How to prove Big Theta on polynomial function?
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Im working on the Big-O notation and struggle to understand whether I have done enough to prove the following:
$5n^3+4n^2+4 in Theta (n^3)$
So based on the definition of $Θ(g(n)):$
Step $1$: $0 ≤ c_1 n^3 leq 5n^3+4n^2+4 leq c_2 n^3$
Divide the inequality by the largest order n-term
Step $2$: $0 ≤ c_1 leq 5+(4/n+4/n^3) leq c_2$
Find constant $c_2$ that will satisfy:
Step $3$: $0 leq 5+(4/n+4/n^3) leq c_2$
For $n=1$, $0 leq 5+(4/1+4/1^3)=13$
For $n=2$, $0 leq 5+(4/2+4/2^3)=7,5$
For $n=3$, $0 leq 5+(4/3+4/3^3)=6,7$
For $n=4$, $0 leq 5+(4/4+4/4^3)=6,06$
Since $c_2$ approaches $5$ when $n to infty$, I pick 5 for $c_2$ as it satisfies the equation in Step $3$
Step $4$: Is to find $c_1$
Since $c_1$ can be $leq 5$
I can say that $5n^3+4n^2+4 in Theta (n^3)$ for $c_1$ $leq 5$ and $c_2 geq 5$. However not sure how to find $n_0$ in that regard.
Appreciate any insights!
discrete-mathematics asymptotics computer-science
asked Dec 10 '17 at 18:19
GoogmeGoogme
23818
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add a comment |
$begingroup$
Im working on the Big-O notation and struggle to understand whether I have done enough to prove the following:
$5n^3+4n^2+4 in Theta (n^3)$
So based on the definition of $Θ(g(n)):$
Step $1$: $0 ≤ c_1 n^3 leq 5n^3+4n^2+4 leq c_2 n^3$
Divide the inequality by the largest order n-term
Step $2$: $0 ≤ c_1 leq 5+(4/n+4/n^3) leq c_2$
Find constant $c_2$ that will satisfy:
Step $3$: $0 leq 5+(4/n+4/n^3) leq c_2$
For $n=1$, $0 leq 5+(4/1+4/1^3)=13$
For $n=2$, $0 leq 5+(4/2+4/2^3)=7,5$
For $n=3$, $0 leq 5+(4/3+4/3^3)=6,7$
For $n=4$, $0 leq 5+(4/4+4/4^3)=6,06$
Since $c_2$ approaches $5$ when $n to infty$, I pick 5 for $c_2$ as it satisfies the equation in Step $3$
Step $4$: Is to find $c_1$
Since $c_1$ can be $leq 5$
I can say that $5n^3+4n^2+4 in Theta (n^3)$ for $c_1$ $leq 5$ and $c_2 geq 5$. However not sure how to find $n_0$ in that regard.
Appreciate any insights!
discrete-mathematics asymptotics computer-science
asked Dec 10 '17 at 18:19
GoogmeGoogme
23818
$endgroup$
add a comment |
$begingroup$
Im working on the Big-O notation and struggle to understand whether I have done enough to prove the following:
$5n^3+4n^2+4 in Theta (n^3)$
So based on the definition of $Θ(g(n)):$
Step $1$: $0 ≤ c_1 n^3 leq 5n^3+4n^2+4 leq c_2 n^3$
Divide the inequality by the largest order n-term
Step $2$: $0 ≤ c_1 leq 5+(4/n+4/n^3) leq c_2$
Find constant $c_2$ that will satisfy:
Step $3$: $0 leq 5+(4/n+4/n^3) leq c_2$
For $n=1$, $0 leq 5+(4/1+4/1^3)=13$
For $n=2$, $0 leq 5+(4/2+4/2^3)=7,5$
For $n=3$, $0 leq 5+(4/3+4/3^3)=6,7$
For $n=4$, $0 leq 5+(4/4+4/4^3)=6,06$
Since $c_2$ approaches $5$ when $n to infty$, I pick 5 for $c_2$ as it satisfies the equation in Step $3$
Step $4$: Is to find $c_1$
Since $c_1$ can be $leq 5$
I can say that $5n^3+4n^2+4 in Theta (n^3)$ for $c_1$ $leq 5$ and $c_2 geq 5$. However not sure how to find $n_0$ in that regard.
Appreciate any insights!
discrete-mathematics asymptotics computer-science
asked Dec 10 '17 at 18:19
GoogmeGoogme
23818
$endgroup$
Im working on the Big-O notation and struggle to understand whether I have done enough to prove the following:
$5n^3+4n^2+4 in Theta (n^3)$
So based on the definition of $Θ(g(n)):$
Step $1$: $0 ≤ c_1 n^3 leq 5n^3+4n^2+4 leq c_2 n^3$
Divide the inequality by the largest order n-term
Step $2$: $0 ≤ c_1 leq 5+(4/n+4/n^3) leq c_2$
Find constant $c_2$ that will satisfy:
Step $3$: $0 leq 5+(4/n+4/n^3) leq c_2$
For $n=1$, $0 leq 5+(4/1+4/1^3)=13$
For $n=2$, $0 leq 5+(4/2+4/2^3)=7,5$
For $n=3$, $0 leq 5+(4/3+4/3^3)=6,7$
For $n=4$, $0 leq 5+(4/4+4/4^3)=6,06$
Since $c_2$ approaches $5$ when $n to infty$, I pick 5 for $c_2$ as it satisfies the equation in Step $3$
Step $4$: Is to find $c_1$
Since $c_1$ can be $leq 5$
I can say that $5n^3+4n^2+4 in Theta (n^3)$ for $c_1$ $leq 5$ and $c_2 geq 5$. However not sure how to find $n_0$ in that regard.
Appreciate any insights!
discrete-mathematics asymptotics computer-science
discrete-mathematics asymptotics computer-science
asked Dec 10 '17 at 18:19
GoogmeGoogme
23818
asked Dec 10 '17 at 18:19
GoogmeGoogme
23818
edited Dec 10 '17 at 18:40
Googme
asked Dec 10 '17 at 18:19
GoogmeGoogme
23818
asked Dec 10 '17 at 18:19
GoogmeGoogme
23818
asked Dec 10 '17 at 18:19
GoogmeGoogme
23818
23818
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1 Answer
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Hint: $1 le n^2 le n^3$ for $n ge 1,$, so $5n^3 ;le; 5n^3+4n^2+4 ;le; 5n^3+4n^3+4n^3 = 13 n^3,$.
answered Dec 10 '17 at 20:19
dxivdxiv
57.8k648101
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1 Answer
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$begingroup$
Hint: $1 le n^2 le n^3$ for $n ge 1,$, so $5n^3 ;le; 5n^3+4n^2+4 ;le; 5n^3+4n^3+4n^3 = 13 n^3,$.
answered Dec 10 '17 at 20:19
dxivdxiv
57.8k648101
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add a comment |
$begingroup$
Hint: $1 le n^2 le n^3$ for $n ge 1,$, so $5n^3 ;le; 5n^3+4n^2+4 ;le; 5n^3+4n^3+4n^3 = 13 n^3,$.
answered Dec 10 '17 at 20:19
dxivdxiv
57.8k648101
$endgroup$
add a comment |
$begingroup$
Hint: $1 le n^2 le n^3$ for $n ge 1,$, so $5n^3 ;le; 5n^3+4n^2+4 ;le; 5n^3+4n^3+4n^3 = 13 n^3,$.
answered Dec 10 '17 at 20:19
dxivdxiv
57.8k648101
$endgroup$
Hint: $1 le n^2 le n^3$ for $n ge 1,$, so $5n^3 ;le; 5n^3+4n^2+4 ;le; 5n^3+4n^3+4n^3 = 13 n^3,$.
answered Dec 10 '17 at 20:19
dxivdxiv
57.8k648101
answered Dec 10 '17 at 20:19
dxivdxiv
57.8k648101
answered Dec 10 '17 at 20:19
dxivdxiv
57.8k648101
answered Dec 10 '17 at 20:19
dxivdxiv
57.8k648101
57.8k648101
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