Let fmathend000#, gmathend000# et hmathend000# be functions from
mathend000# to
mathend000#
We say that g(n)
mathend000# is in the ORDER OF MAGNITUDE
of f (n)
mathend000# and we write
f (n) (g(n))
mathend000# if
there exist two (strictly) positive constants c1mathend000# and c2mathend000#
such that for nmathend000# big enough we have
0 c1g(n) f (n) c2g(n).
(2)
We say that g(n)
mathend000# is an ASYMPTOTIC UPPER BOUND
of f (n)
mathend000# and we write
f (n) (g(n))
mathend000# if there exists
a (strictly) positive constants c2mathend000#
such that for nmathend000# big enough we have
0 f (n) c2g(n).
(3)
We say that g(n)
mathend000# is an ASYMPTOTIC LOWER BOUND
of f (n)
mathend000# and we write
f (n) (g(n))
mathend000# if there exists
a (strictly) positive constants c1mathend000#
such that for nmathend000# big enough we have
0 c1g(n) f (n).
(4)
Let us give some examples.
With
f (n) = n2 - 3nmathend000# and
g(n) = n2mathend000# we have
f (n) (g(n)).
mathend000#
Indeed we have
c1n2n2 -3nc2n2.
(5)
for n 12
mathend000# with
c1 = mathend000# and
c2 = mathend000#.
Assume that there exists a positive integer n0mathend000#
such that f (n) > 0
mathend000# and g(n) > 0
mathend000# for every
nn0mathend000#.
Then we have
max(f (n), g(n)) (f (n) + g(n)).
(6)
Indeed we have
(f (n) + g(n)) max(f (n), g(n)) (f (n) + g(n)).
(7)
Assume amathend000# and bmathend000# are positive real constants.
Then we have
(n + a)b(nb).
(8)
Indeed for namathend000# we have
0 nb (n + a)b (2n)b.
(9)
Hence we can choose c1 = 1
mathend000# and
c2 = 2bmathend000#.
We have the following properties.
f (n) (g(n))
mathend000# holds iff
f (n) (g(n))
mathend000# and
f (n) (g(n))
mathend000# hold together.
Each of the predicates
f (n) (g(n))
mathend000#,
f (n) (g(n))
mathend000# and
f (n) (g(n))
mathend000#
define a reflexive and transitive binary relation
among the
mathend000#-to-
mathend000# functions.
Moreover
f (n) (g(n))
mathend000# is symmetric.
We have the following TRANSPOSITION FORMULA
f (n) (g(n)) g(n) (f (n)).
(10)
In practice mathend000# is replaced by =
mathend000#
in each of the expressions
f (n) (g(n))
mathend000#,
f (n) (g(n))
mathend000# and
f (n) (g(n))
mathend000#.
Hence, the following
f (n) = h(n) + (g(n))
(11)
means
f (n) - h(n) (g(n)).
(12)
Let us give another fundamental example.
Let p(n)
mathend000# be a (univariate) polynomial with degree d > 0
mathend000#.
Let admathend000# be its leading coefficient and assume ad > 0
mathend000#.
Then we have