Maclaurin series of $1-\left(\frac{\sin x}{x}\right) ^{\frac{2}{5}}$ has all coefficients positive

Question

I had a similar problem posted here, and after experimentation with WA I noticed that the function $1-\left(\frac{\sin x}{x}\right) ^{\frac{2}{5}}$ has a Maclaurin expansion with all coefficients positive.

It seems to hold. A possible approach might using the product expansion for the function $\sin x$. Or maybe a differential equation.

Thank you for your interest!

$\bf{Added:}$ Denote the expression by $y(x)$. It's enough to show that $z\colon=y'$ has a positive expansion. We have $$x\cdot z' = \left(\frac{x^2}{1 - x \cot x} + \frac{3}{5}( 1- x \cot x) - 2 \right)\cdot z$$

If the expression in the brackets has a positive expansion (it seems so) then we can show that $z$ has a positive expasion.

Answer 1

Not an answer, but here's my attempt: the function $f(x) = 1 - \left(\frac{\sin x}{x}\right)^{2/5}$ satisfies a differential equation $$ \frac{y'}{1-y} = -\frac{2}{5x}(x\cot x - 1) $$ so the coefficients of the series $f(x) = 1 + \sum_{k\geq 1} a_{2k}x^{2k}$ ($f$ is an even function) satisfy the recurrence relation $$ 2na_{2n} = \frac{2}{5}\left(b_{2n} - a_{2}b_{2n-2} - a_{4}b_{2n-4} - \cdots - a_{2n-2}b_{2} \right) $$ where $$ b_{2k} = \frac{(-1)^{k+1}B_{2k}2^{2k}}{(2k)!} $$ and $B_{2k}$ is Bernoulli number. Note that $b_{2k} > 0$ for all $k\geq 1$. Assume that the following inequality holds for all $n\geq 2$: $$ \sum_{1\leq k \leq n-1} \frac{1}{5k}b_{k}b_{n-k} \leq b_{n} $$ then one can prove $0\leq a_{n} \leq \frac{b_{2n}}{5n}$ for all $n$, by induction on $n$. However, I failed to prove the above inequaltiy, and I don't even know whether the inequality is true or not. I tried to use an identity $$ \sum_{1\leq k \leq n} b_{k}b_{n-k} = (2n+1)b_{n} $$ (for $n\geq 2$) which comes from $$ \coth x = \sum_{n \geq 0} \frac{2^{2n}B_{2n}x^{2n-1}}{(2n)!} $$ and $\frac{d}{dx}\coth x = 1 - \coth^{2}x$, but failed.

Answer 2

Using the same approach as earlier $$1-\left(\frac{\sin (x)}{x}\right)^a= \frac{a x^2}{6}\Bigg[1+\frac 1{60} \sum_{n=1}^\infty (-1)^n\frac{P_n(a)}{b_n }x^{2n}\Bigg]$$ where the first $b_n$ are $$\{1,126,15120,997920,16345929600,\cdots\}$$ and the first $P_n(a)$ are $$\left( \begin{array}{cc} n & P_n(a) \\ 1 & 5 a-2 \\ 2 & 35 a^2-42 a+16 \\ 3 & (5 a-4) \left(35 a^2-56 a+36\right) \\ 4 & 385 a^4-1540 a^3+2684 a^2-2288 a+768 \\ 5 & 175175 a^5-1051050 a^4+2862860 a^3-4252248 a^2+3327584 a-1061376 \end{array} \right)$$

and all coefficients are positive as long as $a \leq \frac 25$

Answer 3

I think I found a solution to the problem, by proving that the Maclaurin series of $$\frac{x^2}{1- x \cot x} + \frac{3}{5}(1- x \cot x) - 2$$ has all coefficients positive.

Now, from the answer of @Gary: we found out that there is the expansion $$\frac{x^2}{1- x \cot x} = 3 - \frac{1}{5} x^2 + \sum_{n=2}^{\infty} \frac{(-1)^n 3 \cdot 2^n V_{2n}}{(2n)!} x^{2n}$$ where $V_n$ are the van der Pol numbers. Moreover we have the inequality $$|V_{2n}|< \frac{4 (2n)!}{15 (2\pi)^{2n}}$$

(See : Howard -- van der Pol numbers and polynomials).

Moreover, we have the expansion

$$1 - x \cot x = \sum_{n=1}^{\infty} \frac{(-1)^{n-1} 2^{2n} B_{2n}} {(2n)!} x^{2n} $$

Now I think we are able to prove that the series for the above expression has all of the coefficients positive. Details to follow.

Answer 4

In the arXiv preprint [1] below, the following series expansions were established.

When $r\ge0$, the series expansions \begin{equation}\label{recip-sin-ser-closed-eq} \biggl(\frac{\sin z}z\biggr)^r=1+\sum_{q=1}^{\infty}(-1)^q\Biggl[\sum_{k=1}^{2q}\frac{(-r)_k}{k!} \sum_{j=1}^k(-1)^j\binom{k}{j} \frac{T(2q+j,j)}{\binom{2q+j}{j}}\Biggr]\frac{(2z)^{2q}}{(2q)!} \end{equation} and \begin{equation}\label{recip-sin-stirl-closed-eq} \biggl(\frac{\sin z}z\biggr)^r=1+\sum_{q=1}^{\infty}(-1)^q\Biggl[\sum_{k=1}^{2q}\frac{(-r)_k}{k!} \sum_{j=1}^k(-1)^j\binom{k}{j} \sum_{m=0}^{2q}(-1)^{m}\binom{2q}{m} \biggl(\frac{j}{2}\biggr)^{m} \frac{S(2q+j-m,j)} {\binom{2q+j-m}{j}}\Biggr]\frac{(2z)^{2q}}{(2q)!} \end{equation} are convergent in $z\in\mathbb{C}$, where $T(n,k)$ and $S(n,k)$ denote the central factorial numbers and the Stirling numbers of the second kinds, and the rising factorial $(r)_k$ is defined by \begin{equation*}%\label{rising-Factorial} (r)_k=\prod_{\ell=0}^{k-1}(r+\ell) = \begin{cases} r(r+1)\dotsm(r+k-1), & k\ge1;\\ 1, & k=0. \end{cases} \end{equation*} When $r<0$, the above two series expansions are convergent in $|z|<\pi$.

However, by virtue of these two series expansions applied to $r=\frac25$, it is very difficult to verify that the function $1-\bigl(\frac{\sin z}z\bigr)^{2/5}$ has a Maclaurin expansion with all coefficients positive.

Reference

1. F. Qi and P. Taylor, Several series expansions for real powers and several formulas for partial Bell polynomials of sinc and sinhc functions in terms of central factorial and Stirling numbers of second kind, arXiv (2022), available online at https://arxiv.org/abs/2204.05612v4 or https://doi.org/10.48550/arXiv.2204.05612.