# Big O notation question

1. Aug 4, 2014

### ericm1234

I am trying to reconcile the following statement:

" ||u||<=eps*||f|| means ||u||=O(eps) ("||u|| is order eps")... "

with the limit definition of "big O"; considering it's not clear that ||u|| here even depends on eps:

" lim as eps goes to 0 of ||u||/eps, by definition of "big O", should equal some constant "; is it necessarily ||f||?

I understand that, for example, sin(x)=O(x) because lim as x goes to 0 of sin(x)/x is bounded.

So to casually say something is "order epsilon", if it doesn't depend on epsilon, I am not sure how to reconcile that with the definition of "big O" given by the limit one should be able to take, as above.

2. Aug 4, 2014

3. Aug 4, 2014

### ericm1234

I paraphrased for the sake of drawing attention to my (pedantic) question;
The original statement, from a book on finite element, was:
||u-u_s||<=eps*||u-u_s||_E <=eps^2*||f|| (the middle norm is energy norm)

the following text then goes:
"The point of course is that ||u-u_s||_E (energy norm) is of order eps, whereas ||u-u_s|| is of order eps^2."

So maybe you can explain based on this context then?

And sorry for lack of "tex" skills.

4. Aug 4, 2014

### Simon Bridge

It looks like the author is not using regular big-O notation and/or conventions here.

Possibly the author just refers to the order in "epsilon".
What is epsilon supposed to stand for?

Last edited: Aug 4, 2014
5. Aug 4, 2014

### ericm1234

Well, that was sort of what I figured..but anyway, Brenner, Mathematical Theory of Finite Element, p.6 bottom.

Thanks

6. Aug 4, 2014

### Simon Bridge

I don't know that one - the best way to be sure is to look for the same lesson in another text and compare.

If I said that |y| = x|k| while |z|=x^2|k| then you could say that y = O(x) while z=O(x^2).
I'm thinking that the dependent variable in the inequalities above is epsilon.

7. Aug 4, 2014

### ericm1234

Thanks; do you believe (as in, is it plausible) that the author means: ||u-u_s||/||f||<=eps means that thing less than eps is "of order epsilon"?

8. Aug 4, 2014

### Simon Bridge

Really need the context.
But, especially if ||f|| is just a number, it would be fair to say that ||u-u_s||=O(eps)
i.e. it seems one could expand the expressions as $\sum_n a_n\epsilon^n$ ... whatever epsilon is supposed to represent. You are the one with the book in front of you.