1.5 O and o

Definition 1.5.1 f=O(g) when x→a if there exists M>0 and δ0>0 such that for any 0<∥x−a∥<δ0, ∥f(x)∥≤M∥g(x)∥.

Definition 1.5.2 f and g has the same order when x→a if and only if f=O(g) and g=O(f), meaning there exists M>0 and δ0>0 such that for any 0<∥x−a∥<δ0, 1M∥g(x)∥≤∥f(x)∥≤M∥g(x)∥.

Definition 1.5.3 f=o(g) when x→a if for any ϵ>0, there exists δϵ>0 such that for any 0<∥x−a∥<δ0, ∥f(x)∥≤ϵ∥g(x)∥.

Definition 1.5.4 f and g are equivalent if and only if f=g+o(g).

Theorem 1.5.5 Given f,g:E→Rp, where f(x)=(f1(x)⋯fp(x))Tg(x)=(g1(x)⋯gp(x))T

then f=O(g)⇔fk=O(gk), for any k.

Proof Hint.

 1           1
---|fk(x)| < ---∥f(x )∥ ∞ ≤ ∥f (x)∥ ≤ M  ∥f(x)∥1 = M  ∖lef t[|f1(x)| + ...+ |fp(x)|∖right ]
M           M

◻

Example 1.5.1 Several examples for O and o.

1.

Linear (matrix) A. Ax=O(x), since ∥Ax∥≤∥A∥∥x∥.

2.

Quadratic form. ⟨Ax,x⟩=O(∥x∥2), since |⟨Ax,x⟩|≤∥Ax∥∥x∥≤∥A∥∥x∥2.

3.

xy=O(x2+y2),(x,y)→(0,0), since |xy|≤x2+y22. Does xy has the same order as x2+y2 when (x,y)→(0,0) ? No! For any δ-deleted neighborhood of (0,0), there exists a point whose components x≠0 and y=0. Here 0<∥(x,0)−(0,0)∥2=|x|<δ. If x2+y2=O(xy), then there exists M,δ1>0 such that for any (x,0), 0<∥(x,y)−(0,0)∥<δ1, x2+y2≤M|xy|, here 0<x2≤M|x⋅0|=0, contradicted.

4.

x2+y2 has the same order as 2x2+3y2, since 2(x2+y2)≤2x2+3y2≤3(x2+y2).

5.

The difference between xy and x2+y2 is shown in Figure 1.1. They don’t have the same order because their graph near (0,0) is different. Therefore, the definition of an exact order on the textbook is not well-defined, because the relation between x,y may result in various results.

PIC
(a) Difference between xy and x2+y2

PIC
(b) The graph of z=xy
PIC
(c) The graph of z=x2+y2
Figure 1.1: xy and x2+y2