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Big-o notation confused
The Next CEO of Stack OverflowWhat does it mean to select $O(k log k / epsilon^2)$ indices?Big Oh notation/estimationParticular Use of Big O NotationAsymptotic/Big-O notation with multiple variables?Big O notation and double “less than” notationClarifying the definition of big-ohQuestion about the meaning of “exact” in backward stabilityBig-O Notation of an expressionBig O notation question, loglinear/polynomial growth rateStrange big-O notation?
$begingroup$
In error analysis we say the error is of order $epsilon$ if the error is less than or equal constant multiple of $epsilon$.
The question is: what is the benefit we have if we can multiply by any constant we choose?
linear-algebra asymptotics numerical-linear-algebra
$endgroup$
add a comment |
$begingroup$
In error analysis we say the error is of order $epsilon$ if the error is less than or equal constant multiple of $epsilon$.
The question is: what is the benefit we have if we can multiply by any constant we choose?
linear-algebra asymptotics numerical-linear-algebra
$endgroup$
add a comment |
$begingroup$
In error analysis we say the error is of order $epsilon$ if the error is less than or equal constant multiple of $epsilon$.
The question is: what is the benefit we have if we can multiply by any constant we choose?
linear-algebra asymptotics numerical-linear-algebra
$endgroup$
In error analysis we say the error is of order $epsilon$ if the error is less than or equal constant multiple of $epsilon$.
The question is: what is the benefit we have if we can multiply by any constant we choose?
linear-algebra asymptotics numerical-linear-algebra
linear-algebra asymptotics numerical-linear-algebra
edited Mar 20 at 14:34
Omnomnomnom
129k793187
129k793187
asked Mar 20 at 10:25
user8965733user8965733
453
453
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1 Answer
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$begingroup$
The point is that you usually don't care about the constant. If $epsilon = 10^-10$, it's more important to know whether your error bounded by $8 epsilon$ or $8 epsilon^2$ than it is to know whether your error is bounded by $8epsilon$ or $pi^2 epsilon$...
$endgroup$
$begingroup$
Thank you........
$endgroup$
– user8965733
Mar 20 at 11:47
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The point is that you usually don't care about the constant. If $epsilon = 10^-10$, it's more important to know whether your error bounded by $8 epsilon$ or $8 epsilon^2$ than it is to know whether your error is bounded by $8epsilon$ or $pi^2 epsilon$...
$endgroup$
$begingroup$
Thank you........
$endgroup$
– user8965733
Mar 20 at 11:47
add a comment |
$begingroup$
The point is that you usually don't care about the constant. If $epsilon = 10^-10$, it's more important to know whether your error bounded by $8 epsilon$ or $8 epsilon^2$ than it is to know whether your error is bounded by $8epsilon$ or $pi^2 epsilon$...
$endgroup$
$begingroup$
Thank you........
$endgroup$
– user8965733
Mar 20 at 11:47
add a comment |
$begingroup$
The point is that you usually don't care about the constant. If $epsilon = 10^-10$, it's more important to know whether your error bounded by $8 epsilon$ or $8 epsilon^2$ than it is to know whether your error is bounded by $8epsilon$ or $pi^2 epsilon$...
$endgroup$
The point is that you usually don't care about the constant. If $epsilon = 10^-10$, it's more important to know whether your error bounded by $8 epsilon$ or $8 epsilon^2$ than it is to know whether your error is bounded by $8epsilon$ or $pi^2 epsilon$...
answered Mar 20 at 10:28
GlougloubarbakiGlougloubarbaki
5,83511439
5,83511439
$begingroup$
Thank you........
$endgroup$
– user8965733
Mar 20 at 11:47
add a comment |
$begingroup$
Thank you........
$endgroup$
– user8965733
Mar 20 at 11:47
$begingroup$
Thank you........
$endgroup$
– user8965733
Mar 20 at 11:47
$begingroup$
Thank you........
$endgroup$
– user8965733
Mar 20 at 11:47
add a comment |
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