How can I prove $20n = O(n^6)$ [closed]

-1

$begingroup$

I am trying to understand how to prove or disprove this:

$$20n = O(n^6) $$

I think it is false, but I am lost on how to show it.

asymptotics

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edited Jan 21 at 8:36

dkaeae

306311

asked Jan 20 at 22:50

J_StrautonJ_Strauton

1043

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closed as off-topic by max_zorn, Cesareo, user91500, Riccardo.Alestra, mrtaurho Jan 21 at 10:44

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." – max_zorn, Cesareo, user91500, Riccardo.Alestra, mrtaurho

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is not false. Big-O indicates asymptotic upper bound.
  $endgroup$
  – lightxbulb
  Jan 20 at 22:52
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @lightxbulb that means I am more lost than I thought.
  $endgroup$
  – J_Strauton
  Jan 20 at 22:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  en.wikipedia.org/wiki/Big_O_notation#Formal_definition
  $endgroup$
  – lightxbulb
  Jan 20 at 22:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  To be fair, this depends on the implicit limit here ($n to infty$, $n to 0$ or $n to$ whatever...)
  $endgroup$
  – Klaus
  Jan 20 at 23:00
  
  
  

  
  
  
  

add a comment | 

-1

$begingroup$

I am trying to understand how to prove or disprove this:

$$20n = O(n^6) $$

I think it is false, but I am lost on how to show it.

asymptotics

share|cite|improve this question

edited Jan 21 at 8:36

dkaeae

306311

asked Jan 20 at 22:50

J_StrautonJ_Strauton

1043

$endgroup$

closed as off-topic by max_zorn, Cesareo, user91500, Riccardo.Alestra, mrtaurho Jan 21 at 10:44

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." – max_zorn, Cesareo, user91500, Riccardo.Alestra, mrtaurho

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is not false. Big-O indicates asymptotic upper bound.
  $endgroup$
  – lightxbulb
  Jan 20 at 22:52
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @lightxbulb that means I am more lost than I thought.
  $endgroup$
  – J_Strauton
  Jan 20 at 22:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  en.wikipedia.org/wiki/Big_O_notation#Formal_definition
  $endgroup$
  – lightxbulb
  Jan 20 at 22:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  To be fair, this depends on the implicit limit here ($n to infty$, $n to 0$ or $n to$ whatever...)
  $endgroup$
  – Klaus
  Jan 20 at 23:00
  
  
  

  
  
  
  

add a comment | 

-1

-1

-1

$begingroup$

I am trying to understand how to prove or disprove this:

$$20n = O(n^6) $$

I think it is false, but I am lost on how to show it.

asymptotics

share|cite|improve this question

edited Jan 21 at 8:36

dkaeae

306311

asked Jan 20 at 22:50

J_StrautonJ_Strauton

1043

$endgroup$

I am trying to understand how to prove or disprove this:

$$20n = O(n^6) $$

I think it is false, but I am lost on how to show it.

asymptotics

asymptotics

share|cite|improve this question

edited Jan 21 at 8:36

dkaeae

306311

asked Jan 20 at 22:50

J_StrautonJ_Strauton

1043

share|cite|improve this question

edited Jan 21 at 8:36

dkaeae

306311

asked Jan 20 at 22:50

J_StrautonJ_Strauton

1043

share|cite|improve this question

share|cite|improve this question

edited Jan 21 at 8:36

dkaeae

306311

edited Jan 21 at 8:36

dkaeae

306311

edited Jan 21 at 8:36

dkaeae

306311

306311

asked Jan 20 at 22:50

J_StrautonJ_Strauton

1043

asked Jan 20 at 22:50

J_StrautonJ_Strauton

1043

asked Jan 20 at 22:50

J_StrautonJ_Strauton

1043

1043

closed as off-topic by max_zorn, Cesareo, user91500, Riccardo.Alestra, mrtaurho Jan 21 at 10:44

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." – max_zorn, Cesareo, user91500, Riccardo.Alestra, mrtaurho

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

closed as off-topic by max_zorn, Cesareo, user91500, Riccardo.Alestra, mrtaurho Jan 21 at 10:44

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." – max_zorn, Cesareo, user91500, Riccardo.Alestra, mrtaurho

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is not false. Big-O indicates asymptotic upper bound.
  $endgroup$
  – lightxbulb
  Jan 20 at 22:52
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @lightxbulb that means I am more lost than I thought.
  $endgroup$
  – J_Strauton
  Jan 20 at 22:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  en.wikipedia.org/wiki/Big_O_notation#Formal_definition
  $endgroup$
  – lightxbulb
  Jan 20 at 22:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  To be fair, this depends on the implicit limit here ($n to infty$, $n to 0$ or $n to$ whatever...)
  $endgroup$
  – Klaus
  Jan 20 at 23:00
  
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is not false. Big-O indicates asymptotic upper bound.
  $endgroup$
  – lightxbulb
  Jan 20 at 22:52
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @lightxbulb that means I am more lost than I thought.
  $endgroup$
  – J_Strauton
  Jan 20 at 22:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  en.wikipedia.org/wiki/Big_O_notation#Formal_definition
  $endgroup$
  – lightxbulb
  Jan 20 at 22:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  To be fair, this depends on the implicit limit here ($n to infty$, $n to 0$ or $n to$ whatever...)
  $endgroup$
  – Klaus
  Jan 20 at 23:00
  
  
  

  
  
  
  

$begingroup$
It is not false. Big-O indicates asymptotic upper bound.
$endgroup$
– lightxbulb
Jan 20 at 22:52

$begingroup$
It is not false. Big-O indicates asymptotic upper bound.
$endgroup$
– lightxbulb
Jan 20 at 22:52

$begingroup$
@lightxbulb that means I am more lost than I thought.
$endgroup$
– J_Strauton
Jan 20 at 22:54

$begingroup$
@lightxbulb that means I am more lost than I thought.
$endgroup$
– J_Strauton
Jan 20 at 22:54

$begingroup$
en.wikipedia.org/wiki/Big_O_notation#Formal_definition
$endgroup$
– lightxbulb
Jan 20 at 22:56

$begingroup$
en.wikipedia.org/wiki/Big_O_notation#Formal_definition
$endgroup$
– lightxbulb
Jan 20 at 22:56

1

1

$begingroup$
To be fair, this depends on the implicit limit here ($n to infty$, $n to 0$ or $n to$ whatever...)
$endgroup$
– Klaus
Jan 20 at 23:00

$begingroup$
To be fair, this depends on the implicit limit here ($n to infty$, $n to 0$ or $n to$ whatever...)
$endgroup$
– Klaus
Jan 20 at 23:00

add a comment | 

1 Answer
1

active

oldest

votes

4

$begingroup$

Remember that

$$f(x) = O(g(x)) mbox{ as } x rightarrow infty$$

is the same as saying

$$exists M > 0, x_0 in mathbb{R}: |f(x)| leq Mg(x) forall x geq x_0$$

i.e. there is a cut-off for $x$, beyond which $g$ times a constant is always larger than $f$.

In this case, there's an easy pick: let $M = 20$ and $x_0 = 1$. Then for $x geq x_0$ it's pretty simple to see that $20x leq 20x^6$, and hence $20x = O(x^6)$.

share|cite|improve this answer

answered Jan 20 at 22:57

ConManConMan

7,9021324

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add a comment | 

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

4

$begingroup$

Remember that

$$f(x) = O(g(x)) mbox{ as } x rightarrow infty$$

is the same as saying

$$exists M > 0, x_0 in mathbb{R}: |f(x)| leq Mg(x) forall x geq x_0$$

i.e. there is a cut-off for $x$, beyond which $g$ times a constant is always larger than $f$.

In this case, there's an easy pick: let $M = 20$ and $x_0 = 1$. Then for $x geq x_0$ it's pretty simple to see that $20x leq 20x^6$, and hence $20x = O(x^6)$.

share|cite|improve this answer

answered Jan 20 at 22:57

ConManConMan

7,9021324

$endgroup$

add a comment | 

4

$begingroup$

Remember that

$$f(x) = O(g(x)) mbox{ as } x rightarrow infty$$

is the same as saying

$$exists M > 0, x_0 in mathbb{R}: |f(x)| leq Mg(x) forall x geq x_0$$

i.e. there is a cut-off for $x$, beyond which $g$ times a constant is always larger than $f$.

In this case, there's an easy pick: let $M = 20$ and $x_0 = 1$. Then for $x geq x_0$ it's pretty simple to see that $20x leq 20x^6$, and hence $20x = O(x^6)$.

share|cite|improve this answer

answered Jan 20 at 22:57

ConManConMan

7,9021324

$endgroup$

add a comment | 

4

4

4

$begingroup$

Remember that

$$f(x) = O(g(x)) mbox{ as } x rightarrow infty$$

is the same as saying

$$exists M > 0, x_0 in mathbb{R}: |f(x)| leq Mg(x) forall x geq x_0$$

i.e. there is a cut-off for $x$, beyond which $g$ times a constant is always larger than $f$.

In this case, there's an easy pick: let $M = 20$ and $x_0 = 1$. Then for $x geq x_0$ it's pretty simple to see that $20x leq 20x^6$, and hence $20x = O(x^6)$.

share|cite|improve this answer

answered Jan 20 at 22:57

ConManConMan

7,9021324

$endgroup$

Remember that

$$f(x) = O(g(x)) mbox{ as } x rightarrow infty$$

is the same as saying

$$exists M > 0, x_0 in mathbb{R}: |f(x)| leq Mg(x) forall x geq x_0$$

i.e. there is a cut-off for $x$, beyond which $g$ times a constant is always larger than $f$.

In this case, there's an easy pick: let $M = 20$ and $x_0 = 1$. Then for $x geq x_0$ it's pretty simple to see that $20x leq 20x^6$, and hence $20x = O(x^6)$.

share|cite|improve this answer

answered Jan 20 at 22:57

ConManConMan

7,9021324

share|cite|improve this answer

share|cite|improve this answer

answered Jan 20 at 22:57

ConManConMan

7,9021324

answered Jan 20 at 22:57

ConManConMan

7,9021324

answered Jan 20 at 22:57

ConManConMan

7,9021324

7,9021324

add a comment | 

add a comment | 

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