# a new way of defining $$e$$ the Euler number

This is inspired by a post on Quora by Alon Amit. I found this approach of defining e, the Euler number more intuitive than the traditional way($$e=\lim_{n \to \infty} (1+\frac{1}{n})^n$$). In addition, it explains the importance of the Euler number very well.

First of all, we need a function that solves the elementary differential equation:

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

Of course, without specifying an initial value, this equation will have infinite number of solutions. Just for convenience, let’s specify:

$$f(0)=1$$

By Picard–Lindelöf theorem, our equation:

$$\begin{gather*} \frac{d f(x)}{d x} = f(x)\\f(0)=1 \end{gather*}$$

has one and only one solution.

Now, let’s see what kind of property this function $$f(x)$$ has.

Notice that if we ignore the initial value for now, if $$f(x)$$ is a solution to the differential equation, then for any constant $$c$$, $$c*f(x)$$ is also a solution to the differential equation. To simplify our discussion, let’s call $$f(x)$$ “$$exp(x)$$”.

That means, all the solution to the differential equation $$\frac{d f(x)}{d x}=f(x)$$ has the form of $$c*exp(x)$$.

Also notice that for any constant $$a$$, $$exp(x+a)$$ is also a solution to the differential equation. So we know:

$$exp(x+a)=c*exp(x)$$

Now, since we have $$exp(0)=1$$, insert it into the equation above, we get:

$$exp(a)=c$$

That means:

$$exp(x+a)=exp(a)*exp(x)$$

Magic, we’ve proved that the solution to the elementary differential equation has the property of:

$$f(a+b)=f(a)*f(b)$$

So it must be a exponential function. The question now is, what’s its base? Or, what is the value of $$exp(1)$$?

To answer that question, let’s look at another form of $$f(x)$$ that satisfying the elementary differential equation:

$$f(x)=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+…+\frac{x^n}{n!}+…$$

It’s easy to see that this function is the solution to the differential equation. It also satisfies the initial value. So we have:

$$exp(x)=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+…+\frac{x^n}{n!}+…$$

With this equation, we can very easily get:

$$exp(1)=1+1+\frac{1}{2!}+\frac{1}{3!}+…+\frac{1}{n!}+…$$

It’s not hard to prove that this series converges. And the value is:

$$e=exp(1)=1+1+\frac{1}{2!}+\frac{1}{3!}+…+\frac{1}{n!}+…=2.71828182845904523536028747135266249775724709369995…$$

And our solution to the elementary differential equation is:

$$f(x)=exp(x)=e^x$$

So this is how we define the Euler number: We define it as the base of a exponential function that satisfy the elementary differential equation.

Now, why is $$e$$ important? All its importance actually comes from this definition. Let’s see some examples.

In physics, oscillation is a central phenomenon. All oscillations, be it in motion or in electromagnetic, share the same underlying math:

$$f”=-a*f$$, where $$a>0$$ is a constant.

How do we solve this equation? Since we have $$\frac{de^x}{dx}=e^x$$, let $$f=e^{k*x}$$, we get:

$$f”=k^2*e^{k*x}$$

Now, we only have a find $$k$$ so that:

$$k^2=-a$$

That gives us:

$$k=\pm\sqrt{a}*i$$, where $$i=\sqrt{-1}$$. Plug in the Euler equation and then you’ll get the normal solution of oscillation.

In the same post, Alon Amit also pointed out the important role $$e^x$$ plays in Fourier transformation and normal distribution. The role $$e^x$$ plays in Fourier transformation is obvious. The role $$e^x$$ plays in statistics and probability is not that obvious. Alon pointed out that the normal distribution is a solution that satisfy the following differential equation:

$$f’+x*f=0$$

Alon didn’t explain why, but I found a explanation from here. I’ll create another post for it.

In summary, $$e$$ is important because the exponential function $$e^x$$ is the solution to the elementary differential equation, which is essential in different disciplines of modern science.