Euler's Number, e

e is perhaps the most celebrated, most feted number in Maths this side of pi. First described by Euler, it is difficult to define as it's an irrational number, i.e. without an integeric value and unexpressable as a fraction. It is best denoted as the limit of (1+1/n)^n, when n tends to infinity. The mathematical value of e is approximately 2.71828..

You do not begin to appreciate the importance of e until you consider the following problems, provided as examples:

1. A bank offers you 5.6% interest rate, compounded for 5 years. How much would you get back after 5 years?
2. UK has a population of 63 million, Germany has 78 million. UK is growing at the rate of 2 new subjects per 1000 per year, Germany at roughly half that. How many years would it take for UK's population to catch up with that of Germany's?
3. A food item will spoil at a critical bacterial mass of 5 million per gram of foodstaff. If initially, there were only 10,000 bacteria, and if the number of bacteria grow from 10,000 to 20,000 in 1 minute and to 250,000 in the next 10 minutes, how long would it take for the food to spoil?

You need e to solve all the problems. Essentially, if an entity is growing continuously, at a given rate, you can use e to calculate its growth. e is the natural base log (log 10 being the "common" log). OTOH, the natural log of a number when the result is expressed as an exponent of e is called "ln" (l for Lima, n for November, using phonetics.

A simple way to remember this is that e^kt gives you the value of an entity growing at the rate of "k" after a certain time period "t", while ln x gives you the exponent- i.e. the power by which the entity has grown. If the rate of growth is known, ln x will let you find the time taken for the entity to grow from y to x.

Things will become clearer with the above examples.

1. A bank offers you 5.6% interest rate, compounded for 5 years. How much would you get back after 5 years if you invested £5000?

You might think the answer is 5000*(1.056)^5, but you would be wrong. The answer of 6565 worked this way is less than what you'd actually get.

The answer is 5000*e^(0.056*5)= approximately 6616. This is how compound interest is calculated.

On the other hand, you might wonder how long it would take for your money to double at this rate of 5.6%? If time for doubling is denoted as t, you get e^(0.056t)= 2. However, you know that ln 2 = 0.693 from a scientific calculator. Hence ln 2 = 0.693 = 0.056t or t = 12.375 years.

Since interest rates are expressed as percentages, you can simply multiply both numerator and denominator by 100 and get t=69.3/5.6= 12.375 years.

In practice, 69.3 is rounded off to 72 to allow us to take advantage of the fact that 72 is divisible by many small integers (the usual duration of a fixed deposit). However, this gives us an approximate result. Thus 72/5.6 in the previous example would give 12.85 years. This is called the rule of 72.

Similarly, you can calculate the trebling time for your deposit at the same rate of interest. Thus, e^0.056t = 3 . ln 3 = 1.10 approximately. Thus for trebling, you can use the rule of 110. Your money will treble in 110/5.6 = 19.64 years.


2. UK has a population of 63 million, Germany has 78 million. UK is growing at the rate of 2 new subjects per 1000 per year, Germany at roughly half that. How many years would it take for UK's population to catch up with that of Germany's?

In this case, if the projected period for equalisation of numbers is t, you can write the equation as:

63 million*e^(0.002t)= 78 million* (e^0.001t)

Or e^.002t/e^.001t = 78/63.

Here, it is useful to know that since e is a logrithmic function, e^x/e^y = e^(x-y).

Thus, in the above problem, e^(.002t-.001t) = 78/63.

Or e^.001t = 1.238

now, ln 1.238= 0.213 = .001t

or t = .213/.001 = 213 years.


3. A food item will spoil at a critical bacterial mass of 5 million per gram of foodstaff. If initially, there were only 10,000 bacteria, and if the number of bacteria grow from 10,000 to 20,000 in 1 minute and to 250,000 in the next 10 minutes, how long would it take for the food to spoil?

Here, after 1 minute, 10,000* e^(k.1), where k is the rate at which bacteria are multiplying.

Or 10,000*e^k= 20,000- This is equation 1.

After a further 10 minutes, 20,000*e^[k.(1+10)]= 250,000- This is equation 2.

Using logrithmic notation, e^[k(1+10)] can also be written as e^k * e^10k

Dividing equation 2 by equation 1, we get:

(20,000*e^k*e^10k = 250,000)/(10,000*e^k = 20,000)

You get 2e^10k = 12.5

or, e^10k = 6.25.

ln 6.25 = 1.83 = 10k, or k = .183

How long will it take to attain the critical mass of 5 million bacteria per gramme? (call this T)

10000*e^(.183T) = 5 million

or e^.183T = 500

ln 500 = 6.21 = .183 T,

or T= 6.21/.183 = 33.93 minutes.