Help to solve $y'=y$, building exp function

Question

I come to ask for help building the exponential function as the solution to $y'=y$.

This question is different from :

Since I would like help to prove it using the following arguments :

show that the solution should verify : $f(a+b)=f(a)f(b)$
show that $f(x)$ for any $x$ in $\Bbb R$, will write $f(x)=c a^x$.
show that if the function value is $1$ at $0$, using a numerical tool we will be able to find the Euler constant value and not it e.

For the moment here are my ideas :

no idea – this is here that I need the more help
prove it for naturals, rationals then all real numbers using density arguments.
using Euler method,I can show that $a$ is the limit of $f(1) = \lim_{n\to\infty} (1+1/n)^n$ As you can see here, the computation will tend to $e$:

Many thanks, I'll appreciate your help

G

Does this answer your question? Prove that $C\exp(x)$ is the only set of functions for which $f(x) = f'(x)$ — KReiser, Jul 11 '20 at 22:12
No... Not really... I see here that this proof is not complete : https://math.stackexchange.com/a/1175664/608989 . however i really liked the proof using the taylor method. I read deeper the first one and come back to you. I was more looking for a proof closer to my 3 steps methodology, - but have issues using these.. do you follow ? — StudentInFinance, Jul 11 '20 at 22:19
I finally read all. But none of them uses the approach i mention. — StudentInFinance, Jul 11 '20 at 22:28

score 4 · Accepted Answer · answered Jul 11 '20 at 22:47

4

A hint for 1: consider the function $g$ defined as $$g(x)=\frac{f(a+x)}{f(x)}$$ and calculate $g'(x)$. What can you conclude?

answered Jul 11 '20 at 22:47

Bernard

I get g'(x)= [f'(x+h)f(x)-f(x+h)f'(x)]/f(x)^2. Since f'=f, g'(x) =0. Thus g(x) =Constant. This makes me conclude that, for all h, f(x+h) = C f(x)... But i don t get C=f(h)... Where am i going the wrong way ? Moreover choosing h=0 gives C=1. Which make me conclude that f is a constant function – StudentInFinance Jul 12 '20 at 07:37
Why do you replace $a$ with $h$? The latter is mostly used tio denote a small increment of the variable, to obtain the definition of the derivative via the limit of the rate of variation when $h$ tends to $0$. This is not the case here – $a$ is just a (fixed) parameter. To answer your question, you just have set $x=0$ to get the constant. – Bernard Jul 12 '20 at 08:06
Sorry for using another letter. I agree that h is conventionally reserved for small increments and using it might be confusing. Using a and evaluating on x=0, i get C= f(h)/f(0). But, assuming we want the solution that verifies f(0)=1, we end up with C=f(h). Which proves what i needed ! – StudentInFinance Jul 12 '20 at 08:13
It's exactly that! – Bernard Jul 12 '20 at 08:22

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