Proof for upper bound of function using an approximation through Squeeze Theorem. The Next CEO of Stack OverflowFinding lower bound sequence for squeeze theoremHow to use Taylor's Theorem to obtain an upper bound for an error approximationProof verification for the Squeeze Theorem.squeeze theorem proof, using the same epsilonsqueeze theorem proof checkProof using squeeze theorem for convergence of infinite sequenceUsing a Taylor polynomial to calculate the error in using $x$ as an approximation for $sinx$Proof Verification: Squeeze theoremProve that $sqrt[k] frac1k+2 to 1$ by squeeze theorem2 has a square root in $mathbbR$ - proof explanation
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Proof for upper bound of function using an approximation through Squeeze Theorem.
The Next CEO of Stack OverflowFinding lower bound sequence for squeeze theoremHow to use Taylor's Theorem to obtain an upper bound for an error approximationProof verification for the Squeeze Theorem.squeeze theorem proof, using the same epsilonsqueeze theorem proof checkProof using squeeze theorem for convergence of infinite sequenceUsing a Taylor polynomial to calculate the error in using $x$ as an approximation for $sinx$Proof Verification: Squeeze theoremProve that $sqrt[k] frac1k+2 to 1$ by squeeze theorem2 has a square root in $mathbbR$ - proof explanation
$begingroup$
I am trying to turn
$n_0log_2(n_0)$ into an approximation of $nlog_2(n)$ using the following statement:
$$n geq n_0+1 $$
For a bit of context, $n$ is the number of nodes on a minimum spanning tree [MST] and $n_0$ the number of odd degree nodes on MST.
I am having trouble justifying the following. I wish to set an upper bound to the prior statement, and make an approximation of $n_0$ for large numbers of $nuparrowuparrow$ so I state:
$$ n_0 lessapprox n rightarrow n_0log_2(n_0) lessapprox nlog_2(n) $$
Is there any better mathematical way you can think of to prove my reasoning above?
I had thought of doing it through Squeeze Theorem. What do you think?
calculus proof-verification proof-explanation approximation alternative-proof
asked Mar 18 at 10:34
Aidan DeavesAidan Deaves
36
$endgroup$
add a comment |
$begingroup$
I am trying to turn
$n_0log_2(n_0)$ into an approximation of $nlog_2(n)$ using the following statement:
$$n geq n_0+1 $$
For a bit of context, $n$ is the number of nodes on a minimum spanning tree [MST] and $n_0$ the number of odd degree nodes on MST.
I am having trouble justifying the following. I wish to set an upper bound to the prior statement, and make an approximation of $n_0$ for large numbers of $nuparrowuparrow$ so I state:
$$ n_0 lessapprox n rightarrow n_0log_2(n_0) lessapprox nlog_2(n) $$
Is there any better mathematical way you can think of to prove my reasoning above?
I had thought of doing it through Squeeze Theorem. What do you think?
calculus proof-verification proof-explanation approximation alternative-proof
asked Mar 18 at 10:34
Aidan DeavesAidan Deaves
36
$endgroup$
add a comment |
$begingroup$
I am trying to turn
$n_0log_2(n_0)$ into an approximation of $nlog_2(n)$ using the following statement:
$$n geq n_0+1 $$
For a bit of context, $n$ is the number of nodes on a minimum spanning tree [MST] and $n_0$ the number of odd degree nodes on MST.
I am having trouble justifying the following. I wish to set an upper bound to the prior statement, and make an approximation of $n_0$ for large numbers of $nuparrowuparrow$ so I state:
$$ n_0 lessapprox n rightarrow n_0log_2(n_0) lessapprox nlog_2(n) $$
Is there any better mathematical way you can think of to prove my reasoning above?
I had thought of doing it through Squeeze Theorem. What do you think?
calculus proof-verification proof-explanation approximation alternative-proof
asked Mar 18 at 10:34
Aidan DeavesAidan Deaves
36
$endgroup$
I am trying to turn
$n_0log_2(n_0)$ into an approximation of $nlog_2(n)$ using the following statement:
$$n geq n_0+1 $$
For a bit of context, $n$ is the number of nodes on a minimum spanning tree [MST] and $n_0$ the number of odd degree nodes on MST.
I am having trouble justifying the following. I wish to set an upper bound to the prior statement, and make an approximation of $n_0$ for large numbers of $nuparrowuparrow$ so I state:
$$ n_0 lessapprox n rightarrow n_0log_2(n_0) lessapprox nlog_2(n) $$
Is there any better mathematical way you can think of to prove my reasoning above?
I had thought of doing it through Squeeze Theorem. What do you think?
calculus proof-verification proof-explanation approximation alternative-proof
calculus proof-verification proof-explanation approximation alternative-proof
asked Mar 18 at 10:34
Aidan DeavesAidan Deaves
36
asked Mar 18 at 10:34
Aidan DeavesAidan Deaves
36
asked Mar 18 at 10:34
Aidan DeavesAidan Deaves
36
asked Mar 18 at 10:34
Aidan DeavesAidan Deaves
36
asked Mar 18 at 10:34
Aidan DeavesAidan Deaves
36
36
add a comment |
add a comment |
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