How to prove that if i got a series $f \in \mathbb{R}[X,Y]$ that vanishes for all $X,Y \in [-1,1]$ then all his coefficients must be $0$ ?
Thank you for your answers.
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$X\in\mathbb{R}[X,Y]$ vanishes at all the points $(X,Y)=(0,y)$, for $y\in\mathbb{R}$, but it is not zero. – Jul 26 '18 at 03:11
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Not sure if this helps or not but I think you should only consider the case for uncountable zeros since if it is countable, we can construct the sequence $a_{n}=(-1)^{n}$ to obtain the divergence series $S_{n}$ with countably many zeros. – Evan William Chandra Jul 26 '18 at 03:12
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Yes of course you're both right, I edited my question. – Jul 26 '18 at 03:34
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See Zeros of polynomial over an infinite field. This answers the question for polynomials. If you speak about power series, converging for all $(x,y)\in [-1,1] \times [-1,1]$, the same holds true by the identity theorem for analytic functions in several variables.
Jens Schwaiger
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If $f=0$ in $(0,1)^2,$ then all partial derivatives of $f$ at $(0,0)$ are $0.$ Couple that with the following: If $f(x,y) = \sum_{m=0}^{M}\sum_{n=0}^{N} a_{mn}x^my^n,$ then $D_x^mD_y^n f(0,0) = m!n!a_{mn}.$
zhw.
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