Vol. 277

1977

GENERAL EXPRESSIONS FOR THE FUNCTION \(f(x)\) AND ITS DERIVATIVES \(f^{(x)}(x)\), OBTAINED BY THE METHOD OF PARTIAL MATCHING FOR AN ARBITRARY ARRANGEMENT OF THE NODES \(x_s\)

B. F. Krutoi

(Presented by the scientific seminar of the Department of Engineering and Computational Mathematics)

1. By the method of partial matching we shall mean the following method of approximate representation, on the interval \(a \le x \le b\), of a function \(f(x) \in L\) from the linear space \(L\) by means of some other function \(\varphi(x) \in L_{n+1}\) from an \((n+1)\)-dimensional subspace \(L_{n+1} \subset L\), under which, at \(n+1\) prescribed points \(x_s\) of the interval \(a \le x \le b\), the condition

\[ \varepsilon_s=\varphi(x_s)-f(x_s)=0,\qquad (s=0,1,2,\ldots,n). \tag{1} \]

\[ (a\le x\le b) \]

is satisfied.

However, at all other points \(x_j \ne x_s\) of the interval \(a \le x \le b\), condition (1) is generally not satisfied, so that

\[ \varepsilon_j=\varphi(x_j)-f(x_j)\ne 0,\qquad (j\ne s) \tag{2} \]

\[ (a\le x\le b) \]

in the general case. The points \(x_s\) of the interval \(a \le x \le b\) at which condition (1) is satisfied shall be called the nodes of the method of partial matching.

Usually the approximate representation \(\varphi(x) \in L_{n+1}\subset L\) for the function \(f(x)\in L\) on the interval under consideration \(a \le x \le b\) is specified by expanding the representation \(\varphi(x)\) in terms of some \(n+1\) supporting, mutually independent functions \(\varphi_\nu(x)\in L_{n+1}\), so that

\[ f(x)\simeq \varphi(x)=\sum_{\nu=0}^{n} c_\nu \varphi_\nu(x), \tag{3} \]

where \(c_\nu\) are coefficients to be determined. Often the following are used as the \(n+1\) supporting functions \(\varphi_\nu(x)\):

\[ 1)\ \varphi_\nu(x)=x^\nu,\qquad 2)\ \varphi_\nu(x)=\frac{1}{x^\nu},\qquad 3)\ \varphi_\nu(x)=\sin \nu x,\ \cos \nu x, \]

\[ 4)\ \varphi_\nu(x)=T_\nu(x). \tag{4} \]

In the last case \(T_\nu(x)\) are Chebyshev polynomials of the second kind, which are defined as follows:

\[ \begin{gathered} 1)\ T_\nu(x)=\cos(\nu\arccos x),\quad 2)\ \cos x=\theta,\quad 3)\ T_\nu(x)=\cos\nu\theta,\\ 4)\ T_{\nu+1}(x)=2xT_\nu(x)-T_{\nu-1}(x),\quad T_0(x)=1,\quad T_1(x)=x. \end{gathered} \tag{5} \]

For any method of choosing the approximate representation (3) for the function \(f(x)\), it is useful to transform the interval \(a\le x\le b\) into the interval \(-1\le \xi\le +1\), which is done as follows:

\[ \begin{gathered} 1)\ x=\frac{b+a}{2}+\frac{b-a}{2}\xi,\qquad 2)\ \xi=\frac{2x-(a+b)}{b-a},\\ 3)\ (x=a)\to(\xi=-1),\qquad 4)\ (x=b)\to(\xi=+1). \end{gathered} \tag{6} \]

If the representation \(\varphi(x)\) for the function \(f(x)\) is given by the expansion of general form (3), then the coefficients \(c_\nu\), \((\nu=0,1,2,\ldots,n)\), of this expansion are found by solving a set of \(n+1\) linear equations

\[ \sum_{\nu=0}^{n}\varphi_\nu(x_s)c_\nu-\varphi(x_s) =\sum_{\nu=0}^{n}\varphi_\nu(x_s)c_\nu-f(x_s)= \]

\[ =\sum_{\nu=0}^{n}a_{s\nu}c_\nu-f_s=0,\qquad (s=0,1,\ldots,n), \tag{7} \]

which we obtain by substituting in (3) the values \(x=x_0,x_1,\ldots,x_n\) and taking into account condition (1). By virtue of the mutual independence of the supporting functions \(\varphi_\mu(x)\), \(\varphi_\nu(x)\), the determinant of the system of equations (7) is nonzero,

\[ \left|\varphi_\nu(x_s)\right|_0^n=\left|a_{s\nu}\right|_0^n\ne 0. \tag{8} \]

and therefore, from the solution of the system (7), its \(n+1\) unknowns \(c_\nu\) are determined uniquely.

1. Suppose now that the approximate representation \(\varphi(x)\) for the function \(f(x)\) on the interval \(a\le x\le b\) is given in a form somewhat different from (3), namely [1]:

\[ f(x)\approx\varphi(x)=(x-x_0)(x-x_1)\ldots(x-x_n)\sum_{\chi=0}^{n}\frac{A_\chi}{(x-x_\chi)} \]

\[ =\prod_{\nu=0}^{n}(x-x_\nu)\sum_{\chi=0}^{n}\frac{A_\chi}{(x-x_\chi)}. \tag{9} \]

Put here \(x=x_s\). Then we may write that

\[ f(x_s)\approx\varphi(x_s)= \prod_{\nu=0}^{s-1}(x_s-x_\nu)(x_s-x_s) \prod_{\chi=s+1}^{n}(x_s-x_\chi)\times \]

\[ \times\left[ \sum_{\chi=0}^{s-1}\frac{A_\chi}{(x_s-x_\chi)} +\frac{A_s}{(x_s-x_s)} +\sum_{\chi=s+1}^{n}\frac{A_\chi}{(x_s-x_\chi)} \right]= \]

\[ =\prod_{\nu=0}^{n}(x_s-x_\nu)\sum_{\chi=0}^{s-1}\frac{A_\chi}{(x_s-x_\chi)} +\prod_{\substack{\nu=0\\ \nu\ne s}}^{n}(x_s-x_\nu)A_s +\prod_{\nu=0}^{n}(x_s-x_\nu)\sum_{\chi=s+1}^{n}\frac{A_\chi}{(x_s-x_\chi)}. \]

But the first and third terms vanish, since

\[ \prod_{\nu=0}^{n}(x_s-x_\nu) = \prod_{\nu=0}^{s-1}(x_s-x_\nu)(x_s-x_s) \prod_{\nu=s+1}^{n}(x_s-x_\nu)=0 . \]

Hence we conclude that

\[ f(x_s)\simeq \varphi(x_s)=A_s \prod_{\substack{\nu=0\\ \nu\ne s}}^{n}(x_s-x_\nu) \]

and, consequently,

\[ A_s\simeq \frac{f(x_s)} {\displaystyle \prod_{\substack{\nu=0\\ \nu\ne s}}^{n}(x_s-x_\nu)}, \qquad (s=0,1,\ldots,n;\ \nu\ne s). \tag{10} \]

Substitution of the value (10) for \(A_s\) into (9), after replacing \(\nu\) by \(s\), gives

\[ \begin{aligned} f(x)\simeq \varphi(x) &= \prod_{\nu=0}^{n}(x-x_\nu) \sum_{s=0}^{n}\frac{A_s}{(x-x_s)} = \sum_{s=0}^{n} A_s \prod_{\substack{\nu=0\\ \nu\ne s}}^{n}(x-x_\nu) \\ &= \sum_{s=0}^{n} \frac{f(x_s)} {\displaystyle \prod_{\substack{\nu=0\\ \nu\ne s}}^{n}(x_s-x_\nu)} \prod_{\substack{\nu=0\\ \nu\ne s}}^{n}(x-x_\nu) = \sum_{s=0}^{n} B_s \prod_{\substack{\nu=0\\ \nu\ne s}}^{n}(x-x_\nu), \end{aligned} \tag{11} \]

where

\[ B_s= \frac{f(x_s)} {\displaystyle \prod_{\substack{\nu=0\\ \nu\ne s}}^{n}(x_s-x_\nu)}, \qquad (s,\nu=0,1,\ldots,n;\ \nu\ne s). \tag{12} \]

Thus, we have established that if the approximate representation \(\varphi(x)\) for \(f(x)\) on the interval \(a\le x\le b\) is specified according to (9) with coefficients \(A_s\) still undetermined, then the values of these coefficients are found according to (10), and for the function \(f(x)\) we obtain the explicit expression (11).

The approximate representation (11) for \(f(x)\), obtained under condition (1) for an arbitrary arrangement of \(n+1\) nodes \(x_s\) on the interval \(a\le x\le b\), is, obviously, an algebraic polynomial of degree \(n\) and in the partial matching method is called the Lagrange expansion. It is easy to see that the Lagrange expansion corresponds to the case in which, as the supporting functions \(\varphi_\nu(x)\) in the finite series (3), \(\varphi_\nu(x)=x^\nu\) are taken.

1. The list (4) of frequently used supporting functions \(\varphi_\nu(x)\) shows that, for \(f(x)\) in the partial matching method, the following expansions convenient for calculation are possible:

\[ \begin{gathered} 1)\quad f(x)\simeq \sum_{\nu=0}^{n} c_\nu x^\nu, \qquad 2)\quad f(x)=\sum_{\nu=0}^{n} c_\nu x^{-\nu}; \\[6pt] 3)\quad f(x)\simeq \sum_{\nu=0}^{n}\bigl(c_{\nu 1}\sin \nu x+c_{\nu 2}\cos \nu x\bigr), \end{gathered} \tag{13} \]

or, in expanded form,

\[ \begin{aligned} 3)\quad f(x)&=c_{01}\sin 0x+c_{02}\cos 0x+\sum_{\nu=1}^{n}(c_{\nu 1}\sin \nu x+c_{\nu 2}\cos \nu x)\\ &=c_{02}+\sum_{\nu=1}^{n}(c_{\nu 1}\sin \nu x+c_{\nu 2}\cos \nu x); \end{aligned} \tag{13} \]

\[ \begin{aligned} 4)\quad f(x)&=\sum_{\nu=0}^{n} c_\nu T_\nu(x)=c_0+c_1x+c_2(2x^2-1)+c_3x(4x^2-3)+{}\\ &\quad +c_4(8x^4-8x^2+1)+c_5x(16x^4-20x^2+5)+c_6(32x^6-48x^4+18x^2-1)+\cdots \end{aligned} \]

The only difficulty in using expansions (13) for an approximate representation of the function \(f(x)\) by this method is the fact that, for each new set of nodes \(x_s\) on the interval \(a\le x\le b\), it is necessary to recompute the matrix \([a_{s\nu}]_0^n\) of the coefficients \(a_{s\nu}\) and the vector \([f_s]_0^1\) for the corresponding system of equations (7), and then solve it in order to determine all \(n+1\) coefficients \(c_\nu\) entering into expansion (3). However, calculations of this kind, if they are carried out on a digital computer, can be programmed in advance for prescribed limiting values of \(n\) and a prescribed form (4) of the supporting functions \(\varphi_\nu(x)\), which will substantially facilitate and accelerate these calculations.

1. Let us now calculate the successive derivatives \(f'(x)\), \(f''(x),\ldots\) for the function \(f(x)\). This calculation can be performed in two ways:

a) starting from the finished finite series (13) for \(f(x)\),

b) relying on the expression of \(f(x)\) by the Lagrange series (9), which does not yet give the final representation for \(f(x)\) and therefore is not entirely convenient for differentiation.

The first approach makes it possible to find very simply, for \(f(x)\), derivatives \(f^{(\nu)}(x)\) of any order \(\nu\), but requires a large amount of preliminary work to calculate, by the method indicated above, all coefficients \(c_\nu\) of the chosen variety (13) for \(f(x)\).

The second way of calculating the derivatives \(f'(x)\), \(f''(x),\ldots\) is based directly on the Lagrange expansion (9) for \(f(x)\) and requires only knowledge of the coefficients \(B_s\), but leads to cumbersome expressions for \(f'(x)\), \(f''(x),\ldots\) because of the complexity of the initial expression (11).

1. Let us find, in general form, the derivatives \(f'(x)\), \(f''(x)\), and \(f'''(x)\) for \(f(x)\), and also obtain from them the general expressions needed below for \(f'(x_r)\), \(f''(x_r)\), and \(f'''(x_r)\) for these derivatives at the nodal points \(x_r\) of the interval \(a\le x\le b\), relying on the Lagrange expansion (11).

Writing expansion (11) in compact form,

\[ f(x)=\sum_{s=0}^{n}\left[B_s\prod_{\nu=0}^{n}(x-x_\nu)\right],\qquad \nu\ne s, \tag{11} \]

where

\[ B_s=\frac{f(x_s)}{\displaystyle\prod_{\nu=0}^{n}(x_s-x_\nu)},\qquad (s,\nu=0,1\ldots,n;\ \nu\ne s) \tag{10} \]

and differentiating this expansion for \(f(x)\), we find

\[ f'(x)=\sum_{s=0}^{n}\left[B_s\left(\sum_{\substack{\lambda=0\\ \lambda\ne \nu}}^{n}\frac{1}{(x-x_\lambda)}\prod_{\substack{\nu=0\\ \nu\ne s}}^{n}(x-x_\nu)\right)\right]. \]

Hence we obtain the following final expression for

\[ f'(x)=\sum_{s=0}^{n}\left[B_s\left(\sum_{\lambda=0}^{n}\eta_\lambda \prod_{\nu=0}^{n}(x-x_\nu)\right)\right]_{\lambda\ne s\cap\nu}^{s\ne\nu\cap\lambda,\ \nu\ne s\cap\lambda}, \tag{14} \]

where

\[ \eta_\lambda= \begin{cases} +1, & \text{if } \lambda\ne s\cap\nu,\\ 0, & \text{if } \lambda=s\cup\nu. \end{cases} \tag{15} \]

Further differentiating expression (11) for \(f'(x)\), we find

\[ f''(x)=\sum_{s=0}^{n}\left\{ B_s\left[ \sum_{\substack{\lambda=0\\ \lambda\ne\mu}}^{n}\eta_\lambda \left( \sum_{\substack{\mu=0\\ \mu\ne\nu}}^{n} \frac{\displaystyle\prod_{\nu=0}^{n}(x-x_\nu)} {x-x_\mu} \right)_{\nu\ne s} \right]\right\}. \]

Hence we finally obtain

\[ f''(x)=\sum_{s=0}^{n}\left\{ B_s\left[ \sum_{\lambda=0}^{n}\eta_\lambda \left( \sum_{\mu=0}^{n}\eta_\mu \prod_{\nu=0}^{n}(x-x_\nu) \right) \right]\right\}, \tag{16} \]

\[ \left( \begin{array}{ll} s\ne\nu\cap\lambda\cap\mu & \lambda\ne s\cap\nu\cap\mu\\ \nu\ne s\cap\lambda\cap\mu & \mu\ne s\cap\nu\cap\lambda \end{array} \right) \]

where

\[ \eta_\mu= \begin{cases} +1, & \text{if } \mu\ne s\cap\nu\cap\lambda,\\ 0, & \text{if } \mu=s\cup\nu\cup\lambda. \end{cases} \tag{17} \]

In the same way we find that

\[ f'''(x)=\sum_{s=0}^{n}\left\langle B_s\left\{ \sum_{\lambda=0}^{n}\eta_\lambda \left[ \sum_{\mu=0}^{n}\eta_\mu \left( \sum_{\sigma=0}^{n}\eta_\sigma \prod_{\nu=0}^{n}(x-x_\nu) \right) \right]\right\} \right\rangle, \tag{18} \]

\[ \left( \begin{array}{lll} s\ne\nu\cap\lambda\cap\mu\cap\sigma & \lambda\ne s\cap\nu\cap\mu\cap\sigma & \sigma\ne s\cap\nu\cap\lambda\cap\mu\\ \nu\ne s\cap\lambda\cap\mu\cap\sigma & \mu\ne s\cap\nu\cap\lambda\cap\sigma & \end{array} \right) \]

where

\[ \eta_\sigma= \begin{cases} +1, & \text{if } \sigma\ne s\cap\nu\cap\lambda\cap\mu,\\ 0, & \text{if } \sigma=s\cup\nu\cup\lambda\cup\mu. \end{cases} \tag{19} \]

Proceeding now from the expressions obtained by us, (14), (16), and (18), for the derivatives \(f'(x)\), \(f''(x)\), and \(f'''(x)\), we shall derive expressions for the values \(f'(x_i)\), \(f''(x_i)\), and \(f'''(x_i)\) of these derivatives at a prescribed node \(x_i=x_0,x_1,\ldots,x_n\) of the interval \(a\le x\le b\).

First of all, proceeding from expression (14) for \(f'(x)\), we find that

\[ f'(x_i)=\sum_{s=0}^{n}B_s\left[ \sum_{\lambda=0}^{n}\eta_\lambda \prod_{\nu=0}^{n}(x_i-x_\nu) \right]_{\lambda\ne s\cap\nu}^{s\ne\nu\cap\lambda,\ \nu\ne s\cap\lambda}. \]

Hence, in more expanded notation, we obtain

\[ f'(x_i)= \left\{ B_i\left[ \sum_{\lambda=0}^{n}\eta_\lambda \prod_{\nu=0}^{n}(x_i-x_\nu) \right] + \sum_{\substack{s=0\\ s\ne i}}^{n}B_s \left[ \sum_{\lambda=0}^{n}\eta_\lambda \prod_{\nu=0}^{n}(x_i-x_\nu) \right] \right\}_{\lambda\ne s\cap\nu}^{s\ne\nu\cap\lambda,\ \nu\ne s\cap\lambda}, \]

\[ f'(x_i)=\left\{B_i\left[\sum_{\lambda=0}^{n}\eta_\lambda \prod_{\nu=0}^{n}(x_i-x_\nu)\right] +\sum_{\substack{s=0\\ s\ne i}}^{n}B_s\left[ \eta_i\prod_{\nu=0}^{i-1}(x_i-x_\nu) \prod_{\nu=i+1}^{n}(x_i-x_\nu) +\sum_{\substack{\lambda=0\\ \lambda\ne i}}^{n}\eta_\lambda \prod_{\nu=0}^{n}(x_i-x_\nu)\right]\right\} \begin{matrix} s\ne \nu\cap\lambda\\ \nu\ne s\cap\lambda\\ \lambda\ne s\cap\nu \end{matrix} \]

But the terms of the last sum

\[ \eta_\lambda\prod_{\nu=0}^{n}(x_i-x_\nu)=0,\qquad (\lambda\ne i;\ \nu\ne \lambda), \]

since these terms contain the factor \((x_i-x_i)=0\). Therefore, for \(f'(x_i)\) at the prescribed node \(x_i\) of the interval \(a\le x\le b\), we find the following final expression:

\[ f'(x_i)=\left\{ B_i\left[\sum_{\lambda=0}^{n}\eta_\lambda\prod_{\nu=0}^{n}(x_i-x_\nu)\right] +\sum_{s=0}^{n}B_s\prod_{\nu=0}^{n}(x_i-x_\nu) \right\}. \tag{20} \]

\[ \begin{pmatrix} s\ne \nu\cap\lambda\cap i;\ \lambda\ne s\cap\nu\cap i;\\ \nu\ne s\cap\lambda\cap i;\ i\ne s\cap\nu\cap\lambda. \end{pmatrix} \]

In the same way, starting from expression (16) for \(f''(x)\), we establish first of all that at the prescribed node \(x_i\) of the interval \(a\le x\le b\) we shall have

\[ f''(x_i)= \sum_{s=0}^{n}B_s\left\{ \sum_{\lambda=0}^{n}\eta_\lambda \left[ \sum_{\mu=0}^{n}\eta_\mu \prod_{\nu=0}^{n}(x_i-x_\nu) \right]\right\}. \]

\[ \begin{pmatrix} s\ne \nu\cap\lambda\cap\mu;\ \lambda\ne s\cap\nu\cap\mu;\\ \nu\ne s\cap\lambda\cap\mu;\ \mu\ne s\cap\nu\cap\lambda; \end{pmatrix} \]

Hence, in more expanded form, we find that

\[ f''(x_i)=B_i\left\{ \sum_{\lambda=0}^{n}\eta_\lambda \left[ \sum_{\mu=0}^{n}\eta_\mu \prod_{\nu=0}^{n}(x_i-x_\nu) \right]\right\} + \]

\[ \begin{pmatrix} i\ne \nu\cap\lambda\cap\mu;\ \lambda\ne i\cap\nu\cap\mu;\\ \nu\ne i\cap\lambda\cap\mu;\ \mu\ne i\cap\nu\cap\lambda. \end{pmatrix} \]

\[ +\sum_{\substack{s=0\\ s\ne i}}^{n}B_s\left\{ \eta_i\left[ \sum_{\mu=0}^{n}\eta_\mu \prod_{\nu=0}^{i-1}(x_i-x_\nu) \prod_{\nu=i+1}^{n}(x_i-x_\nu) \right]\right. + \]

\[ \left. +\sum_{\lambda=0}^{n}\eta_\lambda \left[ \sum_{\substack{\mu=0\\ \mu\ne i}}^{n}\eta_\mu \prod_{\nu=0}^{n}(x_i-x_\nu) \right]\right\}, \quad \begin{pmatrix} s\ne \nu\cap\lambda\cap\mu;\ \lambda\ne s\cap\nu\cap\mu;\\ \nu\ne s\cap\lambda\cap\mu;\ \mu\ne s\cap\nu\cap\lambda. \end{pmatrix} \]

But the terms of the inner sum in the third term of this expression for \(f''(x_i)\) are equal to zero:

\[ \eta_\mu\prod_{\nu=0}^{n}(x_i-x_\nu)=0,\qquad (\mu\ne i;\ \nu\ne \mu), \]

since all these terms contain the factor \((x_i-x_i)=0\). Therefore, for the second derivative \(f''(x_i)\) at the prescribed node \(x_i\) of the interval \(a\le x\le b\), we obtain the following final expression:

\[ f'''(x_i)=B_i\left\{\sum_{\lambda=0}^{n}\eta_\lambda \left[\sum_{\mu=0}^{n}\eta_\mu \prod_{\nu=0}^{n}(x_i-x_\nu)\right]\right\}+ \tag{21} \]

\[ +\sum_{s=0}^{n}B_s\left\{\left[\sum_{\mu=0}^{n}\eta_\mu \prod_{\nu=0}^{n}(x_i-x_\nu)\right]\right\}. \]

\[ \left( \begin{array}{lll} s\ne \nu\cap\lambda\cap\mu\cap i; & \lambda\ne s\cap\nu\cap\mu\cap i; & i\ne s\cap\nu\cap\lambda\cap\mu.\\ \nu\ne s\cap\lambda\cap\mu\cap i; & \mu\ne s\cap\nu\cap\lambda\cap i;& \end{array} \right) \]

Finally, for the third derivative \(f'''(x_i)\) at the node \(x\) of the interval \(a\le x\le b\), we find in the same way the following final expression:

\[ f'''(x_i)=\left\{B_i\left\{\sum_{\lambda=0}^{n}\eta_\lambda \left\{\sum_{\mu=0}^{n}\eta_\mu \left[\sum_{\chi=0}^{n}\eta_\chi \prod_{\nu=0}^{n}(x_i-x_\nu)\right]\right\}\right\}\right\}+ \]

\[ +\sum_{s=0}^{n}B_s\left\{\sum_{\mu=0}^{n}\eta_\mu \left[\sum_{\chi=0}^{n}\eta_\chi \prod_{\nu=0}^{n}(x_i-x_\nu)\right]\right\}\right\}. \tag{22} \]

\[ \left( \begin{array}{lll} s\ne \nu\cap\lambda\cap\mu\cap\chi\cap i; & \lambda\ne s\cap\nu\cap\mu\cap\chi\cap i; & \chi\ne s\cap\nu\cap\lambda\cap\mu\cap i;\\ \nu\ne s\cap\lambda\cap\mu\cap\chi\cap i; & \mu\ne s\cap\nu\cap\lambda\cap\chi\cap i; & i\ne s\cap\nu\cap\lambda\cap\mu\cap\chi. \end{array} \right) \]

Example. Let \(s=0,1,2,3,4\), and \(i=2\). Then

\[ \text{a) }\quad f(x)= \frac{f(x_0)} {(x_0-x_1)(x_0-x_2)(x_0-x_3)(x_0-x_4)} (x-x_1)(x-x_2)(x-x_3)(x-x_4)+ \]

\[ +\frac{f(x_1)} {(x_1-x_0)(x_1-x_2)(x_1-x_3)(x_1-x_4)} (x-x_0)(x-x_2)(x-x_3)(x-x_4)+ \]

\[ +\frac{f(x_2)} {(x_2-x_0)(x_2-x_1)(x_2-x_3)(x_2-x_4)} (x-x_0)(x-x_1)(x-x_3)(x-x_4)+ \]

\[ +\frac{f(x_3)} {(x_3-x_0)(x_3-x_1)(x_3-x_2)(x_3-x_4)} (x-x_0)(x-x_1)(x-x_2)(x-x_4)+ \]

\[ +\frac{f(x_4)} {(x_4-x_0)(x_4-x_1)(x_4-x_2)(x_4-x_3)} (x-x_0)(x-x_1)(x-x_2)(x-x_3)= \]

\[ B_0(x-x_1)(x-x_2)(x-x_3)(x-x_4)+B_1(x-x_0)(x-x_2)(x-x_3)(x-x_4)+ \]

\[ B_2(x-x_0)(x-x_1)(x-x_3)(x-x_4)+B_3(x-x_0)(x-x_1)(x-x_2)(x-x_4)+ \]

\[ +B_4(x-x_0)(x-x_1)(x-x_2)(x-x_3). \]

If here we introduce the notation

\[ 1)\quad x_\chi-x_\lambda=x_{\chi\lambda},\qquad 2)\quad x-x_a, \]

then \(f(x)\) is written in the following abbreviated form:

\[ f(x)= \frac{f(x_0)}{x_{01}x_{02}x_{03}x_{04}}x_{a1}x_{a2}x_{a3}x_{a4} + \frac{f(x_1)}{x_{10}x_{12}x_{13}x_{14}}x_{a0}x_{a2}x_{a3}x_{a4} + \]

\[ +\frac{f(x_2)}{x_{20}x_{21}x_{23}x_{24}}x_{a0}x_{a1}x_{a3}x_{a4} + \frac{f(x_3)}{x_{30}x_{31}x_{32}x_{34}}x_{a0}x_{a1}x_{a2}x_{a4} + \]

\[ +\frac{f(x_4)}{x_{40}x_{41}x_{42}x_{43}}x_{a0}x_{a1}x_{a2}x_{a3}. \]

or, in still more abbreviated form,

\[ f(x)=B_0x_{a1}x_{a2}x_{a3}x_{a4} +B_1x_{a0}x_{a2}x_{a3}x_{a4} +B_2x_{a0}x_{a1}x_{a3}x_{a4} + \]
\[ +B_3x_{a0}x_{a1}x_{a2}x_{a4} +B_4x_{a0}x_{a1}x_{a2}x_{a3}. \]

b)

\[ f'(x_2)=B_0x_{21}x_{23}x_{24} +B_1x_{20}x_{23}x_{24} +B_3x_{20}x_{21}x_{24} +B_4x_{20}x_{21}x_{23} + \]
\[ +B_2(\eta_0x_{21}x_{23}x_{24} +\eta_1x_{20}x_{23}x_{24} +\eta_3x_{20}x_{21}x_{24} +\eta_4x_{20}x_{21}x_{23}) \]

or otherwise

\[ f'(x_2)=(B_0x_{21}+B_1x_{20})x_{23}x_{24} +(B_3x_{24}+B_4x_{23})x_{20}x_{21} + \]
\[ +B_2[(x_{20}+x_{21})x_{23}x_{24} +(x_{23}+x_{24})x_{20}x_{21}]. \]

c)

\[ f''(x_2)=B_0(\eta_1x_{23}x_{24} +\eta_3x_{21}x_{24} +\eta_4x_{21}x_{23})+ \]
\[ +B_1(\eta_0x_{23}x_{24} +\eta_3x_{20}x_{24} +\eta_4x_{20}x_{23})+ \]
\[ +B_3(\eta_0x_{21}x_{24} +\eta_1x_{20}x_{24} +\eta_4x_{20}x_{21})+ \]
\[ +B_4(\eta_0x_{21}x_{23} +\eta_1x_{20}x_{23} +\eta_3x_{20}x_{21})+ \]
\[ +B_2[\eta_0(\eta_1x_{23}x_{24} +\eta_3x_{21}x_{24} +\eta_4x_{21}x_{23})+ \]
\[ +\eta_1(\eta_0x_{23}x_{24} +\eta_3x_{20}x_{24} +\eta_4x_{20}x_{23})+ \]
\[ +\eta_3(\eta_0x_{21}x_{24} +\eta_1x_{20}x_{24} +\eta_4x_{20}x_{21})+ \]
\[ +\eta_4(\eta_0x_{21}x_{23} +\eta_1x_{20}x_{23} +\eta_3x_{20}x_{21})]. \]

Hence, after collecting like terms, we find that

\[ f''(x_2)=(B_0+B_1)x_{23}x_{24} +(B_0+B_3)x_{21}x_{24} +(B_0+B_4)x_{21}x_{23}+ \]
\[ +(B_1+B_3)x_{20}x_{24} +(B_1+B_4)x_{20}x_{23} +(B_3+B_4)x_{20}x_{21}+ \]
\[ +2B_2(x_{20}x_{21}+x_{20}x_{23}+x_{20}x_{24} +x_{21}x_{23}+x_{21}x_{24}+x_{23}x_{24}). \]

d)

\[ f'''(x_2)=B_0[\eta_1(\eta_3x_{24}+\eta_4x_{23}) +\eta_3(\eta_1x_{24}+\eta_4x_{21}) +\eta_4(\eta_1x_{23}+\eta_3x_{21})]+ \]
\[ +B_1[\eta_0(\eta_3x_{24}+\eta_4x_{23}) +\eta_3(\eta_0x_{24}+\eta_4x_{20}) +\eta_4(\eta_0x_{23}+\eta_3x_{20})]+ \]
\[ +B_3[\eta_0(\eta_1x_{24}+\eta_4x_{21}) +\eta_1(\eta_0x_{24}+\eta_4x_{20}) +\eta_4(\eta_0x_{21}+\eta_1x_{20})]+ \]
\[ +B_4[\eta_0(\eta_1x_{23}+\eta_3x_{21}) +\eta_1(\eta_0x_{23}+\eta_3x_{20}) +\eta_3(\eta_0x_{21}+\eta_1x_{20})]+ \]
\[ +6B_2(x_{20}+x_{21}+x_{23}+x_{24}). \]

Carrying out here one more collection of like terms, we obtain

\[ \frac{1}{2}f'''(x_2)=(B_0+B_1)(x_{23}+x_{24}) +(B_0+B_3)(x_{21}+x_{24})+ \]
\[ +(B_0+B_4)(x_{21}+x_{23}) +(B_1+B_3)(x_{20}+x_{24})+ \]
\[ +(B_1+B_4)(x_{20}+x_{23}) +(B_3+B_4)(x_{20}+x_{21})+ \]
\[ +3B_2(x_{20}+x_{21}+x_{23}+x_{24}). \]

e) If, in the expansion for \(\frac{1}{2}f'''(x_2)\), we replace \(x_2\) by \(x=x_a\), and then differentiate the resulting expression with respect to \(x=x_a\), then, after simplification, we find that

\[ \frac{1}{6}f''''(x_2)=B_0+B_1+B_3+B_4+4B_2, \]

\[ \text{e)}\quad f'''''(x_2)=0. \]

The example presented shows that numerical computation according to (20)—(22) for the derivatives \(f^{(x)}(x_i)\) at arbitrarily specified nodes \(x_i\) of the interval \(a \le x \le b\) is expedient only for \(n \le 5\), and then \(f^{(n)}(x_i)=0\). If, however, it is required to find numerical values of derivatives of higher order, one must proceed from the representation of \(f(x)\) by Newton’s expansion with divided differences [2]

\[ \begin{aligned} f(x) &= f(x_0)+f(x_0x_1)(x-x_0)+f(x_0x_1x_2)(x-x_0)(x-x_1)+\ldots = \\ &= f(x_0)+\sum_{\chi=1}^{n} f(x_0x_1\ldots x_\chi)\prod_{\nu=0}^{\chi-1}(x-x_\nu) + f(xx_0x_1\ldots x_n)\prod_{\nu=0}^{n}(x-x_\nu). \end{aligned} \tag{23} \]

Here \(f(x_0x_1\ldots x_\chi)\) and \(f(xx_0x_1\ldots x_n)\) are divided differences of orders \(\chi-1\) and \(n\), which are determined as follows:

\[ \begin{aligned} 1)\quad f(x_0x_1\ldots x_\chi) &= \frac{f(x_0x_2\ldots x_{\chi-1})-f(x_1x_2\ldots x_\chi)} {x_0-x_\chi}, \\[6pt] 2)\quad f(xx_0x_1\ldots x_n) &= \frac{f(xx_0x_1\ldots x_{n-1})-f(x_0x_1x_2\ldots x_n)} {x-x_n}, \end{aligned} \tag{24} \]

so that

\[ 1)\quad f(x_0x_1)=\frac{f(x_0)-f(x_1)}{x_0-x_1},\qquad 2)\quad f(xx_0)=\frac{f(x)-f(x_0)}{x-x_0}. \]

However, if general expressions are required for the derivatives \(f^{(x)}(x_i)\) at nonuniform nodes \(x_i\) of the interval \(a \le x \le b\), then recourse to the corresponding expressions of the form (20)—(22) is unavoidable, despite their complication as the number of nodes \(n\) taken in the calculation of \(f^{(x)}(x_i)\) increases.

1. In the present article, various general expressions have been considered for the functions \(f(x)\) and its derivatives \(f^{(x)}(x)\), obtained by the method of partial matching for an arbitrary arrangement of the nodes \(x_s\). From these we formed the corresponding general expressions for the values \(f^{(x)}(x_i)\) of the derivatives \(f^{(x)}(x)\) at specified nodes \(x_i\) of a certain interval \(a \le x \le b\). Subsequently it will be shown how these nodal values \(f^{(x)}(x_i)\) of the derivatives \(f^{(x)}(x)\) can be used for the exact solution, by the method of partial matching, of a boundary-value problem in an arbitrary closed domain \(D^{(2)}=D^{(2)}+\Gamma\) with boundary \(\Gamma\)

\[ 0)\quad x=(x^1,x^2),\qquad 1)\quad L(x)u(x)=f(x), \]

\[ 2)\quad \mathcal{L}(x)u(x)\big|_{\Gamma}=h(x,y), \tag{25} \]

if \(L(x)\) and \(\mathcal{L}(x)\) are linear differential operators in partial derivatives, independent of the time \(t\).

REFERENCES

1. E. D. But. Numerical Methods. Moscow, Fizmatgiz, 1959.

2. E. Whittaker and G. Robinson. Mathematical Processing of Observational Results. Moscow–Leningrad, ONTI, 1935.