Taylor series remainder equaling zero

Question

I asked this question a while ago and I figured out my initial problem. I basically learned that convergence of series doesn't mean that it converges to that function since there could be error

My question now is, if there is no error between a function and Taylor series, then does that mean that Taylor series is equal to that function (at least on that interval?)

And, for example, for $e^x$ or $\sin(x)$. if I prove the error is zero, then do I need to prove the radius of convergence of that series (perhaps by ratio test?)?

Answer 1

Look at the case of $e^x$. The $n$th Taylor polynomial is $$ T_n (x) = \sum\limits_{k = 0}^n {\frac{{x^k }}{{k!}}} . $$ It is easy to show, using the Lagrange form of the remainder, that $$ \left| {e^x - T_n (x)} \right| \le \frac{{x^{n + 1} }}{{(n + 1)!}}\max (1,e^x ) $$ for any real $x$. The right-hand side converges point-wise to zero as $n\to +\infty$. Thus, the function sequence $T_n(x)$ converges to $e^x$ for any real $x$, i.e., $\lim_{n\to +\infty}T_n(x)=e^x$ for any real $x$. By the definition of an infinite series $$ \mathop {\lim }\limits_{n \to + \infty } T_n (x) =: \sum\limits_{k = 0}^\infty {\frac{{x^k }}{{k!}}} . $$ Consequently, $$ e^x = \sum\limits_{k = 0}^\infty {\frac{{x^k }}{{k!}}} $$ for any real $x$. You do not need to look at the radius of convergence, because we have just shown that this equality holds for all real $x$, i.e., the series on the right-hand side converges for all real $x$ (and its sum is $e^x$). Does this answer your question?