Let's look at an unusual one coming from linguistics.

If we read English poetry from antiquity, we'll see some unfamiliar words.

And if you look at Chaucer, the Canterbury Tales,

many of the words are unfamiliar to us.

But if we move the clock ahead to Shakespeare's time,

then some of that poetry is familiar.

Some is unfamiliar.

Moving ahead a little bit further to Milton,

things are still a little obtuse but more,

the words make sense, and we could keep going with Blake and others.

What we might guess is the following linguistic model.

Namely, that the number of words in English remaining in common use

decreases at a rate proportional to the remaining words.

That means that words fall out of use.

But this model states that words fall out of use according to a linear

differential equation where W is the amount of words remaining in

usage and alpha is a decay constant.

Assuming this model, let's answer the following question.

If we find 20% of the words in Milton's poem to be unusual,

then what fraction of Chaucer's poetry did Shakespeare's audience recognize?

Whew, that sounds difficult.

What can we do with that?

Well, let's say that t corresponds to the year.

We're going to be interested in two functions of t.

The first, W sub M, is the number of

words in Milton's usage currently used at time t.

That is, according to the solution to this differential equation,

W sub M at 1667, the initial condition, times e to the minus

alpha times the number of years that have elapsed since 1667.

Likewise, we'll be interested in W c of t,

the number of words from Chaucer's time still in common usage.

This has the same solution with 1667 being replaced by 1400.

Most importantly, the alphas are assumed same because this is all English.

Now what is it that we are given?

We are told that the number of words from Milton's time in common use today,

let's say in 2012, is 80% of those originally available.

Plugging in the solution to the differential equation for W sub M when t

equals 2012, and then dividing by the initial condition

gives us a single equation that has only alpha as an unknown.

We can therefore solve for alpha and

we get a quantity that is about 6.5 times 10 to the -4.

Now, if we use that alpha in our solution for

W sub C of t, then what can we get?

Well, our goal is to find the fraction of words from

Chaucer's time that people in Shakespeare's time would have understood.

That is, W c of 1600 divided by W c of 1400.

Plugging in our solution to the differential equation for W sub C