Little o(h) limit about h=0
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I understand that generally if a function $f(h)$ is described as $o(h)$ that $f(h)$ has a smaller rate of growth than $h$ (like it would have to be $sqrth$). i.e. $sqrth = o(h)$, just like (for example) $4h=o(h^2)$. The notes I'm reading (CT3), however, states that:
A function $f(h)$ is described as $o(h)$ if:
$$lim_h to 0 fracf(h)h = 0 $$
but if I use, for example, $f(h)= sqrth$ which does have a slower growth rate then $h$ then the limit doesn't go to $0$. Is there a different meaning to little $0$ when its approaching $0$ compared to when it goes to infinity cause the only way that limit holds is if $f(h)$ goes to $0$ faster then $h$ does which I guess means $f(h)$ decreases faster then $h$ as $h to 0$.
Anyway you can ignore my thoughts on the question but an explanation of the text I quoted would be greatly appreciated.
calculus
edited Jan 31 '15 at 6:55
Daniel W. Farlow
17.2k114187
asked Jan 31 '15 at 6:42
Mauro Augusto
296
 |Â
show 2 more comments
up vote
0
down vote
favorite
I understand that generally if a function $f(h)$ is described as $o(h)$ that $f(h)$ has a smaller rate of growth than $h$ (like it would have to be $sqrth$). i.e. $sqrth = o(h)$, just like (for example) $4h=o(h^2)$. The notes I'm reading (CT3), however, states that:
A function $f(h)$ is described as $o(h)$ if:
$$lim_h to 0 fracf(h)h = 0 $$
but if I use, for example, $f(h)= sqrth$ which does have a slower growth rate then $h$ then the limit doesn't go to $0$. Is there a different meaning to little $0$ when its approaching $0$ compared to when it goes to infinity cause the only way that limit holds is if $f(h)$ goes to $0$ faster then $h$ does which I guess means $f(h)$ decreases faster then $h$ as $h to 0$.
Anyway you can ignore my thoughts on the question but an explanation of the text I quoted would be greatly appreciated.
calculus
edited Jan 31 '15 at 6:55
Daniel W. Farlow
17.2k114187
asked Jan 31 '15 at 6:42
Mauro Augusto
296
1
It is a definition - I don't understand what you mean by explanation? You could interpret it as saying $f(0) = 0$ and $f'(0) = 0$.
– copper.hat
Jan 31 '15 at 6:51
Im talking about little o notation?
– Mauro Augusto
Jan 31 '15 at 7:06
1
I understand. I don't understand what you are asking.
– copper.hat
Jan 31 '15 at 7:09
I just need an explanation of the grey bit. Like how does o(h) work when h->0 instead of going to infinity
– Mauro Augusto
Jan 31 '15 at 7:24
I don't know what you mean, the definition only involves $h to 0$? It is exactly equivalent to $f(0) =0$ and $f'(0) = 0$.
– copper.hat
Jan 31 '15 at 7:28
 |Â
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I understand that generally if a function $f(h)$ is described as $o(h)$ that $f(h)$ has a smaller rate of growth than $h$ (like it would have to be $sqrth$). i.e. $sqrth = o(h)$, just like (for example) $4h=o(h^2)$. The notes I'm reading (CT3), however, states that:
A function $f(h)$ is described as $o(h)$ if:
$$lim_h to 0 fracf(h)h = 0 $$
but if I use, for example, $f(h)= sqrth$ which does have a slower growth rate then $h$ then the limit doesn't go to $0$. Is there a different meaning to little $0$ when its approaching $0$ compared to when it goes to infinity cause the only way that limit holds is if $f(h)$ goes to $0$ faster then $h$ does which I guess means $f(h)$ decreases faster then $h$ as $h to 0$.
Anyway you can ignore my thoughts on the question but an explanation of the text I quoted would be greatly appreciated.
calculus
edited Jan 31 '15 at 6:55
Daniel W. Farlow
17.2k114187
asked Jan 31 '15 at 6:42
Mauro Augusto
296
I understand that generally if a function $f(h)$ is described as $o(h)$ that $f(h)$ has a smaller rate of growth than $h$ (like it would have to be $sqrth$). i.e. $sqrth = o(h)$, just like (for example) $4h=o(h^2)$. The notes I'm reading (CT3), however, states that:
A function $f(h)$ is described as $o(h)$ if:
$$lim_h to 0 fracf(h)h = 0 $$
but if I use, for example, $f(h)= sqrth$ which does have a slower growth rate then $h$ then the limit doesn't go to $0$. Is there a different meaning to little $0$ when its approaching $0$ compared to when it goes to infinity cause the only way that limit holds is if $f(h)$ goes to $0$ faster then $h$ does which I guess means $f(h)$ decreases faster then $h$ as $h to 0$.
Anyway you can ignore my thoughts on the question but an explanation of the text I quoted would be greatly appreciated.
calculus
edited Jan 31 '15 at 6:55
Daniel W. Farlow
17.2k114187
asked Jan 31 '15 at 6:42
Mauro Augusto
296
edited Jan 31 '15 at 6:55
Daniel W. Farlow
17.2k114187
edited Jan 31 '15 at 6:55
Daniel W. Farlow
17.2k114187
edited Jan 31 '15 at 6:55
Daniel W. Farlow
17.2k114187
17.2k114187
asked Jan 31 '15 at 6:42
Mauro Augusto
296
asked Jan 31 '15 at 6:42
Mauro Augusto
296
asked Jan 31 '15 at 6:42
Mauro Augusto
296
296
1
It is a definition - I don't understand what you mean by explanation? You could interpret it as saying $f(0) = 0$ and $f'(0) = 0$.
– copper.hat
Jan 31 '15 at 6:51
Im talking about little o notation?
– Mauro Augusto
Jan 31 '15 at 7:06
1
I understand. I don't understand what you are asking.
– copper.hat
Jan 31 '15 at 7:09
I just need an explanation of the grey bit. Like how does o(h) work when h->0 instead of going to infinity
– Mauro Augusto
Jan 31 '15 at 7:24
I don't know what you mean, the definition only involves $h to 0$? It is exactly equivalent to $f(0) =0$ and $f'(0) = 0$.
– copper.hat
Jan 31 '15 at 7:28
 |Â
show 2 more comments
1
It is a definition - I don't understand what you mean by explanation? You could interpret it as saying $f(0) = 0$ and $f'(0) = 0$.
– copper.hat
Jan 31 '15 at 6:51
Im talking about little o notation?
– Mauro Augusto
Jan 31 '15 at 7:06
1
I understand. I don't understand what you are asking.
– copper.hat
Jan 31 '15 at 7:09
I just need an explanation of the grey bit. Like how does o(h) work when h->0 instead of going to infinity
– Mauro Augusto
Jan 31 '15 at 7:24
I don't know what you mean, the definition only involves $h to 0$? It is exactly equivalent to $f(0) =0$ and $f'(0) = 0$.
– copper.hat
Jan 31 '15 at 7:28
1
1
It is a definition - I don't understand what you mean by explanation? You could interpret it as saying $f(0) = 0$ and $f'(0) = 0$.
– copper.hat
Jan 31 '15 at 6:51
It is a definition - I don't understand what you mean by explanation? You could interpret it as saying $f(0) = 0$ and $f'(0) = 0$.
– copper.hat
Jan 31 '15 at 6:51
Im talking about little o notation?
– Mauro Augusto
Jan 31 '15 at 7:06
Im talking about little o notation?
– Mauro Augusto
Jan 31 '15 at 7:06
1
1
I understand. I don't understand what you are asking.
– copper.hat
Jan 31 '15 at 7:09
I understand. I don't understand what you are asking.
– copper.hat
Jan 31 '15 at 7:09
I just need an explanation of the grey bit. Like how does o(h) work when h->0 instead of going to infinity
– Mauro Augusto
Jan 31 '15 at 7:24
I just need an explanation of the grey bit. Like how does o(h) work when h->0 instead of going to infinity
– Mauro Augusto
Jan 31 '15 at 7:24
I don't know what you mean, the definition only involves $h to 0$? It is exactly equivalent to $f(0) =0$ and $f'(0) = 0$.
– copper.hat
Jan 31 '15 at 7:28
I don't know what you mean, the definition only involves $h to 0$? It is exactly equivalent to $f(0) =0$ and $f'(0) = 0$.
– copper.hat
Jan 31 '15 at 7:28
 |Â
show 2 more comments
2 Answers
2
active
oldest
votes
up vote
2
down vote
Yes, the little-o notation (and Landau symbols in general) behaves differently for $xto 0$ and for $xtoinfty$.
When we're considering $xtoinfty$ (as you may be familiar with from analysis of algorithms), $sqrt x$ grows slower than $x$ -- because for large $x$, the square root of $x$ is smaller than $x$ by a ratio that becomes ever more lopsided. Therefore in this context we say that $sqrt x = o(x)$.
On the other hand, if we're considering $xto 0$ (which is more common in analysis), then when $x$ is close to zero, $sqrt x$ is larger than $x$ by a ratio that tends to infinity. Therefore in that context we say that $x = o(sqrt x)$.
If would be less confusing to make it explicit which limit we're working with, and write something like
$$ sqrt x = mathop o_xtoinfty(x) qquadqquadqquad x = mathop o_xto 0(sqrt x) $$
However, in most practical uses of the notation, the limit is the same throughout the entire calculation, so repeating it for every $o$ would be tedious and distracting. So generally it is left implicit, though one should make it clear before one starts using asymptotic notation which limit one is considering.
answered Aug 14 at 9:16
Henning Makholm
228k16293524
add a comment |Â
up vote
0
down vote
Your starting point is wrong. $h^2=o(h)$ and not the reverse. $h=o(sqrth)$ and not the reverse.The definition of the Landau notation in the vicinity of a point $x_0$ is $$f=o(g)Leftrightarrowlim_xto x_0fracf(x)g(x)=0$$
answered Jan 31 '15 at 8:03
marwalix
13.1k12338
"$sqrth$ has slower growth than $h$" is for $h to infty$, and not for $h to 0$ as we have here.
– GEdgar
Oct 4 '17 at 23:46
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Yes, the little-o notation (and Landau symbols in general) behaves differently for $xto 0$ and for $xtoinfty$.
When we're considering $xtoinfty$ (as you may be familiar with from analysis of algorithms), $sqrt x$ grows slower than $x$ -- because for large $x$, the square root of $x$ is smaller than $x$ by a ratio that becomes ever more lopsided. Therefore in this context we say that $sqrt x = o(x)$.
On the other hand, if we're considering $xto 0$ (which is more common in analysis), then when $x$ is close to zero, $sqrt x$ is larger than $x$ by a ratio that tends to infinity. Therefore in that context we say that $x = o(sqrt x)$.
If would be less confusing to make it explicit which limit we're working with, and write something like
$$ sqrt x = mathop o_xtoinfty(x) qquadqquadqquad x = mathop o_xto 0(sqrt x) $$
However, in most practical uses of the notation, the limit is the same throughout the entire calculation, so repeating it for every $o$ would be tedious and distracting. So generally it is left implicit, though one should make it clear before one starts using asymptotic notation which limit one is considering.
answered Aug 14 at 9:16
Henning Makholm
228k16293524
add a comment |Â
up vote
2
down vote
Yes, the little-o notation (and Landau symbols in general) behaves differently for $xto 0$ and for $xtoinfty$.
When we're considering $xtoinfty$ (as you may be familiar with from analysis of algorithms), $sqrt x$ grows slower than $x$ -- because for large $x$, the square root of $x$ is smaller than $x$ by a ratio that becomes ever more lopsided. Therefore in this context we say that $sqrt x = o(x)$.
On the other hand, if we're considering $xto 0$ (which is more common in analysis), then when $x$ is close to zero, $sqrt x$ is larger than $x$ by a ratio that tends to infinity. Therefore in that context we say that $x = o(sqrt x)$.
If would be less confusing to make it explicit which limit we're working with, and write something like
$$ sqrt x = mathop o_xtoinfty(x) qquadqquadqquad x = mathop o_xto 0(sqrt x) $$
However, in most practical uses of the notation, the limit is the same throughout the entire calculation, so repeating it for every $o$ would be tedious and distracting. So generally it is left implicit, though one should make it clear before one starts using asymptotic notation which limit one is considering.
answered Aug 14 at 9:16
Henning Makholm
228k16293524
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Yes, the little-o notation (and Landau symbols in general) behaves differently for $xto 0$ and for $xtoinfty$.
When we're considering $xtoinfty$ (as you may be familiar with from analysis of algorithms), $sqrt x$ grows slower than $x$ -- because for large $x$, the square root of $x$ is smaller than $x$ by a ratio that becomes ever more lopsided. Therefore in this context we say that $sqrt x = o(x)$.
On the other hand, if we're considering $xto 0$ (which is more common in analysis), then when $x$ is close to zero, $sqrt x$ is larger than $x$ by a ratio that tends to infinity. Therefore in that context we say that $x = o(sqrt x)$.
If would be less confusing to make it explicit which limit we're working with, and write something like
$$ sqrt x = mathop o_xtoinfty(x) qquadqquadqquad x = mathop o_xto 0(sqrt x) $$
However, in most practical uses of the notation, the limit is the same throughout the entire calculation, so repeating it for every $o$ would be tedious and distracting. So generally it is left implicit, though one should make it clear before one starts using asymptotic notation which limit one is considering.
answered Aug 14 at 9:16
Henning Makholm
228k16293524
Yes, the little-o notation (and Landau symbols in general) behaves differently for $xto 0$ and for $xtoinfty$.
When we're considering $xtoinfty$ (as you may be familiar with from analysis of algorithms), $sqrt x$ grows slower than $x$ -- because for large $x$, the square root of $x$ is smaller than $x$ by a ratio that becomes ever more lopsided. Therefore in this context we say that $sqrt x = o(x)$.
On the other hand, if we're considering $xto 0$ (which is more common in analysis), then when $x$ is close to zero, $sqrt x$ is larger than $x$ by a ratio that tends to infinity. Therefore in that context we say that $x = o(sqrt x)$.
If would be less confusing to make it explicit which limit we're working with, and write something like
$$ sqrt x = mathop o_xtoinfty(x) qquadqquadqquad x = mathop o_xto 0(sqrt x) $$
However, in most practical uses of the notation, the limit is the same throughout the entire calculation, so repeating it for every $o$ would be tedious and distracting. So generally it is left implicit, though one should make it clear before one starts using asymptotic notation which limit one is considering.
answered Aug 14 at 9:16
Henning Makholm
228k16293524
edited Aug 14 at 12:35
answered Aug 14 at 9:16
Henning Makholm
228k16293524
answered Aug 14 at 9:16
Henning Makholm
228k16293524
answered Aug 14 at 9:16
Henning Makholm
228k16293524
228k16293524
add a comment |Â
add a comment |Â
up vote
0
down vote
Your starting point is wrong. $h^2=o(h)$ and not the reverse. $h=o(sqrth)$ and not the reverse.The definition of the Landau notation in the vicinity of a point $x_0$ is $$f=o(g)Leftrightarrowlim_xto x_0fracf(x)g(x)=0$$
answered Jan 31 '15 at 8:03
marwalix
13.1k12338
"$sqrth$ has slower growth than $h$" is for $h to infty$, and not for $h to 0$ as we have here.
– GEdgar
Oct 4 '17 at 23:46
add a comment |Â
up vote
0
down vote
Your starting point is wrong. $h^2=o(h)$ and not the reverse. $h=o(sqrth)$ and not the reverse.The definition of the Landau notation in the vicinity of a point $x_0$ is $$f=o(g)Leftrightarrowlim_xto x_0fracf(x)g(x)=0$$
answered Jan 31 '15 at 8:03
marwalix
13.1k12338
"$sqrth$ has slower growth than $h$" is for $h to infty$, and not for $h to 0$ as we have here.
– GEdgar
Oct 4 '17 at 23:46
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Your starting point is wrong. $h^2=o(h)$ and not the reverse. $h=o(sqrth)$ and not the reverse.The definition of the Landau notation in the vicinity of a point $x_0$ is $$f=o(g)Leftrightarrowlim_xto x_0fracf(x)g(x)=0$$
answered Jan 31 '15 at 8:03
marwalix
13.1k12338
Your starting point is wrong. $h^2=o(h)$ and not the reverse. $h=o(sqrth)$ and not the reverse.The definition of the Landau notation in the vicinity of a point $x_0$ is $$f=o(g)Leftrightarrowlim_xto x_0fracf(x)g(x)=0$$
answered Jan 31 '15 at 8:03
marwalix
13.1k12338
answered Jan 31 '15 at 8:03
marwalix
13.1k12338
answered Jan 31 '15 at 8:03
marwalix
13.1k12338
answered Jan 31 '15 at 8:03
marwalix
13.1k12338
13.1k12338
"$sqrth$ has slower growth than $h$" is for $h to infty$, and not for $h to 0$ as we have here.
– GEdgar
Oct 4 '17 at 23:46
add a comment |Â
"$sqrth$ has slower growth than $h$" is for $h to infty$, and not for $h to 0$ as we have here.
– GEdgar
Oct 4 '17 at 23:46
"$sqrth$ has slower growth than $h$" is for $h to infty$, and not for $h to 0$ as we have here.
– GEdgar
Oct 4 '17 at 23:46
"$sqrth$ has slower growth than $h$" is for $h to infty$, and not for $h to 0$ as we have here.
– GEdgar
Oct 4 '17 at 23:46
add a comment |Â
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1
It is a definition - I don't understand what you mean by explanation? You could interpret it as saying $f(0) = 0$ and $f'(0) = 0$.
– copper.hat
Jan 31 '15 at 6:51
Im talking about little o notation?
– Mauro Augusto
Jan 31 '15 at 7:06
1
I understand. I don't understand what you are asking.
– copper.hat
Jan 31 '15 at 7:09
I just need an explanation of the grey bit. Like how does o(h) work when h->0 instead of going to infinity
– Mauro Augusto
Jan 31 '15 at 7:24
I don't know what you mean, the definition only involves $h to 0$? It is exactly equivalent to $f(0) =0$ and $f'(0) = 0$.
– copper.hat
Jan 31 '15 at 7:28