# New PDF release: Calculus of Finite Differences (AMS Chelsea Publishing) By Charles Jordan

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Example text

L-~5)x-(1+ 1/ p'l , and finally (1) We could have obtained the same result by solving the difference equation deduced in Sfirling’s method (Method II, 6 12). ;-1 15” = FL gives the interesting relation ,,z (“-F-l). Example 2. Decomposition of the reciprocal factorial We u(t) = 1 /t (t-l) (t-2) . , . iave ai = therefore mDt(t-I) u = 1 ... wig (t-m)lz wi(~) m! (t-i) In the same manner we should have r=O 1-l m - (-l)m’i -\$, 37 m+1 w,‘(y) 1 ’ = t (f+l) (t+2) . , . (t+m) - 20 ml (t+i) B. g are real: therefore if rl = a+@ is a root of y(t) = 0, rs = a-pi will be a root too.

1 have du -du dydr = d y df unction of function. [See Schliimilch’s us write u = u(y) and y = y(f), We &U ;i’Ti where the functions Y,t are ; determine them by choosing Let du - d2yd y df2 of u. , + . . J emurJ, From this we deduce that d” e-w’ dm ewY w=. 32 Moreover, hence from this it follows by aid of (7’) that so that Finally we have For every particular function y we may determine once for al1 the quantities Y,, , Y,, , , . , Y,,. Example 1. Let y = il. We may write [pfls = esr (e”-lp = esr 21 (-1) i ( f ) ehls-il hence -\$[ +d]s = esr Putting 21 (-1)’ (s) (s-i)” eh(ri) .

G\$=e’ 3. --log ( l - f ) = j,; f” - - G; 4. (lff)” = Tsl (;) f” A’=0 G (;) = Il+fl’: 5. f (f-l) (f-2) . , . (f--n+l) = therefore et+ 1 G 1 = & I: s: /=l = - log (l-f) f” G S:,= f(f-1) . (f--n+l), Where the numbers Si are the Stirling Numbers of the first kind. We shall see them later ((j 50). 6. f(2r) = 0 , 2 fG5+1) = 2”+* l+f u(f) = log l - - f 7. f(2v) E 0 , f(2v+l) u ( f ) = arctan f 8. I--11’ f(2vfl) x0, f(2e) = (2s) 9. f (2v) =o, fWil1 10. f(%+l)=o, f(h) 11. = = (3 Y = (;)yl;)! (2;,,! f(O)xl, f(21~+1)=0, f(2v)=Elr (2r) !