This observation follows from the definition of the derivative of a function of onevariable, namely
f′(x) ≡ lim f-(x+-h)−-f-(x).
h→0 h
Thus
|f (x+ h)− f (x)− f′(x) h| ||f (x + h)− f (x) ′ ||
hli→m0 -----------|h|-----------= lhim→0||-------h------ − f (x)|| = 0
Definition 21.1.2A vector valued function of a vector v is called o
(v)
(referred to as “little o of v”) if
lim o-(v-)= 0. (21.1)
|v|→0 |v|
(21.1)
Thus for a function of one variable, the function f
(x+ h)
− f
(x)
− f^{′}
(x )
h is o
(h)
.
When we say a function is o
(h)
, it is used like an adjective. It is like saying the function
is white or black or green or fat or thin. The term is used very imprecisely. Thus in
general,
o(v) = o (v )+ o(v), o (v ) = 45× o(v), o(v) = o (v )− o(v),etc.
When you add two functions with the property of the above definition, you get another
one having that same property. When you multiply by 45, the property is also retained,
as it is when you subtract two such functions. How could something so sloppy be useful?
The notation is useful precisely because it prevents you from obsessing over things which
are not relevant and should be ignored.
Theorem 21.1.3Let f :
(a,b)
→ ℝ be a function of one variable. Then f^{′}
(x)
exists if and only if there exists p such that
f (x +h )− f (x) = ph + o(h) (21.2)
(21.2)
In this case, p = f^{′}
(x)
.
Proof:From the above observation it follows that if f^{′}
sin (x + h)− sin (x) = sin(x)cos(h)+ cos(x) sin(h)− sin (x )
(cos(h)− 1)
= cos(x)sin (h)+ sin(x) ----h-----h
(sin(h)− h) (cos(h)− 1)
= cos(x)h + cos(x) ----h-----h+ sinx-----h-----h.
Now
cos(x) (sin(h)−-h)h + sinx (cos(h)−-1)h = o(h). (21.3)
h h
(21.3)
Remember the fundamental limits which allowed you to find the derivative of sin
(x)
were
lim sin(h)-= 1, lim cos(h)−-1 = 0. (21.4)
h→0 h h→0 h
(21.4)
These same limits are what is needed to verify 21.3.
How can you tell whether a function of two variables
(u,v)
is o
( )
u
v
? In general,
there is no substitute for the definition, but you can often identify this property by
observing that the expression involves only “higher order terms”. These are terms like
u^{2}v,uv,v^{4}, etc. If you sum the exponents on the u and the v you get something larger
than 1. For example,
|| vu || 1 ( 2 2) 1 1∘ -------
||√--2---2|| ≤ 2 u + v √--2---2-= 2 u2 + v2
u + v u + v
and this converges to 0 as
(u,v)
→
(0,0)
. This follows from the inequality
|uv|
≤
1
2
( 2 2)
u + v
which you can verify from
(u − v)
^{2}≥ 0. Similar considerations apply
in higher dimensions also. In general, this is a hard question because it involves a limit
of a function of many variables. Furthermore, there is really no substitute for
answering this question, because its resolution involves the definition of whether a
function is differentiable. That may be why we spend most of our time on one
dimensional considerations which involve taking the partial derivatives. The following
exercises should help give you an idea of how to determine whether something is
o.