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 Post subject: function and Little o
PostPosted: Sat, 18 Sep 2010 21:49:04 UTC 
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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).


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PostPosted: Sat, 18 Sep 2010 23:02:50 UTC 
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The easy answer is any constant function.

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


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 Post subject: Thanks
PostPosted: Sat, 18 Sep 2010 23:58:52 UTC 
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Thanks for the reply Matt that does work but how would I prove it


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PostPosted: Sun, 19 Sep 2010 00:53:08 UTC 
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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?


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PostPosted: Sun, 19 Sep 2010 01:25:27 UTC 
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I graphed both and log(x) was increasing greater than √log(x) [/list]


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PostPosted: Sun, 19 Sep 2010 01:45:46 UTC 
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Okay, do you have an answer to my second question?
If so, then please provide the definition here.


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PostPosted: Sun, 19 Sep 2010 02:07:48 UTC 
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This is what I understand as x is increasing to infinity the function √log(x) is decreasing less that the log(x).


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PostPosted: Sun, 19 Sep 2010 06:12:20 UTC 
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f(x)=o(\log x) means that $\lim_{x\to\infty}\frac{f(x)}{\log x}=0


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PostPosted: Sun, 19 Sep 2010 16:57:19 UTC 
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Thanks Matt I got it now


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