Function representation of power series convergent at x = a

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).

1 Answer

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.

• 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. – Julia Kim Jan 29 at 3:16
• Yes, that is correct. – José Carlos Santos Jan 29 at 7:55