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By Jordan C.

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To prove that formula (1) holds too for any values of x it is necessary to show that the operation E is identical with that corresponding to the series E: = l+(;jA+[;)Y+ . . +{;r]6’+ I..... This is obviously impossible if the series f(O) + (~]AW + . . + 1;) A”fIo) +-a / I is divergent. On the other hand if f(x) is a polynomial of degree n then A*+I f(x) = 0 and the corresponding series is finite. Steffensen 110~. cit. 1. p. 1841 has. shown that in such cases the expansions are justified. If we limit the use of the symbolical expansions to these cases, their application becomes somewhat restricted; but as Steffensen remarked, these expansions are nevertheless of considerable use, since the form of an 1 2 interpolation or summation formula does not depend on whether the function is a polynomial or not (except the remainder term) If certain conditions are satisfied, the expansion of operation symbols into infinite series may be permitted even if the function to which the operations are applied, is not a polynomial when the corresponding series is convergent.

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. First multiplying sf (x,) by w. (x,,J, then B2f (x0) by o1 (x,) and so on: jDvf (x,) by wIW1 (x,,) and finally Bmf (x,,) by CO,,,-, (x,) we obtain + m+1 $1 cd”-1 = i=l v=i D (x,,,) wy (Xi) f(xi)W . =~ D toy (x,) = -’ ’ so that the coefficient of f(x,,) is equal to -1. Moreover it can be demonstrated that VI+1 WY-l (&) ,zl D (tiv (Xi) = O if m ’ i ’ O therefore for these values the term f(xi) will vanish from the equation and we have f M = f (x0) + h---x,) w (4 + brr-x0) (x,,,--xl) 6’f(X”) + (4 + bi---x,1 hr--XI) (w---x,) %3f(X,,) + ’ * * ’ + hr-4 I&c--xl) * * - (xm--x,-l) 9”f(X”).

F,,. (DY,,) ‘I (F )“” ’ . , , RnD”vo ( n! 1 - 0 13. Expansion of functions by aid of decomposition into partial fractions. /p(t). We may always suppose that ,, (t) and (,1(f) have no roots in common; since if they had, it would always be possible to simplify the fraction, dividing by f-r,,, , if r,,, i s . the root. A. Let us suppose that the roots t,, r3, , , , , r,, of t/*(f) are all real and unequal. We have u= tt+l ai z a=, f--c 35 Reducing the fractions to a common denominator we obtain u if for every value of t we have *+1 44 Y(t) = iz, Qi-ri therefore this is an identity: so that the coefficients of P’, in both members, must be equal, This gives R equations of the first degree which determine the coefficients oi.

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Calculus of Finite Differences by Jordan C.

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