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why $ln(n!)=nln(n)-n+frac12ln(2pi n)+o(1)$?
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I know that $$n!sim_infty left(fracneright)^nsqrt2pi n,$$
and thus $$ln(n!)sim_infty lnleftleft(fracneright)^nsqrt2pi nright.tag*$$
But why does it implies
$$ln(n!)=nln(n)-n+frac12ln(2pi n)+o(1) ?$$
For me $(*)$ gives us $$ln(n!)=nln(n)-n+frac12ln(2pi n)+oleft(lnleftleft(fracneright)^nsqrt2pi nrightright),$$
and since $$lim_nto infty lnleftleft(fracneright)^nsqrt2pi nrightneq 0,$$
why do we have this $o(1)$ at the end ?
asymptotics factorial
asked Jul 22 at 11:56
Peter
348112
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up vote
0
down vote
favorite
I know that $$n!sim_infty left(fracneright)^nsqrt2pi n,$$
and thus $$ln(n!)sim_infty lnleftleft(fracneright)^nsqrt2pi nright.tag*$$
But why does it implies
$$ln(n!)=nln(n)-n+frac12ln(2pi n)+o(1) ?$$
For me $(*)$ gives us $$ln(n!)=nln(n)-n+frac12ln(2pi n)+oleft(lnleftleft(fracneright)^nsqrt2pi nrightright),$$
and since $$lim_nto infty lnleftleft(fracneright)^nsqrt2pi nrightneq 0,$$
why do we have this $o(1)$ at the end ?
asymptotics factorial
asked Jul 22 at 11:56
Peter
348112
Forget $(*)$
– Lord Shark the Unknown
Jul 22 at 12:00
@LordSharktheUnknown: what do you mean ?
– Peter
Jul 22 at 12:01
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I know that $$n!sim_infty left(fracneright)^nsqrt2pi n,$$
and thus $$ln(n!)sim_infty lnleftleft(fracneright)^nsqrt2pi nright.tag*$$
But why does it implies
$$ln(n!)=nln(n)-n+frac12ln(2pi n)+o(1) ?$$
For me $(*)$ gives us $$ln(n!)=nln(n)-n+frac12ln(2pi n)+oleft(lnleftleft(fracneright)^nsqrt2pi nrightright),$$
and since $$lim_nto infty lnleftleft(fracneright)^nsqrt2pi nrightneq 0,$$
why do we have this $o(1)$ at the end ?
asymptotics factorial
asked Jul 22 at 11:56
Peter
348112
I know that $$n!sim_infty left(fracneright)^nsqrt2pi n,$$
and thus $$ln(n!)sim_infty lnleftleft(fracneright)^nsqrt2pi nright.tag*$$
But why does it implies
$$ln(n!)=nln(n)-n+frac12ln(2pi n)+o(1) ?$$
For me $(*)$ gives us $$ln(n!)=nln(n)-n+frac12ln(2pi n)+oleft(lnleftleft(fracneright)^nsqrt2pi nrightright),$$
and since $$lim_nto infty lnleftleft(fracneright)^nsqrt2pi nrightneq 0,$$
why do we have this $o(1)$ at the end ?
asymptotics factorial
asked Jul 22 at 11:56
Peter
348112
asked Jul 22 at 11:56
Peter
348112
asked Jul 22 at 11:56
Peter
348112
asked Jul 22 at 11:56
Peter
348112
348112
Forget $(*)$
– Lord Shark the Unknown
Jul 22 at 12:00
@LordSharktheUnknown: what do you mean ?
– Peter
Jul 22 at 12:01
add a comment |Â
Forget $(*)$
– Lord Shark the Unknown
Jul 22 at 12:00
@LordSharktheUnknown: what do you mean ?
– Peter
Jul 22 at 12:01
Forget $(*)$
– Lord Shark the Unknown
Jul 22 at 12:00
Forget $(*)$
– Lord Shark the Unknown
Jul 22 at 12:00
@LordSharktheUnknown: what do you mean ?
– Peter
Jul 22 at 12:01
@LordSharktheUnknown: what do you mean ?
– Peter
Jul 22 at 12:01
add a comment |Â
3 Answers
3
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3
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Your calculus is incorrect. The asymptotic equality $f(n)sim g(n)$ is equivalent to $f(n)= g(n)cdot (1+o(1))$.
Taking logarithm yields the equivalent condition $log f(n) = log g(n) + log (1+o(1))$. To finish the proof, note that $log (1+o(1))= o(1)$. So to sum up: $log f(n) = log g(n) + o(1)$.
answered Jul 22 at 12:04
A. Pongrácz
2,159221
Nice answer +1 :)
– Peter
Jul 22 at 12:09
add a comment |Â
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0
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Taking the logarithm on both sides of $n!sim_infty left(fracneright)^nsqrt2pi n$ and keeping the $sim_infty$ is an awful waste of information.
$sim_infty$ means that the ratio between the expressions on either side goes to $1$ as $ntoinfty$. That means that the difference between the logarithm of the two terms goes to $0$, which is much stronger (at least since the terms both go to $infty$). That's exactly $o(1)$.
answered Jul 22 at 12:06
Arthur
98.5k793174
add a comment |Â
up vote
0
down vote
The original asymptote is stronger than you think it is; $sim$ means the expressions on either side have a ratio of the form $1+o(1)$. In fact, one can prove $n!$ is exactly the usual approximation for it multiplied by a Taylor series in $1/n$; the result is called the Stirling series.
answered Jul 22 at 12:06
J.G.
13.2k11424
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Your calculus is incorrect. The asymptotic equality $f(n)sim g(n)$ is equivalent to $f(n)= g(n)cdot (1+o(1))$.
Taking logarithm yields the equivalent condition $log f(n) = log g(n) + log (1+o(1))$. To finish the proof, note that $log (1+o(1))= o(1)$. So to sum up: $log f(n) = log g(n) + o(1)$.
answered Jul 22 at 12:04
A. Pongrácz
2,159221
Nice answer +1 :)
– Peter
Jul 22 at 12:09
add a comment |Â
up vote
3
down vote
accepted
Your calculus is incorrect. The asymptotic equality $f(n)sim g(n)$ is equivalent to $f(n)= g(n)cdot (1+o(1))$.
Taking logarithm yields the equivalent condition $log f(n) = log g(n) + log (1+o(1))$. To finish the proof, note that $log (1+o(1))= o(1)$. So to sum up: $log f(n) = log g(n) + o(1)$.
answered Jul 22 at 12:04
A. Pongrácz
2,159221
Nice answer +1 :)
– Peter
Jul 22 at 12:09
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Your calculus is incorrect. The asymptotic equality $f(n)sim g(n)$ is equivalent to $f(n)= g(n)cdot (1+o(1))$.
Taking logarithm yields the equivalent condition $log f(n) = log g(n) + log (1+o(1))$. To finish the proof, note that $log (1+o(1))= o(1)$. So to sum up: $log f(n) = log g(n) + o(1)$.
answered Jul 22 at 12:04
A. Pongrácz
2,159221
Your calculus is incorrect. The asymptotic equality $f(n)sim g(n)$ is equivalent to $f(n)= g(n)cdot (1+o(1))$.
Taking logarithm yields the equivalent condition $log f(n) = log g(n) + log (1+o(1))$. To finish the proof, note that $log (1+o(1))= o(1)$. So to sum up: $log f(n) = log g(n) + o(1)$.
answered Jul 22 at 12:04
A. Pongrácz
2,159221
answered Jul 22 at 12:04
A. Pongrácz
2,159221
answered Jul 22 at 12:04
A. Pongrácz
2,159221
answered Jul 22 at 12:04
A. Pongrácz
2,159221
2,159221
Nice answer +1 :)
– Peter
Jul 22 at 12:09
add a comment |Â
Nice answer +1 :)
– Peter
Jul 22 at 12:09
Nice answer +1 :)
– Peter
Jul 22 at 12:09
Nice answer +1 :)
– Peter
Jul 22 at 12:09
add a comment |Â
up vote
0
down vote
Taking the logarithm on both sides of $n!sim_infty left(fracneright)^nsqrt2pi n$ and keeping the $sim_infty$ is an awful waste of information.
$sim_infty$ means that the ratio between the expressions on either side goes to $1$ as $ntoinfty$. That means that the difference between the logarithm of the two terms goes to $0$, which is much stronger (at least since the terms both go to $infty$). That's exactly $o(1)$.
answered Jul 22 at 12:06
Arthur
98.5k793174
add a comment |Â
up vote
0
down vote
Taking the logarithm on both sides of $n!sim_infty left(fracneright)^nsqrt2pi n$ and keeping the $sim_infty$ is an awful waste of information.
$sim_infty$ means that the ratio between the expressions on either side goes to $1$ as $ntoinfty$. That means that the difference between the logarithm of the two terms goes to $0$, which is much stronger (at least since the terms both go to $infty$). That's exactly $o(1)$.
answered Jul 22 at 12:06
Arthur
98.5k793174
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Taking the logarithm on both sides of $n!sim_infty left(fracneright)^nsqrt2pi n$ and keeping the $sim_infty$ is an awful waste of information.
$sim_infty$ means that the ratio between the expressions on either side goes to $1$ as $ntoinfty$. That means that the difference between the logarithm of the two terms goes to $0$, which is much stronger (at least since the terms both go to $infty$). That's exactly $o(1)$.
answered Jul 22 at 12:06
Arthur
98.5k793174
Taking the logarithm on both sides of $n!sim_infty left(fracneright)^nsqrt2pi n$ and keeping the $sim_infty$ is an awful waste of information.
$sim_infty$ means that the ratio between the expressions on either side goes to $1$ as $ntoinfty$. That means that the difference between the logarithm of the two terms goes to $0$, which is much stronger (at least since the terms both go to $infty$). That's exactly $o(1)$.
answered Jul 22 at 12:06
Arthur
98.5k793174
answered Jul 22 at 12:06
Arthur
98.5k793174
answered Jul 22 at 12:06
Arthur
98.5k793174
answered Jul 22 at 12:06
Arthur
98.5k793174
98.5k793174
add a comment |Â
add a comment |Â
up vote
0
down vote
The original asymptote is stronger than you think it is; $sim$ means the expressions on either side have a ratio of the form $1+o(1)$. In fact, one can prove $n!$ is exactly the usual approximation for it multiplied by a Taylor series in $1/n$; the result is called the Stirling series.
answered Jul 22 at 12:06
J.G.
13.2k11424
add a comment |Â
up vote
0
down vote
The original asymptote is stronger than you think it is; $sim$ means the expressions on either side have a ratio of the form $1+o(1)$. In fact, one can prove $n!$ is exactly the usual approximation for it multiplied by a Taylor series in $1/n$; the result is called the Stirling series.
answered Jul 22 at 12:06
J.G.
13.2k11424
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The original asymptote is stronger than you think it is; $sim$ means the expressions on either side have a ratio of the form $1+o(1)$. In fact, one can prove $n!$ is exactly the usual approximation for it multiplied by a Taylor series in $1/n$; the result is called the Stirling series.
answered Jul 22 at 12:06
J.G.
13.2k11424
The original asymptote is stronger than you think it is; $sim$ means the expressions on either side have a ratio of the form $1+o(1)$. In fact, one can prove $n!$ is exactly the usual approximation for it multiplied by a Taylor series in $1/n$; the result is called the Stirling series.
answered Jul 22 at 12:06
J.G.
13.2k11424
answered Jul 22 at 12:06
J.G.
13.2k11424
answered Jul 22 at 12:06
J.G.
13.2k11424
answered Jul 22 at 12:06
J.G.
13.2k11424
13.2k11424
add a comment |Â
add a comment |Â
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Forget $(*)$
– Lord Shark the Unknown
Jul 22 at 12:00
@LordSharktheUnknown: what do you mean ?
– Peter
Jul 22 at 12:01