Does the dimension converge? [closed]
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I'm doing an exercise in a textbook and I need to calculate the effective dimension of a possibly non-Euclidean network lattice. I have $r^d=2r^2+2r+1$ where $d$ is the dimension of the network and I need to count the number of nodes up to $r$ connections away from a given node. I'm trying to figure out if $d$ converges as $r$ goes to infinity. From my calculations it looks like $d$ might converge to $2,$ or something close to it.
general-topology limits lattice-orders network
asked Jan 2 at 22:52
UltradarkUltradark
2241517
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closed as off-topic by amWhy, Leucippus, Lord Shark the Unknown, Paul Frost, user91500 Jan 3 at 12:08
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I'm doing an exercise in a textbook and I need to calculate the effective dimension of a possibly non-Euclidean network lattice. I have $r^d=2r^2+2r+1$ where $d$ is the dimension of the network and I need to count the number of nodes up to $r$ connections away from a given node. I'm trying to figure out if $d$ converges as $r$ goes to infinity. From my calculations it looks like $d$ might converge to $2,$ or something close to it.
general-topology limits lattice-orders network
asked Jan 2 at 22:52
UltradarkUltradark
2241517
$endgroup$
closed as off-topic by amWhy, Leucippus, Lord Shark the Unknown, Paul Frost, user91500 Jan 3 at 12:08
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." – amWhy, Leucippus, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
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Would the downvoter care to explain their point of view?
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– Ultradark
Jan 2 at 23:03
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I'm doing an exercise in a textbook and I need to calculate the effective dimension of a possibly non-Euclidean network lattice. I have $r^d=2r^2+2r+1$ where $d$ is the dimension of the network and I need to count the number of nodes up to $r$ connections away from a given node. I'm trying to figure out if $d$ converges as $r$ goes to infinity. From my calculations it looks like $d$ might converge to $2,$ or something close to it.
general-topology limits lattice-orders network
asked Jan 2 at 22:52
UltradarkUltradark
2241517
$endgroup$
I'm doing an exercise in a textbook and I need to calculate the effective dimension of a possibly non-Euclidean network lattice. I have $r^d=2r^2+2r+1$ where $d$ is the dimension of the network and I need to count the number of nodes up to $r$ connections away from a given node. I'm trying to figure out if $d$ converges as $r$ goes to infinity. From my calculations it looks like $d$ might converge to $2,$ or something close to it.
general-topology limits lattice-orders network
general-topology limits lattice-orders network
asked Jan 2 at 22:52
UltradarkUltradark
2241517
asked Jan 2 at 22:52
UltradarkUltradark
2241517
asked Jan 2 at 22:52
UltradarkUltradark
2241517
asked Jan 2 at 22:52
UltradarkUltradark
2241517
asked Jan 2 at 22:52
UltradarkUltradark
2241517
2241517
closed as off-topic by amWhy, Leucippus, Lord Shark the Unknown, Paul Frost, user91500 Jan 3 at 12:08
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." – amWhy, Leucippus, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, Leucippus, Lord Shark the Unknown, Paul Frost, user91500 Jan 3 at 12:08
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." – amWhy, Leucippus, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
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Would the downvoter care to explain their point of view?
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– Ultradark
Jan 2 at 23:03
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Would the downvoter care to explain their point of view?
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– Ultradark
Jan 2 at 23:03
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Would the downvoter care to explain their point of view?
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– Ultradark
Jan 2 at 23:03
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Would the downvoter care to explain their point of view?
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– Ultradark
Jan 2 at 23:03
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1 Answer
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For an upper bound, we have $d = log_r(2r^2 + 2r + 1) leq log_r(5r^2) = 2log_r(r) + log_r(5) = 2 + log_r(5) to 2$ as $r to infty$. On the lower side, we have $d = log_r(2r^2 + 2r + 1) geq log_r(2r^2) = 2 + log_r(2) to 2$ as $r to infty$. So yes, indeed, $d$ converges to $2$.
answered Jan 2 at 22:57
user3482749user3482749
4,296919
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For an upper bound, we have $d = log_r(2r^2 + 2r + 1) leq log_r(5r^2) = 2log_r(r) + log_r(5) = 2 + log_r(5) to 2$ as $r to infty$. On the lower side, we have $d = log_r(2r^2 + 2r + 1) geq log_r(2r^2) = 2 + log_r(2) to 2$ as $r to infty$. So yes, indeed, $d$ converges to $2$.
answered Jan 2 at 22:57
user3482749user3482749
4,296919
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add a comment |
$begingroup$
For an upper bound, we have $d = log_r(2r^2 + 2r + 1) leq log_r(5r^2) = 2log_r(r) + log_r(5) = 2 + log_r(5) to 2$ as $r to infty$. On the lower side, we have $d = log_r(2r^2 + 2r + 1) geq log_r(2r^2) = 2 + log_r(2) to 2$ as $r to infty$. So yes, indeed, $d$ converges to $2$.
answered Jan 2 at 22:57
user3482749user3482749
4,296919
$endgroup$
add a comment |
$begingroup$
For an upper bound, we have $d = log_r(2r^2 + 2r + 1) leq log_r(5r^2) = 2log_r(r) + log_r(5) = 2 + log_r(5) to 2$ as $r to infty$. On the lower side, we have $d = log_r(2r^2 + 2r + 1) geq log_r(2r^2) = 2 + log_r(2) to 2$ as $r to infty$. So yes, indeed, $d$ converges to $2$.
answered Jan 2 at 22:57
user3482749user3482749
4,296919
$endgroup$
For an upper bound, we have $d = log_r(2r^2 + 2r + 1) leq log_r(5r^2) = 2log_r(r) + log_r(5) = 2 + log_r(5) to 2$ as $r to infty$. On the lower side, we have $d = log_r(2r^2 + 2r + 1) geq log_r(2r^2) = 2 + log_r(2) to 2$ as $r to infty$. So yes, indeed, $d$ converges to $2$.
answered Jan 2 at 22:57
user3482749user3482749
4,296919
answered Jan 2 at 22:57
user3482749user3482749
4,296919
answered Jan 2 at 22:57
user3482749user3482749
4,296919
answered Jan 2 at 22:57
user3482749user3482749
4,296919
4,296919
add a comment |
add a comment |
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Would the downvoter care to explain their point of view?
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– Ultradark
Jan 2 at 23:03