Maclaurin series of $1- \cos^{2/3} x$ has all coefficients positive

Question

Experimenting with WA I noticed that the function $1- \cos^{\frac{2}{3}}x$ has the Maclaurin expansion with all coefficients positive ( works for any exponent in $[0, \frac{2}{3}]$). A trivial conclusion from this is $|\cos x|\le 1$, but it implies more than that, for instance see this. Maybe some ''natural'' proofs are available. Thank you for your interest!

Note: an attempt used a differential equation satisfied by the function. But the answer by @metamorphy just solved it the right way.

$\bf{Added:}$ Some comments about series with positive coefficients.

By $P$ we denote a series with positive coefficients ( no free term),

1. If $a>0$ then $\frac{1}{(1-P)^a} = 1+P$ (moreover, the positive expression on RHS is a polynomial in $a$ with positive coefficients

2. If $0<a < 1$ then $(1-P)^a = 1-P$. Similarly the expressions for $a = \frac{t}{t+1}$ are positive in $t$.

2'. If $1-f= P$ then $1- f^{a} =P$ for any $0 < a < 1$, and similar with above.

$\bf{Added:}$ It turns out that the function $\cos^{2/3} x$ has a continued fraction (an $S$-fraction, from Stieltjes) that is "positive" ( similar to the continued fraction for $\tan x$). This is a stronger statement than the one before. Maybe there is some approach using hypergeometric functions.

Answer 1

(A proof, not very "natural" though.) $f(x)=1-\cos^{2/3}x$ satisfies $xf''(x)=g(x)f'(x)$, where $$g(x)=x\cot x+\frac{x}3\tan x=1+\sum_{n=1}^\infty g_n x^n$$ with $g_n=0$ for odd $n$, and $3g_{2n}=(-1)^n 2^{2n}(4-2^{2n})B_{2n}/(2n)!\geqslant 0$ using Bernoulli numbers (and the alternating-sign property of these).

Now $f'(x)=\sum_{n=1}^\infty f_n x^n$ implies $(n-1)f_n=\sum_{k=1}^{n-1}g_{n-k}f_k$, giving $f_n\geqslant 0$ by induction.

Answer 2

This is not an answer but it is too long for a comment.

This is an interesting observation. I think that we just need to look at the coefficient of $x^4$; as soon as it is positive, all the next coefficients are positive too.

There are amazing patterns.

Let $a=\frac {2 \times 10^k+1}{3 \times 10^k}$. The coefficient of $x^4$ in the expansion of $\big[1-\cos^a(x)\big]$ write $$c_4(k)=-\frac {\alpha_k}{24\times 10^{2k}}$$ and the $\alpha_k$ form the sequence $$\{1,7,67,667,6667,66667,666667,\cdots\}$$

Let $b=\frac {2 \times 10^k-1}{3 \times 10^k}$. The coefficient of $x^4$ in the expansion of $\big[1-\cos^b(x)\big]$ write $$c_4(k)=\frac {\beta_k}{72\times 10^{2k}}$$ and the $\beta_k$ form the sequence $$\{1,19,199,1999,19999,199999,1999999,\cdots\}$$

Answer 3

Start with the positive series $$\csc x = \frac{1}{x} + \frac{1}{6} x + \frac{7}{360} x^3 + \frac{31}{15120} x^5 + \cdots $$

Take the square and obtain the positive series $$\csc^2 x =\frac{1}{x^2} + \frac{1}{3} + \frac{1}{15} x^2 + \frac{2}{189} x^4 + \cdots $$

Now take the derivative and get the series $$- \frac{2\cos x}{\sin^3 x} = -\frac{2}{x^3} + \frac{2}{15} x + \frac{8}{189} x^3 + \cdots $$ where all the coefficients after the first are positive. We conclude that the series $$ 1 - \frac{x^3 \cos x}{ \sin ^3 x}=\frac{1}{15}x^4 + \frac{4}{189}x^6 + \frac{1}{225}x^8 + \cdots $$ is positive. Now, since $(1-t)^{-1/3}$ is a positive series in $t$, if we substitute for $t$ a positive series in $x$ without free term we get again a positive series in $x$. We conclude that $$\frac{\sin x}{x \cos^{1/3} x}= 1 + \frac{1}{45}x^4 + \frac{4}{567}x^6 + \frac{1}{405} x^8+ \cdots $$ is a positive series. Now multiply by $\frac{2}{3} x$ and integrate and get the positive $1- \cos^{2/3} x$ ( all coefficients are, except perhaps the free term, which turns out to be $0$).

$\bf{Note:}$ In fact we can prove that the series for $$\sin x - \frac{x^3 \cos x}{ \sin^2 x}$$ is positive. For this, we start with the expansion $$\csc x= x^{-1} + \frac{1}{6} x + \frac{7}{360}x^3 + \frac{31}{15120} x^5 + \cdots = \sum_{n=0}^{\infty} \frac{(-1)^{n+1} 2(2^{2n-1} -1) B_{2n}}{(2n)!}x^{2n-1}$$

Taking the derivative with respect to $x$ we get the expansion of $\frac{\cos x}{\sin^2 x}$. From here we conclude that $\sin x- \frac{x^3 \cos x}{\sin^2 x}$ is positive. Multiplying by the positive $\csc x$ we conclude that $1 - \frac{x^3 \cos x}{\sin^3 x}$ is positive.

Answer 4

As a series, we have $$1-\cos^a(x)=a\sum_{n=1}^\infty(-1)^{n+1}\,\frac {P_n(a)}{(2n)!}\, x^{2n}$$ and the first polynomials are $$\left( \begin{array}{cc} n & P_n(a) \\ 1 & 1 \\ 2 & 3 a-2 \\ 3 & 15 a^2-30 a+16 \\ 4 & 105 a^3-420 a^2+588 a-272 \\ 5 & 945 a^4-6300 a^3+16380 a^2-18960 a+7936 \\ 6 & 10395 a^5-103950 a^4+429660 a^3-893640 a^2+911328 a-353792 \end{array} \right)$$

Explored up to $n=100$, none of the $P_n(a)$ is factorable and none of them shows rational solution (except for $n=2$ !)

Edit

After @orangeskid's comment, let $a=\frac{2 t}{3 (t+1)}$ $$1-\cos ^{\frac{2 t}{3 (t+1)}}(x)=\frac t 6\sum_{n=1}^\infty \frac {Q_n(t)} {b^n\,(1+t)^n}\,x^{2n}$$ where the $b_n$ form the sequence $$\{1,72,97200,457228800,617258880000,16132678087680000,\cdots\}$$ which is quite interesting; for example, looking at $\frac{b_{n+1}}{b_n}$ gives $$\{72,1350,4704,1350,26136,49686,28800,140454,216600,106722\}$$

and the first $Q_n(t)$ are $$\left( \begin{array}{cc} n & Q_n(t) \\ 1 & 1 \\ 2 & 1 \\ 3 & 2 t^2+9 t+12 \\ 4 & 40 t^3+246 t^2+477 t+306 \\ 5 & 168 t^4+1222 t^3+3183 t^2+3582 t+1488 \\ 6 & 9920 t^5+83304 t^4+269106 t^3+423249 t^2+326646 t+99504 \end{array} \right)$$

Continuing for $$1-\cos ^{\frac{2 a}{3 (a+1)}}\left(x\sqrt{a+1} \right)=\frac a 6\sum_{n=1}^\infty \frac {R_n(a)} {b^n\,(1+t)^n}\,x^{2n}$$ the first $R_n(a)$ being $$\left( \begin{array}{cc} 1 & 1 \\ 2 & 1 \\ 3 & 2 a^2+9 a+12 \\ 4 & 40 a^3+246 a^2+477 a+306 \\ 5 & 168 a^4+1222 a^3+3183 a^2+3582 a+1488 \\ 6 & 9920 a^5+83304 a^4+269106 a^3+423249 a^2+326646 a+99504 \end{array} \right)$$ Compare $Q_n(t)$ and $R_n(a)$.

Amazing problem !

Update

$$P_3(a)=15 a^2-30 a+16=0 \quad \implies \quad a_\pm=1 \pm \frac{i}{\sqrt{15}}$$

$$1-\cos^{a_-}(x)=\frac{x^2}2-$$ $$\frac{x^4}{30}\Bigg[1+\frac{x^4}{315}+\frac{x^6}{945}+\frac{37 x^8}{111375}+\frac{22996 x^{10}}{212837625}+\frac{139 x^{12}}{3869775}+\frac{3109 x^{14}}{255605625}+O\left(x^{16}\right) \Bigg]-$$ $$-i\frac{x^2}{2 \sqrt{15}}+$$ $$i\frac{x^4}{6 \sqrt{15}}\Bigg[1+\frac{x^4}{105}+\frac{53 x^6}{23625}+\frac{23 x^8}{37125}+\frac{63964 x^{10}}{354729375}+\frac{353 x^{12}}{6449625}+\frac{153547 x^{14}}{8946196875}+O\left(x^{16}\right) \Bigg]$$

Answer 5

Some thoughts:

Let $f(x) := \cos^{2/3} x$.

We have $$f'(x) = \frac23 \cos^{-1/3} x\, (-\sin x) = -\frac23 f(x)\tan x $$ and $$f''(x) = -\frac23 f'(x) \tan x - \frac23 f(x) (\tan x)' = - \frac23 f(x) - \frac29 f(x)\tan^2 x .$$

Conjecture 1: For $n = 1, 2, \cdots$, $$f^{(2n - 1)}(x) = f(x) \sum_{k=1}^n c_{2n-1, k} \tan^{2k-1} x$$ and $$f^{(2n)}(x) = f(x) c_{2n, 0} + f(x) \sum_{k=1}^n c_{2n, k} \tan^{2k} x$$ where $c_{2n-1, k} \le 0$ for all $n\ge 1,\, 1\le k \le n$, and $c_{2n, k} \le 0$ for all $n\ge 1,\, 0 \le k \le n$.

I think we can use Mathematical Induction or Strong Induction to prove it.