Vtyjkyuk

Suppose a function $f(t)$ is defined only on $[t_0,infty)$. Suppose all "right'' derivatives $f^(n)(t)$ exist, that is,
$$f^(1)(t_0)=lim_deltarightarrow 0+fracf(t_0)-f(t_0+delta)delta<infty,$$ and in general,

$$f^(n)(t_0)=lim_deltarightarrow 0+fracf^(n-1)(t_0)-f^(n-1)(t_0+delta)delta<infty,$$ for every $ngeq 1$.

Is Taylor's expansion,
$$
f(t_0)+sum_n=1^infty fracf^(n)(t_0)n!(t-t_0)^n,
$$ 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.

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 $Ysubseteq X$ 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.