Question

Equivalence of taylor series and its corresponding function

Answer 1

Is it valid to say that:
\[f(x)=\sum_{n=0}^{\infty}\frac{ 1 }{ n! }f ^{(n)}(a)(x-a)^n\]
for all values of x irrespective of whether the taylor series converges for all x? In other words, if we were to sum up all the infinite terms of the taylor series all at once, should we end up with f(x), irrespective of whether the series converges for that particular value of x?

Answer 2

Try it:)

Answer 3

So in order to sum an infinite number of things all at once is similar to the Axiom of Choice, in the respect that it is doing an infinite action since the Axiom of Choice is about taking something from each of infinite number of things. So assuming we use an axiom which allows us to sum an infinite number of things all at once (does such an axiom exist? is actually related to the AoC?), should we end up with the equivalence?

Answer 4

At first glance, it seems like the question is "Will the Taylor series converge for all values of \(x \) for any function \(f\), provided it converges on some interval?" If that is the question, unfortunately the answer is no. This idea is not unfounded though. This is true for all complex analytic functions.