Reports of the Academy of Sciences of the USSR
1961. Volume 141, No. 6

MATHEMATICS

D. V. IONESCU

REPRESENTATION OF THE DIVIDED DIFFERENCE OF ORDER \((m,n)\) OF A FUNCTION OF TWO VARIABLES BY A DOUBLE INTEGRAL. II

(Presented by Academician A. N. Kolmogorov on 12 V 1961)

1. In the preceding note \((^1)\) we proved that the divided difference of order \((m,n)\) of a function \(f(x,y)\) at the nodes \((x_i,y_k)\), where \(x_0 < x_1 < \cdots < x_m\), \(y_0 < y_1 < \cdots < y_n\), can be represented by the formula

\[ \left[ \begin{array}{c} x_0, x_1,\ldots,x_m; f \\ y_0, y_1,\ldots,y_n \end{array} \right] = \iint_D \Phi(x,y)\, \frac{\partial^{m+n} f}{\partial x^m \partial y^n}\, dx\,dy, \tag{1} \]

where \(D\) is the rectangle defined by the inequalities \(x_0 \leq x \leq x_m\), \(y_0 \leq y \leq y_n\), and the function \(\Phi(x,y)\) coincides in each rectangle \(D_i^k\), defined by the inequalities \(x_i \leq x \leq x_{i+1}\), \(y_k \leq y \leq y_{k+1}\), with the function \(\varphi_i^k(x,y)\), defined by the formula

\[ \varphi_i^k(x,y) = \frac{(-1)^{m-n}}{(m-1)!(n-1)!} \sum_{\alpha=0}^{i}\sum_{\beta=0}^{k} C_\alpha^\beta (x-x_\alpha)^{m-1}(y-y_\beta)^{n-1}; \tag{2} \]

the coefficients \(C_\alpha^\beta\) are given by the formula

\[ C_\alpha^\beta = (-1)^{\alpha+\beta} \frac{ V(x_0,x_1,\ldots,x_{\alpha-1},x_{\alpha+1},\ldots,x_m) }{ V(x_0,x_1,\ldots,x_m) } \, \frac{ V(y_0,y_1,\ldots,y_{\beta-1},y_{\beta+1},\ldots,y_n) }{ V(y_0,y_1,\ldots,y_n) } . \tag{3} \]

The expansions

\[ \frac{1}{(x-x_0)(x-x_1)\cdots(x-x_m)} = \sum_{\alpha=0}^{m}\frac{A_\alpha}{x-x_\alpha}, \]

\[ \frac{1}{(y-y_0)(y-y_1)\cdots(y-y_n)} = \sum_{\beta=0}^{n}\frac{B_\beta}{y-y_\beta} \tag{4} \]

show that one may also write

\[ C_\alpha^\beta = (-1)^{m+n} A_\alpha B_\beta. \tag{5} \]

In the present note we shall prove that the function \(\Phi(x,y)\), which vanishes on the sides of the rectangle \(D\), has constant sign inside this rectangle.

2. Consider the functions \(\varphi_i^k(x,y)\) associated with the rectangles \(D_i^k\) lying between the lines \(y=y_k\) and \(y=y_{k+1}\), and study the functions of \(x\), \(\chi_i(x)=\varphi_i^k(x,y)\), where \(y\) has a fixed value from the interval \((y_k,y_{k+1})\), if \(k=0,1,\ldots,n-2\), or from the interval \((y_{n-1},y_n)\), if \(k=n-1\). It is proved that the functions \(\chi_i(x)\) satisfy the boundary conditions

\[ \chi_0^{(j)}(x_0)=0,\qquad \chi_i^{(j)}(x_i)=\chi_{i-1}^{(j)}(x_i),\qquad \chi_{m-1}^{(j)}(x_m)=0 \tag{6} \]

for \(i=1,2,\ldots,m-1\) and \(j=0,1,\ldots,m-2\).

Similarly, as functions of \(y\), \(\theta_k(y)=\varphi_i^k(x,y)\), where \(x\) has a fixed value from the interval \((x_i,x_{i+1})\), if \(i=0,1,\ldots,m-2\), or from the interval \((x_{m-1},x_m)\), if \(i=m-1\), satisfy the boundary conditions

\[ \theta_0^{(j)}(y_0)=0,\qquad \theta_k^{(j)}(y_k)=\theta_{k-1}^{(j)}(y_k),\qquad \theta_{n-1}^{(j)}(y_n)=0 \tag{7} \]

for \(k=1,2,\ldots,n-1\) and \(j=0,1,\ldots,n-2\).

1. We shall assume that the derivative \(\chi_i^{(m-1)}(x)\) does not vanish on the interval \((x_0,x_m)\). This derivative is constant with respect to \(x\), and we have

\[ \chi_i^{(m-1)}(x)= \frac{A_0+A_1+\cdots+A_i}{(n-1)!} \left[ B_0(y-y_0)^{n-1}+B_1(y-y_1)^{n-1}+\cdots \right. \]

\[ \left. \cdots+B_k(y-y_k)^{n-1} \right]. \tag{8} \]

We proved in (2) that \(A_0+A_1+\cdots+A_i\ne0\), and it remains for us to prove that the sum

\[ h_k=B_0(y-y_0)^{n-1}+B_1(y-y_1)^{n-1}+\cdots+B_k(y-y_k)^{n-1} \tag{9} \]

is not equal to zero for \(k=0,1,\ldots,n-1\), where \(y\) has a fixed value from the interval \((y_k,y_{k+1})\), if \(k=0,1,\ldots,n-2\), or from \((y_{n-1},y_n)\), if \(k=n-1\). We assume that \(n>1\).

Considering the expansion

\[ \frac{(y-Y)^{n-1}}{(Y-y_0)(Y-y_1)\cdots(Y-y_n)} = \sum_{j=0}^{n}\frac{B_j'}{Y-y_j}, \tag{10} \]

one may write

\[ h_k=B_0'+B_1'+\cdots+B_k'. \tag{11} \]

Introducing the function

\[ f(Y)=\frac{(y-Y)^{n-1}}{(Y-y_0)(Y-y_1)\cdots(Y-y_k)}, \tag{12} \]

we prove that

\[ h_k=-[y_{k+1},y_{k+2},\ldots,y_n;f], \tag{13} \]

and, by virtue of a known theorem, we have

\[ h_k=-\frac{f^{(n-k-1)}(\bar Y)}{(n-k-1)!}, \tag{14} \]

where \(y_{k+1}<\bar Y<y_n\).

1. We have

\[ f^{(n-k-1)}(Y) = (-1)^n(n-k-1)!\sum_{i=0}^{k} A_i \frac{(y_i-y)^{n-1}}{(y_i-Y)^{n-k}}, \]

and, introducing a function of \(\eta\)

\[ g(\eta)=\frac{(\eta-y)^{n-1}}{(\eta-Y)^{n-k}}, \tag{15} \]

we obtain

\[ f^{(n-k-1)}(Y) = (-1)^n(n-k-1)![y_0,y_1,\ldots,y_k;g] = \]

\[ = (-1)^n\frac{(n-k-1)!}{k!}\,g^{(k)}(\bar\eta), \tag{16} \]

where \(y_0<\bar\eta<y_k\).

It can be proved that

\[ g^{(k)}(\eta) = (-1)^k\frac{(n-1)!}{(n-k-1)!}\, \frac{(Y-y)^k(\eta-y)^{\,n-k-1}}{(\eta-Y)^n}. \tag{17} \]

Let us return to formulas (14), (16), (17) and denote by \(\bar{\eta}^{*}\) the number corresponding to \(\bar{Y}\) in formula (16). We have

\[ h_k=(-1)^{n+k+1}\binom{n-1}{k} \frac{(\bar{Y}-y)^k(\bar{\eta}^{*}-y)^{\,n-k-1}}{(\bar{\eta}^{*}-\bar{Y})^n}, \tag{18} \]

and this proves that \(h_k\neq 0\), since

\[ \bar{\eta}^{*}<y_k<y\leq y_{k+1}<\bar{Y}. \tag{19} \]

The preceding arguments were valid for \(k=1,2,\ldots,n-2\), but one can verify directly that we also have \(h_0\neq 0\) and \(h_{n-1}\neq 0\).

1. It can be proved in the same way that the derivative \(\theta_k^{(n-1)}(y)\) does not vanish on the interval \((y_k,y_{k+1})\). We shall restrict ourselves to considering the functions \(\theta_k(y)\) corresponding to \(i=0\), i.e.

\[ \theta_k(y)=\frac{(-1)^{m-n}}{(m-1)!(n-1)!} \left[C_0^0(y-y_0)^{n-1}+C_0^1(y-y_1)^{n-1}+\cdots\right. \]
\[ \left.\cdots+C_0^k\times(y-y_k)^{n-1}\right](x-x_0)^{m-1}. \tag{20} \]

For these functions we have

\[ \theta_k^{(n-1)}(y)= \frac{(-1)^{m-n}}{(m-1)!}A_0(B_0+B_1+\cdots+B_k)(x-x_0)^{m-1}, \tag{21} \]

and, consequently, \(\theta_k^{(n-1)}(y)\neq 0\) on the interval \((y_k,y_{k+1})\), since \(B_0+B_1+\cdots+B_k\neq 0\).

1. We can now prove that the function \(\Phi(x,y)\) has the same sign as \((-1)^{m-n}\) inside the rectangle \(D\).

First of all, we have

\[ \varphi_0^0(x,y)= \frac{(-1)^{m-n}}{(m-1)!(n-1)!} C_0^0(x-x_0)^{m-1}(y-y_0)^{n-1}, \tag{22} \]

where the coefficient \(C_0^0\) is positive. Hence \(\varphi_0^0(x,y)\) has the sign \((-1)^{m-n}\) for \(x_0<x\leq x_1,\ y_0<y\leq y_1\).

Next we shall prove that the function \(\Phi(x,y)\) has the sign \((-1)^{m-n}\) in the strip \(x_0<x\leq x_1,\ y_0<y<y_n\). To this end consider the function \(\theta(y)\), which coincides on each interval \([y_k,y_{k+1}],\ k=0,1,\ldots,n-1\), with the functions \(\theta_k(y)=\varphi_0^k(x,y)\), where \(x\) has a fixed value in the interval \((x_0,x_1]\). We have proved that \(\theta(y)\) is a function continuous on the interval \([y_0,y_n]\), together with its derivatives up to order \(n-2\), and that it satisfies the conditions \(\theta^j(x_0)=0,\ \theta^{(j)}(x_n)=0\) for \(j=0,1,\ldots,n-2\). By Rolle’s theorem, applied to the function \(\theta(y)\) and to the interval \([y_0,y_n]\), the derivative \(\theta'(y)\) has at least one zero in the interval \((y_0,y_n)\). This zero is unique.

Indeed, suppose that the derivative \(\theta'(y)\) has two zeros in the interval \((y_0,y_n)\). Then, applying Rolle’s theorem successively and taking into account the conditions which the function \(\theta(y)\) satisfies at the points \(y_0\) and \(y_n\), we conclude that the derivative \(\theta^{(n-2)}(y)\) has \(n-1\) zeros in the interval \((y_0,y_n)\). In the intervals \((y_0,y_1]\), \([y_{n-1},y_n)\) there is not a single zero of \(\theta^{(n-2)}(y)\), since we have proved that \(\theta_0^{(n-1)}(y)\) and \(\theta_n^{(n-1)}(y)\) do not vanish in the intervals \((y_0,y_1)\), \((y_{n-1},y_n)\). Consequently, the \(n-1\) zeros of the derivative \(\theta^{(n-2)}(y)\) lie in the interval \((y_1,y_{n-1})\), but in each interval \((y_k,y_{k+1}]\), where \(k=1,2,\ldots,n-2\), there lies only one zero of the function \(\theta^{(n-2)}(y)\), since we have proved that the derivative \(\theta^{(n-1)}(y)\) does not vanish on the interval \((y_k,y_{k+1})\). Thus the interval \((y_1,y_{n-1})\) can contain only \(n-2\) zeros of the function \(\theta^{(n-2)}(y)\), and we have arrived at a contradiction, which shows that the derivative \(\theta'(y)\) has only one zero in the interval \((y_0,y_n)\).

It follows from this that the function \(\theta(y)\) has only one extremum in the interval \((y_0,y_n)\), and, since it has sign \((-1)^{m-n}\) in the interval \((y_0,y_1)\), it has sign \((-1)^{m-n}\) in the interval \((y_0,y_n)\). Hence we conclude that the function \(\Phi(x,y)\) has sign \((-1)^{m-n}\) in the strip \(x_0<x\le x_1,\ y_0<y<y_n\).

By the same method, using the properties of the functions \(\chi_i(x)\), one can prove that the function \(\Phi(x,y)\) has sign \((-1)^{m-n}\) in the strip \(x_0<x<x_m,\ y_k<y\le y_{k+1}\) for each value \(k=0,1,\ldots,n-2\), or in the strip \(x_0<x<x_n,\ y_{n-1}<y<y_n\).

The function \(\Phi(x,y)\), therefore, has sign \((-1)^{m-n}\) inside the rectangle \(D\).

1. From the preceding property it follows that

\[ \left[ \begin{array}{c} x_0,x_1,\ldots,x_m\\ y_0,y_1,\ldots,y_n \end{array} ; f \right] = \frac{\partial^{m+n} f(\xi,\eta)}{\partial x^m \partial y^n} \iint_D \Phi(x,y)\,dx\,dy, \tag{23} \]

where \(x_0<\xi<x_n,\ y_0<\eta<y_n\). We have

\[ \iint_D \Phi(x,y)\,dx\,dy = \frac{(-1)^{m-n}}{m!n!} \tag{24} \]

and, consequently, the divided difference of order \((m,n)\) of the function \(f(x,y)\) can be represented by the formula

\[ \left[ \begin{array}{c} x_0,x_1,\ldots,x_m\\ y_0,y_1,\ldots,y_n \end{array} ; f \right] = \frac{(-1)^{m-n}}{m!n!}\, \frac{\partial^{m+n} f(\xi,\eta)}{\partial x^m \partial y^n}, \tag{25} \]

where \(x_0<\xi<x_n,\ y_0<y<y_n\).

Hence the estimate follows:

\[ \left| \left[ \begin{array}{c} x_0,x_1,\ldots,x_m\\ y_0,y_1,\ldots,y_n \end{array} ; f \right] \right| \le \frac{M_{m,n}}{m!n!}, \tag{26} \]

where \(M_{m,n}\) is the least upper bound of the absolute value of

\[ \frac{\partial^{m+n}f}{\partial x^m \partial y^n} \]

in the rectangle \(x_0<x<x_m,\ y_0<y<y_n\).

Cluj
Romanian People’s Republic

Received
25 IV 1961

References

\({}^{1}\) D. V. Ionescu, DAN, 141, No. 5 (1961). \({}^{2}\) D. V. Ionescu, Cuadraturi numerice, București, 1957, chap. III.