Can the remainder of a Taylor expansion be estimated from the coefficients?

Question

Given a formula for the coefficients $c_n\in\mathbb C$ of a not analytically known function $f:\mathbb C\to\mathbb C, z\mapsto f(z)$'s Taylor series, is there any way to estimate the remainder term of the order $N$

$$r_N(z) := f(z) - \sum_{n=0}^N c_n z^n$$

within a given radius $|z|\le\rho$ (truly smaller than the convergence radius) that only depends on a finite amount of the $c_n$ or some limit thereof? In other words, is there a way to obtain a $R_N$ such that $|r_n(z)|\le R_n(\rho)\ \forall |z|\le\rho$?

Answer 1

Let's give this a shot, also to clarify the question and hopefully attract better answers...

The convergence radius $R$ is given by

$$\frac 1R = \limsup_{n\to\infty} |c_n|^{\frac1n}.$$

Therefore,

$$\forall N\in\mathbb N\,\exists \epsilon_N>0\forall n>N:|c_n|^{\frac1n} < \frac{1+\epsilon_N}R \tag{*}\label{*}$$

(and $\epsilon_{n>N}\le\epsilon_N$ and $\lim\limits_{N\to\infty}\epsilon_N = 0$).

Thus for $|z|<R/(1+\epsilon_N)$,

$$\begin{align*} |r_{N-1}(z)| &= \Bigg|\sum_{n=N}^\infty c_n z^n\Bigg| \\ &\le \sum_{n=N}^\infty |c_n|\cdot |z|^n \\ &\stackrel{\eqref{*}}< \sum_{n=N}^\infty \underbrace{(1+\epsilon_N)^n \left|\frac zR\right|^n}_{=:(\zeta_N)^n} \tag{#}\label{#} \\ &= \frac{(\zeta_N)^N}{1-\zeta_N} < \frac{R}{R-(1+\epsilon_N)|z|} \end{align*}$$

The final inequality is probably too generous... Note that $\eqref{#}$ converges due to $|z|<R/(1+\epsilon_N) \Leftrightarrow \zeta_N<1$. The same procedure can probably be applied to estimate the remainder of a Laurent series' principal part using $r = \limsup\limits_{n\to\infty}|c_{-n}|^{\frac1n}$.

It's probably not a spectacular boundary, so I hope someone else knows a better one...

Answer 2

It seems to depend strongly upon what knowledge made you conclude that the radius of convergence is at least $\rho$? If for example, you happen to know that the function is analytic on a disk $D(0,\rho)$ and is uniformly bounded on that disk, say $|f(z)|\leq M<+\infty$ for all $|z|<\rho$ then indeed you can estimate the remainder. From Cauchy, integrating over a circle of radius $\rho-\epsilon$ and for $|z|<\rho$: $$ f(z) = \oint_{\partial B} \frac{f(u)}{u-z}\frac{du}{2\pi i} = \oint_{\partial B} \frac{f(u)}{u} \left( 1+ \frac{z}{u} + \cdots \right) \frac{du}{2\pi i}=\sum_{n=0}^{N} c_nz^n + R_N(z) $$ where (after taking the limit $\epsilon\rightarrow 0$): $$ |R_N(z)| \leq \left|\oint_{\partial B} \frac{f(u)}{u} \frac{z^{N+1}}{u^{N+1}}\frac{1}{u-z} \frac{du}{2\pi i}\right| \leq M \left(\frac{|z|}{\rho}\right)^{N+1} \frac{1}{\rho-|z|}$$ Without any uniform bounds on $|f|$ (or the sequence $c_n$), I don't have any reasonable suggestion. After all you may always add $10^{266} z^{N+1}$ without changing the radius of convergence.