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$f(n)=fracn^2+21+2^-n$ and $g(n)=n+3$

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0

$begingroup$

Given

beginaligned
f(n)&=fracn^2+2; 1+2^-n\[1em] g(n)&= n+3
endaligned
I have to calculate the constant $c$ and the $n_0$ of $g(n)=O(f(n))$.

$$n+3≤n^2+3≤3n^2+3≤3n^2+6≤3(n^2+2)$$

At this point I don't know how to go on because if I divide $3(n^2+2)$ by something I get something less than $3(n^2+2)$.

functions inequality algorithms computer-science

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edited yesterday

Jennifer

8,45721837

asked yesterday

user649882user649882

133

$endgroup$

add a comment | 

0

$begingroup$

Given

beginaligned
f(n)&=fracn^2+2; 1+2^-n\[1em] g(n)&= n+3
endaligned
I have to calculate the constant $c$ and the $n_0$ of $g(n)=O(f(n))$.

$$n+3≤n^2+3≤3n^2+3≤3n^2+6≤3(n^2+2)$$

At this point I don't know how to go on because if I divide $3(n^2+2)$ by something I get something less than $3(n^2+2)$.

functions inequality algorithms computer-science

share|cite|improve this question

edited yesterday

Jennifer

8,45721837

asked yesterday

user649882user649882

133

$endgroup$

add a comment | 

$begingroup$

Given

beginaligned
f(n)&=fracn^2+2; 1+2^-n\[1em] g(n)&= n+3
endaligned
I have to calculate the constant $c$ and the $n_0$ of $g(n)=O(f(n))$.

$$n+3≤n^2+3≤3n^2+3≤3n^2+6≤3(n^2+2)$$

At this point I don't know how to go on because if I divide $3(n^2+2)$ by something I get something less than $3(n^2+2)$.

functions inequality algorithms computer-science

share|cite|improve this question

edited yesterday

Jennifer

8,45721837

asked yesterday

user649882user649882

133

$endgroup$

share|cite|improve this question

edited yesterday

Jennifer

8,45721837

asked yesterday

user649882user649882

133

share|cite|improve this question

edited yesterday

Jennifer

8,45721837

asked yesterday

user649882user649882

133

edited yesterday

Jennifer

8,45721837

edited yesterday

Jennifer

8,45721837

asked yesterday

user649882user649882

133

asked yesterday

user649882user649882

133

1 Answer
1

active

oldest

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0

$begingroup$

HINT of a possible method: Note that $$frac11+2^-n ge frac11+ 2^-1 = frac23$$ for all $n ge 1$. So $f(n) ge frac23left(n^2+2right)$ for all $nge 1$. Therefore, it is sufficient for you to find $c$ and $n_0$ such that $$g(n) le c times frac23left(n^2+2right)$$ for all positive integers $nge n_0$. Are you able to do this (note: you have actually already done most of the work for it!)?

share|cite|improve this answer

edited yesterday

answered yesterday

Minus One-TwelfthMinus One-Twelfth

2,07219

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't understand
  $endgroup$
  – user649882
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you know that $g(n)=n+3le colorblue3left(n^2 + 2right)$, then what happens if you choose $c$ such that $ctimes frac23 = colorblue3$? Also note that if $g(n) le ctimes frac23left(n^2+2right)$ for some positive constant $c$, then $g(n) le cf(n)$, because as we said, $ frac23left(n^2+2right)le f(n)$.
  $endgroup$
  – Minus One-Twelfth
  yesterday
  
  

  
  
  
  

add a comment | 

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0

$begingroup$

HINT of a possible method: Note that $$frac11+2^-n ge frac11+ 2^-1 = frac23$$ for all $n ge 1$. So $f(n) ge frac23left(n^2+2right)$ for all $nge 1$. Therefore, it is sufficient for you to find $c$ and $n_0$ such that $$g(n) le c times frac23left(n^2+2right)$$ for all positive integers $nge n_0$. Are you able to do this (note: you have actually already done most of the work for it!)?

share|cite|improve this answer

edited yesterday

answered yesterday

Minus One-TwelfthMinus One-Twelfth

2,07219

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't understand
  $endgroup$
  – user649882
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you know that $g(n)=n+3le colorblue3left(n^2 + 2right)$, then what happens if you choose $c$ such that $ctimes frac23 = colorblue3$? Also note that if $g(n) le ctimes frac23left(n^2+2right)$ for some positive constant $c$, then $g(n) le cf(n)$, because as we said, $ frac23left(n^2+2right)le f(n)$.
  $endgroup$
  – Minus One-Twelfth
  yesterday
  
  

  
  
  
  

add a comment | 

0

$begingroup$

HINT of a possible method: Note that $$frac11+2^-n ge frac11+ 2^-1 = frac23$$ for all $n ge 1$. So $f(n) ge frac23left(n^2+2right)$ for all $nge 1$. Therefore, it is sufficient for you to find $c$ and $n_0$ such that $$g(n) le c times frac23left(n^2+2right)$$ for all positive integers $nge n_0$. Are you able to do this (note: you have actually already done most of the work for it!)?

share|cite|improve this answer

edited yesterday

answered yesterday

Minus One-TwelfthMinus One-Twelfth

2,07219

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't understand
  $endgroup$
  – user649882
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you know that $g(n)=n+3le colorblue3left(n^2 + 2right)$, then what happens if you choose $c$ such that $ctimes frac23 = colorblue3$? Also note that if $g(n) le ctimes frac23left(n^2+2right)$ for some positive constant $c$, then $g(n) le cf(n)$, because as we said, $ frac23left(n^2+2right)le f(n)$.
  $endgroup$
  – Minus One-Twelfth
  yesterday
  
  

  
  
  
  

add a comment | 

$begingroup$

HINT of a possible method: Note that $$frac11+2^-n ge frac11+ 2^-1 = frac23$$ for all $n ge 1$. So $f(n) ge frac23left(n^2+2right)$ for all $nge 1$. Therefore, it is sufficient for you to find $c$ and $n_0$ such that $$g(n) le c times frac23left(n^2+2right)$$ for all positive integers $nge n_0$. Are you able to do this (note: you have actually already done most of the work for it!)?

share|cite|improve this answer

edited yesterday

answered yesterday

Minus One-TwelfthMinus One-Twelfth

2,07219

$endgroup$

share|cite|improve this answer

edited yesterday

answered yesterday

Minus One-TwelfthMinus One-Twelfth

2,07219

answered yesterday

Minus One-TwelfthMinus One-Twelfth

2,07219

answered yesterday

Minus One-TwelfthMinus One-Twelfth

2,07219