Concrete mathematics : a foundation for computer science

Infinite calculus
avoids E here by
letting 1 -3 0.
1 guess ex = 2”) for
small values of 1
Table 55 What’s the difference?
f = zg Af = g
x0 = 1 0
x1 = x 1
x2=x(x-l) 2 x
mxti
XB
xmf'/(m+l) x=
HX x-‘= l/(x+1)
2.6 FINITE AND INFINITE CALCULUS 55
f=Lg Af = g
2" 2"
CX (c - 1 )cX
c"/(c-1) cx
cf cAf
f+g Af+Ag
f g
fAg + EgAf
after integration and rearranging terms; we can do a similar thing in finite
calculus.
We start by applying the difference operator to the product of two functions
u(x) and v(x):
A@(x) v(x)) = u(x+l) v(x+l) - u(x) v(x)
= u(x+l)v(x+l)-u(x)v(x+l)
+u(x)v(x+l)-u(x)v(x)
= u(x) Av(x) + v(x+l) Au(x). (2.54)
This formula can be put into a convenient form using the shij?! operator E,
defined by
Ef(x) = f(x+l).
Substituting this for v(x+l) yields a compact rule for the difference of a
product:
A(uv) = uAv + EvAu. (2.55)
(The E is a bit of a nuisance, but it makes the equation correct.) Taking
the indefinite sum on both sides of this equation, and rearranging its terms,
yields the advertised rule for summation by parts:
ix uAv = uv- t EvAu. (2.56)
As with infinite calculus, limits can be placed on all three terms, making the
indefinite sums definite.
This rule is useful when the sum on the left is harder to evaluate than the
one on the right. Let’s look at an example. The function s xe’ dx is typically
integrated by parts; its discrete analog is t x2’ 6x, which we encountered
earlier this chapter in the form xt=, k2k. To sum this by parts, we let