logarithm

[Lisa91]

How to prove that [tex] n^{\alpha} > \ln(n) [/tex] for [tex] \alpha>0 [/tex]?

 

[MarkFL]

How are the variables defined?

For instance if $\alpha\in\mathbb{R}$ and $n\in\mathbb{N}$ then one example of a counter-example to your inequality occurs for:

$n=3,\,\alpha=0.01$

This leads me to believe there in some missing information.

 

[ZaidAlyafey]

we have $n^{\alpha}> \ln(n) $ take ln to both sides :

I will separate the problem in two steps :

1-For 0<n<1 this is trivial since the right hand will be always negtative while
and the left hand side is always positive , also for n=1 the inequality holds .

2-Now for n>1 :

${\alpha}\ln(n)> \ln(\ln(n)) \Rightarrow \,\, \alpha> \frac{\ln(\ln(n))}{\ln(n) } $

Now this is only true iff $\frac{\ln(\ln(n))}{\ln(n) } \leq 0$

which holds iff $0<\ln(n)\leq 1\,\, \Rightarrow \,\, 1< n \leq e $

The inequality is true for all $\alpha $ iff $0<n \leq e$

 

[Klaas van Aarsen]

2-Now for n>1 :

${\alpha}\ln(n)> \ln(\ln(n)) \Rightarrow \,\, \alpha> \frac{\ln(\ln(n))}{\ln(n) } $

Since the right hand side approaches zero for large n, this means that for any $\alpha>0$ there is a number N such that the inequality is true for any n > N.


Hey Lisa91!

Can it be there is a condition missing from your problem?
The extra condition that it holds for any n > N for some N?

 

[ZaidAlyafey]

Since the right hand side approaches zero for large n, this means that for any $\alpha>0$ there is a number N such that the inequality is true for any n > N.

since $\alpha $ is an independent variable of n I can choose it as small as possible so that
it becomes lesser than the right-hand side .
Can you give a counter example for $\alpha$ and n that disproves my argument ?

 

[Klaas van Aarsen]

since $\alpha $ is an independent variable of n I can choose it as small as possible so that
it becomes lesser than the right-hand side .
Can you give a counter example for $\alpha$ and n that disproves my argument ?

Your argument is flawless.
It's just that you have assumed that the inequality should hold for specific n and all $\alpha$'s.
Whereas I have assumed it's not for all n.

In other words, you have solved:

Find n such that $n^\alpha > \ln n$ for all $\alpha > 0$.

​

Whereas I have use the first half of your argument to follow up with:

Prove that $n^\alpha > \ln n$ for $\alpha > 0$ if n is big enough given a certain $\alpha$.
​​

That's why we need clarification on what the actual problem is.