Concrete mathematics : a foundation for computer science

Setting E = 1 tells us that
Af(x) = f(x+ 1) -f(x)
= f/(x)/l! + f”(X)/2! + f”‘(X)/3! + “.
9.5 EULER’S SUMMATION FORMULA 457
= (D/l!+D2/2!+D3/3!+...)f(x) = (eD-l)f(x). b69)
Here eD stands for the differential operation 1 + D/l ! + D2/2! + D3/3! + . . . .
Since A = eD - 1, the inverse operator t = l/A should be l/(eD - 1); and
we know from Table 337 that z/(e’ - 1) = &c Bk.zk/k! is a power series
involving Bernoulli numbers. Thus
1 = ~+!?+$D+$D~+... = J+&$Dkpl.
Applying this operator equation to f(x) and attaching limits yields
~;f(x)hx = Jb f(x) dx + x sf’kpl’(x) b
a k>l k! a'
(9.70)
(9.71)
which is exactly Euler’s summation formula (9.67) without the remainder
term. (Euler did not, in fact, consider the remainder, nor did anybody else
until S. D. Poisson [:236] published an important memoir about approximate
summation in 1823. The remainder term is important, because the infinite
sum xk>, (Bk/k!)fCk--‘)(x)li often diverges. Our derivation of (9.71) has been
purely formal, without regard to convergence.)
Now let’s prove (g.67), with the remainder included. It suffices to prove
the case a = 0 and b = 1, namely
J
1
’ Bm(x) (ml
f(0) = f(x) tix + f 3 wyx)~’ - (-l)m J -
0 k=, k! o 0 m! f (x) dx ’
because we can then replace f(x) by f (x + 1) for any integer 1, getting
J
f(l) = 1
lfl
f(x)dx+f ~f(kpl)(x)lL+‘- (-l)m J
k=, k! 1
1
I+’ Bm()’ f’“’ (x) dx
~
The general formula (9.67) is just the sum of this identity over the range
a 6 1 < b, because intermediate terms telescope nicely.
The proof when a = 0 and b = 1 is by induction on m, starting with
m= 1:
f(0) = J’f(x)dx-f(f(l)-f(o))+J’(x-;)f’(x)dx.
0 0