Mixing
Strictly less then $o$ noation.
$begingroup$
We know that $f in o(n)$ if $f(n)$ is strictly less than $n$, i.e., $lim_{n}f(n)/n = 0$.
What do we mean by saying that $f(n) < o(n)$?
Does it means that $f(n) in o(sqrt{n})$ for example?
real-analysis functions cryptography
asked Jan 17 at 13:57
C.S.C.S.
235
$endgroup$
add a comment |
$begingroup$
We know that $f in o(n)$ if $f(n)$ is strictly less than $n$, i.e., $lim_{n}f(n)/n = 0$.
What do we mean by saying that $f(n) < o(n)$?
Does it means that $f(n) in o(sqrt{n})$ for example?
real-analysis functions cryptography
asked Jan 17 at 13:57
C.S.C.S.
235
$endgroup$
1
$begingroup$
"strictly less than" is not the right way to put it (for example, $n/2$ is strictly less than $n$ but the limit is finite).
$endgroup$
– Yves Daoust
Jan 17 at 14:00
add a comment |
$begingroup$
We know that $f in o(n)$ if $f(n)$ is strictly less than $n$, i.e., $lim_{n}f(n)/n = 0$.
What do we mean by saying that $f(n) < o(n)$?
Does it means that $f(n) in o(sqrt{n})$ for example?
real-analysis functions cryptography
asked Jan 17 at 13:57
C.S.C.S.
235
$endgroup$
We know that $f in o(n)$ if $f(n)$ is strictly less than $n$, i.e., $lim_{n}f(n)/n = 0$.
What do we mean by saying that $f(n) < o(n)$?
Does it means that $f(n) in o(sqrt{n})$ for example?
real-analysis functions cryptography
real-analysis functions cryptography
asked Jan 17 at 13:57
C.S.C.S.
235
asked Jan 17 at 13:57
C.S.C.S.
235
asked Jan 17 at 13:57
C.S.C.S.
235
asked Jan 17 at 13:57
C.S.C.S.
235
asked Jan 17 at 13:57
C.S.C.S.
235
235
1
$begingroup$
"strictly less than" is not the right way to put it (for example, $n/2$ is strictly less than $n$ but the limit is finite).
$endgroup$
– Yves Daoust
Jan 17 at 14:00
add a comment |
1
$begingroup$
"strictly less than" is not the right way to put it (for example, $n/2$ is strictly less than $n$ but the limit is finite).
$endgroup$
– Yves Daoust
Jan 17 at 14:00
1
1
$begingroup$
"strictly less than" is not the right way to put it (for example, $n/2$ is strictly less than $n$ but the limit is finite).
$endgroup$
– Yves Daoust
Jan 17 at 14:00
$begingroup$
"strictly less than" is not the right way to put it (for example, $n/2$ is strictly less than $n$ but the limit is finite).
$endgroup$
– Yves Daoust
Jan 17 at 14:00
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I've never seen this notation.
What seems like a sensible interpretation would be that $f<o(g)$ if "$f$ is even smaller than $o(g)$" (or perhaps "$f=o(o((g))$"), meaning that there exists $h$ such that $f=o(h)$ and $h=o(g)$.
Except that that's a little silly, because it's not hard to see that with that definition we have $f<o(g)$ if and only if $f=o(g)$.
Edit: It's been stated that the equivalence mentioned above is false. So here's a proof:
Suppose $f=o(g)$. Wlog $gge0$. This says that $$lim_{ntoinfty}frac{|f(n)|}{g(n)}=0.$$
Since $|f(n)|/g(n)ge0$ there exists a function $phi(n)ge0$ with $$phi(n)^2=frac{|f(n)|}{g(n)}.$$So of course $phi(n)to0$.
Let $h(n)=phi(n)g(n)$. Then $$frac{|f(n)|}{h(n)}=frac{h(n)}{g(n)}=phi(n),$$so $f=o(h)$ and $h=o(g)$.
Oops: It's possible that there is some division by $0$ above. Of course the values of $n$ for which we divided by $0$ don't matter; we leave it to the reader to construct a formally correct proof. (Instead of saying $f=o(g)$ means $f/gto0$ one should say that for every $epsilon>0$ there exists $N$ so that $|f(n)|le epsilon g(n)$ for all $nge N$.)
answered Jan 17 at 14:10
David C. UllrichDavid C. Ullrich
61k43994
$endgroup$
add a comment |
$begingroup$
$f(n)<o(n)$ makes no sense.
Any function that grows slower than $n$ is denoted with $f(n)=o(n)$ (or $f(n)in o(n)$).
If you want to say "grows slower than a function that grows slower than $n$", $f(n)=o(g(n))$, where $g(n)=o(n)$ or $g$ is given explictly.
answered Jan 17 at 14:03
Yves DaoustYves Daoust
128k675227
$endgroup$
$begingroup$
In fact "$f=o(g)$ where $g=o(n)$" is equivalent to "$f=o(n)$".
$endgroup$
– David C. Ullrich
Jan 17 at 14:11
$begingroup$
@DavidC.Ullrich: it is not equivalent. It implies it, but not conversely.
$endgroup$
– Yves Daoust
Jan 17 at 14:34
$begingroup$
Really? Counterexample?
$endgroup$
– David C. Ullrich
Jan 17 at 14:37
$begingroup$
@DavidC.Ullrich Take $g(n)=sqrt n=o(n)$. $f=o(n)nRightarrow f=o(g)$.
$endgroup$
– Yves Daoust
Jan 17 at 14:46
1
$begingroup$
When I wrote "$f=o(g)$ where $g=o(n)$" I meant "there exists $g$ such that $f=o(g)$ and $g=o(n)$." A counterexample to that cannnot begin by specifying $g$.
$endgroup$
– David C. Ullrich
Jan 17 at 14:49
|
show 1 more comment
Your Answer
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2 Answers
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2 Answers
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$begingroup$
I've never seen this notation.
What seems like a sensible interpretation would be that $f<o(g)$ if "$f$ is even smaller than $o(g)$" (or perhaps "$f=o(o((g))$"), meaning that there exists $h$ such that $f=o(h)$ and $h=o(g)$.
Except that that's a little silly, because it's not hard to see that with that definition we have $f<o(g)$ if and only if $f=o(g)$.
Edit: It's been stated that the equivalence mentioned above is false. So here's a proof:
Suppose $f=o(g)$. Wlog $gge0$. This says that $$lim_{ntoinfty}frac{|f(n)|}{g(n)}=0.$$
Since $|f(n)|/g(n)ge0$ there exists a function $phi(n)ge0$ with $$phi(n)^2=frac{|f(n)|}{g(n)}.$$So of course $phi(n)to0$.
Let $h(n)=phi(n)g(n)$. Then $$frac{|f(n)|}{h(n)}=frac{h(n)}{g(n)}=phi(n),$$so $f=o(h)$ and $h=o(g)$.
Oops: It's possible that there is some division by $0$ above. Of course the values of $n$ for which we divided by $0$ don't matter; we leave it to the reader to construct a formally correct proof. (Instead of saying $f=o(g)$ means $f/gto0$ one should say that for every $epsilon>0$ there exists $N$ so that $|f(n)|le epsilon g(n)$ for all $nge N$.)
answered Jan 17 at 14:10
David C. UllrichDavid C. Ullrich
61k43994
$endgroup$
add a comment |
$begingroup$
I've never seen this notation.
What seems like a sensible interpretation would be that $f<o(g)$ if "$f$ is even smaller than $o(g)$" (or perhaps "$f=o(o((g))$"), meaning that there exists $h$ such that $f=o(h)$ and $h=o(g)$.
Except that that's a little silly, because it's not hard to see that with that definition we have $f<o(g)$ if and only if $f=o(g)$.
Edit: It's been stated that the equivalence mentioned above is false. So here's a proof:
Suppose $f=o(g)$. Wlog $gge0$. This says that $$lim_{ntoinfty}frac{|f(n)|}{g(n)}=0.$$
Since $|f(n)|/g(n)ge0$ there exists a function $phi(n)ge0$ with $$phi(n)^2=frac{|f(n)|}{g(n)}.$$So of course $phi(n)to0$.
Let $h(n)=phi(n)g(n)$. Then $$frac{|f(n)|}{h(n)}=frac{h(n)}{g(n)}=phi(n),$$so $f=o(h)$ and $h=o(g)$.
Oops: It's possible that there is some division by $0$ above. Of course the values of $n$ for which we divided by $0$ don't matter; we leave it to the reader to construct a formally correct proof. (Instead of saying $f=o(g)$ means $f/gto0$ one should say that for every $epsilon>0$ there exists $N$ so that $|f(n)|le epsilon g(n)$ for all $nge N$.)
answered Jan 17 at 14:10
David C. UllrichDavid C. Ullrich
61k43994
$endgroup$
add a comment |
$begingroup$
I've never seen this notation.
What seems like a sensible interpretation would be that $f<o(g)$ if "$f$ is even smaller than $o(g)$" (or perhaps "$f=o(o((g))$"), meaning that there exists $h$ such that $f=o(h)$ and $h=o(g)$.
Except that that's a little silly, because it's not hard to see that with that definition we have $f<o(g)$ if and only if $f=o(g)$.
Edit: It's been stated that the equivalence mentioned above is false. So here's a proof:
Suppose $f=o(g)$. Wlog $gge0$. This says that $$lim_{ntoinfty}frac{|f(n)|}{g(n)}=0.$$
Since $|f(n)|/g(n)ge0$ there exists a function $phi(n)ge0$ with $$phi(n)^2=frac{|f(n)|}{g(n)}.$$So of course $phi(n)to0$.
Let $h(n)=phi(n)g(n)$. Then $$frac{|f(n)|}{h(n)}=frac{h(n)}{g(n)}=phi(n),$$so $f=o(h)$ and $h=o(g)$.
Oops: It's possible that there is some division by $0$ above. Of course the values of $n$ for which we divided by $0$ don't matter; we leave it to the reader to construct a formally correct proof. (Instead of saying $f=o(g)$ means $f/gto0$ one should say that for every $epsilon>0$ there exists $N$ so that $|f(n)|le epsilon g(n)$ for all $nge N$.)
answered Jan 17 at 14:10
David C. UllrichDavid C. Ullrich
61k43994
$endgroup$
I've never seen this notation.
What seems like a sensible interpretation would be that $f<o(g)$ if "$f$ is even smaller than $o(g)$" (or perhaps "$f=o(o((g))$"), meaning that there exists $h$ such that $f=o(h)$ and $h=o(g)$.
Except that that's a little silly, because it's not hard to see that with that definition we have $f<o(g)$ if and only if $f=o(g)$.
Edit: It's been stated that the equivalence mentioned above is false. So here's a proof:
Suppose $f=o(g)$. Wlog $gge0$. This says that $$lim_{ntoinfty}frac{|f(n)|}{g(n)}=0.$$
Since $|f(n)|/g(n)ge0$ there exists a function $phi(n)ge0$ with $$phi(n)^2=frac{|f(n)|}{g(n)}.$$So of course $phi(n)to0$.
Let $h(n)=phi(n)g(n)$. Then $$frac{|f(n)|}{h(n)}=frac{h(n)}{g(n)}=phi(n),$$so $f=o(h)$ and $h=o(g)$.
Oops: It's possible that there is some division by $0$ above. Of course the values of $n$ for which we divided by $0$ don't matter; we leave it to the reader to construct a formally correct proof. (Instead of saying $f=o(g)$ means $f/gto0$ one should say that for every $epsilon>0$ there exists $N$ so that $|f(n)|le epsilon g(n)$ for all $nge N$.)
answered Jan 17 at 14:10
David C. UllrichDavid C. Ullrich
61k43994
edited Jan 17 at 15:03
answered Jan 17 at 14:10
David C. UllrichDavid C. Ullrich
61k43994
answered Jan 17 at 14:10
David C. UllrichDavid C. Ullrich
61k43994
answered Jan 17 at 14:10
David C. UllrichDavid C. Ullrich
61k43994
61k43994
add a comment |
add a comment |
$begingroup$
$f(n)<o(n)$ makes no sense.
Any function that grows slower than $n$ is denoted with $f(n)=o(n)$ (or $f(n)in o(n)$).
If you want to say "grows slower than a function that grows slower than $n$", $f(n)=o(g(n))$, where $g(n)=o(n)$ or $g$ is given explictly.
answered Jan 17 at 14:03
Yves DaoustYves Daoust
128k675227
$endgroup$
$begingroup$
In fact "$f=o(g)$ where $g=o(n)$" is equivalent to "$f=o(n)$".
$endgroup$
– David C. Ullrich
Jan 17 at 14:11
$begingroup$
@DavidC.Ullrich: it is not equivalent. It implies it, but not conversely.
$endgroup$
– Yves Daoust
Jan 17 at 14:34
$begingroup$
Really? Counterexample?
$endgroup$
– David C. Ullrich
Jan 17 at 14:37
$begingroup$
@DavidC.Ullrich Take $g(n)=sqrt n=o(n)$. $f=o(n)nRightarrow f=o(g)$.
$endgroup$
– Yves Daoust
Jan 17 at 14:46
1
$begingroup$
When I wrote "$f=o(g)$ where $g=o(n)$" I meant "there exists $g$ such that $f=o(g)$ and $g=o(n)$." A counterexample to that cannnot begin by specifying $g$.
$endgroup$
– David C. Ullrich
Jan 17 at 14:49
|
show 1 more comment
$begingroup$
$f(n)<o(n)$ makes no sense.
Any function that grows slower than $n$ is denoted with $f(n)=o(n)$ (or $f(n)in o(n)$).
If you want to say "grows slower than a function that grows slower than $n$", $f(n)=o(g(n))$, where $g(n)=o(n)$ or $g$ is given explictly.
answered Jan 17 at 14:03
Yves DaoustYves Daoust
128k675227
$endgroup$
$begingroup$
In fact "$f=o(g)$ where $g=o(n)$" is equivalent to "$f=o(n)$".
$endgroup$
– David C. Ullrich
Jan 17 at 14:11
$begingroup$
@DavidC.Ullrich: it is not equivalent. It implies it, but not conversely.
$endgroup$
– Yves Daoust
Jan 17 at 14:34
$begingroup$
Really? Counterexample?
$endgroup$
– David C. Ullrich
Jan 17 at 14:37
$begingroup$
@DavidC.Ullrich Take $g(n)=sqrt n=o(n)$. $f=o(n)nRightarrow f=o(g)$.
$endgroup$
– Yves Daoust
Jan 17 at 14:46
1
$begingroup$
When I wrote "$f=o(g)$ where $g=o(n)$" I meant "there exists $g$ such that $f=o(g)$ and $g=o(n)$." A counterexample to that cannnot begin by specifying $g$.
$endgroup$
– David C. Ullrich
Jan 17 at 14:49
|
show 1 more comment
$begingroup$
$f(n)<o(n)$ makes no sense.
Any function that grows slower than $n$ is denoted with $f(n)=o(n)$ (or $f(n)in o(n)$).
If you want to say "grows slower than a function that grows slower than $n$", $f(n)=o(g(n))$, where $g(n)=o(n)$ or $g$ is given explictly.
answered Jan 17 at 14:03
Yves DaoustYves Daoust
128k675227
$endgroup$
$f(n)<o(n)$ makes no sense.
Any function that grows slower than $n$ is denoted with $f(n)=o(n)$ (or $f(n)in o(n)$).
If you want to say "grows slower than a function that grows slower than $n$", $f(n)=o(g(n))$, where $g(n)=o(n)$ or $g$ is given explictly.
answered Jan 17 at 14:03
Yves DaoustYves Daoust
128k675227
answered Jan 17 at 14:03
Yves DaoustYves Daoust
128k675227
answered Jan 17 at 14:03
Yves DaoustYves Daoust
128k675227
answered Jan 17 at 14:03
Yves DaoustYves Daoust
128k675227
128k675227
$begingroup$
In fact "$f=o(g)$ where $g=o(n)$" is equivalent to "$f=o(n)$".
$endgroup$
– David C. Ullrich
Jan 17 at 14:11
$begingroup$
@DavidC.Ullrich: it is not equivalent. It implies it, but not conversely.
$endgroup$
– Yves Daoust
Jan 17 at 14:34
$begingroup$
Really? Counterexample?
$endgroup$
– David C. Ullrich
Jan 17 at 14:37
$begingroup$
@DavidC.Ullrich Take $g(n)=sqrt n=o(n)$. $f=o(n)nRightarrow f=o(g)$.
$endgroup$
– Yves Daoust
Jan 17 at 14:46
1
$begingroup$
When I wrote "$f=o(g)$ where $g=o(n)$" I meant "there exists $g$ such that $f=o(g)$ and $g=o(n)$." A counterexample to that cannnot begin by specifying $g$.
$endgroup$
– David C. Ullrich
Jan 17 at 14:49
|
show 1 more comment
$begingroup$
In fact "$f=o(g)$ where $g=o(n)$" is equivalent to "$f=o(n)$".
$endgroup$
– David C. Ullrich
Jan 17 at 14:11
$begingroup$
@DavidC.Ullrich: it is not equivalent. It implies it, but not conversely.
$endgroup$
– Yves Daoust
Jan 17 at 14:34
$begingroup$
Really? Counterexample?
$endgroup$
– David C. Ullrich
Jan 17 at 14:37
$begingroup$
@DavidC.Ullrich Take $g(n)=sqrt n=o(n)$. $f=o(n)nRightarrow f=o(g)$.
$endgroup$
– Yves Daoust
Jan 17 at 14:46
1
$begingroup$
When I wrote "$f=o(g)$ where $g=o(n)$" I meant "there exists $g$ such that $f=o(g)$ and $g=o(n)$." A counterexample to that cannnot begin by specifying $g$.
$endgroup$
– David C. Ullrich
Jan 17 at 14:49
$begingroup$
In fact "$f=o(g)$ where $g=o(n)$" is equivalent to "$f=o(n)$".
$endgroup$
– David C. Ullrich
Jan 17 at 14:11
$begingroup$
In fact "$f=o(g)$ where $g=o(n)$" is equivalent to "$f=o(n)$".
$endgroup$
– David C. Ullrich
Jan 17 at 14:11
$begingroup$
@DavidC.Ullrich: it is not equivalent. It implies it, but not conversely.
$endgroup$
– Yves Daoust
Jan 17 at 14:34
$begingroup$
@DavidC.Ullrich: it is not equivalent. It implies it, but not conversely.
$endgroup$
– Yves Daoust
Jan 17 at 14:34
$begingroup$
Really? Counterexample?
$endgroup$
– David C. Ullrich
Jan 17 at 14:37
$begingroup$
Really? Counterexample?
$endgroup$
– David C. Ullrich
Jan 17 at 14:37
$begingroup$
@DavidC.Ullrich Take $g(n)=sqrt n=o(n)$. $f=o(n)nRightarrow f=o(g)$.
$endgroup$
– Yves Daoust
Jan 17 at 14:46
$begingroup$
@DavidC.Ullrich Take $g(n)=sqrt n=o(n)$. $f=o(n)nRightarrow f=o(g)$.
$endgroup$
– Yves Daoust
Jan 17 at 14:46
1
1
$begingroup$
When I wrote "$f=o(g)$ where $g=o(n)$" I meant "there exists $g$ such that $f=o(g)$ and $g=o(n)$." A counterexample to that cannnot begin by specifying $g$.
$endgroup$
– David C. Ullrich
Jan 17 at 14:49
$begingroup$
When I wrote "$f=o(g)$ where $g=o(n)$" I meant "there exists $g$ such that $f=o(g)$ and $g=o(n)$." A counterexample to that cannnot begin by specifying $g$.
$endgroup$
– David C. Ullrich
Jan 17 at 14:49
|
show 1 more comment
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1
$begingroup$
"strictly less than" is not the right way to put it (for example, $n/2$ is strictly less than $n$ but the limit is finite).
$endgroup$
– Yves Daoust
Jan 17 at 14:00