View source for Informal discussion of the Euler's constant

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I'm going to resort to the concept of the '''integral''' without the formalism. If you derive the '''natural log''' the result is the inverse of <math>x</math>. The other direction, if you calculate the '''area under the function's curve''' given by the inverse of <math>x</math> you get the natural log. The reason for this specific function <math>f(x) = 1/x</math> is that deriving logs in any base that isn't <math>e</math> yields <math>1/(xb)</math>, where <math>b \neq 1</math> and is some log. With the natural log we have that <math>\ln(e) = 1 \iff e^1 = e</math>, which implies that if we integrate <math>1/x</math> from 1 to <math>e</math> the result is 1. Why from 1? Because <math>e^0 = 1 \iff \ln(1) = 0</math>. For now it suffices to say that the concept of the integral that we first learn is a sum of rectangles under the function's curve. The base of the rectangle is <math>x_2 - x_1</math> and the height is <math>f(x_2)</math> itself. There is a matter about error in this computation regarding a small area between a rectangle and the function, but I'm disregarding the '''Riemann integral''''s formalism for now.

You can try to calculate the area under <math>1/x</math>, from 1 to <math>e</math>, with a calculator and brute force. Try a rectangle where the base is <math>1.2 - 1.1</math> and the height is <math>f(1.2)</math>. Sum the area of this rectangle with the next, base <math>1.3 - 1.2</math> and the height is <math>f(1.3)</math>. Repeat until you reach a number under 1 and close to it. If we deform the graph of <math>1/x</math> by multiplying it by any <math>c > 0</math> and <math>c \neq 1</math>, the area from 1 to <math>e</math> won't be equal to 1. Which also implies that <math>c \cdot \ln(e) \neq 1</math>. Using composition to deform the graph of <math>1/x</math> by applying a <math>g(x)</math> also has the effect of modifying the area from 1 to <math>e</math>.

A formal way to define the <math>e</math> requires computing a limit of a sequence. Let's skip the theory for now. To compute the sum of areas of rectangles is equivalent to a sum of numbers. With this in mind there must be some sum of numbers such that the result is <math>e</math>. Yes, it's this one:

<math>e =  \sum\limits_{n = 0}^{\infty} \frac{1}{n!} = 1 + \frac{1}{1} + \frac{1}{1\cdot 2} + \frac{1}{1\cdot 2\cdot 3} + \cdots.</math>