ON HERMITE'S INTERPOLATION FORMULA AND SOME APPLICATIONS OF IT*
Ch. Hermite [1] formulated the following general interpolation problem:
To construct a polynomial, of minimal degree, which together with its successive derivatives - up to and including certain orders
R
1
− 1 , … ,
R
S
− 1
R
1
− 1 , … ,
R
S
− 1 r_(1)-1,dots,r_(s)-1 r_{1}-1, \ldots, r_{s}-1 R 1 − 1 , … , R S − 1 , to take on distinct points
(1)
x
1
,
x
2
, … ,
x
S
(1)
x
1
,
x
2
, … ,
x
S
{:(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 values given in advance.
It is known
1
1
^(1) { }^{1} 1 that this polynomial exists and is unique, and its degree is
n
n n n n , where
n =
R
1
+ … +
R
S
− 1
n =
R
1
+ … +
R
S
− 1 n=r_(1)+dots+r_(s)-1 n=r_{1}+\ldots+r_{s}-1 n = R 1 + … + R S − 1 . Let us denote this polynomial by
H
n
( x )
H
n
( x ) H_(n)(x) H_{n}(x) H n ( x ) and suppose that the values given before are the values taken, on the points
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 , of a function
f ( x )
f ( x ) f(x) f(x) f ( x ) and its derivatives. Then the Hermite interpolation polynomial relative to the function
f ( x )
f ( x ) f(x) f(x) f ( x ) and points (1), to which are attached the multiplicity orders respectively
R
1
,
R
2
, … ,
R
S
R
1
,
R
2
, … ,
R
S
r_(1),r_(2),dots,r_(s) r_{1}, r_{2}, \ldots, r_{s} R 1 , R 2 , … , R S ,
(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 )
(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 )
{:(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 ) will be the polynomial that verifies the conditions
(3)
H
n
(
l
)
(
x
k
)
=
f
n
(
l
)
(
x
k
)
(
l
=
0
,
1
,
…
,
r
k
−
1
k
=
1
,
2
,
…
,
s
)
(3)
H
n
(
l
)
x
k
=
f
n
(
l
)
x
k
(
l
=
0
,
1
,
…
,
r
k
−
1
k
=
1
,
2
,
…
,
s
)
{:(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 ( it ) ( x k ) = f n ( it ) ( x k ) ( it = 0 , 1 , … , R k − 1 k = 1 , 2 , … , S )
If
(4)
H
n
(
x
)
=
∑
j
=
0
n
a
j
x
j
,
(4)
H
n
(
x
)
=
∑
j
=
0
n
a
j
x
j
,
{:(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 ,
the previous conditions lead us to the system of
n
+
1
n
+
1
n+1 n+1 n + 1 equations with
n
+
1
n
+
1
n+1 n+1 n + 1 unknown:
a
0
,
a
1
,
…
,
a
n
a
0
,
a
1
,
…
,
a
n
a_(0),a_(1),dots,a_(n) a_{0}, a_{1}, \ldots, a_{n} A 0 , A 1 , … , A n
(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
)
.
(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
.
{:[(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 = and n j ( j − 1 ) … ( j − it + 1 ) A j x k j − it = f ( it ) ( x k ) ( it = 0 , 1 , … , R k − 1 ; k = 1 , 2 , … , S ) . The determinant of this system - as is known - is different from zero, because the points (1) are distinct.
To find the expression of the polynomial
H
n
(
x
)
H
n
(
x
)
H_(n)(x) H_{n}(x) H n ( x ) we could eliminate the unknowns
a
0
,
a
1
,
…
,
a
n
a
0
,
a
1
,
…
,
a
n
a_(0),a_(1),dots,a_(n) a_{0}, a_{1}, \ldots, a_{n} A 0 , A 1 , … , A n between equations (4) and (5). This path is, however, very difficult. G. Z emp 1 en [3] nevertheless tried to find the explicit expression of the polynomial
H
n
(
x
)
H
n
(
x
)
H_(n)(x) H_{n}(x) H n ( x ) . We note, however, that the results obtained by this author are inaccurate; the formula he gave does not correspond to the solution of the problem if
s
≧
2
s
≧
2
s >= 2 s \geqq 2 S ≧ 2 and the numbers
r
i
r
i
r_(i) r_{i} R and are greater than 2 . Of course, the results he gave in the second part of his work, relating to the decomposition of a rational function into simple fractions, are also erroneous. In the review made by We1tzien Zeh1endorf [4] and in the memoir of E. Netto in [5], in which this work is mentioned, no observations were made on the results obtained by G. Zemplen.
3. From the above-mentioned elimination of the unknowns
a
0
,
a
1
,
…
,
a
n
a
0
,
a
1
,
…
,
a
n
a_(0),a_(1),dots,a_(n) a_{0}, a_{1}, \ldots, a_{n} A 0 , A 1 , … , A n and from the formal solution in relation to
H
n
(
x
)
H
n
(
x
)
H_(n)(x) H_{n}(x) H n ( x ) of the equation that was obtained, we can realize that the polynomial sought is of the form
(6)
H
n
(
x
)
=
∑
i
=
1
s
∑
k
=
0
r
i
−
1
l
i
,
k
(
x
)
f
(
k
)
(
x
i
)
,
(6)
H
n
(
x
)
=
∑
i
=
1
s
∑
k
=
0
r
i
−
1
l
i
,
k
(
x
)
f
(
k
)
x
i
,
{:(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 ) = ∑ and = 1 S ∑ k = 0 R and − 1 it and , k ( x ) f ( k ) ( x and ) , where
l
l
,
k
(
x
)
l
l
,
k
(
x
)
l_(l,k)(x) l_{l, k}(x) it it , k ( x ) are polynomials of degree
n
n
n n n From the formal calculation above it is immediately observed that the polynomials
l
l
,
k
(
x
)
l
l
,
k
(
x
)
l_(l,k)(x) l_{l, k}(x) it it , k ( x ) are independent of
f
(
x
)
f
(
x
)
f(x) f(x) f ( x ) and that, taking into account (3), they must verify the following conditions:
(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
)
(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
{:[(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) it and , k ( p ) ( x j ) = 0 ( j ≠ and ; p = 0 , 1 , … , R j − 1 ) (8) it and , k ( p ) ( x and ) = { 0 , for p ≠ k 1 , for p = k ( p = 0 , 1 , … , R and − 1 ) These conditions will completely determine our fundamental interpolation polynomials
l
i
,
k
(
x
)
l
i
,
k
(
x
)
l_(i,k)(x) l_{i, k}(x) it and , k ( x ) Indeed, based on (7) it is observed that
l
i
,
k
(
x
)
l
i
,
k
(
x
)
l_(i,k)(x) l_{i, k}(x) it and , k ( x ) contains as a factor the product
(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
.
(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
.
{:(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 and ( x ) = ( x − x 1 ) γ 1 … ( x − x and − 1 ) γ and − 1 ( x − x and + 1 ) γ and + 1 … ( x − x S ) γ S . Taking into account conditions (8), we conclude that
(
x
−
x
i
)
k
x
−
x
i
k
(x-x_(i))^(k) \left(x-x_{i}\right)^{k} ( x − x and ) k is also a factor of
l
l
,
k
(
x
)
l
l
,
k
(
x
)
l_(l,k)(x) l_{l, k}(x) it it , k ( x ) It follows that this is of the form
(10)
l
i
,
k
(
x
)
=
g
i
(
x
)
(
x
−
x
i
)
k
h
i
,
k
(
x
)
,
(10)
l
i
,
k
(
x
)
=
g
i
(
x
)
x
−
x
i
k
h
i
,
k
(
x
)
,
{:(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) it and , k ( x ) = g and ( x ) ( x − x and ) k h and , k ( x ) , where
h
i
,
k
(
x
)
h
i
,
k
(
x
)
h_(i,k)(x) h_{i, k}(x) h and , k ( x ) is a polynomial of degree
r
i
−
k
−
1
r
i
−
k
−
1
r_(i)-k-1 r_{i}-k-1 R and − k − 1 . It remains to determine this last polynomial. If we expand it according to Taylor's formula in the neighborhood of
x
=
x
i
x
=
x
i
x=x_(i) x=x_{i} x = x and , it is obtained
(11)
h
i
,
k
(
x
)
=
∑
m
=
0
r
i
−
k
−
1
(
x
−
x
i
)
m
m
!
h
i
,
k
(
m
)
(
x
i
)
.
(11)
h
i
,
k
(
x
)
=
∑
m
=
0
r
i
−
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 and , k ( x ) = ∑ m = 0 R and − k − 1 ( x − x and ) m m ! h and , k ( m ) ( x and ) . From (10) we have
(
x
−
x
i
)
k
h
i
,
k
(
x
)
=
l
i
,
k
(
x
)
1
g
i
(
x
)
x
−
x
i
k
h
i
,
k
(
x
)
=
l
i
,
k
(
x
)
1
g
i
(
x
)
(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 and ) k h and , k ( x ) = it and , k ( x ) 1 g and ( x ) If we calculate, according to Leibniz's formula, the derivative of the order
q
q
q q q of both members of this equality, we obtain
∑
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
)
∑
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
)
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 and ) k ] ( q − j ) h and , k ( j ) ( x ) = ∑ j = 0 q ( q j ) it and , k ( q − j ) ( x ) ( 1 g and ( x ) ) ( j ) If it is taken
q
=
k
+
m
q
=
k
+
m
q=k+m q=k+m q = k + m , it is done
x
=
x
i
x
=
x
i
x=x_(i) x=x_{i} x = x and and taking into account (7) and (8), we obtain the relations
k
!
h
i
,
k
(
m
)
(
x
i
)
=
(
1
g
i
(
x
)
)
x
=
x
i
(
m
)
(
m
=
0
,
1
,
…
,
r
i
−
k
−
1
)
k
!
h
i
,
k
(
m
)
x
i
=
1
g
i
(
x
)
x
=
x
i
(
m
)
m
=
0
,
1
,
…
,
r
i
−
k
−
1
{:[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 and , k ( m ) ( x and ) = ( 1 g and ( x ) ) x = x and ( m ) ( m = 0 , 1 , … , R and − k − 1 ) In this way we arrive at the following expressions for the fundamental interpolation polynomials
(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
)
,
(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
)
,
{:(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) it and , k ( x ) = ∑ R = 0 R and − k − 1 [ ( x − x and ) R R ! ( 1 g and ( x ) ) x and ( R ) ] g and ( x ) , and the polynomial (6) becomes
(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
)
.
(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
.
{:(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 ) = ∑ and = 1 S ∑ k = 0 R and − 1 ∑ R = 0 R and − k − 2 ( x − x and ) k k ! [ ( x − x and ) R R ! ( 1 g and ( x ) ) x and ( R ) ] g and ( x ) f ( k ) ( x and ) . To this expression of the polynomial
H
n
(
x
)
H
n
(
x
)
H_(n)(x) H_{n}(x) H n ( x ) VL Gonciarov arrived at this, in a slightly different way [6].
4. If the derivative of the order is calculated
j
j
j j j , at the point
x
i
x
i
x_(i) x_{i} x and , his
1
g
i
(
x
)
1
g
i
(
x
)
(1)/(g_(i)(x)) \frac{1}{g_{i}(x)} 1 g and ( x ) , we obtain
(
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
,
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
,
{:[((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 and ( x ) ) x and ( 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 and − x 1 ) R 1 + α 1 … / … ( x and − x S ) R S + α S , dash indicating the absence of factors with all indices
i
i
i i and Here the sum extends to all systems of non-negative integers that verify the relation
α
1
+
α
2
+
…
+
α
i
−
1
+
α
i
+
1
+
…
+
α
s
=
j
α
1
+
α
2
+
…
+
α
i
−
1
+
α
i
+
1
+
…
+
α
s
=
j
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 and − 1 + α and + 1 + … + α S = j Taking this into account, the fundamental interpolation polynomial
l
i
,
k
(
x
)
l
i
,
k
(
x
)
l_(i,k)(x) l_{i, k}(x) it and , k ( x ) can be put in the form
(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
}
(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
{:[(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) it and , k ( x ) = g and ( x ) g and ( x and ) { ( x − x and ) k k ! ∑ j = 0 R and − k − 1 ( x − x and ) j × × ∑ A 1 + … + A and − 1 + α and + 1 + … + α S = j ( R 1 + α 1 − 1 α 1 ) … / … ( R S + α S − 1 α S ) ( x 1 − x and ) α 1 … / … ( x S − x and ) α S } In this way we arrived at the definitive expression of Hermite's interpolation polynomial (2)
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
)
}
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
{:[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) = ∑ and = 1 S g and ( x ) g and ( x and ) { ∑ k = 0 R and − 1 ( x − x and ) k k ! [ ∑ j = 0 R and − k − 1 A α 1 , … , α and − 1 , α and + 1 , … , α S ( j ) ( x − x and ) j ] f ( k ) ( x and ) } where
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
.
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
.
{:[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 and − 1 , α and + 1 , … , α S ( j ) = (16) = ∑ α 1 + … + α and − 1 + α and − 1 + … + α S = j ( γ 1 + α 1 − 1 α 1 ) … / … ( γ S + α S − 1 α S ) ( x 1 − x and ) α 1 … / … ( x S − x and ) α S .
If we consider Hermite's interpolation formula
(17)
f
(
x
)
=
H
n
(
x
)
+
R
n
+
1
(
x
)
,
(17)
f
(
x
)
=
H
n
(
x
)
+
R
n
+
1
(
x
)
,
{:(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 ) , REST
R
n
+
1
(
x
)
R
n
+
1
(
x
)
R_(n+1)(x) R_{n+1}(x) R n + 1 ( x ) has, as is known, the expression
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
]
,
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
]
,
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 ] , where
[
α
1
,
α
2
,
…
,
α
m
+
1
;
f
]
α
1
,
α
2
,
…
,
α
m
+
1
;
f
[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 ] is the difference divided by the order
m
m
m m m of the function
f
(
x
)
f
(
x
)
f(x) f(x) f ( x ) on the points
α
1
,
α
2
,
…
,
α
m
+
1
α
1
,
α
2
,
…
,
α
m
+
1
alpha_(1),alpha_(2),dots,alpha_(m+1) \alpha_{1}, \alpha_{2}, \ldots, \alpha_{m+1} α 1 , α 2 , … , α m + 1 .
Assuming that
f
(
x
)
f
(
x
)
f(x) f(x) f ( x ) has a derivative of the order
n
+
1
n
+
1
n+1 n+1 n + 1 in the smallest interval containing the values
x
1
,
…
,
x
s
,
x
x
1
,
…
,
x
s
,
x
x_(1),dots,x_(s),x x_{1}, \ldots, x_{s}, x x 1 , … , x S , x , the rest can be expressed, as is known, by the formula
(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
)
,
(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
)
,
{:(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 ) , where
ξ
ξ
xi \xi ξ belongs to the smallest interval containing the values
x
i
x
i
x_(i) x_{i} x and and
x
x
x x x .
6. Example. Find the interpolation polynomial of minimum degree relative to the function
f
(
x
)
f
(
x
)
f(x) f(x) f ( x ) and at the nodes
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 . Based on the previous formula we find
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. Particular cases.
If
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 , (15) reduces to the Lagrange interpolation polynomial
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 ) = ∑ and = 1 S g and ( x ) g and ( x and ) f ( x and ) .
2
∘
2
∘
2^(@) 2^{\circ} 2 ∘ If
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 , from (13) we obtain
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 ) = ∑ and = 1 S g and ( x ) g and ( x and ) [ 1 − ( x − x and ) g and ′ ( x and ) g and ( x and ) ] f ( x and ) + ∑ and = 1 S ( x − x and ) g and ( x ) g and ( x and ) f ′ ( x and ) , and from (14) it follows
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 ) = ∑ and = 1 S g and ( x ) g and ( x and ) ( x − x and ) [ 1 x − x and + 1 x 1 − x and + … + 1 x S − x and ] + + ∑ and = 1 S g and ( x ) g it ( x and ) ( x − x and ) f ′ ( x and ) . This formula was also found by A. Markoff [7].
3
∘
3
∘
3^(@) 3^{\circ} 3 ∘ In the case of
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 , we have
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 ) = ∑ and = 1 S [ it and , 0 ( x ) f ( x and ) + it and , 1 ( x ) f ′ ( x and ) + it and , 2 ( x ) f ′ ′ ( x and ) ] where
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] it and , 0 ( x ) = g and ( x ) g and ( x and ) [ 1 − ( x − x and ) g and ′ ( x and ) g and ( x and ) − ( x − x and ) 2 g and ′ ′ ( x and ) g and ( x and ) − 2 g and ′ 2 ( x and ) 2 g and 2 ( x and ) ]
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} it and , 1 ( x ) = ( x − x and ) g and ( x ) g and ( x and ) [ 1 − ( x − x and ) g and ′ ( x and ) g and ( x and ) ] it and , 2 ( x ) = ( x − x and ) 2 2 g and ( x ) g and ( x and ) .
Let's consider the particular case
s
=
2
s
=
2
s=2 s=2 S = 2 and let's note
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 The interpolation formula (15) reduces in this case to
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 ! [ ∑ and = 0 m − k − 1 ( n + and − 1 and ) ( x − A b − A ) and ] 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 ) . In the corresponding interpolation formula
(21)
f
(
x
)
=
H
m
+
n
−
1
(
x
)
+
R
m
+
n
(
x
)
(21)
f
(
x
)
=
H
m
+
n
−
1
(
x
)
+
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 ) the rest has the expression
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
)
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
)
{:[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 )
An important application of the interpolation formula (15) is that concerning the decomposition of a rational function into simple fractions, in the general case when the denominator of the rational function has multiple roots. Moreover, there is an equivalence between these two problems, as can be easily seen.
Suppose we are given the rational function
R
(
x
)
=
f
(
x
)
ω
(
x
)
R
(
x
)
=
f
(
x
)
ω
(
x
)
R(x)=(f(x))/(omega(x)) R(x)=\frac{f(x)}{\omega(x)} R ( x ) = f ( x ) ω ( x ) where
ω
(
x
)
=
∏
p
=
1
s
(
x
−
x
p
)
r
p
=
g
i
(
x
)
(
x
−
x
i
)
r
i
ω
(
x
)
=
∏
p
=
1
s
x
−
x
p
r
p
=
g
i
(
x
)
x
−
x
i
r
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 and ( x ) ( x − x and ) R and and
f
(
x
)
f
(
x
)
f(x) f(x) f ( x ) is a polynomial of degree
m
<
n
m
<
n
m < n m<n m < n (a case that is always of interest), where we noted as before
n
=
r
1
+
r
2
+
…
+
r
s
−
1
n
=
r
1
+
r
2
+
…
+
r
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 . It is known that
R
(
x
)
R
(
x
)
R(x) R(x) R ( x ) can be represented in
mod
mod
mod \bmod mode unique in form
(23)
R
(
x
)
=
∑
i
=
1
s
∑
p
=
0
r
i
−
1
A
i
p
(
x
−
x
i
)
r
i
−
p
,
(23)
R
(
x
)
=
∑
i
=
1
s
∑
p
=
0
r
i
−
1
A
i
p
x
−
x
i
r
i
−
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 ) = ∑ and = 1 S ∑ p = 0 R and − 1 A and p ( x − x and ) R and − p , where
A
i
p
A
i
p
A_(ip) A_{i p} A and p are constants.
If the interpolation polynomial is written relative to the polynomial
f
(
x
)
f
(
x
)
f(x) f(x) f ( x ) and at the nodes
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 − of orders of multiplicity equal respectively to
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 - the polynomial is obtained
H
n
(
x
)
H
n
(
x
)
H_(n)(x) H_{n}(x) H n ( x ) from (13) or (15). Considering that
f
(
x
)
f
(
x
)
f(x) f(x) f ( x ) is a polynomial of degree
m
<
n
m
<
n
m < n m<n m < n , we have
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 ) , so we will have
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
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
(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 ) = ∑ and = 1 S ∑ p = 0 R and − 1 A and p ( x − x and ) R and − p where, based on formula (13)
(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
)
.
(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
.
{:[(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 and p = ∑ k = 0 p f ( k ) ( x and ) k ! ( p − k ) ! ( 1 g and ( x ) ) x = x and ( p − k ) = 1 p ! ( f ( x ) g and ( x ) ) x = x and ( p ) ( p = 0 , 1 , … , R and − 1 ; and = 1 , 2 , … , S ) . Thus we have
again
l
2
,
k
(
x
)
,
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) it 2 , k ( x ) , it 3 , k ( x ) are obtained from here by circular permutations of the indices
1
,
2
,
3
1
,
2
,
3
1,2,3 1,2,3 1 , 2 , 3 .
assuming of course that in the interval (
a
,
b
a
,
b
a,b a, b A , b ) function
f
(
x
)
f
(
x
)
f(x) f(x) f ( x ) has a derivative of the order
m
+
n
m
+
n
m+n m+n m + n .
9. In the case of
s
=
3
s
=
3
s=3 s=3 S = 3 , (15) reduces to 1a
where
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 it 1 , k ( x ) f ( k ) ( x 1 ) + ∑ k = 0 R 2 − 1 it 2 , k ( x ) f ( k ) ( x 2 ) + ∑ k = 0 R 3 − 1 it 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} it 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 ] } , If formula (15) is used, the explicit formulas are obtained
(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
!
(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
!
{:(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 and p = 1 g and ( x and ) ∑ 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 and ) α 1 … / ⋯ ( x S − x and ) you S ] f ( k ) ( x and ) k !
In the particular case
(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 the decomposition into simple fractions is of the form
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 Based on the previous formulas, we have
(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 ∑ and = 0 p ( − 1 ) p − and ( n + p − and − 1 n − 1 ) ( A − b ) p − and f ( and ) ( A ) and ! 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 !
Let us now assume that we have to calculate the definite integral
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 and = ∫ A b f ( x ) d x If the interpolation formula (17) is used, the quadrature formula is obtained
(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 = ∑ and = 1 S ∑ k = 0 R and − 1 A it , k f ( k ) ( x and ) + ρ ( f ) where
again
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 it , k = ∫ A b it it , 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 The quadrature formula (28) generally has the degree of accuracy
n
n
n n n , that is, the remainder is zero if
f
(
x
)
f
(
x
)
f(x) f(x) f ( x ) is a polynomial of degree at most
n
n
n n n However, it may happen that through a convenient choice of nodes, the degree of accuracy may be higher. Examples.
1
∘
1
∘
1^(@) 1^{\circ} 1 ∘ . Choosing the nodes
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 and writing the corresponding interpolation formula, we have
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 ) where
H
6
(
x
)
H
6
(
x
)
H_(6)(x) H_{6}(x) H 6 ( x ) is the interpolation polynomial (19) and
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 ] Integrating from -1 to +1, we arrive at the quadrature formula of degree of accuracy 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 ) For the rest
ρ
(
f
)
ρ
(
f
)
rho(f) \rho(f) ρ ( f ) the expression has been established
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 ∘ If nodes are used
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 and
−
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 , write the corresponding Hermite interpolation formula, multiply by
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 and integrating from -1 to +1, we obtain the quadrature formula with accuracy degree 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 ( and V ) ( 0 ) + ( ′ ) + 15360 [ f ( − 7 8 ) + f ( | 7 8 ) ] } + π 530841600 f ( 10 ) ( ξ ) Observation. The quadrature formulas (with remainder) from (29) and (29́) were arrived at by doubling the node in the first case
x
=
0
x
=
0
x=0 x=0 x = 0 , and in the second case doubling the nodes
x
6
x
6
x_(6) x_{6} x 6 and
x
7
x
7
x_(7) x_{7} x 7 and taking the
x
=
0
x
=
0
x=0 x=0 x = 0 multiple of order 6.
13. Let's see what the quadrature formula (28) becomes in the particular case from no. 8. We will use the interpolation polynomial (20). Using the generalized integration by parts formula, after systematically performing the calculations, we obtain
∫
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 ) ! [ ∑ and = 0 m − k − 1 ( k + and and ) ( n + k + and + 1 n + and − 1 ) ] f ( k ) ( A ) + + m ∑ R = 0 n − 1 ( A − b ) k ( R + 2 ) ! [ ∑ R = 0 n − R − 1 ( R + j and ) ( m + R + j + 1 R + 2 ) ] f ( j ) ( b ) To obtain a simpler expression of the coefficients of this formula, we will use the identity
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 ∑ and = 0 m − k − 1 ( k + and and ) ( n + k + and + 1 k + 2 ) = ( m k + 1 ) ( m + n k + 1 ) which is proven without difficulty.
In this way we arrive at the quadrature formula of degree of accuracy
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 ) where
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 or
(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 making
m
=
n
m
=
n
m=n m=n m = n , this reduces to the 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 ) where
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 ) Formulas (20) and (32) are due to
2
2
^(2) { }^{2} 2 Hermite [1]. He deduced them with a special method, without using the interpolation formula (20), which he gave up establishing because it seemed to him too complicated. These formulas have recently been generalized, in a certain sense, by Prof. DV Ionescu [12].
14. Using the interpolation formula (15), one can also construct formulas for the numerical calculation of derivatives of different orders of a function
f
(
x
)
f
(
x
)
f(x) f(x) f ( x ) at a point
x
0
x
0
x_(0) x_{0} x 0 . Now we will deal with a concrete case that seems important to us. Namely, we will use the interpolation formula (20) to establish formulas for calculating derivatives
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 ) , where
p
p
p p p is a natural number smaller than
n
n
n n n . For this, we will seek to calculate
(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 Considering formula (20), the coefficient of
f
(
k
)
(
a
)
f
(
k
)
(
a
)
f^((k))(a) f^{(k)}(a) f ( k ) ( A ) from (33) it is found that
(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) ∑ and = 0 m − k − 1 ( − 1 ) and ( A − b ) n + and ⋅ 1 k ! ( n + and − 1 and ) d m + p d x m + p [ ( x − b ) n ( x − A ) k + and ] Since based on Leibniz's formula it is found that
[
(
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 expression (34) can be written
(
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 ∑ and = 0 m − k − 1 ( − 1 ) and ( n + and − 1 and ) ( n m + p − k − and ) If we use the identity
∑
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 ) which is easily proven, is ultimately found for the coefficient of
f
(
k
)
(
a
)
f
(
k
)
(
a
)
f^((k))(a) f^{(k)}(a) f ( k ) ( A ) from (33) the following expression
(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 )
Let's now find the coefficient of
f
(
r
)
(
b
)
f
(
r
)
(
b
)
f(r)(b) f(r)(b) f ( R ) ( b ) from (34).
This is
(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 ) Let's then deal with the evaluation of the amount
(37)
C
r
=
∑
j
=
0
n
−
r
−
1
(
m
−
1
+
j
j
)
(
r
+
j
p
)
(37)
C
r
=
∑
j
=
0
n
−
r
−
1
(
m
−
1
+
j
j
)
(
r
+
j
p
)
{:(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 ) Here some terms at the beginning are null, because in the development according to Leibniz's formula, which we used above, it must be assumed
r
+
j
≧
p
r
+
j
≧
p
r+j >= p r+j \geqq p R + j ≧ p , so that
C
r
C
r
C_(r) C_{r} C R is reduced to 1
(38)
C
r
=
∑
j
=
p
−
r
n
−
r
−
1
(
m
−
1
+
j
j
)
(
r
+
j
r
−
p
+
j
)
(38)
C
r
=
∑
j
=
p
−
r
n
−
r
−
1
(
m
−
1
+
j
j
)
(
r
+
j
r
−
p
+
j
)
{:(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 ) This can also be put in the form
(39)
C
r
=
∑
i
=
0
n
−
p
−
1
(
m
+
p
−
r
−
1
+
i
p
−
r
+
i
)
(
p
+
i
i
)
(39)
C
r
=
∑
i
=
0
n
−
p
−
1
(
m
+
p
−
r
−
1
+
i
p
−
r
+
i
)
(
p
+
i
i
)
{:(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 = ∑ and = 0 n − p − 1 ( m + p − R − 1 + and p − R + and ) ( p + and and ) or
(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
)
(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
)
{:(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 ) ∑ and = 0 n − p − 1 ( p + and and ) ( m + p − R + and and ) ( p − R + and and ) Applying the formula
(
a
b
)
=
∑
j
=
0
l
(
l
j
)
(
a
−
l
b
−
j
)
(
a
b
)
=
∑
j
=
0
l
(
l
j
)
(
a
−
l
b
−
j
)
((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 it ( it j ) ( A − it b − j ) get
(
p
+
i
i
)
=
∑
j
=
0
r
(
r
j
)
(
p
+
i
−
v
i
−
j
)
.
(
p
+
i
i
)
=
∑
j
=
0
r
(
r
j
)
(
p
+
i
−
v
i
−
j
)
.
((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 + and and ) = ∑ j = 0 R ( R j ) ( p + and − V and − j ) . With this
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
)
,
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
)
,
{:[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 = ∑ and = 0 n − p − 1 ∑ j = 0 R ( R j ) ( p + and − R and − j ) ( m + p − R − 1 + and p − R + and ) = = ∑ j = 0 R ( R j ) ∑ and = j n − p − 1 ( p + and − R and − j ) ( m + p − R − 1 + and p − R + and ) , based on the usual convention that
(
m
n
)
(
m
n
)
((m)/(n)) \binom{m}{n} ( m n ) be null if
m
m
m m m or
n
n
n n n would be negative.
Noting
β
=
i
−
j
,
m
=
p
+
j
−
r
,
n
=
m
+
p
+
j
−
r
−
1
β
=
i
−
j
,
m
=
p
+
j
−
r
,
n
=
m
+
p
+
j
−
r
−
1
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 = and − j , m = p + j − R , n = m + p + j − R − 1 HAVE
C
r
=
∑
j
=
0
r
(
r
j
)
n
−
p
−
j
−
1
∑
β
=
0
−
1
(
m
+
β
β
)
(
n
+
β
m
+
β
)
.
C
r
=
∑
j
=
0
r
(
r
j
)
n
−
p
−
j
−
1
∑
β
=
0
−
1
(
m
+
β
β
)
(
n
+
β
m
+
β
)
.
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 ) . Applying the identity
∑
p
=
0
k
(
m
+
β
β
)
(
n
+
β
m
+
β
)
=
(
n
+
k
+
1
k
)
(
n
m
)
∑
p
=
0
k
(
m
+
β
β
)
(
n
+
β
m
+
β
)
=
(
n
+
k
+
1
k
)
(
n
m
)
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 ) we finally get
(41)
C
r
=
∑
j
=
0
r
(
r
j
)
(
m
+
n
−
r
−
1
n
−
p
−
j
−
1
)
(
m
+
p
+
j
−
r
−
1
p
+
j
−
r
)
(41)
C
r
=
∑
j
=
0
r
(
r
j
)
(
m
+
n
−
r
−
1
n
−
p
−
j
−
1
)
(
m
+
p
+
j
−
r
−
1
p
+
j
−
r
)
{:(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 ) For example, for
k
=
0
,
1
,
2
k
=
0
,
1
,
2
k=0,1,2 k=0,1,2 k = 0 , 1 , 2 HAVE
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
)
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
)
{:[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, Taking into account (35), (36), (37), (41), it is seen that we were able to construct the following numerical derivation formula of degree of accuracy
m
+
n
−
1
m
+
n
−
1
m+n-1 m+n-1 m + n − 1
(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
)
(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
)
{:(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 ) where
(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
)
(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
)
{:(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 ) and
(44)
B
r
=
(
−
1
)
r
+
p
(
m
+
p
)
!
r
!
(
b
−
a
)
m
+
p
−
r
C
r
(44)
B
r
=
(
−
1
)
r
+
p
(
m
+
p
)
!
r
!
(
b
−
a
)
m
+
p
−
r
C
r
{:(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 and
(45)
C
r
=
∑
j
−
0
r
(
r
j
)
(
m
+
n
−
r
−
1
n
−
p
−
j
−
1
)
(
m
+
p
+
j
−
r
−
1
p
+
j
−
r
)
(45)
C
r
=
∑
j
−
0
r
(
r
j
)
(
m
+
n
−
r
−
1
n
−
p
−
j
−
1
)
(
m
+
p
+
j
−
r
−
1
p
+
j
−
r
)
{:(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 ) or with
C
r
C
r
C_(r) C_{r} C R given by (38), (39) or (40).
For the rest of this formula, the expression was obtained
(46)
ρ
(
f
)
=
(
m
+
p
)
!
(
m
+
n
)
!
(
n
p
)
(
a
−
b
)
n
−
p
f
(
m
+
n
)
(
ξ
)
,
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 .
Particular cases of the numerical derivation formula (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
)
2
[
f
′
(
a
)
+
f
′
(
b
)
]
+
a
−
b
2
f
(
4
)
(
ξ
)
.
f
′
′
′
(
a
)
=
−
12
(
b
−
a
)
3
[
f
(
a
)
−
f
(
b
)
]
+
6
(
b
−
a
)
2
f
′
(
a
)
+
f
′
(
b
)
+
a
−
b
2
f
(
4
)
(
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
)
=
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
)
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
)
{:[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
′
′
(
b
)
+
12
b
−
a
f
′
′
′
(
b
)
−
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 ( and 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 and ) ( 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 ( and 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} 
