Mixing
Compute using big-o
$begingroup$
Assume that $a<0$ and that for $N$ large we have $a_N= frac{exp(Na)}{N} (1+o(1)) + O(frac{1}{N})$.
Can I write that $a_N$ is equal to $frac{exp(Na)}{N} (1+o(1))$ ? since both terms are of order $O(frac{1}{N})$.
My question must be easy but I am new to this notation and still it's very confusing.
asymptotics
asked Jan 14 at 15:52
vl.athvl.ath
699
$endgroup$
add a comment |
$begingroup$
Assume that $a<0$ and that for $N$ large we have $a_N= frac{exp(Na)}{N} (1+o(1)) + O(frac{1}{N})$.
Can I write that $a_N$ is equal to $frac{exp(Na)}{N} (1+o(1))$ ? since both terms are of order $O(frac{1}{N})$.
My question must be easy but I am new to this notation and still it's very confusing.
asymptotics
asked Jan 14 at 15:52
vl.athvl.ath
699
$endgroup$
add a comment |
$begingroup$
Assume that $a<0$ and that for $N$ large we have $a_N= frac{exp(Na)}{N} (1+o(1)) + O(frac{1}{N})$.
Can I write that $a_N$ is equal to $frac{exp(Na)}{N} (1+o(1))$ ? since both terms are of order $O(frac{1}{N})$.
My question must be easy but I am new to this notation and still it's very confusing.
asymptotics
asked Jan 14 at 15:52
vl.athvl.ath
699
$endgroup$
Assume that $a<0$ and that for $N$ large we have $a_N= frac{exp(Na)}{N} (1+o(1)) + O(frac{1}{N})$.
Can I write that $a_N$ is equal to $frac{exp(Na)}{N} (1+o(1))$ ? since both terms are of order $O(frac{1}{N})$.
My question must be easy but I am new to this notation and still it's very confusing.
asymptotics
asymptotics
asked Jan 14 at 15:52
vl.athvl.ath
699
asked Jan 14 at 15:52
vl.athvl.ath
699
asked Jan 14 at 15:52
vl.athvl.ath
699
asked Jan 14 at 15:52
vl.athvl.ath
699
asked Jan 14 at 15:52
vl.athvl.ath
699
699
add a comment |
add a comment |
1 Answer
1
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$begingroup$
According to definition $$a_n=O(b_n)iff limsup_{nto infty} left|{a_nover b_n}right|<infty\a_n=o(b_n)iff lim_{nto infty} {a_nover b_n}=0$$according to Big O notation
Your conclusion is wrong since ${exp(aN)over N}$ is an exponential decreasing term that vanishes much faster that $O({1over N})$ (i.e. $O({1over N})$ becomes dominant for $N$ sufficiently large) because $1+o(1)$ can't compensate ${exp(aN)over N}$ (you can think of $1over sqrt n$ or $1over ln n$ as $o(1)$ and $1over n^2sin{1over n}$ as $O({1over n})$). Then it is suitable to say that $$a_n=O({1over n})$$here is an example:
answered Jan 14 at 16:22
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
$begingroup$
the definition I use for big-o is : $a_N=O(b_N)$ iff there exists $c>0$ such that $|a_N|leqslant c |b_N|$ , for $N$ large enough .. i thought this is a general definition.. I agree with your little-o definition
$endgroup$
– vl.ath
Jan 14 at 16:33
$begingroup$
In fact I should have used $limsup$. I will edit my answer
$endgroup$
– Mostafa Ayaz
Jan 14 at 16:45
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
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votes
$begingroup$
According to definition $$a_n=O(b_n)iff limsup_{nto infty} left|{a_nover b_n}right|<infty\a_n=o(b_n)iff lim_{nto infty} {a_nover b_n}=0$$according to Big O notation
Your conclusion is wrong since ${exp(aN)over N}$ is an exponential decreasing term that vanishes much faster that $O({1over N})$ (i.e. $O({1over N})$ becomes dominant for $N$ sufficiently large) because $1+o(1)$ can't compensate ${exp(aN)over N}$ (you can think of $1over sqrt n$ or $1over ln n$ as $o(1)$ and $1over n^2sin{1over n}$ as $O({1over n})$). Then it is suitable to say that $$a_n=O({1over n})$$here is an example:
answered Jan 14 at 16:22
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
$begingroup$
the definition I use for big-o is : $a_N=O(b_N)$ iff there exists $c>0$ such that $|a_N|leqslant c |b_N|$ , for $N$ large enough .. i thought this is a general definition.. I agree with your little-o definition
$endgroup$
– vl.ath
Jan 14 at 16:33
$begingroup$
In fact I should have used $limsup$. I will edit my answer
$endgroup$
– Mostafa Ayaz
Jan 14 at 16:45
add a comment |
$begingroup$
According to definition $$a_n=O(b_n)iff limsup_{nto infty} left|{a_nover b_n}right|<infty\a_n=o(b_n)iff lim_{nto infty} {a_nover b_n}=0$$according to Big O notation
Your conclusion is wrong since ${exp(aN)over N}$ is an exponential decreasing term that vanishes much faster that $O({1over N})$ (i.e. $O({1over N})$ becomes dominant for $N$ sufficiently large) because $1+o(1)$ can't compensate ${exp(aN)over N}$ (you can think of $1over sqrt n$ or $1over ln n$ as $o(1)$ and $1over n^2sin{1over n}$ as $O({1over n})$). Then it is suitable to say that $$a_n=O({1over n})$$here is an example:
answered Jan 14 at 16:22
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
$begingroup$
the definition I use for big-o is : $a_N=O(b_N)$ iff there exists $c>0$ such that $|a_N|leqslant c |b_N|$ , for $N$ large enough .. i thought this is a general definition.. I agree with your little-o definition
$endgroup$
– vl.ath
Jan 14 at 16:33
$begingroup$
In fact I should have used $limsup$. I will edit my answer
$endgroup$
– Mostafa Ayaz
Jan 14 at 16:45
add a comment |
$begingroup$
According to definition $$a_n=O(b_n)iff limsup_{nto infty} left|{a_nover b_n}right|<infty\a_n=o(b_n)iff lim_{nto infty} {a_nover b_n}=0$$according to Big O notation
Your conclusion is wrong since ${exp(aN)over N}$ is an exponential decreasing term that vanishes much faster that $O({1over N})$ (i.e. $O({1over N})$ becomes dominant for $N$ sufficiently large) because $1+o(1)$ can't compensate ${exp(aN)over N}$ (you can think of $1over sqrt n$ or $1over ln n$ as $o(1)$ and $1over n^2sin{1over n}$ as $O({1over n})$). Then it is suitable to say that $$a_n=O({1over n})$$here is an example:
answered Jan 14 at 16:22
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
According to definition $$a_n=O(b_n)iff limsup_{nto infty} left|{a_nover b_n}right|<infty\a_n=o(b_n)iff lim_{nto infty} {a_nover b_n}=0$$according to Big O notation
Your conclusion is wrong since ${exp(aN)over N}$ is an exponential decreasing term that vanishes much faster that $O({1over N})$ (i.e. $O({1over N})$ becomes dominant for $N$ sufficiently large) because $1+o(1)$ can't compensate ${exp(aN)over N}$ (you can think of $1over sqrt n$ or $1over ln n$ as $o(1)$ and $1over n^2sin{1over n}$ as $O({1over n})$). Then it is suitable to say that $$a_n=O({1over n})$$here is an example:
answered Jan 14 at 16:22
Mostafa AyazMostafa Ayaz
15.6k3939
edited Jan 14 at 16:47
answered Jan 14 at 16:22
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 14 at 16:22
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 14 at 16:22
Mostafa AyazMostafa Ayaz
15.6k3939
15.6k3939
$begingroup$
the definition I use for big-o is : $a_N=O(b_N)$ iff there exists $c>0$ such that $|a_N|leqslant c |b_N|$ , for $N$ large enough .. i thought this is a general definition.. I agree with your little-o definition
$endgroup$
– vl.ath
Jan 14 at 16:33
$begingroup$
In fact I should have used $limsup$. I will edit my answer
$endgroup$
– Mostafa Ayaz
Jan 14 at 16:45
add a comment |
$begingroup$
the definition I use for big-o is : $a_N=O(b_N)$ iff there exists $c>0$ such that $|a_N|leqslant c |b_N|$ , for $N$ large enough .. i thought this is a general definition.. I agree with your little-o definition
$endgroup$
– vl.ath
Jan 14 at 16:33
$begingroup$
In fact I should have used $limsup$. I will edit my answer
$endgroup$
– Mostafa Ayaz
Jan 14 at 16:45
$begingroup$
the definition I use for big-o is : $a_N=O(b_N)$ iff there exists $c>0$ such that $|a_N|leqslant c |b_N|$ , for $N$ large enough .. i thought this is a general definition.. I agree with your little-o definition
$endgroup$
– vl.ath
Jan 14 at 16:33
$begingroup$
the definition I use for big-o is : $a_N=O(b_N)$ iff there exists $c>0$ such that $|a_N|leqslant c |b_N|$ , for $N$ large enough .. i thought this is a general definition.. I agree with your little-o definition
$endgroup$
– vl.ath
Jan 14 at 16:33
$begingroup$
In fact I should have used $limsup$. I will edit my answer
$endgroup$
– Mostafa Ayaz
Jan 14 at 16:45
$begingroup$
In fact I should have used $limsup$. I will edit my answer
$endgroup$
– Mostafa Ayaz
Jan 14 at 16:45
add a comment |
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