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SOME BACKGROUND INFO: Analytic functions may be (locally) represented by a convergent power/Taylor series. The domain is given by the interval in which the power series represents this function. For example, $f(x) = e^x$ has $domf = (-∞,∞)$.

In addition, it has also been said that every power series is the Taylor series of some $C^∞$ function.

My question is thus: suppose we had a power series centered at $a$, whose radius of convergence $R = a$. By the above, it must have a Taylor series representation of some analytic, $C^∞$ function. But, also by the above, the domain of this function must be ${a}$, a collapsed interval. How are these ideas compatible? (I feel as if there is some contradiction: a function defined only at one point cannot be differentiated an infinite number of times, and moreover, there would be an infinite number of functions that could be represented by this $R =a $ power series).

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Actually, what happens is that every power series with non-zero radius of convergence is the Taylor series of some $C^\infty$ function.

On the other hand, if a Taylor series is centered at $a(>0)$ and if its radius of convergence is $a$, then it converges on the interval $(0,2a)$ and its sum defines a $C^\infty$ function there.

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  • $\begingroup$ Much obliged for your help. Another question: If a power series has $R = 0$, then its coefficient terms, $c_n$, can be in function of anything, correct? That is, they cannot be $f^{(n)}(a)/n!$, as per the Taylor series definition, since such an $f(x)$ does not exist. $\endgroup$ – Julia Kim Jan 29 at 3:16
  • $\begingroup$ Yes, that is correct. $\endgroup$ – José Carlos Santos Jan 29 at 7:55

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