show this inequality with $\frac{d^i}{dx^i}\left(1-\left(\frac{-x}{\ln(1-x)}\right)^{1/K}\right) \Bigg|_{x=0}>0, ~~~\forall i\in N^{+}$

Question

I am trying to solve this Komal problem 661:

Let $K$ be a fixed positive integer. Let $(a_{0},a_{1},\cdots )$ be the sequence of real numbers that satisfies $a_{0}=-1$ and $$\sum_{i_{0},i_{1},\cdots,i_{K}\ge 0,i_{0}+i_{1}+\cdots+i_{K}=n}\dfrac{a_{i_{1}}a_{i_{2}}\cdots a_{i_{K}}}{i_{0}+1}=0$$ for every postive integer $n$. Show that $a_{n}>0$ for $n\ge 1$.

Add edit:For the Iosif Pinelis point out,$b_{0}=(-1)^K$,Now I have known where my wrong,and Ira Gessel point that, Now How to prove

Let $\displaystyle f(x)\triangleq\sum_{i\geq 0} a_i x^i$ and $\displaystyle g(x)\triangleq \sum_{i\geq 0} \cfrac{x^{i}}{i+1}$.

Then, we get $$ f(x)^Kg(x) = \sum_{n\geq 0}b_nx^n \text{ with } b_n=\sum_{\substack{i_{0},i_{1},\cdots,i_{K}\ge 0\\i_{0}+i_{1}+\cdots+i_{K}=n}}\dfrac{a_{i_{1}}a_{i_{2}}\cdots a_{i_{K}}}{i_{0}+1}.$$ Since $b_n=0$ for $n\geq 1$, we get \begin{align} f(x)^Kg(x)&=b_0=(-1)^K\\ \implies \left(\sum_{i\geq 0} a_i x^i\right)^K&=\frac{(-1)^K}{g(x)}=\frac{-(-1)^K\cdot x}{\sum_{i\geq 1} -\cfrac{x^{i}}{i}}=\frac{-(-1)^K\cdot x}{\ln(1-x)}\\ \implies\sum_{i\geq 1} a_i x^i &=-a_{0}+ \left(\frac{-(-1)^Kx}{\ln(1-x)}\right)^{1/K} =1-\left(\dfrac{-x}{\ln{(1-x)}}\right)^{1/K}\end{align} Thus, using the Tyalor series expansion,

$$\dfrac{d^i}{dx^i}\left(1-\left(-\dfrac{x}{\ln{(1-x)}}\right)^{\frac{1}{K}}\right)|_{x=0}>0\tag{1}$$

But the last maybe it not easy prove it,can help me to prove $(1)$?Thanks

Answer 1

The function $f(x) = (x - 1) / \log x$ extends to a holomorphic function on $\mathbb C \setminus (-\infty, 0]$. Clearly, $f(x) > 0$ when $x > 0$. We claim that $\operatorname{Im} f(z) \geqslant 0$ when $\operatorname{Im} z > 0$, that is, $f$ is a complete Bernstein function. For the proof of this claim, see the last two paragraphs of this answer.

Similarly, $g(x) = x^{1/K}$ is a complete Bernstein function: $g(x) > 0$ for $x > 0$, and $g$ extends to a holomorphic function on $\mathbb C \setminus (-\infty, 0]$, and we have $\operatorname{Im} g(z) \geqslant 0$ when $\operatorname{Im} z > 0$.

It follows that $h(x) = g(f(x))$ is a complete Bernstein function. In particular, $h$ is a Bernstein function: $(-1)^{n - 1} h^{(n)}(x) \geqslant 0$ for $x > 0$ and $n \geqslant 1$. We conclude that $F(x) = -h(1 - x) = -(-x / \log(1 - x))^{1/K}$ has all derivatives nonnegative for $x < 1$: $F^{(n)}(x) \geqslant 0$ when $x < 1$ and $n \geqslant 0$, as desired.

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Proof of the claim: My favourite way to prove this kind of results is to observe that $\operatorname{Arg} \log z$ is a bounded harmonic function in the upper complex half-plane with boundary values $0$ on $(1, \infty)$, $\pi$ on $(0, 1)$ and something in $[0, \pi]$ on $(-\infty, 0)$. Similarly, $\operatorname{Arg} (z - 1)$ is a bounded harmonic function in the upper complex half-plane with boundary values $0$ on $(1, \infty)$ and $\pi$ on $(-\infty, 1)$. Thus, $$\operatorname{Arg} (z - 1) - \operatorname{Arg} \log z = \operatorname{Arg} f(z)$$ is a bounded harmonic function in the upper half-plane with boundary values $0$ on $(0, \infty)$ and something in $[0, \pi]$ on $(-\infty, 0)$. This proves that $\operatorname{Arg} f(z)$ is well-defined and belongs to $[0, \pi]$ in the upper complex half-plane.

An alternative approach would be to write the representation of the logarithm as a extended complete Bernstein function (or simply a Nevanlinna–Pick function, to give another keyword): $$ \log z = \int_0^\infty \biggl( \frac{z}{z + s} - \frac{1}{1 + s} \biggr) \frac{1}{s} \, ds . $$ It follows that $$ \tilde f(z) = \frac{z \log z}{z - 1} = \int_0^\infty \frac{z}{z + s} \, \frac{s}{1 + s} \, \frac{1}{s} \, ds $$ is a complete Bernstein function, and consequently $f(z) = z / \tilde f(z)$ is a complete Bernstein function, too.

For more on Bernstein functions, complete Bernstein functions and related topics, see the excellent book Bernstein Functions. Theory and Applications by René Schilling, Renming Song and Zoran Vondraček.

Answer 2

A mistake in your reasoning is that $b_0=(-1)^K$, rather than $b_0=-1$.