function and Little o

mebigp · **Joined:** Sat, 18 Sep 2010 21:28:18 UTC **Posts:** 15

Find a function f(x) that grows at a slower rate that log x and prove that it does. and prove that f(x)=o(log x).

Matt · **Joined:** Wed, 1 Oct 2003 04:45:43 UTC **Posts:** 9632

The easy answer is any constant function.

If it has to be strictly increasing, then how about $f(x)=\sqrt{\log x}$ ?

mebigp · **Joined:** Sat, 18 Sep 2010 21:28:18 UTC **Posts:** 15

Thanks for the reply Matt that does work but how would I prove it

Matt · **Joined:** Wed, 1 Oct 2003 04:45:43 UTC **Posts:** 9632

mebigp wrote:

that does work but how would I prove it

How do you know that it works?

Do you know the definition of Little-o?

mebigp · **Joined:** Sat, 18 Sep 2010 21:28:18 UTC **Posts:** 15

I graphed both and log(x) was increasing greater than √log(x) [/list]

Matt · **Joined:** Wed, 1 Oct 2003 04:45:43 UTC **Posts:** 9632

Okay, do you have an answer to my second question?
If so, then please provide the definition here.

mebigp · **Joined:** Sat, 18 Sep 2010 21:28:18 UTC **Posts:** 15

This is what I understand as x is increasing to infinity the function √log(x) is decreasing less that the log(x).

Matt · **Joined:** Wed, 1 Oct 2003 04:45:43 UTC **Posts:** 9632

$f(x)=o(\log x)$ means that $$\lim_{x\to\infty}\frac{f(x)}{\log x}=0$

mebigp · **Joined:** Sat, 18 Sep 2010 21:28:18 UTC **Posts:** 15

Thanks Matt I got it now

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