Mathematics

Intuition For The Constant e=2.71828…

We will develop an intuition for the constant e = 2.71828.., and its importance in calculus. While introductory calculus courses teach what is true about e, we will explain why it is true. We will be as brief as possible for those not wishing to be overwhelmed by meticulous proofs, while not shying away from the mathematics that describe this constant. We will derive the value for e from literally nothing (zero). We will show why e to the power of x, that is eˣ, is the only function family that is also its own derivative and integral.

Recursive Integration

Suppose we start with 0 (zero) and continuously apply integration to that value:

… ∫ ∫ ∫ ∫ ∫ ∫ ∫ 0 dx dx dx dx dx dx dx …

Let us try to evaluate the whole expression, starting in the middle.

… ∫ ∫ ∫ ∫ ∫ ∫ 1 dx dx dx dx dx dx …

… ∫ ∫ ∫ ∫ ∫ x dx dx dx dx dx …

… ∫ ∫ ∫ ∫ x² / 2 dx dx dx dx …

Continuing for higher numbers ..

… ∫ ∫ ∫ 1/2 ⋅ x³ / 3 dx dx dx …

… ∫ ∫ 1/2 ⋅ 1/3 ⋅ x⁴ / 4 dx dx …

… ∫ 1/2 ⋅ 1/3 ⋅ 1/4 ⋅ x⁵ / 5 dx …

… 1/2 ⋅ 1/3 ⋅ 1/4 ⋅ 1/5 ⋅ 1/6 ⋅ x⁶ …

We carried out n+1 steps and find a pattern emerges

xⁿ/n!

We see this type of expression many times when looking at infinite series. It appears in binomial theorem, Taylor series, series definitions of sin and cosine etc. We now have some intuition where the factorial in these formulae come from, and why the factorial is related to the exponent. It arises from repeating integration on itself in the form of recursion.

Taking it to the Limit … To Infinity and Beyond

What is the value of xⁿ/n! if we extend n to infinity?

As n → ∞, lim ( xⁿ/n! ) = 0

It is zero because n! grows much much faster than xⁿ. So we can never actually reach infinity, but we can start at n=0 and add up each xⁿ/n! in a series called exp(x):

exp(x) = 1 + x + x²/2! + x³/3! + x⁴/4! + x⁵/5! + x⁶/6! + …

You can also build this same series with binomial theorem on (1+1/n)ⁿ, or by applying Taylor’s Theorem to e to the power of x.

Why is exp(x) a function that is its own derivative?

An important theoretical question in calculus is the existence of a function that is its derivative.

y = f(x) = f ′(x) = y′

We must also demand that f(0) = 1. Let’s see if exp(x) is such a function. We take the derivative of exp(x).

exp′(x) = [1 + x + x²/2! + x³/3! + x⁴/4! + x⁵/5! + x⁶/6! + …]′

= [1]′ + [x]′ + [x²/2!]′ + [x³/3!]′ + [x⁴/4!]′ + [x⁵/5!]′ + [x⁶/6!]′ + …

= 0 + 1 + x + x²/2! + x³/3! + x⁴/4! + x⁵/5! + …

= exp(x)

Thus exp′(x)=exp(x) and exp(0)=1, so exp(x) is our special function for calculus that is its own derivative! As a corollary, exp(x) is also its own integral.

We can see how our definition of recursive integrals is critical to understanding the proof of y = y′ = exp(x). Since each term is built from a recursive integration, then a term in exp(x) can be equal to the one before it when it is differentiated. Since there are infinite terms, the sum of the whole series is unchanged and y = y′ = exp(x).

Why the value of the constant e is 2.71828…

Saving the reader a very long proof, it can be shown using binomial theorem that

exp(x + y) = exp(x)⋅ exp(y)

Thus

exp(1 + 1) = exp(1)⋅ exp(1) = exp²(1)

exp(1 + 1 + 1) = exp³(1)

exp(1 + 1 + 1 + 1) = exp⁴(1)

and thus

exp(x) = expˣ(1)

So we can find exp(x) for any x simply by knowing exp(1) and bringing it to the power of x. So lets find exp(1) and name it e.

e = exp(1) = 1 + 1 + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! + …

≈ 2.71828182845904523536028747

We can get as many decimal places as we like just including more computed terms, like 1/9999! or beyond.

Now we can state that

exp(x) = expˣ(1) = (2.718281..)ˣ = eˣ

So we no longer need that clunky infinite series anymore, we just use exponents to find eˣ.

Conclusion

This article describes how I came up with this recursive integration construction to prove these concepts to myself. I hope that it may help students of calculus understand the importance of the constant e = 2.71828...