Landau notation division

  • #1
190
0
Why is it that,

##
\frac{a+\mathcal{O}(h^2)}{b+\mathcal{O}(h^2)} = \frac{a}{b}+\mathcal{O}(h^2)
##

as ##h\rightarrow 0##? It seems like the ##\mathcal{O}(h^2)## term should become ##\mathcal{O}(1)##.
 

Answers and Replies

  • #2
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8,367
Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
 
  • #3
statdad
Homework Helper
1,495
35
Why is it that,

a+(h2)b+(h2)=ab+(h2) \frac{a+\mathcal{O}(h^2)}{b+\mathcal{O}(h^2)} = \frac{a}{b}+\mathcal{O}(h^2)

as h→0h\rightarrow 0? It seems like the (h2)\mathcal{O}(h^2) term should become (1)\mathcal{O}(1).

Think about this very simple illustration. Consider the specific case

[tex]
c = \frac{a + k1h^2}{b + k2h^2}
[/tex]

with k1, k2 positive (specific choices for the second terms in numerator and denominator, but since h-squared is the key term it's find. Bear with me.)

Suppose that fraction is expanded into a power series in h. The first few terms are

[tex]
\frac a b −\frac{((ak2−bk1)h^2)}{b^2}+\frac{((ak2^2−bk1k2)h^4)}{b^3}−\frac{((ak2^3−bk1k2^2)h^6)}{b^4}+\frac{((ak2^4−bk1k2^3)h^8)}{b^5}+... = \frac a b + O(h^2)
[/tex]
 

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