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 paper [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. Feng Qi and Peter Taylor, Series expansions for powers of sinc function and closed-form expressions for specific partial Bell polynomials, Applicable Analysis and Discrete Mathematics 18 (2024), no. 1, 92–115; available online at https://doi.org/10.2298/AADM230902020Q.