Proving Little o Notation: f(x) = o(g(x)) as x → 0

  • Thread starter Thread starter zjhok2004
  • Start date Start date
  • Tags Tags
    Notation
zjhok2004
Messages
8
Reaction score
0

Homework Statement


Given two functions f and g with derivatives in some interval containing 0, where g is positive. Assume also f(x) = o(g(x)) as x → 0. Prove or disprove each of the following statements:


a)∫f(t) dt = o(∫g(t)dt) as x → 0 (Both integrals goes from 0 to x)
b)derivative of f(x) = o( derivative of g(x)) as x → -

Can anyone show me how to prove this? thanks
 
Physics news on Phys.org
Recall, ##f = o(g)## implies :

##\forall k>0, \exists a \space | \space f(x) < kg(x), \forall x>a## where 'k' and 'a' are arbitrary constants.

That's what you meant by "Assume also f(x) = o(g(x)) as x → 0" right?
 
f=o(g) as x -> 0 means lim f/g ->0 as x ->0
 
zjhok2004 said:
f=o(g) as x -> 0 means lim f/g ->0 as x ->0


Not true, what about f = x2, then x2 = o(x3) ( For example ).

Then x2/x3 = 1/x → ±∞ as x → 0.
 
Zondrina said:
Not true, what about f = x2, then x2 = o(x3) ( For example ).

Then x2/x3 = 1/x → ±∞ as x → 0.

That is the definition of the little o notation
 
zjhok2004 said:
That is the definition of the little o notation

No, the definition is what I've given you.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top