Despre precizia calculului numeric în interpolare prin polinoame

Tiberiu Popoviciu
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T. Popoviciu
Institutul de Calcul

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T. Popoviciu, Despre precizia calculului numeric în interpolare prin polinoame, Buletinul științific, Secția de științe matematice și fizice (Academia R.P.R.), 7 (1955) no. 4, pp. 953-961.

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Buletinul Stiintific

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Academia Republicii S.R.

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1955 c -Popoviciu- Bul. St. Sect. St. Mat. Fiz. - Despre precizia calculului numeric in interpolarea
SECTIA DE STIINŢE MATEMATICE ŞI FIZICE
'Tomul VJI, ni. 4, 1955

DESPRE PRECIZIA CALCULULUI NUMERIC IN INTERPOLAREA PRIN POLINOAME

DE

TIBERIU POPOVICIUMEMBRU CORESPONDENT AL ACADEMEI R.P.R.

Comunicare prezentală în şedinţa din 30 iunie 1955
  1. Pentru a calcula o valoare aproximativă a funcției f ( x ) f ( x ) f(x)f(x)f(x), reală și de variabila reală x x xxx, pe punctul x 0 x 0 x_(0)x_{0}x0, cînd se cunose valorile f ( x i ) , i = 1 , 2 , … , n + 1 f x i , i = 1 , 2 , … , n + 1 f(x_(i)),i=1,2,dots,n+1f\left(x_{i}\right), i=1,2, \ldots, n+1f(xi),i=1,2,…,n+1, ale acestei funcți pe n + 1 n + 1 n+1n+1n+1 puncte (distincte) date x i , i = 1 , 2 , … , n + 1 x i , i = 1 , 2 , … , n + 1 x_(i),i=1,2,dots,n+1x_{i}, i=1,2, \ldots, n+1xi,i=1,2,…,n+1, se poate folosi formula de interpolare
(1) f ( x 0 ) = L ( x 1 , x 2 , … , x n + 1 ; f ∣ x 0 ) , (1) f x 0 = L x 1 , x 2 , … , x n + 1 ; f ∣ x 0 , {:(1)f(x_(0))=L(x_(1),x_(2),dots,x_(n+1);f∣x_(0))",":}\begin{equation*} f\left(x_{0}\right)=L\left(x_{1}, x_{2}, \ldots, x_{n+1} ; f \mid x_{0}\right), \tag{1} \end{equation*}(1)f(x0)=L(x1,x2,…,xn+1;f∣x0),
unde L ( x 1 , x 2 , … , x n + 1 ; f ∣ x 0 ) L x 1 , x 2 , … , x n + 1 ; f ∣ x 0 L(x_(1),x_(2),dots,x_(n+1);f∣x_(0))L\left(x_{1}, x_{2}, \ldots, x_{n+1} ; f \mid x_{0}\right)L(x1,x2,…,xn+1;f∣x0) este polinomul lui Lagrange de gradul n n nnn şi care ia valorile f ( x i ) f x i f(x_(i))f\left(x_{i}\right)f(xi) pe punctele sau nodurile de interpolare x i x i x_(i)x_{i}xi.
Vom presupunevĕ nodurile nu sînt neapărat echidistante și ne yom ocupa de calculul membrului al doilea al formulei (1) cu ajutorul formulci lui Newton, L ( x 1 , x 2 , … , x n + 1 ; f ∣ x 0 ) = ∑ i = 0 n ( x 0 − x 1 ) ( x 0 − x 2 ) … ( x 0 − x i ) D 1 j ) L x 1 , x 2 , … , x n + 1 ; f ∣ x 0 = ∑ i = 0 n   x 0 − x 1 x 0 − x 2 … x 0 − x i D 1 j {:L(x_(1),x_(2),dots,x_(n+1);f∣x_(0))=sum_(i=0)^(n)(x_(0)-x_(1))(x_(0)-x_(2))dots(x_(0)-x_(i))D_(1)^(j))\left.L\left(x_{1}, x_{2}, \ldots, x_{n+1} ; f \mid x_{0}\right)=\sum_{i=0}^{n}\left(x_{0}-x_{1}\right)\left(x_{0}-x_{2}\right) \ldots\left(x_{0}-x_{i}\right) D_{1}^{j}\right)L(x1,x2,…,xn+1;f∣x0)=∑i=0n(x0−x1)(x0−x2)…(x0−xi)D1j), unde coeficienții D i i , i = 0 , 1 , … , n D i i , i = 0 , 1 , … , n D_(i)^(i),i=0,1,dots,nD_{i}^{i}, i=0,1, \ldots, nDii,i=0,1,…,n, se obțin din tabloul nr. 1 al diferențelor divizate în care avem, în general,
D i j = D i j [ f ] = [ x i , x i + 1 , … , x i + j ; f ] (3) D i 0 = f i = f ( x i ) D i j = D i j [ f ] = x i , x i + 1 , … , x i + j ; f (3) D i 0 = f i = f x i {:[D_(i)^(j)=D_(i)^(j)[f]=[x_(i),x_(i+1),dots,x_(i+j);f]],[(3)D_(i)^(0)=f_(i)=f(x_(i))]:}\begin{gather*} D_{i}^{j}=D_{i}^{j}[f]=\left[x_{i}, x_{i+1}, \ldots, x_{i+j} ; f\right] \\ D_{i}^{0}=f_{i}=f\left(x_{i}\right) \tag{3} \end{gather*}Dij=Dij[f]=[xi,xi+1,…,xi+j;f](3)Di0=fi=f(xi)
folosind notațile obişuite pentru diferențele divizate ale funcției f ( x ) f ( x ) f(x)f(x)f(x).
Formarea tabloului nr. 1 al diferențelor divizate se face calculînd succesiv coloanele cu ajutorul formulei de recurență
(4) D i j [ f ] = D i + 1 i − 1 [ f ] − D i j − 1 [ f ] x i + j − x i ( D i 0 [ f ] = f i ) i = 1 , 2 , … , n + 1 − j , j = 1 , 2 , … , n (4) D i j [ f ] = D i + 1 i − 1 [ f ] − D i j − 1 [ f ] x i + j − x i D i 0 [ f ] = f i i = 1 , 2 , … , n + 1 − j , j = 1 , 2 , … , n {:[(4)D_(i)^(j)[f]=(D_(i+1)^(i-1)[f]-D_(i)^(j-1)[f])/(x_(i+j)-x_(i))quadquadquad(D_(i)^(0)[f]=f_(i))],[i=1","2","dots","n+1-j","quad j=1","2","dots","n]:}\begin{align*} D_{i}^{j}[f] & =\frac{D_{i+1}^{i-1}[f]-D_{i}^{j-1}[f]}{x_{i+j}-x_{i}} \quad \quad \quad\left(D_{i}^{0}[f]=f_{i}\right) \tag{4}\\ i & =1,2, \ldots, n+1-j, \quad j=1,2, \ldots, n \end{align*}(4)Dij[f]=Di+1i−1[f]−Dij−1[f]xi+j−xi(Di0[f]=fi)i=1,2,…,n+1−j,j=1,2,…,n
Se vede că acest calcul necesită împărțiri (cu diferențe de noduri) din care cauză, în general, diferențele divizate nu se pot calcula, de exemplu, sub forma practică de fracții zecimale limitate, decît cu o anumită aproximație, chiar
Tabloul m. 1
x x xxx f ( x ) f ( x ) f(x)f(x)f(x) D 1 D 1 D^(1)D^{1}D1 D 2 D 2 D^(2)D^{2}D2 D 3 D 3 D^(3)D^{3}D3 ... D n − 1 D n − 1 D^(n-1)D^{n-1}Dn−1
x 1 x 1 x_(1)x_{1}x1 h 1 h 1 h_(1)h_{1}h1 D 1 1 D 1 1 D_(1)^(1)D_{1}^{1}D11
x 2 x 2 x_(2)x_{2}x2 / 2 / 2 //_(2)/_{2}/2 D 2 1 D 2 1 D_(2)^(1)D_{2}^{1}D21 D 1 2 D 1 2 D_(1)^(2)D_{1}^{2}D12 D 1 3 D 1 3 D_(1)^(3)D_{1}^{3}D13 -
r 3 r 3 r_(3)r_{3}r3 f 3 f 3 f_(3)f_{3}f3 D 2 2 D 2 2 D_(2)^(2)D_{2}^{2}D22 .
⋮ ⋮ vdots\vdots⋮ : ⋮ ⋮ vdots\vdots⋮ ⋮ ⋮ vdots\vdots⋮ D 1 n D 1 n D_(1)^(n)D_{1}^{n}D1n
⋮ n − 1 ⋮ n − 1 ^(vdots)_(n-1){ }^{\vdots}{ }_{n-1}⋮n−1 I n − 1 I n − 1 I_(n-1)I_{n-1}In−1 D n − 1 1 D n − 1 1 D_(n-1)^(1)D_{n-1}^{1}Dn−11 D n − 2 2 D n − 2 2 D_(n-2)^(2)D_{n-2}^{2}Dn−22 D n − 2 3 D n − 2 3 D_(n-2)^(3)D_{n-2}^{3}Dn−23
x n x n x_(n)x_{n}xn i n i n i_(n)i_{n}in D n D n D_(n)D_{n}Dn D n − 1 2 D n − 1 2 D_(n-1)^(2)D_{n-1}^{2}Dn−12
x n − 1 x n − 1 x_(n-1)x_{n-1}xn−1 i n + 1 i n + 1 i_(n+1)i_{n+1}in+1
x f(x) D^(1) D^(2) D^(3) ... D^(n-1) x_(1) h_(1) D_(1)^(1) x_(2) //_(2) D_(2)^(1) D_(1)^(2) D_(1)^(3) - r_(3) f_(3) D_(2)^(2) . vdots : vdots vdots D_(1)^(n) ^(vdots)_(n-1) I_(n-1) D_(n-1)^(1) D_(n-2)^(2) D_(n-2)^(3) x_(n) i_(n) D_(n) D_(n-1)^(2) x_(n-1) i_(n+1) | $x$ | $f(x)$ | $D^{1}$ | $D^{2}$ | $D^{3}$ | ... | $D^{n-1}$ | | :--- | :--- | :--- | :--- | :--- | :--- | :--- | | $x_{1}$ | $h_{1}$ | $D_{1}^{1}$ | | | | | | $x_{2}$ | $/_{2}$ | $D_{2}^{1}$ | $D_{1}^{2}$ | $D_{1}^{3}$ | - | | | $r_{3}$ | $f_{3}$ | | $D_{2}^{2}$ | | . | | | | $\vdots$ | : | $\vdots$ | $\vdots$ | | $D_{1}^{n}$ | | ${ }^{\vdots}{ }_{n-1}$ | $I_{n-1}$ | $D_{n-1}^{1}$ | $D_{n-2}^{2}$ | $D_{n-2}^{3}$ | | | | | | | | | | | | $x_{n}$ | $i_{n}$ | $D_{n}$ | $D_{n-1}^{2}$ | | | | | $x_{n-1}$ | $i_{n+1}$ | | | | | |
dacă nodurile și valorile luncţiei pe aceste noduri sîut date sub această formă. Dacă însă elementele tabloului nr. I se calculează aproximativ, erorile comise se tränspun asupra polinomului de interpolare. In cele ce-urmează ne propunem să examinăm efectul acestor erori asupra calculului electiv al formulei (2). Pentru simplificare, vom presupune că nodurile şi punctul x 0 x 0 x_(0)x_{0}x0, unde se interpolează nu sînt afectate de erori.
2. Coloanele tabloului nr. 1. se calculează succesiv cu anumite aproximatii.
Fie f ~ i , i = 1 , 2 , … n + 1 f ~ i , i = 1 , 2 , … n + 1 widetilde(f)_(i),i=1,2,dots n+1\widetilde{f}_{i}, i=1,2, \ldots n+1f~i,i=1,2,…n+1, nişte valori aproximative ale valorilor funcției pe noduri și c i ( 0 ) , i = 1 , 2 , … , n + 1 c i ( 0 ) , i = 1 , 2 , … , n + 1 c_(i)^((0)),i=1,2,dots,n+1c_{i}^{(0)}, i=1,2, \ldots, n+1ci(0),i=1,2,…,n+1, corectiile respective, deci f i = f ~ i + c i ( 0 ) f i = f ~ i + c i ( 0 ) f_(i)= widetilde(f)_(i)+c_(i)^((0))f_{i}=\widetilde{f}_{i}+c_{i}^{(0)}fi=f~i+ci(0), i = 1 , 2 , … , n + 1 i = 1 , 2 , … , n + 1 i=1,2,dots,n+1i=1,2, \ldots, n+1i=1,2,…,n+1. Fie apoi, în general, D ~ i j = D ~ i i [ t ] , i = 1 , 2 , … , n + 1 − j D ~ i j = D ~ i i [ t ] , i = 1 , 2 , … , n + 1 − j widetilde(D)_(i)^(j)= widetilde(D)_(i)^(i)[t],i=1,2,dots,n+1-j\widetilde{D}_{i}^{j}=\widetilde{D}_{i}^{i}[t], i=1,2, \ldots, n+1-jD~ij=D~ii[t],i=1,2,…,n+1−j, valorile aproximative ale elementelor coloanei j j jjj, calculate cu ajutorul elementelor deja calculate ale coloanei j − 1 j − 1 j-1j-1j−1 şi c i ( j ) , i = 1 , 2 , … , n + 1 − j c i ( j ) , i = 1 , 2 , … , n + 1 − j c_(i)^((j)),i=1,2,dots,n+1-jc_{i}^{(j)}, i=1,2, \ldots, n+1-jci(j),i=1,2,…,n+1−j, corecţiile respective. Vom ayea atunci
(5) D ~ i j + c i ( j ) == D ~ i + 1 j − 1 − D ~ i j − 1 x i + j − x i ( D ~ i 0 = f ~ i ) (5) D ~ i j + c i ( j ) == D ~ i + 1 j − 1 − D ~ i j − 1 x i + j − x i D ~ i 0 = f ~ i {:(5) widetilde(D)_(i)^(j)+c_(i)^((j))==( widetilde(D)_(i+1)^(j-1)- widetilde(D)_(i)^(j-1))/(x_(i+j)-x_(i))( widetilde(D)_(i)^(0)= widetilde(f)_(i)):}\begin{equation*} \widetilde{D}_{i}^{j}+c_{i}^{(j)}==\frac{\widetilde{D}_{i+1}^{j-1}-\widetilde{D}_{i}^{j-1}}{x_{i+j}-x_{i}}\left(\widetilde{D}_{i}^{0}=\widetilde{f}_{i}\right) \tag{5} \end{equation*}(5)D~ij+ci(j)==D~i+1j−1−D~ij−1xi+j−xi(D~i0=f~i)
i = 1 , 2 , … , n + 1 − j , j = 1 , 2 , … , n . i = 1 , 2 , … , n + 1 − j , j = 1 , 2 , … , n . i=1,2,dots,n+1-j,j=1,2,dots,n.quadi=1,2, \ldots, n+1-j, j=1,2, \ldots, n . \quadi=1,2,…,n+1−j,j=1,2,…,n. opnoil an
Tabloul m. 1 al diferentelog divizate se inlocuieste atunci eu tabloul mr. 2 a] valorilor aproximative calculate al diferentelor divizate.
Tabloul m. 2
x x xxx I ~ ( x ) I ~ ( x ) tilde(I)(x)\tilde{I}(x)I~(x) D ~ 1 D ~ 1 widetilde(D)^(1)\widetilde{D}^{1}D~1 m ~ 2 m ~ 2 widetilde(m)^(2)\widetilde{m}^{2}m~2 p ~ 3 p ~ 3 tilde(p)^(3)\tilde{p}^{3}p~3 1 ~ n − 1 1 ~ n − 1 tilde(1)^(n-1)\tilde{1}^{n-1}1~n−1
x 1 x 1 x_(1)x_{1}x1 l ~ 1 l ~ 1 tilde(l)_(1)\tilde{l}_{1}l~1 D ~ 1 1 D ~ 1 1 widetilde(D)_(1)^(1)\widetilde{D}_{1}^{1}D~11 ν ~ 1 2 ν ~ 1 2 tilde(nu)_(1)^(2)\tilde{\nu}_{1}^{2}ν~12
x,2 τ 2 τ 2 tau_(2)\tau_{2}τ2 D ~ 2 1 D ~ 2 1 widetilde(D)_(2)^(1)\widetilde{D}_{2}^{1}D~21 . J ~ 1 n 1 J ~ 1 n 1 widetilde(J)_(1)^(n)^(1)\widetilde{J}_{1}^{n}{ }^{1}J~1n1
x 3 x 3 x_(3)x_{3}x3 l ^ 3 l ^ 3 hat(l)_(3)\hat{l}_{3}l^3 . D ~ 2 2 D ~ 2 2 widetilde(D)_(2)^(2)\widetilde{D}_{2}^{2}D~22
⋯ x 2 ⋮ x n − 1 ⋯ x 2 ⋮ x n − 1 cdotsquad{:[x_(2)],[vdots],[x_(n-1)]:}\cdots \quad \begin{gathered}x_{2} \\ \vdots \\ x_{n-1}\end{gathered}⋯x2⋮xn−1 ⋮ ⋮ vdots\vdots⋮ ⋮ ⋮ vdots\vdots⋮ D ~ 2 n − 1 D ~ 2 n − 1 widetilde(D)_(2)^(n-1)\widetilde{D}_{2}^{n-1}D~2n−1
x n − 1 x n − 1 x_(n-1)x_{n-1}xn−1 ⋮ T ¯ n − 1 ⋮ T ¯ n − 1 vdots bar(T)_(n-1)\vdots \bar{T}_{n-1}⋮T¯n−1 D ~ n − 1 1 D ~ n − 1 1 widetilde(D)_(n-1)^(1)\widetilde{D}_{n-1}^{1}D~n−11 D ¯ ― n − 2 2 D ¯ ¯ n − 2 2 bar(bar(D))_(n-2)^(2)\overline{\bar{D}}_{n-2}^{2}D¯―n−22 D ¯ n − 2 3 D ¯ n − 2 3 bar(D)_(n-2)^(3)\bar{D}_{n-2}^{3}D¯n−23
/ n / n //_(n)/_{n}/n
x n x n x_(n)x_{n}xn I ~ n I ~ n tilde(I)_(n)\tilde{I}_{n}I~n D ~ n 1 D ~ n 1 widetilde(D)_(n)^(1)\widetilde{D}_{n}^{1}D~n1 D ~ n − 1 2 D ~ n − 1 2 widetilde(D)_(n-1)^(2)\widetilde{D}_{n-1}^{2}D~n−12
x n + 1 x n + 1 x_(n+1)x_{n+1}xn+1 T ~ n + 1 T ~ n + 1 widetilde(T)_(n+1)\widetilde{T}_{n+1}T~n+1
x tilde(I)(x) widetilde(D)^(1) widetilde(m)^(2) tilde(p)^(3) tilde(1)^(n-1) x_(1) tilde(l)_(1) widetilde(D)_(1)^(1) tilde(nu)_(1)^(2) x,2 tau_(2) widetilde(D)_(2)^(1) . widetilde(J)_(1)^(n)^(1) x_(3) hat(l)_(3) . widetilde(D)_(2)^(2) cdotsquad{:[x_(2)],[vdots],[x_(n-1)]:} vdots vdots widetilde(D)_(2)^(n-1) x_(n-1) vdots bar(T)_(n-1) widetilde(D)_(n-1)^(1) bar(bar(D))_(n-2)^(2) bar(D)_(n-2)^(3) //_(n) x_(n) tilde(I)_(n) widetilde(D)_(n)^(1) widetilde(D)_(n-1)^(2) x_(n+1) widetilde(T)_(n+1) | $x$ | $\tilde{I}(x)$ | $\widetilde{D}^{1}$ | $\widetilde{m}^{2}$ | $\tilde{p}^{3}$ | $\tilde{1}^{n-1}$ | | :--- | :--- | :--- | :--- | :--- | :--- | | $x_{1}$ | $\tilde{l}_{1}$ | $\widetilde{D}_{1}^{1}$ | $\tilde{\nu}_{1}^{2}$ | | | | x,2 | $\tau_{2}$ | $\widetilde{D}_{2}^{1}$ | | . $\widetilde{J}_{1}^{n}{ }^{1}$ | | | $x_{3}$ | $\hat{l}_{3}$ | . | $\widetilde{D}_{2}^{2}$ | | | | $\cdots \quad \begin{gathered}x_{2} \\ \vdots \\ x_{n-1}\end{gathered}$ | | $\vdots$ | | $\vdots$ | $\widetilde{D}_{2}^{n-1}$ | | $x_{n-1}$ | $\vdots \bar{T}_{n-1}$ | $\widetilde{D}_{n-1}^{1}$ | $\overline{\bar{D}}_{n-2}^{2}$ | $\bar{D}_{n-2}^{3}$ | | | | $/_{n}$ | | | | | | $x_{n}$ | $\tilde{I}_{n}$ | $\widetilde{D}_{n}^{1}$ | $\widetilde{D}_{n-1}^{2}$ | | | | $x_{n+1}$ | $\widetilde{T}_{n+1}$ | | | | |
O valoare aproximativă L ~ L ~ tilde(L)\tilde{L}L~ a expresiei (2) se obtine atunci cu ajutorul tabloului nr. 2 cu formula
(6) L ~ = ∑ i = 0 n ( x 0 − x 1 ) ( x 0 − x 2 ) … ( x 0 − x i ) D ~ 1 j (6) L ~ = ∑ i = 0 n   x 0 − x 1 x 0 − x 2 … x 0 − x i D ~ 1 j {:(6) widetilde(L)=sum_(i=0)^(n)(x_(0)-x_(1))(x_(0)-x_(2))dots(x_(0)-x_(i)) widetilde(D)_(1)^(j):}\begin{equation*} \widetilde{L}=\sum_{i=0}^{n}\left(x_{0}-x_{1}\right)\left(x_{0}-x_{2}\right) \ldots\left(x_{0}-x_{i}\right) \widetilde{D}_{1}^{j} \tag{6} \end{equation*}(6)L~=∑i=0n(x0−x1)(x0−x2)…(x0−xi)D~1j
şi vom avea
(7) L ( x 1 , x 2 , … , x n + 1 ; / ∣ x 0 ) = L ~ + λ , (7) L x 1 , x 2 , … , x n + 1 ; / ∣ x 0 = L ~ + λ , {:(7)L(x_(1),x_(2),dots,x_(n+1);//∣x_(0))= widetilde(L)+lambda",":}\begin{equation*} L\left(x_{1}, x_{2}, \ldots, x_{n+1} ; / \mid x_{0}\right)=\widetilde{L}+\lambda, \tag{7} \end{equation*}(7)L(x1,x2,…,xn+1;/∣x0)=L~+λ,
  1. fiiitd corecţia respectivă.
In vedcrea delimitării corectiei λ λ lambda\lambdaλ, vom face cîteva observatii asupraformulei de recurență a diferentelor divizate și asupra delimitării diferențelor divizate, care vor interveni.
3. Find dat şirul de noduri (distincte)
(8) x 1 , x 2 , … , x n + 1 (8) x 1 , x 2 , … , x n + 1 {:(8)x_(1)","x_(2)","dots","x_(n+1):}\begin{equation*} x_{1}, x_{2}, \ldots, x_{n+1} \tag{8} \end{equation*}(8)x1,x2,…,xn+1
orice șir de n + 1 n + 1 n+1n+1n+1 numore
(9) f 1 , f 2 , … , f n + 1 (9) f 1 , f 2 , … , f n + 1 {:(9)f_(1)","f_(2)","dots","f_(n+1):}\begin{equation*} f_{1}, f_{2}, \ldots, f_{n+1} \tag{9} \end{equation*}(9)f1,f2,…,fn+1
poate fi consideral ca find format de valorile f i = f ( x i ) , i = 1 , 2 , … , n + 1 f i = f x i , i = 1 , 2 , … , n + 1 f_(i)=f(x_(i)),i=1,2,dots,n+1f_{i}=f\left(x_{i}\right), i=1,2, \ldots, n+1fi=f(xi),i=1,2,…,n+1, ale unei funcții f ( a ) f ( a ) f(a)f(a)f(a) definite pe punctele (8). Această observație simplă clarifică diferite notații, care vor urma.
Fie k k kkk un număr intreg nenegativ < n < n < n<n<n și să considerăm, pentru j = 0 , 1 , … … , n − k j = 0 , 1 , … … , n − k j=0,1,dots dots,n-kj=0,1, \ldots \ldots, n-kj=0,1,……,n−k, şirul
(10) D 1 j , k [ f ] , D 2 j , k [ f ] , … , D n + 1 − k − j j , k [ f ] (10) D 1 j , k [ f ] , D 2 j , k [ f ] , … , D n + 1 − k − j j , k [ f ] {:(10)D_(1)^(j,k)[f]","D_(2)^(j,k)[f]","dots","D_(n+1-k-j)^(j,k)[f]:}\begin{equation*} D_{1}^{j, k}[f], D_{2}^{j, k}[f], \ldots, D_{n+1-k-j}^{j, k}[f] \tag{10} \end{equation*}(10)D1j,k[f],D2j,k[f],…,Dn+1−k−jj,k[f]
format eu ajutorul formulei de recurență
D i i , h [ f ] = D i + 1 i − 1 , h [ f ] − D i j − 1 , h [ f ] x i + j + h − x i , (11) i = 1 , 2 , … , n + 1 − k − i D i i , h [ f ] = D i + 1 i − 1 , h [ f ] − D i j − 1 , h [ f ] x i + j + h − x i , (11) i = 1 , 2 , … , n + 1 − k − i {:[D_(i)^(i,h)[f]=(D_(i+1)^(i-1,h)[f]-D_(i)^(j-1,h)[f])/(x_(i+j+h)-x_(i))","],[(11)i=1","2","dots","n+1-k-i]:}\begin{gather*} D_{i}^{i, h}[f]=\frac{D_{i+1}^{i-1, h}[f]-D_{i}^{j-1, h}[f]}{x_{i+j+h}-x_{i}}, \\ i=1,2, \ldots, n+1-k-i \tag{11} \end{gather*}Dii,h[f]=Di+1i−1,h[f]−Dij−1,h[f]xi+j+h−xi,(11)i=1,2,…,n+1−k−i
unde D i 0 , k [ f ] = f i , i = 1 , 2 , … , n + 1 − k D i 0 , k [ f ] = f i , i = 1 , 2 , … , n + 1 − k D_(i)^(0,k)[f]=f_(i),i=1,2,dots,n+1-kD_{i}^{0, k}[f]=f_{i}, i=1,2, \ldots, n+1-kDi0,k[f]=fi,i=1,2,…,n+1−k. Astel, pentru j = 0 j = 0 j=0j=0j=0 şirul (10) se reduce la şirul format din primii n + 1 − k n + 1 − k n+1-kn+1-kn+1−k termeni ai şirului (9).
Este clar că termenii şirului (10) se exprimă liniar şi omogen în raport cu primii n + 1 − k n + 1 − k n+1-kn+1-kn+1−k termeni ai șirului (9) și, în general, termenii şirului (10), pentru un j j jjj dat, se exprimă liniar şi omogen cụ termenii şirului (10) pentru valoarea imediat inferioară a lui j j jjj.
Pentru k = 0 k = 0 k=0k=0k=0, avem D i i , 0 [ f ] = D i i , i = 1 , 2 , … , n + 1 − j D i i , 0 [ f ] = D i i , i = 1 , 2 , … , n + 1 − j D_(i)^(i,0)[f]=D_(i)^(i),i=1,2,dots,n+1-jD_{i}^{i, 0}[f]=D_{i}^{i}, i=1,2, \ldots, n+1-jDii,0[f]=Dii,i=1,2,…,n+1−j, formulele (11) se reduc la formulele (4), iar şirul (10) se reduce la givul elementelor coloanci de ordinul j j jjj din tabloul nr. 1 al diferențelor divizate ale suncției f ( x ) f ( x ) f(x)f(x)f(x) pe nodurile (8).
Formula (11) care face legătura între şirurile (10) pentru valorile succesive ale lui j j jjj, se mai poate scrie
(12) D i j , k [ f ] = D i 1 ^ , k + j − 1 [ D j − 1 , k [ f ] ] (12) D i j , k [ f ] = D i 1 ^ , k + j − 1 D j − 1 , k [ f ] {:(12)D_(i)^(j,k)[f]=D_(i)^( hat(1),k+j-1)[D^(j-1,k)[f]]:}\begin{equation*} D_{i}^{j, k}[f]=D_{i}^{\hat{1}, k+j-1}\left[D^{j-1, k}[f]\right] \tag{12} \end{equation*}(12)Dij,k[f]=Di1^,k+j−1[Dj−1,k[f]]
Mai general, avem formula
(13) D i j , k [ / ] = D i r , k + j − r [ D j − r , k [ / ] ] , (13) D i j , k [ / ] = D i r , k + j − r D j − r , k [ / ] , {:(13)D_(i)^(j,k)[//]=D_(i)^(r,k+j-r)[D^(j-r,k)[//]]",":}\begin{equation*} D_{i}^{j, k}[/]=D_{i}^{r, k+j-r}\left[D^{j-r, k}[/]\right], \tag{13} \end{equation*}(13)Dij,k[/]=Dir,k+j−r[Dj−r,k[/]],
unde r r rrr este un număr întreg; 0 ⩽ r ⩽ j 0 ⩽ r ⩽ j 0 <= r <= j0 \leqslant r \leqslant j0⩽r⩽j.
9 - c. 1540
Într-adevăr, se observă că pentru r = 0 r = 0 r=0r=0r=0 și r = j r = j r=jr=jr=j, formula (13) este evidentă, iar pentru r = 1 r = 1 r=1r=1r=1 ea se reduce la (12). Pentru a demonstra deci formula (13) este destul a arăta că, dacă ea este adevărată pentru r = s ( s < j ) r = s ( s < j ) r=s(s < j)r=s(s<j)r=s(s<j), ea va fi adevărată și pentru r = s + 1 r = s + 1 r=s+1r=s+1r=s+1. Acest lucru se arată astfel. Prin ipoteză avem:
D i j − 1 , k [ f ] = D i s , k + j − s − 1 [ D j − s − 1 , k [ f ] ] , D i j − 1 , k [ f ] = D i s , k + j − s − 1 D j − s − 1 , k [ f ] , D_(i)^(j-1,k)[f]=D_(i)^(s,k+j-s-1)[D^(j-s-1,k)[f]],D_{i}^{j-1, k}[f]=D_{i}^{s, k+j-s-1}\left[D^{j-s-1, k}[f]\right],Dij−1,k[f]=Dis,k+j−s−1[Dj−s−1,k[f]],
de unde, dacă ținem seama de formula (12), deducem
D i j , k [ f ] = D i 1 , k + j − 1 [ D s , k − j − s − 1 [ D j − s − 1 , k [ f ] ] = [ L i s + 1 , k + j − s − 1 [ D j − s − 1 , k [ f ] ] . D i j , k [ f ] = D i 1 , k + j − 1 D s , k − j − s − 1 D j − s − 1 , k [ f ] = L i s + 1 , k + j − s − 1 D j − s − 1 , k [ f ] . D_(i)^(j,k)[f]=D_(i)^(1,k+j-1)[D^(s,k-j-s-1)[D^(j-s-1,k)[f]]=[L_(i)^(s+1,k+j-s-1)[D^(j-s-1,k)[f]].:}D_{i}^{j, k}[f]=D_{i}^{1, k+j-1}\left[D^{s, k-j-s-1}\left[D^{j-s-1, k}[f]\right]=\left[L_{i}^{s+1, k+j-s-1}\left[D^{j-s-1, k}[f]\right] .\right.\right.Dij,k[f]=Di1,k+j−1[Ds,k−j−s−1[Dj−s−1,k[f]]=[Lis+1,k+j−s−1[Dj−s−1,k[f]].
De aici rezultă imediat că formula (13) este adevărată pentru r = s + 1 r = s + 1 r=s+1r=s+1r=s+1.
Simbolic, proprietatea exprimată de formula (13) se poate serie
D α + β , k = D α , β + k D β , h . D α + β , k = D α , β + k D β , h . D^(alpha+beta,k)=D^(alpha,beta+k)D^(beta,h).D^{\alpha+\beta, k}=D^{\alpha, \beta+k} D^{\beta, h} .Dα+β,k=Dα,β+kDβ,h.
  1. În cele ce urmcază vom avea nevoie de o delimitare convenabilă a zermenilor şirului (9).
Diferența divizată D 1 n = [ x 1 , x 2 , … , x n + 1 ; f ] D 1 n = x 1 , x 2 , … , x n + 1 ; f D_(1)^(n)=[x_(1),x_(2),dots,x_(n+1);f]D_{1}^{n}=\left[x_{1}, x_{2}, \ldots, x_{n+1} ; f\right]D1n=[x1,x2,…,xn+1;f] se delimiteacă uşor, ținînd seama de formula bine cunoscută
[ x 1 , x 2 , … , x n + 1 ; f ] = ∑ i + 1 n + 1 f i l ′ ( x i ) x 1 , x 2 , … , x n + 1 ; f = ∑ i + 1 n + 1   f i l ′ x i [x_(1),x_(2),dots,x_(n+1);f]=sum_(i+1)^(n+1)(f_(i))/(l^(')(x_(i)))\left[x_{1}, x_{2}, \ldots, x_{n+1} ; f\right]=\sum_{i+1}^{n+1} \frac{f_{i}}{l^{\prime}\left(x_{i}\right)}[x1,x2,…,xn+1;f]=∑i+1n+1fil′(xi)
unde i i _(i){ }_{i}i
l ( x ) = ( x − x 1 ) ( x − x 2 ) … ( x − x n + 1 ) . Avem | D 1 n | ≦ N ( x 1 , x 2 , … , x n + 1 ) M , l ( x ) = x − x 1 x − x 2 … x − x n + 1 .  Avem  D 1 n ≦ N x 1 , x 2 , … , x n + 1 M , {:[l(x)=(x-x_(1))(x-x_(2))dots(x-x_(n+1))." Avem "],[|D_(1)^(n)| <= N(x_(1),x_(2),dots,x_(n+1))M","]:}\begin{array}{r} l(x)=\left(x-x_{1}\right)\left(x-x_{2}\right) \ldots\left(x-x_{n+1}\right) . \text { Avem } \\ \left|D_{1}^{n}\right| \leqq N\left(x_{1}, x_{2}, \ldots, x_{n+1}\right) M, \end{array}l(x)=(x−x1)(x−x2)…(x−xn+1). Avem |D1n|≦N(x1,x2,…,xn+1)M,
unde
(14) N i ( x 1 , x 2 , … , x n + 1 ) = ∑ i = 1 n + 1 1 | l ′ ( x i ) | (14) N i x 1 , x 2 , … , x n + 1 = ∑ i = 1 n + 1   1 l ′ x i {:(14)N_(i)(x_(1),x_(2),dots,x_(n+1))=sum_(i=1)^(n+1)(1)/(|l^(')(x_(i))|):}\begin{equation*} N_{i}\left(x_{1}, x_{2}, \ldots, x_{n+1}\right)=\sum_{i=1}^{n+1} \frac{1}{\left|l^{\prime}\left(x_{i}\right)\right|} \tag{14} \end{equation*}(14)Ni(x1,x2,…,xn+1)=∑i=1n+11|l′(xi)|
şi M = max i = 1 , 2 , … , n + 1 ( | f i | ) M = max i = 1 , 2 , … , n + 1   f i M=max_(i=1,2,dots,n+1)(|f_(i)|)M=\max _{i=1,2, \ldots, n+1}\left(\left|f_{i}\right|\right)M=maxi=1,2,…,n+1(|fi|).
O delimitare analogă a lui D 1 n − k , k [ f ] D 1 n − k , k [ f ] D_(1)^(n-k,k)[f]D_{1}^{n-k, k}[f]D1n−k,k[f] se poate scrie
(15) | D 1 n − k , k [ f ] | ≦ N k ( x 1 , x 2 , … , x n + 1 ) M , (15) D 1 n − k , k [ f ] ≦ N k x 1 , x 2 , … , x n + 1 M , {:(15)|D_(1)^(n-k,k)[f]| <= N_(k)(x_(1),x_(2),dots,x_(n+1))M",":}\begin{equation*} \left|D_{1}^{n-k, k}[f]\right| \leqq N_{k}\left(x_{1}, x_{2}, \ldots, x_{n+1}\right) M, \tag{15} \end{equation*}(15)|D1n−k,k[f]|≦Nk(x1,x2,…,xn+1)M,
unde, de altfel, M M MMM se poate înlocui cu max i = 1 , 2 , … , n + 1 − k ( | / i | ) max i = 1 , 2 , … , n + 1 − k   ( | / i | ) max_(i=1,2,dots,n+1-k)(|//i|)\max _{i=1,2, \ldots, n+1-k}(|/ i|)maxi=1,2,…,n+1−k(|/i|).
In formula (15) avem
N k ( x 1 , x 2 , … , x n + 1 ) = sup | f i | ≦ 1 i = 1 , 2 , … , n + 1 − k | D 1 n − k , k [ f ] | indice fic D N k x 1 , x 2 , … , x n + 1 = sup f i ≦ 1 i = 1 , 2 , … , n + 1 − k   D 1 n − k , k [ f ]  indice fic  D N_(k)(x_(1),x_(2),dots,x_(n+1))=s u p_({:[|f_(i)| <= 1],[i=1","2","dots","n+1-k]:})|D_(1)^(n-k,k)[f]|" indice fic "DN_{k}\left(x_{1}, x_{2}, \ldots, x_{n+1}\right)=\sup _{\substack{\left|f_{i}\right| \leqq 1 \\ i=1,2, \ldots, n+1-k}}\left|D_{1}^{n-k, k}[f]\right| \text { indice fic } DNk(x1,x2,…,xn+1)=sup|fi|≦1i=1,2,…,n+1−k|D1n−k,k[f]| indice fic D
şi deci
N k ( x 1 , x 2 , … , x n + 1 ) = η 1 n − k , k [ f ] N k x 1 , x 2 , … , x n + 1 = η 1 n − k , k [ f ] N_(k)(x_(1),x_(2),dots,x_(n+1))=eta_(1)^(n-k,k)[f]N_{k}\left(x_{1}, x_{2}, \ldots, x_{n+1}\right)=\eta_{1}^{n-k, k}[f]Nk(x1,x2,…,xn+1)=η1n−k,k[f]
cu condiția să alegem toate numerele f i , i = 1 , 2 , … , n + 1 − k f i , i = 1 , 2 , … , n + 1 − k f_(i),i=1,2,dots,n+1-kf_{i}, i=1,2, \ldots, n+1-kfi,i=1,2,…,n+1−k egale cu 1 sau -1 , în mod convenabil.
Avem
N 0 ( x 1 , x 2 , … , x n + 1 ) = N ( x 1 , x 2 , … , x n + 1 ) N 0 x 1 , x 2 , … , x n + 1 = N x 1 , x 2 , … , x n + 1 N_(0)(x_(1),x_(2),dots,x_(n+1))=N(x_(1),x_(2),dots,x_(n+1))N_{0}\left(x_{1}, x_{2}, \ldots, x_{n+1}\right)=N\left(x_{1}, x_{2}, \ldots, x_{n+1}\right)N0(x1,x2,…,xn+1)=N(x1,x2,…,xn+1)
membrul al doilea fiind dat de (1/4). Pentru k > 0 k > 0 k > 0k>0k>0, expresia lui N k ( x 1 , x 2 , … , x n + 1 ) N k x 1 , x 2 , … , x n + 1 N_(k)(x_(1),x_(2),dots,x_(n+1))N_{k}\left(x_{1}, x_{2}, \ldots, x_{n+1}\right)Nk(x1,x2,…,xn+1) este însă mai complicată.
În formarea lui L 1 n − k , k [ f ] L 1 n − k , k [ f ] L_(1)^(n-k,k)[f]L_{1}^{n-k, k}[f]L1n−k,k[f] intervin efectiv numai nodurile x i x i x_(i)x_{i}xi pentru i = 1 , 2 , … , n − k , k + 2 , k + 3 , … , n + 1 i = 1 , 2 , … , n − k , k + 2 , k + 3 , … , n + 1 i=1,2,dots,n-k,k+2,k+3,dots,n+1i=1,2, \ldots, n-k, k+2, k+3, \ldots, n+1i=1,2,…,n−k,k+2,k+3,…,n+1. Dacă deci 2 k ≦ n − 1 2 k ≦ n − 1 2k <= n-12 k \leqq n-12k≦n−1, vor
interveni toate nodurile. Dacă însă 2 k > n − 1 2 k > n − 1 2k > n-12 k>n-12k>n−1 nu vor interveni decît primele n − k n − k n-kn-kn−k și ultimele n − k n − k n-kn-kn−k noduri (8). Deducem de aici următoarea formulă:
N k ( x 1 , x 2 , … , x n + 1 ) = N n − k − 1 ( x 1 , x 2 , … , x n − k , x k + 2 , x k + 3 , … , x n + 1 ) N k x 1 , x 2 , … , x n + 1 = N n − k − 1 x 1 , x 2 , … , x n − k , x k + 2 , x k + 3 , … , x n + 1 N_(k)(x_(1),x_(2),dots,x_(n+1))=N_(n-k-1)(x_(1),x_(2),dots,x_(n-k),x_(k+2),x_(k+3),dots,x_(n+1))N_{k}\left(x_{1}, x_{2}, \ldots, x_{n+1}\right)=N_{n-k-1}\left(x_{1}, x_{2}, \ldots, x_{n-k}, x_{k+2}, x_{k+3}, \ldots, x_{n+1}\right)Nk(x1,x2,…,xn+1)=Nn−k−1(x1,x2,…,xn−k,xk+2,xk+3,…,xn+1) dacă 2 k > n − 1 2 k > n − 1 2k > n-12 k>n-12k>n−1. Este suficient să presupunem k < n k < n k < nk<nk<n, deoarece avem evident
(17) N n ( x 1 , x 2 , … , x n − 1 ) = 1 (17) N n x 1 , x 2 , … , x n − 1 = 1 {:(17)N_(n)(x_(1),x_(2),dots,x_(n-1))=1:}\begin{equation*} N_{n}\left(x_{1}, x_{2}, \ldots, x_{n-1}\right)=1 \tag{17} \end{equation*}(17)Nn(x1,x2,…,xn−1)=1
-5. In particular, să presupunem că
(18) x 1 < x 2 < … < x n + 1 (18) x 1 < x 2 < … < x n + 1 {:(18)x_(1) < x_(2) < dots < x_(n+1):}\begin{equation*} x_{1}<x_{2}<\ldots<x_{n+1} \tag{18} \end{equation*}(18)x1<x2<…<xn+1
Formula de recurență ( 1 i. , no arată că dacă termenii çirului ( 9 ) sînt alternativ pozitivi și negativi, şirurile (10) se bucură de aceeaşi proprietate. Avem atuuci, pe de o parte,
| D i i , k [ f ] | = 1 x i + j + k − x i [ | D i + 1 i − 1 , k [ f ] | + | D i i − 1 , k [ f ] | ] D i i , k [ f ] = 1 x i + j + k − x i D i + 1 i − 1 , k [ f ] + D i i − 1 , k [ f ] |D_(i)^(i,k)[f]|=(1)/(x_(i+j+k)-x_(i))[|D_(i+1)^(i-1,k)[f]|+|D_(i)^(i-1,k)[f]|]\left|D_{i}^{i, k}[f]\right|=\frac{1}{x_{i+j+k}-x_{i}}\left[\left|D_{i+1}^{i-1, k}[f]\right|+\left|D_{i}^{i-1, k}[f]\right|\right]|Dii,k[f]|=1xi+j+k−xi[|Di+1i−1,k[f]|+|Dii−1,k[f]|]
Deoarece, pe de altă parte, avem totdeauna
| D i j , k [ f ] | ≤ 1 | x i + j + k − x i | [ | D L l − k , k [ f ] | + | D i j − 1 , k [ f ] | ] D i j , k [ f ] ≤ 1 x i + j + k − x i D L l − k , k [ f ] + D i j − 1 , k [ f ] |D_(i)^(j,k)[f]| <= (1)/(|x_(i+j+k)-x_(i)|)[|D_(L)^(l-k,k)[f]|+|D_(i)^(j-1,k)[f]|]\left|D_{i}^{j, k}[f]\right| \leq \frac{1}{\left|x_{i+j+k}-x_{i}\right|}\left[\left|D_{L}^{l-k, k}[f]\right|+\left|D_{i}^{j-1, k}[f]\right|\right]|Dij,k[f]|≤1|xi+j+k−xi|[|DLl−k,k[f]|+|Dij−1,k[f]|]
rezultă că, în cazul (18), numărul N k ( x 1 , x 2 , … , x n + 1 ) N k x 1 , x 2 , … , x n + 1 N_(k)(x_(1),x_(2),dots,x_(n+1))N_{k}\left(x_{1}, x_{2}, \ldots, x_{n+1}\right)Nk(x1,x2,…,xn+1) este dat de formula
(19) N k ( x 1 , x 2 , … , x n + 1 ) = α 1 n − k , k (19) N k x 1 , x 2 , … , x n + 1 = α 1 n − k , k {:(19)N_(k)(x_(1),x_(2),dots,x_(n+1))=alpha_(1)^(n-k,k):}\begin{equation*} N_{k}\left(x_{1}, x_{2}, \ldots, x_{n+1}\right)=\alpha_{1}^{n-k, k} \tag{19} \end{equation*}(19)Nk(x1,x2,…,xn+1)=α1n−k,k
unde numerele α i j , i ℏ α i j , i ℏ alpha_(i)^(j,iℏ)\alpha_{i}^{j, i \hbar}αij,iℏ se pot calcula cu ajutorul formulelor de recurenţă
(20) α i j , k = α i + 1 j − 1 , k + α i j − 1 , k x i + j + k − x i , ( α i 0 , k = 1 ) i = 1 , 2 , … , n + 1 − k − j , j = 1 , 2 , … , n + 1 − k (20) α i j , k = α i + 1 j − 1 , k + α i j − 1 , k x i + j + k − x i , α i 0 , k = 1 i = 1 , 2 , … , n + 1 − k − j , j = 1 , 2 , … , n + 1 − k {:[(20)alpha_(i)^(j,k)=(alpha_(i+1)^(j-1,k)+alpha_(i)^(j-1,k))/(x_(i+j+k)-x_(i))","quad(alpha_(i)^(0,k)=1)],[i=1","2","dots","n+1-k-j","j=1","2","dots","n+1-k]:}\begin{gather*} \alpha_{i}^{j, k}=\frac{\alpha_{i+1}^{j-1, k}+\alpha_{i}^{j-1, k}}{x_{i+j+k}-x_{i}}, \quad\left(\alpha_{i}^{0, k}=1\right) \tag{20}\\ i=1,2, \ldots, n+1-k-j, j=1,2, \ldots, n+1-k \end{gather*}(20)αij,k=αi+1j−1,k+αij−1,kxi+j+k−xi,(αi0,k=1)i=1,2,…,n+1−k−j,j=1,2,…,n+1−k
Formula de recurenţă (20) ne mai arată că, în cazul (18), avem
(21) N k ( x 1 , x 2 , … , x i + k + 1 ) = N k ( x 1 , x 2 , … , x j + k ) + N k ( x 2 , x 3 , … , x j + k + 1 ) x j + k + 1 − x 1 (21) N k x 1 , x 2 , … , x i + k + 1 = N k x 1 , x 2 , … , x j + k + N k x 2 , x 3 , … , x j + k + 1 x j + k + 1 − x 1 {:(21)N_(k)(x_(1),x_(2),dots,x_(i+k+1))=(N_(k)(x_(1),x_(2),dots,x_(j+k))+N_(k)(x_(2),x_(3),dots,x_(j+k+1)))/(x_(j+k+1)-x_(1)):}\begin{equation*} N_{k}\left(x_{1}, x_{2}, \ldots, x_{i+k+1}\right)=\frac{N_{k}\left(x_{1}, x_{2}, \ldots, x_{j+k}\right)+N_{k}\left(x_{2}, x_{3}, \ldots, x_{j+k+1}\right)}{x_{j+k+1}-x_{1}} \tag{21} \end{equation*}(21)Nk(x1,x2,…,xi+k+1)=Nk(x1,x2,…,xj+k)+Nk(x2,x3,…,xj+k+1)xj+k+1−x1
Astfel, Lolosind formulele (16), (17), (21) deducem, in particular,
N 0 ( x 1 , x 2 ) = 2 x 2 − x 1 N 0 ( x 1 , x 2 , x 3 ) = 2 ( x 2 − x 1 ) ( x 3 − x 2 ) , N 1 ( x 1 , x 2 , x 3 ) = 2 x 3 − x 1 N 0 ( x 1 , x 2 , x 3 , x 4 ) = 2 ( x 4 + x 2 − x 1 − x 3 ) ( x 2 − x 1 ) ( x 3 − x 2 ) ( x 4 − x 3 ) ( x 4 − x 1 ) N 1 ( x 1 , x 2 , x 3 , x 4 ) = 2 ( x 4 + x 3 − x 1 − x 2 ) ( x 3 − x 1 ) ( x 4 − x 2 ) ( x 4 − x 1 ) , N 2 ( x 1 , x 2 , x 3 , x 4 ) = 2 x 4 − x 1 N 0 x 1 , x 2 = 2 x 2 − x 1 N 0 x 1 , x 2 , x 3 = 2 x 2 − x 1 x 3 − x 2 , N 1 x 1 , x 2 , x 3 = 2 x 3 − x 1 N 0 x 1 , x 2 , x 3 , x 4 = 2 x 4 + x 2 − x 1 − x 3 x 2 − x 1 x 3 − x 2 x 4 − x 3 x 4 − x 1 N 1 x 1 , x 2 , x 3 , x 4 = 2 x 4 + x 3 − x 1 − x 2 x 3 − x 1 x 4 − x 2 x 4 − x 1 , N 2 x 1 , x 2 , x 3 , x 4 = 2 x 4 − x 1 {:[N_(0)(x_(1),x_(2))=(2)/(x_(2)-x_(1))],[N_(0)(x_(1),x_(2),x_(3))=(2)/((x_(2)-x_(1))(x_(3)-x_(2)))","N_(1)(x_(1),x_(2),x_(3))=(2)/(x_(3)-x_(1))],[N_(0)(x_(1),x_(2),x_(3),x_(4))=(2(x_(4)+x_(2)-x_(1)-x_(3)))/((x_(2)-x_(1))(x_(3)-x_(2))(x_(4)-x_(3))(x_(4)-x_(1)))],[N_(1)(x_(1),x_(2),x_(3),x_(4))=(2(x_(4)+x_(3)-x_(1)-x_(2)))/((x_(3)-x_(1))(x_(4)-x_(2))(x_(4)-x_(1)))","N_(2)(x_(1),x_(2),x_(3),x_(4))=(2)/(x_(4)-x_(1))]:}\begin{gathered} N_{0}\left(x_{1}, x_{2}\right)=\frac{2}{x_{2}-x_{1}} \\ N_{0}\left(x_{1}, x_{2}, x_{3}\right)=\frac{2}{\left(x_{2}-x_{1}\right)\left(x_{3}-x_{2}\right)}, N_{1}\left(x_{1}, x_{2}, x_{3}\right)=\frac{2}{x_{3}-x_{1}} \\ N_{0}\left(x_{1}, x_{2}, x_{3}, x_{4}\right)=\frac{2\left(x_{4}+x_{2}-x_{1}-x_{3}\right)}{\left(x_{2}-x_{1}\right)\left(x_{3}-x_{2}\right)\left(x_{4}-x_{3}\right)\left(x_{4}-x_{1}\right)} \\ N_{1}\left(x_{1}, x_{2}, x_{3}, x_{4}\right)=\frac{2\left(x_{4}+x_{3}-x_{1}-x_{2}\right)}{\left(x_{3}-x_{1}\right)\left(x_{4}-x_{2}\right)\left(x_{4}-x_{1}\right)}, N_{2}\left(x_{1}, x_{2}, x_{3}, x_{4}\right)=\frac{2}{x_{4}-x_{1}} \end{gathered}N0(x1,x2)=2x2−x1N0(x1,x2,x3)=2(x2−x1)(x3−x2),N1(x1,x2,x3)=2x3−x1N0(x1,x2,x3,x4)=2(x4+x2−x1−x3)(x2−x1)(x3−x2)(x4−x3)(x4−x1)N1(x1,x2,x3,x4)=2(x4+x3−x1−x2)(x3−x1)(x4−x2)(x4−x1),N2(x1,x2,x3,x4)=2x4−x1
N 0 ( x 1 , x 2 , x 3 , x 4 , x 5 ) = N 0 x 1 , x 2 , x 3 , x 4 , x 5 = N_(0)(x_(1),x_(2),x_(3),x_(4),x_(5))=N_{0}\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right)=N0(x1,x2,x3,x4,x5)=
= 2 [ ( x 2 − x 1 ) ( x 3 − x 2 ) + ( x 5 − x 4 ) ( x 4 − x 3 ) + ( x 2 − x 1 ) ( x 5 − x 4 ) ] ( x 2 − x 1 ) ( x 3 − x 2 ) ( x 4 − x 3 ) ( x 5 − x 4 ) ( x 4 − x 1 ) ( x 5 − x 2 ) = N 1 ( x 1 : x 2 , x 3 , x 4 , x 5 ) = 2 [ ( x 4 − x 1 ) ( x 5 − x 2 ) ( x 5 − x 1 ) + ( x 3 − x 1 ) ( x 4 − x 3 ) ( x 4 − x 4 ) + ( x 5 − x 2 ) ( x 3 − x 2 ) ( x 5 − x 3 ) ] ( x 2 − x 1 ) ( x 4 − x 2 ) ( x 5 − x 3 ) ( x 4 − x 1 ) ( x 5 − x 2 ) ( x 5 − x 1 ) N 2 ( x 1 , x 2 , x 3 , x 4 , x 5 ) = 2 ( x 5 + x 4 − x 1 − x 2 ) ( x 4 − x 1 ) ( x 5 − x 2 ) ( x 5 − x 1 ) N 3 ( x 1 , x 2 , x 3 , x 4 , x 5 ) = 2 x 5 − x 1 = 2 x 2 − x 1 x 3 − x 2 + x 5 − x 4 x 4 − x 3 + x 2 − x 1 x 5 − x 4 x 2 − x 1 x 3 − x 2 x 4 − x 3 x 5 − x 4 x 4 − x 1 x 5 − x 2 = N 1 x 1 : x 2 , x 3 , x 4 , x 5 = 2 x 4 − x 1 x 5 − x 2 x 5 − x 1 + x 3 − x 1 x 4 − x 3 x 4 − x 4 + x 5 − x 2 x 3 − x 2 x 5 − x 3 x 2 − x 1 x 4 − x 2 x 5 − x 3 x 4 − x 1 x 5 − x 2 x 5 − x 1 N 2 x 1 , x 2 , x 3 , x 4 , x 5 = 2 x 5 + x 4 − x 1 − x 2 x 4 − x 1 x 5 − x 2 x 5 − x 1 N 3 x 1 , x 2 , x 3 , x 4 , x 5 = 2 x 5 − x 1 {:[=(2[(x_(2)-x_(1))(x_(3)-x_(2))+(x_(5)-x_(4))(x_(4)-x_(3))+(x_(2)-x_(1))(x_(5)-x_(4))])/((x_(2)-x_(1))(x_(3)-x_(2))(x_(4)-x_(3))(x_(5)-x_(4))(x_(4)-x_(1))(x_(5)-x_(2)))],[=(N_(1)(x_(1):x_(2),x_(3),x_(4),x_(5))=)/(2[(x_(4)-x_(1))(x_(5)-x_(2))(x_(5)-x_(1))+(x_(3)-x_(1))(x_(4)-x_(3))(x_(4)-x_(4))+(x_(5)-x_(2))(x_(3)-x_(2))(x_(5)-x_(3))])],[(x_(2)-x_(1))(x_(4)-x_(2))(x_(5)-x_(3))(x_(4)-x_(1))(x_(5)-x_(2))(x_(5)-x_(1))],[N_(2)(x_(1),x_(2),x_(3),x_(4),x_(5))=(2(x_(5)+x_(4)-x_(1)-x_(2)))/((x_(4)-x_(1))(x_(5)-x_(2))(x_(5)-x_(1)))],[N_(3)(x_(1),x_(2),x_(3),x_(4),x_(5))=(2)/(x_(5)-x_(1))]:}\begin{gathered} =\frac{2\left[\left(x_{2}-x_{1}\right)\left(x_{3}-x_{2}\right)+\left(x_{5}-x_{4}\right)\left(x_{4}-x_{3}\right)+\left(x_{2}-x_{1}\right)\left(x_{5}-x_{4}\right)\right]}{\left(x_{2}-x_{1}\right)\left(x_{3}-x_{2}\right)\left(x_{4}-x_{3}\right)\left(x_{5}-x_{4}\right)\left(x_{4}-x_{1}\right)\left(x_{5}-x_{2}\right)} \\ =\frac{N_{1}\left(x_{1}: x_{2}, x_{3}, x_{4}, x_{5}\right)=}{2\left[\left(x_{4}-x_{1}\right)\left(x_{5}-x_{2}\right)\left(x_{5}-x_{1}\right)+\left(x_{3}-x_{1}\right)\left(x_{4}-x_{3}\right)\left(x_{4}-x_{4}\right)+\left(x_{5}-x_{2}\right)\left(x_{3}-x_{2}\right)\left(x_{5}-x_{3}\right)\right]} \\ \left(x_{2}-x_{1}\right)\left(x_{4}-x_{2}\right)\left(x_{5}-x_{3}\right)\left(x_{4}-x_{1}\right)\left(x_{5}-x_{2}\right)\left(x_{5}-x_{1}\right) \\ N_{2}\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right)=\frac{2\left(x_{5}+x_{4}-x_{1}-x_{2}\right)}{\left(x_{4}-x_{1}\right)\left(x_{5}-x_{2}\right)\left(x_{5}-x_{1}\right)} \\ N_{3}\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right)=\frac{2}{x_{5}-x_{1}} \end{gathered}=2[(x2−x1)(x3−x2)+(x5−x4)(x4−x3)+(x2−x1)(x5−x4)](x2−x1)(x3−x2)(x4−x3)(x5−x4)(x4−x1)(x5−x2)=N1(x1:x2,x3,x4,x5)=2[(x4−x1)(x5−x2)(x5−x1)+(x3−x1)(x4−x3)(x4−x4)+(x5−x2)(x3−x2)(x5−x3)](x2−x1)(x4−x2)(x5−x3)(x4−x1)(x5−x2)(x5−x1)N2(x1,x2,x3,x4,x5)=2(x5+x4−x1−x2)(x4−x1)(x5−x2)(x5−x1)N3(x1,x2,x3,x4,x5)=2x5−x1
Se vede acum uşor cum se pot. delimita termenii șirului (10) eu ajutorul rezultatelor prevedente. Avem
| D i j , k [ f ] | ≦ N k ( x i , x i + 1 , … , x i + j + k ) M D i j , k [ f ] ≦ N k x i , x i + 1 , … , x i + j + k M |D_(i)^(j,k)[f]| <= N_(k)(x_(i),x_(i+1),dots,x_(i+j+k))M\left|D_{i}^{j, k}[f]\right| \leqq N_{k}\left(x_{i}, x_{i+1}, \ldots, x_{i+j+k}\right) M|Dij,k[f]|≦Nk(xi,xi+1,…,xi+j+k)M
unde M = max r = i , i + 1 , … , i + j ( | f r | ) M = max r = i , i + 1 , … , i + j   f r M=max_(r=i,i+1,dots,i+j)(|f_(r)|)M=\max _{r=i, i+1, \ldots, i+j}\left(\left|f_{r}\right|\right)M=maxr=i,i+1,…,i+j(|fr|)
6. Să revenim la delimitarea corceției λ λ lambda\lambdaλ din formula (7). Formula (5) se mai poate scrie
ques
D i 1 , j − 1 [ D ~ j − 1 ] = D ~ i j + c i ( j ) D i 1 , j − 1 D ~ j − 1 = D ~ i j + c i ( j ) D_(i)^(1,j-1)[ widetilde(D)^(j-1)]= widetilde(D)_(i)^(j)+c_(i)^((j))D_{i}^{1, j-1}\left[\widetilde{D}^{j-1}\right]=\widetilde{D}_{i}^{j}+c_{i}^{(j)}Di1,j−1[D~j−1]=D~ij+ci(j)
din care deducem şi
D i 1 , j − r − 1 [ D ~ j − r − 1 ] = D ~ i j − r + c i ( j − r ) , 0 ≦ r ≦ j − 1 D i 1 , j − r − 1 D ~ j − r − 1 = D ~ i j − r + c i ( j − r ) , 0 ≦ r ≦ j − 1 D_(i)^(1,j-r-1)[ widetilde(D)^(j-r-1)]= widetilde(D)_(i)^(j-r)+c_(i)^((j-r)),0 <= r <= j-1D_{i}^{1, j-r-1}\left[\widetilde{D}^{j-r-1}\right]=\widetilde{D}_{i}^{j-r}+c_{i}^{(j-r)}, 0 \leqq r \leqq j-1Di1,j−r−1[D~j−r−1]=D~ij−r+ci(j−r),0≦r≦j−1
De aici rezultă că
D i r , j − r [ D j , j − r − 1 [ D ~ j − r − 1 ] ] = D i r , j − r [ D ~ j − r + c ( j − r ) ] = D i r , j − r [ D ~ j − r ] + D i r , j − r [ c ( j − r ) ] . D i r , j − r D j , j − r − 1 D ~ j − r − 1 = D i r , j − r D ~ j − r + c ( j − r ) = D i r , j − r D ~ j − r + D i r , j − r c ( j − r ) . {:[D_(i)^(r,j-r)[D^(j,j-r-1)[ widetilde(D)^(j-r-1)]]=D_(i)^(r,j-r)[ widetilde(D)^(j-r)+c^((j-r))]=],[D_(i)^(r,j-r)[ widetilde(D)^(j-r)]+D_(i)^(r,j-r)[c^((j-r))].]:}\begin{gathered} D_{i}^{r, j-r}\left[D^{j, j-r-1}\left[\widetilde{D}^{j-r-1}\right]\right]=D_{i}^{r, j-r}\left[\widetilde{D}^{j-r}+c^{(j-r)}\right]= \\ D_{i}^{r, j-r}\left[\widetilde{D}^{j-r}\right]+D_{i}^{r, j-r}\left[c^{(j-r)}\right] . \end{gathered}Dir,j−r[Dj,j−r−1[D~j−r−1]]=Dir,j−r[D~j−r+c(j−r)]=Dir,j−r[D~j−r]+Dir,j−r[c(j−r)].
Dar, pe baza formulei (13),
D i r , j − r [ D 1 , j − r − 1 [ D ~ j − r − 1 ] ] = D i r + 1 , j − r − 1 [ D ~ j − r − 1 ] D i r , j − r D 1 , j − r − 1 D ~ j − r − 1 = D i r + 1 , j − r − 1 D ~ j − r − 1 D_(i)^(r,j-r)[D^(1,j-r-1)[ widetilde(D)^(j-r-1)]]=D_(i)^(r+1,j-r-1)[ widetilde(D)^(j-r-1)]D_{i}^{r, j-r}\left[D^{1, j-r-1}\left[\widetilde{D}^{j-r-1}\right]\right]=D_{i}^{r+1, j-r-1}\left[\widetilde{D}^{j-r-1}\right]Dir,j−r[D1,j−r−1[D~j−r−1]]=Dir+1,j−r−1[D~j−r−1]
şi avem, prin urmare,
D i r + 1 : j − r − 1 [ D ~ j − r − 1 ] = D i r , j − r [ D ~ j − r ] + D i r , j − r [ c ( j − r ) ] , r = 0 , 1 , … , j − 1 . D i r + 1 : j − r − 1 D ~ j − r − 1 = D i r , j − r D ~ j − r + D i r , j − r c ( j − r ) , r = 0 , 1 , … , j − 1 . D_(i)^(r+1:j-r-1)[ widetilde(D)^(j-r-1)]=D_(i)^(r,j-r)[ widetilde(D)^(j-r)]+D_(i)^(r,j-r)[c^((j-r))],r=0,1,dots,j-1.D_{i}^{r+1: j-r-1}\left[\widetilde{D}^{j-r-1}\right]=D_{i}^{r, j-r}\left[\widetilde{D}^{j-r}\right]+D_{i}^{r, j-r}\left[c^{(j-r)}\right], r=0,1, \ldots, j-1 .Dir+1:j−r−1[D~j−r−1]=Dir,j−r[D~j−r]+Dir,j−r[c(j−r)],r=0,1,…,j−1.
Adunînd rnembru cu membru aceste egalități, avem
D i j , 0 [ D ~ 0 ] = D i 0 , j [ D ~ j ] + ∑ r = 0 i − 1 D i r , j − r [ c ( j − k ) ] . D i j , 0 D ~ 0 = D i 0 , j D ~ j + ∑ r = 0 i − 1   D i r , j − r c ( j − k ) . D_(i)^(j,0)[ widetilde(D)^(0)]=D_(i)^(0,j)[ widetilde(D)^(j)]+sum_(r=0)^(i-1)D_(i)^(r,j-r)[c^((j-k))].D_{i}^{j, 0}\left[\widetilde{D}^{0}\right]=D_{i}^{0, j}\left[\widetilde{D}^{j}\right]+\sum_{r=0}^{i-1} D_{i}^{r, j-r}\left[c^{(j-k)}\right] .Dij,0[D~0]=Di0,j[D~j]+∑r=0i−1Dir,j−r[c(j−k)].
İnsă avem şi
D i j , 0 [ D ~ 0 ] = D i j , 0 [ f ] − D i j , 0 [ c ( 0 ) ] = D i j [ f ] − D i j , 0 [ c ( 0 ) ] , D i 0 , j [ D ~ j ] = D ~ i i D i j , 0 D ~ 0 = D i j , 0 [ f ] − D i j , 0 c ( 0 ) = D i j [ f ] − D i j , 0 c ( 0 ) , D i 0 , j D ~ j = D ~ i i D_(i)^(j,0)[ widetilde(D)^(0)]=D_(i)^(j,0)[f]-D_(i)^(j,0)[c^((0))]=D_(i)^(j)[f]-D_(i)^(j,0)[c^((0))],D_(i)^(0,j)[ widetilde(D)^(j)]= widetilde(D)_(i)^(i)D_{i}^{j, 0}\left[\widetilde{D}^{0}\right]=D_{i}^{j, 0}[f]-D_{i}^{j, 0}\left[c^{(0)}\right]=D_{i}^{j}[f]-D_{i}^{j, 0}\left[c^{(0)}\right], D_{i}^{0, j}\left[\widetilde{D}^{j}\right]=\widetilde{D}_{i}^{i}Dij,0[D~0]=Dij,0[f]−Dij,0[c(0)]=Dij[f]−Dij,0[c(0)],Di0,j[D~j]=D~ii
aşa că deducem formula
D i j [ f ] = D i j = D ~ i j + ∑ r = 0 j D i r , i − r [ c ( j − r ) ] D i j [ f ] = D i j = D ~ i j + ∑ r = 0 j   D i r , i − r c ( j − r ) D_(i)^(j)[f]=D_(i)^(j)= widetilde(D)_(i)^(j)+sum_(r=0)^(j)D_(i)^(r,i-r)[c^((j-r))]D_{i}^{j}[f]=D_{i}^{j}=\widetilde{D}_{i}^{j}+\sum_{r=0}^{j} D_{i}^{r, i-r}\left[c^{(j-r)}\right]Dij[f]=Dij=D~ij+∑r=0jDir,i−r[c(j−r)]
Tyinînd seama de această formulă, din (6), (7) deducem
(22) λ = ∑ i = 0 n ( x 0 − x 1 ) ( x 0 − x 2 ) … ( x 0 − x i ) [ ∑ r = 0 i D 1 r , i − r [ c ( i − r ) ] ] (22) λ = ∑ i = 0 n   x 0 − x 1 x 0 − x 2 … x 0 − x i ∑ r = 0 i   D 1 r , i − r c ( i − r ) {:(22)lambda=sum_(i=0)^(n)(x_(0)-x_(1))(x_(0)-x_(2))dots(x_(0)-x_(i))[sum_(r=0)^(i)D_(1)^(r,i-r)[c^((i-r))]]:}\begin{equation*} \lambda=\sum_{i=0}^{n}\left(x_{0}-x_{1}\right)\left(x_{0}-x_{2}\right) \ldots\left(x_{0}-x_{i}\right)\left[\sum_{r=0}^{i} D_{1}^{r, i-r}\left[c^{(i-r)}\right]\right] \tag{22} \end{equation*}(22)λ=∑i=0n(x0−x1)(x0−x2)…(x0−xi)[∑r=0iD1r,i−r[c(i−r)]]
În particular, dacă toate corectuile ci si rămin, în valoare absolută, cel mult egale cu ε ε epsi\varepsilonε, pe baza formulei (15) avem
şi putem scric
unde
(23) V ( x ) = ∑ i = − 0 n | ( x − x 1 ) ( x − x 2 ) … ( x − x i ) | [ ∑ r = 0 i N i − r ( x 1 , x 2 , … , x i + 1 ) ] . (23) V ( x ) = ∑ i = − 0 n   x − x 1 x − x 2 … x − x i ∑ r = 0 i   N i − r x 1 , x 2 , … , x i + 1 . {:(23)V(x)=sum_(i=-0)^(n)|(x-x_(1))(x-x_(2))dots(x-x_(i))|[sum_(r=0)^(i)N_(i-r)(x_(1),x_(2),dots,x_(i+1))].:}\begin{equation*} V(x)=\sum_{i=-0}^{n}\left|\left(x-x_{1}\right)\left(x-x_{2}\right) \ldots\left(x-x_{i}\right)\right|\left[\sum_{r=0}^{i} N_{i-r}\left(x_{1}, x_{2}, \ldots, x_{i+1}\right)\right] . \tag{23} \end{equation*}(23)V(x)=∑i=−0n|(x−x1)(x−x2)…(x−xi)|[∑r=0iNi−r(x1,x2,…,xi+1)].
Cînd valorile functici sînt exacte, deci cînd c i ( j ) = 0 , i = 1 , 2 , … , n + 1 c i ( j ) = 0 , i = 1 , 2 , … , n + 1 c_(i)^((j))=0,i=1,2,dots,n+1c_{i}^{(j)}=0, i=1,2, \ldots, n+1ci(j)=0,i=1,2,…,n+1, se poate lua chiar
(25) V ( x ) = ∑ i = 1 n | ( x − x 1 ) ( x − x 2 ) … ( x − x i ) | [ ∑ r = 0 i − 1 N i − r ( x 1 , x 2 , … , x i + 1 ) ] (25) V ( x ) = ∑ i = 1 n   x − x 1 x − x 2 … x − x i ∑ r = 0 i − 1   N i − r x 1 , x 2 , … , x i + 1 {:(25)V(x)=sum_(i=1)^(n)|(x-x_(1))(x-x_(2))dots(x-x_(i))|[sum_(r=0)^(i-1)N_(i-r)(x_(1),x_(2),dots,x_(i+1))]:}\begin{equation*} V(x)=\sum_{i=1}^{n}\left|\left(x-x_{1}\right)\left(x-x_{2}\right) \ldots\left(x-x_{i}\right)\right|\left[\sum_{r=0}^{i-1} N_{i-r}\left(x_{1}, x_{2}, \ldots, x_{i+1}\right)\right] \tag{25} \end{equation*}(25)V(x)=∑i=1n|(x−x1)(x−x2)…(x−xi)|[∑r=0i−1Ni−r(x1,x2,…,xi+1)]
  1. Vom incheia aceste consideratii cu un exemplu numerie. Fie functia f ( x ) f ( x ) f(x)f(x)f(x) pe nodurile x 1 = 14 , x 2 = 17 , x 3 = 31 , x 4 = 35 x 1 = 14 , x 2 = 17 , x 3 = 31 , x 4 = 35 x_(1)=14,x_(2)=17,x_(3)=31,x_(4)=35x_{1}=14, x_{2}=17, x_{3}=31, x_{4}=35x1=14,x2=17,x3=31,x4=35, data cu valorile ci exacte:
x 1 / 4 17 31 35 f ( x ) . 68 , 7 64 , 0 44 , 0 39 , 1 . . x 1 / 4 17 31 35 f ( x ) . 68 , 7 64 , 0 44 , 0 39 , 1 . . {:[x,1//4,17,31,35],[f(x).,68","7,64","0,44","0,39","1.]:}.\begin{array}{c|cccc} x & 1 / 4 & 17 & 31 & 35 \\ \hline f(x) . & 68,7 & 64,0 & 44,0 & 39,1 . \end{array} .x1/4173135f(x).68,764,044,039,1..
Avem aici n = 3 n = 3 n=3n=3n=3 şi sîntem în cazul (18) prin alegerea convenabilă a notațiilor.
Formulele precedente ne dau, in acest caz,
N 1 ( x 1 , x 2 ) = 1 N 1 x 1 , x 2 = 1 N_(1)(x_(1),x_(2))=1N_{1}\left(x_{1}, x_{2}\right)=1N1(x1,x2)=1
N 2 ( x 1 , x 2 , x 3 ) + N 1 ( x 1 , x 2 , x 3 ) = 1 + 2 x 3 − x 2 = 1 + 2 17 < 1 , 12 N 2 x 1 , x 2 , x 3 + N 1 x 1 , x 2 , x 3 = 1 + 2 x 3 − x 2 = 1 + 2 17 < 1 , 12 N_(2)(x_(1),x_(2),x_(3))+N_(1)(x_(1),x_(2),x_(3))=1+(2)/(x_(3)-x_(2))=1+(2)/(17) < 1,12N_{2}\left(x_{1}, x_{2}, x_{3}\right)+N_{1}\left(x_{1}, x_{2}, x_{3}\right)=1+\frac{2}{x_{3}-x_{2}}=1+\frac{2}{17}<1,12N2(x1,x2,x3)+N1(x1,x2,x3)=1+2x3−x2=1+217<1,12
N 3 ( x 1 , x 2 , x 3 , x 4 ) + N 2 ( x 1 , x 2 , x 3 , x 4 ) + N 1 ( x 1 , x 2 , x 3 , x 4 ) = N 3 x 1 , x 2 , x 3 , x 4 + N 2 x 1 , x 2 , x 3 , x 4 + N 1 x 1 , x 2 , x 3 , x 4 = N_(3)(x_(1),x_(2),x_(3),x_(4))+N_(2)(x_(1),x_(2),x_(3),x_(4))+N_(1)(x_(1),x_(2),x_(3),x_(4))=N_{3}\left(x_{1}, x_{2}, x_{3}, x_{4}\right)+N_{2}\left(x_{1}, x_{2}, x_{3}, x_{4}\right)+N_{1}\left(x_{1}, x_{2}, x_{3}, x_{4}\right)=N3(x1,x2,x3,x4)+N2(x1,x2,x3,x4)+N1(x1,x2,x3,x4)=
= 1 + 2 x 4 − x 1 + 2 ( x 4 + x 3 − x 1 − x 2 ) ( x 3 − x 1 ) ( x 4 − x 2 ) ( x 4 − x 1 ) = 1 + 2 21 + 5 459 < 1 , 11 . = 1 + 2 x 4 − x 1 + 2 x 4 + x 3 − x 1 − x 2 x 3 − x 1 x 4 − x 2 x 4 − x 1 = 1 + 2 21 + 5 459 < 1 , 11 . =1+(2)/(x_(4)-x_(1))+(2(x_(4)+x_(3)-x_(1)-x_(2)))/((x_(3)-x_(1))(x_(4)-x_(2))(x_(4)-x_(1)))=1+(2)/(21)+(5)/(459) < 1,11.=1+\frac{2}{x_{4}-x_{1}}+\frac{2\left(x_{4}+x_{3}-x_{1}-x_{2}\right)}{\left(x_{3}-x_{1}\right)\left(x_{4}-x_{2}\right)\left(x_{4}-x_{1}\right)}=1+\frac{2}{21}+\frac{5}{459}<1,11 .=1+2x4−x1+2(x4+x3−x1−x2)(x3−x1)(x4−x2)(x4−x1)=1+221+5459<1,11.
Avem prin urmare, interpolind în punctul x 0 = 27 x 0 = 27 x_(0)=27x_{0}=27x0=27,
V ( x 0 ) < 13 + 13 × 10 × 1 , 12 + 13 × 10 × 4 × 1 , 11 < 736 . V x 0 < 13 + 13 × 10 × 1 , 12 + 13 × 10 × 4 × 1 , 11 < 736 . V(x_(0)) < 13+13 xx10 xx1,12+13 xx10 xx4xx1,11 < 736.V\left(x_{0}\right)<13+13 \times 10 \times 1,12+13 \times 10 \times 4 \times 1,11<736 .V(x0)<13+13×10×1,12+13×10×4×1,11<736.
Dacă în vederea formării tabloului nr. 2, calculăm diferențele divizate prescurtat la k k kkk zecimale, vom avea
| λ | < 736 2 ⋅ 10 k = 368 10 k | λ | < 736 2 ⋅ 10 k = 368 10 k |lambda| < (736)/(2*10^(k))=(368)/(10^(k))|\lambda|<\frac{736}{2 \cdot 10^{k}}=\frac{368}{10^{k}}|λ|<7362⋅10k=36810k
Pentru a putea obţine cel puțin o zecimală exactă în valoarca polinomului (2), trebuie să avem cel puțin 368 ⋅ 10 − k < 0 , 01 368 ⋅ 10 − k < 0 , 01 368*10^(-k) < 0,01368 \cdot 10^{-k}<0,01368⋅10−k<0,01, deci cel puțin k = 5 k = 5 k=5k=5k=5.
Luînd k = 5 k = 5 k=5k=5k=5, cu datele din tabloul nr. 2 se obțin datele din tabloul nr. 3.
Tabloul nr. 3
x x xxx f(x) ν ~ 1 ν ~ 1 widetilde(nu)^(1)\widetilde{\nu}^{1}ν~1 ν ~ 2 ν ~ 2 tilde(nu)^(2)\tilde{\nu}^{2}ν~2 ν ~ 3 ν ~ 3 widetilde(nu)^(3)\widetilde{\nu}^{3}ν~3
14 68,7 -1,56667 0,00015
17 64,0 0,00812
---1,42857 0,01431
31 44.0
35 39,1 -- 1,22500
x f(x) widetilde(nu)^(1) tilde(nu)^(2) widetilde(nu)^(3) 14 68,7 -1,56667 0,00015 17 64,0 0,00812 ---1,42857 0,01431 31 44.0 35 39,1 -- 1,22500 | $x$ | f(x) | $\widetilde{\nu}^{1}$ | $\tilde{\nu}^{2}$ | $\widetilde{\nu}^{3}$ | | :--- | :--- | :--- | :--- | :--- | | 14 | 68,7 | -1,56667 | | 0,00015 | | 17 | 64,0 | | 0,00812 | | | | | ---1,42857 | 0,01431 | | | 31 | 44.0 | | | | | 35 | 39,1 | -- 1,22500 | | |
Avem atunce
L ~ = 68 , 7 − 13 × 1 , 5667 + 130 × 0 , 00812 − 520 × 0 , 00015 = 49 , 31089 L ~ = 68 , 7 − 13 × 1 , 5667 + 130 × 0 , 00812 − 520 × 0 , 00015 = 49 , 31089 widetilde(L)=68,7-13 xx1,5667+130 xx0,00812-520 xx0,00015=49,31089\widetilde{L}=68,7-13 \times 1,5667+130 \times 0,00812-520 \times 0,00015=49,31089L~=68,7−13×1,5667+130×0,00812−520×0,00015=49,31089
si | λ | < 0 , 00368 | λ | < 0 , 00368 |lambda| < 0,00368|\lambda|<0,00368|λ|<0,00368,
deci
49 , 37021 < L ( 14 , 17 , 31 , 35 ; | | 27 ) < 49 , 31457 . 49 , 37021 < L ( 14 , 17 , 31 , 35 ; | | 27 ) < 49 , 31457 . 49,37021 < L(14,17,31,35;quad|quad|27) < 49,31457.49,37021<L(14,17,31,35 ; \quad|\quad| 27)<49,31457 .49,37021<L(14,17,31,35;||27)<49,31457.
In delul acesta polinomul (2) este obtinut cu trei zecimale exacte, ceea faţă de datele problemei este suficient.
1955

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