You may also like

problem icon

Ball Bearings

If a is the radius of the axle, b the radius of each ball-bearing, and c the radius of the hub, why does the number of ball bearings n determine the ratio c/a? Find a formula for c/a in terms of n.

problem icon

Air Routes

Find the distance of the shortest air route at an altitude of 6000 metres between London and Cape Town given the latitudes and longitudes. A simple application of scalar products of vectors.

problem icon

Epidemic Modelling

Use the computer to model an epidemic. Try out public health policies to control the spread of the epidemic, to minimise the number of sick days and deaths.

Stirling Work

Stage: 5 Challenge Level: Challenge Level:2 Challenge Level:2

Part 1: Estimate of n


$\sqrt{n} > 100$
$n> 10^4$

$\sqrt{n} > 1000\ 000$
$n> 10^{12}$

$\sqrt{n} > 1000\ 000\ 000\ 000$
$n> 10^{24}$

$n^n> 100$
$n\log n> \log 100$

Using the interval bisection method: $n$ is greater than 3.59728502354042

 

$n^n> 1000\ 000$

$n\log n> \log 1\ 000\ 000$

Using the interval bisection method: $n$ is greater than 7.06579672829962

 

$n^n> 1000\ 000\ 000\ 000$

$n\log n> \log 1000\ 000\ 000\ 000$

Using the interval bisection method: $n$ is greater than 11.3673178030006

Interval Bisection tables
 

Part 2:

$n!=1 \times 2 \times 3 \times 4 \times \dots (n-1) \times n$

Taking logs:

$$
\begin{align*} \ln(n!) &= \ln (1 \times 2 \times 3 \times 4 \times \dots (n-1) \times n) \\&= \ln 1 + \ln 2 +\ln 3 +...\ln(n-1) +\ln(n) \\&= \sum_{i=1}^n\ln(i) \approx\int^n_1\ln(x)\,dx
\end{align*}$$

where we've assumed n is large when replacing the sum by an integral. We can evaluate this integral using integrating by parts: 

$$ \int \ln(x) \,dx = x \ln(x) - \int 1 \,dx = x\ln(x) - x = \ln(x^x) - x$$
Substituting the limits of integration gives: $$\ln(n!) = \ln(n^n) -n - (\ln(1^1) - 1)$$
Taking exponentials: $$n!= e^{\ln(n^n) -n + 1}=n^n e^{1-n}$$
Comparing with the form of the approximation given (for large n) it can be seen that we should use the approximation:
$$n! \sim A n^{ +n\pm 0.5} e^{-n} $$
We now need to find the value of the constant A and work out the remaining sign in the term $n^{\pm0.5}$.

If we temporarily let the constant $= 1$ and then compare the values obtained from $An^{n-0.5}e^{-n}$ and $An^{n+0.5}e^{-n}$ with $n!$ for a range of $n$, the results are listed in the spreadsheet below.

Spreadsheet

From the spreadsheet it can be seen that $An^{n+0.5}e^{-n}$ with a constant $A$ greater than unity, will provide the appropriate form. If we were to use the other possible form, $An^{n-0.5}e^{-n}$, we would need an extremely large value of $A$.

Estimate of $A$:

By simply dividing column E by coumn D in the above spreadsheet it can be seen the value of $A$ starts at around 2.6 (for small $n$) and tends to around 2.5 (for large $n$). The exact value of $A$ is actually $\sqrt{2\pi} \approx 2.506628275$
 

Part 3:


There appears to be no limit to the factorial which may be calculated if the answer is left in an algebraic form. Using the formula in Excel, $142!$ was the largest number that I was able to compute. The value of the real and approximated results are listed below.

$142!= 2.695364137888160 \times 10^{245}$


$\sqrt{2\pi} \times 142^{142 +0.5}e^{-142} =2.693782818480670 \times 10^{245}$

Percentage difference = 0.058668118 %