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

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- Apr 5th 2008, 07:34 PM #1

- Apr 5th 2008, 10:30 PM #2
Hello,

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

And what does f(x) ~ g(x) mean ? It means that there exists $\displaystyle \epsilon (x)$ such as $\displaystyle f(x)=g(x)+\epsilon (x)$, where $\displaystyle \epsilon (x)$ 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)...

- Apr 6th 2008, 06:59 AM #3
## That is not what I meant

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

- Apr 6th 2008, 07:02 AM #4
well, $\displaystyle f(x) = g(x)$ certainly satisfies that definition, and $\displaystyle f(x) > g(x)$ for all $\displaystyle x$ 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.

- Apr 6th 2008, 07:30 AM #5
## I dont know if this is a good example

but what relating to the classic $\displaystyle \pi(x)$....I am pretty sure that $\displaystyle Li(x)>\frac{x}{\ln(x)} \forall{x}\in\mathbb{R}$...but assuming the limit exists that then $\displaystyle Li(x)\sim\frac{x}{\ln(x)}$...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?

- Apr 6th 2008, 07:33 AM #6

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Another example is Stirling approximation.