Solve the differential equation that define exp(x)

Question

In the wikipedia page for the exponential function in the "formal definition" section I found this statement:

Solving the ordinary differential equation $y'(x)=y(x)$ with the [initial condition](https://en.wikipedia.org/wiki: Initial_value_problem) $y'(0)=1$ using Euler's method gives the product limit formula, valid for all complex values of $x$: $\exp x=\lim_{n\to\infty}\Bigl(1+\frac{x}{n}\Bigr)^n.$

I watched some Youtube videos about the Euler method to try to reproduce it and obtain this same product limit formula but the Euler method as it is described in the videos I found can only give an approximation of a curve, point after point and doesn't give a limit formula.

So, I'm searching documentation about the method to reproduce and verify the Wikipedia statement or if you have any clue to help me to understand this method?

Answer 1

Imagine using Forward Euler to solve the ODE $y' = y$, $y(0) = 1$ on the interval $0$ to $x$, using $n$ steps. The step size is thus $h = x/n$. The Euler step is $$ y_{k+1} = y_k + hy_k = (1+h) y_k, $$ so by induction, applying all $n$ steps from the initial condition to the end: $$ e^x \approx y_n = (1+h)^n = \left(1 + \frac{x}{n}\right)^n. $$ Taking the limit $n\to\infty$ is equivalent to (loosely speaking) an infinite number of infinitesimal steps.

Answer 2

We set $f(t_n,y_n) = y_n$ to solve for $y_{n+1} = y_n + hf(t_n,y_n)$:

We first notice that $y_{n+2} = y_{n+1}+hy_{n+1} = y_n+hy_n+h(y_n+hy_n) = y_n+2hy_n+h^2y_n$

And of course $y_{n+3}=y_n+2hy_n+h^2y_n+h(y_n+2hy_n+h^2y_n) = y_n+3hy_n+3h^2y_n+h^3y_n$

By induction we get that:

$$y_n=\sum_{k=0}^{n} {n \choose k}h^{(k)}y_0$$

We want to find $y(x)$, and we know $x = hn$, and of course we know $y(0) = 1$ so we simplify:

$$y_n=\sum_{k=0}^{n} {n \choose k}(\frac{x}{n})^{(k)}$$

We take the limit as $n$ tends to $\infty$, from this we get the classic result:

$$\lim_\limits{n \to \infty} y_n = \lim_\limits{n \to \infty} \sum_{k=0}^{n} \frac{n!}{k!(n-k)!n^k}x^k$$

$$\lim_\limits{n \to \infty} y_n = \lim_\limits{n \to \infty} \sum_{k=0}^{n} \frac{n \times (n-1) \times (n-2) \times \dots (n-k+1)}{n!n^k}x^k $$

$$ \lim_\limits{n \to \infty} y_n = \lim_\limits{n \to \infty} \sum_{k=0}^{n} \frac{n^k}{k!n^k}x^k =\lim_\limits{n \to \infty} \sum_{k=0}^{n} \frac{1}{k!}x^k= e^x \ \ \ \square $$

Answer 3

[...] but the Euler method as it is described in the videos I found can only give an approximation of a curve, point after point and doesn't give a limit formula.

That's exactly correct. The Euler method doesn't give us a limit formula; if we want a limit, we need to put it in ourselves.

Suppose that we want to find a formula for $\exp x$. The process we're going to use has two steps. In step 1, we apply the Euler method to the differential equation. I'm omitting the details, but the Euler method gives us the formula

$$\left (1 + \frac{x}{n} \right)^n,$$

where $n$ is the desired number of steps. This is only an approximation to $\exp x$, but the approximation gets better as $n$ gets larger.

Now we're done with the Euler method. In step 2, we start with the formula that we got, and we take the limit as $n$ approaches infinity:

$$\lim_{x \to \infty} \left (1 + \frac{x}{n} \right)^n.$$

Why did we do that? Well, what we're hoping is that if we simply take the limit as $n$ approaches infinity, the result will be the exactly correct answer. It turns out that this is, indeed, what happens, so this is a correct formula for $\exp x$.