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I'd like to know what I've done wrong when I tried to solve the functional equation below for any domain and codomain where $f$ exists and its derivate as well:

$f(x+c)=f(x)+ce^{x}$

$f(x+c)-f(x)=ce^{x}$

$\displaystyle\frac{f(x+c)-f(x)}{c}=e^{x}$

$lim_{c\rightarrow0}\displaystyle\frac{f(x+c)-f(x)}{c}=lim_{c\rightarrow0}e^{x}$

$df(x)/dx=e^{x}$

$f(x)=e^{x}$

but $e^{x+c}=(c+1)e^{x}$ doesn't hold, just take $c=1$ and $x=0$ then we have $e=2$ xD.

Davi Américo
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Assuming $f$ is a differentiable function such that for all $x,c\in\Bbb{R}$, $f(x+c)=f(x)+ce^x$, you've correctly shown that $f'(x)=e^x$ for all $x\in\Bbb{R}$. From here, it follows that there is come constant $A\in\Bbb{R}$ such that for all $x\in\Bbb{R}$, $f(x)=e^x+A$. So $f(x+c)=f(x)+ce^x$ now implies $e^{x+c}+A=e^x+A+ce^x$, which implies $e^ce^x=(1+c)e^x$ for all $c,x\in\Bbb{R}$. By choosing $x=0$, we then get $e^c=1+c$ for all $c\in\Bbb{R}$, which is absurd. The exponential cannot equal an affine function.

What this shows is that the original functional equation you started with has no solutions.

peek-a-boo
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  • where did i go wrong in the calculations? (ignoring integral's constant) – Davi Américo Jun 13 '22 at 19:35
  • Except the case where $c$ satisfies $e^c=c+1$. Of course, this equation has only one real root at $c=0$ which is just the trivial equation $f(x+0)=f(x)+0$. – QC_QAOA Jun 13 '22 at 19:35
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    @DaviAmérico your calculations are mostly right. One minor (inconsequential) mistake is that $f(x)=e^x+A$ (you forgot the constant of integration). The real issue is one of logic. You started by assuming $f$ is a function which satisfies certain functional equation, and then you proceeded to derive consequences. At the end, you arrived at some false statements while making perfectly correct deductions. This means the initial premise (the existence of a solution) was wrong (think of rules of implication: if you start with a false statement, the implication is true). – peek-a-boo Jun 13 '22 at 19:38
  • @QC_QAOA I'm saying $e^c=c+1$ cannot hold for all values of $c$. The fact that there exists $c\in\Bbb{R}$ (namely $0$) such that $e^c=c+1$ is not relevant here. – peek-a-boo Jun 13 '22 at 19:38
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    @DaviAmérico imagine the following line of reasoning. "Suppose $x$ is a real number such that $x+1=x+2$. Then, by subtracting $x+1$ from both sides, we get $1=0$." This is a correct implication. It just so happens that the premise (existence of $x$ such that $x+1=x+2$) is false (there is no such $x$), so the conclusion (that $1=0$) is also false. But the implication itself is true. If this is still not clear, you should look up some basic logic about truth tables and implications and false premises and vacuous truths etc. – peek-a-boo Jun 13 '22 at 19:42
  • @peek-a-boo I think I get it I thought I was finded out a property generator xD – Davi Américo Jun 13 '22 at 19:45