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The definition that $e^x$ is the unique function $f(x)$ such that $f(x)=f'(x)$ and $f(0)=1$ has two problems for me:

  1. How is $e^x$ the unique function that satisfies this property? $ke^x$ also has this property:$\frac{d}{dx}(ke^x)=ke^x$. It would appear that infinitely many functions satisfy this property.
  2. Why does the definition specify that $f(0)=1$? What functions is this part of the definition trying to exclude?
Joe
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First notice that $k e^x $ is equal to $k$ when $x=0$ so it doesn't satisfy $f(0)=1$ when $k \neq 1$. The property $f' = f$ assures you that $f$ is a multiple of $e^x$ and the value of $f$ at $0$ enables you to say which multiple of $e^x$ $f$ is. Here's a proof of that :

If there is another differentiable function $f$ such that $f'(x) = f(x)$ and $f(0) = 1$ then since $e^x \neq 0$ for all $x$ we have

$$ \frac{d}{dx}\frac{f(x)}{e^x} = \frac{f'(x)e^x - (e^x)'f}{e^x} = \frac{f(x)e^x-f(x)e^x}{e^x} = 0.$$ This means that $\frac{f(x)}{e^x}$ is constant and $f(x) = k e^x$ for some $k \in \mathbb R$.

Plugging in $x = 0$ we get

$$ f(0) = k$$

and $f(x) = f(0)e^x = e^x$.

Therefore any other value than $1$ for $f(0)$ would give another different multiple of $e^x.$

Edit:

To prove that $e^x$ is the unique function such that $f' = f$ and $f(0) = 1$ one must of course first show that $(e^x)' = e^x$ and $e^0 = 1$. In the proof above I assume this therefore this proof only shows that if $f' = f, f(0) = 1$ has a solution then it is is unique (you can see this by replacing $e^x$ by $g(x)$ in the proof).

Proving that $e^x$ is a solution of $f' = f,f(0) = 1$ depends heavily on how you define $e^x.$

Using the Picard-Lindelöf theorem one can easily show that $f' =f, f(0)=1$ has a unique (maximal) solution. Define $e^x$ as this unique solution and then by definition $e^x$ is the only function such that $f'=f, f(0) =1.$ Obviously this isn't very insightful since you still want to be sure that this $e^x$ is the one you know as $ \sum_k \frac{x^k}{k!}$ or $\lim_{n \to \infty}(1 + x/n)^n.$

This all comes down to showing that $(e^x)' = e^x,e^0 =1$ with your preferred definition of $e^x.$ You can find this proof here on the website (Derivative of exponential function proof). I'm sure you can find a few other proofs here by using the search functionality.

Digitallis
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  • Thank you very much. This is an extremely good answer as not only does it clear up my confusion, it also proves that $f(x) = ke^x$ are the only family of functions that have the property $f’(x) = f(x)$, which is very enlightening. – Joe May 31 '20 at 17:08
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    @Joe The only thing that remains to be proved, in order to be sure that this is a good definition of $e^x$ is that a function satisfying $f'=f$ and $f(0)=1$ exists! – Brightsun Jun 01 '20 at 07:35
  • @Brightsun Ah I see. So it doesn’t really make sense to define $e$ in this way if we haven’t proven that such a number exists. Is that correct? – Joe Jun 01 '20 at 08:47
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    @Joe To define $e^x$ as the unique solution to $f' = f, f(0) = 1$ one must prove two things 1. that there is a solution to $f' = f, f(0) = 1$ and 2. that the solution is unique. In this answer I've only proved the unicity and not the existence. – Digitallis Jun 01 '20 at 09:20
  • @Digitallissimo Thank you for clarifying. So is it possible to prove the existence of such a function in the following way: define $e^x$ using a different method (e.g. as an infinite series), and show that $e^x$ satisfies the property $f'=f,f(0)=1$? Otherwise, we run the risk of 'defining something into existence' when it hasn't been shown that such a function does exist! – Joe Jun 01 '20 at 09:32
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    @Joe see my edit. – Digitallis Jun 01 '20 at 12:17