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.


1 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.

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    $\begingroup$ I think you are saying the opposite of what I am asking. If a function is defined on a larger set and has a convergent Taylor's series then of course the series is convergent on a restricted smaller set. My problem is that, when $t<t_{0}$, $f(t)$ is not defined, can we still talk about the series' convergence. The radius of convergence implies convergence in an open neighborhood of $t_0$. Maybe I am asking a non-sensical question, but I am thinking about convergence in something like $(t_0-\varepsilon, t_0+ε)\cap [t_0,\infty)$, avoiding the fact that $f(t)$ has no definition for $t<t_{0}$. $\endgroup$ Sep 4, 2018 at 8:25
  • $\begingroup$ The answer to the question you ask in the third sentence is yes. To put it another way, if you have some function $f$ defined on a set $Y$ in a way where insights can be gained from a natural extension to a larger set $X\supseteq Y$, then doing so with a larger function $F$ on $X$ and analysing the behaviour of $F$ on $Y$ alone will often produce equivalent true statements about $f$. To be fair, usually a bit of extra work is required to show that $f$ inherits true statements in a certain way from $F$, and you can view that as an exercise in this case. $\endgroup$ Sep 4, 2018 at 18:27

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