91Ó°ÊÓ

Stirling's approximation:

The approximate value for a factorial function (n!) is obtained using the Stirling formula, also known as Stirling's approximation formula. Gamma function can also be performed with this. Applied mathematics also makes use of Stirling's formula. It simplifies calculating the factorial of higher values.

In this problem, we are to consider the ratio: the number of ways N particles can be arranged in such a way that both sections of a two-sided room have$\frac{N}{2}$ particles each $\left(W_1\right)$to the number of ways N particles can be arranged in such a way that the right side has of the total particles and the left has the remaining .$40\%\left(W_2\right)$

a)

Calling$W_1$ the number of ways N particles can be arranged such that both region contain half of the total number of particles is given by the binomial coefficient:

$W_1=\frac{N!}{(0.5N)!(N-0.5N)!}=\frac{N!}{(0.5N){!}^2}$

Contrastingly, calling $W_2$the number of ways$N$ particles can be arranged such that the right region has $60\%$of the total number of particles and the left region has the remaining $40\%$is given by:

$\begin{array}{rcl}{W}_2& =& \frac{N!}{(0.6N)!(N-0.6N)!}\\ & =& \frac{N!}{(0.6N)!(0.4N)!}\\ & & \end{array}$

Taking their ratio we obtain:

$\begin{array}{rcl}\frac{W_1}{W_2}& =& \frac{\frac{N!}{(0.5N){!}^2}}{\frac{N!}{(0.6N)!(0.4N)!}}\\ & =& \frac{(0.6N)!(0.4N)!}{(0.5N){!}^2}\\ & & \end{array}$

Apply Stirling's approximation to each coefficient with the factorial operation, and simplify the expression:

$\begin{array}{rcl}\frac{W_2}{W_1}& ≈& \frac{\sqrt{2\mathrm{π}}{(0.6N)}^{0.6N+1/2}{e}^{-0.6N}\sqrt{2\mathrm{π}}{(0.4N)}^{0.64+1/2}{e}^{-0.4N}}{{\left(\sqrt{2\mathrm{π}}{(0.5N)}^{0.5N+1/2}{e}^{-0.5N}\right)}^2}\\ & ≈& \frac{{(0.6N)}^{0.6N+1/2}{(0.4N)}^{0.4N+1/2}{e}^{-0.6N}{e}^{-0.4N}}{{\left({(0.5N)}^{0.5N+1/2}\right)}^2}\frac{{\left(e^{-0.5N}\right)}^2}{{\left({(0.5N)}^{0.5N+1/2}\right)}^2}\\ & ≈& \frac{{(0.6N)}^{0.6N+1/2}{(0.4N)}^{0.4N+1/2}}{\left(\right(0.0.6N-0.4N+N}\\ & ≈& \frac{{(0.6N)}^{0.6N+1/2}{(0.4N)}^{0.4N+1/2}}{{\left({(0.5N)}^{0.5N+1/2}\right)}^2}\\ & ≈& \frac{{0.6}^{0.6N+1/2}{0.4}^{0.4N+1/2}}{0.N^{0.6N+1/2}{N}^{0.4N+1/2}}\\ & ≈& \frac{{0.6}^{0.6N+1/2}{0.4}^{0.4N+1/2}}{{0.5}^{NN+1}}{N}^{0.6N+0.4N-N+1/2+1/2-1}\\ & ≈& \frac{{0.6}^{0.6N+1/2}{0.4}^{0.4N+1/2}}{{0.5}^{N+1}}.\end{array}$

Assuming N is very large, we can approximate the above expression further by getting rid of the additive terms in the exponents (i.e. the$1/2$ 's for the 0.6and 0.4, and the 1 for the 0.5. Doing so yields:

$\frac{W_1}{W_2}≈\frac{{0.6}^{0.6N}{0.4}^{0.4N}}{{0.5}^N}$

Making some minor algebraic manipulations we obtain:

$\begin{array}{rcl}\backslash \frac{W_1}{W_2}& ≈& \frac{{\left(0.6\frac{10}{10}\right)}^{0.6N}{\left(0.4\frac{10}{10}\right)}^{0.4N}}{{\left(0.5\frac{10}{10}\right)}^N}\\ & ≈& \frac{4^{0.4N}{6}^{0.6N}}{5^N}\frac{{10}^{-0.6N}{10}^{-0.4N}}{{10}^{-N}}\\ & ≈& \frac{4^{0.4N}{6}^{0.6N}}{5^N}{10}^{N-0.6N-0.4N}\\ & ≈& \frac{4^{0.4N}{6}^{0.6N}}{5^N}\\ & ≈& {\left(\frac{4^{0.4}{6}^{0.6}}{5}\right)}^N·\\ & & \end{array}$

b)

Consider the number N to be a small one. The numerical value of the ratio is around unity based on our calculated expression for the ratio, meaning that both circumstances describing and are roughly equally likely to occur. The ratio is considerably raised by making the room very large, meaning that the condition where both parts of the room contain roughly the same number of particles is heavily favoured. This backs with the theory that for very large numbers of people, average behaviours become relatively predictable.