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I have just started learning about differential equations, as a result I started to think about this question but couldn't get anywhere. So I googled and wasn't able to find any particularly helpful results. I am more interested in the reason or method rather than the actual answer. Also I do not know if there even is a solution to this but if there isn't I am just as interested to hear why not.

Is there a solution to the differential equation:

$$f(x)=\sum_{n=1}^\infty f^{(n)}(x)$$

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    $\begingroup$ Assume everything works well, and differentiate the equation. Looking sharply, you will see a first order ODE emerging. $\endgroup$ – Daniel Fischer Feb 6 '16 at 11:45
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    $\begingroup$ If you differentiate the equation, and assume that everything behaves nicely, what equation do you obtain? $\endgroup$ – Daniel Fischer Feb 6 '16 at 11:50
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    $\begingroup$ Right. And now look, how does that right hand side differ from the original right hand side? $\endgroup$ – Daniel Fischer Feb 6 '16 at 11:56
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    $\begingroup$ @DanielFischer By $f'(x)$. Which gives $f'(x)=f(x)-f'(x)$ therefore $f(x)=2 f'(x)$. Ah Ok got it. Thank you very much!! $\endgroup$ – o.comp Feb 6 '16 at 11:58
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    $\begingroup$ And don't forget the boring solution (for all homogeneous linear differential equations): $0$. $\endgroup$ – Eric Towers Feb 6 '16 at 21:04
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$f(x)=\exp(\frac{1}{2}x)$ is such a function, since $f^{(n)}=2^{-n} f(x)$, you have

$$\sum_{n=1}^\infty f^{(n)}(x)=\sum_{n=1}^\infty 2^{-n}f(x)=(2-1)f(x)=f(x)$$

This is the only function (up to a constant prefactor) for which $\sum_{n}f^{(n)}$ and its derivatives converge uniformly (on compacta), as $$f'=\sum_{n=1}^\infty f^{(n+1)}=f-f'$$ follows from this assumption. But this is the same as $f-2 f'=0$, of which the only (real) solutions are $f(x)= C \exp{\frac{x}{2}}$ for some $C \in \mathbb R$.

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    $\begingroup$ Being little pedantic: the constant zero function would work too. $\endgroup$ – DBS Feb 6 '16 at 14:20
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    $\begingroup$ @DBS this is covered by $C=0$. $\endgroup$ – s.harp Feb 6 '16 at 14:20
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    $\begingroup$ Oops you are right. Sorry my bad. $\endgroup$ – DBS Feb 6 '16 at 14:42
  • $\begingroup$ Does it remain to determine whether there are some more solutions where the convergence of the series is (point-wise but) not uniform, then? $\endgroup$ – Jeppe Stig Nielsen Feb 7 '16 at 9:24
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    $\begingroup$ It remains to be determined :) $\endgroup$ – s.harp Feb 8 '16 at 10:30
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Differentiate both sides to get $f'(x)=f''(x)+f^{(3)}(x)+...$

So the starting equation becomes $f(x)=f'(x)+f'(x)\Rightarrow f(x)=2f'(x)$

Multiply now both sides by $e^{-\frac{x}{2}}$ and this becomes
$$[e^{-\frac{x}{2}}f(x)]'=0$$
So $f(x)=ce^{\frac{x}{2}}$
Done.

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the question means: $$y-y'-y''-y'''-......y^n=0$$ the differential equation is homogeneous and the characteristics equation is $$1-r-r^2-r^3-.......r^n=0$$ $$r(1+r+r^2+r^3+....)=1$$ by using the geometric series (r<1) $$\frac{r}{1-r}=1$$ $$r=1-r$$ $$r=\frac{1}{2}$$ so the function is $$y=Ce^{\frac{x}{2}}$$

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