Why when we try to solve $f = f '$, do we set the $f(0) = 1$ condition? This is regarding finding Euler's number.

Question

I was trying to get a better understanding for e and pi, and came across Alon Amit's explanation here: https://www.quora.com/q/bzxvjykyriufyfio/What-is-math-pi-math-and-while-were-at-it-whats-math-e-math

What I don't understand is this 'normalization' process. Why no set any number of arbitrary conditions? $$f(1) = 0 \text{ or } f(0) = 2?$$

Answer 1

You don't really have to set $f(0)=1$ to retrieve $e$. Any nonzero value of $f(0)$ works quite well if you then render $e$ as the ratio $f(1)/f(0)$.

Answer 2

When you "solve" $f' = f$, you almost get an answer. It's a little like solving $x^2 = 4$. In that case, you get two answers, which are negatives of each other $\pm 2$. And if you have some additional information, like $ x > 0$, then you know your answer is $x = 2$.

For $f' = f$, there's a whole family of solutions, but they have an interesting property: any two of them are proportional. In particular, the function $$ f(x) = 0 $$ is a solution, but a particularly uninteresting one: it's a multiple of any other solution, but no other solution is a multiple of it. So let's pick some particular nonzero solution of the equation, and call it $g$. It turns out (although this requires proof) that merely knowing that $g(x) \ne 0$ for some $x$ actually means that $g(0) \ne 0$. So at this point we have

1. $g$ is a solution of $f' = f$.

2. $g$ is nonzero somewhere, and hence (by a circuitous route) $g(0) = A \ne 0$.

3. Any other solution $f$ of the equation can be written as $f = cg$, a scalar multiple of $g$.

So (as in the $x^2 = 4$ problem), which value of $c$ do we choose? The answer is that it's arbitrary, but in any field (the real numbers form an algebraic structure called a "field") there's an additive identity $0$, and a multiplicative identity, $1$, and these tend to come up a lot. In choosing the value we want for $g(0)$, we could choose $0$...but that doesn't work because of item 2. So instead, we choose $1$, which means that when we multiply by $g(0)$ in some problem, we'll end up multiplying by $1$, which is easy to simplify.

There's another argument: As it happens, we can prove (with some work involving calculus and the inverse function theorem) that there's a number $e$ with the property that for every integer $n$, we have $$ g(n) = A e^n $$

[NB: it's actually true that for any positive rational number $g(n/k) = Ae^{n/k}$, and for irrational numbers $s$, we can then define $Ae^s$ to be the number $g(s)$. In other words: we use this special solution $g$ to define irrational exponentiation.]

Of all possible choices for $A$, the "simplest" one (in our usual notation) is $A = 1$, so that we can write $g(n) = e^n$ (because we can drop any "multiply by 1"). So...that's another good reason to make that choice.

Answer 3

$f(x)=f'(x)$ has for solution $f(x)=ce^x$ for any c. So c=1 is just one of the solutions you can arbitrarily pick. Any arbitrary selection will result in one solution however you cannot pick more than one arbitrary choices for the same solution. That violates the definition of function for f