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- Thread starter Lepros
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- #1

- Jan 30, 2012

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Affirmative. (The forum requires a 5-character minimum reply, so I could not just answer "yes".)

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- Jan 26, 2012

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Changed to 3. Now you can write "No." or "Yes."Affirmative. (The forum requires a 5-character minimum reply, so I could not just answer "yes".)

- Feb 12, 2012

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Why ?Affirmative. (The forum requires a 5-character minimum reply, so I could not just answer "yes".)

- Jan 30, 2012

- 2,589

- Feb 12, 2012

- 26

But how can you write the double inequality ? We know it's okay if we consider absolute values, but if for example $\forall x>x_0~,~f(x)<0~\text{and}~g(x)>0,~\text{while}~C|f(x)|>g(x)$, what would happen ?

Another equivalent definition (or almost) is that limsup |f/g| < infinity, but if it's 0, couldn't there be a problem ?

I'm basing all this on the wikipedia, so please excuse my ignorance

This document ( http://courses.cs.vt.edu/~cs2604/Summer2000/Notes/C03.Asymptotics.pdf ) covers quite a lot of properties related to the big theta. But none of them talk about the equivalence...

Another equivalent definition (or almost) is that limsup |f/g| < infinity, but if it's 0, couldn't there be a problem ?

I'm basing all this on the wikipedia, so please excuse my ignorance

This document ( http://courses.cs.vt.edu/~cs2604/Summer2000/Notes/C03.Asymptotics.pdf ) covers quite a lot of properties related to the big theta. But none of them talk about the equivalence...

Last edited:

- Jan 30, 2012

- 2,589

For functions that can be negative, the answer depends on the details of the definition of $\Theta$. If it means $C_1|g(x)|\le |f(x)|\le C_2|g(x)|$, then again the relation is symmetric. If it means $C_1g(x)\le |f(x)|\le C_2g(x)$, then, as you say, it does not follow that $|g(x)|\le 1/C_1\cdot f(x)$.

- Feb 12, 2012

- 26

Sorry, I was considering big O and not big theta I thought there were the same