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

This question is different from :

Prove that $C\exp(x)$ is the only set of functions for which $f(x) = f'(x)$

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

  1. show that the solution should verify : $f(a+b)=f(a)f(b)$
  2. show that $f(x)$ for any $x$ in $\Bbb R$, will write $f(x)=c a^x$.
  3. 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 :

  1. no idea – this is here that I need the more help
  2. prove it for naturals, rationals then all real numbers using density arguments.
  3. 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 $e$:

https://www.freecodecamp.org/news/eulers-method-explained-with-examples/

Many thanks, I'll appreciate your help

G

Bernard
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StudentInFinance
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    Does this answer your question? [Prove that $C\exp(x)$ is the only set of functions for which $f(x) = f'(x)$](https://math.stackexchange.com/questions/58097/prove-that-c-expx-is-the-only-set-of-functions-for-which-fx-fx) – 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

1 Answers1

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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?

Bernard
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  • 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