Understanding 'e' – the Euler’s constant

This post is going to be very different from my previous posts. This post tries to present an intuitive understanding of a very interesting mathematical constant 'e'.

All introductions to ‘e’ in mathematical texts typically allude to other mathematical constructs which themselves are not very easy or intuitive to understand.

For eg. here is the page on Wolfram Mathworld on’e’

http://mathworld.wolfram.com/e.html

This page first introduces ‘e’ as the base of natural logarithm. ‘Natural logarithm’ itself is a fairly difficult construct to understand. Why makes ‘natural logarithm’ natural? Other definitions don’t help either.

One of the definitions that one typically comes across presents ‘e’ as the following infinite sum

e = ∑(1/k!) where k ranges from 0 to ∞

Such definitions make it extremely difficult to understand why do we even study such numbers ‘e’, where does ‘e’ show up?

Let’s step aside from this track and first study how growth rates work. Growth rates abound in natural phenomena, whether it is population growth rates, interest on principal, bacterial growth etc.

So let’s say today you start with $1 (A). And let’s say you get a 100% annual return. This is to say you are doubling every year.

At time t=0, A = 1

At time t = 1, A = 1 + 1*100% = 1* (1+ 100%)

At time t = 2, A = (1+100%) + (1+100%)*100% = (1+100%)2

In the above example, the time cadence of money doubling is every year. In this case growth happens at discrete steps of time, you get interest credit every year which is when your principal amount grows.

Now let’s say the cadence is not yearly but half yearly. With 100% annual return and money growing half-yearly.

At time t = 0, A = 1

At time t=0.5, A = 1 + 1*100%/2 = 1* (1+100%/2)

At time t = 1, A = 1*(1+100%2) + 1* (1+100%/2)*100%/2 = 1*(1+100%/2)2

What if the compounding happens 3 times a year.

A = 1*(1+100%/3)3

What if compounding happens continuously, you break the entire year into really really small time segments, you earn interest every such segment and it adds up

A = 1* (1+100%/n)n where n is a very large number, n -> ∞

Or A = 1* (1+1/n)n where n is a very large number, n -> ∞

So (1+1/n)n is the scaling factor when your earn interest every infinitesimal amount

If we start with $1 and let’s say split 1 year into 100 time segments which is fairly high number of segments, and our annual interest rate is 100%.

This is like earning 1% interest (100%/100 time segments) every time segment.

When n = 100, (1+100%/100)100 = 2.70481…

When n = 1000, (1+100%/1000)1000 = 2.7169…

This is like earning 0.1% interest for 1000 time segments

When n = 100000, (1+100%/100000)100000 = 2.718…

So you see as we increase the number of time segments we move towards continuous compounding- the scaling factor starts converging towards 2.718…

This is the factor by which your money grows is 2.718 ie $1 become $2.718, so this constant in some sense is the final epoch – you can’t grow more than 2.718 times even if you are growing even pico, femto whatever second.

This is one of the inspirations of ‘e’. ‘e’ appears when you are continuously growing/compounding.

One of the definitions of e is n->∞ (1+1/n)n

And the reason we are interested in understanding this expression is because of the above discussion. Intuitively the reason that above limit converges is because as you go to infinitesimal time segments the compounding effect is offset by the fact that you earn super teeny weeny interest in that time segment.

Now, let’s say the annual rate of growth is not 100% but some other percent, let’s try to get an intuitive feel for how we can map it back to ‘e’

First let’s pick a percentage <100%, say 70% annual interest rate. We start with $1 and do continuous compounding like above

What happens in this case

A1 = 1*(1+70%/n)n

We know to mimic continuous compounding we can pick a large enough n, let’s smartly pick n = 700

So, A1 = 1*(1+70%/700)700 = 1*(1+.1%)70

This is like saying that the growth is equivalent to 0.1% growth rate across 700 time segments

From the previous discussion when we had rate of growth of 100%, one way to mimic continuous compounding was by setting a large enough n = 1000

In this case A = 1*(1+100%/1000)1000 = 1*(1+0.1%)1000 .

We know (1+0.1%)1000 the scaling factor is equal to ‘e’

So if (1+0.1%)1000 is ‘e’,

(1+0.1%)700 is nothing but ((1+0.1%)1000)700/1000 = e0.7

Similarly if the rate of growth is >100%, say 300%, applying the same principles, you can argue this is equivalent to .1% growth rate for 3000 time segments => (1+0.1%)3000 = ((1+0.1%)1000)3 => e3

So, now we know e1 is the scaling factor for continuous growth when rate of growth is 100%, other rates of growth can be easily handled by appropriate exponents, sub 100% growth rates are handled by <1 exponent and greater than 100% growth rates are handled by > 1 exponent (obvious because 100% translates to exponent 1 and becomes the baseline)

Now let’s take the case of changing time, we go back to the original example

With 100% annual return and money growing half-yearly.

At time t = 0, A = 1

At time t=0.5 yr, A = 1 + 1*100%/2 = 1* (1+100%/2)

At time t = 1 yr, A = 1*(1+100%2) + 1* (1+100%/2)*100%/2 = 1*(1+100%/2)2

At time t=1.5 yrs, A = 1*(1+100%/2)3

At time t=2 yrs, A = 1*(1+100%/2)4

With continuous compounding

A (at t=1 yr) = 1*(1+100%/n)n n -> ∞ = e

A (at t=2 yrs) = 1*(1+100%/n)2n n -> ∞ = e2

So, whether it is a change of growth rate or time scale it always translates in an exponent change.

So for eg. continuous growth at 70% rate for 2 years, growth scaling factor = e0.7*2 = e1.4

So, this is the key takeaway, total growth on unit base whether due to growth rate or due to time can easily be modelled by ex

Where x = r*t (r=growth %age and t = time)