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I have a function $f$ infinitely differentiable on $\mathbb{R}$ such as :

  • $f^{(n)}\geq 0$ for every $n$
  • $f(0)=f'(0)=f''(0)=1$

I have to show that $f = \exp$

With the first condition I remarked that $f$ must be analytical. But I can't go further.

(It's an exercise a friend of mine gave to me)

badinmaths
  • 486
  • What context did this arise in? Is this a contest problem of sorts, or an exercise, or something else? Including context elements is advised on this forum, so that users can potentially give more adequate answers or comments. – Bruno B Feb 09 '24 at 07:16
  • There are many more solutions than just $f(x)=e^x$, as it turns out. And it's not necessarily the case that functions satisfying these hypotheses must be analytic. – Greg Martin Feb 09 '24 at 07:28
  • Greg Martin of course $f$ is analytical. You can see https://math.stackexchange.com/questions/1193121/bernsteins-theorem-of-analytic-function-proof – badinmaths Feb 09 '24 at 07:34
  • Why can't you just use the Taylor series expansion? – Ekene E. Feb 09 '24 at 08:19
  • I tried but with no success. – badinmaths Feb 09 '24 at 08:35

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