## Wednesday, 21 December 2011

### Musings on Euler's number and natural logarithms

In other words: a proof that

$\int_{1}^{x}\frac{1}{x}dx=\log_{e}x$

where e=2.718281828... (Euler's number)

Mainly a note to self, just so I don't have to remember it again...

Prompted by a couple of posts by Adrian McMenamin on his blog

Not entirely sure this is entirely robust or complete – but I think it’s basically the way I learned it and have remembered and rehearsed it in the past. Any comments/corrections from those more in the know welcome.

Define the function f(x) as follows:

$f(x)=\int_{1}^{x}\frac{1}{x}dx$

Then f(x) is uniquely defined (for all real non-zero positive values of x) and

$\frac{d(f(x))}{dx}=\frac{1}{x}$

and

$f(1)=0$

Also, for a variable y (independent of x but on the same domain)

$\frac{d(f(xy))}{dx}=\frac{1}{xy}\frac{d(xy)}{dx}=\frac{1}{xy}y=\frac{1}{x}$

and

$\frac{d(f(x)+f(y))}{dx}=\frac{d(f(x))}{dx}+\frac{d(f(y))}{dx}=\frac{d(f(x))}{dx}=\frac{1}{x}$

$\therefore \frac{d(f(xy))}{dx}=\frac{d(f(x)+f(y))}{dx}$

$\therefore f(xy)=f(x)+f(y)+C$

where C is some constant value.

To find the value of C let x = 1. Then, since f(1) = 0

$f(y)=f(y)+C\Rightarrow C=0$

$\therefore f(xy)=f(x)+f(y)$

Now define the function g(x) to be the inverse of the function f(x). i.e.

$f(g(x))=x$

and

$g(f(x))=x$

So

$x+y=f(g(x+y))$

And from the previous result that

$f(xy)=f(x)+f(y)$

we get

$x+y=f(g(x))+f(g(y))=f(g(x)g(y))$

$\therefore f(g(x+y))=f(g(x)g(y))$

$\therefore g(x+y)=g(x)g(y)$

i.e. the function g satisfies the basic exponential identity

$a^{m+n}=a^{m}a^{n}$

So, g(x) can be expressed as some number (go on, let's call it e) raised to the power of x.

$g(x)=e^{x}$

Since f(x) is the inverse of g(x), f(x) can be expressed as the logarithm to the base e of x. i.e.

$f(x)=\log_{e}(x)$

So

$\log_{e}(x)=f(x)=\int_{1}^{x}\frac{1}{x}dx$

That's the first bit done (I think)...

...but what's the value of e?

First, look at the properties of the derivative of the function*

$g(x)=e^{x}$

i.e.

$\frac{d(g(x))}{dx}=\frac{d(e^{x})}{dx}$

Remember (from above)

$x=f(g(x))$

So

$\frac{dx}{d(g(x))}=\frac{d(f(g(x)))}{d(g(x))}$

$\therefore \frac{d(e^{x})}{dx}=\frac{d(g(x))}{dx}=\frac{1}{(\frac{dx}{d(g(x))})}=\frac{1}{(\frac{d(f(g(x)))}{d(g(x))})}$

Remember too that

$\frac{d(f(x))}{dx}=\frac{1}{x}$

$\therefore \frac{d(e^{x})}{dx}=\frac{1}{(\frac{d(f(g(x)))}{d(g(x))})}=\frac{1}{(\frac{1}{g(x)})}=g(x)=e^{x}$

[This result allows for the derivation of both the sum of a series and limit definitions of Euler's number. What follows here is the derivation of the sum of a series definition - the limit definition is derived below]

It follows directly from the result

$\frac{d(e^{x})}{dx}=e^{x}$

that all higher derivatives of

$g(x)=e^{x}$

can be expressed as follows:

$\frac{d^{n}(e^{x})}{dx^{n}}=e^{x}$

for any positive integer n.

So

$g(x)=e^{x}$

is infinitely differentiable and

$g^{(n)}(0)=e^{0}=1$

where

$g^{(n)}(0)$

represents the value of the nth derivative of

$g(x)=e^{x}$

with respect to x at x = 0.

So

$g(x)=e^{x}$

can be expressed as a Maclaurin series:

$g(x)=e^{x}=\sum_{n=0}^{\infty }\frac{g^{(n)}(0)}{n!}x^{n}=\sum_{n=0}^{\infty }\frac{1}{n!}x^{n}$

So set x = 1 to calculate e to arbitrary precision using

$e=e^{1}=\sum_{n=0}^{\infty }\frac{1}{n!}1^{n}=\sum_{n=0}^{\infty }\frac{1}{n!}= 2.718281828...$

LIMIT DEFINITION OF EULER'S NUMBER

Consider the limit

$\lim_{y \to 0} \frac{y}{e^{y}-1}$

Now

$\lim_{y \to 0} y=0$

and

$\lim_{y \to 0} e^{y}-1=0$

and

$\frac{d(e^{y}-1)}{dy}=e^{y}\neq 0$

for any y and

$\lim_{y \to 0} (\frac{d(y)}{dy}}/{\frac{d(e^{y}-1)}{dy})=\lim_{y \to 0}\frac{1}{e^{y}}=1$

is defined, therefore L'Hopital's rule applies. i.e.

$\lim_{y \to 0} \frac{y}{e^{y}-1}=\lim_{y \to 0}(\frac{d(y)}{dy}}/{\frac{d(e^{y}-1)}{dy})=\lim_{y \to 0}\frac{1}{e^{y}}=1$

Substitute

$y=\log_{e}(x+1)$

Then

$\frac{y}{e^{y}-1}=\frac{\log_{e}(x+1)}{x}$

and as

$y \to 0, x \to 0$

Therefore

$\lim_{x \to 0}\frac{\log_{e}(x+1)}{x}=\lim_{y \to 0}\frac{y}{e^{y}-1}=1$

Now

$\lim_{x \to 0}\frac{\log_{e}(x+1)}{x}=\lim_{x \to 0}\log_{e}((x+1)^{1/x})=1$

therefore (since the exponential function is continuous*)

$\lim_{x \to 0}(x+1)^{1/x}=e$

Or alternatively if

$n=\frac{1}{x}$

then

$\lim_{n \to \infty }(1+\frac{1}{n})^{n}=e$

*POSTSCRIPT
I've used g(x) to denote the exponential function here to try and avoid preconceptions but it's obviously more commonly denoted by

$exp(x)=e^{x}$

I think I'm right in saying that, without a definition of the exponential function:

$a^{x}$

and exponential identities such as

$a^{x+y}=a^xa^y$

can only be defined where x (and y) is rational.

However, because

$\log_{e}x=f(x)=\int_{1}^{x}\frac{1}{x}dx$

is (by definition) continuous and differentiable (with non-zero derivative) at all points on its domain of all non-zero positive numbers (range is all real numbers), its inverse

$g(x)=exp(x)=e^{x}$

is also continuous and differentiable at all points on its domain of all real numbers (its range is all non-zero positive numbers). Therefore e can be raised to the power of any real number, not just rationals. This leads on to the definition of values for non-zero positive real numbers raised to the power of any real number - and eventually can be completely extended to include negative and complex numbers. A few steps beyond that and ultimately comes the famous and beautiful (and immensely useful) Euler function and identity:

$e^{i\theta }=\cos \theta +i\sin \theta$

$e^{i\pi }+1=0$

I think... I might post a proof of that too at some point. Weather, time and memory permitting.