Concrete mathematics : a foundation for computer science

474 ASYMPTOTICS
This is our approximation, with
22n+l/2
bk(n) = ~ e -k2/n
e”fi ’
ck(n) = 22nnp1+3e e-k2/n.
Notice that k enters bk(n) and ck(n) in a very simple way. We’re in luck,
because we will be summing over k.
The tail-exchange trick tells us that tk ok(n) will be approximately
tk bk(n) if we have done a good job of estimation. Let us therefore evaluate
xbk(n) = g I- e-k*/n
k k-
=
(Another stroke of luck: We get to use the sum 0, from the previous example.)
This is encouraging, because we know that the original sum is actually
A, = x ‘k”
( )
k
= (1 f 1)2, = 22n.
Therefore it looks as if we will have e“ = 6, as advertised.
But there’s a catch: We still need to prove that our estimates are good
enough. So let’s look first at the error contributed by C&(n):
L(n) = x 22nn~1+3ee~k2/n < 22nn~1+3c@
\ n= O(22”npt+3’).
~kl$n’/2+~
Good; this is asymptotically smaller than the previous sum, if 3~ < i.
Next we must check the tails. We have
x epk2/n < exp(--Ln’/2+eJ2/n) (1 + e-‘/n + e-21” + . )
k>n’/ZiC
= O(ep1L2F) . O(n),
which is O(nPM) for all M; XI Eke,,, bk(n) is asymptotically negligible. (We
chose the cutoff at n’/2fe just so that eek21n would be exponentially small
outside of D,. Other choices like n ‘/2 logn would have been good enough
too, and the resulting estimates would have been slightly sharper, but the
formulas would have come out more complicated. We need not make the
strongest possible estimates, since our main goal is to establish the value of
the constant o.) Similarly, the other tail
What an amazing
coincidence.
I’m tired of getting
to the end of long,
hard books and not
even getting a word
of good wishes from
the author. It would
be nice to read a
“thanks for reading
this, hope it comes
in handy,” instead
of just running into
a hard, cold, cardboard
cover at the
end of a long, dry
proof You know?