By Jordan C.

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Extra resources for Calculus of Finite Differences

Sample text

A,. vo . (R! , . cl,,‘, (DY,,)“I D”y,, an n! +ff,, II . a, + 2~2, + . , . + ncf,, = R. Remark. Comparing formulae (8) and (9) we find (IO) -& I& [Ay(~)jj=~ = x ,I,! “! , . , . ‘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.

We find G{(x) (1) = PI + 1 “=O X A”f(0) “’ (l--fJY+’ 3. Given the generating function of f(x) G f(x) = u(t) = ,z, f(x) t’ we may easily deduce the generating function of f (x+1) ; indeed from the preceding equation we may obtain u(f) -f(O) t (4 % =k” f(xf1) i” = Gf(x+l) Therefore the generating fanction of the difierencc of f(x) will be GAfIx) (2’) = f [(l--t)u(f) -f(o)) In the same manner we have ( 3 ) Gf(x-i- II) = + [U(l) ---f(O) --~ tf(1) - . - t”-‘f(n-1) i. \nlf(x) or of Mlprf(x) and so on, Example.

F”)” but for x