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Which algorithm is more efficient? [closed]

-1

0

$begingroup$

Which algorithm is more efficient. Is it a or b?

a. $O(n^{2.81})$

b. $O((n^3 )/ ln n)$

I want to solve this problem using l'Hospital's rule, $lim(f(n)/g(n)) = lim(f'(n)/g'(n))$

L'Hospital's Rule (1696): If $f(n)$ and $g(n)$ are both differentiable with derivatives $f'(n)$ and $g'(n)$, respectively, and if $lim f(n) = lim g(n) = infty$, then $lim frac{f(n)}{g(n)} = lim frac{f'(n)}{g'(n)}$, whenever the limit on the right exists.

Example (1) $f(n) = n^3$ and $g(n) = ln n$? $f'(n)=3n^2$ and $g'(n)=1/n$ so $lim frac{n^3}{ln n}=lim frac{3n^2}{1/n}=infty$. therefore, $g(n)$ is more efficient than $f(n)$.

calculus asymptotics

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edited Mar 24 at 16:00

Robert Howard

2,3033935

asked Mar 23 at 12:36

bitmappbitmapp

12

$endgroup$

closed as off-topic by user21820, Saad, José Carlos Santos, Shailesh, RRL Mar 25 at 0:20

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Saad, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Deleting information from your question doesn't help anyone, so please don't do it in the future. Other people might find your question weeks, months, or years from now and be helped by it and the answers other users have left, but vandalizing your own question prevents there from being any chance of that happening.
  $endgroup$
  – Robert Howard
  Mar 24 at 16:02
  

  
  
  
  

add a comment | 

-1

0

$begingroup$

Which algorithm is more efficient. Is it a or b?

a. $O(n^{2.81})$

b. $O((n^3 )/ ln n)$

I want to solve this problem using l'Hospital's rule, $lim(f(n)/g(n)) = lim(f'(n)/g'(n))$

L'Hospital's Rule (1696): If $f(n)$ and $g(n)$ are both differentiable with derivatives $f'(n)$ and $g'(n)$, respectively, and if $lim f(n) = lim g(n) = infty$, then $lim frac{f(n)}{g(n)} = lim frac{f'(n)}{g'(n)}$, whenever the limit on the right exists.

Example (1) $f(n) = n^3$ and $g(n) = ln n$? $f'(n)=3n^2$ and $g'(n)=1/n$ so $lim frac{n^3}{ln n}=lim frac{3n^2}{1/n}=infty$. therefore, $g(n)$ is more efficient than $f(n)$.

calculus asymptotics

share|cite|improve this question

edited Mar 24 at 16:00

Robert Howard

2,3033935

asked Mar 23 at 12:36

bitmappbitmapp

12

$endgroup$

closed as off-topic by user21820, Saad, José Carlos Santos, Shailesh, RRL Mar 25 at 0:20

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Saad, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Deleting information from your question doesn't help anyone, so please don't do it in the future. Other people might find your question weeks, months, or years from now and be helped by it and the answers other users have left, but vandalizing your own question prevents there from being any chance of that happening.
  $endgroup$
  – Robert Howard
  Mar 24 at 16:02
  

  
  
  
  

add a comment | 

$begingroup$

Which algorithm is more efficient. Is it a or b?

a. $O(n^{2.81})$

b. $O((n^3 )/ ln n)$

I want to solve this problem using l'Hospital's rule, $lim(f(n)/g(n)) = lim(f'(n)/g'(n))$

L'Hospital's Rule (1696): If $f(n)$ and $g(n)$ are both differentiable with derivatives $f'(n)$ and $g'(n)$, respectively, and if $lim f(n) = lim g(n) = infty$, then $lim frac{f(n)}{g(n)} = lim frac{f'(n)}{g'(n)}$, whenever the limit on the right exists.

Example (1) $f(n) = n^3$ and $g(n) = ln n$? $f'(n)=3n^2$ and $g'(n)=1/n$ so $lim frac{n^3}{ln n}=lim frac{3n^2}{1/n}=infty$. therefore, $g(n)$ is more efficient than $f(n)$.

calculus asymptotics

share|cite|improve this question

edited Mar 24 at 16:00

Robert Howard

2,3033935

asked Mar 23 at 12:36

bitmappbitmapp

12

$endgroup$

share|cite|improve this question

edited Mar 24 at 16:00

Robert Howard

2,3033935

asked Mar 23 at 12:36

bitmappbitmapp

12

share|cite|improve this question

edited Mar 24 at 16:00

Robert Howard

2,3033935

asked Mar 23 at 12:36

bitmappbitmapp

12

edited Mar 24 at 16:00

Robert Howard

2,3033935

edited Mar 24 at 16:00

Robert Howard

2,3033935

asked Mar 23 at 12:36

bitmappbitmapp

12

asked Mar 23 at 12:36

bitmappbitmapp

12

2 Answers
2

active

oldest

votes

1

$begingroup$

$$lim_{ntoinfty} frac{(frac{n^3}{ln{n}})}{n^{2.81}}=lim_{ntoinfty} frac{n^{0.19}}{ln{n}}=lim_{ntoinfty} frac{0.19cdot n^{-0.81}}{n^{-1}}=lim_{ntoinfty} 0.19cdot n^{0.19}=infty$$
So the more efficient algorithm is the one which is $O(n^{2.81})$ as $O(frac{n^3}{ln{n}})$ has an infinitely greater complexity.

share|cite|improve this answer

answered Mar 23 at 12:43

Peter ForemanPeter Foreman

7,2411319

$endgroup$

add a comment | 

0

$begingroup$

L'Hospital's rule is for quotient of functions with limit $0$. Here all you have to do is divide and simplify :
$$frac{n^{2.81}}{frac{n^3}{ln n}} = n^{2.81}frac{ln n}{n^3} = frac{ln n}{n^{0.19}} xrightarrow[ntoinfty]{} 0$$
so $Oleft(n^{2.81}right)$ is more efficient than $Oleft(frac{n^3}{ln n}right)$.

share|cite|improve this answer

edited Mar 23 at 13:07

answered Mar 23 at 12:44

Nicolas FRANCOISNicolas FRANCOIS

3,7921516

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'd like to see a proof of this statement... We never use it in France (just not very useful).
  $endgroup$
  – Nicolas FRANCOIS
  Mar 24 at 18:30
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

$$lim_{ntoinfty} frac{(frac{n^3}{ln{n}})}{n^{2.81}}=lim_{ntoinfty} frac{n^{0.19}}{ln{n}}=lim_{ntoinfty} frac{0.19cdot n^{-0.81}}{n^{-1}}=lim_{ntoinfty} 0.19cdot n^{0.19}=infty$$
So the more efficient algorithm is the one which is $O(n^{2.81})$ as $O(frac{n^3}{ln{n}})$ has an infinitely greater complexity.

share|cite|improve this answer

answered Mar 23 at 12:43

Peter ForemanPeter Foreman

7,2411319

$endgroup$

add a comment | 

1

$begingroup$

$$lim_{ntoinfty} frac{(frac{n^3}{ln{n}})}{n^{2.81}}=lim_{ntoinfty} frac{n^{0.19}}{ln{n}}=lim_{ntoinfty} frac{0.19cdot n^{-0.81}}{n^{-1}}=lim_{ntoinfty} 0.19cdot n^{0.19}=infty$$
So the more efficient algorithm is the one which is $O(n^{2.81})$ as $O(frac{n^3}{ln{n}})$ has an infinitely greater complexity.

share|cite|improve this answer

answered Mar 23 at 12:43

Peter ForemanPeter Foreman

7,2411319

$endgroup$

add a comment | 

$begingroup$

$$lim_{ntoinfty} frac{(frac{n^3}{ln{n}})}{n^{2.81}}=lim_{ntoinfty} frac{n^{0.19}}{ln{n}}=lim_{ntoinfty} frac{0.19cdot n^{-0.81}}{n^{-1}}=lim_{ntoinfty} 0.19cdot n^{0.19}=infty$$
So the more efficient algorithm is the one which is $O(n^{2.81})$ as $O(frac{n^3}{ln{n}})$ has an infinitely greater complexity.

share|cite|improve this answer

answered Mar 23 at 12:43

Peter ForemanPeter Foreman

7,2411319

$endgroup$

share|cite|improve this answer

answered Mar 23 at 12:43

Peter ForemanPeter Foreman

7,2411319

answered Mar 23 at 12:43

Peter ForemanPeter Foreman

7,2411319

answered Mar 23 at 12:43

Peter ForemanPeter Foreman

7,2411319

0

$begingroup$

L'Hospital's rule is for quotient of functions with limit $0$. Here all you have to do is divide and simplify :
$$frac{n^{2.81}}{frac{n^3}{ln n}} = n^{2.81}frac{ln n}{n^3} = frac{ln n}{n^{0.19}} xrightarrow[ntoinfty]{} 0$$
so $Oleft(n^{2.81}right)$ is more efficient than $Oleft(frac{n^3}{ln n}right)$.

share|cite|improve this answer

edited Mar 23 at 13:07

answered Mar 23 at 12:44

Nicolas FRANCOISNicolas FRANCOIS

3,7921516

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'd like to see a proof of this statement... We never use it in France (just not very useful).
  $endgroup$
  – Nicolas FRANCOIS
  Mar 24 at 18:30
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

L'Hospital's rule is for quotient of functions with limit $0$. Here all you have to do is divide and simplify :
$$frac{n^{2.81}}{frac{n^3}{ln n}} = n^{2.81}frac{ln n}{n^3} = frac{ln n}{n^{0.19}} xrightarrow[ntoinfty]{} 0$$
so $Oleft(n^{2.81}right)$ is more efficient than $Oleft(frac{n^3}{ln n}right)$.

share|cite|improve this answer

edited Mar 23 at 13:07

answered Mar 23 at 12:44

Nicolas FRANCOISNicolas FRANCOIS

3,7921516

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'd like to see a proof of this statement... We never use it in France (just not very useful).
  $endgroup$
  – Nicolas FRANCOIS
  Mar 24 at 18:30
  
  
  

  
  
  
  

add a comment | 

$begingroup$

L'Hospital's rule is for quotient of functions with limit $0$. Here all you have to do is divide and simplify :
$$frac{n^{2.81}}{frac{n^3}{ln n}} = n^{2.81}frac{ln n}{n^3} = frac{ln n}{n^{0.19}} xrightarrow[ntoinfty]{} 0$$
so $Oleft(n^{2.81}right)$ is more efficient than $Oleft(frac{n^3}{ln n}right)$.

share|cite|improve this answer

edited Mar 23 at 13:07

answered Mar 23 at 12:44

Nicolas FRANCOISNicolas FRANCOIS

3,7921516

$endgroup$

share|cite|improve this answer

edited Mar 23 at 13:07

answered Mar 23 at 12:44

Nicolas FRANCOISNicolas FRANCOIS

3,7921516

answered Mar 23 at 12:44

Nicolas FRANCOISNicolas FRANCOIS

3,7921516

answered Mar 23 at 12:44

Nicolas FRANCOISNicolas FRANCOIS

3,7921516