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

In summary, the conversation discusses finding a bound for log n < O(n^epsilon) for n sufficiently large, using limits and the definition of big-O. Various methods are suggested, including l'hopital's rule and a pure algebra method, with the final solution involving evaluating the limit using a substitution method.
  • #1
ti89fr33k
13
0
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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  • #2
Did you try looking at

[tex]\lim_{n\rightarrow\infty}\frac{\log n}{n^\epsilon}[/tex]

If you can evaluate this limit, the bound you're after follows.
 
  • #3
I know the limit is 0...but how do you evaluate it?
 
  • #4
The answer is quite intuitive...but i want a rigorous method
 
  • #5
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.
 
  • #6
yes, but my instructor specifically mentioned not to use l'hopitals rule
 
  • #7
there is a pure algebra method
 
  • #8
If I write

[tex]\log n=\frac{2}{\epsilon}\log n^\frac{\epsilon}{2}[/tex]

does that help you evaluate the limit?
 
  • #9
log n^epsilon/2 has to grow slower than n^epsilon?
 
  • #10
oh...i see...
 

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

1. What does the notation "log n" mean in this context?

The notation "log n" represents the logarithmic function, which is the inverse of the exponential function. In this context, it is used to measure the growth rate of a function in terms of the input size n.

2. How does the function "log n" compare to the function "O(n^ε)" for large values of n?

For large values of n, the function "log n" grows much slower than the function "O(n^ε)". This means that the value of "log n" will be much smaller than the value of "O(n^ε)" for the same input size n.

3. What is the significance of the inequality "log n < O(n^ε)" for large n?

This inequality indicates that the function "log n" is an upper bound on the function "O(n^ε)" for large values of n. In other words, the growth rate of "log n" is always smaller than the growth rate of "O(n^ε)" for large input sizes.

4. Can you provide an example of a function that satisfies the inequality "log n < O(n^ε)" for large n?

One example of a function that satisfies this inequality is "log n" itself. For any value of ε greater than 1, the function "log n" will have a smaller growth rate than the function "O(n^ε)" for large input sizes n.

5. How does the growth rate of "log n" compare to other common function growth rates?

The growth rate of "log n" is considered to be slower than linear growth (O(n)), but faster than constant growth (O(1)). It is also slower than polynomial growth (O(n^c)) for any constant c, but faster than exponential growth (O(2^n)).

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