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 $|r_n(z)|\le R_n(\rho)\ \forall |z|\le\rho$?

  • 1
    $\begingroup$ No, it can't. Remember the (in)famous example $f(x)=\exp(-1/x^2)$ for $x\neq0$ and $f(0)=0$: all coefficients of the Taylor series around $x=0$ are zero, and the remainder is small for small $x$, but it can't be estimated from the coefficients. $\endgroup$
    – user436658
    Oct 6, 2017 at 19:20
  • $\begingroup$ @ProfessorVector Silly me, always forgetting to mention what Physicists implicitly assume, e.g. actual convergence and absence of singularities... $\endgroup$ Oct 6, 2017 at 19:37
  • $\begingroup$ Sorry, but my $f(x)$ is in no way singular as a real function, we know a formula for the coefficients ($c_n=0$) , the series actually converges (to $0$), and since we assumed we don't know $f(x)$ analytically, we don't even know the series converges to another value. $\endgroup$
    – user436658
    Oct 6, 2017 at 19:53
  • 1
    $\begingroup$ Well, in that case, you'd be well advised to include somewhere in your question that $z$ is a complex variable, and that the series has some (known?) positive radius of convergence. That radius also gives you the asymptotic behavior of the remainder, but hardly a rigorous estimate. $\endgroup$
    – user436658
    Oct 6, 2017 at 20:14
  • 1
    $\begingroup$ I still have difficulty understanding this question. $\endgroup$ Oct 9, 2017 at 19:34

2 Answers 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.

  • $\begingroup$ Thank you, that's an interesting approach. Of course by adding $10^{266}z^{N+1}$, you'd increase $M$ by up to $10^{266}\rho^{N+1}$ as well. Maybe a restriction to $c_n$ that are related via a finite analytical equation (e.g. a recurrence equation) would help here, since then such an irregular additional term would be excluded $\endgroup$ Dec 7, 2017 at 11:41
  • $\begingroup$ A recurrence relation for $f$ indeed helps. But often by reducing to the above estimate. For example, counting binary trees you wind up looking at a generating fct $f$ for which: $f(z)=1+zf(z)^2$ and you may show that for $|z|\leq \rho<1/4$ there is $M=M(\rho)<+\infty$ for which my suggested estimate holds, and it gives result pretty close to the real ones. When recurrence is in the coeffs $c_n$ there are undoubtedly similar estimates, but I don't have any (natural) example. $\endgroup$
    – H. H. Rugh
    Dec 7, 2017 at 17:59

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

The convergence radius $R$ is given by

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


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

(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...


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .