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Sometimes we use $$ f(x)\equiv\sum_{-\infty}^{\infty} C_n x^n=0 \quad \forall x \quad \Rightarrow \quad C_n=0 \quad \forall_{n} $$ If the summation is $\sum_0^{\infty}$, its proof is easy.
$$ f^{(n)}(0)=0 \quad \forall_{n\ge0} \quad \Rightarrow \quad C_n =0 \quad \forall_{n\ge0} $$ But when the summation is $\sum_{-\infty}^{\infty}$, how can we prove?
Or is this a kind of an assumption?
Thanks.

GotchaP
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  • If you're able to consider complex x then the zero function $f(x) = 0$ has zero residue, as so all $x^{m}f(x)$ for $m \in \mathbb{Z}$. If you're required to stay in $\mathbb{R}$ then if $C_{n} \neq 0$ for $n \in \mathbb{Z}^{-}$ what would $f(0)$ look like? – AlphaNumeric Dec 21 '16 at 14:39
  • You need to specify some region of convergence. – copper.hat Dec 21 '16 at 15:25
  • @copper.hat: If I choose the region in the vicinity of $x=0$, is it a bad choice? – GotchaP Dec 21 '16 at 15:58
  • @AlphaNumeric: When $x\in \mathbb{C}$, unless we show $C_n=0$ for all ${n\le m<0}$, I think we cannot calculate residue... Why $C_n=0$ for all ${n\le m<0}$? And when $x\in \mathbb{R}$, $|f(x)|\to \infty,(x\to 0)$. But how can we exclude the possibility that $\sum_{-\infty}^{-1} C_n x^n=0$ for $C_n \ne 0$? – GotchaP Dec 22 '16 at 11:41

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