Approximation $ln frac(h+f)!(h-g)!$ using Stirling’s when $f+g=o(h)$How best to explain the $sqrt2pi n$ term in Stirling's?Approximation of integral using series expansion of the integrand.Approximating $sqrt101$ using Taylor series methodsUse stirlings approximation to prove inequality.best approximation of $sqrt2$Factorial of a large number and Stirling approximationMeaure-theoretic induction: Why dyadic approximation?Sum ApproximationUsing the normal approximation to a binomial distributionDeriving the fact that the approximation $log(n!) approx nlog(n) - n + frac12log(2pi n)$ is $O(1/n)$.
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Approximation $ln frac(h+f)!(h-g)!$ using Stirling’s when $f+g=o(h)$
How best to explain the $sqrt2pi n$ term in Stirling's?Approximation of integral using series expansion of the integrand.Approximating $sqrt101$ using Taylor series methodsUse stirlings approximation to prove inequality.best approximation of $sqrt2$Factorial of a large number and Stirling approximationMeaure-theoretic induction: Why dyadic approximation?Sum ApproximationUsing the normal approximation to a binomial distributionDeriving the fact that the approximation $log(n!) approx nlog(n) - n + frac12log(2pi n)$ is $O(1/n)$.
$begingroup$
Stirling’s approximation can be extended to a very well known inequality -
$$sqrt2pi nleft(fracneright)^n leq n! leq esqrtnleft(fracneright)^n$$
How can we use this to prove,
$$ln frac(h+f)!(h-g)! = (f+g)ln h pm Oleft(frac(f+g)^2hright) $$
when $f+g=o(h)$
My Attempt
$$
beginalign
ln frac(h+f)!(h-g)! &= ln fracleft(h+fright)^h+f+1/2left(h-gright)^h-g+1/2 + ln frace^h-ge^h+f\ &= (h+f+1/2)ln(h+f) - (h-g+1/2)ln(h-g) - (f+g)
endalign
$$
At this point I can use property of logarithms $ln(h+f)=ln h+ln(1+f/h)$ and $ln(h-g)=ln h+ln(1-g/h)$ to extract $ln h$ terms. But that would leave a lot of other terms with itself which I'm not able to collate.
How should I move ahead from hear? If this is not the right approach then I hope to be pointed towards right direction.
approximation factorial
asked Mar 22 at 10:38
PrateekPrateek
1377
$endgroup$
add a comment |
$begingroup$
Stirling’s approximation can be extended to a very well known inequality -
$$sqrt2pi nleft(fracneright)^n leq n! leq esqrtnleft(fracneright)^n$$
How can we use this to prove,
$$ln frac(h+f)!(h-g)! = (f+g)ln h pm Oleft(frac(f+g)^2hright) $$
when $f+g=o(h)$
My Attempt
$$
beginalign
ln frac(h+f)!(h-g)! &= ln fracleft(h+fright)^h+f+1/2left(h-gright)^h-g+1/2 + ln frace^h-ge^h+f\ &= (h+f+1/2)ln(h+f) - (h-g+1/2)ln(h-g) - (f+g)
endalign
$$
At this point I can use property of logarithms $ln(h+f)=ln h+ln(1+f/h)$ and $ln(h-g)=ln h+ln(1-g/h)$ to extract $ln h$ terms. But that would leave a lot of other terms with itself which I'm not able to collate.
How should I move ahead from hear? If this is not the right approach then I hope to be pointed towards right direction.
approximation factorial
asked Mar 22 at 10:38
PrateekPrateek
1377
$endgroup$
$begingroup$
I never saw the very well known inequality :-)
$endgroup$
– Yves Daoust
Mar 22 at 10:41
$begingroup$
I'm sorry, I should not have presumed too much. But the inequality is there on wiki, and quite straight forward to prove as well.
$endgroup$
– Prateek
Mar 22 at 11:05
add a comment |
$begingroup$
Stirling’s approximation can be extended to a very well known inequality -
$$sqrt2pi nleft(fracneright)^n leq n! leq esqrtnleft(fracneright)^n$$
How can we use this to prove,
$$ln frac(h+f)!(h-g)! = (f+g)ln h pm Oleft(frac(f+g)^2hright) $$
when $f+g=o(h)$
My Attempt
$$
beginalign
ln frac(h+f)!(h-g)! &= ln fracleft(h+fright)^h+f+1/2left(h-gright)^h-g+1/2 + ln frace^h-ge^h+f\ &= (h+f+1/2)ln(h+f) - (h-g+1/2)ln(h-g) - (f+g)
endalign
$$
At this point I can use property of logarithms $ln(h+f)=ln h+ln(1+f/h)$ and $ln(h-g)=ln h+ln(1-g/h)$ to extract $ln h$ terms. But that would leave a lot of other terms with itself which I'm not able to collate.
How should I move ahead from hear? If this is not the right approach then I hope to be pointed towards right direction.
approximation factorial
asked Mar 22 at 10:38
PrateekPrateek
1377
$endgroup$
Stirling’s approximation can be extended to a very well known inequality -
$$sqrt2pi nleft(fracneright)^n leq n! leq esqrtnleft(fracneright)^n$$
How can we use this to prove,
$$ln frac(h+f)!(h-g)! = (f+g)ln h pm Oleft(frac(f+g)^2hright) $$
when $f+g=o(h)$
My Attempt
$$
beginalign
ln frac(h+f)!(h-g)! &= ln fracleft(h+fright)^h+f+1/2left(h-gright)^h-g+1/2 + ln frace^h-ge^h+f\ &= (h+f+1/2)ln(h+f) - (h-g+1/2)ln(h-g) - (f+g)
endalign
$$
At this point I can use property of logarithms $ln(h+f)=ln h+ln(1+f/h)$ and $ln(h-g)=ln h+ln(1-g/h)$ to extract $ln h$ terms. But that would leave a lot of other terms with itself which I'm not able to collate.
How should I move ahead from hear? If this is not the right approach then I hope to be pointed towards right direction.
approximation factorial
approximation factorial
asked Mar 22 at 10:38
PrateekPrateek
1377
asked Mar 22 at 10:38
PrateekPrateek
1377
asked Mar 22 at 10:38
PrateekPrateek
1377
asked Mar 22 at 10:38
PrateekPrateek
1377
asked Mar 22 at 10:38
PrateekPrateek
1377
1377
$begingroup$
I never saw the very well known inequality :-)
$endgroup$
– Yves Daoust
Mar 22 at 10:41
$begingroup$
I'm sorry, I should not have presumed too much. But the inequality is there on wiki, and quite straight forward to prove as well.
$endgroup$
– Prateek
Mar 22 at 11:05
add a comment |
$begingroup$
I never saw the very well known inequality :-)
$endgroup$
– Yves Daoust
Mar 22 at 10:41
$begingroup$
I'm sorry, I should not have presumed too much. But the inequality is there on wiki, and quite straight forward to prove as well.
$endgroup$
– Prateek
Mar 22 at 11:05
$begingroup$
I never saw the very well known inequality :-)
$endgroup$
– Yves Daoust
Mar 22 at 10:41
$begingroup$
I never saw the very well known inequality :-)
$endgroup$
– Yves Daoust
Mar 22 at 10:41
$begingroup$
I'm sorry, I should not have presumed too much. But the inequality is there on wiki, and quite straight forward to prove as well.
$endgroup$
– Prateek
Mar 22 at 11:05
$begingroup$
I'm sorry, I should not have presumed too much. But the inequality is there on wiki, and quite straight forward to prove as well.
$endgroup$
– Prateek
Mar 22 at 11:05
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
To continue, you can use the Taylor expansion of logarithm, just the beginning:
$$ ln(1+x) = x + mathcalO(x^2)$$
That gives you
$$ (h+f+1/2)ln(1+f/h) = f + mathcalOleft(fracf^2hright) $$
and $$-(h-g+1/2)ln(1+g/h) = g + mathcalOleft(fracg^2hright) $$
However, note that Stirling's approxiamtion won't give you the exact equality as you're writing it. What you can get from them is
$$ ln (n!) = (n+1/2)ln n - n + xi(n)$$
where $xi(n) in [lnsqrt2pi,1]$. This last term is negligible compared to the leading term, but it doesn't vanish. In the end, the best you can get this with this method is
$$ lnfrac(h+f)!(h-g)! = (f+g)ln h + xi + mathcalOleft(fracf^2h,fracg^2hright) $$
where $xi = xi(h+f)-xi(h-g)$, $|xi| le 1-lnsqrt2pi$
answered Mar 22 at 12:54
Adam LatosińskiAdam Latosiński
4887
$endgroup$
add a comment |
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1 Answer
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$begingroup$
To continue, you can use the Taylor expansion of logarithm, just the beginning:
$$ ln(1+x) = x + mathcalO(x^2)$$
That gives you
$$ (h+f+1/2)ln(1+f/h) = f + mathcalOleft(fracf^2hright) $$
and $$-(h-g+1/2)ln(1+g/h) = g + mathcalOleft(fracg^2hright) $$
However, note that Stirling's approxiamtion won't give you the exact equality as you're writing it. What you can get from them is
$$ ln (n!) = (n+1/2)ln n - n + xi(n)$$
where $xi(n) in [lnsqrt2pi,1]$. This last term is negligible compared to the leading term, but it doesn't vanish. In the end, the best you can get this with this method is
$$ lnfrac(h+f)!(h-g)! = (f+g)ln h + xi + mathcalOleft(fracf^2h,fracg^2hright) $$
where $xi = xi(h+f)-xi(h-g)$, $|xi| le 1-lnsqrt2pi$
answered Mar 22 at 12:54
Adam LatosińskiAdam Latosiński
4887
$endgroup$
add a comment |
$begingroup$
To continue, you can use the Taylor expansion of logarithm, just the beginning:
$$ ln(1+x) = x + mathcalO(x^2)$$
That gives you
$$ (h+f+1/2)ln(1+f/h) = f + mathcalOleft(fracf^2hright) $$
and $$-(h-g+1/2)ln(1+g/h) = g + mathcalOleft(fracg^2hright) $$
However, note that Stirling's approxiamtion won't give you the exact equality as you're writing it. What you can get from them is
$$ ln (n!) = (n+1/2)ln n - n + xi(n)$$
where $xi(n) in [lnsqrt2pi,1]$. This last term is negligible compared to the leading term, but it doesn't vanish. In the end, the best you can get this with this method is
$$ lnfrac(h+f)!(h-g)! = (f+g)ln h + xi + mathcalOleft(fracf^2h,fracg^2hright) $$
where $xi = xi(h+f)-xi(h-g)$, $|xi| le 1-lnsqrt2pi$
answered Mar 22 at 12:54
Adam LatosińskiAdam Latosiński
4887
$endgroup$
add a comment |
$begingroup$
To continue, you can use the Taylor expansion of logarithm, just the beginning:
$$ ln(1+x) = x + mathcalO(x^2)$$
That gives you
$$ (h+f+1/2)ln(1+f/h) = f + mathcalOleft(fracf^2hright) $$
and $$-(h-g+1/2)ln(1+g/h) = g + mathcalOleft(fracg^2hright) $$
However, note that Stirling's approxiamtion won't give you the exact equality as you're writing it. What you can get from them is
$$ ln (n!) = (n+1/2)ln n - n + xi(n)$$
where $xi(n) in [lnsqrt2pi,1]$. This last term is negligible compared to the leading term, but it doesn't vanish. In the end, the best you can get this with this method is
$$ lnfrac(h+f)!(h-g)! = (f+g)ln h + xi + mathcalOleft(fracf^2h,fracg^2hright) $$
where $xi = xi(h+f)-xi(h-g)$, $|xi| le 1-lnsqrt2pi$
answered Mar 22 at 12:54
Adam LatosińskiAdam Latosiński
4887
$endgroup$
To continue, you can use the Taylor expansion of logarithm, just the beginning:
$$ ln(1+x) = x + mathcalO(x^2)$$
That gives you
$$ (h+f+1/2)ln(1+f/h) = f + mathcalOleft(fracf^2hright) $$
and $$-(h-g+1/2)ln(1+g/h) = g + mathcalOleft(fracg^2hright) $$
However, note that Stirling's approxiamtion won't give you the exact equality as you're writing it. What you can get from them is
$$ ln (n!) = (n+1/2)ln n - n + xi(n)$$
where $xi(n) in [lnsqrt2pi,1]$. This last term is negligible compared to the leading term, but it doesn't vanish. In the end, the best you can get this with this method is
$$ lnfrac(h+f)!(h-g)! = (f+g)ln h + xi + mathcalOleft(fracf^2h,fracg^2hright) $$
where $xi = xi(h+f)-xi(h-g)$, $|xi| le 1-lnsqrt2pi$
answered Mar 22 at 12:54
Adam LatosińskiAdam Latosiński
4887
answered Mar 22 at 12:54
Adam LatosińskiAdam Latosiński
4887
answered Mar 22 at 12:54
Adam LatosińskiAdam Latosiński
4887
answered Mar 22 at 12:54
Adam LatosińskiAdam Latosiński
4887
4887
add a comment |
add a comment |
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$begingroup$
I never saw the very well known inequality :-)
$endgroup$
– Yves Daoust
Mar 22 at 10:41
$begingroup$
I'm sorry, I should not have presumed too much. But the inequality is there on wiki, and quite straight forward to prove as well.
$endgroup$
– Prateek
Mar 22 at 11:05