$f(n)=fracn^2+21+2^-n$ and $g(n)=n+3$Algorithm analysis, finding a constant c and a point n?Ceiling to Floor Function Conversion ProofSolving InequalitiesTime complexity functionProve $ frac x+yx^2+y^2 + frac y+zy^2+z^2 + frac z+xz^2+x^2 leq frac 1x + frac 1y + frac 1z . $Help understanding notation and proofDistributing a set of integers where set follows a pattternRecursion Tree - recurrence problemGrowth of Functions proofNotation for Markov chain
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Signed and unsigned numbers
$f(n)=fracn^2+21+2^-n$ and $g(n)=n+3$
Algorithm analysis, finding a constant c and a point n?Ceiling to Floor Function Conversion ProofSolving InequalitiesTime complexity functionProve $ frac x+yx^2+y^2 + frac y+zy^2+z^2 + frac z+xz^2+x^2 leq frac 1x + frac 1y + frac 1z . $Help understanding notation and proofDistributing a set of integers where set follows a pattternRecursion Tree - recurrence problemGrowth of Functions proofNotation for Markov chain
$begingroup$
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
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
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
edited yesterday
Jennifer
8,45721837
asked yesterday
user649882user649882
133
$endgroup$
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
functions inequality algorithms computer-science
edited yesterday
Jennifer
8,45721837
asked yesterday
user649882user649882
133
edited yesterday
Jennifer
8,45721837
asked yesterday
user649882user649882
133
edited yesterday
Jennifer
8,45721837
edited yesterday
Jennifer
8,45721837
edited yesterday
Jennifer
8,45721837
8,45721837
asked yesterday
user649882user649882
133
asked yesterday
user649882user649882
133
asked yesterday
user649882user649882
133
133
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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!)?
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 |
Your Answer
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1 Answer
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1 Answer
1
active
oldest
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$begingroup$
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!)?
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!)?
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!)?
answered yesterday
Minus One-TwelfthMinus One-Twelfth
2,07219
$endgroup$
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!)?
answered yesterday
Minus One-TwelfthMinus One-Twelfth
2,07219
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
2,07219
$begingroup$
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 |
$begingroup$
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
$begingroup$
I don't understand
$endgroup$
– user649882
yesterday
$begingroup$
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
$begingroup$
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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