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

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 $$0 then $$(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.

• Nice and interesting observation. $\to +1$ May 5 at 6:15
• I'm sure there's a clever proof that obtains for $n\ge1$ results of the form$$\frac{d^n}{dx^n}(1-\cos^{2/3}x)=\sum_kf_{n,\,k}(\cos x)\sin^kx,\,f_{n,\,0}(1)\ge0.$$
– J.G.
May 5 at 7:23
• 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.
– J.G.
May 5 at 9:00
• 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. May 13 at 9:49
• 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 $a$. May 13 at 11:15

(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.

• Very nice! A clean approach. Thank you for your answer! 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 $$\{1,19,199,1999,19999,199999,1999999,\cdots\}$$

• Interesting sequences that pop out. I like this. May 5 at 9:23
• Very interesting patterns indeed! 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 $$n=2$$ !)

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

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

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

• Interestingly, the OP's choice $a=2/3$ is the root of $P_2$.
– Gary
May 7 at 3:11
• @Gary. I was hpoing to find another one. May 7 at 3:22
• 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... May 7 at 3:27
• 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 May 7 at 4:16
• @orangeskid. Look at my edit after your first comment. There are interesting things. 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.