By Jordan C.

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**Extra resources for Calculus of Finite Differences **

**Sample text**

Since the expansion of u,(f) is simpler than that of u(f), we will use the former for the determination of f(x). We get u, (f) zz f”’ Finally we have i f x57. , (-1)” [ :,“I fY = 9, [ m+zy-‘] tmtv. , 22 To determine the generating functions there are several methods. First method. In Infinitesimal Calculus a great number of expansions in power series are known. Each function expanded may be considered as a generating function. For instance we found: 1. 1 s f” l--t= a=0 2. e’ = So -$ f” . , G$=e’ 3.

By the first divided difference of f(xi), denoted by Ff (xi) the following quantity is understood: Zf (xi) = f (xi:! Q+ -j- fo x1-x0 ’ W(4 = (x f(X”l ! fk) 0 --x,1 h---x,) -i- (x’--xJ (XI--XJ + f(x2) + (q--x,) (x2-q ’ I and so on. Putting W,,,(X) = (x-x,,) (x-x,) . , . pf(x,) = &k-g + -w-- -f . . ,. fkol X0 x0 m-l Xl2 . . 1 Xl fkl Xl . , . , . , . I p7f(x,) = l 1x. 1 xm2 ':' x0 xo2 ... Xn Xl Xl ... ymrn-l 2 m f(XfJ m ,.. * . . 1 I . . x,’ xm X”, m NOW we shall deduce an expression forf(x,) by aid of the divided differences.

The roots rr , r2, . . , r, of the polynomial q(f) are all real, but there are multiple roots. ; where m,. may be equal td 1, 2, 3, * * . , and so on. The general formula of decomposition into partial fractions is the following: If we multiply this expression by (f-ry)m~ we get (4) 233 (f---ry)Nlt, = t&l (f--fTy%-l + . . a,em,, + R(f) , (f-r,)“‘v WI We have l)(f) = C(f-rp (f--r2)m~*, . (t), A,(r,) will be different from zero: A”(f”) = am, In the same manner the first m,,--s derivatives give D”~-~dv(rJ = (m,-s)!