvarphi_(k)(x)=((x-x_(k-1))^('))/(2)-(h)/(2)(x-x_(k-1))++(h^(2))/(12)(k=4,5,dots) \varphi_{2}(x)=\frac{\left(x-x_{2}\right)^{\prime}}{2}-\frac{19 h}{36}\left(x-x_{2}\right)+\frac{4 h^{\prime}}{36}, \quad \varphi_{k}(x)=\frac{\left(x-x_{k-1}\right)^{\prime}}{2}-\frac{h}{2}\left(x-x_{k-1}\right)+ +\frac{h^{2}}{12}(k=4,5, \ldots) f 2 ( x ) = ( x − x 2 ) ′ 2 − 19 h 36 ( x − x 2 ) + 4 h ′ 36 , f k ( x ) = ( x − x k − 1 ) ′ 2 − h 2 ( x − x k − 1 ) + + h 2 12 ( k = 4 , 5 , … ) .
If you find cå
∫
x
1
α
1
φ
1
(
x
)
d
x
=
−
h
3
72
,
∫
x
1
x
1
φ
3
(
x
)
d
x
=
3
h
3
72
,
∫
x
2
y
2
φ
3
(
x
)
d
x
=
h
8
72
,
∫
x
1
α
1
φ
1
(
x
)
d
x
=
−
h
3
72
,
∫
x
1
x
1
φ
3
(
x
)
d
x
=
3
h
3
72
,
∫
x
2
y
2
φ
3
(
x
)
d
x
=
h
8
72
,
int_(x_(1))^(alpha_(1))varphi_(1)(x)dx=-(h^(3))/(72),quadint_(x_(1))^(x_(1))varphi_(3)(x)dx=(3h^(3))/(72),quadint_(x_(2))^(y_(2))varphi_(3)(x)dx=(h^(8))/(72), \int_{x_{1}}^{\alpha_{1}} \varphi_{1}(x) \mathrm{d} x=-\frac{h^{3}}{72}, \quad \int_{x_{1}}^{x_{1}} \varphi_{3}(x) \mathrm{d} x=\frac{3 h^{3}}{72}, \quad \int_{x_{2}}^{y_{2}} \varphi_{3}(x) \mathrm{d} x=\frac{h^{8}}{72}, ∫ x 1 a 1 f 1 ( x ) d x = − h 3 72 , ∫ x 1 x 1 f 3 ( x ) d x = 3 h 3 72 , ∫ x 2 and 2 f 3 ( x ) d x = h 8 72 ,
∫
k
−
1
x
k
φ
k
(
x
)
d
x
=
0
(
k
=
4
,
5
,
…
)
.
∫
k
−
1
x
k
φ
k
(
x
)
d
x
=
0
(
k
=
4
,
5
,
…
)
.
int_(k-1)^(x_(k))varphi_(k)(x)dx=0quad(k=4,5,dots). \int_{k-1}^{x_{k}} \varphi_{k}(x) \mathrm{d} x=0 \quad(k=4,5, \ldots) . ∫ k − 1 x k f k ( x ) d x = 0 ( k = 4 , 5 , … ) . The degree of accuracy of formula (C) is 1, because in formula (1) we have
∫
−
4
x
n
φ
(
x
)
d
x
=
h
3
12
.
∫
−
4
x
n
φ
(
x
)
d
x
=
h
3
12
.
int_(-4)^(x_(n))varphi(x)dx=(h^(3))/(12). \int_{-4}^{x_{n}} \varphi(x) \mathrm{d} x=\frac{h^{3}}{12} . ∫ − 4 x n f ( x ) d x = h 3 12 . From formulas (1) an evaluation of s is deduced
R
c
R
c
R_(c) R_{c} R c and name
(2)
|
R
c
|
⩽
K
c
M
2
h
3
,
(2)
R
c
⩽
K
c
M
2
h
3
,
{:(2)|R_(c)| <= K_(c)M_(2)h^(3)",":} \begin{equation*}
\left|R_{c}\right| \leqslant K_{c} M_{2} h^{3}, \tag{2}
\end{equation*} (2) | R c | ⩽ K c M 2 h 3 , where 8-noted
(3)
M
2
=
sup
[
x
n
=
x
n
]
|
f
′
′
(
x
)
|
(3)
M
2
=
sup
x
n
=
x
n
f
′
′
(
x
)
{:(3)M_(2)=s u p_([x_(n)=x_(n)])|f^('')(x)|:} \begin{equation*}
M_{2}=\sup _{\left[x_{n}=x_{n}\right]}\left|f^{\prime \prime}(x)\right| \tag{3}
\end{equation*} (3) M 2 = sup [ x n = x n ] | f ′ ′ ( x ) | and
(4)
R
c
h
s
=
∫
α
0
x
n
|
φ
(
x
)
|
d
ω
(4)
R
c
h
s
=
∫
α
0
x
n
|
φ
(
x
)
|
d
ω
{:(4)R_(c)h^(s)=int_(alpha_(0))^(x_(n))|varphi(x)|d omega:} \begin{equation*}
R_{c} h^{s}=\int_{\alpha_{0}}^{x_{n}}|\varphi(x)| d \omega \tag{4}
\end{equation*} (4) R c h s = ∫ a 0 x n | f ( x ) | d oh some practical formulas for quadrature
Wind calculations can be found
∫
A
0
x
1
|
φ
1
(
x
)
|
d
x
=
1
34
711
992
h
8
,
∫
x
4
x
2
|
φ
e
(
x
)
|
d
x
=
1459
34992
h
2
∫
x
4
x
2
|
φ
1
(
x
)
|
d
x
=
486
+
73
73
34992
h
3
,
∫
x
3
z
4
|
φ
4
(
x
)
|
d
x
=
3
54
h
2
.
∫
A
0
x
1
φ
1
(
x
)
d
x
=
1
34
711
992
h
8
,
∫
x
4
x
2
φ
e
(
x
)
d
x
=
1459
34992
h
2
∫
x
4
x
2
φ
1
(
x
)
d
x
=
486
+
73
73
34992
h
3
,
∫
x
3
z
4
φ
4
(
x
)
d
x
=
3
54
h
2
.
{:[int_(A_(0))^(x_(1))|varphi_(1)(x)|dx=(1)/(34)(711)/(992)h^(8)","quadint_(x_(4))^(x_(2))|varphi_(e)(x)|dx=(1459)/(34992)h^(2)],[int_(x_(4))^(x_(2))|varphi_(1)(x)|dx=(486+73sqrt73)/(34992)h^(3)","quadint_(x_(3))^(z_(4))|varphi_(4)(x)|dx=(sqrt3)/(54)h^(2).]:} \begin{aligned}
& \int_{A_{0}}^{x_{1}}\left|\varphi_{1}(x)\right| \mathrm{d} x=\frac{1}{34} \frac{711}{992} h^{8}, \quad \int_{x_{4}}^{x_{2}}\left|\varphi_{\mathrm{e}}(x)\right| \mathrm{d} x=\frac{1459}{34992} h^{2} \\
& \int_{x_{4}}^{x_{2}}\left|\varphi_{1}(x)\right| \mathrm{d} x=\frac{486+73 \sqrt{73}}{34992} h^{3}, \quad \int_{x_{3}}^{z_{4}}\left|\varphi_{4}(x)\right| \mathrm{d} x=\frac{\sqrt{3}}{54} h^{2} .
\end{aligned} ∫ A 0 x 1 | f 1 ( x ) | d x = 1 34 711 992 h 8 , ∫ x 4 x 2 | f and ( x ) | d x = 1459 34992 h 2 ∫ x 4 x 2 | f 1 ( x ) | d x = 486 + 73 73 34992 h 3 , ∫ x 3 With 4 | f 4 ( x ) | d x = 3 54 h 2 . Taking into account the symmetry of the curve
y
=
φ
(
x
)
y
=
φ
(
x
)
y=varphi(x) y=\varphi(x) and = f ( x ) right side
x
=
x
=
x= x= x = got
x
2
x
2
x_(2) x_{2} x 2 , we have
∫
π
∗
π
∗
|
φ
(
x
)
|
d
x
=
[
2
1711
34992
+
2
′
1459
24
θ
92
+
2
−
486
+
73
/
7
3
―
34982
+
(
n
−
6
)
3
54
|
A
∫
π
∗
π
∗
|
φ
(
x
)
|
d
x
=
2
1711
34992
+
2
′
1459
24
θ
92
+
2
−
486
+
73
/
7
3
¯
34982
+
(
n
−
6
)
3
54
A
int_(pi_(**))^(pi_(**))|varphi(x)|dx=[2(1711)/(34992)+2^(')(1459)/(24 theta92)+2-(486+73//7 bar(3))/(34982)+(n-6)(sqrt3)/(54)|_(A):} \int_{\pi_{*}}^{\pi_{*}}|\varphi(x)| \mathrm{d} x=\left[2 \frac{1711}{34992}+2^{\prime} \frac{1459}{24 \theta 92}+2-\frac{486+73 / 7 \overline{3}}{34982}+\left.(n-6) \frac{\sqrt{3}}{54}\right|_{\mathrm{A}}\right. ∫ p ∗ p ∗ | f ( x ) | d x = [ 2 1711 34992 + 2 ′ 1459 24 i 92 + 2 − 486 + 73 / 7 3 ― 34982 + ( n − 6 ) 3 54 | A and therefore
(b)
K
c
=
3650
+
73
73
−
1944
3
17496
+
n
ψ
3
―
54
(b)
K
c
=
3650
+
73
73
−
1944
3
17496
+
n
ψ
3
¯
54
{:(b)K_(c)=(3650+73sqrt73-1944sqrt3)/(17496)+(n psi bar(3))/(54):} \begin{equation*}
K_{c}=\frac{3650+73 \sqrt{73}-1944 \sqrt{3}}{17496}+\frac{n \psi \overline{3}}{54} \tag{b}
\end{equation*} (b) K c = 3650 + 73 73 − 1944 3 17496 + n ψ 3 ― 54
HESTUL IN FORMULA (D)
Assuming that the function
f
(
x
)
f
(
x
)
f(x) f(x) f ( x ) it is annoying
C
2
[
x
01
x
n
]
C
2
x
01
x
n
C^(2)[x_(01)x_(n)] C^{2}\left[x_{01} x_{n}\right] C 2 [ x 01 x n ] the rest is deducted
R
D
R
D
R_(D) R_{D} R D has the form
(
θ
)
R
D
=
∫
n
0
x
m
ψ
(
x
)
f
′
′
(
x
)
d
x
,
(
θ
)
R
D
=
∫
n
0
x
m
ψ
(
x
)
f
′
′
(
x
)
d
x
,
{:(theta")"R_(D)=int_(n_(0))^(x_(m))psi(x)f^('')(x)dx",":} \begin{equation*}
R_{D}=\int_{n_{0}}^{x_{m}} \psi(x) f^{\prime \prime}(x) \mathrm{d} x, \tag{$\theta$}
\end{equation*} ( i ) R D = ∫ n 0 x m ψ ( x ) f ′ ′ ( x ) d x , where the function
ψ
(
x
)
ψ
(
x
)
psi(x) \psi(x) ψ ( x ) coincides on the intervals
[
x
0
,
x
1
]
,
[
x
,
x
1
]
…
x
0
,
x
1
,
x
,
x
1
…
[x_(0),x_(1)],[x,x_(1)]dots \left[x_{0}, x_{1}\right],\left[x, x_{1}\right] \ldots [ x 0 , x 1 ] , [ x , x 1 ] … I polynomials
ψ
1
(
x
)
,
ψ
2
(
x
)
,
…
ψ
1
(
x
)
,
ψ
2
(
x
)
,
…
psi_(1)(x),psi_(2)(x),dots \psi_{1}(x), \psi_{2}(x), \ldots ψ 1 ( x ) , ψ 2 ( x ) , … given by the equations
ψ
1
(
x
)
=
(
x
−
x
0
)
1
2
−
5
h
12
(
x
−
x
0
)
,
ψ
k
(
x
)
=
(
x
−
x
k
−
1
)
1
2
−
h
2
(
x
−
x
1
−
1
)
+
h
1
12
ψ
1
(
x
)
=
x
−
x
0
1
2
−
5
h
12
x
−
x
0
,
ψ
k
(
x
)
=
x
−
x
k
−
1
1
2
−
h
2
x
−
x
1
−
1
+
h
1
12
psi_(1)(x)=((x-x_(0))^(1))/(2)-(5h)/(12)(x-x_(0)),quadpsi_(k)(x)=((x-x_(k-1))^(1))/(2)-(h)/(2)(x-x_(1-1))+(h^(1))/(12) \psi_{1}(x)=\frac{\left(x-x_{0}\right)^{1}}{2}-\frac{5 h}{12}\left(x-x_{0}\right), \quad \psi_{k}(x)=\frac{\left(x-x_{k-1}\right)^{1}}{2}-\frac{h}{2}\left(x-x_{1-1}\right)+\frac{h^{1}}{12} ψ 1 ( x ) = ( x − x 0 ) 1 2 − 5 h 12 ( x − x 0 ) , ψ k ( x ) = ( x − x k − 1 ) 1 2 − h 2 ( x − x 1 − 1 ) + h 1 12
(
k
=
2
,
3
,
⋯
)
.
(
k
=
2
,
3
,
⋯
)
.
(k=2,3,cdots). (k=2,3, \cdots) . ( k = 2 , 3 , ⋯ ) . It is found that
∫
x
1
x
1
ψ
1
(
x
)
d
x
=
−
h
3
24
,
∫
k
+
1
x
k
ψ
1
(
x
)
d
x
=
0
(
k
=
2
,
3
,
…
)
∫
x
1
x
1
ψ
1
(
x
)
d
x
=
−
h
3
24
,
∫
k
+
1
x
k
ψ
1
(
x
)
d
x
=
0
(
k
=
2
,
3
,
…
)
int_(x_(1))^(x_(1))psi_(1)(x)dx=-(h^(3))/(24),quadint_(k+1)^(x_(k))psi_(1)(x)dx=0quad(k=2,3,dots) \int_{x_{1}}^{x_{1}} \psi_{1}(x) \mathrm{d} x=-\frac{h^{3}}{24}, \quad \int_{k+1}^{x_{k}} \psi_{1}(x) \mathrm{d} x=0 \quad(k=2,3, \ldots) ∫ x 1 x 1 ψ 1 ( x ) d x = − h 3 24 , ∫ k + 1 x k ψ 1 ( x ) d x = 0 ( k = 2 , 3 , … ) and therefore the result is
∫
x
0
x
n
ψ
(
x
)
d
x
=
−
h
3
12
,
∫
x
0
x
n
ψ
(
x
)
d
x
=
−
h
3
12
,
int_(x_(0))^(x_(n))psi(x)dx=-(h^(3))/(12), \int_{x_{0}}^{x_{n}} \psi(x) \mathrm{d} x=-\frac{h^{3}}{12}, ∫ x 0 x n ψ ( x ) d x = − h 3 12 , which means that the degree of accuracy of formula (D) is 1.
From formula (0) it is deduced that
(T)
|
R
D
|
⩽
K
D
M
B
h
3
(T)
R
D
⩽
K
D
M
B
h
3
{:(T)|R_(D)| <= K_(D)M_(B)h^(3):} \begin{equation*}
\left|R_{D}\right| \leqslant K_{D} M_{B} h^{3} \tag{T}
\end{equation*} (T) | R D | ⩽ K D M B h 3
ande
K
b
A
a
=
∫
M
0
x
0
|
Ψ
(
x
)
|
d
x
.
ande
K
b
A
a
=
∫
M
0
x
0
|
Ψ
(
x
)
|
d
x
.
" ande "K_(b)A^(a)=int_(M_(0))^(x_(0))|Psi(x)|dx. \text { ande } K_{b} A^{\mathrm{a}}=\int_{M_{0}}^{x_{0}}|\Psi(x)| d x . Spirit K b A a = ∫ M 0 x 0 | P ( x ) | d x . Doing the calculations, we find
∫
x
0
π
1
|
ψ
1
(
x
)
|
d
x
=
71
k
8
1296
,
∫
α
1
π
2
|
ψ
8
(
x
)
|
d
x
=
V
3
―
54
h
0
∫
x
0
π
1
ψ
1
(
x
)
d
x
=
71
k
8
1296
,
∫
α
1
π
2
ψ
8
(
x
)
d
x
=
V
3
¯
54
h
0
int_(x_(0))^(pi_(1))|psi_(1)(x)|dx=(71k^(8))/(1296),quadint_(alpha_(1))^(pi_(2))|psi_(8)(x)|dx=(V bar(3))/(54)h^(0) \int_{x_{0}}^{\pi_{1}}\left|\psi_{1}(x)\right| d x=\frac{71 k^{8}}{1296}, \quad \int_{\alpha_{1}}^{\pi_{2}}\left|\psi_{8}(x)\right| d x=\frac{V \overline{3}}{54} h^{0} ∫ x 0 p 1 | ψ 1 ( x ) | d x = 71 k 8 1296 , ∫ a 1 p 2 | ψ 8 ( x ) | d x = In 3 ― 54 h 0 and taking into account the symmetry of the curve
y
=
ψ
(
x
)
y
=
ψ
(
x
)
y=psi(x) y=\psi(x) and = ψ ( x ) right side
x
=
x
0
+
x
0
2
x
=
x
0
+
x
0
2
x=(x_(0)+x_(0))/(2) x=\frac{x_{0}+x_{0}}{2} x = x 0 + x 0 2 HAVE
∫
x
0
x
0
[
ψ
(
x
)
|
d
x
=
[
2
71
1296
+
(
π
−
2
)
V
3
―
54
]
A
3
=
(
71
−
12
V
3
648
+
n
V
3
―
54
)
h
3
∫
x
0
x
0
ψ
(
x
)
d
x
=
2
71
1296
+
(
π
−
2
)
V
3
¯
54
A
3
=
71
−
12
V
3
648
+
n
V
3
¯
54
h
3
int_(x_(0))^(x_(0))[psi(x)|dx=[2(71)/(1296)+(pi-2)(V bar(3))/(54)]A^(3)=((71-12 V3)/(648)+(nV bar(3))/(54))h^(3):} \int_{x_{0}}^{x_{0}}\left[\psi(x) \left\lvert\, \mathrm{d} x=\left[2 \frac{71}{1296}+(\pi-2) \frac{V \overline{3}}{54}\right] A^{3}=\left(\frac{71-12 V 3}{648}+\frac{n V \overline{3}}{54}\right) h^{3}\right.\right. ∫ x 0 x 0 [ ψ ( x ) | d x = [ 2 71 1296 + ( p − 2 ) In 3 ― 54 ] A 3 = ( 71 − 12 In 3 648 + n In 3 ― 54 ) h 3
(8)
K
D
=
71
−
12
3
648
+
n
3
54
.
(8)
K
D
=
71
−
12
3
648
+
n
3
54
.
{:(8)K_(D)=(71-12sqrt3)/(648)+(nsqrt3)/(54).:} \begin{equation*}
K_{D}=\frac{71-12 \sqrt{3}}{648}+\frac{n \sqrt{3}}{54} . \tag{8}
\end{equation*} (8) K D = 71 − 12 3 648 + n 3 54 . Presuming that it works
f
(
x
)
f
(
x
)
f(x) f(x) f ( x ) it is classy
C
2
[
x
01
x
p
]
C
2
x
01
x
p
C^(2)[x_(01)x_(p)] C^{2}\left[x_{01} x_{p}\right] C 2 [ x 01 x p ] we can serialize her crime in the form
(9)
R
L
=
∫
x
v
x
n
θ
(
x
)
∫
′
′
(
x
)
d
x
(9)
R
L
=
∫
x
v
x
n
θ
(
x
)
∫
′
′
(
x
)
d
x
{:(9)R_(L)=int_(x_(v))^(x_(n))theta(x)int^('')(x)dx:} \begin{equation*}
R_{L}=\int_{x_{v}}^{x_{n}} \theta(x) \int^{\prime \prime}(x) \mathrm{d} x \tag{9}
\end{equation*} (9) R L = ∫ x in x n i ( x ) ∫ ′ ′ ( x ) d x where the function
θ
(
x
)
θ
(
x
)
theta(x) \theta(x) i ( x ) coincides on the intervals
[
x
0
,
x
1
]
[
x
1
,
x
1
]
,
…
x
0
,
x
1
x
1
,
x
1
,
…
[x_(0),x_(1)][x_(1),x_(1)],dots \left[x_{0}, x_{1}\right]\left[x_{1}, x_{1}\right], \ldots [ x 0 , x 1 ] [ x 1 , x 1 ] , … I polynomials
θ
1
(
x
)
=
(
x
−
x
0
)
2
2
−
3
h
8
(
x
−
x
0
)
,
θ
3
(
x
)
=
(
x
−
x
1
)
2
2
−
13
h
24
(
x
−
x
1
)
+
h
2
8
θ
k
(
x
)
=
(
x
−
x
k
−
1
)
2
2
−
h
2
(
x
−
x
k
−
1
)
+
h
2
12
(
k
=
3
,
4
,
…
)
θ
1
(
x
)
=
x
−
x
0
2
2
−
3
h
8
x
−
x
0
,
θ
3
(
x
)
=
x
−
x
1
2
2
−
13
h
24
x
−
x
1
+
h
2
8
θ
k
(
x
)
=
x
−
x
k
−
1
2
2
−
h
2
x
−
x
k
−
1
+
h
2
12
(
k
=
3
,
4
,
…
)
{:[theta_(1)(x)=((x-x_(0))^(2))/(2)-(3h)/(8)(x-x_(0))","quadtheta_(3)(x)=((x-x_(1))^(2))/(2)-(13 h)/(24)(x-x_(1))+(h^(2))/(8)],[theta_(k)(x)=((x-x_(k-1))^(2))/(2)-(h)/(2)(x-x_(k-1))+(h^(2))/(12)quad(k=3","4","dots)]:} \begin{gathered}
\theta_{1}(x)=\frac{\left(x-x_{0}\right)^{2}}{2}-\frac{3 h}{8}\left(x-x_{0}\right), \quad \theta_{3}(x)=\frac{\left(x-x_{1}\right)^{2}}{2}-\frac{13 h}{24}\left(x-x_{1}\right)+\frac{h^{2}}{8} \\
\theta_{k}(x)=\frac{\left(x-x_{k-1}\right)^{2}}{2}-\frac{h}{2}\left(x-x_{k-1}\right)+\frac{h^{2}}{12} \quad(k=3,4, \ldots)
\end{gathered} i 1 ( x ) = ( x − x 0 ) 2 2 − 3 h 8 ( x − x 0 ) , i 3 ( x ) = ( x − x 1 ) 2 2 − 13 h 24 ( x − x 1 ) + h 2 8 i k ( x ) = ( x − x k − 1 ) 2 2 − h 2 ( x − x k − 1 ) + h 2 12 ( k = 3 , 4 , … ) we
∫
x
0
x
1
θ
1
(
x
)
d
x
=
−
h
8
48
,
∫
x
1
x
1
θ
2
(
x
)
d
x
=
h
8
48
⋅
∫
x
1
x
1
θ
3
(
x
)
d
x
=
0
.
∫
x
0
x
1
θ
1
(
x
)
d
x
=
−
h
8
48
,
∫
x
1
x
1
θ
2
(
x
)
d
x
=
h
8
48
⋅
∫
x
1
x
1
θ
3
(
x
)
d
x
=
0
.
int_(x_(0))^(x_(1))theta_(1)(x)dx=-(h^(8))/(48),quadint_(x_(1))^(x_(1))theta_(2)(x)dx=(h^(8))/(48)*int_(x_(1))^(x_(1))theta_(3)(x)dx=0. \int_{x_{0}}^{x_{1}} \theta_{1}(x) \mathrm{d} x=-\frac{h^{8}}{48}, \quad \int_{x_{1}}^{x_{1}} \theta_{2}(x) \mathrm{d} x=\frac{h^{8}}{48} \cdot \int_{x_{1}}^{x_{1}} \theta_{3}(x) \mathrm{d} x=0 . ∫ x 0 x 1 i 1 ( x ) d x = − h 8 48 , ∫ x 1 x 1 i 2 ( x ) d x = h 8 48 ⋅ ∫ x 1 x 1 i 3 ( x ) d x = 0 . Result horse
∫
x
0
z
u
θ
(
x
)
d
x
=
0
∫
x
0
z
u
θ
(
x
)
d
x
=
0
int_(x_(0))^(z_(u))theta(x)dx=0 \int_{x_{0}}^{z_{u}} \theta(x) \mathrm{d} x=0 ∫ x 0 With in i ( x ) d x = 0 which means that the degree of enactity of the formula (L) is much greater than 1. It is easily seen that the degree of enactity of the formula (L) is ib.
SOME PRACTICAL FORMULAS OF SQUAREDNESS
603
From formula (9) it follows that
|
R
t
|
⩽
K
L
M
2
A
∗
K
t
h
3
=
∫
x
0
x
∗
|
θ
(
x
)
|
d
x
R
t
⩽
K
L
M
2
A
∗
K
t
h
3
=
∫
x
0
x
∗
|
θ
(
x
)
|
d
x
{:[|R_(t)| <= K_(L)M_(2)A^(**)],[K_(t)h^(3)=int_(x_(0))^(x_(**))|theta(x)|dx]:} \begin{gathered}
\left|R_{t}\right| \leqslant K_{L} M_{2} A^{*} \\
K_{t} h^{3}=\int_{x_{0}}^{x_{*}}|\theta(x)| d x
\end{gathered} | R t | ⩽ K L M 2 A ∗ K t h 3 = ∫ x 0 x ∗ | i ( x ) | d x Doing the calculations we find
∫
x
1
x
1
|
0
1
(
x
)
|
d
x
=
19
384
h
8
,
∫
α
1
x
1
|
θ
2
(
x
)
|
d
x
=
341
10368
h
2
,
∫
α
0
α
1
|
θ
1
(
x
)
|
d
x
=
3
54
h
1
∫
x
1
x
1
0
1
(
x
)
d
x
=
19
384
h
8
,
∫
α
1
x
1
θ
2
(
x
)
d
x
=
341
10368
h
2
,
∫
α
0
α
1
θ
1
(
x
)
d
x
=
3
54
h
1
int_(x_(1))^(x_(1))|0_(1)(x)|dx=(19)/(384)h^(8),quadint_(alpha_(1))^(x_(1))|theta_(2)(x)|dx=(341)/(10368)h^(2),int_(alpha_(0))^(alpha_(1))|theta_(1)(x)|dx=(sqrt3)/(54)h^(1) \int_{x_{1}}^{x_{1}}\left|0_{1}(x)\right| \mathrm{d} x=\frac{19}{384} h^{8}, \quad \int_{\alpha_{1}}^{x_{1}}\left|\theta_{2}(x)\right| \mathrm{d} x=\frac{341}{10368} h^{2}, \int_{\alpha_{0}}^{\alpha_{1}}\left|\theta_{1}(x)\right| \mathrm{d} x=\frac{\sqrt{3}}{54} h^{1} ∫ x 1 x 1 | 0 1 ( x ) | d x = 19 384 h 8 , ∫ a 1 x 1 | i 2 ( x ) | d x = 341 10368 h 2 , ∫ a 0 a 1 | i 1 ( x ) | d x = 3 54 h 1 .
It is a rule that
∫
x
n
x
n
|
θ
(
x
)
|
d
x
=
2
(
19
384
+
341
10368
)
h
3
+
(
n
−
4
)
3
54
h
2
=
427
2592
h
2
+
+
(
n
−
4
)
3
54
h
2
∫
x
n
x
n
|
θ
(
x
)
|
d
x
=
2
19
384
+
341
10368
h
3
+
(
n
−
4
)
3
54
h
2
=
427
2592
h
2
+
+
(
n
−
4
)
3
54
h
2
{:[int_(x_(n))^(x_(n))|theta(x)|dx=2((19)/(384)+(341)/(10368))h^(3)+(n-4)(sqrt3)/(54)h^(2)=(427)/(2592)h^(2)+],[+(n-4)(sqrt3)/(54)h^(2)]:} \begin{gathered}
\int_{x_{n}}^{x_{n}}|\theta(x)| d x=2\left(\frac{19}{384}+\frac{341}{10368}\right) h^{3}+(n-4) \frac{\sqrt{3}}{54} h^{2}=\frac{427}{2592} h^{2}+ \\
+(n-4) \frac{\sqrt{3}}{54} h^{2}
\end{gathered} ∫ x n x n | i ( x ) | d x = 2 ( 19 384 + 341 10368 ) h 3 + ( n − 4 ) 3 54 h 2 = 427 2592 h 2 + + ( n − 4 ) 3 54 h 2 I declare
(11)
K
L
=
427
−
192
3
2
―
592
+
n
3
54
.
(11)
K
L
=
427
−
192
3
2
¯
592
+
n
3
54
.
{:(11)K_(L)=(427-192sqrt3)/( bar(2)592)+n(sqrt3)/(54).:} \begin{equation*}
K_{L}=\frac{427-192 \sqrt{3}}{\overline{2} 592}+n \frac{\sqrt{3}}{54} . \tag{11}
\end{equation*} (11) K L = 427 − 192 3 2 ― 592 + n 3 54 .
COMPARISON OF NUMBERS
K
C
,
K
D
,
K
L
K
C
,
K
D
,
K
L
K_(C),K_(D),K_(L) K_{C}, K_{D}, K_{L} K C , K D , K L , WITH THE NUMBER
K
,
K
,
K_(", ") K_{\text {, }} K , CORRESPONDING TO THE FORMULA (T)
As is known in formula (T), assuming that the function aff(x) is of class
C
1
[
x
0
,
x
n
]
C
1
x
0
,
x
n
C^(1)[x_(0),x_(n)] C^{1}\left[x_{0}, x_{n}\right] C 1 [ x 0 , x n ] , evem
(12)
R
T
=
∫
x
0
a
x
x
(
x
)
f
′
′
(
x
)
d
x
(12)
R
T
=
∫
x
0
a
x
x
(
x
)
f
′
′
(
x
)
d
x
{:(12)R_(T)=int_(x_(0))^(a_(x))x(x)f^('')(x)dx:} \begin{equation*}
R_{T}=\int_{x_{0}}^{a_{x}} x(x) f^{\prime \prime}(x) d x \tag{12}
\end{equation*} (12) R T = ∫ x 0 a x x ( x ) f ′ ′ ( x ) d x where the function
χ
(
x
)
χ
(
x
)
chi(x) \chi(x) x ( x ) coincides on the intervals
[
x
0
,
x
1
]
,
[
x
1
,
x
0
]
,
…
x
0
,
x
1
,
x
1
,
x
0
,
…
[x_(0),x_(1)],[x_(1),x_(0)],dots \left[x_{0}, x_{1}\right],\left[x_{1}, x_{0}\right], \ldots [ x 0 , x 1 ] , [ x 1 , x 0 ] , … with polynomials
χ
1
(
x
)
=
(
x
−
x
0
)
(
x
−
x
1
)
2
,
χ
1
(
x
)
=
(
x
−
x
1
)
(
x
−
x
1
)
2
,
…
χ
1
(
x
)
=
x
−
x
0
x
−
x
1
2
,
χ
1
(
x
)
=
x
−
x
1
x
−
x
1
2
,
…
chi_(1)(x)=((x-x_(0))(x-x_(1)))/(2),quadchi_(1)(x)=((x-x_(1))(x-x_(1)))/(2),dots \chi_{1}(x)=\frac{\left(x-x_{0}\right)\left(x-x_{1}\right)}{2}, \quad \chi_{1}(x)=\frac{\left(x-x_{1}\right)\left(x-x_{1}\right)}{2}, \ldots x 1 ( x ) = ( x − x 0 ) ( x − x 1 ) 2 , x 1 ( x ) = ( x − x 1 ) ( x − x 1 ) 2 , … It is noted that
(13)
(
R
T
)
⩽
K
T
M
2
h
a
(13)
R
T
⩽
K
T
M
2
h
a
{:(13)(R_(T)) <= K_(T)M_(2)h^(a):} \begin{equation*}
\left(R_{T}\right) \leqslant K_{T} M_{2} h^{\mathrm{a}} \tag{13}
\end{equation*} (13) ( R T ) ⩽ K T M 2 h a where
((14)
K
r
=
n
12
.
((14)
K
r
=
n
12
.
{:((14)K_(r)=(n)/( 12).:} \begin{equation*}
K_{r}=\frac{n}{12} . \tag{(14}
\end{equation*} ((14) K r = n 12 . Let's see what we have.
K
C
−
K
L
=
1095
+
292
|
73
−
2592
|
3
69984
=
0
,
0157
…
,
K
p
−
K
c
=
1620
+
3
−
73
/
73
−
1739
=
0
,
0253
.
K
C
−
K
L
=
1095
+
292
|
73
−
2592
|
3
69984
=
0
,
0157
…
,
K
p
−
K
c
=
1620
+
3
−
73
/
73
−
1739
=
0
,
0253
.
{:[K_(C)-K_(L)=(1095+292|73-2592|3)/(69984)=0","0157 dots","],[K_(p)-K_(c)=1620+3-73//73-1739=0","0253.]:} \begin{gathered}
K_{C}-K_{L}=\frac{1095+292|73-2592| 3}{69984}=0,0157 \ldots, \\
K_{p}-K_{c}=1620+3-73 / 73-1739=0,0253 .
\end{gathered} K C − K L = 1095 + 292 | 73 − 2592 | 3 69984 = 0 , 0157 … , K p − K c = 1620 + 3 − 73 / 73 − 1739 = 0 , 0253 . It is also found that we have
K
r
−
K
D
=
(
54
−
12
∣
3
)
(
n
−
2
)
+
(
37
−
12
∣
3
―
)
648
>
0
(
n
⩾
2
)
K
r
−
K
D
=
(
54
−
12
∣
3
)
(
n
−
2
)
+
(
37
−
12
∣
3
¯
)
648
>
0
(
n
⩾
2
)
K_(r)-K_(D)=((54-12∣3)(n-2)+(37-12∣ bar(3)))/(648) > 0quad(n >= 2) K_{r}-K_{D}=\frac{(54-12 \mid 3)(n-2)+(37-12 \mid \overline{3})}{648}>0 \quad(n \geqslant 2) K r − K D = ( 54 − 12 ∣ 3 ) ( n − 2 ) + ( 37 − 12 ∣ 3 ― ) 648 > 0 ( n ⩾ 2 ) from which it follows that we have
(15)
K
t
<
K
r
<
K
D
<
K
T
.
(15)
K
t
<
K
r
<
K
D
<
K
T
.
{:(15)K_(t) < K_(r) < K_(D) < K_(T).:} \begin{equation*}
\boldsymbol{K}_{t}<\boldsymbol{K}_{r}<\boldsymbol{K}_{\mathrm{D}}<\boldsymbol{K}_{T} . \tag{15}
\end{equation*} (15) K t < K r < K D < K T . In conclusion, we have determined and evaluated the residues in formulas (C), (D), (L). From inequalities (15) it follows that the more advantageous is formula (L), which immediately follows, with a very small difference for the coefficient
K
K
K K K , formula (C). The most disadvantageous of all the formulas considered in this note is formula (T). We will now introduce a method for constructing formulas (C), (D), (L) and other analogous ones.
RPR Academy Cluj Branch,
Computing Institute
SUMMARY
The remainders of the quadrature formulas (C), (D), (L), (T) are determined by the formulas (1), (6), (9), (12), and the evaluations (2), (7), (10), (13), characterized by the numbers
K
G
,
K
D
,
K
L
,
K
T
K
G
,
K
D
,
K
L
,
K
T
K_(G),K_(D),K_(L),K_(T) K_{G}, K_{D}, K_{L}, K_{T} K G , K D , K L , K T given by the formulas Cateva formula practice of cuadratura
895
(1), (14). We demonstrate the inequalities, (15), from which it follows that (1). Ja (1) is the most advantageous, followed by formula (C), and that formula (T) is the most disadvantageous.
BIBLIOGRAPHY
G Goblarl Operalions sur les commes experimeniales, CF Aced. Sel. Fatii, 1058, e46, 1799-1800.
Minkull, Technique de Caleut namérique, Libralite Polylechnique Ch. Béranger, Paris, 1052, 244.
3. DV IONFBCT, CDOdraturi numerice, Edit. Lehnica, Bucurest, 1957.
