Mixing
Logarithmic asymptotic formula using o(1)
$begingroup$
Let's say that we have $limlimits_{nrightarrowinfty} frac{1}{n}log a_n , = , a$.
My professor said that this is equivalent to $a_n sim exp(na)$ for $n$ large. Intuitively it makes sense but I cannot really do the algebra to get the result. I must show that $a_n = exp(na)(1+o(1))$ for $n$ large, right ? Could someone help me with this ?
What I can do so far is to say that the limit above is equivalent to $frac{1}{n}log a_n , - , a = o(1)$ for $n$ large, which then gives $a_n= exp(n(a+o(1))$ for $n$ large. How is my result related to the one my prof said ? is it weaker or is it the same ?
asymptotics
asked Jan 13 at 11:22
vl.athvl.ath
649
$endgroup$
add a comment |
$begingroup$
Let's say that we have $limlimits_{nrightarrowinfty} frac{1}{n}log a_n , = , a$.
My professor said that this is equivalent to $a_n sim exp(na)$ for $n$ large. Intuitively it makes sense but I cannot really do the algebra to get the result. I must show that $a_n = exp(na)(1+o(1))$ for $n$ large, right ? Could someone help me with this ?
What I can do so far is to say that the limit above is equivalent to $frac{1}{n}log a_n , - , a = o(1)$ for $n$ large, which then gives $a_n= exp(n(a+o(1))$ for $n$ large. How is my result related to the one my prof said ? is it weaker or is it the same ?
asymptotics
asked Jan 13 at 11:22
vl.athvl.ath
649
$endgroup$
3
$begingroup$
I believe this is what your prof meant. By saying $a_nsim exp (na)$ he was saying that $a_n = exp(na + no(1))$ and not that $a_n=exp(na)(1+o(1))$. The latter is a stronger assertion and it's false. If you take $a_n= e^{na+sqrt{n}}$ then $frac{1}{n}log a_nrightarrow a$ but it's not satisfying that $a_n=exp(na)(1+o(1))$.
$endgroup$
– Yanko
Jan 13 at 11:30
add a comment |
$begingroup$
Let's say that we have $limlimits_{nrightarrowinfty} frac{1}{n}log a_n , = , a$.
My professor said that this is equivalent to $a_n sim exp(na)$ for $n$ large. Intuitively it makes sense but I cannot really do the algebra to get the result. I must show that $a_n = exp(na)(1+o(1))$ for $n$ large, right ? Could someone help me with this ?
What I can do so far is to say that the limit above is equivalent to $frac{1}{n}log a_n , - , a = o(1)$ for $n$ large, which then gives $a_n= exp(n(a+o(1))$ for $n$ large. How is my result related to the one my prof said ? is it weaker or is it the same ?
asymptotics
asked Jan 13 at 11:22
vl.athvl.ath
649
$endgroup$
Let's say that we have $limlimits_{nrightarrowinfty} frac{1}{n}log a_n , = , a$.
My professor said that this is equivalent to $a_n sim exp(na)$ for $n$ large. Intuitively it makes sense but I cannot really do the algebra to get the result. I must show that $a_n = exp(na)(1+o(1))$ for $n$ large, right ? Could someone help me with this ?
What I can do so far is to say that the limit above is equivalent to $frac{1}{n}log a_n , - , a = o(1)$ for $n$ large, which then gives $a_n= exp(n(a+o(1))$ for $n$ large. How is my result related to the one my prof said ? is it weaker or is it the same ?
asymptotics
asymptotics
asked Jan 13 at 11:22
vl.athvl.ath
649
asked Jan 13 at 11:22
vl.athvl.ath
649
asked Jan 13 at 11:22
vl.athvl.ath
649
asked Jan 13 at 11:22
vl.athvl.ath
649
asked Jan 13 at 11:22
vl.athvl.ath
649
649
3
$begingroup$
I believe this is what your prof meant. By saying $a_nsim exp (na)$ he was saying that $a_n = exp(na + no(1))$ and not that $a_n=exp(na)(1+o(1))$. The latter is a stronger assertion and it's false. If you take $a_n= e^{na+sqrt{n}}$ then $frac{1}{n}log a_nrightarrow a$ but it's not satisfying that $a_n=exp(na)(1+o(1))$.
$endgroup$
– Yanko
Jan 13 at 11:30
add a comment |
3
$begingroup$
I believe this is what your prof meant. By saying $a_nsim exp (na)$ he was saying that $a_n = exp(na + no(1))$ and not that $a_n=exp(na)(1+o(1))$. The latter is a stronger assertion and it's false. If you take $a_n= e^{na+sqrt{n}}$ then $frac{1}{n}log a_nrightarrow a$ but it's not satisfying that $a_n=exp(na)(1+o(1))$.
$endgroup$
– Yanko
Jan 13 at 11:30
3
3
$begingroup$
I believe this is what your prof meant. By saying $a_nsim exp (na)$ he was saying that $a_n = exp(na + no(1))$ and not that $a_n=exp(na)(1+o(1))$. The latter is a stronger assertion and it's false. If you take $a_n= e^{na+sqrt{n}}$ then $frac{1}{n}log a_nrightarrow a$ but it's not satisfying that $a_n=exp(na)(1+o(1))$.
$endgroup$
– Yanko
Jan 13 at 11:30
$begingroup$
I believe this is what your prof meant. By saying $a_nsim exp (na)$ he was saying that $a_n = exp(na + no(1))$ and not that $a_n=exp(na)(1+o(1))$. The latter is a stronger assertion and it's false. If you take $a_n= e^{na+sqrt{n}}$ then $frac{1}{n}log a_nrightarrow a$ but it's not satisfying that $a_n=exp(na)(1+o(1))$.
$endgroup$
– Yanko
Jan 13 at 11:30
add a comment |
1 Answer
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Considering that $a_n sim exp(na) iff lim_{nto infty} frac{a_n }{exp(na)}=1 $, the assertion is wrong:
$$ frac{a_n}{exp(na)}=expleft(n left(frac{log a_n}{n}-aright)right)=e^{ng(n)}$$
with $g(n)=frac{log a_n}{n}-a$. You are told that $frac{1}{n}log a_n , to , a implies g(n)to 0$, but you need more than that, you need $g(n)=o(1/n)$.
answered Jan 13 at 18:38
leonbloyleonbloy
41.1k645107
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$begingroup$
Considering that $a_n sim exp(na) iff lim_{nto infty} frac{a_n }{exp(na)}=1 $, the assertion is wrong:
$$ frac{a_n}{exp(na)}=expleft(n left(frac{log a_n}{n}-aright)right)=e^{ng(n)}$$
with $g(n)=frac{log a_n}{n}-a$. You are told that $frac{1}{n}log a_n , to , a implies g(n)to 0$, but you need more than that, you need $g(n)=o(1/n)$.
answered Jan 13 at 18:38
leonbloyleonbloy
41.1k645107
$endgroup$
add a comment |
$begingroup$
Considering that $a_n sim exp(na) iff lim_{nto infty} frac{a_n }{exp(na)}=1 $, the assertion is wrong:
$$ frac{a_n}{exp(na)}=expleft(n left(frac{log a_n}{n}-aright)right)=e^{ng(n)}$$
with $g(n)=frac{log a_n}{n}-a$. You are told that $frac{1}{n}log a_n , to , a implies g(n)to 0$, but you need more than that, you need $g(n)=o(1/n)$.
answered Jan 13 at 18:38
leonbloyleonbloy
41.1k645107
$endgroup$
add a comment |
$begingroup$
Considering that $a_n sim exp(na) iff lim_{nto infty} frac{a_n }{exp(na)}=1 $, the assertion is wrong:
$$ frac{a_n}{exp(na)}=expleft(n left(frac{log a_n}{n}-aright)right)=e^{ng(n)}$$
with $g(n)=frac{log a_n}{n}-a$. You are told that $frac{1}{n}log a_n , to , a implies g(n)to 0$, but you need more than that, you need $g(n)=o(1/n)$.
answered Jan 13 at 18:38
leonbloyleonbloy
41.1k645107
$endgroup$
Considering that $a_n sim exp(na) iff lim_{nto infty} frac{a_n }{exp(na)}=1 $, the assertion is wrong:
$$ frac{a_n}{exp(na)}=expleft(n left(frac{log a_n}{n}-aright)right)=e^{ng(n)}$$
with $g(n)=frac{log a_n}{n}-a$. You are told that $frac{1}{n}log a_n , to , a implies g(n)to 0$, but you need more than that, you need $g(n)=o(1/n)$.
answered Jan 13 at 18:38
leonbloyleonbloy
41.1k645107
edited Jan 13 at 21:42
answered Jan 13 at 18:38
leonbloyleonbloy
41.1k645107
answered Jan 13 at 18:38
leonbloyleonbloy
41.1k645107
answered Jan 13 at 18:38
leonbloyleonbloy
41.1k645107
41.1k645107
add a comment |
add a comment |
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I believe this is what your prof meant. By saying $a_nsim exp (na)$ he was saying that $a_n = exp(na + no(1))$ and not that $a_n=exp(na)(1+o(1))$. The latter is a stronger assertion and it's false. If you take $a_n= e^{na+sqrt{n}}$ then $frac{1}{n}log a_nrightarrow a$ but it's not satisfying that $a_n=exp(na)(1+o(1))$.
$endgroup$
– Yanko
Jan 13 at 11:30