How Big Must N Be for Stirling's Formula to Achieve 2% Accuracy?

[Physgeek64]

Homework Statement

How big must N be for the simple version of stirlings formula to be accurate to within 2%

Homework Equations

The Attempt at a Solution

So I think the starting point is

$\frac{N lnN-N}{lnN!} =\alpha$ where $\alpha=0.98$

But i have no idea how to solve this expression for N

Many thanks

 

[Mark44]

Homework Statement

How big must N be for the simple version of stirlings formula to be accurate to within 2%

Homework Equations

The Attempt at a Solution

So I think the starting point is

$\frac{N lnN-N}{lnN!} =\alpha$ where $\alpha=0.98$

But i have no idea how to solve this expression for N

You can't solve it algebraically, but you can substitute values of N. With N = 100, $\alpha \approx 0.991$

 

[Charles Link]

One thing that can aid in the computation is $\ln{ N!}=\ln{N}+\ln(N-1)+... +\ln{2}+\ln{1}$, but otherwise, this simply needs to be computed with the computer doing the work.

 

[Ray Vickson]

Homework Statement

How big must N be for the simple version of stirlings formula to be accurate to within 2%

Homework Equations

The Attempt at a Solution

So I think the starting point is

$\frac{N lnN-N}{lnN!} =\alpha$ where $\alpha=0.98$

But i have no idea how to solve this expression for N

Many thanks

What, exactly, do you regard as the "simple version" of Stirling's formula? The simplest version that actually works is
$$N! \sim \sqrt{2 \pi} \, N^{N+ (1/2)} \, e^{-N},$$
and without the factors $\sqrt{2 \pi}$ and $\sqrt{N}$ the formula can never, ever get within 2% of $N!$, no matter how large is $N$.

However, in pp. 52--54 of his beautiful probability book, Feller gives a surprisingly simple derivation of the following:
$$\sqrt{2 \pi} N^{N+ (1/2)} e^{-N} < N! < \sqrt{2 \pi} \, N^{N+ (1/2)} \, e^{-N\:+\: 1/(12N)},$$
and this works for all $N \geq 1$. (For clarity:the exponent is $-N + \frac{1}{12N}$.) For example, even when $N$ is as small as $N=1$ the inequalities read as $0.9221< 1 < 1.002$.

The inequalities should allow you to figure out when $\sqrt{2 \pi}\, N^{N+ (1/2)} \, e^{-N}$ comes within 2% of $N!$.

See, eg., https://www.amazon.com/dp/0471257087/?tag=pfamazon01-20 for more information on Feller's book. I understand that free versions are now available on-line as pdf files.

 

[Physgeek64]

What, exactly, do you regard as the "simple version" of Stirling's formula? The simplest version that actually works is
$$N! \sim \sqrt{2 \pi} \, N^{N+ (1/2)} \, e^{-N},$$
and without the factors $\sqrt{2 \pi}$ and $\sqrt{N}$ the formula can never, ever get within 2% of $N!$, no matter how large is $N$.

However, in pp. 52--54 of his beautiful probability book, Feller gives a surprisingly simple derivation of the following:
$$\sqrt{2 \pi} N^{N+ (1/2)} e^{-N} < N! < \sqrt{2 \pi} \, N^{N+ (1/2)} \, e^{-N\:+\: 1/(12N)},$$
and this works for all $N \geq 1$. (For clarity:the exponent is $-N + \frac{1}{12N}$.) For example, even when $N$ is as small as $N=1$ the inequalities read as $0.9221< 1 < 1.002$.

The inequalities should allow you to figure out when $\sqrt{2 \pi}\, N^{N+ (1/2)} \, e^{-N}$ comes within 2% of $N!$.

See, eg., https://www.amazon.com/dp/0471257087/?tag=pfamazon01-20 for more information on Feller's book. I understand that free versions are now available on-line as pdf files.

I think the 'simplest formula' is meant to be $NlnN-N$ in this case

 

[Ray Vickson]

I think the 'simplest formula' is meant to be $NlnN-N$ in this case

It is simple for sure; it is also wrong. You must beware of assuming that an accurate formula for $\ln (f(n)$ produces an accurate formula for $f(n)$ via exponentiation. It generally does not, and Stirling's formula is a perfect example of that.

 

[Physgeek64]

It is simple for sure; it is also wrong. You must beware of assuming that an accurate formula for $\ln (f(n)$ produces an accurate formula for $f(n)$ via exponentiation. It generally does not, and Stirling's formula is a perfect example of that.

Okay, but how would i go about working out for which N gives an answer to within 2% of the true value? Thanks :)

 

[Ray Vickson]

Okay, but how would i go about working out for which N gives an answer to within 2% of the true value? Thanks :)

Use the two inequalities I wrote in #4.

 

[Physgeek64]

Use the two inequalities I wrote in #4.

Can you solve for N in these, or just plug in values of N?

 

[FactChecker]

You must beware of assuming that an accurate formula for $\ln (f(n)$ produces an accurate formula for $f(n)$ via exponentiation. It generally does not, and Stirling's formula is a perfect example of that.

If the accuracy of ln( f(n) ) is in terms of abs( trueValue - estimatedValue ) and the desired accuracy is in terms of percentage, I think this should be possible.

 

[Ray Vickson]

If the accuracy of ln( f(n) ) is in terms of abs( trueValue - estimatedValue ) and the desired accuracy is in terms of percentage, I think this should be possible.

Unfortunately, not. If $f(n) \sim g(n) \to \infty$ as well as $\ln f(n), \ln g(n) \to \infty$, then we have $\ln f(n) \sim \ln g(n)$. However, we also have $\ln f(n) \sim \ln 2 + \ln g(n)$, and yet we do not have $2 g(n)$ asymptotic to $f(n)$. And that is the problem right there: an additive constant in the logarithm (whose importance goes to zero in the limit) becomes a constant mutliplicative factor in the original function, whose importance never goes away.

 

[Ray Vickson]

Can you solve for N in these, or just plug in values of N?

You can do it either way, but solving directly for N is easier.