MATHS BITE: Shoelace Theorem

The Shoelace theorem is a useful formula for finding the area of a polygon when we know the coordinates of its vertices. The formula was described by Meister in 1769, and then by Gauss in 1795.


Let’s suppose that a polygon P has vertices (a1, b1), (a2, b2), …, (an, bn), in clockwise order. Then the area of P is given by

\[\dfrac{1}{2} |(a_1b_2 + a_2b_3 + \cdots + a_nb_1) - (b_1a_2 + b_2a_3 + \cdots + b_na_1)|\]

The name of this theorem comes from the fact that if you were to list the coordinates in a column and mark the pairs to be multiplied, then the image looks like laced-up shoes.

Screen Shot 2017-08-04 at 11.59.29 AM.png


(Note: this proof is taken from artofproblemsolving.)

Let $\Omega$ be the set of points that belong to the polygon. Then


where $\alpha=dx\wedge dy$.

Note that the volume form $\alpha$ is an exact form since $d\omega=\alpha$, where


Substitute this in to give us


and then use Stokes’ theorem (a key theorem in vector calculus) to obtain



$\partial \Omega=\bigcup A(i)$

and $A(i)$ is the line segment from $(x_i,y_i)$ to $(x_{i+1},y_{i+1})$, i.e. Screen Shot 2017-08-04 at 12.05.20 PM.png is the boundary of the polygon.

Next we substitute for $\omega$:


Parameterising this expression gives us


Then, by integrating this we obtain

\[\frac{1}{2}\sum_{i=1}^n\frac{1}{2}[(x_i+x_{i+1})(y_{i+1}-y_i)- (y_{i}+y_{i+1})(x_{i+1}-x_i)].\]

This then yields, after further manipulation, the shoelace formula:


M x


Stirling’s Formula

Today I wanted to discuss something I learnt last week in my Probability course: Stirling’s Formula. Stirling’s Formula is an approximation for factorials, and leads to quite accurate results even for small values of n.

The formula can be written in two ways:

{\displaystyle \ln n!=n\ln n-n+O(\ln n)}


{\displaystyle n!\sim {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n},}

where the ~ sign means that the two quantities are asymptotic (i.e. their ratios tend to 1 as n tends to infinity).

File:Mplwp factorial gamma stirling.svg
Comparison of Factorial with Stirling’s Approximation | Source: Wikipedia

Proof of Stirling’s Formula

The following identity arises using integration by parts:

Screen Shot 2017-02-06 at 8.11.00 AM.png

Taking f(x) = log x, we obtain

Screen Shot 2017-02-06 at 8.11.48 AM.png

Next, sum over n, and by recalling that log x + log y = log xy we get the following expression:

Screen Shot 2017-02-06 at 8.12.23 AM.png


Screen Shot 2017-02-06 at 8.13.23 AM.png

Next, define

Screen Shot 2017-02-06 at 8.13.38 AM.png

which allows us to rearrange the above expression to:

Screen Shot 2017-02-06 at 8.13.41 AM.png

So as n tends to infinity we get

Screen Shot 2017-02-06 at 8.13.46 AM.png

How do we show that Screen Shot 2017-02-06 at 8.15.19 AM.png ?

Firstly, note that from (*) it follows that

Screen Shot 2017-02-06 at 8.16.31 AM.png

So, we need to show that

Screen Shot 2017-02-06 at 8.16.57 AM.png

Let’s set

Screen Shot 2017-02-06 at 8.17.25 AM.png

Note that I0=π/2 and I1 = 1. Then for n≥2, we can integrate by parts to see that

Screen Shot 2017-02-06 at 8.19.17 AM.png

And so, we obtain the following two expressions:

Screen Shot 2017-02-06 at 8.19.36 AM.png

In is decreasing in n and In/In-2 → 1, so it follows that I2n/I2n+1 → 1. Therefore,

Screen Shot 2017-02-06 at 8.21.18 AM.png

as required.

Although the end result is satisfying, I find that some steps in this proof are like ‘pulling-a-rabbit-out-of-a-hat’! What do you think? Mx