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Given the ODE $f(x)=f'(x)$ one obtains that its only solution is $ce^x$ for some constant $c$.

So if one adds the constraint $f(0) = 0$, it's easy to check that there's no way to bend the particular solution to generate this value besides setting $c=0$.

Technically, we should have enough information to reason this without any use of our knowledge about ODEs, simply by proving that the function can't be not constant.

So by basic method, I mean a method as simple as possible, but at least one that's not using knowledge about the possible functions solving the ODE.

A few examples of possible answers, which in my opinion still use too much machinery for the simplicity of the problem:

1) If the solution were analytical, its Taylor polynomial simply would be 0.

2) We can make an iterative approximation by splitting the interval $[0,b]$ into $n$ pieces, and for every piece add the increase caused by $f$'s gradient to the function (which trivially means $f$ will stay constant throughout the whole iteration). We can then show that if we let $n\to\infty$, the error of the approximation goes to 0.

I really feel like there should be a simply and easy solution to this: After all, we know no less than
$0=f(0)= f'(0)= f''(0)= ...$

Which intuitively, combined with $f$'s continuity already seems to force $0$ to be the only solution.

Sudix
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    $0=f(0)=f'(0)=f''(0)=\ldots$ does not ensure that $f$ is $0$ (there is the famous example $e^{-\frac{1}{x^2}}$). – Delta-u Jul 12 '18 at 11:28
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    Compare https://math.stackexchange.com/questions/58097/proof-that-c-expx-is-the-only-set-of-functions-for-which-fx-fx: The derivative of $f(x)e^{-x}$ is zero ... – Martin R Jul 12 '18 at 11:28
  • @Delta-u That's a nice counter example - thanks for proving my intuition wrong! – Sudix Jul 12 '18 at 11:41
  • @MartinR A few of the answers to that question are truly simple and aesthetic. The only thing there I dislike is that (of cause) all solutions show that the solution has to have a particular form, whereas here I hope to find a reasoning "The function can not be not constant, as this would induce a contradiction with the definition of continuity/differentiability" – Sudix Jul 12 '18 at 11:44
  • Another one: https://math.stackexchange.com/questions/406686/fx-fx-and-f0-0-implies-that-fx-0-formal-proof. – Martin R Jul 12 '18 at 11:49

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