One-sided Taylor's expansion

Question

Suppose a function $f(t)$ is defined only on $[t_{0},\infty)$. Suppose all "right'' derivatives $f^{(n)}(t)$ exist, that is, $$f^{(1)}(t_{0})=\lim_{\delta\rightarrow 0+}\frac{f(t_{0})-f(t_{0}+\delta)}{\delta}<\infty,$$ and in general,
$$f^{(n)}(t_{0})=\lim_{\delta\rightarrow 0+}\frac{f^{(n-1)}(t_{0})-f^{(n-1)}(t_{0}+\delta)}{\delta}<\infty,$$ for every $n\geq 1$.

Is Taylor's expansion, $$ f(t_{0})+\sum_{n=1}^{\infty} \frac{f^{(n)}(t_{0})}{n!}(t-t_{0})^n, $$ defined on $[t_{0},t_{0}+\varepsilon)$ for some $\varepsilon$?

I am not asking whether the expansion converges to $f(t)$ (that is a different question) in a neighborhood of $t_{0}$. I am just asking if $f(t)$ can be expanded only to the right hand side without the left hand side having been defined.

Intuitively one would think so, but I cannot find any literature items specifically on this point. All Taylor's expansions seem to assume all derivatives exist in an open neighborhood of $t_{0}$. Please enlighten me.

Answer 1

Your intuition is correct, in the sense that you may generally take a function which is defined on some set $X$ and restrict it to a smaller subset $Y\subseteq X$ unconditionally.

If this series has nonzero radius of convergence, then there is a function defined on a neighbourhood of $t_0$ which corresponds to the function given in your question on some $(t_0-\epsilon,t_0+\epsilon)$ with those (two-sided) derivatives. Then what you have is just a restriction of the function defined on that interval to the right side, which is perfectly consistent.

Admittedly I don't know if a function can even exist which defines such a series that has zero radius of convergence, but given what you're asking I don't think it's relevant anyway.