Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2005 / Paper 1 / 7

The logarithm turns the product defining the factorial into $\ln (N!)={\sum }_{k=2}^N\ln k$. Since $\ln x$ is increasing, each term bounds integrals on its adjacent unit intervals:Summing from $k=2$ to $N$ proves the integral bounds for a logarithmic factorial:For $N=1$, both sides and the empty sum are zero. For $N\ge 2$, the rectangle diagram gives the same inequalities geometrically. These bounds establish the leading $N\ln N-N$ growth, while Stirling's formula gives finer corrections.

Write $L(N)=(N+a)\ln N-N+b$. The two data points give the linear equations $a\ln 10+b=15.104413-10\ln 10+10$ and $a\ln 50+b=148.477767-50\ln 50+50$. Subtraction and back-substitution giveor $a=0.4959$, $b=0.9368$ to four significant figures. The six-decimal-place value $148.477767$ here is the one printed in the original PDF. Retain guard digits for the estimates below. The fitted coefficients are close to the Stirling's formula values $1/2$ and $\frac{1}{2}\ln (2\pi )$ but need not equal them because the fit absorbs finite-$N$ corrections.

Substituting into the fitted factorial logarithm givesThis is an estimate from the requested two-point fit, not an exact factorial evaluation.

The same fitted factorial approximation gives

For the stated $N_A=6.022137\times 10^{23}$, the leading term $N_A(\ln N_A-1)$ dominates the logarithmic correction $a\ln N_A+b$. ThusThis uses the numerical value specified in the question; such a large logarithm can be computed without constructing the factorial itself.

Separate the even factors from $1000!$:The double factorial therefore satisfies $\ln (999!!)=\ln (1000!)-500\ln 2-\ln (500!)$. Applying the same fit consistently gives