In real analysis and complex analysis, branches of mathematics, the identity theorem for analytic functions states: given functions f and g analytic on a domain D (open and connected subset of or ), if f = g on some , where has an accumulation point in D, then f = g on D. Thus an analytic function is completely determined by its values on a single open neighborhood in D, or even a countable subset of D (provided this contains a converging sequence together with its limit). This is not true in general for real-differentiable functions, even infinitely real-differentiable functions. In comparison, analytic functions are a much more rigid notion. Informally, one sometimes summarizes the theorem by saying analytic functions are "hard" (as opposed to, say, continuous functions which are "soft"). The underpinning fact from which the theorem is established is the expandability of a holomorphic function into its Taylor series. The connectedness assumption on the domain D is necessary. For example, if D consists of two disjoint open sets, can be on one open set, and on another, while is on one, and on another. If two holomorphic functions and on a domain D agree on a set S which has an accumulation point in , then on a disk in centered at . To prove this, it is enough to show that for all . If this is not the case, let be the smallest nonnegative integer with . By holomorphy, we have the following Taylor series representation in some open neighborhood U of : By continuity, is non-zero in some small open disk around . But then on the punctured set . This contradicts the assumption that is an accumulation point of . This lemma shows that for a complex number , the fiber is a discrete (and therefore countable) set, unless . Define the set on which and have the same Taylor expansion: We'll show is nonempty, open, and closed. Then by connectedness of , must be all of , which implies on . By the lemma, in a disk centered at in , they have the same Taylor series at , so , is nonempty.

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