Power series converging at $1$ with all derivatives zero

Question

It is well-known that there are non zero functions with derivatives of all orders at $1$ equal to zero, like $x \mapsto \exp\big(-\frac{1}{(x-1)^2}\big)$.

I'm trying to construct an explicit non zero series with a similar property, that is $F : x \mapsto \sum \limits_{k=0}^{\infty} a_k x^k$ converging for $|x| \le 1$ (including $x=1$), such that for all $k \ge 0$, $F^{(k)}(1) = 0$. Here $F^{(k)}(1)$ is defined as the limit (if it exists) of $F^{k}(z)$ when $z \to 1$ for $|z|<1$.

Is this possible with this interpretation? Is it possible if we take a derivative in the radial sense by taking $z$ to be real in the previous limit?

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A similar question was asked some years ago, but I do not know whether the given solution converges for $|x|=1$, since it isn't quite explicit: $$\exp\Big(\frac{-1}{(x-1)^2}\Big) \cdot e = 1 - 2 x - x^2 + \dfrac{2}{3} x^3 + \dfrac{13}{6} x^4 + \dfrac{41}{15} x^5 + \ldots$$

Some numerical tests show that the coefficients are not even converging to zero, but I have no way of checking this.

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Edit 2022/10/11

One additional comment on what I tried for $\exp\big(\frac{1}{x-1}\big)$, in order to get its Taylor series at $0$ and study its convergence at $1$. I am replicating Feng Qi's approach: with $B_{n,k}$ the Bell polynomials of the second kind, for $|x|<1$, \begin{align*} \bigl[e^{1/(x-1)}\bigr]^{(n)} &=\sum_{k=1}^ne^{1/(x-1)}B_{n,k}\biggl(\frac{-1!}{(x-1)^2}, \frac{2!}{(x-1)^3}, \dotsc,\frac{(-1)^{n-k+1}(n-k+1)!}{(x-1)^{n-k+2}}\biggr)\\ &=\sum_{k=1}^ne^{1/(x-1)}B_{n,k}\biggl(\frac{-1!}{(1-x)^2}, \frac{-2!}{(1-x)^3}, \dotsc,\frac{-(n-k+1)!}{(1-x)^{n-k+2}}\biggr)\\ &=e^{1/(x-1)}\sum_{k=1}^n\frac{(-1)^k}{(1-x)^{n+k}}B_{n,k}(1!, 2!, \dotsc,(n-k+1)!)\\ &=e^{1/(x-1)}\sum_{k=1}^n\frac{(-1)^k}{(1-x)^{n+k}}\binom{n}{k}\binom{n-1}{k-1}(n-k)!\\ &\to \frac{1}{e}\sum_{k=1}^n (-1)^k\binom{n}{k}\binom{n-1}{k-1}(n-k)!, \quad x\to0, \end{align*}

Consequently, the series expansion of $\exp\big(\frac{1}{x-1}\big)$ is $$\sum \limits_{n=0}^{\infty} \Biggl[\frac{1}{e} \sum_{k=1}^n \frac{(-1)^k}{k!}\binom{n-1}{k-1} \Biggl] x^n$$

Now the last step is to show that the sum of these coefficients converges (for $x=1$, in order to apply Abel's theorem). This doesn't seem obvious at all, considering that in the general term for the coefficients, the first terms of the sum grow arbitrarily large as $n$ increases.

Answer 1

Consider the singular inner function $H(z)=\exp{\frac{1+z}{z-1}}=\sum_{n \ge 0}a_nz^n, |z|<1$; since $\Re \frac{1+z}{z-1}<0, |z|<1$ it follows that $H$ is bounded by $1$ in the unit disc hence $\sum |a_n|^2\le 1$ (by Parseval).

In particular $\sum_{n \ge 0}{|a_n|/{n+1}} < \infty$ so if we take $F(z)=b_0+\sum_{n \ge 0}\frac{a_nz^{n+1}}{n+1}$ with $b_0$ chosen so $F(1)=0$ we have that $F$ converges absolutely on $|z| \le 1$

But $F^{(k)}=H^{(k-1)}, k \ge 1, |z|<1$ and it is very easy to see that for $r \to 1^-$ we have $H^{(m)}(r) \to 0, m \ge 0$ (and more generally $H^{(m)}(r)/(1-r)^k \to 0, k,m \ge 0$) so we can define $H^{(m)}(1)= 0, m \ge 1$ and we have $\frac{H^{(m)}(1)-H^{(m)}(r)}{1-r} \to 0, r \to 1^-$ so the extension of $F$ and all its derivatives at $1$ exists and is $0$ in the radial sense

(note that $H$ and all its derivatives are discontinuous at $1$ for more general convergence $z \to 1$ than radial)

Not sure that this is what OP wants as while the series of $F$ converges absolutely, the series of its derivatives do not and we can define their values only by radial limits at $1$, as they are discontinuous there in a general sense on the unit disc, but they are continuous and differentiable if we restrict to $[0,1]$

Edit later: Polya and Szego PTA I Pt III 4, Pr 250 has a class of examples $f_{\mu}(z), 0<\mu <1/2$ for which $f^{(k)}$ is continuous on the closed unit disc and analytic inside, while $f^{(k)}(1)=0$

For example for $\mu=1/4$ one can take:

$$f(z)=\int_0^{\infty}e^{-x^{1/4}/\sqrt 2}\sin (x^{1/4}/\sqrt 2)e^{-x(1-z)}dx$$

It is obvious that the integral is absolutely convergent for $\Re z \le 1$ so in particular on the closed unit disc and since $$f^{(k)}(z)=\int_0^{\infty}e^{-x^{1/4}/\sqrt 2}\sin (x^{1/4}/\sqrt 2)x^ke^{-x(1-z)}dx$$ same is true for all its derivatives, hence they are analytic on the open unit disc and continuous on the closed unit disc.

Now it is shown in the same book cited above,Problem 153 Pt III 1, (and follows by change of variables) that

$$\int_0^{\infty}e^{-x^{1/4}\cos \alpha}\sin (x^{1/4}\sin \alpha)x^kdx=4(4n+3)!\sin (4(k+1)\alpha), 0<\alpha <\pi/2, k \ge 0$$

so with $\alpha=\pi/4$ above we get that $f^{(k)}(1)=0$ for all $k \ge 0$ since $\sin (4(k+1)\pi/4)=0$ (!), while $|e^{-x^{1/4}/\sqrt 2}\sin (x^{1/4}/\sqrt 2)|<1$ so in particular if we let $z=-N$ for a large $N$ we have $$|f(-N-1)|\ge \int_0^{1}e^{-x^{1/4}/\sqrt 2}\sin (x^{1/4}/\sqrt 2)e^{-xN}dx-|\int_1^{\infty}e^{-x^{1/4}/\sqrt 2}\sin (x^{1/4}/\sqrt 2)e^{-xN}dx|$$ or $$|f(-N-1)| \ge \int_0^{1}e^{-x^{1/4}/\sqrt 2}\sin (x^{1/4}/\sqrt 2)e^{-xN}dx-\int_1^{\infty}e^{-xN}dx$$

But $$\int_0^{1}e^{-x^{1/4}/\sqrt 2}\sin (x^{1/4}/\sqrt 2)e^{-xN}dx \ge \int_0^{1/N}e^{-x^{1/4}/\sqrt 2}\sin (x^{1/4}/\sqrt 2)e^{-xN}dx \ge CN^{-5/4}$$

(where $C=8\sqrt 2e^{-2}/(5\pi)$ since the exponentials are at least $e^{-1}$ each and $\sin (x^{1/4}/\sqrt 2) \ge 2\sqrt 2 x^{1/4}/\pi $ when $0 \le x \le 1/N$ and the integrand is positive on $[0,1]$ as the sine term is)

Now $\int_1^{\infty}e^{-xN}=e^{-N}/N$ so $|f(-N-1)|>0$ for large $N$ hence $f$ doesn't vanish identically on $\Re z <1$, hence in the unit disc and while the same proof works for all derivatives we do not need it since, clearly $f$ cannot be a polynomial because of the vanishing at $1$ of all derivatives, so we are done!