How do I prove $log(n^3+n^-3$ is $O(n^2)$? [on hold] The 2019 Stack Overflow Developer Survey Results Are InHow many edges does a full binary tree with 1000 internal vertices have?The sequence T is defined by $T_n = n! + 2, n geq 1$How to convert to disjunctive normal form?How many different ways of choosing a committee of 7 individuals from 10 officersDeMorgan's lawsProve $forall x ; forall y ; (x + y = y + x)$Prove log(log(n)) is Big-O (log(n))State True or False and explain. If 2a ≡ 4b mod 8, then a ≡ 2b mod 8Show that $sum_k=0^n binomnkGamma(k+frac 1 2)Gamma(n-k+frac 12) = pi n!$Prove $frac1n+ 3;log(log(n))+ frac5;log(n+1)2$ is $O(n)$
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How do I prove $log(n^3+n^-3$ is $O(n^2)$? [on hold]
The 2019 Stack Overflow Developer Survey Results Are InHow many edges does a full binary tree with 1000 internal vertices have?The sequence T is defined by $T_n = n! + 2, n geq 1$How to convert to disjunctive normal form?How many different ways of choosing a committee of 7 individuals from 10 officersDeMorgan's lawsProve $forall x ; forall y ; (x + y = y + x)$Prove log(log(n)) is Big-O (log(n))State True or False and explain. If 2a ≡ 4b mod 8, then a ≡ 2b mod 8Show that $sum_k=0^n binomnkGamma(k+frac 1 2)Gamma(n-k+frac 12) = pi n!$Prove $frac1n+ 3;log(log(n))+ frac5;log(n+1)2$ is $O(n)$
$begingroup$
Provide a full derivation explaining your answer.
discrete-mathematics
asked Apr 6 at 22:16
someonehelpmesomeonehelpme
11
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someonehelpme is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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put on hold as off-topic by Saucy O'Path, Wojowu, davidlowryduda♦ Apr 6 at 22:45
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." – Saucy O'Path, Wojowu, davidlowryduda
add a comment |
$begingroup$
Provide a full derivation explaining your answer.
discrete-mathematics
asked Apr 6 at 22:16
someonehelpmesomeonehelpme
11
New contributor
someonehelpme is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
put on hold as off-topic by Saucy O'Path, Wojowu, davidlowryduda♦ Apr 6 at 22:45
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." – Saucy O'Path, Wojowu, davidlowryduda
$begingroup$
A parenthesis needed in title. Also bad way to ask a question (as a simple demand...)
$endgroup$
– coffeemath
Apr 6 at 22:44
$begingroup$
An intuition is that for large $n$, $n^-3$ is close to $0$, so $$logleft(n^3 + n^-3right) approx logleft(n^3right) = 3log n,$$ which you should know is less than $n^2$ for all large $n$. See if you can fashion a proof out of this idea.
$endgroup$
– Minus One-Twelfth
Apr 6 at 23:45
add a comment |
$begingroup$
Provide a full derivation explaining your answer.
discrete-mathematics
asked Apr 6 at 22:16
someonehelpmesomeonehelpme
11
New contributor
someonehelpme is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Provide a full derivation explaining your answer.
discrete-mathematics
discrete-mathematics
asked Apr 6 at 22:16
someonehelpmesomeonehelpme
11
New contributor
someonehelpme is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 6 at 22:16
someonehelpmesomeonehelpme
11
New contributor
someonehelpme is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 6 at 22:19
someonehelpme
asked Apr 6 at 22:16
someonehelpmesomeonehelpme
11
New contributor
someonehelpme is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 6 at 22:16
someonehelpmesomeonehelpme
11
asked Apr 6 at 22:16
someonehelpmesomeonehelpme
11
11
New contributor
someonehelpme is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
someonehelpme is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
someonehelpme is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Saucy O'Path, Wojowu, davidlowryduda♦ Apr 6 at 22:45
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." – Saucy O'Path, Wojowu, davidlowryduda
put on hold as off-topic by Saucy O'Path, Wojowu, davidlowryduda♦ Apr 6 at 22:45
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." – Saucy O'Path, Wojowu, davidlowryduda
$begingroup$
A parenthesis needed in title. Also bad way to ask a question (as a simple demand...)
$endgroup$
– coffeemath
Apr 6 at 22:44
$begingroup$
An intuition is that for large $n$, $n^-3$ is close to $0$, so $$logleft(n^3 + n^-3right) approx logleft(n^3right) = 3log n,$$ which you should know is less than $n^2$ for all large $n$. See if you can fashion a proof out of this idea.
$endgroup$
– Minus One-Twelfth
Apr 6 at 23:45
add a comment |
$begingroup$
A parenthesis needed in title. Also bad way to ask a question (as a simple demand...)
$endgroup$
– coffeemath
Apr 6 at 22:44
$begingroup$
An intuition is that for large $n$, $n^-3$ is close to $0$, so $$logleft(n^3 + n^-3right) approx logleft(n^3right) = 3log n,$$ which you should know is less than $n^2$ for all large $n$. See if you can fashion a proof out of this idea.
$endgroup$
– Minus One-Twelfth
Apr 6 at 23:45
$begingroup$
A parenthesis needed in title. Also bad way to ask a question (as a simple demand...)
$endgroup$
– coffeemath
Apr 6 at 22:44
$begingroup$
A parenthesis needed in title. Also bad way to ask a question (as a simple demand...)
$endgroup$
– coffeemath
Apr 6 at 22:44
$begingroup$
An intuition is that for large $n$, $n^-3$ is close to $0$, so $$logleft(n^3 + n^-3right) approx logleft(n^3right) = 3log n,$$ which you should know is less than $n^2$ for all large $n$. See if you can fashion a proof out of this idea.
$endgroup$
– Minus One-Twelfth
Apr 6 at 23:45
$begingroup$
An intuition is that for large $n$, $n^-3$ is close to $0$, so $$logleft(n^3 + n^-3right) approx logleft(n^3right) = 3log n,$$ which you should know is less than $n^2$ for all large $n$. See if you can fashion a proof out of this idea.
$endgroup$
– Minus One-Twelfth
Apr 6 at 23:45
add a comment |
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$begingroup$
A parenthesis needed in title. Also bad way to ask a question (as a simple demand...)
$endgroup$
– coffeemath
Apr 6 at 22:44
$begingroup$
An intuition is that for large $n$, $n^-3$ is close to $0$, so $$logleft(n^3 + n^-3right) approx logleft(n^3right) = 3log n,$$ which you should know is less than $n^2$ for all large $n$. See if you can fashion a proof out of this idea.
$endgroup$
– Minus One-Twelfth
Apr 6 at 23:45