A definite integral can be evaluated approximately by means of the Trapezium rule:
\[
\int_{x_{0}}^{x_{N}}\mathrm{f}(x)\,\mathrm{d}x\approx\tfrac{1}{2}h\left\{ \mathrm{f}\left(x_{0}\right)+2\mathrm{f}\left(x_{1}\right)+\ldots+2\mathrm{f}\left(x_{N-1}\right)+\mathrm{f}\left(x_{N}\right)\right\} ,
\]
where the interval length \(h\) is given by \(Nh=x_{N}-x_{0}\), and \(x_{r}=x_{0}+rh\). Justify briefly this approximation.
Use the Trapezium rule with intervals of unit length to evaluate approximately the integral
\[
\int_{1}^{n}\ln x\,\mathrm{d}x,
\]
where \(n(>2)\) is an integer. Deduce that \(n!\approx\mathrm{g}(n)\), where
\[
\mathrm{g}(n)=n^{n+\frac{1}{2}}\mathrm{e}^{1-n},
\]
and show by means of a sketch, or otherwise, that
\[
n!<\mathrm{g}(n).
\]
By using the Trapezium rule on the above integral with intervals of width \(k^{-1}\), where \(k\) is a positive integer, show that
\[
\left(kn\right)!\approx k!n^{kn+\frac{1}{2}}\left(\frac{\mathrm{e}}{k}\right)^{k\left(1-n\right)}.
\]
Determine whether this approximation or \(\mathrm{g}(kn)\) is closer to \(\left(kn\right)!\).
Solution:
We can approximate the integral by \(N\) trapeziums, each with height \(x_{i+1}-x_{i} = \frac{x_N-x_0}{N} = \frac{h}{N}\). The will have area \(\frac{(f(x_i)+f(x_{i+1}))h}{2}\) and summing all these areas we will get:
\[\frac12 h \l f(x_0) + f(x_1) + f(x_1)+f(x_2) + \cdots + f(x_{N-1})+f(x_N) \r = \frac12 h \l f(x_0) +2 f(x_1) + + \cdots +2f(x_{N-1})+f(x_N) \r\]
But this is approximately the integral \(\displaystyle \int_{x_0}^{x_N} f(x) \d x\)
\begin{align*}
&& \int_1^n \ln x \d x &= [x \ln x]_1^n - \int_1^n x \cdot \frac{1}{x} \d x \\
&&&= n \ln n - n+1 \\
&&&\approx \frac12 \l \ln 1 + 2\sum_{k=2}^{n-1} \ln k + \ln n \r \\
&&&= \ln (n!) - \frac12 \ln n \\
\Rightarrow && \ln (n!) &\approx n \ln n + \frac12 \ln n - n + 1 \\
\Rightarrow && n! &\approx \exp(n \ln n + \frac12 \ln n - n + 1) \\
&&&=n^{n+\frac12}e^{1-n}
\end{align*}
Since \(\ln x\) is a concave function, we should expect all the trapeziums to all lie under the curve, therefore this is always an underestimate for the integral, ie \(n! < g(n)\)
\begin{align*}
&& \int_1^n \ln x \d x &= n \ln n - n+1 \\
&&&\approx \frac12 k^{-1} \l \ln 1 + 2\sum_{r=1}^{k(n-1)-1} \ln \l 1+\frac{r}{k} \r + \ln n \r \\
&&&=\frac{1}{2k} \l 2\sum_{r=1}^{k(n-1)-1} \l \ln(k+r) - \ln k)\r + \ln n\r \\
&&&=\frac1{k} \l \ln ((k+k(n-1)-1)!) - \ln(k!) - (k(n-1)-1) \ln k+\frac12\ln n \r \\
&&&=\frac1{k} \l \ln ((kn-1)!) - \ln(k!) - (k(n-1)-1) \ln k+\frac12 \ln n \r \\
&&&=\frac1{k} \l \ln ((kn)! ) -\ln k -\ln n - \ln(k!) - (k(n-1)-1) \ln k+\frac12\ln n \r \\
&&&= \frac1{k} \l \ln ((kn)! ) - \ln(k!) - (k(n-1)) \ln k - \frac12 \ln n\r \\
\Rightarrow && \ln ((kn)!) &\approx kn \ln n - kn + k + \ln(k!) + (k(n-1)) \ln k + \frac12 \ln n\\
\Rightarrow && (kn)! &\approx n^{kn+\frac12}e^{-k(n-1)}k!k^{k(n-1)} \\
&&&= n^{kn+\frac12} k! \l \frac{e}{k} \r^{k(1-n)}
\end{align*}
I would expect this approximation to be a better approximation for \((kn)!\) since it is created using a finer mesh.
