ASUPRA FORMULEI DE INTERPOLARE A LUI HERMITE ȘI A UNOR APLICAȚII ALE ACESTEIA*
Ch. Hermite [1] a formulat următoarea problemă generală de interpolare:
Să se construiască un polinom, de grad minim, care împreună cu derivatele sale succesive - pînă inclusiv la anumite ordine
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,
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r_(1)-1,dots,r_(s)-1 r_{1}-1, \ldots, r_{s}-1 r 1 − 1 , … , r s − 1 , să ia pe punctele distincte
(1)
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(1)
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{:(1)x_(1)","x_(2)","dots","x_(s):} \begin{equation*}
x_{1}, x_{2}, \ldots, x_{s} \tag{1}
\end{equation*} (1) x 1 , x 2 , … , x s valori date dinainte.
Se știe
1
1
^(1) { }^{1} 1 că acest polinom există și că e unic, iar gradul său este
n
n
n n n , unde
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1
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n=r_(1)+dots+r_(s)-1 n=r_{1}+\ldots+r_{s}-1 n = r 1 + … + r s − 1 .
Să notăm acest polinom cu
H
n
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x
)
H
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H_(n)(x) H_{n}(x) H n ( x ) și să presupunem că valorile date dinainte sînt valorile luate, pe punctele
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x_(1),x_(2),dots,x_(s) x_{1}, x_{2}, \ldots, x_{s} x 1 , x 2 , … , x s , de o funcție
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f
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f(x) f(x) f ( x ) şi derivatele ei. Atunci polinomul de interpolare al lui Hermite relativ la funcția
f
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f
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f(x) f(x) f ( x ) și punctele (1), cărora li se ataşează respectiv ordinele de multiplicitate
r
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s
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r_(1),r_(2),dots,r_(s) r_{1}, r_{2}, \ldots, r_{s} r 1 , r 2 , … , r s ,
(2)
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⏟
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⏟
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∣
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(2)
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{:(2)H_(n)(x)=H_(n)(ubrace(x_(1),dots,x_(1))_(r_(1))","ubrace(x_(2),dots,x_(2))_(r_(2))","dots","ubrace(x_(s),dots,x_(s))_(r_(s));f∣x):} \begin{equation*}
H_{n}(x)=H_{n}(\underbrace{x_{1}, \ldots, x_{1}}_{r_{1}}, \underbrace{x_{2}, \ldots, x_{2}}_{r_{2}}, \ldots, \underbrace{x_{s}, \ldots, x_{s}}_{r_{s}} ; f \mid x) \tag{2}
\end{equation*} (2) H n ( x ) = H n ( x 1 , … , x 1 ⏟ r 1 , x 2 , … , x 2 ⏟ r 2 , … , x s , … , x s ⏟ r s ; f ∣ x ) va fi polinomul care verifică condițiile
(3)
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(3)
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{:(3)H_(n)^((l))(x_(k))=f_(n)^((l))(x_(k))((l=0,1,dots,r_(k)-1)/(k=1,2,dots,s)):} \begin{equation*}
H_{n}^{(l)}\left(x_{k}\right)=f_{n}^{(l)}\left(x_{k}\right)\binom{l=0,1, \ldots, r_{k}-1}{k=1,2, \ldots, s} \tag{3}
\end{equation*} (3) H n ( l ) ( x k ) = f n ( l ) ( x k ) ( l = 0 , 1 , … , r k − 1 k = 1 , 2 , … , s )
Dacă
(4)
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(4)
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{:(4)H_(n)(x)=sum_(j=0)^(n)a_(j)x^(j)",":} \begin{equation*}
H_{n}(x)=\sum_{j=0}^{n} a_{j} x^{j}, \tag{4}
\end{equation*} (4) H n ( x ) = ∑ j = 0 n a j x j ,
condițiile precedente ne conduc la sistemul de
n
+
1
n
+
1
n+1 n+1 n + 1 ecuații cu
n
+
1
n
+
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n+1 n+1 n + 1 necunoscute :
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a_(0),a_(1),dots,a_(n) a_{0}, a_{1}, \ldots, a_{n} a 0 , a 1 , … , a n
(5)
∑
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…
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.
(5)
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{:[(5)sum_(j=i)^(n)j(j-1)dots(j-l+1)a_(j)x_(k)^(j-l)=f^((l))(x_(k))],[(l=0,1,dots,r_(k)-1;k=1,2,dots,s).]:} \begin{gather*}
\sum_{j=i}^{n} j(j-1) \ldots(j-l+1) a_{j} x_{k}^{j-l}=f^{(l)}\left(x_{k}\right) \tag{5}\\
\left(l=0,1, \ldots, r_{k}-1 ; k=1,2, \ldots, s\right) .
\end{gather*} (5) ∑ j = i n j ( j − 1 ) … ( j − l + 1 ) a j x k j − l = f ( l ) ( x k ) ( l = 0 , 1 , … , r k − 1 ; k = 1 , 2 , … , s ) . Determinantul acestui sistem - se știe - e diferit de zero, căci punctele (1) sînt distincte.
Pentru a găsi expresia polinomului
H
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H
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H_(n)(x) H_{n}(x) H n ( x ) am putea elimina necunoscutele
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a_(0),a_(1),dots,a_(n) a_{0}, a_{1}, \ldots, a_{n} a 0 , a 1 , … , a n între ecuațiile (4) și (5). Calea aceasta e însă foarte anevoioasă. G. Z e m p 1 e n [3] a încercat totuşi pe această cale să găsească expresia explicită a polinomului
H
n
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x
)
H
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H_(n)(x) H_{n}(x) H n ( x ) . Menționăm însă că rezultatele pe care le-a obținut acest autor sînt inexacte; formula pe care a dat-o nu corespunde soluției problemei dacă
s
≧
2
s
≧
2
s >= 2 s \geqq 2 s ≧ 2 și numerele
r
i
r
i
r_(i) r_{i} r i sînt mai mari decît 2 . Bineînțeles că și rezultatele pe care le-a dat în partea a doua a lucrării sale, relative la descompunerea unei funcții raționale în fracții simple, sînt de asemenea eronate. În recenzia făcută de We1tzien Zeh1endorf [4] şi în memoriul lui E. Netto din [5], în care se aminteşte de această lucrare, nu s-au făcut observații asupra rezultatelor obținute de G. Zemplen.
3. Din eliminarea amintită mai sus a necunoscutelor
a
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n
a
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a_(0),a_(1),dots,a_(n) a_{0}, a_{1}, \ldots, a_{n} a 0 , a 1 , … , a n și din rezolvarea formală în raport cu
H
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H
n
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H_(n)(x) H_{n}(x) H n ( x ) a ecuației care s-a obținut, ne putem da seama că polinomul căutat este de forma
(6)
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(6)
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{:(6)H_(n)(x)=sum_(i=1)^(s)sum_(k=0)^(r_(i)-1)l_(i,k)(x)f^((k))(x_(i))",":} \begin{equation*}
H_{n}(x)=\sum_{i=1}^{s} \sum_{k=0}^{r_{i}-1} l_{i, k}(x) f^{(k)}\left(x_{i}\right), \tag{6}
\end{equation*} (6) H n ( x ) = ∑ i = 1 s ∑ k = 0 r i − 1 l i , k ( x ) f ( k ) ( x i ) , unde
l
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l
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l_(l,k)(x) l_{l, k}(x) l l , k ( x ) sînt polinoame de grad
n
n
n n n . Din calculul formal de mai sus se observă imediat că polinoamele
l
l
,
k
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l
l
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k
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l_(l,k)(x) l_{l, k}(x) l l , k ( x ) sînt independente de
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f
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f(x) f(x) f ( x ) și că, t, tinînd seama de (3), ele trebuie să verifice următoarele condiții :
(7)
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(8)
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pentru
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pentru
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(7)
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(8)
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{:[(7)l_(i,k)^((p))(x_(j))=0(j!=i;p=0,1,dots,r_(j)-1)],[(8)l_(i,k)^((p))(x_(i))={[0","" pentru "p!=k],[1","" pentru "p=k]quad(p=0,1,dots,r_(i)-1):}]:} \begin{gather*}
l_{i, k}^{(p)}\left(x_{j}\right)=0\left(j \neq i ; p=0,1, \ldots, r_{j}-1\right) \tag{7}\\
l_{i, k}^{(p)}\left(x_{i}\right)=\left\{\begin{array}{l}
0, \text { pentru } p \neq k \\
1, \text { pentru } p=k
\end{array} \quad\left(p=0,1, \ldots, r_{i}-1\right)\right. \tag{8}
\end{gather*} (7) l i , k ( p ) ( x j ) = 0 ( j ≠ i ; p = 0 , 1 , … , r j − 1 ) (8) l i , k ( p ) ( x i ) = { 0 , pentru p ≠ k 1 , pentru p = k ( p = 0 , 1 , … , r i − 1 ) Aceste conditii ne vor determina complet polinoamele fundamentale de interpolare
l
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l_(i,k)(x) l_{i, k}(x) l i , k ( x ) . Intr-adevăr, în baza lui (7) se observă că
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i
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l
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l_(i,k)(x) l_{i, k}(x) l i , k ( x ) conține ca factor produsul
(9)
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γ
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…
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γ
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.
(9)
g
i
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γ
1
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γ
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.
{:(9)g_(i)(x)=(x-x_(1))^(gamma_(1))dots(x-x_(i-1))^(gamma_(i-1))(x-x_(i+1))^(gamma_(i+1))dots(x-x_(s))^(gamma_(s)).:} \begin{equation*}
g_{i}(x)=\left(x-x_{1}\right)^{\gamma_{1}} \ldots\left(x-x_{i-1}\right)^{\gamma_{i-1}}\left(x-x_{i+1}\right)^{\gamma_{i+1}} \ldots\left(x-x_{s}\right)^{\gamma_{s}} . \tag{9}
\end{equation*} (9) g i ( x ) = ( x − x 1 ) γ 1 … ( x − x i − 1 ) γ i − 1 ( x − x i + 1 ) γ i + 1 … ( x − x s ) γ s . Ținînd seama și de condițiile (8), se ajunge la concluzia că
(
x
−
x
i
)
k
x
−
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i
k
(x-x_(i))^(k) \left(x-x_{i}\right)^{k} ( x − x i ) k este de asemenea un factor al lui
l
l
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k
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)
l
l
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k
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l_(l,k)(x) l_{l, k}(x) l l , k ( x ) . Rezultă că acesta este de forma
(10)
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(10)
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{:(10)l_(i,k)(x)=g_(i)(x)(x-x_(i))^(k)h_(i,k)(x)",":} \begin{equation*}
l_{i, k}(x)=g_{i}(x)\left(x-x_{i}\right)^{k} h_{i, k}(x), \tag{10}
\end{equation*} (10) l i , k ( x ) = g i ( x ) ( x − x i ) k h i , k ( x ) , unde
h
i
,
k
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x
)
h
i
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k
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x
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h_(i,k)(x) h_{i, k}(x) h i , k ( x ) este un polinom de gradul
r
i
−
k
−
1
r
i
−
k
−
1
r_(i)-k-1 r_{i}-k-1 r i − k − 1 . Rămîne să determinăm acest ultim polinom. Dacă îl dezvoltăm după formula lui Taylor în vecinătatea lui
x
=
x
i
x
=
x
i
x=x_(i) x=x_{i} x = x i , se obtine
(11)
h
i
,
k
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=
∑
m
=
0
r
i
−
k
−
1
(
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−
x
i
)
m
m
!
h
i
,
k
(
m
)
(
x
i
)
.
(11)
h
i
,
k
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=
∑
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0
r
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k
−
1
x
−
x
i
m
m
!
h
i
,
k
(
m
)
x
i
.
{:(11)h_(i,k)(x)=sum_(m=0)^(r_(i)-k-1)((x-x_(i))^(m))/(m!)h_(i,k)^((m))(x_(i)).:} \begin{equation*}
h_{i, k}(x)=\sum_{m=0}^{r_{i}-k-1} \frac{\left(x-x_{i}\right)^{m}}{m!} h_{i, k}^{(m)}\left(x_{i}\right) . \tag{11}
\end{equation*} (11) h i , k ( x ) = ∑ m = 0 r i − k − 1 ( x − x i ) m m ! h i , k ( m ) ( x i ) . Din (10) avem
(
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−
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)
k
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1
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1
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(x-x_(i))^(k)h_(i,k)(x)=l_(i,k)(x)(1)/(g_(i)(x)) \left(x-x_{i}\right)^{k} h_{i, k}(x)=l_{i, k}(x) \frac{1}{g_{i}(x)} ( x − x i ) k h i , k ( x ) = l i , k ( x ) 1 g i ( x ) Dacă se calculează, după formula lui Leibniz, derivata de ordinul
q
q
q q q a ambilor membri ai acestei egalități, se obține
∑
j
=
0
q
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q
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)
[
(
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k
]
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g
i
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∑
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sum_(j=0)^(q)((q)/(j))[(x-x_(i))^(k)]^((q-j))h_(i,k)^((j))(x)=sum_(j=0)^(q)((q)/(j))l_(i,k)^((q-j))(x)((1)/(g_(i)(x)))^((j)) \sum_{j=0}^{q}\binom{q}{j}\left[\left(x-x_{i}\right)^{k}\right]^{(q-j)} h_{i, k}^{(j)}(x)=\sum_{j=0}^{q}\binom{q}{j} l_{i, k}^{(q-j)}(x)\left(\frac{1}{g_{i}(x)}\right)^{(j)} ∑ j = 0 q ( q j ) [ ( x − x i ) k ] ( q − j ) h i , k ( j ) ( x ) = ∑ j = 0 q ( q j ) l i , k ( q − j ) ( x ) ( 1 g i ( x ) ) ( j ) Dacă se ia
q
=
k
+
m
q
=
k
+
m
q=k+m q=k+m q = k + m , se face
x
=
x
i
x
=
x
i
x=x_(i) x=x_{i} x = x i şi se ține seama de (7) și (8), se obţin relațiile
k
!
h
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k
!
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x
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x
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{:[k!h_(i,k)^((m))(x_(i))=((1)/(g_(i)(x)))_(x=x_(i))^((m))],[(m=0,1,dots,r_(i)-k-1)]:} \begin{aligned}
& k!h_{i, k}^{(m)}\left(x_{i}\right)=\left(\frac{1}{g_{i}(x)}\right)_{x=x_{i}}^{(m)} \\
& \left(m=0,1, \ldots, r_{i}-k-1\right)
\end{aligned} k ! h i , k ( m ) ( x i ) = ( 1 g i ( x ) ) x = x i ( m ) ( m = 0 , 1 , … , r i − k − 1 ) În felul acesta se ajunge la următoarele expresii pentru polinoa mele fundamentale de interpolare
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{:(12)l_(i,k)(x)=sum_(r=0)^(r_(i)-k-1)[((x-x_(i))^(r))/(r!)((1)/(g_(i)(x)))_(x_(i))^((r))]g_(i)(x)",":} \begin{equation*}
l_{i, k}(x)=\sum_{r=0}^{r_{i}-k-1}\left[\frac{\left(x-x_{i}\right)^{r}}{r!}\left(\frac{1}{g_{i}(x)}\right)_{x_{i}}^{(r)}\right] g_{i}(x), \tag{12}
\end{equation*} (12) l i , k ( x ) = ∑ r = 0 r i − k − 1 [ ( x − x i ) r r ! ( 1 g i ( x ) ) x i ( r ) ] g i ( x ) , iar polinomul (6) devine
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{:(13)H_(n)(x)=sum_(i=1)^(s)sum_(k=0)^(r_(i)-1)sum_(r=0)^(r_(i)-k-2)((x-x_(i))^(k))/(k!)[((x-x_(i))^(r))/(r!)((1)/(g_(i)(x)))_(x_(i))^((r))]g_(i)(x)f^((k))(x_(i)).:} \begin{equation*}
H_{n}(x)=\sum_{i=1}^{s} \sum_{k=0}^{r_{i}-1} \sum_{r=0}^{r_{i}-k-2} \frac{\left(x-x_{i}\right)^{k}}{k!}\left[\frac{\left(x-x_{i}\right)^{r}}{r!}\left(\frac{1}{g_{i}(x)}\right)_{x_{i}}^{(r)}\right] g_{i}(x) f^{(k)}\left(x_{i}\right) . \tag{13}
\end{equation*} (13) H n ( x ) = ∑ i = 1 s ∑ k = 0 r i − 1 ∑ r = 0 r i − k − 2 ( x − x i ) k k ! [ ( x − x i ) r r ! ( 1 g i ( x ) ) x i ( r ) ] g i ( x ) f ( k ) ( x i ) . La această expresie a polinomului
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H_(n)(x) H_{n}(x) H n ( x ) a ajuns, pe o cale puţin deosebită, V. L. Gonciarov [6].
4. Dacă se calculează derivata de ordinul
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j j j , în punctul
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x_(i) x_{i} x i , a lui
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(1)/(g_(i)(x)) \frac{1}{g_{i}(x)} 1 g i ( x ) , se obține
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{:[((1)/(g_(i)(x)))_(x_(i))^((j))=(-1)^(j)sum(j!)/(alpha_(1)!dots//dotsalpha_(s)!)xx],[xx(r_(1)(r_(1)+1)dots(r_(1)+alpha_(1)-1)dots//dotsr_(s)(r_(s)+1)dots(r_(s)+alpha_(s)-1))/((x_(i)-x_(1))^(r_(1)+alpha_(1))dots//dots(x_(i)-x_(s))^(r_(s)+alpha_(s)))","]:} \begin{gathered}
\left(\frac{1}{g_{i}(x)}\right)_{x_{i}}^{(j)}=(-1)^{j} \sum \frac{j!}{\alpha_{1}!\ldots / \ldots \alpha_{s}!} \times \\
\times \frac{r_{1}\left(r_{1}+1\right) \ldots\left(r_{1}+\alpha_{1}-1\right) \ldots / \ldots r_{s}\left(r_{s}+1\right) \ldots\left(r_{s}+\alpha_{s}-1\right)}{\left(x_{i}-x_{1}\right)^{r_{1}+\alpha_{1}} \ldots / \ldots\left(x_{i}-x_{s}\right)^{r_{s}+\alpha_{s}}},
\end{gathered} ( 1 g i ( x ) ) x i ( j ) = ( − 1 ) j ∑ j ! α 1 ! … / … α s ! × × r 1 ( r 1 + 1 ) … ( r 1 + α 1 − 1 ) … / … r s ( r s + 1 ) … ( r s + α s − 1 ) ( x i − x 1 ) r 1 + α 1 … / … ( x i − x s ) r s + α s , liniuța indicînd absența factorilor cu toți indicii
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i i i . Aici suma se extinde la toate sistemele de întregi nenegativi care verifică relaţia
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alpha_(1)+alpha_(2)+dots+alpha_(i-1)+alpha_(i+1)+dots+alpha_(s)=j \alpha_{1}+\alpha_{2}+\ldots+\alpha_{i-1}+\alpha_{i+1}+\ldots+\alpha_{s}=j α 1 + α 2 + … + α i − 1 + α i + 1 + … + α s = j Ținînd seama de aceasta, polinomul fundamental de interpolare
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l_(i,k)(x) l_{i, k}(x) l i , k ( x ) se poate pune sub forma
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{:[(14)l_(i,k)(x)=(g_(i)(x))/(g_(i)(x_(i))){((x-x_(i))^(k))/(k!)sum_(j=0)^(r_(i)-k-1)(x-x_(i))^(j)xx:}],[{: xxsum_(a_(1)+dots+a_(i-1)+alpha_(i+1)+dots+alpha_(s)=j)(((r_(1)+alpha_(1)-1)/(alpha_(1)))dots//dots((r_(s)+alpha_(s)-1)/(alpha_(s))))/((x_(1)-x_(i))^(alpha_(1))dots//dots(x_(s)-x_(i))^(alpha_(s)))}]:} \begin{gather*}
l_{i, k}(x)=\frac{g_{i}(x)}{g_{i}\left(x_{i}\right)}\left\{\frac{\left(x-x_{i}\right)^{k}}{k!} \sum_{j=0}^{r_{i}-k-1}\left(x-x_{i}\right)^{j} \times\right. \tag{14}\\
\left.\times \sum_{a_{1}+\ldots+a_{i-1}+\alpha_{i+1}+\ldots+\alpha_{s}=j} \frac{\binom{r_{1}+\alpha_{1}-1}{\alpha_{1}} \ldots / \ldots\binom{r_{s}+\alpha_{s}-1}{\alpha_{s}}}{\left(x_{1}-x_{i}\right)^{\alpha_{1}} \ldots / \ldots\left(x_{s}-x_{i}\right)^{\alpha_{s}}}\right\}
\end{gather*} (14) l i , k ( x ) = g i ( x ) g i ( x i ) { ( x − x i ) k k ! ∑ j = 0 r i − k − 1 ( x − x i ) j × × ∑ a 1 + … + a i − 1 + α i + 1 + … + α s = j ( r 1 + α 1 − 1 α 1 ) … / … ( r s + α s − 1 α s ) ( x 1 − x i ) α 1 … / … ( x s − x i ) α s } În felul acesta am ajuns la expresia definitivă a polinomului de interpolare (2) al lui Hermite
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{:[H_(n)(x)=],[(15)=sum_(i=1)^(s)(g_(i)(x))/(g_(i)(x_(i))){sum_(k=0)^(r_(i)-1)((x-x_(i))^(k))/(k!)[sum_(j=0)^(r_(i)-k-1)A_(alpha_(1),dots,alpha_(i-1),alpha_(i+1),dots,alpha_(s))^((j))(x-x_(i))^(j)]f^((k))(x_(i))}]:} \begin{align*}
& H_{n}(x)= \\
& =\sum_{i=1}^{s} \frac{g_{i}(x)}{g_{i}\left(x_{i}\right)}\left\{\sum_{k=0}^{r_{i}-1} \frac{\left(x-x_{i}\right)^{k}}{k!}\left[\sum_{j=0}^{r_{i}-k-1} A_{\alpha_{1}, \ldots, \alpha_{i-1}, \alpha_{i+1}, \ldots, \alpha_{s}}^{(j)}\left(x-x_{i}\right)^{j}\right] f^{(k)}\left(x_{i}\right)\right\} \tag{15}
\end{align*} H n ( x ) = (15) = ∑ i = 1 s g i ( x ) g i ( x i ) { ∑ k = 0 r i − 1 ( x − x i ) k k ! [ ∑ j = 0 r i − k − 1 A α 1 , … , α i − 1 , α i + 1 , … , α s ( j ) ( x − x i ) j ] f ( k ) ( x i ) } unde
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{:[A_(alpha_(1),dots,alpha_(i-1),alpha_(i+1),dots,alpha_(s))^((j))=],[(16)=sum_(alpha_(1)+dots+alpha_(i-1)+alpha_(i-1)+dots+alpha_(s)=j)(((gamma_(1)+alpha_(1)-1)/(alpha_(1)))dots//dots((gamma_(s)+alpha_(s)-1)/(alpha_(s))))/((x_(1)-x_(i))^(alpha_(1))dots//dots(x_(s)-x_(i))^(alpha_(s))).]:} \begin{gather*}
A_{\alpha_{1}, \ldots, \alpha_{i-1}, \alpha_{i+1}, \ldots, \alpha_{s}}^{(j)}= \\
=\sum_{\alpha_{1}+\ldots+\alpha_{i-1}+\alpha_{i-1}+\ldots+\alpha_{s}=j} \frac{\binom{\gamma_{1}+\alpha_{1}-1}{\alpha_{1}} \ldots / \ldots\binom{\gamma_{s}+\alpha_{s}-1}{\alpha_{s}}}{\left(x_{1}-x_{i}\right)^{\alpha_{1}} \ldots / \ldots\left(x_{s}-x_{i}\right)^{\alpha_{s}}} . \tag{16}
\end{gather*} A α 1 , … , α i − 1 , α i + 1 , … , α s ( j ) = (16) = ∑ α 1 + … + α i − 1 + α i − 1 + … + α s = j ( γ 1 + α 1 − 1 α 1 ) … / … ( γ s + α s − 1 α s ) ( x 1 − x i ) α 1 … / … ( x s − x i ) α s .
Dacă se consideră formula de interpolare a lui Hermite
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{:(17)f(x)=H_(n)(x)+R_(n+1)(x)",":} \begin{equation*}
f(x)=H_{n}(x)+R_{n+1}(x), \tag{17}
\end{equation*} (17) f ( x ) = H n ( x ) + R n + 1 ( x ) , restul
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R_(n+1)(x)=(x-x_(1))^(r_(1))dots(x-x_(s))^(r_(s))[x,ubrace(x_(1),dots,x_(1))_(r_(1)),ubrace(x_(2),dots,x_(2))_(r_(2)),dots,ubrace(x_(s),dots,x_(s))_(r_(s));f]", " R_{n+1}(x)=\left(x-x_{1}\right)^{r_{1}} \ldots\left(x-x_{s}\right)^{r_{s}}[x, \underbrace{x_{1}, \ldots, x_{1}}_{r_{1}}, \underbrace{x_{2}, \ldots, x_{2}}_{r_{2}}, \ldots, \underbrace{x_{s}, \ldots, x_{s}}_{r_{s}} ; f] \text {, } R n + 1 ( x ) = ( x − x 1 ) r 1 … ( x − x s ) r s [ x , x 1 , … , x 1 ⏟ r 1 , x 2 , … , x 2 ⏟ r 2 , … , x s , … , x s ⏟ r s ; f ] , unde
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[alpha_(1),alpha_(2),dots,alpha_(m+1);f] \left[\alpha_{1}, \alpha_{2}, \ldots, \alpha_{m+1} ; f\right] [ α 1 , α 2 , … , α m + 1 ; f ] este diferența divizată de ordinul
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m m m a funcției
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alpha_(1),alpha_(2),dots,alpha_(m+1) \alpha_{1}, \alpha_{2}, \ldots, \alpha_{m+1} α 1 , α 2 , … , α m + 1 .
În ipoteza că
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f(x) f(x) f ( x ) are derivată de ordinul
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n+1 n+1 n + 1 în intervalul cel mai mic care conține valorile
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x_(1),dots,x_(s),x x_{1}, \ldots, x_{s}, x x 1 , … , x s , x , restul se poate exprima, după cum se cunoaşte, prin formula
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,
{:(18)R_(n+1)(x)=((x-x_(1))^(r_(1))(x-x_(2))^(r_(2))dots(x-x_(s))^(r_(s)))/((n+1)!)f^((n+1))(xi)",":} \begin{equation*}
R_{n+1}(x)=\frac{\left(x-x_{1}\right)^{r_{1}}\left(x-x_{2}\right)^{r_{2}} \ldots\left(x-x_{s}\right)^{r_{s}}}{(n+1)!} f^{(n+1)}(\xi), \tag{18}
\end{equation*} (18) R n + 1 ( x ) = ( x − x 1 ) r 1 ( x − x 2 ) r 2 … ( x − x s ) r s ( n + 1 ) ! f ( n + 1 ) ( ξ ) , unde
ξ
ξ
xi \xi ξ aparține celui mai mic interval care conţine valorile
x
i
x
i
x_(i) x_{i} x i și
x
x
x x x .
6. Exemplu. Să se găsească polinomul de interpolare de grad minim relativ la funcţia
f
(
x
)
f
(
x
)
f(x) f(x) f ( x ) și la nodurile
x
1
=
x
2
=
x
3
=
−
1
,
x
4
=
0
x
1
=
x
2
=
x
3
=
−
1
,
x
4
=
0
x_(1)=x_(2)=x_(3)=-1,x_(4)=0 x_{1}=x_{2}=x_{3}=-1, x_{4}=0 x 1 = x 2 = x 3 = − 1 , x 4 = 0 ,
x
5
=
x
6
=
x
7
=
1
x
5
=
x
6
=
x
7
=
1
x_(5)=x_(6)=x_(7)=1 x_{5}=x_{6}=x_{7}=1 x 5 = x 6 = x 7 = 1 .
Pe baza formulei precedente găsim
H
6
(
x
)
=
H
6
(
−
1
,
−
1
,
−
1
,
0
,
1
,
1
,
1
;
f
∣
x
)
=
(
1
−
x
2
)
3
f
(
0
)
+
+
x
(
x
−
1
)
3
[
1
8
+
5
(
x
+
1
)
16
+
(
x
+
1
)
2
2
]
f
(
−
1
)
+
x
(
x
−
1
)
3
(
x
+
1
)
[
1
8
+
5
(
x
+
1
)
16
]
f
′
(
−
1
)
+
+
x
(
x
−
1
)
3
(
x
+
1
)
2
16
f
′
′
(
−
1
)
+
x
(
x
+
1
)
3
[
1
8
−
5
16
(
x
−
1
)
+
(
x
−
1
)
2
2
]
f
(
1
)
+
(19)
+
x
(
x
+
1
)
3
(
x
−
1
)
[
1
8
−
5
(
x
−
1
)
16
]
f
′
(
1
)
+
x
(
x
+
1
)
3
(
x
−
1
)
2
16
f
′
′
(
1
)
H
6
(
x
)
=
H
6
(
−
1
,
−
1
,
−
1
,
0
,
1
,
1
,
1
;
f
∣
x
)
=
1
−
x
2
3
f
(
0
)
+
+
x
(
x
−
1
)
3
1
8
+
5
(
x
+
1
)
16
+
(
x
+
1
)
2
2
f
(
−
1
)
+
x
(
x
−
1
)
3
(
x
+
1
)
1
8
+
5
(
x
+
1
)
16
f
′
(
−
1
)
+
+
x
(
x
−
1
)
3
(
x
+
1
)
2
16
f
′
′
(
−
1
)
+
x
(
x
+
1
)
3
1
8
−
5
16
(
x
−
1
)
+
(
x
−
1
)
2
2
f
(
1
)
+
(19)
+
x
(
x
+
1
)
3
(
x
−
1
)
1
8
−
5
(
x
−
1
)
16
f
′
(
1
)
+
x
(
x
+
1
)
3
(
x
−
1
)
2
16
f
′
′
(
1
)
{:[H_(6)(x)=H_(6)(-1","-1","-1","0","1","1","1;f∣x)=(1-x^(2))^(3)f(0)+],[+x(x-1)^(3)[(1)/(8)+(5(x+1))/(16)+((x+1)^(2))/(2)]f(-1)+x(x-1)^(3)(x+1)[(1)/(8)+(5(x+1))/(16)]f^(')(-1)+],[+(x(x-1)^(3)(x+1)^(2))/(16)f^('')(-1)+x(x+1)^(3)[(1)/(8)-(5)/(16)(x-1)+((x-1)^(2))/(2)]f(1)+],[(19)+x(x+1)^(3)(x-1)[(1)/(8)-(5(x-1))/(16)]f^(')(1)+(x(x+1)^(3)(x-1)^(2))/(16)f^('')(1)]:} \begin{gather*}
H_{6}(x)=H_{6}(-1,-1,-1,0,1,1,1 ; f \mid x)=\left(1-x^{2}\right)^{3} f(0)+ \\
+x(x-1)^{3}\left[\frac{1}{8}+\frac{5(x+1)}{16}+\frac{(x+1)^{2}}{2}\right] f(-1)+x(x-1)^{3}(x+1)\left[\frac{1}{8}+\frac{5(x+1)}{16}\right] f^{\prime}(-1)+ \\
+\frac{x(x-1)^{3}(x+1)^{2}}{16} f^{\prime \prime}(-1)+x(x+1)^{3}\left[\frac{1}{8}-\frac{5}{16}(x-1)+\frac{(x-1)^{2}}{2}\right] f(1)+ \\
+x(x+1)^{3}(x-1)\left[\frac{1}{8}-\frac{5(x-1)}{16}\right] f^{\prime}(1)+\frac{x(x+1)^{3}(x-1)^{2}}{16} f^{\prime \prime}(1) \tag{19}
\end{gather*} H 6 ( x ) = H 6 ( − 1 , − 1 , − 1 , 0 , 1 , 1 , 1 ; f ∣ x ) = ( 1 − x 2 ) 3 f ( 0 ) + + x ( x − 1 ) 3 [ 1 8 + 5 ( x + 1 ) 16 + ( x + 1 ) 2 2 ] f ( − 1 ) + x ( x − 1 ) 3 ( x + 1 ) [ 1 8 + 5 ( x + 1 ) 16 ] f ′ ( − 1 ) + + x ( x − 1 ) 3 ( x + 1 ) 2 16 f ′ ′ ( − 1 ) + x ( x + 1 ) 3 [ 1 8 − 5 16 ( x − 1 ) + ( x − 1 ) 2 2 ] f ( 1 ) + (19) + x ( x + 1 ) 3 ( x − 1 ) [ 1 8 − 5 ( x − 1 ) 16 ] f ′ ( 1 ) + x ( x + 1 ) 3 ( x − 1 ) 2 16 f ′ ′ ( 1 )
7. Cazuri particulare.
Dacă
r
1
=
r
2
=
…
=
r
3
=
1
r
1
=
r
2
=
…
=
r
3
=
1
r_(1)=r_(2)=dots=r_(3)=1 r_{1}=r_{2}=\ldots=r_{3}=1 r 1 = r 2 = … = r 3 = 1
H
s
−
1
(
x
)
=
∑
i
=
1
s
g
i
(
x
)
g
i
(
x
i
)
f
(
x
i
)
.
H
s
−
1
(
x
)
=
∑
i
=
1
s
g
i
(
x
)
g
i
x
i
f
x
i
.
H_(s-1)(x)=sum_(i=1)^(s)(g_(i)(x))/(g_(i)(x_(i)))f(x_(i)). H_{s-1}(x)=\sum_{i=1}^{s} \frac{g_{i}(x)}{g_{i}\left(x_{i}\right)} f\left(x_{i}\right) . H s − 1 ( x ) = ∑ i = 1 s g i ( x ) g i ( x i ) f ( x i ) .
2
∘
2
∘
2^(@) 2^{\circ} 2 ∘ . Dacă
r
1
=
r
2
=
…
=
r
s
=
2
r
1
=
r
2
=
…
=
r
s
=
2
r_(1)=r_(2)=dots=r_(s)=2 r_{1}=r_{2}=\ldots=r_{s}=2 r 1 = r 2 = … = r s = 2 , din (13) obținem
H
2
s
−
1
(
x
)
=
∑
i
=
1
s
g
i
(
x
)
g
i
(
x
i
)
[
1
−
(
x
−
x
i
)
g
i
′
(
x
i
)
g
i
(
x
i
)
]
f
(
x
i
)
+
∑
i
=
1
s
(
x
−
x
i
)
g
i
(
x
)
g
i
(
x
i
)
f
′
(
x
i
)
,
H
2
s
−
1
(
x
)
=
∑
i
=
1
s
g
i
(
x
)
g
i
x
i
1
−
x
−
x
i
g
i
′
x
i
g
i
x
i
f
x
i
+
∑
i
=
1
s
x
−
x
i
g
i
(
x
)
g
i
x
i
f
′
x
i
,
H_(2s-1)(x)=sum_(i=1)^(s)(g_(i)(x))/(g_(i)(x_(i)))[1-(x-x_(i))(g_(i)^(')(x_(i)))/(g_(i)(x_(i)))]f(x_(i))+sum_(i=1)^(s)(x-x_(i))(g_(i)(x))/(g_(i)(x_(i)))f^(')(x_(i)), H_{2 s-1}(x)=\sum_{i=1}^{s} \frac{g_{i}(x)}{g_{i}\left(x_{i}\right)}\left[1-\left(x-x_{i}\right) \frac{g_{i}^{\prime}\left(x_{i}\right)}{g_{i}\left(x_{i}\right)}\right] f\left(x_{i}\right)+\sum_{i=1}^{s}\left(x-x_{i}\right) \frac{g_{i}(x)}{g_{i}\left(x_{i}\right)} f^{\prime}\left(x_{i}\right), H 2 s − 1 ( x ) = ∑ i = 1 s g i ( x ) g i ( x i ) [ 1 − ( x − x i ) g i ′ ( x i ) g i ( x i ) ] f ( x i ) + ∑ i = 1 s ( x − x i ) g i ( x ) g i ( x i ) f ′ ( x i ) , iar din (14) rezultă
H
2
s
−
1
(
x
)
=
∑
i
=
1
s
g
i
(
x
)
g
i
(
x
i
)
(
x
−
x
i
)
[
1
x
−
x
i
+
1
x
1
−
x
i
+
…
+
1
x
s
−
x
i
]
+
+
∑
i
=
1
s
g
i
(
x
)
g
l
(
x
i
)
(
x
−
x
i
)
f
′
(
x
i
)
.
H
2
s
−
1
(
x
)
=
∑
i
=
1
s
g
i
(
x
)
g
i
x
i
x
−
x
i
1
x
−
x
i
+
1
x
1
−
x
i
+
…
+
1
x
s
−
x
i
+
+
∑
i
=
1
s
g
i
(
x
)
g
l
x
i
x
−
x
i
f
′
x
i
.
{:[H_(2s-1)(x)=sum_(i=1)^(s)(g_(i)(x))/(g_(i)(x_(i)))(x-x_(i))[(1)/(x-x_(i))+(1)/(x_(1)-x_(i))+dots+(1)/(x_(s)-x_(i))]+],[+sum_(i=1)^(s)(g_(i)(x))/(g_(l)(x_(i)))(x-x_(i))f^(')(x_(i)).]:} \begin{gathered}
H_{2 s-1}(x)=\sum_{i=1}^{s} \frac{g_{i}(x)}{g_{i}\left(x_{i}\right)}\left(x-x_{i}\right)\left[\frac{1}{x-x_{i}}+\frac{1}{x_{1}-x_{i}}+\ldots+\frac{1}{x_{s}-x_{i}}\right]+ \\
+\sum_{i=1}^{s} \frac{g_{i}(x)}{g_{l}\left(x_{i}\right)}\left(x-x_{i}\right) f^{\prime}\left(x_{i}\right) .
\end{gathered} H 2 s − 1 ( x ) = ∑ i = 1 s g i ( x ) g i ( x i ) ( x − x i ) [ 1 x − x i + 1 x 1 − x i + … + 1 x s − x i ] + + ∑ i = 1 s g i ( x ) g l ( x i ) ( x − x i ) f ′ ( x i ) . Această formulă a fost găsită și de A. Markoff [7].
3
∘
3
∘
3^(@) 3^{\circ} 3 ∘ . In cazul
r
1
=
r
2
=
…
=
r
s
=
3
r
1
=
r
2
=
…
=
r
s
=
3
r_(1)=r_(2)=dots=r_(s)=3 r_{1}=r_{2}=\ldots=r_{s}=3 r 1 = r 2 = … = r s = 3 , avem
H
3
s
−
1
(
x
)
=
∑
i
=
1
s
[
l
i
,
0
(
x
)
f
(
x
i
)
+
l
i
,
1
(
x
)
f
′
(
x
i
)
+
l
i
,
2
(
x
)
f
′
′
(
x
i
)
]
H
3
s
−
1
(
x
)
=
∑
i
=
1
s
l
i
,
0
(
x
)
f
x
i
+
l
i
,
1
(
x
)
f
′
x
i
+
l
i
,
2
(
x
)
f
′
′
x
i
H_(3s-1)(x)=sum_(i=1)^(s)[l_(i,0)(x)f(x_(i))+l_(i,1)(x)f^(')(x_(i))+l_(i,2)(x)f^('')(x_(i))] H_{3 s-1}(x)=\sum_{i=1}^{s}\left[l_{i, 0}(x) f\left(x_{i}\right)+l_{i, 1}(x) f^{\prime}\left(x_{i}\right)+l_{i, 2}(x) f^{\prime \prime}\left(x_{i}\right)\right] H 3 s − 1 ( x ) = ∑ i = 1 s [ l i , 0 ( x ) f ( x i ) + l i , 1 ( x ) f ′ ( x i ) + l i , 2 ( x ) f ′ ′ ( x i ) ] unde
l
i
,
0
(
x
)
=
g
i
(
x
)
g
i
(
x
i
)
[
1
−
(
x
−
x
i
)
g
i
′
(
x
i
)
g
i
(
x
i
)
−
(
x
−
x
i
)
2
g
i
′
′
(
x
i
)
g
i
(
x
i
)
−
2
g
i
′
2
(
x
i
)
2
g
i
2
(
x
i
)
]
l
i
,
0
(
x
)
=
g
i
(
x
)
g
i
x
i
1
−
x
−
x
i
g
i
′
x
i
g
i
x
i
−
x
−
x
i
2
g
i
′
′
x
i
g
i
x
i
−
2
g
i
′
2
x
i
2
g
i
2
x
i
l_(i,0)(x)=(g_(i)(x))/(g^(i)(x_(i)))[1-(x-x_(i))(g_(i)^(')(x_(i)))/(g_(i)(x_(i)))-(x-x_(i))^(2)(g_(i)^('')(x_(i))g_(i)(x_(i))-2g_(i)^('2)(x_(i)))/(2g_(i)^(2)(x_(i)))] l_{i, 0}(x)=\frac{g_{i}(x)}{g^{i}\left(x_{i}\right)}\left[1-\left(x-x_{i}\right) \frac{g_{i}^{\prime}\left(x_{i}\right)}{g_{i}\left(x_{i}\right)}-\left(x-x_{i}\right)^{2} \frac{g_{i}^{\prime \prime}\left(x_{i}\right) g_{i}\left(x_{i}\right)-2 g_{i}^{\prime 2}\left(x_{i}\right)}{2 g_{i}^{2}\left(x_{i}\right)}\right] l i , 0 ( x ) = g i ( x ) g i ( x i ) [ 1 − ( x − x i ) g i ′ ( x i ) g i ( x i ) − ( x − x i ) 2 g i ′ ′ ( x i ) g i ( x i ) − 2 g i ′ 2 ( x i ) 2 g i 2 ( x i ) ]
l
i
,
1
(
x
)
=
(
x
−
x
i
)
g
i
(
x
)
g
i
(
x
i
)
[
1
−
(
x
−
x
i
)
g
i
′
(
x
i
)
g
i
(
x
i
)
]
l
i
,
2
(
x
)
=
(
x
−
x
i
)
2
2
g
i
(
x
)
g
i
(
x
i
)
.
l
i
,
1
(
x
)
=
x
−
x
i
g
i
(
x
)
g
i
x
i
1
−
x
−
x
i
g
i
′
x
i
g
i
x
i
l
i
,
2
(
x
)
=
x
−
x
i
2
2
g
i
(
x
)
g
i
x
i
.
{:[l_(i,1)(x)=(x-x_(i))(g_(i)(x))/(g_(i)(x_(i)))[1-(x-x_(i))(g_(i)^(')(x_(i)))/(g_(i)(x_(i)))]],[l_(i,2)(x)=((x-x_(i))^(2))/(2)(g_(i)(x))/(g_(i)(x_(i))).]:} \begin{aligned}
& l_{i, 1}(x)=\left(x-x_{i}\right) \frac{g_{i}(x)}{g_{i}\left(x_{i}\right)}\left[1-\left(x-x_{i}\right) \frac{g_{i}^{\prime}\left(x_{i}\right)}{g_{i}\left(x_{i}\right)}\right] \\
& l_{i, 2}(x)=\frac{\left(x-x_{i}\right)^{2}}{2} \frac{g_{i}(x)}{g_{i}\left(x_{i}\right)} .
\end{aligned} l i , 1 ( x ) = ( x − x i ) g i ( x ) g i ( x i ) [ 1 − ( x − x i ) g i ′ ( x i ) g i ( x i ) ] l i , 2 ( x ) = ( x − x i ) 2 2 g i ( x ) g i ( x i ) .
Să considerăm cazul particular
s
=
2
s
=
2
s=2 s=2 s = 2
x
1
=
a
,
x
2
=
b
,
r
1
=
m
x
1
=
a
,
x
2
=
b
,
r
1
=
m
x_(1)=a,x_(2)=b,r_(1)=m x_{1}=a, x_{2}=b, r_{1}=m x 1 = a , x 2 = b , r 1 = m
r
2
=
n
r
2
=
n
r_(2)=n r_{2}=n r 2 = n
H
m
+
n
−
1
(
x
)
=
H
m
+
n
−
1
(
a
,
…
,
a
⏟
m
,
b
,
…
,
b
⏟
n
;
f
(
x
)
=
(22)
=
(
x
−
b
a
−
b
)
n
∑
k
=
0
m
−
1
(
x
−
a
)
k
k
!
[
∑
i
=
0
m
−
k
−
1
(
n
+
i
−
1
i
)
(
x
−
a
b
−
a
)
i
]
f
(
k
)
(
a
)
+
(20)
+
(
x
−
a
b
−
a
)
m
∑
r
=
0
n
−
1
(
x
−
b
)
r
r
!
[
∑
j
=
0
n
−
r
−
1
(
m
+
j
−
1
j
)
(
x
−
b
a
−
b
)
j
]
f
(
r
)
(
b
)
.
H
m
+
n
−
1
(
x
)
=
H
m
+
n
−
1
(
a
,
…
,
a
⏟
m
,
b
,
…
,
b
⏟
n
;
f
(
x
)
=
(22)
=
x
−
b
a
−
b
n
∑
k
=
0
m
−
1
(
x
−
a
)
k
k
!
∑
i
=
0
m
−
k
−
1
(
n
+
i
−
1
i
)
x
−
a
b
−
a
i
f
(
k
)
(
a
)
+
(20)
+
x
−
a
b
−
a
m
∑
r
=
0
n
−
1
(
x
−
b
)
r
r
!
∑
j
=
0
n
−
r
−
1
(
m
+
j
−
1
j
)
x
−
b
a
−
b
j
f
(
r
)
(
b
)
.
{:[H_(m+n-1)(x)=H_(m+n-1)(ubrace(a,dots,a)_(m)","ubrace(b,dots,b)_(n);f(x)=],[(22)=((x-b)/(a-b))^(n)sum_(k=0)^(m-1)((x-a)^(k))/(k!)[sum_(i=0)^(m-k-1)((n+i-1)/(i))((x-a)/(b-a))^(i)]f^((k))(a)+],[(20)+((x-a)/(b-a))^(m)sum_(r=0)^(n-1)((x-b)^(r))/(r!)[sum_(j=0)^(n-r-1)((m+j-1)/(j))((x-b)/(a-b))^(j)]f^((r))(b).]:} \begin{gather*}
H_{m+n-1}(x)=H_{m+n-1}(\underbrace{a, \ldots, a}_{m}, \underbrace{b, \ldots, b}_{n} ; f(x)= \\
=\left(\frac{x-b}{a-b}\right)^{n} \sum_{k=0}^{m-1} \frac{(x-a)^{k}}{k!}\left[\sum_{i=0}^{m-k-1}\binom{n+i-1}{i}\left(\frac{x-a}{b-a}\right)^{i}\right] f^{(k)}(a)+ \tag{22}\\
+\left(\frac{x-a}{b-a}\right)^{m} \sum_{r=0}^{n-1} \frac{(x-b)^{r}}{r!}\left[\sum_{j=0}^{n-r-1}\binom{m+j-1}{j}\left(\frac{x-b}{a-b}\right)^{j}\right] f^{(r)}(b) . \tag{20}
\end{gather*} H m + n − 1 ( x ) = H m + n − 1 ( a , … , a ⏟ m , b , … , b ⏟ n ; f ( x ) = (22) = ( x − b a − b ) n ∑ k = 0 m − 1 ( x − a ) k k ! [ ∑ i = 0 m − k − 1 ( n + i − 1 i ) ( x − a b − a ) i ] f ( k ) ( a ) + (20) + ( x − a b − a ) m ∑ r = 0 n − 1 ( x − b ) r r ! [ ∑ j = 0 n − r − 1 ( m + j − 1 j ) ( x − b a − b ) j ] f ( r ) ( b ) . În formula de interpolare corespunzătoare
(21)
f
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x
)
=
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m
+
n
−
1
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x
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+
R
m
+
n
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(21)
f
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=
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m
+
n
−
1
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+
R
m
+
n
(
x
)
{:(21)f(x)=H_(m+n-1)(x)+R_(m+n)(x):} \begin{equation*}
f(x)=H_{m+n-1}(x)+R_{m+n}(x) \tag{21}
\end{equation*} (21) f ( x ) = H m + n − 1 ( x ) + R m + n ( x ) restul are expresia
R
m
+
n
(
x
)
=
(
x
−
a
)
m
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b
)
n
[
x
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a
,
…
,
a
,
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,
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b
;
f
]
=
=
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m
(
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)
n
(
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+
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)
!
f
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(
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,
(
a
<
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<
b
)
R
m
+
n
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=
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−
a
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n
[
x
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a
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,
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,
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;
f
]
=
=
(
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n
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!
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)
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,
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<
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<
b
)
{:[R_(m+n)(x)=(x-a)^(m)(x-b)^(n)[x","a","dots","a","b","dots","b;f]=],[quad=((x-a)^(m)(x-b)^(n))/((m+n)!)f^((m+n))(xi)","(a < xi < b)]:} \begin{gathered}
R_{m+n}(x)=(x-a)^{m}(x-b)^{n}[x, a, \ldots, a, b, \ldots, b ; f]= \\
\quad=\frac{(x-a)^{m}(x-b)^{n}}{(m+n)!} f^{(m+n)}(\xi),(a<\xi<b)
\end{gathered} R m + n ( x ) = ( x − a ) m ( x − b ) n [ x , a , … , a , b , … , b ; f ] = = ( x − a ) m ( x − b ) n ( m + n ) ! f ( m + n ) ( ξ ) , ( a < ξ < b )
O importantă aplicație a formulei de interpolare (15) este aceea referitoare la descompunerea unei funcții raționale în fracții simple, în cazul general cînd numitorul funcției raționale are rădăcini multiple. De altfel există o echivalență între aceste două probleme, după cum se poate vedea uşor.
Să presupunem că se dă funcția rațională
R
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x
)
=
f
(
x
)
ω
(
x
)
R
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x
)
=
f
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x
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ω
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)
R(x)=(f(x))/(omega(x)) R(x)=\frac{f(x)}{\omega(x)} R ( x ) = f ( x ) ω ( x ) unde
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=
∏
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s
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i
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ω
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i
omega(x)=prod_(p=1)^(s)(x-x_(p))^(r)_(p)=g_(i)(x)(x-x_(i))^(r_(i)) \omega(x)=\prod_{p=1}^{s}\left(x-x_{p}\right)^{r}{ }_{p}=g_{i}(x)\left(x-x_{i}\right)^{r_{i}} ω ( x ) = ∏ p = 1 s ( x − x p ) r p = g i ( x ) ( x − x i ) r i iar
f
(
x
)
f
(
x
)
f(x) f(x) f ( x ) este un polinom de grad
m
<
n
m
<
n
m < n m<n m < n (caz care interesează totdeauna), unde am notat ca și înainte
n
=
r
1
+
r
2
+
…
+
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s
−
1
n
=
r
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+
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+
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s
−
1
n=r_(1)+r_(2)+dots+r_(s)-1 n=r_{1}+r_{2}+\ldots+r_{s}-1 n = r 1 + r 2 + … + r s − 1 .
Se știe că
R
(
x
)
R
(
x
)
R(x) R(x) R ( x ) se poate reprezenta în
mod
mod
mod \bmod mod unic sub forma
(23)
R
(
x
)
=
∑
i
=
1
s
∑
p
=
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r
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A
i
p
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x
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)
r
i
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p
,
(23)
R
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∑
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∑
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A
i
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x
i
r
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p
,
{:(23)R(x)=sum_(i=1)^(s)sum_(p=0)^(r_(i)-1)(A_(ip))/((x-x_(i))^(r_(i)-p))",":} \begin{equation*}
R(x)=\sum_{i=1}^{s} \sum_{p=0}^{r_{i}-1} \frac{A_{i p}}{\left(x-x_{i}\right)^{r_{i}-p}}, \tag{23}
\end{equation*} (23) R ( x ) = ∑ i = 1 s ∑ p = 0 r i − 1 A i p ( x − x i ) r i − p , unde
A
i
p
A
i
p
A_(ip) A_{i p} A i p sînt constante.
Dacă se scrie polinomul de interpolare relativ la polinomul
f
(
x
)
f
(
x
)
f(x) f(x) f ( x ) și la nodurile
x
1
,
x
2
,
…
,
x
s
−
x
1
,
x
2
,
…
,
x
s
−
x_(1),x_(2),dots,x_(s)- x_{1}, x_{2}, \ldots, x_{s}- x 1 , x 2 , … , x s − de ordine de multiplicitate egale respectiv cu
r
1
,
r
2
,
…
,
x
s
r
1
,
r
2
,
…
,
x
s
r_(1),r_(2),dots,x_(s) r_{1}, r_{2}, \ldots, x_{s} r 1 , r 2 , … , x s - se obține polinomul
H
n
(
x
)
H
n
(
x
)
H_(n)(x) H_{n}(x) H n ( x ) de la (13) sau (15). Avînd în vedere că
f
(
x
)
f
(
x
)
f(x) f(x) f ( x ) e un polinom de grad
m
<
n
m
<
n
m < n m<n m < n , avem
H
n
(
x
)
≡
f
(
x
)
,
H
n
(
x
)
≡
f
(
x
)
,
H_(n)(x)-=f(x), H_{n}(x) \equiv f(x), H n ( x ) ≡ f ( x ) , astfel că vom avea
f
(
x
)
ω
(
x
)
≡
H
n
(
x
)
ω
(
x
)
=
∑
i
=
1
s
∑
p
=
0
r
i
−
1
A
i
p
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x
−
x
i
)
r
i
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p
f
(
x
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ω
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x
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≡
H
n
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x
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ω
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=
∑
i
=
1
s
∑
p
=
0
r
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−
1
A
i
p
x
−
x
i
r
i
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p
(f(x))/(omega(x))-=(H_(n)(x))/(omega(x))=sum_(i=1)^(s)sum_(p=0)^(r_(i)-1)(A_(ip))/((x-x_(i))^(r_(i)-p)) \frac{f(x)}{\omega(x)} \equiv \frac{H_{n}(x)}{\omega(x)}=\sum_{i=1}^{s} \sum_{p=0}^{r_{i}-1} \frac{A_{i p}}{\left(x-x_{i}\right)^{r_{i}-p}} f ( x ) ω ( x ) ≡ H n ( x ) ω ( x ) = ∑ i = 1 s ∑ p = 0 r i − 1 A i p ( x − x i ) r i − p unde, pe baza formulei (13)
(21))
A
i
p
=
∑
k
=
0
p
f
(
k
)
(
x
i
)
k
!
(
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!
(
1
g
i
(
x
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)
x
=
x
i
(
p
−
k
)
=
1
p
!
(
f
(
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g
i
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)
x
=
x
i
(
p
)
(
p
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0
,
1
,
…
,
r
i
−
1
;
i
=
1
,
2
,
…
,
s
)
.
(21))
A
i
p
=
∑
k
=
0
p
f
(
k
)
x
i
k
!
(
p
−
k
)
!
1
g
i
(
x
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x
=
x
i
(
p
−
k
)
=
1
p
!
f
(
x
)
g
i
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x
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x
=
x
i
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p
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p
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0
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1
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r
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;
i
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1
,
2
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s
.
{:[(21))A_(ip)=sum_(k=0)^(p)(f^((k))(x_(i)))/(k!(p-k)!)((1)/(g_(i)(x)))_(x=x_(i))^((p-k))=(1)/(p!)((f(x))/(g_(i)(x)))_(x=x_(i))^((p))],[(p=0,1,dots,r_(i)-1;i=1,2,dots,s).]:} \begin{gather*}
A_{i p}=\sum_{k=0}^{p} \frac{f^{(k)}\left(x_{i}\right)}{k!(p-k)!}\left(\frac{1}{g_{i}(x)}\right)_{x=x_{i}}^{(p-k)}=\frac{1}{p!}\left(\frac{f(x)}{g_{i}(x)}\right)_{x=x_{i}}^{(p)} \tag{21)}\\
\left(p=0,1, \ldots, r_{i}-1 ; i=1,2, \ldots, s\right) .
\end{gather*} (21)) A i p = ∑ k = 0 p f ( k ) ( x i ) k ! ( p − k ) ! ( 1 g i ( x ) ) x = x i ( p − k ) = 1 p ! ( f ( x ) g i ( x ) ) x = x i ( p ) ( p = 0 , 1 , … , r i − 1 ; i = 1 , 2 , … , s ) . Astfel avem
iar
l
2
,
k
(
x
)
,
l
3
,
k
(
x
)
l
2
,
k
(
x
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,
l
3
,
k
(
x
)
l_(2,k)(x),l_(3,k)(x) l_{2, k}(x), l_{3, k}(x) l 2 , k ( x ) , l 3 , k ( x ) se obțin de aici prin permutări circulare ale indicilor
1
,
2
,
3
1
,
2
,
3
1,2,3 1,2,3 1 , 2 , 3 .
în ipoteza bineînţeles că în intervalul (
a
,
b
a
,
b
a,b a, b a , b ) funcția
f
(
x
)
f
(
x
)
f(x) f(x) f ( x ) are derivată de ordinul
m
+
n
m
+
n
m+n m+n m + n .
9. In cazul
s
=
3
s
=
3
s=3 s=3 s = 3 , (15) se reduce 1a
unde
H
r
1
+
r
2
+
r
3
−
1
(
x
)
=
H
r
1
+
r
2
+
r
3
−
1
(
x
1
,
…
,
x
1
⏟
r
1
,
x
2
,
…
,
x
2
⏟
r
2
,
x
3
,
…
,
x
3
⏟
r
3
;
f
∣
x
)
=
=
∑
k
=
0
r
1
−
1
l
1
,
k
(
x
)
f
(
k
)
(
x
1
)
+
∑
k
=
0
r
2
−
1
l
2
,
k
(
x
)
f
(
k
)
(
x
2
)
+
∑
k
=
0
r
3
−
1
l
3
,
k
(
x
)
f
(
k
)
(
x
3
)
H
r
1
+
r
2
+
r
3
−
1
(
x
)
=
H
r
1
+
r
2
+
r
3
−
1
(
x
1
,
…
,
x
1
⏟
r
1
,
x
2
,
…
,
x
2
⏟
r
2
,
x
3
,
…
,
x
3
⏟
r
3
;
f
∣
x
)
=
=
∑
k
=
0
r
1
−
1
l
1
,
k
(
x
)
f
(
k
)
x
1
+
∑
k
=
0
r
2
−
1
l
2
,
k
(
x
)
f
(
k
)
x
2
+
∑
k
=
0
r
3
−
1
l
3
,
k
(
x
)
f
(
k
)
x
3
{:[H_(r_(1)+r_(2)+r_(3)-1)(x)=H_(r_(1)+r_(2)+r_(3)-1)(ubrace(x_(1),dots,x_(1))_(r_(1))","ubrace(x_(2),dots,x_(2))_(r_(2))","ubrace(x_(3),dots,x_(3))_(r_(3));f∣x)=],[quad=sum_(k=0)^(r_(1)-1)l_(1,k)(x)f^((k))(x_(1))+sum_(k=0)^(r_(2)-1)l_(2,k)(x)f^((k))(x_(2))+sum_(k=0)^(r_(3)-1)l_(3,k)(x)f^((k))(x_(3))]:} \begin{aligned}
& H_{r_{1}+r_{2}+r_{3}-1}(x)=H_{r_{1}+r_{2}+r_{3}-1}(\underbrace{x_{1}, \ldots, x_{1}}_{r_{1}}, \underbrace{x_{2}, \ldots, x_{2}}_{r_{2}}, \underbrace{x_{3}, \ldots, x_{3}}_{r_{3}} ; f \mid x)= \\
& \quad=\sum_{k=0}^{r_{1}-1} l_{1, k}(x) f^{(k)}\left(x_{1}\right)+\sum_{k=0}^{r_{2}-1} l_{2, k}(x) f^{(k)}\left(x_{2}\right)+\sum_{k=0}^{r_{3}-1} l_{3, k}(x) f^{(k)}\left(x_{3}\right)
\end{aligned} H r 1 + r 2 + r 3 − 1 ( x ) = H r 1 + r 2 + r 3 − 1 ( x 1 , … , x 1 ⏟ r 1 , x 2 , … , x 2 ⏟ r 2 , x 3 , … , x 3 ⏟ r 3 ; f ∣ x ) = = ∑ k = 0 r 1 − 1 l 1 , k ( x ) f ( k ) ( x 1 ) + ∑ k = 0 r 2 − 1 l 2 , k ( x ) f ( k ) ( x 2 ) + ∑ k = 0 r 3 − 1 l 3 , k ( x ) f ( k ) ( x 3 )
l
1
,
k
(
x
)
=
(
x
−
x
2
x
1
−
x
2
)
r
2
(
x
−
x
3
x
1
−
x
3
)
r
3
×
×
{
(
x
−
x
1
)
k
k
!
∑
α
=
0
r
1
−
k
−
1
(
x
−
x
1
x
2
−
x
1
)
α
[
∑
j
=
0
α
(
r
2
+
α
−
j
−
1
r
2
−
1
)
(
r
3
+
j
−
1
j
)
(
x
1
−
x
2
x
1
−
x
3
)
j
]
}
,
l
1
,
k
(
x
)
=
x
−
x
2
x
1
−
x
2
r
2
x
−
x
3
x
1
−
x
3
r
3
×
×
x
−
x
1
k
k
!
∑
α
=
0
r
1
−
k
−
1
x
−
x
1
x
2
−
x
1
α
∑
j
=
0
α
(
r
2
+
α
−
j
−
1
r
2
−
1
)
(
r
3
+
j
−
1
j
)
x
1
−
x
2
x
1
−
x
3
j
,
{:[l_(1,k)(x)=((x-x_(2))/(x_(1)-x_(2)))^(r_(2))((x-x_(3))/(x_(1)-x_(3)))^(r_(3))xx],[xx{((x-x_(1))^(k))/(k!)sum_(alpha=0)^(r_(1)-k-1)((x-x_(1))/(x_(2)-x_(1)))^(alpha)[sum_(j=0)^(alpha)((r_(2)+alpha-j-1)/(r_(2)-1))((r_(3)+j-1)/(j))((x_(1)-x_(2))/(x_(1)-x_(3)))^(j)]}","]:} \begin{gathered}
l_{1, k}(x)=\left(\frac{x-x_{2}}{x_{1}-x_{2}}\right)^{r_{2}}\left(\frac{x-x_{3}}{x_{1}-x_{3}}\right)^{r_{3}} \times \\
\times\left\{\frac{\left(x-x_{1}\right)^{k}}{k!} \sum_{\alpha=0}^{r_{1}-k-1}\left(\frac{x-x_{1}}{x_{2}-x_{1}}\right)^{\alpha}\left[\sum_{j=0}^{\alpha}\binom{r_{2}+\alpha-j-1}{r_{2}-1}\binom{r_{3}+j-1}{j}\left(\frac{x_{1}-x_{2}}{x_{1}-x_{3}}\right)^{j}\right]\right\},
\end{gathered} l 1 , k ( x ) = ( x − x 2 x 1 − x 2 ) r 2 ( x − x 3 x 1 − x 3 ) r 3 × × { ( x − x 1 ) k k ! ∑ α = 0 r 1 − k − 1 ( x − x 1 x 2 − x 1 ) α [ ∑ j = 0 α ( r 2 + α − j − 1 r 2 − 1 ) ( r 3 + j − 1 j ) ( x 1 − x 2 x 1 − x 3 ) j ] } , Dacă se folosește formula (15), se obțin formulele explicite
(25)
A
i
p
=
1
g
i
(
x
i
)
∑
k
=
0
k
[
∑
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1
+
…
/
…
+
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k
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r
1
+
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1
a
1
)
⋯
/
⋯
(
r
s
+
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s
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1
α
s
)
(
x
1
−
x
i
)
α
1
…
/
⋯
(
x
s
−
x
i
)
u
s
]
f
(
k
)
(
x
i
)
k
!
(25)
A
i
p
=
1
g
i
x
i
∑
k
=
0
k
∑
α
1
+
…
/
…
+
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s
=
p
−
k
(
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1
+
α
1
−
1
a
1
)
⋯
/
⋯
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s
+
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)
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1
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−
x
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u
s
f
(
k
)
x
i
k
!
{:(25)A_(ip)=(1)/(g_(i)(x_(i)))sum_(k=0)^(k)[sum_(alpha_(1)+dots//dots+alpha_(s)=p-k)(((r_(1)+alpha_(1)-1)/(a_(1)))cdots//cdots((r_(s)+alpha_(s)-1)/(alpha_(s))))/((x_(1)-x_(i))^(alpha_(1))dots//cdots(x_(s)-x_(i))^(u_(s)))](f^((k))(x_(i)))/(k!):} \begin{equation*}
A_{i p}=\frac{1}{g_{i}\left(x_{i}\right)} \sum_{k=0}^{k}\left[\sum_{\alpha_{1}+\ldots / \ldots+\alpha_{s}=p-k} \frac{\binom{r_{1}+\alpha_{1}-1}{a_{1}} \cdots / \cdots\binom{r_{s}+\alpha_{s}-1}{\alpha_{s}}}{\left(x_{1}-x_{i}\right)^{\alpha_{1}} \ldots / \cdots\left(x_{s}-x_{i}\right)^{u_{s}}}\right] \frac{f^{(k)}\left(x_{i}\right)}{k!} \tag{25}
\end{equation*} (25) A i p = 1 g i ( x i ) ∑ k = 0 k [ ∑ α 1 + … / … + α s = p − k ( r 1 + α 1 − 1 a 1 ) ⋯ / ⋯ ( r s + α s − 1 α s ) ( x 1 − x i ) α 1 … / ⋯ ( x s − x i ) u s ] f ( k ) ( x i ) k !
În cazul particular
(26)
R
(
x
)
=
f
(
x
)
(
x
−
a
)
m
(
x
−
b
)
n
(26)
R
(
x
)
=
f
(
x
)
(
x
−
a
)
m
(
x
−
b
)
n
{:(26)R(x)=(f(x))/((x-a)^(m)(x-b)^(n)):} \begin{equation*}
R(x)=\frac{f(x)}{(x-a)^{m}(x-b)^{n}} \tag{26}
\end{equation*} (26) R ( x ) = f ( x ) ( x − a ) m ( x − b ) n descompunerea în fracții simple este de forma
R
(
x
)
=
∑
p
=
0
m
−
1
A
p
(
x
−
a
)
p
+
∑
q
=
0
n
−
1
B
q
(
x
−
b
)
q
R
(
x
)
=
∑
p
=
0
m
−
1
A
p
(
x
−
a
)
p
+
∑
q
=
0
n
−
1
B
q
(
x
−
b
)
q
R(x)=sum_(p=0)^(m-1)(A_(p))/((x-a)^(p))+sum_(q=0)^(n-1)(B_(q))/((x-b)^(q)) R(x)=\sum_{p=0}^{m-1} \frac{A_{p}}{(x-a)^{p}}+\sum_{q=0}^{n-1} \frac{B_{q}}{(x-b)^{q}} R ( x ) = ∑ p = 0 m − 1 A p ( x − a ) p + ∑ q = 0 n − 1 B q ( x − b ) q Pe baza formulelor precedente, avem
(27)
A
p
=
1
(
a
−
b
)
n
∑
i
=
0
p
(
−
1
)
p
−
i
(
n
+
p
−
i
−
1
n
−
1
)
(
a
−
b
)
p
−
i
f
(
i
)
(
a
)
i
!
B
q
=
1
(
b
−
a
)
m
∑
j
=
0
q
(
−
1
)
q
−
j
(
m
+
q
−
j
−
1
m
−
1
)
(
b
−
a
)
q
−
j
f
(
j
)
(
b
)
j
!
(27)
A
p
=
1
(
a
−
b
)
n
∑
i
=
0
p
(
−
1
)
p
−
i
(
n
+
p
−
i
−
1
n
−
1
)
(
a
−
b
)
p
−
i
f
(
i
)
(
a
)
i
!
B
q
=
1
(
b
−
a
)
m
∑
j
=
0
q
(
−
1
)
q
−
j
(
m
+
q
−
j
−
1
m
−
1
)
(
b
−
a
)
q
−
j
f
(
j
)
(
b
)
j
!
{:[(27)A_(p)=(1)/((a-b)^(n))sum_(i=0)^(p)(-1)^(p-i)(((n+p-i-1)/(n-1)))/((a-b)^(p-i))(f^((i))(a))/(i!)],[B_(q)=(1)/((b-a)^(m))sum_(j=0)^(q)(-1)^(q-j)(((m+q-j-1)/(m-1)))/((b-a)^(q-j))(f^((j))(b))/(j!)]:} \begin{align*}
& A_{p}=\frac{1}{(a-b)^{n}} \sum_{i=0}^{p}(-1)^{p-i} \frac{\binom{n+p-i-1}{n-1}}{(a-b)^{p-i}} \frac{f^{(i)}(a)}{i!} \tag{27}\\
& B_{q}=\frac{1}{(b-a)^{m}} \sum_{j=0}^{q}(-1)^{q-j} \frac{\binom{m+q-j-1}{m-1}}{(b-a)^{q-j}} \frac{f^{(j)}(b)}{j!}
\end{align*} (27) A p = 1 ( a − b ) n ∑ i = 0 p ( − 1 ) p − i ( n + p − i − 1 n − 1 ) ( a − b ) p − i f ( i ) ( a ) i ! B q = 1 ( b − a ) m ∑ j = 0 q ( − 1 ) q − j ( m + q − j − 1 m − 1 ) ( b − a ) q − j f ( j ) ( b ) j !
Să presupunem acum că avem de calculat integrala definită
I
=
∫
a
b
f
(
x
)
d
x
I
=
∫
a
b
f
(
x
)
d
x
I=int_(a)^(b)f(x)dx I=\int_{a}^{b} f(x) d x I = ∫ a b f ( x ) d x Dacă se foloseste formula de interpolare (17), se obține formula de cuadratură
(28)
∫
a
b
f
(
x
)
d
x
=
∑
i
=
1
s
∑
k
=
0
r
i
−
1
A
l
,
k
f
(
k
)
(
x
i
)
+
ρ
(
f
)
(28)
∫
a
b
f
(
x
)
d
x
=
∑
i
=
1
s
∑
k
=
0
r
i
−
1
A
l
,
k
f
(
k
)
x
i
+
ρ
(
f
)
{:(28)int_(a)^(b)f(x)dx=sum_(i=1)^(s)sum_(k=0)^(r_(i)-1)A_(l,k)f^((k))(x_(i))+rho(f):} \begin{equation*}
\int_{a}^{b} f(x) d x=\sum_{i=1}^{s} \sum_{k=0}^{r_{i}-1} A_{l, k} f^{(k)}\left(x_{i}\right)+\rho(f) \tag{28}
\end{equation*} (28) ∫ a b f ( x ) d x = ∑ i = 1 s ∑ k = 0 r i − 1 A l , k f ( k ) ( x i ) + ρ ( f ) unde
A
l
,
k
=
∫
a
b
l
l
,
k
(
x
)
d
x
A
l
,
k
=
∫
a
b
l
l
,
k
(
x
)
d
x
A_(l,k)=int_(a)^(b)l_(l,k)(x)dx A_{l, k}=\int_{a}^{b} l_{l, k}(x) d x A l , k = ∫ a b l l , k ( x ) d x
ρ
(
f
)
=
∫
a
b
R
(
x
)
d
x
ρ
(
f
)
=
∫
a
b
R
(
x
)
d
x
rho(f)=int_(a)^(b)R(x)dx \rho(f)=\int_{a}^{b} R(x) d x ρ ( f ) = ∫ a b R ( x ) d x Formula de cuadratură (28) are în general gradul de exactitate
n
n
n n n , adică restul e nul dacă
f
(
x
)
f
(
x
)
f(x) f(x) f ( x ) este un polinom de grad cel mult
n
n
n n n . Se poate însă întîmpla ca printr-o alegere convenabilă a nodurilor, gradul de exactitate să fie mai înalt.
Exemple.
1
∘
1
∘
1^(@) 1^{\circ} 1 ∘ . Alegînd nodurile
x
1
=
x
2
=
x
3
=
−
1
,
x
4
=
0
,
x
5
=
x
6
==
x
7
=
1
x
1
=
x
2
=
x
3
=
−
1
,
x
4
=
0
,
x
5
=
x
6
==
x
7
=
1
x_(1)=x_(2)=x_(3)=-1,x_(4)=0,x_(5)=x_(6)==x_(7)=1 x_{1}=x_{2}=x_{3}=-1, x_{4}=0, x_{5}=x_{6}= =x_{7}=1 x 1 = x 2 = x 3 = − 1 , x 4 = 0 , x 5 = x 6 == x 7 = 1 şi scriind formula de interpolare corespunzătoare, avem
f
(
x
)
=
H
6
(
x
)
+
R
7
(
x
)
f
(
x
)
=
H
6
(
x
)
+
R
7
(
x
)
f(x)=H_(6)(x)+R_(7)(x) f(x)=H_{6}(x)+R_{7}(x) f ( x ) = H 6 ( x ) + R 7 ( x ) unde
H
6
(
x
)
H
6
(
x
)
H_(6)(x) H_{6}(x) H 6 ( x ) e polinomul de interpolare (19) iar
R
η
(
x
)
=
x
(
x
2
−
1
)
3
[
x
,
−
1
,
−
1
,
−
1
,
0
,
1
,
1
,
1
;
f
]
R
η
(
x
)
=
x
x
2
−
1
3
[
x
,
−
1
,
−
1
,
−
1
,
0
,
1
,
1
,
1
;
f
]
R_(eta)(x)=x(x^(2)-1)^(3)[x,-1,-1,-1,0,1,1,1;f] R_{\eta}(x)=x\left(x^{2}-1\right)^{3}[x,-1,-1,-1,0,1,1,1 ; f] R η ( x ) = x ( x 2 − 1 ) 3 [ x , − 1 , − 1 , − 1 , 0 , 1 , 1 , 1 ; f ] Integrînd de la -1 la +1 , se ajunge la formula de cuadratură de grad de exactitate 7 :
(29)
∫
−
1
+
1
f
(
x
)
d
x
=
1
105
[
57
f
(
−
1
)
+
12
f
′
(
−
1
)
+
f
′
′
(
−
1
)
+
96
(
0
)
+
f
′
′
(
1
)
−
−
12
f
′
(
1
)
+
57
f
(
1
)
]
+
ρ
(
f
)
(29)
∫
−
1
+
1
f
(
x
)
d
x
=
1
105
57
f
(
−
1
)
+
12
f
′
(
−
1
)
+
f
′
′
(
−
1
)
+
96
(
0
)
+
f
′
′
(
1
)
−
−
12
f
′
(
1
)
+
57
f
(
1
)
+
ρ
(
f
)
{:[(29)int_(-1)^(+1)f(x)dx=(1)/(105)[57 f(-1)+12f^(')(-1)+f^('')(-1)+96(0)+f^('')(1)-:}],[{:-12f^(')(1)+57 f(1)]+rho(f)]:} \begin{gather*}
\int_{-1}^{+1} f(x) d x=\frac{1}{105}\left[57 f(-1)+12 f^{\prime}(-1)+f^{\prime \prime}(-1)+96(0)+f^{\prime \prime}(1)-\right. \tag{29}\\
\left.-12 f^{\prime}(1)+57 f(1)\right]+\rho(f)
\end{gather*} (29) ∫ − 1 + 1 f ( x ) d x = 1 105 [ 57 f ( − 1 ) + 12 f ′ ( − 1 ) + f ′ ′ ( − 1 ) + 96 ( 0 ) + f ′ ′ ( 1 ) − − 12 f ′ ( 1 ) + 57 f ( 1 ) ] + ρ ( f ) Pentru restul
ρ
(
f
)
ρ
(
f
)
rho(f) \rho(f) ρ ( f ) s-a stabilit expresia
ρ
(
f
)
=
−
1
396900
f
(
8
)
(
ξ
)
,
−
1
<
ξ
<
+
1
ρ
(
f
)
=
−
1
396900
f
(
8
)
(
ξ
)
,
−
1
<
ξ
<
+
1
rho(f)=(-1)/(396900)f^((8))(xi),quad-1 < xi < +1 \rho(f)=\frac{-1}{396900} f^{(8)}(\xi), \quad-1<\xi<+1 ρ ( f ) = − 1 396900 f ( 8 ) ( ξ ) , − 1 < ξ < + 1
2
∘
2
∘
2^(@) 2^{\circ} 2 ∘ . Dacă se folosesc nodurile
x
1
=
x
2
=
x
3
=
x
4
=
x
5
=
0
x
1
=
x
2
=
x
3
=
x
4
=
x
5
=
0
x_(1)=x_(2)=x_(3)=x_(4)=x_(5)=0 x_{1}=x_{2}=x_{3}=x_{4}=x_{5}=0 x 1 = x 2 = x 3 = x 4 = x 5 = 0 și
−
x
6
=
x
7
=
7
8
−
x
6
=
x
7
=
7
8
-x_(6)=x_(7)=sqrt((7)/(8)) -x_{6}=x_{7}=\sqrt{\frac{7}{8}} − x 6 = x 7 = 7 8 , se scrie formula de interpolare corespunzătoare a lui Hermite, se înmulțeşte cu
p
(
x
)
=
(
1
−
x
2
)
−
1
2
p
(
x
)
=
1
−
x
2
−
1
2
p(x)=(1-x^(2))^(-(1)/(2)) p(x)=\left(1-x^{2}\right)^{-\frac{1}{2}} p ( x ) = ( 1 − x 2 ) − 1 2 şi se integrează de la -1 la +1 , se obţine formula de cuadratură de grad de exactitate 9
∫
−
1
+
1
f
(
x
)
1
−
x
2
d
x
=
π
65856
{
35136
f
(
0
)
+
3024
f
′
′
(
0
)
+
49
f
(
I
V
)
(
0
)
+
(
′
)
+
15360
[
f
(
−
7
8
)
+
f
(
|
7
8
)
]
}
+
π
530841600
f
(
10
)
(
ξ
)
∫
−
1
+
1
f
(
x
)
1
−
x
2
d
x
=
π
65856
35136
f
(
0
)
+
3024
f
′
′
(
0
)
+
49
f
(
I
V
)
(
0
)
+
(
′
)
+
15360
f
−
7
8
+
f
7
8
+
π
530841600
f
(
10
)
(
ξ
)
{:[int_(-1)^(+1)(f(x))/(sqrt(1-x^(2)))dx=(pi)/(65856){35136 f(0)+3024f^('')(0)+49f^((IV))(0)+:}],[('")"{: quad+15360[f(-sqrt((7)/(8)))+f(|sqrt((7)/(8)))]}+(pi)/(530841600)f^((10))(xi)]:} \begin{align*}
& \int_{-1}^{+1} \frac{f(x)}{\sqrt{1-x^{2}}} d x=\frac{\pi}{65856}\left\{35136 f(0)+3024 f^{\prime \prime}(0)+49 f^{(I V)}(0)+\right. \\
& \left.\quad+15360\left[f\left(-\sqrt{\frac{7}{8}}\right)+f\left(\left\lvert\, \sqrt{\frac{7}{8}}\right.\right)\right]\right\}+\frac{\pi}{530841600} f^{(10)}(\xi) \tag{$\prime$}
\end{align*} ∫ − 1 + 1 f ( x ) 1 − x 2 d x = π 65856 { 35136 f ( 0 ) + 3024 f ′ ′ ( 0 ) + 49 f ( I V ) ( 0 ) + ( ′ ) + 15360 [ f ( − 7 8 ) + f ( | 7 8 ) ] } + π 530841600 f ( 10 ) ( ξ ) Observatie. La formulele de cuadratură (cu rest) de la (29) și (29́) s-a ajuns dublînd în primul caz nodul
x
=
0
x
=
0
x=0 x=0 x = 0 , iar în al doilea caz dublînd nodurile
x
6
x
6
x_(6) x_{6} x 6 și
x
7
x
7
x_(7) x_{7} x 7 și luînd pe
x
=
0
x
=
0
x=0 x=0 x = 0 multiplu de ordinul 6 .
13. Să vedem ce devine formula de cuadratură (28) în cazul particular de la nr. 8 . Ne vom folosi de polinomul de interpolare (20).
Folosind formula de integrare prin părt, generalizată, după efectuarea sistematiçă a calculelor se obţine
∫
a
b
H
m
+
n
=
1
(
x
)
d
x
=
n
∑
k
=
0
m
−
1
(
b
−
a
)
k
+
1
(
k
+
2
)
!
[
∑
i
=
0
m
−
k
−
1
(
k
+
i
i
)
(
n
+
k
+
i
+
1
n
+
i
−
1
)
]
f
(
k
)
(
a
)
+
+
m
∑
r
=
0
n
−
1
(
a
−
b
)
k
(
r
+
2
)
!
[
∑
r
=
0
n
−
r
−
1
(
r
+
j
i
)
(
m
+
r
+
j
+
1
r
+
2
)
]
f
(
j
)
(
b
)
∫
a
b
H
m
+
n
=
1
(
x
)
d
x
=
n
∑
k
=
0
m
−
1
(
b
−
a
)
k
+
1
(
k
+
2
)
!
∑
i
=
0
m
−
k
−
1
(
k
+
i
i
)
(
n
+
k
+
i
+
1
n
+
i
−
1
)
f
(
k
)
(
a
)
+
+
m
∑
r
=
0
n
−
1
(
a
−
b
)
k
(
r
+
2
)
!
∑
r
=
0
n
−
r
−
1
(
r
+
j
i
)
(
m
+
r
+
j
+
1
r
+
2
)
f
(
j
)
(
b
)
{:[int_(a)^(b)H_(m+n=1)(x)dx=nsum_(k=0)^(m-1)((b-a)^(k+1))/((k+2)!)[sum_(i=0)^(m-k-1)(((k+i)/(i)))/(((n+k+i+1)/(n+i-1)))]f^((k))(a)+],[+msum_(r=0)^(n-1)((a-b)^(k))/((r+2)!)[sum_(r=0)^(n-r-1)(((r+j)/(i)))/(((m+r+j+1)/(r+2)))]f^((j))(b)]:} \begin{gathered}
\int_{a}^{b} H_{m+n=1}(x) d x=n \sum_{k=0}^{m-1} \frac{(b-a)^{k+1}}{(k+2)!}\left[\sum_{i=0}^{m-k-1} \frac{\binom{k+i}{i}}{\binom{n+k+i+1}{n+i-1}}\right] f^{(k)}(a)+ \\
+m \sum_{r=0}^{n-1} \frac{(a-b)^{k}}{(r+2)!}\left[\sum_{r=0}^{n-r-1} \frac{\binom{r+j}{i}}{\binom{m+r+j+1}{r+2}}\right] f^{(j)}(b)
\end{gathered} ∫ a b H m + n = 1 ( x ) d x = n ∑ k = 0 m − 1 ( b − a ) k + 1 ( k + 2 ) ! [ ∑ i = 0 m − k − 1 ( k + i i ) ( n + k + i + 1 n + i − 1 ) ] f ( k ) ( a ) + + m ∑ r = 0 n − 1 ( a − b ) k ( r + 2 ) ! [ ∑ r = 0 n − r − 1 ( r + j i ) ( m + r + j + 1 r + 2 ) ] f ( j ) ( b ) Pentru a obține o expresie mai simplă a coeficientilor acestei formule, ne vom folosi de identitatea
n
k
+
3
∑
i
=
0
m
−
k
−
1
(
k
+
i
i
)
(
n
+
k
+
i
+
1
k
+
2
)
=
(
m
k
+
1
)
(
m
+
n
k
+
1
)
n
k
+
3
∑
i
=
0
m
−
k
−
1
(
k
+
i
i
)
(
n
+
k
+
i
+
1
k
+
2
)
=
(
m
k
+
1
)
(
m
+
n
k
+
1
)
(n)/(k+3)sum_(i=0)^(m-k-1)(((k+i)/(i)))/(((n+k+i+1)/(k+2)))=(((m)/(k+1)))/(((m+n)/(k+1))) \frac{n}{k+3} \sum_{i=0}^{m-k-1} \frac{\binom{k+i}{i}}{\binom{n+k+i+1}{k+2}}=\frac{\binom{m}{k+1}}{\binom{m+n}{k+1}} n k + 3 ∑ i = 0 m − k − 1 ( k + i i ) ( n + k + i + 1 k + 2 ) = ( m k + 1 ) ( m + n k + 1 ) care se demonstrează fără greutate.
In felul acesta se ajunge la formula de cuadratură de grad de exactitate
m
+
n
−
1
m
+
n
−
1
m+n-1 m+n-1 m + n − 1 :
(30)
∫
a
b
f
(
x
)
d
x
=
∑
k
=
0
m
−
1
(
b
−
a
)
k
+
1
(
k
+
1
)
!
⋅
(
m
k
+
1
)
(
m
+
n
k
+
1
)
f
(
k
)
(
a
)
−
−
∑
r
=
0
n
−
1
(
a
−
b
)
r
+
1
(
r
+
1
)
!
⋅
(
n
r
+
1
)
(
m
+
n
r
+
1
)
f
(
r
)
(
b
)
+
ρ
m
+
n
(
f
)
(30)
∫
a
b
f
(
x
)
d
x
=
∑
k
=
0
m
−
1
(
b
−
a
)
k
+
1
(
k
+
1
)
!
⋅
(
m
k
+
1
)
(
m
+
n
k
+
1
)
f
(
k
)
(
a
)
−
−
∑
r
=
0
n
−
1
(
a
−
b
)
r
+
1
(
r
+
1
)
!
⋅
(
n
r
+
1
)
(
m
+
n
r
+
1
)
f
(
r
)
(
b
)
+
ρ
m
+
n
(
f
)
{:[(30)int_(a)^(b)f(x)dx=sum_(k=0)^(m-1)((b-a)^(k+1))/((k+1)!)*(((m)/(k+1)))/(((m+n)/(k+1)))f^((k))(a)-],[-sum_(r=0)^(n-1)((a-b)^(r+1))/((r+1)!)*(((n)/(r+1)))/(((m+n)/(r+1)))f^((r))(b)+rho_(m+n)(f)]:} \begin{align*}
& \int_{a}^{b} f(x) d x=\sum_{k=0}^{m-1} \frac{(b-a)^{k+1}}{(k+1)!} \cdot \frac{\binom{m}{k+1}}{\binom{m+n}{k+1}} f^{(k)}(a)- \tag{30}\\
& -\sum_{r=0}^{n-1} \frac{(a-b)^{r+1}}{(r+1)!} \cdot \frac{\binom{n}{r+1}}{\binom{m+n}{r+1}} f^{(r)}(b)+\rho_{m+n}(f)
\end{align*} (30) ∫ a b f ( x ) d x = ∑ k = 0 m − 1 ( b − a ) k + 1 ( k + 1 ) ! ⋅ ( m k + 1 ) ( m + n k + 1 ) f ( k ) ( a ) − − ∑ r = 0 n − 1 ( a − b ) r + 1 ( r + 1 ) ! ⋅ ( n r + 1 ) ( m + n r + 1 ) f ( r ) ( b ) + ρ m + n ( f ) unde
ρ
m
+
n
(
f
)
=
∫
a
b
(
x
−
a
)
m
(
x
−
b
)
n
[
x
,
a
,
…
,
a
,
b
,
…
,
b
;
f
]
d
x
ρ
m
+
n
(
f
)
=
∫
a
b
(
x
−
a
)
m
(
x
−
b
)
n
[
x
,
a
,
…
,
a
,
b
,
…
,
b
;
f
]
d
x
rho_(m+n)(f)=int_(a)^(b)(x-a)^(m)(x-b)^(n)[x,a,dots,a,b,dots,b;f]dx \rho_{m+n}(f)=\int_{a}^{b}(x-a)^{m}(x-b)^{n}[x, a, \ldots, a, b, \ldots, b ; f] d x ρ m + n ( f ) = ∫ a b ( x − a ) m ( x − b ) n [ x , a , … , a , b , … , b ; f ] d x sau
(31)
ρ
m
+
n
(
f
)
=
(
−
1
)
n
(
b
−
a
)
m
+
n
+
1
(
m
+
n
+
1
)
!
(
m
+
n
n
)
f
(
m
+
n
)
(
ξ
)
,
a
<
ξ
<
b
(31)
ρ
m
+
n
(
f
)
=
(
−
1
)
n
(
b
−
a
)
m
+
n
+
1
(
m
+
n
+
1
)
!
(
m
+
n
n
)
f
(
m
+
n
)
(
ξ
)
,
a
<
ξ
<
b
{:(31)rho_(m+n)(f)=((-1)^(n)(b-a)^(m+n+1))/((m+n+1)!((m+n)/(n)))f^((m+n))(xi)","quad a < xi < b:} \begin{equation*}
\rho_{m+n}(f)=\frac{(-1)^{n}(b-a)^{m+n+1}}{(m+n+1)!\binom{m+n}{n}} f^{(m+n)}(\xi), \quad a<\xi<b \tag{31}
\end{equation*} (31) ρ m + n ( f ) = ( − 1 ) n ( b − a ) m + n + 1 ( m + n + 1 ) ! ( m + n n ) f ( m + n ) ( ξ ) , a < ξ < b Făcînd
m
=
n
m
=
n
m=n m=n m = n , aceasta se reduce la formula
(32)
∫
a
b
f
(
x
)
d
x
=
∑
k
=
0
m
−
1
(
b
−
a
)
k
+
1
(
k
+
1
)
!
(
m
k
+
1
)
(
2
m
k
+
1
)
[
f
(
k
)
(
a
)
+
(
−
1
)
k
f
(
k
)
(
b
)
]
+
ρ
2
m
(
f
)
(32)
∫
a
b
f
(
x
)
d
x
=
∑
k
=
0
m
−
1
(
b
−
a
)
k
+
1
(
k
+
1
)
!
(
m
k
+
1
)
(
2
m
k
+
1
)
f
(
k
)
(
a
)
+
(
−
1
)
k
f
(
k
)
(
b
)
+
ρ
2
m
(
f
)
{:(32)int_(a)^(b)f(x)dx=sum_(k=0)^(m-1)((b-a)^(k+1))/((k+1)!)(((m)/(k+1)))/(((2m)/(k+1)))[f^((k))(a)+(-1)^(k)f^((k))(b)]+rho_(2m)(f):} \begin{equation*}
\int_{a}^{b} f(x) d x=\sum_{k=0}^{m-1} \frac{(b-a)^{k+1}}{(k+1)!} \frac{\binom{m}{k+1}}{\binom{2 m}{k+1}}\left[f^{(k)}(a)+(-1)^{k} f^{(k)}(b)\right]+\rho_{2 m}(f) \tag{32}
\end{equation*} (32) ∫ a b f ( x ) d x = ∑ k = 0 m − 1 ( b − a ) k + 1 ( k + 1 ) ! ( m k + 1 ) ( 2 m k + 1 ) [ f ( k ) ( a ) + ( − 1 ) k f ( k ) ( b ) ] + ρ 2 m ( f ) unde
ρ
2
m
(
f
)
=
(
−
1
)
n
(
b
−
a
)
2
m
+
1
(
2
m
+
1
)
!
(
2
m
m
)
f
(
2
m
)
(
ξ
)
ρ
2
m
(
f
)
=
(
−
1
)
n
(
b
−
a
)
2
m
+
1
(
2
m
+
1
)
!
(
2
m
m
)
f
(
2
m
)
(
ξ
)
rho_(2m)(f)=(-1)^(n)((b-a)^(2m+1))/((2m+1)!((2m)/(m)))f^((2m))(xi) \rho_{2 m}(f)=(-1)^{n} \frac{(b-a)^{2 m+1}}{(2 m+1)!\binom{2 m}{m}} f^{(2 m)}(\xi) ρ 2 m ( f ) = ( − 1 ) n ( b − a ) 2 m + 1 ( 2 m + 1 ) ! ( 2 m m ) f ( 2 m ) ( ξ ) Formulele (20) și (32) se datoresc
2
2
^(2) { }^{2} 2 lui Hermite [1]. El le-a dedus cu o metodă deosebită, fără să folosească formula de interpolare (20), pe care a renunțat să o stabilească căci i se părea că ar fi prea complicată.
Aceste formule au fost recent generalizate, într-un anumit sens, de către prof. D. V. Ionescu [12].
14. Folosind formula de interpolare (15), se pot construi de asemenea și formule pentru calculul numeric al derivatelor de diferite ordine ale unei funcţii
f
(
x
)
f
(
x
)
f(x) f(x) f ( x ) într-un punct
x
0
x
0
x_(0) x_{0} x 0 .
Acum ne vom ocupa de un caz concret care ni se pare important. Anume, ne vom folosi de formula de interpolare (20) pentru a stabili formule pentru calculul derivatelor
f
′
(
a
)
,
f
′
′
(
a
)
,
…
,
f
(
m
+
p
)
f
′
(
a
)
,
f
′
′
(
a
)
,
…
,
f
(
m
+
p
)
f^(')(a),f^('')(a),dots,f^((m+p)) f^{\prime}(a), f^{\prime \prime}(a), \ldots, f^{(m+p)} f ′ ( a ) , f ′ ′ ( a ) , … , f ( m + p ) , unde
p
p
p p p e un număr natural mai mic ca
n
n
n n n .
Pentru aceasta, vom căuta să calculăm
(33)
d
m
+
p
d
x
m
+
p
H
m
+
n
−
1
(
x
)
|
x
=
a
(33)
d
m
+
p
d
x
m
+
p
H
m
+
n
−
1
(
x
)
x
=
a
{:(33)(d^(m+p))/(dx^(m+p))H_(m+n-1)(x)|_(x=a):} \begin{equation*}
\left.\frac{d^{m+p}}{d x^{m+p}} H_{m+n-1}(x)\right|_{x=a} \tag{33}
\end{equation*} (33) d m + p d x m + p H m + n − 1 ( x ) | x = a Avînd în vedere formula (20), coeficientul lui
f
(
k
)
(
a
)
f
(
k
)
(
a
)
f^((k))(a) f^{(k)}(a) f ( k ) ( a ) din (33) se găseşte că este
(34)
∑
i
=
0
m
−
k
−
1
(
−
1
)
i
(
a
−
b
)
n
+
i
⋅
1
k
!
(
n
+
i
−
1
i
)
d
m
+
p
d
x
m
+
p
[
(
x
−
b
)
n
(
x
−
a
)
k
+
i
]
(34)
∑
i
=
0
m
−
k
−
1
(
−
1
)
i
(
a
−
b
)
n
+
i
⋅
1
k
!
(
n
+
i
−
1
i
)
d
m
+
p
d
x
m
+
p
(
x
−
b
)
n
(
x
−
a
)
k
+
i
{:(34)sum_(i=0)^(m-k-1)((-1)^(i))/((a-b)^(n+i))*(1)/(k!)((n+i-1)/(i))(d^(m+p))/(dx^(m+p))[(x-b)^(n)(x-a)^(k+i)]:} \begin{equation*}
\sum_{i=0}^{m-k-1} \frac{(-1)^{i}}{(a-b)^{n+i}} \cdot \frac{1}{k!}\binom{n+i-1}{i} \frac{d^{m+p}}{d x^{m+p}}\left[(x-b)^{n}(x-a)^{k+i}\right] \tag{34}
\end{equation*} (34) ∑ i = 0 m − k − 1 ( − 1 ) i ( a − b ) n + i ⋅ 1 k ! ( n + i − 1 i ) d m + p d x m + p [ ( x − b ) n ( x − a ) k + i ] Intrucît pe baza formulei lui Leibniz se găseşte că
[
(
x
−
b
)
n
(
x
−
a
)
k
]
x
=
a
(
m
+
p
)
=
=
n
(
n
−
1
)
…
(
n
−
m
−
p
+
k
+
1
)
!
k
!
(
m
+
p
k
)
(
a
−
b
)
n
−
m
−
p
+
k
(
x
−
b
)
n
(
x
−
a
)
k
x
=
a
(
m
+
p
)
=
=
n
(
n
−
1
)
…
(
n
−
m
−
p
+
k
+
1
)
!
k
!
(
m
+
p
k
)
(
a
−
b
)
n
−
m
−
p
+
k
{:[[(x-b)^(n)(x-a)^(k)]_(x=a)^((m+p))=],[=n(n-1)dots(n-m-p+k+1)!k!((m+p)/(k))(a-b)^(n-m-p+k)]:} \begin{gathered}
{\left[(x-b)^{n}(x-a)^{k}\right]_{x=a}^{(m+p)}=} \\
=n(n-1) \ldots(n-m-p+k+1)!k!\binom{m+p}{k}(a-b)^{n-m-p+k}
\end{gathered} [ ( x − b ) n ( x − a ) k ] x = a ( m + p ) = = n ( n − 1 ) … ( n − m − p + k + 1 ) ! k ! ( m + p k ) ( a − b ) n − m − p + k expresia (34) se poate scrie
(
m
+
p
)
!
k
!
(
a
−
b
)
m
+
p
−
k
∑
i
=
0
m
−
k
−
1
(
−
1
)
i
(
n
+
i
−
1
i
)
(
n
m
+
p
−
k
−
i
)
(
m
+
p
)
!
k
!
(
a
−
b
)
m
+
p
−
k
∑
i
=
0
m
−
k
−
1
(
−
1
)
i
(
n
+
i
−
1
i
)
(
n
m
+
p
−
k
−
i
)
((m+p)!)/(k!(a-b)^(m+p-k))sum_(i=0)^(m-k-1)(-1)^(i)((n+i-1)/(i))((n)/(m+p-k-i)) \frac{(m+p)!}{k!(a-b)^{m+p-k}} \sum_{i=0}^{m-k-1}(-1)^{i}\binom{n+i-1}{i}\binom{n}{m+p-k-i} ( m + p ) ! k ! ( a − b ) m + p − k ∑ i = 0 m − k − 1 ( − 1 ) i ( n + i − 1 i ) ( n m + p − k − i ) Dacă ne folosim de identitatea
∑
α
=
0
k
(
−
1
)
α
(
p
+
α
α
)
(
p
+
1
j
−
α
)
=
(
−
1
)
k
(
j
−
1
k
)
(
p
+
1
+
k
j
)
∑
α
=
0
k
(
−
1
)
α
(
p
+
α
α
)
(
p
+
1
j
−
α
)
=
(
−
1
)
k
(
j
−
1
k
)
(
p
+
1
+
k
j
)
sum_(alpha=0)^(k)(-1)^(alpha)((p+alpha)/(alpha))((p+1)/(j-alpha))=(-1)^(k)((j-1)/(k))((p+1+k)/(j)) \sum_{\alpha=0}^{k}(-1)^{\alpha}\binom{p+\alpha}{\alpha}\binom{p+1}{j-\alpha}=(-1)^{k}\binom{j-1}{k}\binom{p+1+k}{j} ∑ α = 0 k ( − 1 ) α ( p + α α ) ( p + 1 j − α ) = ( − 1 ) k ( j − 1 k ) ( p + 1 + k j ) care se demonstrează uşor, se găseşte în definitiv pentru coeficientul lui
f
(
k
)
(
a
)
f
(
k
)
(
a
)
f^((k))(a) f^{(k)}(a) f ( k ) ( a ) din (33) următoarea expresie
(35)
(
−
1
)
p
+
1
(
m
+
p
)
!
k
!
(
b
−
a
)
m
+
p
−
k
(
m
+
p
−
k
−
1
p
)
(
m
+
n
−
k
−
1
n
−
p
−
1
)
(35)
(
−
1
)
p
+
1
(
m
+
p
)
!
k
!
(
b
−
a
)
m
+
p
−
k
(
m
+
p
−
k
−
1
p
)
(
m
+
n
−
k
−
1
n
−
p
−
1
)
{:(35)(-1)^(p+1)((m+p)!)/(k!(b-a)^(m+p-k))((m+p-k-1)/(p))((m+n-k-1)/(n-p-1)):} \begin{equation*}
(-1)^{p+1} \frac{(m+p)!}{k!(b-a)^{m+p-k}}\binom{m+p-k-1}{p}\binom{m+n-k-1}{n-p-1} \tag{35}
\end{equation*} (35) ( − 1 ) p + 1 ( m + p ) ! k ! ( b − a ) m + p − k ( m + p − k − 1 p ) ( m + n − k − 1 n − p − 1 )
Să căutăm acum coeficientul lui
f
(
r
)
(
b
)
f
(
r
)
(
b
)
f(r)(b) f(r)(b) f ( r ) ( b )
Acesta este
(36)
1
(
b
−
a
)
m
⋅
1
r
!
∑
j
=
0
n
−
r
−
1
(
m
+
j
−
1
j
)
1
(
b
−
a
)
j
[
(
x
−
a
)
m
(
x
−
b
)
r
+
j
]
x
=
a
(
m
+
p
)
=
=
(
−
1
)
r
−
p
(
m
+
p
)
!
r
!
1
(
b
−
a
)
m
+
p
−
r
∑
j
=
0
n
−
r
−
1
(
r
+
j
p
)
(
m
+
j
−
1
j
)
(36)
1
(
b
−
a
)
m
⋅
1
r
!
∑
j
=
0
n
−
r
−
1
(
m
+
j
−
1
j
)
1
(
b
−
a
)
j
(
x
−
a
)
m
(
x
−
b
)
r
+
j
x
=
a
(
m
+
p
)
=
=
(
−
1
)
r
−
p
(
m
+
p
)
!
r
!
1
(
b
−
a
)
m
+
p
−
r
∑
j
=
0
n
−
r
−
1
(
r
+
j
p
)
(
m
+
j
−
1
j
)
{:[(36)(1)/((b-a)^(m))*(1)/(r!)sum_(j=0)^(n-r-1)((m+j-1)/(j))(1)/((b-a)^(j))[(x-a)^(m)(x-b)^(r+j)]_(x=a)^((m+p))=],[quad=(-1)^(r-p)((m+p)!)/(r!)(1)/((b-a)^(m+p-r))sum_(j=0)^(n-r-1)((r+j)/(p))((m+j-1)/(j))]:} \begin{gather*}
\frac{1}{(b-a)^{m}} \cdot \frac{1}{r!} \sum_{j=0}^{n-r-1}\binom{m+j-1}{j} \frac{1}{(b-a)^{j}}\left[(x-a)^{m}(x-b)^{r+j}\right]_{x=a}^{(m+p)}= \tag{36}\\
\quad=(-1)^{r-p} \frac{(m+p)!}{r!} \frac{1}{(b-a)^{m+p-r}} \sum_{j=0}^{n-r-1}\binom{r+j}{p}\binom{m+j-1}{j}
\end{gather*} (36) 1 ( b − a ) m ⋅ 1 r ! ∑ j = 0 n − r − 1 ( m + j − 1 j ) 1 ( b − a ) j [ ( x − a ) m ( x − b ) r + j ] x = a ( m + p ) = = ( − 1 ) r − p ( m + p ) ! r ! 1 ( b − a ) m + p − r ∑ j = 0 n − r − 1 ( r + j p ) ( m + j − 1 j ) Să ne ocupăm apoi cu evaluarea sumei
(37)
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(37)
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{:(37)C_(r)=sum_(j=0)^(n-r-1)((m-1+j)/(j))((r+j)/(p)):} \begin{equation*}
C_{r}=\sum_{j=0}^{n-r-1}\binom{m-1+j}{j}\binom{r+j}{p} \tag{37}
\end{equation*} (37) C r = ∑ j = 0 n − r − 1 ( m − 1 + j j ) ( r + j p ) Aici unii termeni de la început sînt nuli, căci în dezvoltarea după formula lui Leibniz, de care ne-am folosit mai sus, trebuie presupus
r
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≧
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r+j >= p r+j \geqq p r + j ≧ p , astfel că
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C_(r) C_{r} C r se reduce 1a
(38)
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(38)
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{:(38)C_(r)=sum_(j=p-r)^(n-r-1)((m-1+j)/(j))((r+j)/(r-p+j)):} \begin{equation*}
C_{r}=\sum_{j=p-r}^{n-r-1}\binom{m-1+j}{j}\binom{r+j}{r-p+j} \tag{38}
\end{equation*} (38) C r = ∑ j = p − r n − r − 1 ( m − 1 + j j ) ( r + j r − p + j ) Acesta se poate pune şi sub forma
(39)
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(39)
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{:(39)C_(r)=sum_(i=0)^(n-p-1)((m+p-r-1+i)/(p-r+i))((p+i)/(i)):} \begin{equation*}
C_{r}=\sum_{i=0}^{n-p-1}\binom{m+p-r-1+i}{p-r+i}\binom{p+i}{i} \tag{39}
\end{equation*} (39) C r = ∑ i = 0 n − p − 1 ( m + p − r − 1 + i p − r + i ) ( p + i i ) sau
(40)
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(40)
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{:(40)C_(r)=((m+p-r-1)/(m-1))sum_(i=0)^(n-p-1)((p+i)/(i))(((m+p-r+i)/(i)))/(((p-r+i)/(i))):} \begin{equation*}
C_{r}=\binom{m+p-r-1}{m-1} \sum_{i=0}^{n-p-1}\binom{p+i}{i} \frac{\binom{m+p-r+i}{i}}{\binom{p-r+i}{i}} \tag{40}
\end{equation*} (40) C r = ( m + p − r − 1 m − 1 ) ∑ i = 0 n − p − 1 ( p + i i ) ( m + p − r + i i ) ( p − r + i i ) Aplicînd formula
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((a)/(b))=sum_(j=0)^(l)((l)/(j))((a-l)/(b-j)) \binom{a}{b}=\sum_{j=0}^{l}\binom{l}{j}\binom{a-l}{b-j} ( a b ) = ∑ j = 0 l ( l j ) ( a − l b − j ) obţinem
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((p+i)/(i))=sum_(j=0)^(r)((r)/(j))((p+i-v)/(i-j)). \binom{p+i}{i}=\sum_{j=0}^{r}\binom{r}{j}\binom{p+i-v}{i-j} . ( p + i i ) = ∑ j = 0 r ( r j ) ( p + i − v i − j ) . Cu aceasta
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{:[C_(r)=sum_(i=0)^(n-p-1)sum_(j=0)^(r)((r)/(j))((p+i-r)/(i-j))((m+p-r-1+i)/(p-r+i))=],[=sum_(j=0)^(r)((r)/(j))sum_(i=j)^(n-p-1)((p+i-r)/(i-j))((m+p-r-1+i)/(p-r+i))","]:} \begin{aligned}
C_{r} & =\sum_{i=0}^{n-p-1} \sum_{j=0}^{r}\binom{r}{j}\binom{p+i-r}{i-j}\binom{m+p-r-1+i}{p-r+i}= \\
& =\sum_{j=0}^{r}\binom{r}{j} \sum_{i=j}^{n-p-1}\binom{p+i-r}{i-j}\binom{m+p-r-1+i}{p-r+i},
\end{aligned} C r = ∑ i = 0 n − p − 1 ∑ j = 0 r ( r j ) ( p + i − r i − j ) ( m + p − r − 1 + i p − r + i ) = = ∑ j = 0 r ( r j ) ∑ i = j n − p − 1 ( p + i − r i − j ) ( m + p − r − 1 + i p − r + i ) , în baza obişnuitei convenții ca
(
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n
)
(
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((m)/(n)) \binom{m}{n} ( m n ) să fie nul dacă
m
m
m m m sau
n
n
n n n ar fi negativi.
Notînd
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beta=i-j,quad m=p+j-r,quad n=m+p+j-r-1 \beta=i-j, \quad m=p+j-r, \quad n=m+p+j-r-1 β = i − j , m = p + j − r , n = m + p + j − r − 1 avem
C
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C_(r)=sum_(j=0)^(r)((r)/(j))^(n-p-j-1)sum_(beta=0)^(-1)((m+beta)/(beta))((n+beta)/(m+beta)). C_{r}=\sum_{j=0}^{r}\binom{r}{j}^{n-p-j-1} \sum_{\beta=0}^{-1}\binom{m+\beta}{\beta}\binom{n+\beta}{m+\beta} . C r = ∑ j = 0 r ( r j ) n − p − j − 1 ∑ β = 0 − 1 ( m + β β ) ( n + β m + β ) . Aplicînd identitatea
∑
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sum_(p=0)^(k)((m+beta)/(beta))((n+beta)/(m+beta))=((n+k+1)/(k))((n)/(m)) \sum_{p=0}^{k}\binom{m+\beta}{\beta}\binom{n+\beta}{m+\beta}=\binom{n+k+1}{k}\binom{n}{m} ∑ p = 0 k ( m + β β ) ( n + β m + β ) = ( n + k + 1 k ) ( n m ) obținem în definitiv
(41)
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(41)
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{:(41)C_(r)=sum_(j=0)^(r)((r)/(j))((m+n-r-1)/(n-p-j-1))((m+p+j-r-1)/(p+j-r)):} \begin{equation*}
C_{r}=\sum_{j=0}^{r}\binom{r}{j}\binom{m+n-r-1}{n-p-j-1}\binom{m+p+j-r-1}{p+j-r} \tag{41}
\end{equation*} (41) C r = ∑ j = 0 r ( r j ) ( m + n − r − 1 n − p − j − 1 ) ( m + p + j − r − 1 p + j − r ) De exemplu, pentru
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k=0,1,2 k=0,1,2 k = 0 , 1 , 2 avem
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+
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{:[C_(0)=((m+p-1)/(m-1))((m+n-1)/(m+p))","],[C_(1)=((m+n-2)/(m+p-1))((m+p-2)/(m-1))+((m+n-2)/(m+p))((m+p-1)/(m-1))","],[C_(2)=((m+n-3)/(n-p-1))((m+p-3)/(p-2))+2((m+n-3)/(n-p-2))((m+p-2)/(p-1))+((m+n-3)/(n-p-3))((m+p-1)/(p))]:} \begin{gathered}
C_{0}=\binom{m+p-1}{m-1}\binom{m+n-1}{m+p}, \\
C_{1}=\binom{m+n-2}{m+p-1}\binom{m+p-2}{m-1}+\binom{m+n-2}{m+p}\binom{m+p-1}{m-1}, \\
C_{2}=\binom{m+n-3}{n-p-1}\binom{m+p-3}{p-2}+2\binom{m+n-3}{n-p-2}\binom{m+p-2}{p-1}+\binom{m+n-3}{n-p-3}\binom{m+p-1}{p}
\end{gathered} C 0 = ( m + p − 1 m − 1 ) ( m + n − 1 m + p ) , C 1 = ( m + n − 2 m + p − 1 ) ( m + p − 2 m − 1 ) + ( m + n − 2 m + p ) ( m + p − 1 m − 1 ) , C 2 = ( m + n − 3 n − p − 1 ) ( m + p − 3 p − 2 ) + 2 ( m + n − 3 n − p − 2 ) ( m + p − 2 p − 1 ) + ( m + n − 3 n − p − 3 ) ( m + p − 1 p )
T, Tinînd seama de (35), (36), (37), (41), se vede că am putut construi următoarea formulă de derivare numerică de grad de exactitate
m
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m+n-1 m+n-1 m + n − 1
(42)
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(42)
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{:(42)f^((m+p))(a)=sum_(k=0)^(m-1)A_(k)f^((k))(a)+sum_(r=0)^(n-1)B_(r)f^((r))(b)+rho(f):} \begin{equation*}
f^{(m+p)}(a)=\sum_{k=0}^{m-1} A_{k} f^{(k)}(a)+\sum_{r=0}^{n-1} B_{r} f^{(r)}(b)+\rho(f) \tag{42}
\end{equation*} (42) f ( m + p ) ( a ) = ∑ k = 0 m − 1 A k f ( k ) ( a ) + ∑ r = 0 n − 1 B r f ( r ) ( b ) + ρ ( f ) unde
(43)
A
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(43)
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{:(43)A_(k)=(-1)^(p+1)((m+p)!)/(k!(b-a)^(m+p-k))((m+p-k-1)/(p))((n+m-k-1)/(n-p-1)):} \begin{equation*}
A_{k}=(-1)^{p+1} \frac{(m+p)!}{k!(b-a)^{m+p-k}}\binom{m+p-k-1}{p}\binom{n+m-k-1}{n-p-1} \tag{43}
\end{equation*} (43) A k = ( − 1 ) p + 1 ( m + p ) ! k ! ( b − a ) m + p − k ( m + p − k − 1 p ) ( n + m − k − 1 n − p − 1 ) iar
(44)
B
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(44)
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{:(44)B_(r)=(-1)^(r+p)((m+p)!)/(r!(b-a)^(m+p-r))C_(r):} \begin{equation*}
B_{r}=(-1)^{r+p} \frac{(m+p)!}{r!(b-a)^{m+p-r}} C_{r} \tag{44}
\end{equation*} (44) B r = ( − 1 ) r + p ( m + p ) ! r ! ( b − a ) m + p − r C r și
(45)
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(45)
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{:(45)C_(r)=sum_(j-0)^(r)((r)/(j))((m+n-r-1)/(n-p-j-1))((m+p+j-r-1)/(p+j-r)):} \begin{equation*}
C_{r}=\sum_{j-0}^{r}\binom{r}{j}\binom{m+n-r-1}{n-p-j-1}\binom{m+p+j-r-1}{p+j-r} \tag{45}
\end{equation*} (45) C r = ∑ j − 0 r ( r j ) ( m + n − r − 1 n − p − j − 1 ) ( m + p + j − r − 1 p + j − r ) sau cu
C
r
C
r
C_(r) C_{r} C r dat de (38), (39) sau (40).
Pentru restul acestei formule s-a obţinut expresia
(46)
ρ
(
f
)
=
(
m
+
p
)
!
(
m
+
n
)
!
(
n
p
)
(
a
−
b
)
n
−
p
f
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m
+
n
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(
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,
a
<
ξ
<
b
.
(46)
ρ
(
f
)
=
(
m
+
p
)
!
(
m
+
n
)
!
(
n
p
)
(
a
−
b
)
n
−
p
f
(
m
+
n
)
(
ξ
)
,
a
<
ξ
<
b
.
{:(46)rho(f)=((m+p)!)/((m+n)!)((n)/(p))(a-b)^(n-p)f^((m+n))(xi)","a < xi < b.:} \begin{equation*}
\rho(f)=\frac{(m+p)!}{(m+n)!}\binom{n}{p}(a-b)^{n-p} f^{(m+n)}(\xi), a<\xi<b . \tag{46}
\end{equation*} (46) ρ ( f ) = ( m + p ) ! ( m + n ) ! ( n p ) ( a − b ) n − p f ( m + n ) ( ξ ) , a < ξ < b .
Cazuri particulare ale formulei de derivare numerică (42) :
1
∘
.
m
=
n
=
2
,
p
=
1
1
∘
.
m
=
n
=
2
,
p
=
1
1^(@).m=n=2,p=1 1^{\circ} . m=n=2, p=1 1 ∘ . m = n = 2 , p = 1
f
′
′
′
(
a
)
=
−
12
(
b
−
a
)
3
[
f
(
a
)
−
f
(
b
)
]
+
6
(
b
−
a
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2
[
f
′
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a
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+
f
′
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b
)
]
+
a
−
b
2
f
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4
)
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ξ
)
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f
′
′
′
(
a
)
=
−
12
(
b
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a
)
3
[
f
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a
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−
f
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b
)
]
+
6
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a
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2
f
′
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a
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+
f
′
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b
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+
a
−
b
2
f
(
4
)
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ξ
)
.
f^(''')(a)=-(12)/((b-a)^(3))[f(a)-f(b)]+(6)/((b-a)^(2))[f^(')(a)+f^(')(b)]+(a-b)/(2)f^((4))(xi). f^{\prime \prime \prime}(a)=-\frac{12}{(b-a)^{3}}[f(a)-f(b)]+\frac{6}{(b-a)^{2}}\left[f^{\prime}(a)+f^{\prime}(b)\right]+\frac{a-b}{2} f^{(4)}(\xi) . f ′ ′ ′ ( a ) = − 12 ( b − a ) 3 [ f ( a ) − f ( b ) ] + 6 ( b − a ) 2 [ f ′ ( a ) + f ′ ( b ) ] + a − b 2 f ( 4 ) ( ξ ) .
2
∘
.
m
=
3
,
n
=
4
,
p
=
0
2
∘
.
m
=
3
,
n
=
4
,
p
=
0
2^(@).m=3,n=4,p=0 2^{\circ} . m=3, n=4, p=0 2 ∘ . m = 3 , n = 4 , p = 0 :
f
′
′
′
(
a
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=
120
(
b
−
a
)
3
[
f
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b
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−
f
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a
)
]
−
60
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b
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a
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2
[
f
′
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b
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+
f
′
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a
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+
+
12
b
−
a
[
f
′
′
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b
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f
′
′
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a
)
]
−
f
′
′
′
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b
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+
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−
a
)
4
840
f
(
7
)
(
ξ
)
f
′
′
′
(
a
)
=
120
(
b
−
a
)
3
[
f
(
b
)
−
f
(
a
)
]
−
60
(
b
−
a
)
2
f
′
(
b
)
+
f
′
(
a
)
+
+
12
b
−
a
f
′
′
(
b
)
−
f
′
′
(
a
)
−
f
′
′
′
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b
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+
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b
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a
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4
840
f
(
7
)
(
ξ
)
{:[f^(''')(a)=(120)/((b-a)^(3))[f(b)-f(a)]-(60)/((b-a)^(2))[f^(')(b)+f^(')(a)]+],[quad+(12)/(b-a)[f^('')(b)-f^('')(a)]-f^(''')(b)+((b-a)^(4))/(840)f^((7))(xi)]:} \begin{aligned}
& f^{\prime \prime \prime}(a)=\frac{120}{(b-a)^{3}}[f(b)-f(a)]-\frac{60}{(b-a)^{2}}\left[f^{\prime}(b)+f^{\prime}(a)\right]+ \\
& \quad+\frac{12}{b-a}\left[f^{\prime \prime}(b)-f^{\prime \prime}(a)\right]-f^{\prime \prime \prime}(b)+\frac{(b-a)^{4}}{840} f^{(7)}(\xi)
\end{aligned} f ′ ′ ′ ( a ) = 120 ( b − a ) 3 [ f ( b ) − f ( a ) ] − 60 ( b − a ) 2 [ f ′ ( b ) + f ′ ( a ) ] + + 12 b − a [ f ′ ′ ( b ) − f ′ ′ ( a ) ] − f ′ ′ ′ ( b ) + ( b − a ) 4 840 f ( 7 ) ( ξ )
3
∘
.
m
=
3
,
n
=
4
,
p
=
1
3
∘
.
m
=
3
,
n
=
4
,
p
=
1
3^(@).m=3,n=4,p=1 3^{\circ} . m=3, n=4, p=1 3 ∘ . m = 3 , n = 4 , p = 1 :
f
(
I
V
)
(
a
)
=
1080
(
b
−
a
)
4
f
(
a
)
+
480
(
b
−
a
)
3
f
′
(
a
)
+
72
(
b
−
a
)
2
f
′
′
(
a
)
−
1080
(
b
−
a
)
4
f
(
b
)
+
+
600
(
b
−
a
)
3
f
′
(
b
)
−
132
(
b
−
a
)
2
f
′
′
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+
12
b
−
a
f
′
′
′
(
b
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−
2
(
b
−
a
)
3
105
f
(
7
)
(
ξ
)
f
(
I
V
)
(
a
)
=
1080
(
b
−
a
)
4
f
(
a
)
+
480
(
b
−
a
)
3
f
′
(
a
)
+
72
(
b
−
a
)
2
f
′
′
(
a
)
−
1080
(
b
−
a
)
4
f
(
b
)
+
+
600
(
b
−
a
)
3
f
′
(
b
)
−
132
(
b
−
a
)
2
f
′
′
(
b
)
+
12
b
−
a
f
′
′
′
(
b
)
−
2
(
b
−
a
)
3
105
f
(
7
)
(
ξ
)
{:[f^((IV))(a)=(1080)/((b-a)^(4))f(a)+(480)/((b-a)^(3))f^(')(a)+(72)/((b-a)^(2))f^('')(a)-(1080)/((b-a)^(4))f(b)+],[quad+(600)/((b-a)^(3))f^(')(b)-(132)/((b-a)^(2))f^('')(b)+(12)/(b-a)f^(''')(b)-(2(b-a)^(3))/(105)f^((7))(xi)]:} \begin{aligned}
& f^{(I V)}(a)=\frac{1080}{(b-a)^{4}} f(a)+\frac{480}{(b-a)^{3}} f^{\prime}(a)+\frac{72}{(b-a)^{2}} f^{\prime \prime}(a)-\frac{1080}{(b-a)^{4}} f(b)+ \\
& \quad+\frac{600}{(b-a)^{3}} f^{\prime}(b)-\frac{132}{(b-a)^{2}} f^{\prime \prime}(b)+\frac{12}{b-a} f^{\prime \prime \prime}(b)-\frac{2(b-a)^{3}}{105} f^{(7)}(\xi)
\end{aligned} f ( I V ) ( a ) = 1080 ( b − a ) 4 f ( a ) + 480 ( b − a ) 3 f ′ ( a ) + 72 ( b − a ) 2 f ′ ′ ( a ) − 1080 ( b − a ) 4 f ( b ) + + 600 ( b − a ) 3 f ′ ( b ) − 132 ( b − a ) 2 f ′ ′ ( b ) + 12 b − a f ′ ′ ′ ( b ) − 2 ( b − a ) 3 105 f ( 7 ) ( ξ )
4
∘
.
m
==
3
,
n
=
4
,
p
=
2
4
∘
.
m
==
3
,
n
=
4
,
p
=
2
4^(@).m==3,n=4,p=2 4^{\circ} . m==3, n=4, p=2 4 ∘ . m == 3 , n = 4 , p = 2 :
f
(
V
)
(
a
)
=
−
4320
(
b
−
a
)
5
f
(
a
)
−
1800
(
b
−
a
)
4
f
′
(
a
)
−
240
(
b
−
a
)
3
f
′
′
(
a
)
+
4320
(
b
−
a
)
5
f
(
b
)
−
−
2520
(
b
−
a
)
4
f
′
(
b
)
+
600
(
b
−
a
)
3
f
′
′
(
b
)
−
60
(
b
−
a
)
2
f
′
′
′
(
b
)
+
1
7
(
b
−
a
)
2
f
(
7
)
(
ξ
)
f
(
V
)
(
a
)
=
−
4320
(
b
−
a
)
5
f
(
a
)
−
1800
(
b
−
a
)
4
f
′
(
a
)
−
240
(
b
−
a
)
3
f
′
′
(
a
)
+
4320
(
b
−
a
)
5
f
(
b
)
−
−
2520
(
b
−
a
)
4
f
′
(
b
)
+
600
(
b
−
a
)
3
f
′
′
(
b
)
−
60
(
b
−
a
)
2
f
′
′
′
(
b
)
+
1
7
(
b
−
a
)
2
f
(
7
)
(
ξ
)
{:[f^((V))(a)=(-4320)/((b-a)^(5))f(a)-(1800)/((b-a)^(4))f^(')(a)-(240)/((b-a)^(3))f^('')(a)+(4320)/((b-a)^(5))f(b)-],[quad-(2520)/((b-a)^(4))f^(')(b)+(600)/((b-a)^(3))f^('')(b)-(60)/((b-a)^(2))f^(''')(b)+(1)/(7)(b-a)^(2)f^((7))(xi)]:} \begin{aligned}
& f^{(V)}(a)=\frac{-4320}{(b-a)^{5}} f(a)-\frac{1800}{(b-a)^{4}} f^{\prime}(a)-\frac{240}{(b-a)^{3}} f^{\prime \prime}(a)+\frac{4320}{(b-a)^{5}} f(b)- \\
& \quad-\frac{2520}{(b-a)^{4}} f^{\prime}(b)+\frac{600}{(b-a)^{3}} f^{\prime \prime}(b)-\frac{60}{(b-a)^{2}} f^{\prime \prime \prime}(b)+\frac{1}{7}(b-a)^{2} f^{(7)}(\xi)
\end{aligned} f ( V ) ( a ) = − 4320 ( b − a ) 5 f ( a ) − 1800 ( b − a ) 4 f ′ ( a ) − 240 ( b − a ) 3 f ′ ′ ( a ) + 4320 ( b − a ) 5 f ( b ) − − 2520 ( b − a ) 4 f ′ ( b ) + 600 ( b − a ) 3 f ′ ′ ( b ) − 60 ( b − a ) 2 f ′ ′ ′ ( b ) + 1 7 ( b − a ) 2 f ( 7 ) ( ξ )
5
∘
.
m
=
3
,
n
=
4
,
p
=
3
5
∘
.
m
=
3
,
n
=
4
,
p
=
3
5^(@).m=3,n=4,p=3 5^{\circ} . m=3, n=4, p=3 5 ∘ . m = 3 , n = 4 , p = 3 :
f
(
V
I
)
(
a
)
=
7200
(
b
−
a
)
6
f
(
a
)
+
2880
(
b
−
a
)
5
f
′
(
a
)
+
360
(
b
−
a
)
4
f
′
′
(
a
)
−
7200
(
b
−
a
)
6
f
(
b
)
+
+
4320
(
b
−
a
)
5
f
′
(
b
)
−
1080
(
b
−
a
)
4
f
′
′
(
b
)
+
120
(
b
−
a
)
4
f
′
′
′
(
b
)
−
4
(
b
−
a
)
7
f
(
7
)
(
ξ
)
f
(
V
I
)
(
a
)
=
7200
(
b
−
a
)
6
f
(
a
)
+
2880
(
b
−
a
)
5
f
′
(
a
)
+
360
(
b
−
a
)
4
f
′
′
(
a
)
−
7200
(
b
−
a
)
6
f
(
b
)
+
+
4320
(
b
−
a
)
5
f
′
(
b
)
−
1080
(
b
−
a
)
4
f
′
′
(
b
)
+
120
(
b
−
a
)
4
f
′
′
′
(
b
)
−
4
(
b
−
a
)
7
f
(
7
)
(
ξ
)
{:[f^((VI))(a)=(7200)/((b-a)^(6))f(a)+(2880)/((b-a)^(5))f^(')(a)+(360)/((b-a)^(4))f^('')(a)-(7200)/((b-a)^(6))f(b)+],[+(4320)/((b-a)^(5))f^(')(b)-(1080)/((b-a)^(4))f^('')(b)+(120)/((b-a)^(4))f^(''')(b)-(4(b-a))/(7)f^((7))(xi)]:} \begin{aligned}
& f^{(V I)}(a)=\frac{7200}{(b-a)^{6}} f(a)+\frac{2880}{(b-a)^{5}} f^{\prime}(a)+\frac{360}{(b-a)^{4}} f^{\prime \prime}(a)-\frac{7200}{(b-a)^{6}} f(b)+ \\
& +\frac{4320}{(b-a)^{5}} f^{\prime}(b)-\frac{1080}{(b-a)^{4}} f^{\prime \prime}(b)+\frac{120}{(b-a)^{4}} f^{\prime \prime \prime}(b)-\frac{4(b-a)}{7} f^{(7)}(\xi)
\end{aligned} f ( V I ) ( a ) = 7200 ( b − a ) 6 f ( a ) + 2880 ( b − a ) 5 f ′ ( a ) + 360 ( b − a ) 4 f ′ ′ ( a ) − 7200 ( b − a ) 6 f ( b ) + + 4320 ( b − a ) 5 f ′ ( b ) − 1080 ( b − a ) 4 f ′ ′ ( b ) + 120 ( b − a ) 4 f ′ ′ ′ ( b ) − 4 ( b − a ) 7 f ( 7 ) ( ξ )
6
∘
.
m
=
1
,
n
=
6
,
p
=
0
6
∘
.
m
=
1
,
n
=
6
,
p
=
0
6^(@).m=1,n=6,p=0 6^{\circ} . m=1, n=6, p=0 6 ∘ . m = 1 , n = 6 , p = 0 :
f
′
(
a
)
=
6
b
−
a
[
f
(
b
)
−
f
(
a
)
]
−
5
f
′
(
b
)
+
2
(
b
−
a
)
f
′
′
(
b
)
−
1
2
f
′
′
′
(
b
)
+
+
1
12
(
b
−
a
)
3
f
(
I
V
)
(
b
)
−
1
120
(
b
−
a
)
4
f
(
V
)
(
b
)
+
(
b
−
a
)
6
5040
f
(
7
)
(
ξ
)
f
′
(
a
)
=
6
b
−
a
[
f
(
b
)
−
f
(
a
)
]
−
5
f
′
(
b
)
+
2
(
b
−
a
)
f
′
′
(
b
)
−
1
2
f
′
′
′
(
b
)
+
+
1
12
(
b
−
a
)
3
f
(
I
V
)
(
b
)
−
1
120
(
b
−
a
)
4
f
(
V
)
(
b
)
+
(
b
−
a
)
6
5040
f
(
7
)
(
ξ
)
{:[f^(')(a)=(6)/(b-a)[f(b)-f(a)]-5f^(')(b)+2(b-a)f^('')(b)-(1)/(2)f^(''')(b)+],[quad+(1)/(12)(b-a)^(3)f^((IV))(b)-(1)/(120)(b-a)^(4)f^((V))(b)+((b-a)^(6))/(5040)f^((7))(xi)]:} \begin{aligned}
& f^{\prime}(a)=\frac{6}{b-a}[f(b)-f(a)]-5 f^{\prime}(b)+2(b-a) f^{\prime \prime}(b)-\frac{1}{2} f^{\prime \prime \prime}(b)+ \\
& \quad+\frac{1}{12}(b-a)^{3} f^{(I V)}(b)-\frac{1}{120}(b-a)^{4} f^{(V)}(b)+\frac{(b-a)^{6}}{5040} f^{(7)}(\xi)
\end{aligned} f ′ ( a ) = 6 b − a [ f ( b ) − f ( a ) ] − 5 f ′ ( b ) + 2 ( b − a ) f ′ ′ ( b ) − 1 2 f ′ ′ ′ ( b ) + + 1 12 ( b − a ) 3 f ( I V ) ( b ) − 1 120 ( b − a ) 4 f ( V ) ( b ) + ( b − a ) 6 5040 f ( 7 ) ( ξ )
7
∘
.
m
=
1
,
n
=
6
,
p
=
1
7
∘
.
m
=
1
,
n
=
6
,
p
=
1
7^(@).m=1,n=6,p=1 7^{\circ} . m=1, n=6, p=1 7 ∘ . m = 1 , n = 6 , p = 1 :
f
′
′
(
a
)
=
30
(
b
−
a
)
2
[
f
(
a
)
−
f
(
b
)
]
+
30
b
−
a
f
′
(
b
)
−
14
f
′
′
(
b
)
+
4
(
b
−
a
)
f
′
′
′
(
b
)
−
−
3
4
(
b
−
a
)
2
f
(
I
V
)
(
b
)
+
1
12
(
b
−
a
)
3
f
(
V
)
(
b
)
−
(
b
−
a
)
5
420
f
(
7
)
(
ξ
)
f
′
′
(
a
)
=
30
(
b
−
a
)
2
[
f
(
a
)
−
f
(
b
)
]
+
30
b
−
a
f
′
(
b
)
−
14
f
′
′
(
b
)
+
4
(
b
−
a
)
f
′
′
′
(
b
)
−
−
3
4
(
b
−
a
)
2
f
(
I
V
)
(
b
)
+
1
12
(
b
−
a
)
3
f
(
V
)
(
b
)
−
(
b
−
a
)
5
420
f
(
7
)
(
ξ
)
{:[f^('')(a)=(30)/((b-a)^(2))[f(a)-f(b)]+(30)/(b-a)f^(')(b)-14f^('')(b)+4(b-a)f^(''')(b)-],[quad-(3)/(4)(b-a)^(2)f^((IV))(b)+(1)/(12)(b-a)^(3)f^((V))(b)-((b-a)^(5))/(420)f^((7))(xi)]:} \begin{aligned}
& f^{\prime \prime}(a)=\frac{30}{(b-a)^{2}}[f(a)-f(b)]+\frac{30}{b-a} f^{\prime}(b)-14 f^{\prime \prime}(b)+4(b-a) f^{\prime \prime \prime}(b)- \\
& \quad-\frac{3}{4}(b-a)^{2} f^{(I V)}(b)+\frac{1}{12}(b-a)^{3} f^{(V)}(b)-\frac{(b-a)^{5}}{420} f^{(7)}(\xi)
\end{aligned} f ′ ′ ( a ) = 30 ( b − a ) 2 [ f ( a ) − f ( b ) ] + 30 b − a f ′ ( b ) − 14 f ′ ′ ( b ) + 4 ( b − a ) f ′ ′ ′ ( b ) − − 3 4 ( b − a ) 2 f ( I V ) ( b ) + 1 12 ( b − a ) 3 f ( V ) ( b ) − ( b − a ) 5 420 f ( 7 ) ( ξ )
8
∘
.
m
=
1
,
n
=
6
,
p
=
2
8
∘
.
m
=
1
,
n
=
6
,
p
=
2
8^(@).m=1,n=6,p=2 8^{\circ} . m=1, n=6, p=2 8 ∘ . m = 1 , n = 6 , p = 2 :
f
′
′
′
(
a
)
=
120
(
b
−
a
)
3
[
f
(
b
)
−
f
(
a
)
]
−
120
(
b
−
a
)
2
f
′
(
b
)
+
60
b
−
a
f
′
′
(
b
)
−
−
19
f
′
′
′
(
b
)
+
4
(
b
−
a
)
f
(
I
V
)
(
b
)
−
1
2
(
b
−
a
)
2
f
(
V
)
(
b
)
+
(
b
−
a
)
4
56
f
(
7
)
(
ξ
)
f
′
′
′
(
a
)
=
120
(
b
−
a
)
3
[
f
(
b
)
−
f
(
a
)
]
−
120
(
b
−
a
)
2
f
′
(
b
)
+
60
b
−
a
f
′
′
(
b
)
−
−
19
f
′
′
′
(
b
)
+
4
(
b
−
a
)
f
(
I
V
)
(
b
)
−
1
2
(
b
−
a
)
2
f
(
V
)
(
b
)
+
(
b
−
a
)
4
56
f
(
7
)
(
ξ
)
{:[f^(''')(a)=(120)/((b-a)^(3))[f(b)-f(a)]-(120)/((b-a)^(2))f^(')(b)+(60)/(b-a)f^('')(b)-],[-19f^(''')(b)+4(b-a)f^((IV))(b)-(1)/(2)(b-a)^(2)f^((V))(b)+((b-a)^(4))/(56)f^((7))(xi)]:} \begin{aligned}
& f^{\prime \prime \prime}(a)=\frac{120}{(b-a)^{3}}[f(b)-f(a)]-\frac{120}{(b-a)^{2}} f^{\prime}(b)+\frac{60}{b-a} f^{\prime \prime}(b)- \\
& -19 f^{\prime \prime \prime}(b)+4(b-a) f^{(I V)}(b)-\frac{1}{2}(b-a)^{2} f^{(V)}(b)+\frac{(b-a)^{4}}{56} f^{(7)}(\xi)
\end{aligned} f ′ ′ ′ ( a ) = 120 ( b − a ) 3 [ f ( b ) − f ( a ) ] − 120 ( b − a ) 2 f ′ ( b ) + 60 b − a f ′ ′ ( b ) − − 19 f ′ ′ ′ ( b ) + 4 ( b − a ) f ( I V ) ( b ) − 1 2 ( b − a ) 2 f ( V ) ( b ) + ( b − a ) 4 56 f ( 7 ) ( ξ )
9
∘
.
m
=
1
,
n
=
6
,
p
=
3
9
∘
.
m
=
1
,
n
=
6
,
p
=
3
9^(@).m=1,n=6,p=3 9^{\circ} . m=1, n=6, p=3 9 ∘ . m = 1 , n = 6 , p = 3 :
f
(
I
V
)
(
a
)
=
360
(
b
−
a
)
4
[
f
(
a
)
−
f
(
b
)
]
+
360
(
b
−
a
)
3
f
′
(
b
)
−
180
(
b
−
a
)
2
f
′
′
(
b
)
+
+
60
b
−
a
f
′
′
′
(
b
)
−
14
f
(
I
V
)
(
b
)
+
2
(
b
−
a
)
f
(
V
)
(
b
)
−
2
(
b
−
a
)
3
21
f
(
7
)
(
ξ
)
f
(
I
V
)
(
a
)
=
360
(
b
−
a
)
4
[
f
(
a
)
−
f
(
b
)
]
+
360
(
b
−
a
)
3
f
′
(
b
)
−
180
(
b
−
a
)
2
f
′
′
(
b
)
+
+
60
b
−
a
f
′
′
′
(
b
)
−
14
f
(
I
V
)
(
b
)
+
2
(
b
−
a
)
f
(
V
)
(
b
)
−
2
(
b
−
a
)
3
21
f
(
7
)
(
ξ
)
{:[f^((IV))(a)=(360)/((b-a)^(4))[f(a)-f(b)]+(360)/((b-a)^(3))f^(')(b)-(180)/((b-a)^(2))f^('')(b)+],[quad+(60)/(b-a)f^(''')(b)-14f^((IV))(b)+2(b-a)f^((V))(b)-(2(b-a)^(3))/(21)f^((7))(xi)]:} \begin{aligned}
& f^{(I V)}(a)=\frac{360}{(b-a)^{4}}[f(a)-f(b)]+\frac{360}{(b-a)^{3}} f^{\prime}(b)-\frac{180}{(b-a)^{2}} f^{\prime \prime}(b)+ \\
& \quad+\frac{60}{b-a} f^{\prime \prime \prime}(b)-14 f^{(I V)}(b)+2(b-a) f^{(V)}(b)-\frac{2(b-a)^{3}}{21} f^{(7)}(\xi)
\end{aligned} f ( I V ) ( a ) = 360 ( b − a ) 4 [ f ( a ) − f ( b ) ] + 360 ( b − a ) 3 f ′ ( b ) − 180 ( b − a ) 2 f ′ ′ ( b ) + + 60 b − a f ′ ′ ′ ( b ) − 14 f ( I V ) ( b ) + 2 ( b − a ) f ( V ) ( b ) − 2 ( b − a ) 3 21 f ( 7 ) ( ξ )
10
∘
.
m
=
1
,
n
=
6
,
p
=
4
10
∘
.
m
=
1
,
n
=
6
,
p
=
4
10^(@).m=1,n=6,p=4 10^{\circ} . m=1, n=6, p=4 10 ∘ . m = 1 , n = 6 , p = 4 :
f
(
V
)
(
a
)
=
720
(
b
−
a
)
5
[
f
(
b
)
−
f
(
a
)
]
−
720
(
b
−
a
)
4
f
′
(
b
)
+
360
(
b
−
a
)
3
f
′
′
(
b
)
−
−
120
(
b
−
a
)
2
f
′
′
′
(
b
)
+
30
b
−
a
f
(
I
V
)
(
b
)
−
5
f
(
V
)
(
b
)
+
5
(
b
−
a
)
2
14
f
(
7
)
(
ξ
)
f
(
V
)
(
a
)
=
720
(
b
−
a
)
5
[
f
(
b
)
−
f
(
a
)
]
−
720
(
b
−
a
)
4
f
′
(
b
)
+
360
(
b
−
a
)
3
f
′
′
(
b
)
−
−
120
(
b
−
a
)
2
f
′
′
′
(
b
)
+
30
b
−
a
f
(
I
V
)
(
b
)
−
5
f
(
V
)
(
b
)
+
5
(
b
−
a
)
2
14
f
(
7
)
(
ξ
)
{:[f^((V))(a)=(720)/((b-a)^(5))[f(b)-f(a)]-(720)/((b-a)^(4))f^(')(b)+(360)/((b-a)^(3))f^('')(b)-],[-(120)/((b-a)^(2))f^(''')(b)+(30)/(b-a)f^((IV))(b)-5f^((V))(b)+(5(b-a)^(2))/(14)f^((7))(xi)]:} \begin{aligned}
& f^{(V)}(a)=\frac{720}{(b-a)^{5}}[f(b)-f(a)]-\frac{720}{(b-a)^{4}} f^{\prime}(b)+\frac{360}{(b-a)^{3}} f^{\prime \prime}(b)- \\
& -\frac{120}{(b-a)^{2}} f^{\prime \prime \prime}(b)+\frac{30}{b-a} f^{(I V)}(b)-5 f^{(V)}(b)+\frac{5(b-a)^{2}}{14} f^{(7)}(\xi)
\end{aligned} f ( V ) ( a ) = 720 ( b − a ) 5 [ f ( b ) − f ( a ) ] − 720 ( b − a ) 4 f ′ ( b ) + 360 ( b − a ) 3 f ′ ′ ( b ) − − 120 ( b − a ) 2 f ′ ′ ′ ( b ) + 30 b − a f ( I V ) ( b ) − 5 f ( V ) ( b ) + 5 ( b − a ) 2 14 f ( 7 ) ( ξ )
11
∘
.
m
=
1
,
n
=
6
,
p
=
5
11
∘
.
m
=
1
,
n
=
6
,
p
=
5
11^(@).m=1,n=6,p=5 11^{\circ} . m=1, n=6, p=5 11 ∘ . m = 1 , n = 6 , p = 5 :
f
(
V
I
)
(
a
)
=
720
(
b
−
a
)
6
[
f
(
a
)
−
f
(
b
)
]
+
720
(
b
−
a
)
5
f
′
(
b
)
−
360
(
b
−
a
)
4
f
′
′
(
b
)
+
+
120
(
b
−
a
)
3
f
′
′
′
(
b
)
−
30
(
b
−
a
)
2
f
(
I
V
)
(
b
)
+
6
b
−
a
f
(
V
)
(
b
)
−
6
(
b
−
a
)
7
f
(
7
)
(
ξ
)
f
(
V
I
)
(
a
)
=
720
(
b
−
a
)
6
[
f
(
a
)
−
f
(
b
)
]
+
720
(
b
−
a
)
5
f
′
(
b
)
−
360
(
b
−
a
)
4
f
′
′
(
b
)
+
+
120
(
b
−
a
)
3
f
′
′
′
(
b
)
−
30
(
b
−
a
)
2
f
(
I
V
)
(
b
)
+
6
b
−
a
f
(
V
)
(
b
)
−
6
(
b
−
a
)
7
f
(
7
)
(
ξ
)
{:[f^((VI))(a)=(720)/((b-a)^(6))[f(a)-f(b)]+(720)/((b-a)^(5))f^(')(b)-(360)/((b-a)^(4))f^('')(b)+],[quad+(120)/((b-a)^(3))f^(''')(b)-(30)/((b-a)^(2))f^((IV))(b)+(6)/(b-a)f^((V))(b)-(6(b-a))/(7)f^((7))(xi)]:} \begin{aligned}
& f^{(V I)}(a)=\frac{720}{(b-a)^{6}}[f(a)-f(b)]+\frac{720}{(b-a)^{5}} f^{\prime}(b)-\frac{360}{(b-a)^{4}} f^{\prime \prime}(b)+ \\
& \quad+\frac{120}{(b-a)^{3}} f^{\prime \prime \prime}(b)-\frac{30}{(b-a)^{2}} f^{(I V)}(b)+\frac{6}{b-a} f^{(V)}(b)-\frac{6(b-a)}{7} f^{(7)}(\xi)
\end{aligned} f ( V I ) ( a ) = 720 ( b − a ) 6 [ f ( a ) − f ( b ) ] + 720 ( b − a ) 5 f ′ ( b ) − 360 ( b − a ) 4 f ′ ′ ( b ) + + 120 ( b − a ) 3 f ′ ′ ′ ( b ) − 30 ( b − a ) 2 f ( I V ) ( b ) + 6 b − a f ( V ) ( b ) − 6 ( b − a ) 7 f ( 7 ) ( ξ )
Menționăm că primele 22 formule de derivare numerică - construite în mod explicit, ca exemple, de către prof. T. Popoviciu în importantul său memoriu [13] - se pot obtine imediat din formula mai generală (42). Mai sus am dat alte formule decît cele care se găsesc în lucrarea amintită.
BIBLIOGRAFIE
Ch. Hermite, Sur la formule d'interpolation de Lagrange. Journ. f. d. reine n. angew. Math., 1878, tom. LXXXIV, p. 70-79.
I. P. Natanson, Konstruktivnaia leoria funktii. Moscova-Leningrad, 1949, p. 501.
Gy. Zemp1en, Étude sur l'interpolation et la décomposition des fonctions rationnelles en fractions partielles. Archiv der Mathematik und Physik, 1905 (3) 8, p. 214-226.
Jahrbuch über die Fortschritte der Mathematik, 1905, vol. XXXV, p. 283-284.
E. Netto, Les fonctions rationnelles. Encyclopédie des Sci. Math., 1910, tom. I, vol. II. fasc. 2, p. 62.
V. L. Gonciarov, Teoria interpolirovania i priblijenia tunktii. Moscova, 1954, p. 66.
A. Markoff, Sur la méthode de Gauss pour le calcul approché des intégrales. Math. Annalen, 1885, vol. XXV, p. 427.
K. Petr, O jedné formuli pro numerický výpocet určitých integrálu. Čaopis pro péstováni Mathematiky a Fysiky, 1915, tom. XLIV, p. 454-455.
N. Obreschkoff, Neue Quadraturformeln. Abhandl. d. preuss. Akad. d. Wiss., Math. -naturwiss. Kl., 1940, nr. 4, p. 1-20.
Ş. E. Mikeladze, Cislennîe metodî matematiceskogo analiza. Moscova, 1953, p. 394.
Cislennoe integrirovanie. Uspehi Mat. Nauk, t. III, fasc. 6, 1948.
D. V. Ionescu, Generalizarea tormulei de cuadratură a lui N. Obreschkoff. Studii şi cercet. ştiint, -Acad. R.P.R. Filiala Cluj, 1952, nr. 3-4, p. 1-9.
T. Popoviciu, Asupra restului în unele formule de derivare numerică. Studii şi cercet. mat. - Acad. R.P.R., tom. III, nr. 1-2, p. 103.
Об интерполяционной формуле Хермита и о некоторых их применениях
(Краткое содержание)
После того как отмечается тот факт, что явное выражение, данное Г. Земпленом [3], для интерполяционного многочлена Хермита [2] неточно при
s
≧
2
s
≧
2
s >= 2 s \geqq 2 s ≧ 2 и
r
>
2
r
>
2
r > 2 r>2 r > 2 , устанавливается, несколько более простым путем чем в [6], формула (13); потом дается более явную формулу (15) для интерполяционного многочлена Хермита, удовлетворяющего условиям (3). В (20) дается эффективное выражение для интерполяционного многочлена:
H
m
+
n
−
1
(
a
,
…
,
a
⏟
m
,
b
,
…
,
b
⏟
n
;
f
∣
x
)
.
H
m
+
n
−
1
(
a
,
…
,
a
⏟
m
,
b
,
…
,
b
⏟
n
;
f
∣
x
)
.
H_(m+n-1)ubrace((a,dots,a)_(m),ubrace(b,dots,b)_(n);f∣x). H_{m+n-1} \underbrace{(a, \ldots, a}_{m}, \underbrace{b, \ldots, b}_{n} ; f \mid x) . H m + n − 1 ( a , … , a ⏟ m , b , … , b ⏟ n ; f ∣ x ) . В №. 10-11 применяются предыдущие формулы для разложения рациональной функции на простые дроби. Таким образом, для коеффициентов разложения (23) установились формулы (24). В (27) даны эффективные выражения для коэффициентов разложения частной рациональной функции (26).
В №. 12-13 даны некоторые применения к численному интегрированию функций. В качестве примеров, построены в (29) и (29') две квадратурных формулы высокой точности, полученные при подходящем выборе узлов. Воспользовавшись интерполяционной формулой (20), была установлена квадратурная формула (30). В связи с этим замечается, что как формула (30), как и частная формула (32) были даны Хермитом [1].
В (42) установлена формула численного дифференцирования, аналогичная квадратурной формуле (30) Хермита.
В №. 17 даны 11 примеров таких формул.
Après avoir mentionné que l'expression donnée par G. Zemp1en [3] pour le polynôme d'interpolation d' Hermite (2) est inexacte si
s
≧
2
s
≧
2
s >= 2 s \geqq 2 s ≧ 2 et si les nombres
r
i
r
i
r_(i) r_{i} r i sont plus grands que 2 , on établit, par une voie plus simple que celle utilisée dans [6], la formule (13); puis on donne une formule plus explicite (15) pour le polynôme d'interpolation d'Hermite qui vérifie les conditions (3). A (20) on donne l'expression effective pour le polynôme d'interpolation
H
m
+
n
−
1
(
a
,
…
,
a
⏟
m
,
b
,
…
,
b
⏟
n
;
f
∣
x
)
H
m
+
n
−
1
(
a
,
…
,
a
⏟
m
,
b
,
…
,
b
⏟
n
;
f
∣
x
)
H_(m+n-1)ubrace((a,dots,a)_(m),ubrace(b,dots,b)_(n);f∣x) H_{m+n-1} \underbrace{(a, \ldots, a}_{m}, \underbrace{b, \ldots, b}_{n} ; f \mid x) H m + n − 1 ( a , … , a ⏟ m , b , … , b ⏟ n ; f ∣ x ) .
Aux nos. 10-11 on applique les formules précédentes à la décomposition d'une fonction rationnelle en fractions simples. Ainsi, pour les coefficients de la décomposition (23) on a établi les formules (24). A (27) on donne les expressions efectives des coefficients de la décomposition de la fonction rationnelle particulière (26).
Aux nos. 12-13 on fait quelques applications à l'intégration numérique des fonctions. A titre d'exemple, on construit à (29) et (29') deux formules de quadrature d'un haut degré d'exactitude obtenues par un choix convenable des noeuds. En utilisant la formule d'interpolation (20) on établit la formule de quadratre (30). Ici on fait la remarque que la formule (30) ainsi que la formule particulière (32) ont été données pour 1a première fois par Hermite [1].
A (42) on établit une formule de dérivation numérique analogue à la formule de quadrature (30) d'Hermite. Au no. 17 on donne 11 exemples de pareilles formules.
