Big O-Notation and lim sup
Clash Royale CLAN TAG#URR8PPP
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this may sound like a stupid question, so please excuse me for that. From my understanding after quickly reading up on it (only know the definition of general limits before), $limsuplimits_xrightarrowinfty sin(x) = 1$. Because $f in O(g) leftrightarrow limsuplimits_xrightarrowinfty |fracf(x)g(x)| < infty$ I can state that $f(x) = 0.5 in O(sin(x))$.
My problem is when looking at the "regular" Big-O Definition, I am not really sure whether my conclusion $f(x) = 0.5 in O(sin(x))$ is correct as $sin(x)$ oscillates between 0 and 1, thus there is no $c$ for that $0.5 < c cdot sin(x) forall n > n_0$. Those two results contradict and I do not know where my error is...
I would be glad if somebody should tell me where I did something wrong.
asymptotics
asked Aug 26 at 21:54
Maxbit
1203
add a comment |Â
up vote
3
down vote
favorite
this may sound like a stupid question, so please excuse me for that. From my understanding after quickly reading up on it (only know the definition of general limits before), $limsuplimits_xrightarrowinfty sin(x) = 1$. Because $f in O(g) leftrightarrow limsuplimits_xrightarrowinfty |fracf(x)g(x)| < infty$ I can state that $f(x) = 0.5 in O(sin(x))$.
My problem is when looking at the "regular" Big-O Definition, I am not really sure whether my conclusion $f(x) = 0.5 in O(sin(x))$ is correct as $sin(x)$ oscillates between 0 and 1, thus there is no $c$ for that $0.5 < c cdot sin(x) forall n > n_0$. Those two results contradict and I do not know where my error is...
I would be glad if somebody should tell me where I did something wrong.
asymptotics
asked Aug 26 at 21:54
Maxbit
1203
1
Your feeling about this seems right. Where did you get the equivalent formulation for $O(g)$? That does not look right.
– 4-ier
Aug 26 at 21:57
@4-ier I got it from en.wikipedia.org/wiki/… (2. row of the table)
– Maxbit
Aug 26 at 22:01
6
Are you sure that $limsuplimits_xrightarrowinfty |frac0.5sin(x)| < infty$?
– Mjiig
Aug 26 at 22:01
Yup. Completely agree with @Mjiig
– 4-ier
Aug 26 at 22:03
@Mijig Ohh thank you. I visualized $0.5 / sin(x)$ wrongly in my head. When looking at a plot, of course it's not $< infty$. Thank you!
– Maxbit
Aug 26 at 22:06
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
this may sound like a stupid question, so please excuse me for that. From my understanding after quickly reading up on it (only know the definition of general limits before), $limsuplimits_xrightarrowinfty sin(x) = 1$. Because $f in O(g) leftrightarrow limsuplimits_xrightarrowinfty |fracf(x)g(x)| < infty$ I can state that $f(x) = 0.5 in O(sin(x))$.
My problem is when looking at the "regular" Big-O Definition, I am not really sure whether my conclusion $f(x) = 0.5 in O(sin(x))$ is correct as $sin(x)$ oscillates between 0 and 1, thus there is no $c$ for that $0.5 < c cdot sin(x) forall n > n_0$. Those two results contradict and I do not know where my error is...
I would be glad if somebody should tell me where I did something wrong.
asymptotics
asked Aug 26 at 21:54
Maxbit
1203
this may sound like a stupid question, so please excuse me for that. From my understanding after quickly reading up on it (only know the definition of general limits before), $limsuplimits_xrightarrowinfty sin(x) = 1$. Because $f in O(g) leftrightarrow limsuplimits_xrightarrowinfty |fracf(x)g(x)| < infty$ I can state that $f(x) = 0.5 in O(sin(x))$.
My problem is when looking at the "regular" Big-O Definition, I am not really sure whether my conclusion $f(x) = 0.5 in O(sin(x))$ is correct as $sin(x)$ oscillates between 0 and 1, thus there is no $c$ for that $0.5 < c cdot sin(x) forall n > n_0$. Those two results contradict and I do not know where my error is...
I would be glad if somebody should tell me where I did something wrong.
asymptotics
asked Aug 26 at 21:54
Maxbit
1203
asked Aug 26 at 21:54
Maxbit
1203
asked Aug 26 at 21:54
Maxbit
1203
asked Aug 26 at 21:54
Maxbit
1203
1203
1
Your feeling about this seems right. Where did you get the equivalent formulation for $O(g)$? That does not look right.
– 4-ier
Aug 26 at 21:57
@4-ier I got it from en.wikipedia.org/wiki/… (2. row of the table)
– Maxbit
Aug 26 at 22:01
6
Are you sure that $limsuplimits_xrightarrowinfty |frac0.5sin(x)| < infty$?
– Mjiig
Aug 26 at 22:01
Yup. Completely agree with @Mjiig
– 4-ier
Aug 26 at 22:03
@Mijig Ohh thank you. I visualized $0.5 / sin(x)$ wrongly in my head. When looking at a plot, of course it's not $< infty$. Thank you!
– Maxbit
Aug 26 at 22:06
add a comment |Â
1
Your feeling about this seems right. Where did you get the equivalent formulation for $O(g)$? That does not look right.
– 4-ier
Aug 26 at 21:57
@4-ier I got it from en.wikipedia.org/wiki/… (2. row of the table)
– Maxbit
Aug 26 at 22:01
6
Are you sure that $limsuplimits_xrightarrowinfty |frac0.5sin(x)| < infty$?
– Mjiig
Aug 26 at 22:01
Yup. Completely agree with @Mjiig
– 4-ier
Aug 26 at 22:03
@Mijig Ohh thank you. I visualized $0.5 / sin(x)$ wrongly in my head. When looking at a plot, of course it's not $< infty$. Thank you!
– Maxbit
Aug 26 at 22:06
1
1
Your feeling about this seems right. Where did you get the equivalent formulation for $O(g)$? That does not look right.
– 4-ier
Aug 26 at 21:57
Your feeling about this seems right. Where did you get the equivalent formulation for $O(g)$? That does not look right.
– 4-ier
Aug 26 at 21:57
@4-ier I got it from en.wikipedia.org/wiki/… (2. row of the table)
– Maxbit
Aug 26 at 22:01
@4-ier I got it from en.wikipedia.org/wiki/… (2. row of the table)
– Maxbit
Aug 26 at 22:01
6
6
Are you sure that $limsuplimits_xrightarrowinfty |frac0.5sin(x)| < infty$?
– Mjiig
Aug 26 at 22:01
Are you sure that $limsuplimits_xrightarrowinfty |frac0.5sin(x)| < infty$?
– Mjiig
Aug 26 at 22:01
Yup. Completely agree with @Mjiig
– 4-ier
Aug 26 at 22:03
Yup. Completely agree with @Mjiig
– 4-ier
Aug 26 at 22:03
@Mijig Ohh thank you. I visualized $0.5 / sin(x)$ wrongly in my head. When looking at a plot, of course it's not $< infty$. Thank you!
– Maxbit
Aug 26 at 22:06
@Mijig Ohh thank you. I visualized $0.5 / sin(x)$ wrongly in my head. When looking at a plot, of course it's not $< infty$. Thank you!
– Maxbit
Aug 26 at 22:06
add a comment |Â
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1
Your feeling about this seems right. Where did you get the equivalent formulation for $O(g)$? That does not look right.
– 4-ier
Aug 26 at 21:57
@4-ier I got it from en.wikipedia.org/wiki/… (2. row of the table)
– Maxbit
Aug 26 at 22:01
6
Are you sure that $limsuplimits_xrightarrowinfty |frac0.5sin(x)| < infty$?
– Mjiig
Aug 26 at 22:01
Yup. Completely agree with @Mjiig
– 4-ier
Aug 26 at 22:03
@Mijig Ohh thank you. I visualized $0.5 / sin(x)$ wrongly in my head. When looking at a plot, of course it's not $< infty$. Thank you!
– Maxbit
Aug 26 at 22:06