Log n < O(n^ε) for Large n - Simple Function Growth

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Discussion Overview

The discussion revolves around demonstrating that log n is less than O(n^ε) for sufficiently large n. Participants explore various methods to evaluate this relationship, focusing on limits and algebraic approaches.

Discussion Character

  • Exploratory, Technical explanation, Mathematical reasoning

Main Points Raised

  • Mary initiates the discussion by asking how to show that log n < O(n^ε) for large n.
  • One participant suggests evaluating the limit \lim_{n\rightarrow\infty}\frac{\log n}{n^\epsilon} as a means to establish the desired bound.
  • Another participant acknowledges that the limit approaches 0 but seeks clarification on how to evaluate it rigorously.
  • There is a mention of using l'Hôpital's rule to handle the limit, though one participant notes their instructor advised against this method.
  • Alternative approaches are proposed, including a pure algebra method to evaluate the limit.
  • A participant suggests rewriting log n in a specific form to aid in evaluating the limit, questioning whether this helps in demonstrating that log n grows slower than n^ε.
  • One participant expresses realization or understanding in response to the algebraic manipulation presented.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the method to evaluate the limit, with differing opinions on the use of l'Hôpital's rule and the preference for algebraic methods. The discussion remains unresolved regarding the most rigorous approach.

Contextual Notes

Some participants express limitations in their approaches based on instructional guidance, particularly regarding the use of l'Hôpital's rule. There is also an emphasis on the need for a rigorous method, indicating potential gaps in understanding or application of techniques.

Who May Find This Useful

This discussion may be useful for students and individuals interested in mathematical analysis, particularly in understanding limits and big-O notation in the context of function growth.

ti89fr33k
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Show log n < O(n^epsilon) for n sufficiently large.

Not actually calculus, but it has something to do with limits, so there. :smile:

Thanks,

Mary
 
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Did you try looking at

\lim_{n\rightarrow\infty}\frac{\log n}{n^\epsilon}

If you can evaluate this limit, the bound you're after follows.
 
I know the limit is 0...but how do you evaluate it?
 
The answer is quite intuitive...but i want a rigorous method
 
l'hopital's rule will handle the limit.

Are you able to convert from the limit to the big-O bound? This will follow from the definition of the limit.
 
yes, but my instructor specifically mentioned not to use l'hospital's rule
 
there is a pure algebra method
 
If I write

\log n=\frac{2}{\epsilon}\log n^\frac{\epsilon}{2}

does that help you evaluate the limit?
 
log n^epsilon/2 has to grow slower than n^epsilon?
 
  • #10
oh...i see...
 

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