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.

  • 2
    $\begingroup$ Nice and interesting observation. $\to +1$ $\endgroup$ May 5 at 6:15
  • 1
    $\begingroup$ I'm sure there's a clever proof that obtains for $n\ge1$ results of the form$$\frac{d^n}{dx^n}(1-\cos^{2/3}x)=\sum_kf_{n,\,k}(\cos x)\sin^kx,\,f_{n,\,0}(1)\ge0.$$ $\endgroup$
    – J.G.
    May 5 at 7:23
  • 1
    $\begingroup$ Well, we expect a Taylor series for $1-\cos^{2/3}x$ whose $x^n$ coefficient is $1/n!$ times the $n$th derivative at $x=0$. @metamorphy's recent answer is similar to what I'd hoped for: actually, their approach is even nicer. $\endgroup$
    – J.G.
    May 5 at 9:00
  • 1
    $\begingroup$ Although the coefficient of $x^{2m}$ can be written as $c(m) = \frac{(-1)^{m-1}}{2^{a}(2m)!} \sum_{k\ge 0} \binom{a}{k} (2k-a)^{2m}$ it seems hard to make a proof out of it. $\endgroup$ May 13 at 9:49
  • 1
    $\begingroup$ Replacing $cos(x)$ by $1-\frac{x^2}{2}$ gives for the coefficient of $x^{2k}$ the expression $c(k) = -(1/2^k) \binom{k - a - 1}{k}$ which has the sign of $a$. $\endgroup$ May 13 at 11:15

5 Answers 5


(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.

  • $\begingroup$ Very nice! A clean approach. Thank you for your answer! $\endgroup$
    – orangeskid
    May 6 at 11:20

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 $$\{1,19,199,1999,19999,199999,1999999,\cdots\}$$

  • $\begingroup$ Interesting sequences that pop out. I like this. $\endgroup$
    – K.defaoite
    May 5 at 9:23
  • $\begingroup$ Very interesting patterns indeed! $\endgroup$
    – orangeskid
    May 6 at 11:28

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.


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 $n=2$ !)


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 !


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

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

  • $\begingroup$ Interestingly, the OP's choice $a=2/3$ is the root of $P_2$. $\endgroup$
    – Gary
    May 7 at 3:11
  • $\begingroup$ @Gary. I was hpoing to find another one. $\endgroup$ May 7 at 3:22
  • $\begingroup$ it is fascinating. So we know now that the polynomials $P_n$ are positive on $[0, 2/3]$. That would mean $P_n(\frac{2}{3}\cdot \frac{t}{t+1})$ is positive for $t\ge 0$. It seems that the numerator after that substitution again has positive coefficients... $\endgroup$
    – orangeskid
    May 7 at 3:27
  • $\begingroup$ I considered the expansion of $1- \cos( \sqrt{a+1} x )^{\frac{2}{3} \cdot \frac{a}{a+1}}$ in $x$ and got positive coefficients as far as I can see $\endgroup$
    – orangeskid
    May 7 at 4:16
  • $\begingroup$ @orangeskid. Look at my edit after your first comment. There are interesting things. $\endgroup$ May 7 at 4:22

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.