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Apologies if the phrasing is poor..

If we have two functions f and g with the same domain X, is the following statement correct:

If there is exists a point a within X such that $f^n$(a) = $g^n$(a) for all integers n > 0, then f(x) = g(x) for all x in X? Basically if there is a point where you know the derivitaves, of all order, are equal for both functions, does that mean the functions are identical across the domain?

If so - what is the proof? Thanks!

Sam White
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1 Answers1

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If X is $\mathbb R$, then the canonical counter-example is $f(x)=0$ everywhere, and $g(x)=e^{-1/x^2}$ (with $g(0)=0$). Both functions have the property that the derivatives of all orders are zero at $x=0$.

If X is $\mathbb C$, however, this counterexample doesn't work, because $g$ has an essential singularity at $x=0$. And in fact if X is $\mathbb C$, then $f$ and $g$ must be identical. For a proof of this, consult a textbook on complex analysis.

TonyK
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  • Ahh great thanks - this feels counter intuitive but I can't argue with that. Do you know anywhere I can find the proof for the domain in C? – Sam White Sep 03 '22 at 17:29
  • Google "complex analysis" -- there's plenty of material out there. Briefly, all derivatives of the function $h=f-g$ are zero at a; in $\mathbb C$ (but not in $\mathbb R$), this guarantees that $h$ is everywhere zero. This has to do with the fact that in $\mathbb C$, a differentiable function has a unique Taylor series about every point. – TonyK Sep 03 '22 at 18:07