On interpolation with connected polynomials

Ion Păvăloiu
(original), html, Numerical Analysis, paper

Abstract

We consider a function \(f\in C^{n-1}[a,b]\) and the nodes \(a=x_{0}<x_{1}<\ldots<x_{m}=b\). Given the values \begin{align}f\left( x_{i}\right) =&u_{i}, \quad i=1,…,m  \\ f^{\left( k\right) }\left( x_0\right) =&p_{0}^{k},\quad k=1,…,n-1,\end{align} we show that there exists on each interval \(\left[ x_{i-1},x_{i}\right] ,\ i=1,…,m,\) a unique polynomial \(P_{i}\) of degree \(m\) that satisfies \begin{align} P\left( x_{i}\right) =& u_{i},  \\ P_{i}^{\left( k\right) }\left( x_{i}\right) =&P_{i-1}^{\left( k\right) }\left(x_{i}\right) ,\quad k=1,…,n-1,\end{align} i.e. the resulted polynomials are successively joined. In this paper we show how these polynomials may be constructed. The functions \(f\) is therefore approximated by these polynomials.

Original Title (in French)

Sur l’intérpolation à l’aide des polynômes raccordées

Authors

Ion Păvăloiu
Tiberiu Popoviciu Institute of Numerical Analysis

Keywords

approximation by polynomials, joined polynomials, Hermite interpolation

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Cite this paper as:

I. Păvăloiu, Sur l’intérpolations à l’aide des polynômes raccordées, Mathematica, 6(27) (1964) no. 2, pp. 295-299 (in French).

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Mathematica

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Editura Academiei R.S. Române

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References

[1] Boor Carl de, Bicubic spline interpolation. Journal of Mathenatics and Physics. 41 (1962) 3.

[2]                 I959.

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On-interpolation-with-connected-polynomials

ON INTERPOLATION WITH CONNECTED POLYNOMIALS

byION PĂVĂLOIUin Cluj

  1. Either f ( x ) ∈ C n − 1 f ( x ) ∈ C n − 1 f(x)inC^(n-1)f(x) \in C^{n-1}f(x)∈Cn−1 a bounded function, defined in the interval [ a , b ] [ a , b ] [a,b][a, b][has,b] and either a = x 0 < x 1 < … < x m = b a = x 0 < x 1 < … < x m = b a=x_(0) < x_(1) < dots < x_(m)=ba=x_{0}<x_{1}<\ldots<x_{m}=bhas=x0<x1<…<xM=b a division of 1 interval [ a , b ] [ a , b ] [a,b][a, b][has,b].
Suppose that we know the values of 1a function f ( x ) f ( x ) f(x)f(x)f(x) In the m + 1 m + 1 m+1m+1M+1 Interval division nodes [ a , b ] [ a , b ] [a,b][a, b][has,b] as well as the values of the successive derivatives of 1a function f ( x ) f ( x ) f(x)f(x)f(x) in the initial node x 0 x 0 x_(0)x_{0}x0.
We ask:
(1) f ( x i ) = u i , i = 1 , 0 , … , n f ( k ) ( x 0 ) = p 0 k , k = 1 , 2 , … , n − 1 (1) f x i = u i , i = 1 , 0 , … , n f ( k ) x 0 = p 0 k , k = 1 , 2 , … , n − 1 {:[(1)f(x_(i))=u_(i)","quad i=1","0","dots","n],[f^((k))(x_(0))=p_(0)^(k)","quad k=1","2","dots","n-1]:}\begin{align*} f\left(x_{i}\right) & =u_{i}, \quad i=1,0, \ldots, n \tag{1}\\ f^{(k)}\left(x_{0}\right) & =p_{0}^{k}, \quad k=1,2, \ldots, n-1 \end{align*}(1)f(xi)=ui,i=1,0,…,nf(k)(x0)=p0k,k=1,2,…,n−1
Either P n P n P_(n)\mathcal{P}_{n}Pn the class of all functions defined in the interval [ x 0 , x m ] x 0 , x m [x_(0),x_(m)]\left[x_{0}, x_{m}\right][x0,xM] and who in each of the intervals [ x i − 1 , x i ] i = 1 , 2 , … , m x i − 1 , x i i = 1 , 2 , … , m [x_(i-1),x_(i)]i=1,2,dots,m\left[x_{i-1}, x_{i}\right] i=1,2, \ldots, m[xi−1,xi]i=1,2,…,M are identically equal to a polynomial P n i ( x ) P n i ( x ) P_(n_(i))(x)P_{n_{i}}(x)Pni(x) of degree n n nnn.
Definite n. We call a connected polynomial of the order n − 1 n − 1 n-1n-1n−1, each function φ ( x ) φ ( x ) varphi(x)\varphi(x)φ(x) who belongs to the p n p n p_(n)p_{n}pn and for which we have
P n i ( k ) ( x i ) = P n i + 1 ( k ) ( x i ) , (2) i = 1 , 2 , … , m − 1 k = 0 , 1 , … , n − 1 . P n i ( k ) x i = P n i + 1 ( k ) x i , (2) i = 1 , 2 , … , m − 1 k = 0 , 1 , … , n − 1 . P_(n_(i))^((k))(x_(i))=P_(n_(i+1))^((k))(x_(i)),quad{:[(2)i=1","2","dots","m-1],[k=0","1","dots","n-1.]:}P_{n_{i}}^{(k)}\left(x_{i}\right)=P_{n_{i+1}}^{(k)}\left(x_{i}\right), \quad \begin{align*} & i=1,2, \ldots, m-1 \tag{2}\\ & k=0,1, \ldots, n-1 . \end{align*}Pni(k)(xi)=Pni+1(k)(xi),(2)i=1,2,…,M−1k=0,1,…,n−1.
Let be the values given (1) - we ask to determine the polynomial P ( x ) P ( x ) P(x)P(x)P(x) connected of 1 order n − 1 n − 1 n-1n-1n−1 who meets the conditions:
P ( x i ) = u i , i = 0 , 1 , … , m P x i = u i , i = 0 , 1 , … , m P(x_(i))=u_(i),quad i=0,1,dots,mP\left(x_{i}\right)=u_{i}, \quad i=0,1, \ldots, mP(xi)=ui,i=0,1,…,M
(3) P ( k ) ( x 0 ) = p 0 k , k = 1 , 2 , … , n − 1 (3) P ( k ) x 0 = p 0 k , k = 1 , 2 , … , n − 1 {:(3)P^((k))(x_(0))=p_(0)^(k)","quad k=1","2","dots","n-1:}\begin{equation*} P^{(k)}\left(x_{0}\right)=p_{0}^{k}, \quad k=1,2, \ldots, n-1 \tag{3} \end{equation*}(3)P(k)(x0)=p0k,k=1,2,…,n−1
Lemma 1. If we give the values
f ( k ) ( a ) = β k , k = 0 , 1 , … , n − 1 (4) f ( c ) = γ 0 , c ≠ a f ( k ) ( a ) = β k , k = 0 , 1 , … , n − 1 (4) f ( c ) = γ 0 , c ≠ a {:[f^((k))(a)=beta_(k)","quad k=0","1","dots","n-1],[(4)f(c)=gamma_(0)","quad c!=a]:}\begin{align*} & f^{(k)}(a)=\beta_{k}, \quad k=0,1, \ldots, n-1 \\ & f(c)=\gamma_{0}, \quad c \neq a \tag{4} \end{align*}f(k)(has)=βk,k=0,1,…,n−1(4)f(c)=γ0,c≠has
where c is a point in the interval [ a , b ] [ a , b ] [a,b][a, b][has,b] then there exists a single polynomial of degree, Q n ( x ) Q n ( x ) Q_(n)(x)Q_{n}(x)Qn(x) of the form Q n ( x ) = ∑ j = 0 n α j ( x − a ) j Q n ( x ) = ∑ j = 0 n   α j ( x − a ) j Q_(n)(x)=sum_(j=0)^(n)alpha_(j)(x-a)^(j)Q_{n}(x)=\sum_{j=0}^{n} \alpha_{j}(x-a)^{j}Qn(x)=∑j=0nαj(x−has)j which satisfies conditions (4) in points a and c.
Proof: Writing that the polynomial Q n ( x ) Q n ( x ) Q_(n)(x)Q_{n}(x)Qn(x) satisfies the conditions (4) we obtain for the determination of α j , j = 0 , n ― α j , j = 0 , n ¯ alpha_(j),quad j= bar(0,n)\alpha_{j}, \quad j=\overline{0, n}αj,j=0,n― the following system of algebraic linear equations:
(5) l ! α l = β l , l = 0 , 1 , … , n − 1 α 0 + α 1 ( c − a ) + … + α n ( c − a ) n = γ 0 (5) l ! α l = β l , l = 0 , 1 , … , n − 1 α 0 + α 1 ( c − a ) + … + α n ( c − a ) n = γ 0 {:[(5)l!alpha_(l)=beta_(l)","quad l=0","1","dots","n-1],[alpha_(0)+alpha_(1)(c-a)+dots+alpha_(n)(c-a)^(n)=gamma_(0)]:}\begin{align*} & l!\alpha_{l}=\beta_{l}, \quad l=0,1, \ldots, n-1 \tag{5}\\ & \alpha_{0}+\alpha_{1}(c-a)+\ldots+\alpha_{n}(c-a)^{n}=\gamma_{0} \end{align*}(5)l!αl=βl,l=0,1,…,n−1α0+α1(c−has)+…+αn(c−has)n=γ0
We determinant of the system has the form:
(6) Δ = | 1 0 . . 0 0 0 1 ! . . . 0 0 . . . . . . . . . . . . . . . 0 ˙ . . . . . 0 . . . ( n − 1 ) ! 0 1 c − a . . ( c − a ) n − 1 ( c − a ) n | = ∏ i = 0 n − 1 i ! ( c − a ) n (6) Δ = 1 0 . . 0 0 0 1 ! . . . 0 0 . . . . . . . . . . . . . . . 0 ˙ . . . . . 0 . . . ( n − 1 ) ! 0 1 c − a . . ( c − a ) n − 1 ( c − a ) n = ∏ i = 0 n − 1   i ! ( c − a ) n {:(6)Delta=|[1,0,.,.,0,0],[0,1!,.,.,.,0,0],[.,.,.,.,.,.,.],[.,.,.,.,.,.,.],[.,0^(˙),.,.,.,.,.],[0,.,.,.,(n-1)!,0],[1,c-a,.,.,(c-a)^(n-1),(c-a)^(n)]|=prod_(i=0)^(n-1)i!(c-a)^(n):}\Delta=\left|\begin{array}{ccccccc} 1 & 0 & . & . & 0 & 0 \tag{6}\\ 0 & 1! & . & . & . & 0 & 0 \\ . & . & . & . & . & . & . \\ . & . & . & . & . & . & . \\ . & \dot{0} & . & . & . & . & . \\ 0 & . & . & . & (n-1)! & 0 \\ 1 & c-a & . & . & (c-a)^{n-1} & (c-a)^{n} \end{array}\right|=\prod_{i=0}^{n-1} i!(c-a)^{n}(6)Δ=|10..0001!...00...............0˙.....0...(n−1)!01c−has..(c−has)n−1(c−has)n|=∏i=0n−1i!(c−has)n
Given that we have assumed that c ≠ a c ≠ a c!=ac \neq ac≠has, it follows that Δ ≠ 0 Δ ≠ 0 Delta!=0\Delta \neq 0Δ≠0 and consequently, the system (5) has a single solution.
Solving system (5) we obtain for α 0 , … , α n α 0 , … , α n alpha_(0),dots,alpha_(n)\alpha_{0}, \ldots, \alpha_{n}α0,…,αn
(7)
α l = β l l ! , l = 0 , 1 , … , n − 1 α l = β l l ! , l = 0 , 1 , … , n − 1 alpha_(l)=(beta_(l))/(l!),quad l=0,1,dots,n-1\alpha_{l}=\frac{\beta_{l}}{l!}, \quad l=0,1, \ldots, n-1αl=βll!,l=0,1,…,n−1
α n = γ 0 − β 0 ( c − a ) n − ∑ k = 1 n − 1 β k k ! ( c − a ) n − k α n = γ 0 − β 0 ( c − a ) n − ∑ k = 1 n − 1   β k k ! ( c − a ) n − k alpha_(n)=(gamma_(0)-beta_(0))/((c-a)^(n))-sum_(k=1)^(n-1)(beta_(k))/(k!(c-a)^(n-k))\alpha_{n}=\frac{\gamma_{0}-\beta_{0}}{(c-a)^{n}}-\sum_{k=1}^{n-1} \frac{\beta_{k}}{k!(c-a)^{n-k}}αn=γ0−β0(c−has)n−∑k=1n−1βkk!(c−has)n−k
Lemma 2. Let ( x 0 , x 1 , x 2 x 0 , x 1 , x 2 x_(0),x_(1),x_(2)x_{0}, x_{1}, x_{2}x0,x1,x2 ) a given system of three consecutive knots, such that we have,
Δ x 0 = x 1 − x 0 ≠ 0 , Δ x 1 = x 2 − x 1 ≠ 0 . Δ x 0 = x 1 − x 0 ≠ 0 , Δ x 1 = x 2 − x 1 ≠ 0 . Deltax_(0)=x_(1)-x_(0)!=0,Deltax_(1)=x_(2)-x_(1)!=0.\Delta x_{0}=x_{1}-x_{0} \neq 0, \Delta x_{1}=x_{2}-x_{1} \neq 0 .Δx0=x1−x0≠0,Δx1=x2−x1≠0.
If P ( x ) P ( x ) P(x)P(x)P(x) and R ( x ) R ( x ) R(x)R(x)R(x) are two degree polynomials n n nnn who belong to the class P n P n P_(n)P_{n}Pn and who on the knots x 0 , x 1 , x 2 x 0 , x 1 , x 2 x_(0),x_(1),x_(2)x_{0}, x_{1}, x_{2}x0,x1,x2 meet the conditions
(8) R ( k ) ( x 0 ) = p 0 k , k = 1 , n − 1 ― , R ( x 0 ) = u 0 , R ( x 1 ) = u 1 P ( k ) ( x 1 ) = p 1 k , k = 1 , n − 1 ― , P ( x 1 ) = u 1 , P ( x 2 ) = u 2 (8) R ( k ) x 0 = p 0 k , k = 1 , n − 1 ¯ , R x 0 = u 0 , R x 1 = u 1 P ( k ) x 1 = p 1 k , k = 1 , n − 1 ¯ , P x 1 = u 1 , P x 2 = u 2 {:[(8)R^((k))(x_(0))=p_(0)^(k)","quad k= bar(1,n-1)","quad R(x_(0))=u_(0)","quad R(x_(1))=u_(1)],[P^((k))(x_(1))=p_(1)^(k)","quad k= bar(1,n-1)","quad P(x_(1))=u_(1)","quad P(x_(2))=u_(2)]:}\begin{gather*} R^{(k)}\left(x_{0}\right)=p_{0}^{k}, \quad k=\overline{1, n-1}, \quad R\left(x_{0}\right)=u_{0}, \quad R\left(x_{1}\right)=u_{1} \tag{8}\\ P^{(k)}\left(x_{1}\right)=p_{1}^{k}, \quad k=\overline{1, n-1}, \quad P\left(x_{1}\right)=u_{1}, \quad P\left(x_{2}\right)=u_{2} \end{gather*}(8)R(k)(x0)=p0k,k=1,n−1―,R(x0)=u0,R(x1)=u1P(k)(x1)=p1k,k=1,n−1―,P(x1)=u1,P(x2)=u2
then the necessary and sufficient conditions for the point to be x 1 x 1 x_(1)x_{1}x1 we have
(9) P ( k ) ( x 1 ) = R ( k ) ( x 1 ) , k = 1 , n − 1 (9) P ( k ) x 1 = R ( k ) x 1 , k = 1 , n − 1 {:(9)P^((k))(x_(1))=R^((k))(x_(1))","quad k=1","n-1:}\begin{equation*} P^{(k)}\left(x_{1}\right)=R^{(k)}\left(x_{1}\right), \quad k=1, n-1 \tag{9} \end{equation*}(9)P(k)(x1)=R(k)(x1),k=1,n−1
is that between p 1 k , p 0 k , k = 1 , 2 , … , n − 1 , u 0 , u 1 p 1 k , p 0 k , k = 1 , 2 , … , n − 1 , u 0 , u 1 p_(1)^(k),p_(0)^(k),k=1,2,dots,n-1,u_(0),u_(1)p_{1}^{k}, p_{0}^{k}, k=1,2, \ldots, n-1, u_{0}, u_{1}p1k,p0k,k=1,2,…,n−1,u0,u1, we have the relations:
where
(10) p 1 k = ∑ i = 0 n − k − 1 p 0 k + i Δ x 0 i i ! + A ∏ i = n − k + 1 n i Δ x 0 n − k (11) A = u 1 − u 0 Δ x n 0 − ∑ i = 1 n − 1 p 0 i i ! Δ x 0 n − i (10) p 1 k = ∑ i = 0 n − k − 1   p 0 k + i Δ x 0 i i ! + A ∏ i = n − k + 1 n   i Δ x 0 n − k (11) A = u 1 − u 0 Δ x n 0 − ∑ i = 1 n − 1   p 0 i i ! Δ x 0 n − i {:[(10)p_(1)^(k)=sum_(i=0)^(n-k-1)p_(0)^(k+i)(Deltax_(0)^(i))/(i!)+Aprod_(i=n-k+1)^(n)i Deltax_(0)^(n-k)],[(11)A=(u_(1)-u_(0))/(Deltax_(n)^(0))-sum_(i=1)^(n-1)(p_(0)^(i))/(i!Deltax_(0)^(n-i))]:}\begin{gather*} p_{1}^{k}=\sum_{i=0}^{n-k-1} p_{0}^{k+i} \frac{\Delta x_{0}^{i}}{i!}+A \prod_{i=n-k+1}^{n} i \Delta x_{0}^{n-k} \tag{10}\\ A=\frac{u_{1}-u_{0}}{\Delta x_{n}^{0}}-\sum_{i=1}^{n-1} \frac{p_{0}^{i}}{i!\Delta x_{0}^{n-i}} \tag{11} \end{gather*}(10)p1k=∑i=0n−k−1p0k+iΔx0ii!+Has∏i=n−k+1niΔx0n−k(11)Has=u1−u0Δxn0−∑i=1n−1p0ii!Δx0n−i

Demonstration.

Necessity: Polynomials P ( x ) P ( x ) P(x)P(x)P(x) and R ( x ) R ( x ) R(x)R(x)R(x) unambiguously determined by the conditions (8) have the form:
(12) R ( x ) = u 0 + ∑ i = 1 n − 1 p 0 i ( x − x 0 ) i i ! + [ u 1 − u 0 Δ x 0 n − ∑ i = 1 n − 1 p 0 i i ! Δ x 0 n − i ] ( x − x 0 ) n (12) R ( x ) = u 0 + ∑ i = 1 n − 1   p 0 i x − x 0 i i ! + u 1 − u 0 Δ x 0 n − ∑ i = 1 n − 1   p 0 i i ! Δ x 0 n − i x − x 0 n {:(12)R(x)=u_(0)+sum_(i=1)^(n-1)(p_(0)^(i)(x-x_(0))^(i))/(i!)+[(u_(1)-u_(0))/(Deltax_(0)^(n))-sum_(i=1)^(n-1)(p_(0)^(i))/(i!Deltax_(0)^(n-i))](x-x_(0))^(n):}\begin{equation*} R(x)=u_{0}+\sum_{i=1}^{n-1} \frac{p_{0}^{i}\left(x-x_{0}\right)^{i}}{i!}+\left[\frac{u_{1}-u_{0}}{\Delta x_{0}^{n}}-\sum_{i=1}^{n-1} \frac{p_{0}^{i}}{i!\Delta x_{0}^{n-i}}\right]\left(x-x_{0}\right)^{n} \tag{12} \end{equation*}(12)R(x)=u0+∑i=1n−1p0i(x−x0)ii!+[u1−u0Δx0n−∑i=1n−1p0ii!Δx0n−i](x−x0)n
P ( x ) = u 1 + ∑ i = 1 n − 1 p 1 i ( x − x 1 ) i i ! + [ u 2 − u 1 Δ x 1 n − ∑ i = 1 n − 1 p 1 i i ! Δ x 1 n − i ] ( x − x 1 ) n P ( x ) = u 1 + ∑ i = 1 n − 1   p 1 i x − x 1 i i ! + u 2 − u 1 Δ x 1 n − ∑ i = 1 n − 1   p 1 i i ! Δ x 1 n − i x − x 1 n P(x)=u_(1)+sum_(i=1)^(n-1)(p_(1)^(i)(x-x_(1))^(i))/(i!)+[(u_(2)-u_(1))/(Deltax_(1)^(n))-sum_(i=1)^(n-1)(p_(1)^(i))/(i!Deltax_(1)^(n-i))](x-x_(1))^(n)P(x)=u_{1}+\sum_{i=1}^{n-1} \frac{p_{1}^{i}\left(x-x_{1}\right)^{i}}{i!}+\left[\frac{u_{2}-u_{1}}{\Delta x_{1}^{n}}-\sum_{i=1}^{n-1} \frac{p_{1}^{i}}{i!\Delta x_{1}^{n-i}}\right]\left(x-x_{1}\right)^{n}P(x)=u1+∑i=1n−1p1i(x−x1)ii!+[u2−u1Δx1n−∑i=1n−1p1ii!Δx1n−i](x−x1)n
If we now write that these polynomials satisfy conditions (9), we successively obtain relations (10).
Sufficiency. If between p 0 k , p 1 k , k = 1 , 2 , … , n − 1 , u 0 , u 1 p 0 k , p 1 k , k = 1 , 2 , … , n − 1 , u 0 , u 1 p_(0)^(k),p_(1)^(k),k=1,2,dots,n-1,u_(0),u_(1)p_{0}^{k}, p_{1}^{k}, k=1,2, \ldots, n-1, u_{0}, u_{1}p0k,p1k,k=1,2,…,n−1,u0,u1, there are relations (10), so we verify directly by calculation that the polynomials (12) satisfy the relations (9).
Consequence. If the values (1) are given, then there is a unique system of values
p i j , i = 1 ― , m − 1 ― , j = 1 , n − 1 ― p i j , i = 1 ¯ , m − 1 ¯ , j = 1 , n − 1 ¯ p_(i)^(j),i= bar(1),m- bar(1),j= bar(1,n-1)p_{i}^{j}, i=\overline{1}, m-\overline{1}, j=\overline{1, n-1}pij,i=1―,M−1―,j=1,n−1―
which with the values (1) unambiguously determine a connected polynomial of the order n − 1 n − 1 n-1n-1n−1, which satisfies the conditions (3).
Values p i j , i = 1 , m − 1 , j = 1 , n − 1 ― p i j , i = 1 , m − 1 , j = 1 , n − 1 ¯ p_(i)^(j),i=1,m-1,j= bar(1,n-1)p_{i}^{j}, i=1, m-1, j=\overline{1, n-1}pij,i=1,M−1,j=1,n−1― are given by the following recurrence formulas:
(13) p i + 1 j = ∑ k = 0 n − j − 1 p i j + k Δ x i k k ! + A i ∏ k = n − j + 1 n k ⋅ Δ x i n − j j = 1 , 2 , … n − 1 , i = 0 , 1 , … , m − 2 (13) p i + 1 j = ∑ k = 0 n − j − 1   p i j + k Δ x i k k ! + A i ∏ k = n − j + 1 n   k ⋅ Δ x i n − j j = 1 , 2 , … n − 1 , i = 0 , 1 , … , m − 2 {:[(13)p_(i+1)^(j)=sum_(k=0)^(n-j-1)p_(i)^(j+k)(Deltax_(i)^(k))/(k!)+A_(i)prod_(k=n-j+1)^(n)k*Deltax_(i)^(n-j)],[j=1","2","dots n-1","i=0","1","dots","m-2]:}\begin{align*} p_{i+1}^{j} & =\sum_{k=0}^{n-j-1} p_{i}^{j+k} \frac{\Delta x_{i}^{k}}{k!}+A_{i} \prod_{k=n-j+1}^{n} k \cdot \Delta x_{i}^{n-j} \tag{13}\\ j & =1,2, \ldots n-1, i=0,1, \ldots, m-2 \end{align*}(13)pi+1j=∑k=0n−j−1pij+kΔxikk!+Hasi∏k=n−j+1nk⋅Δxin−jj=1,2,…n−1,i=0,1,…,M−2
where
A i = u i + 1 − u i Δ x i n − ∑ k = 1 n − 1 p i k k ! Δ x n − k , i = 0 , m − 2 ― A i = u i + 1 − u i Δ x i n − ∑ k = 1 n − 1   p i k k ! Δ x n − k , i = 0 , m − 2 ¯ A_(i)=(u_(i+1)-u_(i))/(Deltax_(i)^(n))-sum_(k=1)^(n-1)(p_(i)^(k))/(k!Deltax^(n-k)),i= bar(0,m-2)A_{i}=\frac{u_{i+1}-u_{i}}{\Delta x_{i}^{n}}-\sum_{k=1}^{n-1} \frac{p_{i}^{k}}{k!\Delta x^{n-k}}, i=\overline{0, m-2}Hasi=ui+1−uiΔxin−∑k=1n−1pikk!Δxn−k,i=0,M−2―
and
Δ x i = x i + 1 − x i , i = 0 , m − 2 ― Δ x i = x i + 1 − x i , i = 0 , m − 2 ¯ Deltax_(i)=x_(i+1)-x_(i),i= bar(0,m-2)\Delta x_{i}=x_{i+1}-x_{i}, i=\overline{0, m-2}Δxi=xi+1−xi,i=0,M−2―
From lemnes 1 and 2 the following theorem follows:
THEORÈA I i I i I_(i)\mathrm{I}_{\mathrm{i}}Ii. If the values (1) are given, then the function f ( x ) f ( x ) f(x)f(x)f(x) corresponds to a single connected polynomial of order n − 1 n − 1 n-1n-1n−1 which takes these values from the nodes x i , i = 0 , m x i , i = 0 , m x_(i),i=0,mx_{i}, i=0, mxi,i=0,M.
Let us designate by S n ( U , P ) S n ( U , P ) S_(n)(U,P)S_{n}(U, P)Sn(U,P) the linear space of all connected polynomials of order n − 1 n − 1 n-1n-1n−1 of the interval [ x 0 , x m ] x 0 , x m [x_(0),x_(m)]\left[x_{0}, x_{m}\right][x0,xM]. We have the following consequence:
Consequence. Space S n ( U , P ) S n ( U , P ) S_(n)(U,P)S_{n}(U, P)Sn(U,P) is a linear space of dimension m + n m + n m+nm+nM+n.
Demonstration. Theorem 1 corresponds to each vector
( u 0 , … , u m , p 0 1 , … , p 0 n − 1 ) u 0 , … , u m , p 0 1 , … , p 0 n − 1 (u_(0),dots,u_(m),p_(0)^(1),dots,p_(0)^(n-1))\left(u_{0}, \ldots, u_{m}, p_{0}^{1}, \ldots, p_{0}^{n-1}\right)(u0,…,uM,p01,…,p0n−1)
a connected polynomial determined without ambiquity u ( x ) ∈ S n ( U , P ) u ( x ) ∈ S n ( U , P ) u(x)inS_(n)(U,P)u(x) \in S_{n}(U, P)u(x)∈Sn(U,P).
Conversely, the equalities (1) cause that each polynomial U ( x ) ∈ S n ( U , P ) U ( x ) ∈ S n ( U , P ) U(x)inS_(n)(U,P)U(x) \in S_{n}(U, P)U(x)∈Sn(U,P) corresponds to a single vector
( U 0 , … , U m , p 0 1 , … , p 0 n − 1 ) U 0 , … , U m , p 0 1 , … , p 0 n − 1 (U_(0),dots,U_(m),p_(0)^(1),dots,p_(0)^(n-1))\left(U_{0}, \ldots, U_{m}, p_{0}^{1}, \ldots, p_{0}^{n-1}\right)(U0,…,UM,p01,…,p0n−1)
Lemma 2 and its consequence provide a practical procedure for calculating the approximate value of the function f ( x ) f ( x ) f(x)f(x)f(x) at each point,
x ¯ ∈ [ a , b ] , x ¯ ≠ x i , i = 0 , m ― x ¯ ∈ [ a , b ] , x ¯ ≠ x i , i = 0 , m ¯ bar(x)in[a,b],quad bar(x)!=x_(i),quad i= bar(0,m)\bar{x} \in[a, b], \quad \bar{x} \neq x_{i}, \quad i=\overline{0, m}x¯∈[has,b],x¯≠xi,i=0,M―
if we give the values (1).
If u ( x ) u ( x ) u(x)u(x)u(x) is the connected polynomial of order n − 1 n − 1 n-1n-1n−1 which satisfies conditions (3) then we mean by its approximate value of the function f ( x ) f ( x ) f(x)f(x)f(x) in the point x ¯ x ¯ bar(x)\bar{x}x¯ The value of the polynomial u ( x ) u ( x ) u(x)u(x)u(x) in the point x ¯ x ¯ bar(x)\bar{x}x¯ and write f ( x ) ≃ u ( x ) f ( x ) ≃ u ( x ) f(x)≃u(x)f(x) \simeq u(x)f(x)≃u(x).
For the determination of the polynomial U ( x ) U ( x ) U(x)U(x)U(x) we proceed as
follows: Either x k < x < x k + 1 x k < x < x k + 1 x_(k) < x < x_(k+1)x_{k}<x<x_{k+1}xk<x<xk+1. Now using the recurrence formulas (13) we successively calculate the values:
p 1 j , p 2 j , … , p k − 1 j , p k j , j = 1 , n − 1 ― p 1 j , p 2 j , … , p k − 1 j , p k j , j = 1 , n − 1 ¯ p_(1)^(j),p_(2)^(j),dots,p_(k-1)^(j),p_(k)^(j),j= bar(1,n-1)p_{1}^{j}, p_{2}^{j}, \ldots, p_{k-1}^{j}, p_{k}^{j}, j=\overline{1, n-1}p1j,p2j,…,pk−1j,pkj,j=1,n−1―
The polynomial U ( x ) U ( x ) U(x)U(x)U(x) for x k < x < x k + 1 x k < x < x k + 1 x_(k) < x < x_(k+1)x_{k}<x<x_{k+1}xk<x<xk+1 then takes the following form:
(14) U ( x ) = u k + ∑ i = 1 n − 1 p k i i ! ( x − x k ) i + A k ( x − x k ) n (14) U ( x ) = u k + ∑ i = 1 n − 1   p k i i ! x − x k i + A k x − x k n {:(14)U(x)=u_(k)+sum_(i=1)^(n-1)(p_(k)^(i))/(i!)(x-x_(k))^(i)+A_(k)(x-x_(k))^(n):}\begin{equation*} U(x)=u_{k}+\sum_{i=1}^{n-1} \frac{p_{k}^{i}}{i!}\left(x-x_{k}\right)^{i}+A_{k}\left(x-x_{k}\right)^{n} \tag{14} \end{equation*}(14)U(x)=uk+∑i=1n−1pkii!(x−xk)i+Hask(x−xk)n
where
A k = u k + 1 − u k Δ x k n − ∑ i = 1 n − 1 p k i i ! Δ x k n − i A k = u k + 1 − u k Δ x k n − ∑ i = 1 n − 1   p k i i ! Δ x k n − i A_(k)=(u_(k+1)-u_(k))/(Deltax_(k)^(n))-sum_(i=1)^(n-1)(p_(k)^(i))/(i!Deltax_(k)^(n-i))A_{k}=\frac{u_{k+1}-u_{k}}{\Delta x_{k}^{n}}-\sum_{i=1}^{n-1} \frac{p_{k}^{i}}{i!\Delta x_{k}^{n-i}}Hask=uk+1−ukΔxkn−∑i=1n−1pkii!Δxkn−i
and
Δ x k = x k + 1 − x k Δ x k = x k + 1 − x k Deltax_(k)=x_(k+1)-x_(k)\Delta x_{k}=x_{k+1}-x_{k}Δxk=xk+1−xk
Calculation of the coefficients of the polynomial U ( x ) U ( x ) U(x)U(x)U(x) for
x k < x < x k + 1 k = 1 , m ― x k < x < x k + 1 k = 1 , m ¯ x_(k) < x < x_(k+1)k= bar(1,m)x_{k}<x<x_{k+1} k=\overline{1, m}xk<x<xk+1k=1,M―
As we can see, is easily algorithmizable, so the results can be obtained simply with the help of a numerical electronic computing machine.

BIBLIOGRAPHY

[1] Boor Car1 (de), Bicubic spline interpolation. Journal of Mathematics and Physics. 41, 3 (1962).
[2] Березин И. С., Жидков Н. П, Методы вышислении. Государственное издательство Физико-математематическои Литературы, Москва, 1959. Received on the 19th. XI. 1963
1962

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