Find all holomorphic functions at $c=0$ , $f:\mathbb{R}\rightarrow\mathbb{R}$ that satisfy $f(x)=f'(x)$
From my understanding the only function that satisfies that condition is $f(x)=e^x$. (?) How do I show that? I want to show it using its Taylor series obviously but am not sure how to formally prove this. Thank you :)
It's Taylor series at $0$ would be $f(0)\cdot \sum_{n=1}^\infty \frac {x^n}{n!}$ since all the derivatives are $f(x)$...
I think you can do it this way: $$f(x)=f'(x)$$ If $f(x) \neq 0$: $$\frac{f'(x)}{f(x)}=1$$ $$\left[\log(f(x))\right]'=1$$ $$\left[\log(f(x))\right]'=[x+c]'$$ $$\log(f(x))=x+c$$ $$\exp(\log(f(x)))=\exp(x+c)$$ $$f(x)=c\exp(x)$$
There's also a cute trick we can borrow from the differential equations handbook, one that we can use without having to assume $f$ is analytic. Suppose the equation $$f'(x) - f(x) = 0$$ holds true for all $x$. Then this implies that, $$f'(x) e^{-x} - f(x) e^{-x} = 0.$$ But, using the product rule, the left hand side turns into $$(e^{-x} f(x))' = 0.$$ It follows therefore that, for some $c \in \mathbb{R}$, $$e^{-x} f(x) = c \implies f(x) = c e^{-x}.$$ This is just the integrating factor technique from first order linear ODEs!
