The Maclaurin series of $1-(1-\frac{x^2}{2} + \frac{x^4}{24})^{2/3}$ has all coefficients positive

Question

It was shown in a previous post that the Maclaurin series of $1 - \cos^{2/3} x$ has positive coefficients. There @Dr. Wolfgang Hintze: has noticed that the truncation $1- \frac{x^2}{2} + \frac{x^4}{24}$ can be substituted for $\cos x$ ( seems to be true for all the truncations). The proof is escaping me. Thank you for your attention!

$\bf{Added:}$ Thomas Laffey in this paper directs to a proof of the fact that if $a_1$, $\ldots$, $a_n\ge 0$ then $\alpha = \frac{1}{n}$ makes the following series positive:

$$1- (\prod_{i=1}^n (1- a_i x))^{\alpha}$$

Numerical testing suggests that $\alpha = \frac{\sum a_i^2}{(\sum a_i)^2} \ge \frac{1}{n}$ works as well ( see the case $n=2$ tested here). So in our case, instead of $\alpha = \frac{1}{2}$ we can take $\alpha = \frac{2}{3}$. Clearly, this would then be the optimal value. This would be a test case for $n=2$. The result for $\cos x$ used the special properties of the function ( solution of a certain differential equation of second order). Maybe $1- x/2 + x^2/24$ is as general as any quadratic with two positive (distinct) roots.

Answer 1

Hint

Consider $$f(a)=1-\left(1-\frac{x^2}{2}+\frac{x^4}{24}\right)^a$$ and use the binomial expansion.

It should write $$f(a)=\frac a 2 \sum_{n=1}^\infty\frac {P_n(a)}{b_n} x^{2n}$$ You should see it very quickly.

Edit

After your edit and remarks, considering $$f(k)=1-\big[ (1-ax)(1-b x)\big]^k$$ let $b=a c$ to obtain $$f(k)=k\sum_{n=1}^\infty \frac{a^n}{n!}\, Q_n(c)\, x^n$$ with $$Q_1(c)=c+1 \qquad\text{and} \qquad Q_2(c)=(c^2+1)-k (c+1)^2$$ If, as you did, we choose to cancel $Q_2(c)$ that is to say to use $k=\frac{c^2+1}{(c+1)^2}$ we have $$\color{blue}{Q_n(c)= (n-2)\, 2^{\left\lfloor \frac{n-1}{2}\right\rfloor }\, \frac{c\, (c^2-1)^2}{(c+1)^{n}} \,\,R_n(c)}$$ with $$\left( \begin{array}{cc} n & R_n(c) \\ 3 & 1 \\ 4 & c^2+3 c+1 \\ 5 & \left(c^2+c+1\right) \left(c^2+4 c+1\right) \\ 6 & \left(c^2+1\right) \left(c^2+4 c+1\right) \left(3 c^2+8 c+3\right) \\ 7 & \left(c^2+3 c+1\right) \left(c^2+4 c+1\right) \left(6 c^4+7 c^3+6 c^2+7 c+6\right) \end{array} \right)$$

Answer 2

After testing with WolframAlpha I realized we can prove even more.

First, it is enough to theck that the derivative of $f(x) = 1- (1- x/2 + x^2/24)^{2/3}$ has positive coefficients. Even more, it is enough to check that $f'(x) = c \cdot \exp(g(x))$, where $g$ has positive coefficients. But that is equivalent to showing that $\frac{f''(x)}{f'(x)}$ has. The big advantage here is that this fraction is rational, and is relatively easy to deal with. It is in fact better to consider the general case of $$f(x) = 1 - ((1-a x)(1- b x))^p$$ where $p$ is an arbitrary exponent. We may also assume $b=1$. We get $$\frac{f''(x)}{f'(x)} = \frac{1 - p}{1 - x} - \frac{2 a}{1 + a - 2 a x} + \frac{a (1 - p)}{1 - a x}$$

Now for $n\ge 1$ the coefficient of $x^{n-1}$ equals $$(1-p)(1+a^n) - \frac{(2 a)^n}{(a+1)^n}$$

Now choose $p=\frac{a^2+1}{(a+1)^2}$. We need to show that $$\frac{2 a}{(1+a)^2}\ge \frac{ (2 a)^n}{(1+a^n)(1+a)^n}$$ for $n\ge 1$ and $a \ge 0$. Note that for $n=1$ we have equality. (Also note that both sides do not change under $a\mapsto \frac{1}{a}$). To show the inequality for $n\ge 1$ it enough to show that the function $t \mapsto \frac{ (2 a)^t}{(1+a^t)(1+a)^t}$ is decreasing for $a\ge 1$ on $[1, \infty)$. Its derivative equals

$$\frac{ (\frac {2a}{a+1})^t (\log \frac{2a}{a+1} - a^t \log \frac{a+1}{2})}{(a^t+1)^2}$$

Now observe that $a^t\ge 1$ and $\frac{a+1}{2}\ge \frac{2a}{a+1} \ge 1$, so the top large bracket is negative. We are done.

Note: This method might just work for the general case, but the machine proof would be just for a concrete $n$. In any case, for functions of the form $f(x)= 1- \prod_{i=1}^n(1- a_i x)^{\alpha_i}$ the quotient $\frac{f''(x)}{f'(x)}$ is a rational function.

$\bf{Added:}$ Let's see how the general case might work. Consider a function of the form $f(x) = 1- g(x)^{\alpha}$. Then we get $$\frac{f''(x)}{f'(x)} = \frac{(\alpha-1)g'(x)}{g(x)} + \frac{g''(x)}{g'(x)} = \alpha \frac{ g'(x)}{g(x)}+ \frac{g''(x) g(x) - g'(x)^2}{g(x) g'(x)}$$. Note that we have $$(\frac{g'(x)}{g(x)})'= \frac{ g''(x) g(x) - g'(x)^2}{g(x)^2}$$ In the end we get $$\frac{f''}{f'} = \alpha \frac{g'}{g} + \frac{(g'/g)'}{g'/g}$$

Now suppose $g(x) = \prod_{i=1}^n (1- a_i x)$. Then $g'/g = - \sum \frac{a_i}{1- a_i x}$ and $(g'/g)' = -\sum \frac{a_i^2}{(1- a_i x)^2}$ . We get
$$\frac{\sum \frac{a_i^2}{(1-a_i x)^2}}{\sum \frac{a_i}{1-a_i x}}- \alpha \cdot \sum \frac{a_i}{1-a_i x}$$. Each of has coefficients in terms of the Newton sums.

$\bf{Added:}$ In the above expression for $f(x) = 1- g(x)^{\alpha}$ we can iqnore the free term. Therefore, we may consider $g(x)$ any polynomial with real roots and positive free term, say $g(x) = c(r_1-x)(r_2-x) \cdots$. Now $g'(x)/g(x) = - \sum_{k\ge 0} S_{-k-1} x^k$, while $g''(x)/g'(x) = \sum_{k\ge 0} S'_{-k-1} x^k$, where $S_l$, $S'_l$ are the Newton sums for the polynomial $g(x)$, $g'(x)$. Now the optimal $\alpha$ is such that the free term in $(1-\alpha) (-\frac{g'}{g}) + \frac{g''}{g}$ cancels. To show the positivity one needs to show this: $$\frac{S'_{-k}}{S_{-k}}\le \frac{S'_{-1}}{S_{-1}}$$ for all $k$ natural. With some algebra and the Maclaurin inequalities one can show this for $k=2$ ($n$ arbitrary). Perhaps it is worth thinking about general $k$.

Answer 3

The question is equivalent to the claim that all coefficients of the Maclaurin series of the function $1-\bigl(1-\frac{t}{2}+\frac{t^2}{24}\bigr)^{2/3}$ are positive. From \begin{align*} \Biggl[\biggl(1-\frac{t}{2}+\frac{t^2}{24}\biggr)^{2/3}\Biggr]^{(n)} &=\sum_{k=1}^n \biggl\langle\frac23\biggr\rangle_{k}\biggl(1-\frac{t}{2}+\frac{t^2}{24}\biggr)^{2/3-k} B_{n,k}\biggl(-\frac{1}{2}+\frac{t}{12}, \frac{1}{12}, 0,\dotsc,0\biggr)\\ &\to \sum_{k=1}^n \biggl\langle\frac23\biggr\rangle_{k} B_{n,k}\biggl(-\frac{1}{2}, \frac{1}{12}, 0,\dotsc,0\biggr), \quad t\to0\\ &=\sum_{k=1}^n \biggl\langle\frac23\biggr\rangle_{k} \biggl(\frac{1}{12}\biggr)^k B_{n,k}(-6,1, 0,\dotsc,0)\\ &=\sum_{k=1}^n \biggl\langle\frac23\biggr\rangle_{k} \biggl(\frac{1}{12}\biggr)^k \frac{1}{2^{n-k}}\frac{n!}{k!}\binom{k}{n-k}(-6)^{2k-n}\\ &=(-1)^{n+1}\frac{n!}{12^n}\sum_{k=1}^n (-1)^{k}\frac{2^{k+1}}{k!}\binom{k}{n-k} \prod _{j=2}^k (3 j-5) \end{align*} for $n\ge1$, we see that it is sufficient to show the positivity of the sequence $$ (-1)^{n}\sum_{k=1}^n (-1)^{k}\frac{2^{k+1}}{k!}\binom{k}{n-k} \prod _{j=2}^k (3 j-5), \quad n\ge1. $$ However, it is seemingly not easy to confirm the positivity of the above sequence.

Feng Qi, Da-Wei Niu, Dongkyu Lim, and Yong-Hong Yao, Special values of the Bell polynomials of the second kind for some sequences and functions, Journal of Mathematical Analysis and Applications 491 (2020), no. 2, Paper No. 124382, 31 pages; available online at https://doi.org/10.1016/j.jmaa.2020.124382.