What can we say about the limit of : $O((n-1)/n)$
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Let's say I want to compute :
$$lim_n to infty Oleft(fracn-1nright)$$
Then can we say that to infty this is equal to $O(1)$ ?
Because it could also mean that $forall n in mathbbN$ there is a constant $C_n$ such that : $mid fmid leq C_n cdot fracn-1n$
And in the case where the sequence : $C_n$ is such that : $n = o(C_n)$
then we have :
$$lim_n to infty Oleft(fracn-1nright) = O(+infty)$$
That's why I am wondering : when we say that a function $f$ is $O(fracn-1n)$ then does it mean : $mid fmid leq C cdot fracn-1n forall n$, or : $forall n, exists C_n$ such that $mid f mid leq C_n cdot fracn-1n$ ?
real-analysis limits asymptotics
edited Sep 7 at 11:20
G K
1196
asked Sep 7 at 11:00
auhasard
388
add a comment |Â
up vote
2
down vote
favorite
Let's say I want to compute :
$$lim_n to infty Oleft(fracn-1nright)$$
Then can we say that to infty this is equal to $O(1)$ ?
Because it could also mean that $forall n in mathbbN$ there is a constant $C_n$ such that : $mid fmid leq C_n cdot fracn-1n$
And in the case where the sequence : $C_n$ is such that : $n = o(C_n)$
then we have :
$$lim_n to infty Oleft(fracn-1nright) = O(+infty)$$
That's why I am wondering : when we say that a function $f$ is $O(fracn-1n)$ then does it mean : $mid fmid leq C cdot fracn-1n forall n$, or : $forall n, exists C_n$ such that $mid f mid leq C_n cdot fracn-1n$ ?
real-analysis limits asymptotics
edited Sep 7 at 11:20
G K
1196
asked Sep 7 at 11:00
auhasard
388
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let's say I want to compute :
$$lim_n to infty Oleft(fracn-1nright)$$
Then can we say that to infty this is equal to $O(1)$ ?
Because it could also mean that $forall n in mathbbN$ there is a constant $C_n$ such that : $mid fmid leq C_n cdot fracn-1n$
And in the case where the sequence : $C_n$ is such that : $n = o(C_n)$
then we have :
$$lim_n to infty Oleft(fracn-1nright) = O(+infty)$$
That's why I am wondering : when we say that a function $f$ is $O(fracn-1n)$ then does it mean : $mid fmid leq C cdot fracn-1n forall n$, or : $forall n, exists C_n$ such that $mid f mid leq C_n cdot fracn-1n$ ?
real-analysis limits asymptotics
edited Sep 7 at 11:20
G K
1196
asked Sep 7 at 11:00
auhasard
388
Let's say I want to compute :
$$lim_n to infty Oleft(fracn-1nright)$$
Then can we say that to infty this is equal to $O(1)$ ?
Because it could also mean that $forall n in mathbbN$ there is a constant $C_n$ such that : $mid fmid leq C_n cdot fracn-1n$
And in the case where the sequence : $C_n$ is such that : $n = o(C_n)$
then we have :
$$lim_n to infty Oleft(fracn-1nright) = O(+infty)$$
That's why I am wondering : when we say that a function $f$ is $O(fracn-1n)$ then does it mean : $mid fmid leq C cdot fracn-1n forall n$, or : $forall n, exists C_n$ such that $mid f mid leq C_n cdot fracn-1n$ ?
real-analysis limits asymptotics
real-analysis limits asymptotics
edited Sep 7 at 11:20
G K
1196
asked Sep 7 at 11:00
auhasard
388
edited Sep 7 at 11:20
G K
1196
asked Sep 7 at 11:00
auhasard
388
edited Sep 7 at 11:20
G K
1196
edited Sep 7 at 11:20
G K
1196
edited Sep 7 at 11:20
G K
1196
1196
asked Sep 7 at 11:00
auhasard
388
asked Sep 7 at 11:00
auhasard
388
asked Sep 7 at 11:00
auhasard
388
388
add a comment |Â
add a comment |Â
1 Answer
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when we say that a function $f$ is $O(fracn-1n)$ then does it mean : $mid fmid leq C cdot fracn-1n forall n$, or : $forall n, exists C_n$ such that $mid f mid leq C_n cdot fracn-1n$ ?
Neither. It means $exists n_0 exists C$ such that $forall n ge n_0: |f| le C cdot fracn-1n$.
$O$-notation is often abused, but I've never before seen $lim_n to infty O(f(n))$. If that makes any sense at all, it's to disambiguate between the asymptotic behaviour for large $n$ and the asymptotic behaviour for small $n$.
Finally, observe that if $n > 0$ and $|f| le C cdot fracn-1n$ then certainly $|f| le C cdot 1$, and we can indeed say that $f(n) in O(1)$.
answered Sep 7 at 11:09
Peter Taylor
7,99712240
Thank you. Then if I have sequence $(u_n)_n geq 1$ such that : $u_n = O(frac1n)$ it means that $exists C, exists n_O$ such that : $forall n geq n_0 : mid u_n mid leq Ccdot frac1n$ so in particular : $mid u_n mid to 0$. Moreover it also means that the notation $O$ is only true at $infty$, so I can't say anything about $u_1$ ? Is verything I've said so far true ?
– auhasard
Sep 7 at 11:33
1
I wouldn't use the phrase "is only true at $infty$". Talking about the notation being true can give the wrong idea. I would prefer to talk about what the notation means or describes. $O$ notation describes the behaviour for large $n$ (although be careful: one of the more common abuses is to use it for small $n$, and in particular when you see $O(n^-1)$ the context is often this inverted meaning). But you're correct to say that you can't say anything about $u_1$: the notation conceals the cutoff point.
– Peter Taylor
Sep 7 at 12:10
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
when we say that a function $f$ is $O(fracn-1n)$ then does it mean : $mid fmid leq C cdot fracn-1n forall n$, or : $forall n, exists C_n$ such that $mid f mid leq C_n cdot fracn-1n$ ?
Neither. It means $exists n_0 exists C$ such that $forall n ge n_0: |f| le C cdot fracn-1n$.
$O$-notation is often abused, but I've never before seen $lim_n to infty O(f(n))$. If that makes any sense at all, it's to disambiguate between the asymptotic behaviour for large $n$ and the asymptotic behaviour for small $n$.
Finally, observe that if $n > 0$ and $|f| le C cdot fracn-1n$ then certainly $|f| le C cdot 1$, and we can indeed say that $f(n) in O(1)$.
answered Sep 7 at 11:09
Peter Taylor
7,99712240
Thank you. Then if I have sequence $(u_n)_n geq 1$ such that : $u_n = O(frac1n)$ it means that $exists C, exists n_O$ such that : $forall n geq n_0 : mid u_n mid leq Ccdot frac1n$ so in particular : $mid u_n mid to 0$. Moreover it also means that the notation $O$ is only true at $infty$, so I can't say anything about $u_1$ ? Is verything I've said so far true ?
– auhasard
Sep 7 at 11:33
1
I wouldn't use the phrase "is only true at $infty$". Talking about the notation being true can give the wrong idea. I would prefer to talk about what the notation means or describes. $O$ notation describes the behaviour for large $n$ (although be careful: one of the more common abuses is to use it for small $n$, and in particular when you see $O(n^-1)$ the context is often this inverted meaning). But you're correct to say that you can't say anything about $u_1$: the notation conceals the cutoff point.
– Peter Taylor
Sep 7 at 12:10
add a comment |Â
up vote
2
down vote
accepted
when we say that a function $f$ is $O(fracn-1n)$ then does it mean : $mid fmid leq C cdot fracn-1n forall n$, or : $forall n, exists C_n$ such that $mid f mid leq C_n cdot fracn-1n$ ?
Neither. It means $exists n_0 exists C$ such that $forall n ge n_0: |f| le C cdot fracn-1n$.
$O$-notation is often abused, but I've never before seen $lim_n to infty O(f(n))$. If that makes any sense at all, it's to disambiguate between the asymptotic behaviour for large $n$ and the asymptotic behaviour for small $n$.
Finally, observe that if $n > 0$ and $|f| le C cdot fracn-1n$ then certainly $|f| le C cdot 1$, and we can indeed say that $f(n) in O(1)$.
answered Sep 7 at 11:09
Peter Taylor
7,99712240
Thank you. Then if I have sequence $(u_n)_n geq 1$ such that : $u_n = O(frac1n)$ it means that $exists C, exists n_O$ such that : $forall n geq n_0 : mid u_n mid leq Ccdot frac1n$ so in particular : $mid u_n mid to 0$. Moreover it also means that the notation $O$ is only true at $infty$, so I can't say anything about $u_1$ ? Is verything I've said so far true ?
– auhasard
Sep 7 at 11:33
1
I wouldn't use the phrase "is only true at $infty$". Talking about the notation being true can give the wrong idea. I would prefer to talk about what the notation means or describes. $O$ notation describes the behaviour for large $n$ (although be careful: one of the more common abuses is to use it for small $n$, and in particular when you see $O(n^-1)$ the context is often this inverted meaning). But you're correct to say that you can't say anything about $u_1$: the notation conceals the cutoff point.
– Peter Taylor
Sep 7 at 12:10
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
when we say that a function $f$ is $O(fracn-1n)$ then does it mean : $mid fmid leq C cdot fracn-1n forall n$, or : $forall n, exists C_n$ such that $mid f mid leq C_n cdot fracn-1n$ ?
Neither. It means $exists n_0 exists C$ such that $forall n ge n_0: |f| le C cdot fracn-1n$.
$O$-notation is often abused, but I've never before seen $lim_n to infty O(f(n))$. If that makes any sense at all, it's to disambiguate between the asymptotic behaviour for large $n$ and the asymptotic behaviour for small $n$.
Finally, observe that if $n > 0$ and $|f| le C cdot fracn-1n$ then certainly $|f| le C cdot 1$, and we can indeed say that $f(n) in O(1)$.
answered Sep 7 at 11:09
Peter Taylor
7,99712240
when we say that a function $f$ is $O(fracn-1n)$ then does it mean : $mid fmid leq C cdot fracn-1n forall n$, or : $forall n, exists C_n$ such that $mid f mid leq C_n cdot fracn-1n$ ?
Neither. It means $exists n_0 exists C$ such that $forall n ge n_0: |f| le C cdot fracn-1n$.
$O$-notation is often abused, but I've never before seen $lim_n to infty O(f(n))$. If that makes any sense at all, it's to disambiguate between the asymptotic behaviour for large $n$ and the asymptotic behaviour for small $n$.
Finally, observe that if $n > 0$ and $|f| le C cdot fracn-1n$ then certainly $|f| le C cdot 1$, and we can indeed say that $f(n) in O(1)$.
answered Sep 7 at 11:09
Peter Taylor
7,99712240
answered Sep 7 at 11:09
Peter Taylor
7,99712240
answered Sep 7 at 11:09
Peter Taylor
7,99712240
answered Sep 7 at 11:09
Peter Taylor
7,99712240
7,99712240
Thank you. Then if I have sequence $(u_n)_n geq 1$ such that : $u_n = O(frac1n)$ it means that $exists C, exists n_O$ such that : $forall n geq n_0 : mid u_n mid leq Ccdot frac1n$ so in particular : $mid u_n mid to 0$. Moreover it also means that the notation $O$ is only true at $infty$, so I can't say anything about $u_1$ ? Is verything I've said so far true ?
– auhasard
Sep 7 at 11:33
1
I wouldn't use the phrase "is only true at $infty$". Talking about the notation being true can give the wrong idea. I would prefer to talk about what the notation means or describes. $O$ notation describes the behaviour for large $n$ (although be careful: one of the more common abuses is to use it for small $n$, and in particular when you see $O(n^-1)$ the context is often this inverted meaning). But you're correct to say that you can't say anything about $u_1$: the notation conceals the cutoff point.
– Peter Taylor
Sep 7 at 12:10
add a comment |Â
Thank you. Then if I have sequence $(u_n)_n geq 1$ such that : $u_n = O(frac1n)$ it means that $exists C, exists n_O$ such that : $forall n geq n_0 : mid u_n mid leq Ccdot frac1n$ so in particular : $mid u_n mid to 0$. Moreover it also means that the notation $O$ is only true at $infty$, so I can't say anything about $u_1$ ? Is verything I've said so far true ?
– auhasard
Sep 7 at 11:33
1
I wouldn't use the phrase "is only true at $infty$". Talking about the notation being true can give the wrong idea. I would prefer to talk about what the notation means or describes. $O$ notation describes the behaviour for large $n$ (although be careful: one of the more common abuses is to use it for small $n$, and in particular when you see $O(n^-1)$ the context is often this inverted meaning). But you're correct to say that you can't say anything about $u_1$: the notation conceals the cutoff point.
– Peter Taylor
Sep 7 at 12:10
Thank you. Then if I have sequence $(u_n)_n geq 1$ such that : $u_n = O(frac1n)$ it means that $exists C, exists n_O$ such that : $forall n geq n_0 : mid u_n mid leq Ccdot frac1n$ so in particular : $mid u_n mid to 0$. Moreover it also means that the notation $O$ is only true at $infty$, so I can't say anything about $u_1$ ? Is verything I've said so far true ?
– auhasard
Sep 7 at 11:33
Thank you. Then if I have sequence $(u_n)_n geq 1$ such that : $u_n = O(frac1n)$ it means that $exists C, exists n_O$ such that : $forall n geq n_0 : mid u_n mid leq Ccdot frac1n$ so in particular : $mid u_n mid to 0$. Moreover it also means that the notation $O$ is only true at $infty$, so I can't say anything about $u_1$ ? Is verything I've said so far true ?
– auhasard
Sep 7 at 11:33
1
1
I wouldn't use the phrase "is only true at $infty$". Talking about the notation being true can give the wrong idea. I would prefer to talk about what the notation means or describes. $O$ notation describes the behaviour for large $n$ (although be careful: one of the more common abuses is to use it for small $n$, and in particular when you see $O(n^-1)$ the context is often this inverted meaning). But you're correct to say that you can't say anything about $u_1$: the notation conceals the cutoff point.
– Peter Taylor
Sep 7 at 12:10
I wouldn't use the phrase "is only true at $infty$". Talking about the notation being true can give the wrong idea. I would prefer to talk about what the notation means or describes. $O$ notation describes the behaviour for large $n$ (although be careful: one of the more common abuses is to use it for small $n$, and in particular when you see $O(n^-1)$ the context is often this inverted meaning). But you're correct to say that you can't say anything about $u_1$: the notation conceals the cutoff point.
– Peter Taylor
Sep 7 at 12:10
add a comment |Â
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