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T. Popoviciu, Introduction à la théorie des différences divisées, Bull. Math. de la Soc. Roum. des Sci., t. 42 (1940), pp. 65-78 (in French)
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Bulletin mathématique de la Société Roumaine des Sciences
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Societatea de Științe Matematice din România
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INTRODUCTION TO THE THEORY OF DIVIDED DIFFERENCES
BY
TIBERIU POPOVICIU
In what follows we propose to give the demonstration of several formulas which intervene in the theory of divided differences of functions of one variable. Most of these formulas are already exposed, without demonstration, in our Thesis^(1){ }^{1}), These formulas being in common use in the theory of higher order convex functions, it will not be useless to establish them here in all rigor. I decided to return to these questions by noting the appearance of some works where my results are not cited^(2){ }^{2}).
- The Lagrange polynomial. Considern+1n+1distinct pointsx_(1),x_(2),dots,x_(n+1)x_{1}, x_{2}, \ldots, x_{n+1}and eitherf=f(x)f=f(x)a uniform function defined on these points. We will assume that the pointsx_(i)x_{i}are on the real axis and thatffis a real function.
There exists one and only one polynomial of degree n which takes the valuesf(x_(i))\mathrm{f}\left(\mathrm{x}_{i}\right)to the points{:x_(i),i=1,2,dots,n+1^(3))\left.\mathrm{x}_{i}, \mathrm{i}=1,2, \ldots, \mathrm{n}+1{ }^{3}\right).
The existence of at least one polynomial satisfying the imposed conditions can be demonstrated by induction. Forn=0n=0the property is immediate. The constantf(x_(1))f\left(x_{1}\right)(polynomial of degree 0) satisfies the imposed condition. Suppose that the property is true fornnand we will demonstrate it forn+1n+1. There therefore exists, by hypothesis, at least one polynomialP(x)\mathrm{P}(x)of degreen-1n-1taking the valuesf(x_(i))f\left(x_{i}\right)to the pointsx_(i),i=1x_{i}, i=1,2,dots,n2, \ldots, n. We then see that the polynomial
P(x)+[f(x_(n+1))-P(x_(n+1))]((x-x_(1))(x-x_(2))dots(x-x_(n)))/((x_(n+1)-x_(1))(x_(n+1)-x_(2))dots(x_(n+1)-x_(n)))P(x)+\left[f\left(x_{n+1}\right)-P\left(x_{n+1}\right)\right] \frac{\left(x-x_{1}\right)\left(x-x_{2}\right) \ldots\left(x-x_{n}\right)}{\left(x_{n+1}-x_{1}\right)\left(x_{n+1}-x_{2}\right) \ldots\left(x_{n+1}-x_{n}\right)}
is of degreennand takes the valuesf(x_(i))f\left(x_{i}\right)to the pointsx_(i),i=1,2,dots,n+1x_{i}, i=1,2, \ldots, n+1.
Uniqueness is easily obtained. If we had two polynomialsP(x)\mathrm{P}(x)AndQ(x)\mathrm{Q}(x), not identical, of degreennand verifying the imposed conditions, the differenceP(x)-Q(x)\mathrm{P}(x)-\mathrm{Q}(x)would be zero forx=x_(1),x_(2),dots,x_(n+1)x=x_{1}, x_{2}, \ldots, x_{n+1}, which is impossible.
The unique polynomial, determined as we have seen, is called the Lagrange (interpolation) polynomial of the functionffon the pointsx_(i),i=1,2,dots,n+1\mathrm{x}_{i}, \mathrm{i}=1,2, \ldots, \mathrm{n}+1.
We will designate this polynomial by
{:(1)P(x_(1),x_(2),dots,x_(n+1);f∣x).:}\begin{equation*}
\mathrm{P}\left(x_{1}, x_{2}, \ldots, x_{n+1} ; f \mid x\right) . \tag{1}
\end{equation*}
The general form of polynomials taking the valuesf(x_(i))f\left(x_{i}\right)to the pointsx_(i),i=1,2,dots,n+1x_{i}, i=1,2, \ldots, n+1, East
P(x_(1),x_(2),dots,x_(n+1);f∣x)+(x-x_(1))(x-x_(2))dots(x-x_(n+1))Q(x)\mathrm{P}\left(x_{1}, x_{2}, \ldots, x_{n+1} ; f \mid x\right)+\left(x-x_{1}\right)\left(x-x_{2}\right) \ldots\left(x-x_{n+1}\right) \mathrm{Q}(x)
Q(x)\mathrm{Q}(x)being any polynomial^(4){ }^{4}).
The uniqueness of the polynomial (1) allows us to write the following well-known formulas,
The uniqueness of the polynomial (1) allows us to write the following well-known formulas,
P(x_(1),x_(2),dots,x_(n+1);f∣x)=\mathrm{P}\left(x_{1}, x_{2}, \ldots, x_{n+1} ; f \mid x\right)=
=sum_(i=1)^(n+1)f(x_(i))((x-x_(1))dots(x-x_(i-1))(x-x_(i+1))dots(x-x_(n+1)))/((x_(i)-x_(1))dots(x_(i)-x_(i-1))(x_(i)-x_(i+1))dots(x_(i)-x_(n+1)))=sum_(i=1)^(n+1)(f(x_(i))?(x))/(P^(')(x_(i))(x-x_(i)))=\sum_{i=1}^{n+1} f\left(x_{i}\right) \frac{\left(x-x_{1}\right) \ldots\left(x-x_{i-1}\right)\left(x-x_{i+1}\right) \ldots\left(x-x_{n+1}\right)}{\left(x_{i}-x_{1}\right) \ldots\left(x_{i}-x_{i-1}\right)\left(x_{i}-x_{i+1}\right) \ldots\left(x_{i}-x_{n+1}\right)}=\sum_{i=1}^{n+1} \frac{f\left(x_{i}\right) ?(x)}{\mathcal{P}^{\prime}\left(x_{i}\right)\left(x-x_{i}\right)},
P(x_(1),x_(2),dots,x_(n+1);f∣x)=-(|[1,x_(1),x_(1)^(2),dots,x_(1)^(n),f(x_(1))],[1,x_(2),x_(2)^(2),dots,x_(2)^(n),f(x_(2))],[dots dots,dots,dots,dots,dots,dots],[1,x_(n+1),x_(n+1)^(2),dots,x_(n+1)^(n),f(x_(n+1))],[1,x,x^(2),dots,x^(n),0]|)/(|[1,x_(1),x_(1)^(2),dots,x_(1)^(n)],[1,x_(2),x_(2)^(2),dots,x_(2)^(n)],[dots dots dots,dots,dots,dots],[1,x_(n+1),x_(n+1)^(2),dots,x_(n+1)^(n)]|)\mathrm{P}\left(x_{1}, x_{2}, \ldots, x_{n+1} ; f \mid x\right)=-\frac{\left|\begin{array}{cccccc}
1 & x_{1} & x_{1}^{2} & \ldots & x_{1}^{n} & f\left(x_{1}\right) \\
1 & x_{2} & x_{2}^{2} & \ldots & x_{2}^{n} & f\left(x_{2}\right) \\
\ldots \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\
1 & x_{n+1} & x_{n+1}^{2} & \ldots & x_{n+1}^{n} & f\left(x_{n+1}\right) \\
1 & x & x^{2} & \ldots & x^{n} & 0
\end{array}\right|}{\left|\begin{array}{ccccc}
1 & x_{1} & x_{1}^{2} & \ldots & x_{1}^{n} \\
1 & x_{2} & x_{2}^{2} & \ldots & x_{2}^{n} \\
\ldots \ldots \ldots & \ldots & \ldots & \ldots \\
1 & x_{n+1} & x_{n+1}^{2} & \ldots & x_{n+1}^{n}
\end{array}\right|}
where we posed
{:(2)f(x)=(x-x_(1))(x-x_(2))dots(x-x_(n+1)).:}\begin{equation*}
f(x)=\left(x-x_{1}\right)\left(x-x_{2}\right) \ldots\left(x-x_{n+1}\right) . \tag{2}
\end{equation*}
Let's pose
(3)
(3)
[x_(1),x_(2),dots,x_(n+1);f]=sum_(i=1)^(n+1)(f(x_(i)))/(P^(')(x_(i)))\left[x_{1}, x_{2}, \ldots, x_{n+1} ; f\right]=\sum_{i=1}^{n+1} \frac{f\left(x_{i}\right)}{\mathcal{P}^{\prime}\left(x_{i}\right)}
^(4){ }^{4}) The polynomial (1) can also be characterized by the following extremal property. There exists one and only one polynomial of minimum effective degree which takes the valuesf(x_(i))f\left(x_{i}\right), Forx=x_(i),i=1,2,dots,n+1x=x_{i}, i=1,2, \ldots, n+1.
which is the coefficient ofx^(n)x^{n}in the polynomial (1). This polynomial can then be written in the following forms^(5){ }^{5})
which is the coefficient ofx^(n)x^{n}in the polynomial (1). This polynomial can then be written in the following forms^(5){ }^{5})
{:[P(x_(1),x_(2),dots,x_(n);f∣x)+[x_(1),x_(2),dots,x_(n+1);f](varphi(x))/(x-x_(n+1))],[P(x_(2),x_(3),dots,x_(n+1);f∣x)+[x_(1),x_(2),dots,x_(n+1);f](varphi(x))/(x-x_(1))]:}\begin{aligned}
& \mathrm{P}\left(x_{1}, x_{2}, \ldots, x_{n} ; f \mid x\right)+\left[x_{1}, x_{2}, \ldots, x_{n+1} ; f\right] \frac{\varphi(x)}{x-x_{n+1}} \\
& \mathrm{P}\left(x_{2}, x_{3}, \ldots, x_{n+1} ; f \mid x\right)+\left[x_{1}, x_{2}, \ldots, x_{n+1} ; f\right] \frac{\varphi(x)}{x-x_{1}}
\end{aligned}
from which results the recurrence relation of Lagrange polynomials
{:[(4)P(x_(1),x_(2),dots,x_(n+1);f∣x)=],[=((x-x_(1))P(x_(2),x_(3),dots,x_(n+1);f∣x)-(x-x_(n+1))P(x_(1),x_(2),dots,x_(n);f∣x))/(x_(n+1)-x_(1))]:}\begin{gather*}
\mathrm{P}\left(x_{1}, x_{2}, \ldots, x_{n+1} ; f \mid x\right)= \tag{4}\\
=\frac{\left(x-x_{1}\right) \mathrm{P}\left(x_{2}, x_{3}, \ldots, x_{n+1} ; f \mid x\right)-\left(x-x_{n+1}\right) \mathrm{P}\left(x_{1}, x_{2}, \ldots, x_{n} ; f \mid x\right)}{x_{n+1}-x_{1}}
\end{gather*}
- Divided differences. The divided difference of order n of the function f on the pointsx_(1),x_(2),dots,x_(n+1)\mathrm{x}_{1}, \mathrm{x}_{2}, \ldots, \mathrm{x}_{n+1}is characterized by the following properties:
a) It is a linear and homogeneous expression with respect tof(x_(1)),f(x_(2)),dots,f(x_(n+1))\mathrm{f}\left(\mathrm{x}_{1}\right), \mathrm{f}\left(\mathrm{x}_{2}\right), \ldots, \mathrm{f}\left(\mathrm{x}_{n+1}\right), with coefficients independent of the function f .
b) It is zero identically for the functionsf=1,x,x^(2),dots\mathrm{f}=1, \mathrm{x}, \mathrm{x}^{2}, \ldots,x^(n+1)x^{n+1}.
c) It reduces to 1 identically for the functionf=x^(n)\mathrm{f}=\mathrm{x}^{n}.
The expression determined by these conditionsaa),bb),cc), is unique since the system
sum_(i=1)^(n+1)lambda_(i)x_(i)^(m)={[0",",m=0","1","dots","n-1","],[1",",m=n","]:}\sum_{i=1}^{n+1} \lambda_{i} x_{i}^{m}= \begin{cases}0, & m=0,1, \ldots, n-1, \\ 1, & m=n,\end{cases}
is a system ofC_("Ramer ")\mathrm{C}_{\text {Ramer }}inlambda_(1),lambda_(2),dots,lambda_(n+1)\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n+1}The determinant of this system is, in fact, the Vandermonde determinant
V(x_(1),x_(2)dots,x_(n+1))=prod_(i < j)(x_(j)-x_(i))\mathrm{V}\left(x_{1}, x_{2} \ldots, x_{n+1}\right)=\prod_{i<j}\left(x_{j}-x_{i}\right)
distinct numbersx_(i)x_{i}.
Note that expression (3) verifies the propertiesaa),bb),cc) since
Note that expression (3) verifies the propertiesaa),bb),cc) since
P(x_(1),x_(2),dots,x_(n+1);x^(i)∣x)=x^(i),i=0,1,dots,n-1,\mathrm{P}\left(x_{1}, x_{2}, \ldots, x_{n+1} ; x^{i} \mid x\right)=x^{i}, i=0,1, \ldots, n-1,
it is therefore the value of the divided difference of f on the pointsx_(l)\mathrm{x}_{l}.
Let us designate byU(x_(1),x_(2),dots,x_(n+1);f)\mathrm{U}\left(x_{1}, x_{2}, \ldots, x_{n+1} ; f\right)the determinant that we obtain fromV(x_(1),x_(2),dots,x_(n+1))V\left(x_{1}, x_{2}, \ldots, x_{n+1}\right)if we replace the elementsx_(i)^(n)x_{i}^{n}byf(x_(i))f\left(x_{i}\right)respectively. The divided difference (3) is then written
{:(5)[x_(1),x_(2),dots,x_(n+1);f]=(U(x_(1),x_(2),dots,x_(n+1);f))/(V(x_(1),x_(2),dots,x_(n+1))):}\begin{equation*}
\left[x_{1}, x_{2}, \ldots, x_{n+1} ; f\right]=\frac{\mathrm{U}\left(x_{1}, x_{2}, \ldots, x_{n+1} ; f\right)}{\mathrm{V}\left(x_{1}, x_{2}, \ldots, x_{n+1}\right)} \tag{5}
\end{equation*}
in the form of the quotient of two determinants. We see that the divided difference (5) is symmetrical with respect to the pointsx_(i)\mathrm{x}_{i}.
Formula (4) gives us the recurrence formula for divided differences
{:[(6)[x_(1),x_(2),dots,x_(n+1);f]=([x_(2),x_(3),dots,x_(n+1);f]-[x_(1),x_(2),dots,x_(n);f])/(x_(n+1)-x_(1))],[[x_(1);f]=f(x_(1))]:}\begin{gather*}
{\left[x_{1}, x_{2}, \ldots, x_{n+1} ; f\right]=\frac{\left[x_{2}, x_{3}, \ldots, x_{n+1} ; f\right]-\left[x_{1}, x_{2}, \ldots, x_{n} ; f\right]}{x_{n+1}-x_{1}}} \tag{6}\\
{\left[x_{1} ; f\right]=f\left(x_{1}\right)}
\end{gather*}
which can also be used to define them and which justifies their name.
We can therefore write
We can therefore write
{:(7)[x_(1),x_(2),dots,x_(n+1);x^(m)]={[0",",m=0","1","dots","n-1],[1",",m=n]:}:}\left[x_{1}, x_{2}, \ldots, x_{n+1} ; x^{m}\right]= \begin{cases}0, & m=0,1, \ldots, n-1 \tag{7}\\ 1, & m=n\end{cases}
We have
{:(8)[x_(1),x_(2),dots,x_(n+1);Cf]=C[x_(1),x_(2),dots,x_(n+1);f]:}\begin{equation*}
\left[x_{1}, x_{2}, \ldots, x_{n+1} ; \mathrm{C} f\right]=\mathrm{C}\left[x_{1}, x_{2}, \ldots, x_{n+1} ; f\right] \tag{8}
\end{equation*}
(9)quad[x_(1),x_(2)dots,x_(n+1);f+g]=[x_(1),x_(2),dots,x_(n+1);f]+[x_(1),x_(2),dots,x_(n+1);g]\quad\left[x_{1}, x_{2} \ldots, x_{n+1} ; f+g\right]=\left[x_{1}, x_{2}, \ldots, x_{n+1} ; f\right]+\left[x_{1}, x_{2}, \ldots, x_{n+1} ; g\right]
C being a constant andf,gf, gtwo functions defined on the pointsx_(", ")i=1,2,dots,n+1x_{\text {, }} i=1,2, \ldots, n+1.
If{f_(m)}\left\{f_{m}\right\}is a convergent sequence of functions defined on the pointsx_(i)x_{i}, we obviously have
lim_(m rarr oo)[x_(1),x_(2),dots,x_(n+1);f_(m)]=[x_(1),x_(2),dots,x_(n+1);lim_(m rarr oo)f_(m)]\lim _{m \rightarrow \infty}\left[x_{1}, x_{2}, \ldots, x_{n+1} ; f_{m}\right]=\left[x_{1}, x_{2}, \ldots, x_{n+1} ; \lim _{m \rightarrow \infty} f_{m}\right]
so, if the seriessum_(m=0)^(oo)f_(m)\sum_{m=0}^{\infty} f_{m}converges, we have
sum_(m=0)^(oo)[x_(1),x_(2),dots,x_(n+1);f_(m)]=[x_(1),x_(2)dots,x_(n+1);sum_(m=0)^(oo)f_(m)]\sum_{m=0}^{\infty}\left[x_{1}, x_{2}, \ldots, x_{n+1} ; f_{m}\right]=\left[x_{1}, x_{2} \ldots, x_{n+1} ; \sum_{m=0}^{\infty} f_{m}\right]
From the above it follows that the divided difference of ordernnof a polynomial of degreen-1n-1is still zero.
Let us note again the formula
f(x)-P(x_(1),x_(2),dots,x_(n+1);f∣x)=varphi(x)[x_(1),x_(2),dots,x_(n+1),x;f]f(x)-\mathrm{P}\left(x_{1}, x_{2}, \ldots, x_{n+1} ; f \mid x\right)=\varphi(x)\left[x_{1}, x_{2}, \ldots, x_{n+1}, x ; f\right]
which is a simple relationship between the Lagrange polynomial and the divided difference.
Application. Let us calculate the divided differences (7) formmentire> n>n. Without restricting the generality we can assumex_(i)!=0,i=1,2dotsx_{i} \neq 0, i=1,2 \ldots,n+1n+1and either|z| < min((1)/(|x_(1)|),(1)/(|x_(2)|),dots,(1)/(|x_(n+1)|)),z|z|<\min \left(\frac{1}{\left|x_{1}\right|}, \frac{1}{\left|x_{2}\right|}, \ldots, \frac{1}{\left|x_{n+1}\right|}\right), zbeing a parameter independent ofxx. We have
{:[(z^(n))/((1-zx_(1))(1-zx_(2))dots(1-zx_(n+1)))=sum_(i=1)^(n+1)(1)/((1-zx_(i))^(')(o)/(q)^(')(x_(i)))=],[=[x_(1),x_(2),dots,x_(n+1);(1)/(1-zx)]]:}\begin{gathered}
\frac{z^{n}}{\left(1-z x_{1}\right)\left(1-z x_{2}\right) \ldots\left(1-z x_{n+1}\right)}=\sum_{i=1}^{n+1} \frac{1}{\left(1-z x_{i}\right)^{\prime} \frac{o}{q}^{\prime}\left(x_{i}\right)}= \\
=\left[x_{1}, x_{2}, \ldots, x_{n+1} ; \frac{1}{1-z x}\right]
\end{gathered}
But,
(1)/(1-zx)=(z^(n)x^(n))/(1-zx)+1+zx+z^(2)x^(2)+dots+z^(n-1)x^(n-1)\frac{1}{1-z x}=\frac{z^{n} x^{n}}{1-z x}+1+z x+z^{2} x^{2}+\ldots+z^{n-1} x^{n-1}
and taking into account (7), (8), (9),
[x_(1),x_(2),dots,x_(n+1);(1)/(1-zx)]=z^(n)[x_(1),x_(2),dots,x_(n+1);(x^(n))/(1-zx)]\left[x_{1}, x_{2}, \ldots, x_{n+1} ; \frac{1}{1-z x}\right]=z^{n}\left[x_{1}, x_{2}, \ldots, x_{n+1} ; \frac{x^{n}}{1-z x}\right]
If we therefore pose
(10)(1)/((1-zx_(1))(1-zx_(2))dots(1cdots zx_(n+1)))=S_(0)+S_(1)z+S_(2)z^(2)+dots\frac{1}{\left(1-z x_{1}\right)\left(1-z x_{2}\right) \ldots\left(1 \cdots z x_{n+1}\right)}=\mathrm{S}_{0}+\mathrm{S}_{1} z+\mathrm{S}_{2} z^{2}+\ldotsWe have
(10)(1)/((1-zx_(1))(1-zx_(2))dots(1cdots zx_(n+1)))=S_(0)+S_(1)z+S_(2)z^(2)+dots\frac{1}{\left(1-z x_{1}\right)\left(1-z x_{2}\right) \ldots\left(1 \cdots z x_{n+1}\right)}=\mathrm{S}_{0}+\mathrm{S}_{1} z+\mathrm{S}_{2} z^{2}+\ldotsWe have
S_(1)=[x_(1),x_(2),dots,x_(n+1);x^(i+n)]\mathrm{S}_{1}=\left[x_{1}, x_{2}, \ldots, x_{n+1} ; x^{i+n}\right]
If we write that the first member of (10) is equal to the product of the developments
(1)/(1-zx_(i))=1+zx_(i)+z^(2)x_(i)^(2)+dots+z^(m)x_(i)^(m)+dots,i=1,2,dots,n+1\frac{1}{1-z x_{i}}=1+z x_{i}+z^{2} x_{i}^{2}+\ldots+z^{m} x_{i}^{m}+\ldots, i=1,2, \ldots, n+1
we find
S_(i)=[x_(1),x_(2),dots,x_(n+1);x^(i+n)]=Sigmax_(1)^(alpha_(1))x_(2)^(alpha_(2))dotsx_(n+1)^(alpha_(n+1))\mathrm{S}_{i}=\left[x_{1}, x_{2}, \ldots, x_{n+1} ; x^{i+n}\right]=\Sigma x_{1}^{\alpha_{1}} x_{2}^{\alpha_{2}} \ldots x_{n+1}^{\alpha_{n+1}}
the summation being extended to all the whole and positive solutions of the equationalpha_(1)+alpha_(2)+dots+alpha_(n+1)=i\alpha_{1}+\alpha_{2}+\ldots+\alpha_{n+1}=i.
3. The fundamental formula for transforming divided differences. Now consider the functionffset tommdistinct points
3. The fundamental formula for transforming divided differences. Now consider the functionffset tommdistinct points
{:(11)x_(1)","x_(2)","dots","x_(m)quad(m >= n+1):}\begin{equation*}
x_{1}, x_{2}, \ldots, x_{m} \quad(m \geqq n+1) \tag{11}
\end{equation*}
To simplify the notations we will set
{:[(12)Delta_(j)^(i)(f)=[x_(i),x_(i+1),dots,x_(i+j);f]","Delta_(0)^(i)(f)=f(x_(i))","],[(13)xi_(i,j+1)(x)=(x-x_(i))(x-x_(i+1))dots(x-x_(i+j))","quadxi_(i,0)(x)=1","],[i=1","2","dots","m-j","quad j=0","1","dots","m-1.]:}\begin{align*}
\Delta_{j}^{i}(f) & =\left[x_{i}, x_{i+1}, \ldots, x_{i+j} ; f\right], \Delta_{0}^{i}(f)=f\left(x_{i}\right), \tag{12}\\
\xi_{i, j+1}(x) & =\left(x-x_{i}\right)\left(x-x_{i+1}\right) \ldots\left(x-x_{i+j}\right), \quad \xi_{i, 0}(x)=1, \tag{13}\\
i & =1,2, \ldots, m-j, \quad j=0,1, \ldots, m-1 .
\end{align*}
So_(1,n+1)(x){ }_{1, n+1}(x)is precisely the polynomial (2) already considered. With these notations we have
Delta_(j)^(i)(f)=sum_(r=i)^(i⊬j)(f(x_(r)))/(varphi_(i,j+1)^(')(x_(r)))\Delta_{j}^{i}(f)=\sum_{r=i}^{i \nvdash j} \frac{f\left(x_{r}\right)}{\varphi_{i, j+1}^{\prime}\left(x_{r}\right)}
We will now consider a linear and homogeneous expression off(x_(i))f\left(x_{i}\right)of the form
F=sum_(i=1)^(m)lambda_(i)f(x_(i))\mathrm{F}=\sum_{i=1}^{m} \lambda_{i} f\left(x_{i}\right)
THElambda_(i)\lambda_{i}being independent of the functionff.
We can always put such an expression in the form
We can always put such an expression in the form
F=sum_(i=1)^(n)mu_(i)Delta_(i-1)^(1)(f)+sum_(i=1)^(m-n)v_(i)Delta_(n)^(i)(f)\mathrm{F}=\sum_{i=1}^{n} \mu_{i} \Delta_{i-1}^{1}(f)+\sum_{i=1}^{m-n} v_{i} \Delta_{n}^{i}(f)
where themu_(l)\mu_{l}and thenu_(i)\nu_{i}do not depend on the functiontt. These coefficients are completely determined by the coefficientslambda_(r)\lambda_{r}. Indeed, identification leads us to a linear system ofmmequations with respect tommunknownsmu_(l),nu_(l)\mu_{l}, \nu_{l}. The determinant of this system
(1)/(varphi_(1,2)^(')(x_(2))varphi_(1,3)^(')(x_(3))dotsvarphi_(1,n)^(')(x_(n))varphi_(1,n+1)^(')(x_(n+1))varphi_(2,n+1)^(')(x_(n+2))dotsvarphi_(m-n,n+1)^(')(x_(m)))\frac{1}{\varphi_{1,2}^{\prime}\left(x_{2}\right) \varphi_{1,3}^{\prime}\left(x_{3}\right) \ldots \varphi_{1, n}^{\prime}\left(x_{n}\right) \varphi_{1, n+1}^{\prime}\left(x_{n+1}\right) \varphi_{2, n+1}^{\prime}\left(x_{n+2}\right) \ldots \varphi_{m-n, n+1}^{\prime}\left(x_{m}\right)}is different from zero.
(1)/(varphi_(1,2)^(')(x_(2))varphi_(1,3)^(')(x_(3))dotsvarphi_(1,n)^(')(x_(n))varphi_(1,n+1)^(')(x_(n+1))varphi_(2,n+1)^(')(x_(n+2))dotsvarphi_(m-n,n+1)^(')(x_(m)))\frac{1}{\varphi_{1,2}^{\prime}\left(x_{2}\right) \varphi_{1,3}^{\prime}\left(x_{3}\right) \ldots \varphi_{1, n}^{\prime}\left(x_{n}\right) \varphi_{1, n+1}^{\prime}\left(x_{n+1}\right) \varphi_{2, n+1}^{\prime}\left(x_{n+2}\right) \ldots \varphi_{m-n, n+1}^{\prime}\left(x_{m}\right)}is different from zero.
It remains to be seen how the coefficients are determined.mu_(i),y_(i)\mu_{i}, y_{i}We will obtain them by choosing the function appropriately.ff.
To have the coefficientsmu_(i)\mu_{i}, let's take forffthe polynomialvarphi_(1,j-1)=varphi_(1,j-1)(x)\varphi_{1, j-1}=\varphi_{1, j-1}(x). We then have
{:[Delta_(i-1)^(1)(varphi_(1,j-1))={[0",",i=1","2","dots","j-1],[1",",i=j],[0",",i=j+1","j+2","dots","n]:}],[Delta_(n)^(t)(ℑ_(1,j-1))=0","i=1","2","dots","m-n]:}\begin{aligned}
\Delta_{i-1}^{1}\left(\varphi_{1, j-1}\right) & = \begin{cases}0, & i=1,2, \ldots, j-1 \\
1, & i=j \\
0, & i=j+1, j+2, \ldots, n\end{cases} \\
\Delta_{n}^{t}\left(\Im_{1, j-1}\right) & =0, i=1,2, \ldots, m-n
\end{aligned}
SO
{:[(14)mu_(j)=sum_(i=1)^(m)lambda_(i)f_(1,j-1)(x_(l))=sum_(i=j)^(m)lambda_(i)(x_(i)-x_(1))(x_(i)-x_(2))dots(x_(i)-x_(j-1))],[j=1","2","dots","n.]:}\begin{gather*}
\mu_{j}=\sum_{i=1}^{m} \lambda_{i} f_{1, j-1}\left(x_{l}\right)=\sum_{i=j}^{m} \lambda_{i}\left(x_{i}-x_{1}\right)\left(x_{i}-x_{2}\right) \ldots\left(x_{i}-x_{j-1}\right) \tag{14}\\
j=1,2, \ldots, n .
\end{gather*}
To obtain the coefficients%_(i)\%_{i}, we take forffthe following functions
{:(15)f_(j)^(**)=f_(j)^(**)(x)={[0","," pour "x=x_(1)","x_(2)","dots","x_(j+n-1)],[varphi_(j+1,n-1)(x)","," pour "x=x_(j+n)","x_(j+n+1)","dots","x_(m)],[j=1","2","dots",",m-n]:}:}\begin{align*}
f_{j}^{*}=f_{j}^{*}(x)= & \begin{cases}0, & \text { pour } x=x_{1}, x_{2}, \ldots, x_{j+n-1} \\
\varphi_{j+1, n-1}(x), & \text { pour } x=x_{j+n}, x_{j+n+1}, \ldots, x_{m} \\
j=1,2, \ldots, & m-n\end{cases} \tag{15}
\end{align*}
We have
{:[Delta_(i-1)^(1)(f_(j)^(**))=0","i=1","2","dots","n],[Delta_(n)^(i)(f_(j)^(**))={[0",",i=1","2","dots","j-1],[Delta_(n)^(j)(f_(j)^(**))",",i=j],[0",",i=j+1","j+2","dots","m-n]:}]:}\begin{aligned}
\Delta_{i-1}^{1}\left(f_{j}^{*}\right) & =0, i=1,2, \ldots, n \\
\Delta_{n}^{i}\left(f_{j}^{*}\right) & = \begin{cases}0, & i=1,2, \ldots, j-1 \\
\Delta_{n}^{j}\left(f_{j}^{*}\right), & i=j \\
0, & i=j+1, j+2, \ldots, m-n\end{cases}
\end{aligned}
SO
F=y_(j)Delta_(n)^(j)(f_(j)^(**))=(f_(j)^(**)(x_(j+n)))/((x_(j+n)-x_(j))(x_(j+n)-x_(j+1))dots(x_(j+n)-x_(j+n-1)))=(v_(j))/(x_(j+n)-x_(j))\mathrm{F}=y_{j} \Delta_{n}^{j}\left(f_{j}^{*}\right)=\frac{f_{j}^{*}\left(x_{j+n}\right)}{\left(x_{j+n}-x_{j}\right)\left(x_{j+n}-x_{j+1}\right) \ldots\left(x_{j+n}-x_{j+n-1}\right)}=\frac{v_{j}}{x_{j+n}-x_{j}}.
We deduce from this
F=y_(j)Delta_(n)^(j)(f_(j)^(**))=(f_(j)^(**)(x_(j+n)))/((x_(j+n)-x_(j))(x_(j+n)-x_(j+1))dots(x_(j+n)-x_(j+n-1)))=(v_(j))/(x_(j+n)-x_(j))\mathrm{F}=y_{j} \Delta_{n}^{j}\left(f_{j}^{*}\right)=\frac{f_{j}^{*}\left(x_{j+n}\right)}{\left(x_{j+n}-x_{j}\right)\left(x_{j+n}-x_{j+1}\right) \ldots\left(x_{j+n}-x_{j+n-1}\right)}=\frac{v_{j}}{x_{j+n}-x_{j}}.
We deduce from this
{:[(16)y_(j)=(x_(j+n)-x_(j))sum_(i=1)^(m)lambda_(i)f_(j)^(**)(x_(i))=(x_(j+n)-x_(j))sum_(i=j+n)^(m)lambda_(i)varphi_(j+1,n-1)(x_(i))=],[=(x_(j+n)-x_(j))sum_(i=j+n)^(m)lambda_(i)(x_(i)-x_(j+1))(x_(i)-x_(j+2))dots(x_(i)-x_(j+n-1)).]:}\begin{align*}
y_{j} & =\left(x_{j+n}-x_{j}\right) \sum_{i=1}^{m} \lambda_{i} f_{j}^{*}\left(x_{i}\right)=\left(x_{j+n}-x_{j}\right) \sum_{i=j+n}^{m} \lambda_{i} \varphi_{j+1, n-1}\left(x_{i}\right)= \tag{16}\\
& =\left(x_{j+n}-x_{j}\right) \sum_{i=j+n}^{m} \lambda_{i}\left(x_{i}-x_{j+1}\right)\left(x_{i}-x_{j+2}\right) \ldots\left(x_{i}-x_{j+n-1}\right) .
\end{align*}
The fundamental transformation formula is therefore finally written in the following form
{:[(17)sum_(i=1)^(m)lambda_(i)f(x_(i))=sum_(j=1)^(n)[sum_(i=j)^(m)lambda_(i)rho_(1,j-1)(x_(i))]Delta_(j-1)^(1)(f)+],[+sum_(j=1)^(m-n)[(x_(j+n)-x_(j))sum_(i=j+n)^(m)lambda_(i)varphi_(j+1,n-1)(x_(l))]Delta_(n)^(j)(f)]:}\begin{align*}
& \sum_{i=1}^{m} \lambda_{i} f\left(x_{i}\right)=\sum_{j=1}^{n}\left[\sum_{i=j}^{m} \lambda_{i} \rho_{1, j-1}\left(x_{i}\right)\right] \Delta_{j-1}^{1}(f)+ \tag{17}\\
& +\sum_{j=1}^{m-n}\left[\left(x_{j+n}-x_{j}\right) \sum_{i=j+n}^{m} \lambda_{i} \varphi_{j+1, n-1}\left(x_{l}\right)\right] \Delta_{n}^{j}(f)
\end{align*}
For any natural numbernnwe have such a formula. The coefficientsmu_(i),nu_(i)\mu_{i}, \nu_{i}can also be obtained by recurrence relations. If we denote bymu_(i)^((n)),v_(i)^((n))\mu_{i}^{(n)}, v_{i}^{(n)}these coefficients, to highlight the numbernn, We have
{:[mu_(j)^((n))=mu_(j)^((n-1))","j=1","2","dots","n-1],[mu_(n)^((n))=v_(1)^((n-1))+v_(2)^((n-1))+dots dots+v_(m-n+1)^((n-1))],[v_(j)^((n))=(x_(j+n)-x_(j))[v_(j+1)^((n-1))+v_(j+2)^((n-1))+dots+v_(m-n+1)^((n-1))]],[j=1","2","dots","m-n]:}\begin{gathered}
\mu_{j}^{(n)}=\mu_{j}^{(n-1)}, j=1,2, \ldots, n-1 \\
\mu_{n}^{(n)}=v_{1}^{(n-1)}+v_{2}^{(n-1)}+\ldots \ldots+v_{m-n+1}^{(n-1)} \\
v_{j}^{(n)}=\left(x_{j+n}-x_{j}\right)\left[v_{j+1}^{(n-1)}+v_{j+2}^{(n-1)}+\ldots+v_{m-n+1}^{(n-1)}\right] \\
j=1,2, \ldots, m-n
\end{gathered}
4.--Some applications of the fundamental formula.
I. Letf=f(x),g=g(x)f=f(x), g=g(x)two functions defined on the pointsx_(i)x_{i}. Consider the divided difference
F=[x_(1),x_(2),dots,x_(n+1);fg]\mathrm{F}=\left[x_{1}, x_{2}, \ldots, x_{n+1} ; f g\right]
of the productfgf g. In this case we have
lambda_(i)=(g(x_(i)))/(g_(1n+1)^(')(x_(i))),i=1,2,dots,n+1,lambda_(i)=0,i > n+1.\lambda_{i}=\frac{g\left(x_{i}\right)}{g_{1 n+1}^{\prime}\left(x_{i}\right)}, i=1,2, \ldots, n+1, \lambda_{i}=0, i>n+1 .
If we notice that^(')_(1,n+1)=varphi_(1,j-1)*varphi_(j,n-j+2){ }^{\prime}{ }_{1, n+1}=\varphi_{1, j-1} \cdot \varphi_{j, n-j+2}, we deduce from this
varphi_(1,j-1)(x_(i))=(varphi_(1,n+1)^(')(x_(i)))/(varphi_(j,n-j+2)^(')(x_(i)))", pour "j <= i <= n+1\varphi_{1, j-1}\left(x_{i}\right)=\frac{\varphi_{1, n+1}^{\prime}\left(x_{i}\right)}{\varphi_{j, n-j+2}^{\prime}\left(x_{i}\right)} \text {, pour } j \leqq i \leqq n+1
and formula (14) gives us
mu_(j)=sum_(i=j)^(n+1)(g(x_(i)))/(psi_(j,n-j+2)(x_(i)))=[x_(j),x_(j+1),dots,x_(n+1);g],j=1,2,dots,n.\mu_{j}=\sum_{i=j}^{n+1} \frac{g\left(x_{i}\right)}{\psi_{j, n-j+2}\left(x_{i}\right)}=\left[x_{j}, x_{j+1}, \ldots, x_{n+1} ; g\right], j=1,2, \ldots, n .
Formula (16) gives us
v_(1)=g(x_(n+1)),v_(j)=0,j=2,3,dotsv_{1}=g\left(x_{n+1}\right), v_{j}=0, j=2,3, \ldots
As a result, we have the following formula
(18)[x_(1),x_(2),dots,x_(n+1);fg]=sum_(i=1)^(n+1)[x_(1),x_(2),dots,x_(i);f]|__x_(i),x_(i+1),dotsx_(n+1);g]\left[x_{1}, x_{2}, \ldots, x_{n+1} ; f g\right]=\sum_{i=1}^{n+1}\left[x_{1}, x_{2}, \ldots, x_{i} ; f\right]\left\lfloor x_{i}, x_{i+1}, \ldots x_{n+1} ; g\right], which gives the divided difference of a product. This is the generalization of Leibniz's formula.
II. The Lagrange polynomial (1) is of the form F. A simple calculation shows us that in this case
(18)[x_(1),x_(2),dots,x_(n+1);fg]=sum_(i=1)^(n+1)[x_(1),x_(2),dots,x_(i);f]|__x_(i),x_(i+1),dotsx_(n+1);g]\left[x_{1}, x_{2}, \ldots, x_{n+1} ; f g\right]=\sum_{i=1}^{n+1}\left[x_{1}, x_{2}, \ldots, x_{i} ; f\right]\left\lfloor x_{i}, x_{i+1}, \ldots x_{n+1} ; g\right], which gives the divided difference of a product. This is the generalization of Leibniz's formula.
II. The Lagrange polynomial (1) is of the form F. A simple calculation shows us that in this case
lambda_(i)=(S_(1,n+1)(x))/((x-x_(i))P_(1,n+1)^(')(x_(i))),i=1,2,dots,n+1,lambda_(i)=0,i > n+1.\lambda_{i}=\frac{\mathrm{S}_{1, n+1}(x)}{\left(x-x_{i}\right) \mathrm{P}_{1, n+1}^{\prime}\left(x_{i}\right)}, i=1,2, \ldots, n+1, \lambda_{i}=0, i>n+1 .
Formula (14) gives us
mu_(j)=P(x_(1),x_(2),dots,x_(n+1);varphi_(1,j-1)∣x)=varphi_(1,j-1)(x),j=1,2,dots,n\mu_{j}=\mathrm{P}\left(x_{1}, x_{2}, \ldots, x_{n+1} ; \varphi_{1, j-1} \mid x\right)=\varphi_{1, j-1}(x), \mathrm{j}=1,2, \ldots, n
And
{:[y_(1)=(x_(n+1)-x_(1))P(x_(1),x_(2),dots,x_(n+1);f_(1)^(**)∣x)=],[=(x_(n+1)-x_(1))_(**_(1,n))(x)(f_(1)^(**)(x_(n+1)))/(xi_(1,n+1)^(')(x_(n+1)))=xi_(1,n)(x)","y_(i)=0","i > 1]:}\begin{gathered}
y_{1}=\left(x_{n+1}-x_{1}\right) \mathrm{P}\left(x_{1}, x_{2}, \ldots, x_{n+1} ; f_{1}^{*} \mid x\right)= \\
=\left(x_{n+1}-x_{1}\right){ }_{*_{1, n}}(x) \frac{f_{1}^{*}\left(x_{n+1}\right)}{\xi_{1, n+1}^{\prime}\left(x_{n+1}\right)}=\xi_{1, n}(x), y_{i}=0, i>1
\end{gathered}
and we get the well-known interpolation formula
{:[P(x_(1),x_(2),dots,x_(n+1);f∣x)=f(x_(1))+(x-x_(1))[x_(1),x_(2);f]+],[+(x-x_(1))(x-x_(2))[x_(1),x_(2),x_(3);f]+dots],[+(x-x_(1))(x-x_(2))dots(x-x_(n))[x_(1),x_(2),dots,x_(n+1);f]]:}\begin{gathered}
\mathrm{P}\left(x_{1}, x_{2}, \ldots, x_{n+1} ; f \mid x\right)=f\left(x_{1}\right)+\left(x-x_{1}\right)\left[x_{1}, x_{2} ; f\right]+ \\
+\left(x-x_{1}\right)\left(x-x_{2}\right)\left[x_{1}, x_{2}, x_{3} ; f\right]+\ldots \\
+\left(x-x_{1}\right)\left(x-x_{2}\right) \ldots\left(x-x_{n}\right)\left[x_{1}, x_{2}, \ldots, x_{n+1} ; f\right]
\end{gathered}
III. Let us take
F=[x_(n+1),x_(n+2),dots,x_(2n);f]-[x_(1),x_(2),dots,x_(n);f].\mathrm{F}=\left[x_{n+1}, x_{n+2}, \ldots, x_{2 n} ; f\right]-\left[x_{1}, x_{2}, \ldots, x_{n} ; f\right] .
It is easily verified thatmu_(j)=0,j=1,2,dots,n\mu_{j}=0, j=1,2, \ldots, n. We find
{:[y_(j)=(x_(j+n)-x_(j)){[x_(n+1),x_(n+2),dots,x_(2n);f_(j)^(**)]-[x_(1),x_(2),.,x_(n);f_(j)^(**)]}=],[=(x_(j+n)-x_(n))[x_(n+1),x_(n+2),dots,x_(2n);P_(j+1,n-1)]=(x_(j+n)-x_(j))]:}\begin{aligned}
y_{j}= & \left(x_{j+n}-x_{j}\right)\left\{\left[x_{n+1}, x_{n+2}, \ldots, x_{2 n} ; f_{j}^{*}\right]-\left[x_{1}, x_{2}, ., x_{n} ; f_{j}^{*}\right]\right\}= \\
= & \left(x_{j+n}-x_{n}\right)\left[x_{n+1}, x_{n+2}, \ldots, x_{2 n} ; \mathscr{P}_{j+1, n-1}\right]=\left(x_{j+n}-x_{j}\right)
\end{aligned}
and we have, with the notations (12), the following formula
{:(19)Delta_(n-1)^(n+1)(t)-Delta_(n-1)^(1)(f)=sum_(j=1)^(n)(x_(j+n)-x_(j))Delta_(n)^(j)(f):}\begin{equation*}
\Delta_{n-1}^{n+1}(t)-\Delta_{n-1}^{1}(f)=\sum_{j=1}^{n}\left(x_{j+n}-x_{j}\right) \Delta_{n}^{j}(f) \tag{19}
\end{equation*}
which can, moreover, be established just as simply using the recurrence formula (6)
IV. Let
IV. Let
x_(i_(1)),x_(i_(2)),dots,x_(i_(n+1))quad(1 <= i_(1) < i_(2) < dots < i_(n+1) <= m)x_{i_{1}}, x_{i_{2}}, \ldots, x_{i_{n+1}} \quad\left(1 \leqq i_{1}<i_{2}<\ldots<i_{n+1} \leqq m\right)
a partial sequence extracted from the sequence (11). Let us take
F=[x_(i_(1)),x_(i_(2)),dots,x_(i_(n+1));f]\mathrm{F}=\left[x_{i_{1}}, x_{i_{2}}, \ldots, x_{i_{n+1}} ; f\right]
The fordamental formula (17) then gives us
(20)[x_(i_(1)),x_(i_(2)),dots,x_(i_(n+1));f]=sum_(j=4)^(m-n)(x_(j+n)-x_(j))[x_(i_(1)),x_(i_(2)),dots,x_(i_(n+1));f_(j)^(**)]Delta_(n)^(j)(f)\left[x_{i_{1}}, x_{i_{2}}, \ldots, x_{i_{n+1}} ; f\right]=\sum_{j=4}^{m-n}\left(x_{j+n}-x_{j}\right)\left[x_{i_{1}}, x_{i_{2}}, \ldots, x_{i_{n+1}} ; f_{j}^{*}\right] \Delta_{n}^{j}(f).
(20)[x_(i_(1)),x_(i_(2)),dots,x_(i_(n+1));f]=sum_(j=4)^(m-n)(x_(j+n)-x_(j))[x_(i_(1)),x_(i_(2)),dots,x_(i_(n+1));f_(j)^(**)]Delta_(n)^(j)(f)\left[x_{i_{1}}, x_{i_{2}}, \ldots, x_{i_{n+1}} ; f\right]=\sum_{j=4}^{m-n}\left(x_{j+n}-x_{j}\right)\left[x_{i_{1}}, x_{i_{2}}, \ldots, x_{i_{n+1}} ; f_{j}^{*}\right] \Delta_{n}^{j}(f).
Remarks.1^(0)1^{0}. When F is identically zero for any polynomial of degreer-1r-1We havemu_(j)=0,j=1,2,dots,r\mu_{j}=0, j=1,2, \ldots, r.
2^(0)2^{0}. Formula (19) is valid even if thex_(i)x_{i}are not all distinct. It is sufficient that each of the sequences
2^(0)2^{0}. Formula (19) is valid even if thex_(i)x_{i}are not all distinct. It is sufficient that each of the sequences
{:[x_(1)","x_(2)","dots","x_(n);quadx_(n+1)","x_(n+2)","dots","x_(2n)","],[x_(j)","x_(j+1)","dots","x_(j+n)","quad j=1","2","dots","n","]:}\begin{aligned}
& x_{1}, x_{2}, \ldots, x_{n} ; \quad x_{n+1}, x_{n+2}, \ldots, x_{2 n}, \\
& x_{j}, x_{j+1}, \ldots, x_{j+n}, \quad j=1,2, \ldots, n,
\end{aligned}
is formed by distinct points. We deduce that if
alpha_(1),alpha_(2),dots,alpha_(j),beta_(1),beta_(2),dots,beta_(j),alpha_(j+1)=beta_(j+1),alpha_(j+2)=beta_(j+2),dots,alpha_(n)=beta_(n)\alpha_{1}, \alpha_{2}, \ldots, \alpha_{j}, \beta_{1}, \beta_{2}, \ldots, \beta_{j}, \alpha_{j+1}=\beta_{j+1}, \alpha_{j+2}=\beta_{j+2}, \ldots, \alpha_{n}=\beta_{n}
are distinct points, we have the following formula
{:[[alpha_(1),alpha_(2),dots,alpha_(n);f]-[beta_(1),beta_(2),dots,beta_(n);f]=],[quad=sum_(i=1)^(j)(alpha_(i)-beta_(i))[alpha_(1),alpha_(2),dots,alpha_(i),beta_(i),beta_(i+1),dots,beta_(n);f]]:}\begin{aligned}
& {\left[\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n} ; f\right]-\left[\beta_{1}, \beta_{2}, \ldots, \beta_{n} ; f\right]=} \\
& \quad=\sum_{i=1}^{j}\left(\alpha_{i}-\beta_{i}\right)\left[\alpha_{1}, \alpha_{2}, \ldots, \alpha_{i}, \beta_{i}, \beta_{i+1}, \ldots, \beta_{n} ; f\right]
\end{aligned}
- The formula for the mean of the divided differences. We will now assume that the points (11) are ordered, so that
x_(1) < x_(2) < dots < x_(m).x_{1}<x_{2}<\ldots<x_{m} .
The determinantV(x_(1),x_(2),dots,x_(n+1))\mathrm{V}\left(x_{1}, x_{2}, \ldots, x_{n+1}\right)then has a positive value and if we set
sg alpha={[-1","quad alpha < 0","],[0","quad alpha=0","],[1","quad alpha > 0","]:}\operatorname{sg} \alpha=\left\{\begin{array}{r}
-1, \quad \alpha<0, \\
0, \quad \alpha=0, \\
1, \quad \alpha>0,
\end{array}\right.
We have
sg[x_(1),x_(2),dots,x_(n+1);f]=sgU(x_(1),x_(2),dotsx_(n+1);f)s g\left[x_{1}, x_{2}, \ldots, x_{n+1} ; f\right]=\operatorname{sgU}\left(x_{1}, x_{2}, \ldots x_{n+1} ; f\right)
Let us then take up formula (20). The followingx_(l_(1)),x_(i_(2)),dots,x_(i_(n+1))x_{l_{1}}, x_{i_{2}}, \ldots, x_{i_{n+1}}is also ordered and we havei_(n+1)-i_(1)=p >= ni_{n+1}-i_{1}=p \geqq n. We will demonstrate that then all the coefficients ofDelta_(n)^(j)(f)\Delta_{n}^{j}(f)are>= 0\geq 0The demonstration can be done by induction on the numberpp. The property is obvious top=np=n, because in this case[x_(i_(1)),x_(i_(2)),dots,x_(i_(n+1));f]=Delta_(n)^(i_(1))(f)\left[x_{i_{1}}, x_{i_{2}}, \ldots, x_{i_{n+1}} ; f\right]=\Delta_{n}^{i_{1}}(f).
Forp=n+1p=n+1, let's take
{:[i_(1)=i","i_(2)=i+1","dots","i_(j)=i+j-1],[i_(j+1)=i+j+1","i_(j+2)=i+j+2","dots","i_(n+1)=i+n+1]:}\begin{gathered}
i_{1}=i, i_{2}=i+1, \ldots, i_{j}=i+j-1 \\
i_{j+1}=i+j+1, i_{j+2}=i+j+2, \ldots, i_{n+1}=i+n+1
\end{gathered}
and a simple calculation then gives us
(21)
(21)
{:[[x_(i),x_(i+1),dots,x_(i+j-1),x_(i+j+1),dots,x_(i+n+1);f]=],[=((x_(i+j)-x_(i))Delta_(n)^(i)(f)+(x_(i+n+1)-x_(i+j))Delta_(n)^(i+1)(f))/(x_(i+n+1)-x_(i))]:}\begin{gathered}
{\left[x_{i}, x_{i+1}, \ldots, x_{i+j-1}, x_{i+j+1}, \ldots, x_{i+n+1} ; f\right]=} \\
=\frac{\left(x_{i+j}-x_{i}\right) \Delta_{n}^{i}(f)+\left(x_{i+n+1}-x_{i+j}\right) \Delta_{n}^{i+1}(f)}{x_{i+n+1}-x_{i}}
\end{gathered}
In general, forp=n+1,i_(j+1)-i_(j)=2p=n+1, i_{j+1}-i_{j}=2we can write
[x_(i_(1)),x_(i_(2)),dots,x_(i_(n+1));f]=((x_(i_(j)+1)-x_(i_(1)))Delta_(n)^(i_(1))(f)+(x_(i_(n)+1)-x_(i_(j)+1))Delta_(n)^(i_(1)+1)(f))/(x_(i_(n+1))-x_(i_(1)))\left[x_{i_{1}}, x_{i_{2}}, \ldots, x_{i_{n+1}} ; f\right]=\frac{\left(x_{i_{j}+1}-x_{i_{1}}\right) \Delta_{n}^{i_{1}}(f)+\left(x_{i_{n}+1}-x_{i_{j}+1}\right) \Delta_{n}^{i_{1}+1}(f)}{x_{i_{n+1}}-x_{i_{1}}}
Now suppose that the property is true forp=np=n,n+1,dots,n+rn+1, \ldots, n+rand demonstrate that it will also be true forp=n+r+1p=n+r+1.
Ifi_(n+1)-i_(1)=n+r+1i_{n+1}-i_{1}=n+r+1, formula (21) gives us
{:[[x_(i_(1)),x_(i_(2)),dots,x_(i_(n+1));f]=],[=(1)/(x_(i_(n+1))-x_(i_(1))){(x_(i_(j+1))-x_(i_(1)))[x_(i_(1)),x_(i_(2)),dots,x_(i_(j)),x_(i_(j+1)),x_(i_(j+1)),dots,x_(i_(n));f]+:}],[+(x_(i_(n+1))-x_(i_(j+1)))[x_(i_(2)),x_(i_(3)),dots,x_(i_(j)),x_(i_(j+1)),x_(i_(j+1)),dots,x_(i_(n+1));f]]:}\begin{gathered}
{\left[x_{i_{1}}, x_{i_{2}}, \ldots, x_{i_{n+1}} ; f\right]=} \\
=\frac{1}{x_{i_{n+1}}-x_{i_{1}}}\left\{\left(x_{i_{j+1}}-x_{i_{1}}\right)\left[x_{i_{1}}, x_{i_{2}}, \ldots, x_{i_{j}}, x_{i_{j+1}}, x_{i_{j+1}}, \ldots, x_{i_{n}} ; f\right]+\right. \\
+\left(x_{i_{n+1}}-x_{i_{j+1}}\right)\left[x_{i_{2}}, x_{i_{3}}, \ldots, x_{i_{j}}, x_{i_{j+1}}, x_{i_{j+1}}, \ldots, x_{i_{n+1}} ; f\right]
\end{gathered}
Orjjis an index such thati_(j)+1 < i_(j+1)i_{j}+1<i_{j+1}. The property results immediately.
By takingf=x^(n)f=x^{n}in (20), we see that the sum of the coefficients in the second member is equal to 1. We therefore have the following property.
If the sequence (11) is ordered, any divided difference[x_(i_(1)),x_(i_(2)),dots,x_(i_(n+1));f]\left[\mathrm{x}_{i_{1}}, \mathrm{x}_{i_{2}}, \ldots, \mathrm{x}_{i_{n+1}} ; \mathrm{f}\right]onn+1\mathrm{n}+1of these points is an arithmetic (generalized) mean of the differences quotedDelta_(n)^(1)(f),Delta_(n)^(2)(f),dots,Delta_(n)^(m-n)(f)\Delta_{n}^{1}(\mathrm{f}), \Delta_{n}^{2}(\mathrm{f}), \ldots, \Delta_{n}^{m-n}(\mathrm{f}). So we have
{:[{:[x_(i_(1)),x_(i_(2)),dots,x_(i_(n+1));f]=sum_(j=1)^(m-n)A_(j)Delta_(n)^(j)(f):}],[A_(j) >= 0","j=1","2","..","m-n","sum_(j=1)^(m-n)A_(j)=1]:}\begin{aligned}
& {\left[x_{i_{1}}, x_{i_{2}}, \ldots, x_{i_{n+1}} ; f\right]=\sum_{j=1}^{m-n} \mathrm{~A}_{j} \Delta_{n}^{j}(f)} \\
& \mathrm{A}_{j} \geqq 0, j=1,2, . ., m-n, \sum_{j=1}^{m-n} \mathrm{~A}_{j}=1
\end{aligned}
THEA_(j)\mathrm{A}_{j}being independent of the function f .
The coefficientsA_(j)\mathrm{A}_{j}are given by formula (20). The previous demonstration shows us, moreover, that
The coefficientsA_(j)\mathrm{A}_{j}are given by formula (20). The previous demonstration shows us, moreover, that
A_(j)=0,j=1,2,dots,i_(1)-1,i_(n+1)-n+1,i_(n+1)-n+2,dots,m-n,\mathrm{A}_{j}=0, j=1,2, \ldots, i_{1}-1, i_{n+1}-n+1, i_{n+1}-n+2, \ldots, m-n,
which can also be seen in formula (20), taking into account definition (15) of the functionst_(j)^(**)t_{j}^{*}. The coefficientsA_(i_(1)),A_(i_(n+1)-n)\mathrm{A}_{i_{1}}, \mathrm{~A}_{i_{n+1}-n}are surely positive. We can easily calculate their values. The functionf_(i)^(**)f_{i}^{*}can be considered, on the pointsx_(i_(1)),x_(i_(2))dots,x_(i_(n+1))x_{i_{1}}, x_{i_{2}} \ldots, x_{i_{n+1}}, as the sum of the polynomial_(f_(1)+1,n-1){ }_{f_{1}+1, n-1}of degreen-1n-1and a functiong=g(x)g=g(x)equal tovarphi_(i_(1)+1,n-1)(x_(i_(1)))\varphi_{i_{1}+1, n-1}\left(x_{i_{1}}\right)Forx=x_(i_(1))x=x_{i_{1}}and zero outside this point. We then have
{:[A_(i_(1))=(x_(i_(1)+n)-x_(i_(1)))[x_(i_(1)),x_(i_(2)),dots,x_(i_(n+1));f_(i)^(**)]=(x_(i_(1)+n)-x_(i_(1)))[x_(i_(1)),x_(i_(2)),dots,x_(i_(n+1));g]=],[=((-1)^(n+1)(x_(i_(1)+n)-x_(i_(1)))V(x_(i_(2)),x_(i_(3)),dots,x_(i_(n+1)))q_(i_(1)+1,n-1)(x_(i_(1))))/(V(x_(i_(1)),x_(i_(2)),dots,x_(i_(n+1))))=],[=((x_(i_(1)+1)-x_(i_(1)))(x_(i_(1)+2)-x_(i_(1)))dots(x_(i_(1)+n)-x_(i_(1))))/((x_(i_(2))-x_(i_(1)))(x_(i_(3))-x_(i_(1)))dots(x_(i_(n+1))-x_(i_(1))))]:}\begin{gathered}
\mathrm{A}_{i_{1}}=\left(x_{i_{1}+n}-x_{i_{1}}\right)\left[x_{i_{1}}, x_{i_{2}}, \ldots, x_{i_{n+1}} ; f_{i}^{*}\right]=\left(x_{i_{1}+n}-x_{i_{1}}\right)\left[x_{i_{1}}, x_{i_{2}}, \ldots, x_{i_{n+1}} ; \mathrm{g}\right]= \\
=\frac{(-1)^{n+1}\left(x_{i_{1}+n}-x_{i_{1}}\right) \mathrm{V}\left(x_{i_{2}}, x_{i_{3}}, \ldots, x_{i_{n+1}}\right) q_{i_{1}+1, n-1}\left(x_{i_{1}}\right)}{\mathrm{V}\left(x_{i_{1}}, x_{i_{2}}, \ldots, x_{i_{n+1}}\right)}= \\
=\frac{\left(x_{i_{1}+1}-x_{i_{1}}\right)\left(x_{i_{1}+2}-x_{i_{1}}\right) \ldots\left(x_{i_{1}+n}-x_{i_{1}}\right)}{\left(x_{i_{2}}-x_{i_{1}}\right)\left(x_{i_{3}}-x_{i_{1}}\right) \ldots\left(x_{i_{n+1}}-x_{i_{1}}\right)}
\end{gathered}
We find in the same way
A_(i_(n+1)-n)=((x_(i_(n+1))-x_(i_(n+1)-1))(x_(i_(n+1))-x_(i_(n+1)-2))dots(x_(i_(n+1))-x_(i_(n+1)-n)))/((x_(i_(n+1))-x_(i_(1)))(x_(i_(n+1))-x_(i_(2)))dots(x_(i_(n+1))-x_(i_(n))))\mathrm{A}_{i_{n+1}-n}=\frac{\left(x_{i_{n+1}}-x_{i_{n+1}-1}\right)\left(x_{i_{n+1}}-x_{i_{n+1}-2}\right) \ldots\left(x_{i_{n+1}}-x_{i_{n+1}-n}\right)}{\left(x_{i_{n+1}}-x_{i_{1}}\right)\left(x_{i_{n+1}}-x_{i_{2}}\right) \ldots\left(x_{i_{n+1}}-x_{i_{n}}\right)}
Note. The previous demonstration shows us that the functionsf_(j)^(**)f_{j}^{*}are non-concave of ordern-1n-1on points (11). This property can also be established directly.
6. The divided difference of a function of function. We have established formula (18) giving the divided difference of a product of two functions. Let us now consider a function of functionG(f)\mathrm{G}(f)and we propose to find an expression for the divided difference
6. The divided difference of a function of function. We have established formula (18) giving the divided difference of a product of two functions. Let us now consider a function of functionG(f)\mathrm{G}(f)and we propose to find an expression for the divided difference
F=[x_(1),x_(2),dots,x_(n+1);G(f)]\mathrm{F}=\left[x_{1}, x_{2}, \ldots, x_{n+1} ; \mathrm{G}(f)\right]
If we posef_(i)=f(x_(i)),i=1,2,dots,n+1f_{i}=f\left(x_{i}\right), i=1,2, \ldots, n+1, the expression F is linear and homogeneous inG(f_(1)),G(f_(2)),dots,G(f_(n+1))\mathrm{G}\left(f_{1}\right), \mathrm{G}\left(f_{2}\right), \ldots, \mathrm{G}\left(f_{n+1}\right), therefore of the form
F=sum_(i=1)^(n+1)lambda_(l)G(f_(i))\mathrm{F}=\sum_{i=1}^{n+1} \lambda_{l} \mathrm{G}\left(f_{i}\right)
So we will have
F=sum_(i=1)^(n+1)mu_(i)[f_(1),f_(2),dots,f_(i);G]\mathrm{F}=\sum_{i=1}^{n+1} \mu_{i}\left[f_{1}, f_{2}, \ldots, f_{i} ; \mathrm{G}\right]
We have
lambda_(i)=(1)/(j_(i1,n+1)^(')(x_(i))),i=1,2dots,n+1\lambda_{i}=\frac{1}{{\underset{i 1, n+1}{j}}^{\prime}\left(x_{i}\right)}, i=1,2 \ldots, n+1
and formula (14) shows us that if we set
{:[g_(j)=g_(j)(x)=(f-f_(1))(f-f_(2))dots(f-f_(j-1))","quadg_(1)=1],[j=1","2","dots","n+1]:}\begin{gathered}
g_{j}=g_{j}(x)=\left(f-f_{1}\right)\left(f-f_{2}\right) \ldots\left(f-f_{j-1}\right), \quad g_{1}=1 \\
j=1,2, \ldots, n+1
\end{gathered}
We have
{:(22)mu_(j)=[x_(1),x_(2),..x_(n+1);g_(j)]","j=1","2","dots","n+1:}\begin{equation*}
\mu_{j}=\left[x_{1}, x_{2}, . . x_{n+1} ; g_{j}\right], j=1,2, \ldots, n+1 \tag{22}
\end{equation*}
In particular, we havemu_(1)=0\mu_{1}=0. Taking into account (18), formulas[x_(1);f-f_(1)]=0,[x_(2);f-f_(2)]=0\left[x_{1} ; f-f_{1}\right]=0,\left[x_{2} ; f-f_{2}\right]=0and relations (7), (8), (9), we can write
mu_(2)=sum_(i=3)^(n+1)[x_(1),x_(3),dotsx_(i);f][x_(i),x_(i+1),dots,x_(n+1),x_(2);f]\mu_{2}=\sum_{i=3}^{n+1}\left[x_{1}, x_{3}, \ldots x_{i} ; f\right]\left[x_{i}, x_{i+1}, \ldots, x_{n+1}, x_{2} ; f\right]
This coefficient is therefore a sum ofn-1n-1products of divided differences of order>= 1\geq 1offf, each product being formed by two divided differences whose sum of orders is equal tonn. Let us demonstrate that in general
The coefficientmu_(j)\mu_{j}is a sum of products of divided differences of order>= 1\geqq 1off\mathfrak{f}, each product being formed by divided differences whose sum of orders is equal to n.
Each divided difference is taken, of course, at certain points of the sequencex_(1),x_(2),dots,x_(n+1)x_{1}, x_{2}, \ldots, x_{n+1}
The demonstration is done by induction. We have seen that the property is true for the coefficientmu_(2)\mu_{2}, regardless ofnn. Assume the property is true for the coefficientmu_(j)\mu_{j}for all possible values ​​ofnnand let's demonstrate it formu_(j+1)\mu_{j+1}. We have, taking into account (22), formula (18) andg_(j+1)=g_(j)(f-f_(j))g_{j+1}=g_{j}\left(f-f_{j}\right),
{:(23)mu_(j+1)=sum_(i=j+1)^(n+1)[x_(1),x_(2),dots,x_(j-1),x_(j+1),dots,x_(i);g_(j)][x_(i),x_(i+1),dots,x_(n+1),x_(j);f]:}\begin{equation*}
\mu_{j+1}=\sum_{i=j+1}^{n+1}\left[x_{1}, x_{2}, \ldots, x_{j-1}, x_{j+1}, \ldots, x_{i} ; g_{j}\right]\left[x_{i}, x_{i+1}, \ldots, x_{n+1}, x_{j} ; f\right] \tag{23}
\end{equation*}
which demonstrates the property.
This formula also tells us the number of terms of the coefficientmu_(j)\mu_{j}. EitherN_(j)^(n)\mathrm{N}_{j}^{n}the number of terms ofmu_(j)\mu_{j}Forn+1n+1points. We then haveN_(2)^(n)=n-1\mathrm{N}_{2}^{n}=n-1and formula (23) shows us thatN_(j+1)^(n)=N_(j)^(j-1)++N_(j)^(j)+dots+N_(j)^(n-1)\mathrm{N}_{j+1}^{n}=\mathrm{N}_{j}^{j-1}+ +\mathrm{N}_{j}^{j}+\ldots+\mathrm{N}_{j}^{n-1}from which we easily deduce,
This formula also tells us the number of terms of the coefficientmu_(j)\mu_{j}. EitherN_(j)^(n)\mathrm{N}_{j}^{n}the number of terms ofmu_(j)\mu_{j}Forn+1n+1points. We then haveN_(2)^(n)=n-1\mathrm{N}_{2}^{n}=n-1and formula (23) shows us thatN_(j+1)^(n)=N_(j)^(j-1)++N_(j)^(j)+dots+N_(j)^(n-1)\mathrm{N}_{j+1}^{n}=\mathrm{N}_{j}^{j-1}+ +\mathrm{N}_{j}^{j}+\ldots+\mathrm{N}_{j}^{n-1}from which we easily deduce,
N_(j)^(n)=((n-1)(n-2)*(n-j+1))/((j-1)!)\mathrm{N}_{j}^{n}=\frac{(n-1)(n-2) \cdot(n-j+1)}{(j-1)!}
Let us designate byd_(r)^('),d_(r)^(''),dotsd_{r}^{\prime}, d_{r}^{\prime \prime}, \ldotsdivided differences of orderrroffftaken from groups ofr+1r+1points of the sequencex_(1),x_(2),dots,x_(n+1)x_{1}, x_{2}, \ldots, x_{n+1}. We then have
{:(24)mu_(j)=sumd_(1)^(')d_(1)^('')dotsd_(1)^((j_(1)))d_(2)^(')d_(2)^('')dotsd_(2)^((j_(2)))dotsd_(n)^(')d_(n)^('')dotsd_(n)^((j_(n))):}\begin{equation*}
\mu_{j}=\sum d_{1}^{\prime} d_{1}^{\prime \prime} \ldots d_{1}^{\left(j_{1}\right)} d_{2}^{\prime} d_{2}^{\prime \prime} \ldots d_{2}^{\left(j_{2}\right)} \ldots d_{n}^{\prime} d_{n}^{\prime \prime} \ldots d_{n}^{\left(j_{n}\right)} \tag{24}
\end{equation*}
Or
{:(25)j_(1)+j_(2)+dots+j_(n)=j","quadj_(1)+2j_(2)+dots+nj_(n)=n:}\begin{equation*}
j_{1}+j_{2}+\ldots+j_{n}=j, \quad j_{1}+2 j_{2}+\ldots+n j_{n}=n \tag{25}
\end{equation*}
The summation (24) extends to all integer and positive or zero solutions of the system (25) with respect toj_(1),j_(2),dots,j_(n)j_{1}, j_{2}, \ldots, j_{n}. Any solution corresponds to(n!)/(j_(1)!j_(2)!dotsj_(n)!)\frac{n!}{j_{1}!j_{2}!\ldots j_{n}!}terms inmu_(j)\mu_{j}.
7. The case of differentiable functions. We have assumed that the points (11) are all distict. We can also assume the opposite. We then obtain limit formulas from (17) by making several of the points tendx_(i)x_{i}towards each other. The divided differences which are introduced then have the values ​​which we have given in our Thesis^(6){ }^{6}). We
7. The case of differentiable functions. We have assumed that the points (11) are all distict. We can also assume the opposite. We then obtain limit formulas from (17) by making several of the points tendx_(i)x_{i}towards each other. The divided differences which are introduced then have the values ​​which we have given in our Thesis^(6){ }^{6}). We
let us suppose, well extended, that these are now functions defined in an interval containing the pointsx_(i)x_{i} et un nombre suffisant de fois dérivables. Par exemple si tous les points viennent se confondre, la formule (18) devient la formule de Leibniz donnant la dérivée nème d'un produit. La formule de la différence divisée d'une fonction de fonction devient la formule donnant la n^("eme ")n^{\text {eme }} dérivée d'une fonction de fonction.
Bucureşti, le 29 octobre 1940.
- ^(1){ }^{1} ) Tiberiu Popoviciu "Sur quelques propriétés des fonctions d'une ou de deux variables réelles" Thèse, Paris 1933 ou Mathematica, 8, 1-85 (1934).
^(2){ }^{2} ) Voir par ex. J. F. Steffensen „Note on divided differences" Danske Vid. Selsk. Math-fys. Medd. 17, Nr. 3, 1-12 (1939).
^(3){ }^{3} ) Un polynome de degré nn est une expression de la formec_(o)x^(n)+c_(1)x^(n-1)+dots+c_(n)c_{o} x^{n}+c_{1} x^{n-1}+\ldots+c_{n}les c_(i)c_{i} étant des constantes positives, nulles ou négatives (il suffit de nous limiter au cas réel). - ^(5){ }^{5} ) En vertu de l'unicité ơu polynome (1). Nous tenons compte du fait qu'un polynome de degré nn est complètement déterminé par la valeur de son premier coefficient et ses valeurs en nn points distincts.
- ^(6){ }^{6} ) Voir loc. cit. ^(1){ }^{1} ), p. 43.