Thread: Ok

**Mathstud28** · April 5th 2008, 07:34 PM

I will start a new thread here to ask this question...suppose that $f(x)<g(x),\forall{x}\in\mathbb{R}$ and also suppose that $f(x)\sim{g(x)}$ could you argue that the original inequality could be untrue?

**Moo** · April 5th 2008, 10:30 PM

Hello,

It depends on where f(x) ~ g(x)

And what does f(x) ~ g(x) mean ? It means that there exists $\epsilon (x)$ such as $f(x)=g(x)+\epsilon (x)$ , where $\epsilon (x)$ tends to be very small as approaching the domain where f(x) ~ g(x).

This doesn't impeach f(x) to be inferior to g(x)...

**Mathstud28** · April 6th 2008, 06:59 AM

I have always known $f(x)\sim{g(x)}$ to mean $\lim_{x \to {\infty}}\frac{f(x)}{g(x)}=1$ or $\lim_{x \to {\infty}}\frac{g(x)}{f(x)}=1$ ...or in other words they are asymptotically equivalent

**Jhevon** · April 6th 2008, 07:02 AM

Originally Posted by Mathstud28

I have always known $f(x)\sim{g(x)}$ to mean $\lim_{x \to {\infty}}\frac{f(x)}{g(x)}=1$ or $\lim_{x \to {\infty}}\frac{g(x)}{f(x)}=1$ ...or in other words they are asymptotically equivalent

well, $f(x) = g(x)$ certainly satisfies that definition, and $f(x) > g(x)$ for all $x$ could satisfy it as well. so yes, we could argue that the original inequality was false. but of course, to do that, we would need more information that you have provided.

**Mathstud28** · April 6th 2008, 07:30 AM

but what relating to the classic $\pi(x)$ ....I am pretty sure that $Li(x)>\frac{x}{\ln(x)} \forall{x}\in\mathbb{R}$ ...but assuming the limit exists that then $Li(x)\sim\frac{x}{\ln(x)}$ ...I am not sure that is the best example but Its the only one I can think of right now...so would you still agree with the original inequality?

**ThePerfectHacker** · April 6th 2008, 07:33 AM

Another example is Stirling approximation.

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Ok

That is not what I meant

I dont know if this is a good example