Maclaurin series of $1-\left(\frac{\sin x}{x}\right) ^{\frac{2}{5}}$ has all coefficients positive 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.
 A: 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 $a \leq \frac 25$
A: 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.
A: 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.
