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How can one prove without using anything but differentiation, that $e^x$ is the only function with $f'=f$? Clearly I can prove that $(e^x)'=e^x$, and $0'=0$, but how can one show that no other functions exist? Thank you.

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marked as duplicate by Hans Lundmark, Claude Leibovici, Przemysław Scherwentke, Rory Daulton, PhoemueX Jan 7 '15 at 11:26

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    $\begingroup$ There are other functions, like $2 e^x$. $\endgroup$ – Najib Idrissi Jan 7 '15 at 9:09
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For $f \neq 0$ you have

$\dfrac{f'}{f}=1$, that is $(\ln{f})'=1$

$\ln{f}=x+c$, $c \in \mathbb{R}$ being a constant.

$f(x)=\alpha e^x$, with $\alpha=e^c$

Obviously you have to take care of domains for which this is true...

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  • $\begingroup$ thank you very much guys for a quick answer. Martigan, special thanks to you, since it was very helpful $\endgroup$ – Eu2718 Jan 7 '15 at 10:27
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Here it depends strongly on what you allow yourself to use. If "differentiation" implies solving ODEs then you know that since $f'=f$ is a homogeneous linear first-order ODE with constant coefficients, your space of solutions is a one-dimensional vector space. Once you know that $x\mapsto \mathrm e^x$ solves the problem (for example by using the series representation of $\mathrm e^x$) you obtain as your solution-space the space $$ \left\{f(x)=C~ \mathrm e^x, C\in \mathbb R\right\}. $$

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There is a general theorem that says that if $y'=F(y,t)$ and $F$ is Lipshitz in $y$ (that is, for two functions $f,g$, $|F(f,t)-F(g,t)|\leq M||f-g||$ for some $M$), then the ODE $y'=F(y,t), y(a)=c$ has at most one solution for any $a,c$.

In your case, $F(y,t)=y$ and therefore it is Lipshitz. You can therefore replace $M$ in the proof with 1.

The standard proof goes as follows:

Let $f,g$ be two solutions to the problem. Denote $\sigma(t)=(f(t)-g(t))^2$, and note that $\sigma(a)=0$.

We have $|\sigma'(t)|=|2(f'(t)-g'(t))(f(t)-g(t))=2(F(f,t)-F(g,t))(f(t)-g(t))|\leq 2M(f(t)-g(t))^2=2M\sigma(t)$.

We look at $(\sigma(t)e^{-2Mt})'=e^{-2Mt}(\sigma'(t)-2M\sigma(t))\leq 0$, therefore the function is decreasing. $\sigma(t)$ is always non-negative, $e^{-2Mt}$ is always positive, and $\sigma(a)=0$, therefore $\forall t\geq a, \sigma(t)e^{-2Mt}=0$, but that means that for all $t\geq a, \sigma(t)=0$.

For $t<a$, we do the same with $(\sigma(t)e^{2Mt})'\geq 0$, that is, this is a none decreasing function, that reaches 0 at $t=a$, but is non-negative for any $t<a$, and therefore $\sigma(t)\equiv 0$ everywhere.

$\sigma(t)=0$ everywhere means that $f-g=0$ everywhere.

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This is might be true only if $f'(x)=f(x)$ for every $x\in \mathbb{R}$. So lets rewrite it more properly:
$f$ is a differentiable function, $f:\mathbb{R} \rightarrow \mathbb{R}$ with fulfills: $f'(x)=f(x)$ for every $x\in \mathbb{R}$. show that $f(x) = 0$ or $f(x) = e^x$


Since $f$ is differentiable it is also continuous.
Since $f'(x)=f(x)$ ,$f^{(n)}(x)=f(x)$ and $f$ is differentiable $n$ times. Therefore according to Taylor's theorem, $f$ can be represented as:
$f(x) = \sum \frac{f^{(k)}(0)}{k!}x^k$
$\Downarrow $
$f(x) = \sum \frac{f(0)}{k!}x^k$
$\Downarrow $
$f(x) = f(0)\sum \frac{x^k}{k!} = f(0)e^{x}$
Therefore, $f$ must be of the form $Ce^{x}$, $C \in \mathbb{R}$

for $C=0,1$ we have $f = 0, f= e^x$ but also other functions for different values of $C$. For example $f= 2e^x$ and $f= \frac{1}{3}e^x$. Therefore the statement is not correct

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