1

Sometimes I see Euler's number defined the following way:

$$e = \lim_{n \rightarrow + \infty} \biggr ( 1 + \frac 1n \biggr)^n \tag1$$

And sometimes I see the exponential function defined the following way:

$$\exp(x) = e^x = \lim_{n \rightarrow + \infty} \biggr ( 1 + \frac xn \biggr)^n \tag2$$

If we define $e$ as $(1)$, then $e^x$ becomes the following:

$$e^x = \biggr(\lim_{n \rightarrow + \infty} \biggr ( 1 + \frac 1n \biggr)^n\biggr)^x \tag 3$$

And if we define $e^x$ as $(2)$, then it remains to show that such an equation, for all $x$'s, yields a single value for $e$.

I say that because one cannot simply say $(1)$ follows from $(2)$ by definition, because I do not know whether the value for $e$ of which satisfies $(2)$ for a specific $x$ will be the same value for all $x$'s. Given $(1)$ is just the special case where $x=1$, I am not able to derive it from $(2)$ by definition, because I do not know whether the $e$-value that satisfies $(1)$ is the same $e$-value that satisfies $(2)$ for any given $x$-value.

user110391
  • 1,079
  • There are a couple ways to show this, but it heavily depends on what you already know. For instance, if you know $e^x = \sum_{k=0}^\infty \frac{x^k}{k!}$ (the latter infinite series is more commonly written $\exp(x)$), then I think we can get it from the binomial theorem and the monotone convergence theorem. If not, I think you need to know $(1+\frac1x)^x$ is increasing for $x >0$ and $x^y$ is continuous on $x>0$ for any fixed real $y$ – Brian Moehring May 31 '23 at 23:44
  • What is your definition of $e^x$? – LL 3.14 Jun 01 '23 at 01:10

1 Answers1

2

To obtain (1) from (2), you only need to take $x=1$, as you notice.

Now, the most simply way (in my opinion) to obtain (2) from (1), is the following:

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

Fixing $x >0$ and taking $m = \frac{n}{x}$, we have that

$$ \lim_{n \to +\infty} \left(1 + \frac{x}{n}\right)^n = \lim_{m \to +\infty} \left[\left(1 + \frac{1}{m}\right)^m\right]^x = \left[\lim_{m \to +\infty}\left(1 + \frac{1}{m}\right)^m\right]^x = e^x$$

The proof for $x<0$ is similar and for $x=0$ is trivial.

Renato Fernandes
  • 537
  • 2
  • 11
  • this invokes way too many things, without even mentioning them, to be considered a good proof of equivalence. but OP didn’t specify how much they’re already willing to accept so I guess one can’t really give a complete answer. – peek-a-boo Jun 01 '23 at 01:10
  • Yes, I was going to write that too, here if $m$ is not an integer, you are implicitly using $x^m = e^{m\ln x}$. But it is still a good heuristic to understand what is happening, in particular if $m$ is an integer. – LL 3.14 Jun 01 '23 at 01:14
  • I understand your proof of getting $(2)$ from $(1)$, although @peek-a-boo saying it invokes too many things makes me unsure, because to me, this proof does not have all that many moving parts (thus implying I am missing something). However, I disagree that you can simply obtain $(1)$ from $(2)$ by substituting $x$ with $1$. The definition alone cannot show how the $e$-value is the same for all $x$-values. See my edited post for a clarification of my confusion. – user110391 Jun 01 '23 at 17:32