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It's well known that $x > \sin x$ for $x> 0$. The Taylor series of $ x - \sin x$ is also well known, and the coefficients are alternating. However, it appears that the Taylor coefficients of the function $\tan x - \tan (\sin x)$ are all positive ( and this implies that $x > \sin x$ on $(0, \pi/2)$, as it should). It's not clear for me why this is true.

In fact, one can go further as follows. It is known that we have inequalities of the form $$\sum_{k=0}^{2 l} (-1)^k\frac{x^{2k+1}}{(2k+1)!}< \sin x < \sum_{k=0}^{2 m + 1}(-1)^k \frac{x^{2k+1}}{(2k+1)!}$$ for $x > 0$.

Let's consider for instance the inequality $$x< \sin x + \frac{x^3}{6}$$ for $x>0$. Now, it appears again that the function $\tan ( \sin x + x^3/6) - \tan x$ has a "positive" Taylor expansion.

Similarly for $\tan ( x + \frac{x^5}{120}) - \tan( \sin x + x^3/6)$, and so on, for any inequality with positive coefficients obtained from the above by switching sides of terms.

One can substitute the function $\sec$ for the function $\tan$.

I am aware of the Taylor expansions of the functions $\tan$ and $\sec$ ( see the wikipedia article on trigonometric functions), they are all positive, and have a combinatorial interpretation.

One can do some testing with WolframAlpha or any computer algebra system.

orangeskid
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  • Because... all of it's derivatives at 0 are positive? In the Taylor expansion of $x\to f(x)$ around $x_0$ the coefficient to each term is $\frac{f^{(k)}(x_0)}{k!}$, and the factorials are all $>0$ right? Maybe you can make a proof by induction with differentiation and the chain rule. – mathreadler Aug 14 '17 at 10:24
  • @mathreadler: I can definitely show that, say, the $11$th coefficient is positive, by calculation, it's $\frac{6931}{887040}$. But there are infinitely many of them... and no induction in sight at the moment. – orangeskid Aug 14 '17 at 10:34
  • It seems to be: $,\displaystyle 0 < \frac{1}{1.11}[x^{2m+1}]\left(\tan x - \tan(x-\frac{x^3}{6})\right) $$\displaystyle < [x^{2m+1}]\left(\tan x - \tan(\sin x)\right) < $$\displaystyle [x^{2m+1}]\left(\tan x - \tan(x-\frac{x^3}{6})\right) ,$ for $,m>1,$ and $,\displaystyle [x^{2m+1}]\left(\tan x - \tan(\sin x)\right) = $$\displaystyle [x^{2m+1}]\left(\tan x - \tan(\sum\limits_{k=0}^n (-1)^k\frac{x^{2k+1}}{(2k+1)!})\right) ,$ for $,m\in{1,2,3,...,n},$ – user90369 Aug 17 '17 at 10:47

2 Answers2

6

The coefficients of the Taylor series at the origin of $\tan x$ are non-negative since $f(x)=\tan(x)$ is a solution of the differential equation $$f''(x) = 2\,f(x)\left(1+f(x)^2\right)\tag{1} $$ leading to $$ f'''(x) = 2\,f'(x)\left(1+3\,f(x)^2\right)=2\left(1+f(x)^2\right)\left(1+3\,f(x)^2\right)\tag{2} $$ $$ f^{(4)}(x) = 8\,f(x)\left(1+f(x)^2\right)\left(2+3\,f(x)^2\right) \tag{3}$$ by termwise differentiation. By repeating the process we only get positive signs in the involved RHSs, hence $f^{(n)}(0)\geq 0$ simply follows by induction. $\tan(x)$ is an odd function and it is not difficult to refine the previous bound and get $f^{(2n)}(0)=0,\, f^{(2n+1)}(0)>0$.
As an alternative, $\tan(x)=-\frac{d}{dx}\log\cos x$ leads to: $$ \tan(x) = \frac{d}{dx}\sum_{n\geq 0}-\log\left(1-\frac{4x^2}{(2n+1)^2 \pi^2}\right)=\sum_{n\geq 0}\frac{8x}{(2n+1)^2 \pi^2-4x^2}\tag{4} $$ then to: $$ \frac{\tan x}{x}=8\sum_{n\geq 0}\sum_{m\geq 0}\frac{4^m}{(2n+1)^{2m+2}\pi^{2m+2}}x^{2m}=\sum_{m\geq 0}\frac{2\left(2^{2m+2}-1\right)\zeta(2m+2)}{\pi^{2m+2}}x^{2m}\tag{5} $$ from which $[x^{2m+1}]\tan(x)\geq \frac{2^{2m+3}-\tfrac{1}{2}}{\pi^{2m+2}}$, which is not a surprising inequality, since the radius of convergence of the Taylor series of $\tan(x)$ at the origin is $\frac{\pi}{2}$ ($x=\pm\frac{\pi}{2}$ are the closest singularities to the origin, and they are simple poles). The radius of convergence of the Taylor series at the origin of $\tan(\sin x)$ (which still is an odd function) is a bit larger, since the closest singularities are approximately located at $\pm\frac{\pi}{2}\pm i$.
In particular we may expect to have $\left|[x^{2m+1}]\tan\sin x\right|\leq [x^{2m+1}]\tan(x)$ for any $m\geq 0$, proving the claim. By $(5)$ we have:

$$\begin{eqnarray*}\left|[x^{2m+1}]\tan\sin x\right| &=& \left|\sum_{k\geq 0}\frac{2\left(2^{2k+2}-1\right)\zeta(2k+2)}{\pi^{2k+2}}[x^{2m+1}]\left(\sin x\right)^{2k+1}\right|\\ &\leq& \sum_{k\geq 0}\frac{2^{2k+2}-1}{3\pi^{2k}}\left|[x^{2m+1}]\sin(x)^{2k+1}\right|\\&=&\sum_{k\geq 0}\frac{2^{2k+2}-1}{3\pi^{2k}}\left|\frac{1}{2\pi i}\oint_{\gamma_k}\frac{\sin(z)^{2k+1}}{z^{2m+2}}\,dz\right| \tag{6}\end{eqnarray*}$$ where $\gamma_k$ can be chosen as the boundary (counter-clockwise oriented) of a disk centered at the origin with radius $\frac{1}{\sqrt{k+1}}$. By invoking the maximum modulus principle and applying simple inequalities, to prove $\left|[x^{2m+1}]\tan\sin x\right|\leq [x^{2m+1}]\tan(x)$ for any $m\geq 5$ is simple, and the remaining cases can be simply checked by hand by computing a few derivatives at the origin.

Jack D'Aurizio
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  • I wonder what $[x^{2m+1} ] \tan x$ means. As far as I understand, you deal with the size of the coefficients, using the radius of convergence. That would certainly show that infinitely many coefficients are positive. Not clear how the other things work. But it's definitely a start! Nice. – orangeskid Aug 14 '17 at 21:09
  • @orangeskid: $[x^k],f(x)$ stands for the coefficient of $x^k$ in the Taylor series of $f(x)$ at the origin. Which can be computed through Cauchy's integral formula, like in $(6)$. – Jack D'Aurizio Aug 14 '17 at 21:11
  • Ok, I see, it's about the size of the coefficients. Not clear to me right now how you are proving those inequalities – orangeskid Aug 14 '17 at 21:37
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let your function be labelled $f(x)$, then consider

$$f\left( \arcsin \left( x \right) \right)=\frac{x}{\sqrt{1-{{x}^{2}}}}-\tan \left( x \right)$$

Take the series expansion of these functions about $x=0$ and collect coefficients, i.e.

$$f\left( \arcsin \left( x \right) \right)=\sum\limits_{n=1}^{\infty }{\left( \frac{\Gamma \left( n-\tfrac{1}{2} \right)}{\Gamma \left( n \right)\sqrt{\pi }}-\frac{{{\left( -1 \right)}^{n-1}}{{2}^{2n}}\left( {{2}^{2n}}-1 \right){{B}_{2n}}}{\left( 2n \right)!} \right)}{{x}^{2n-1}}$$ Note

${{B}_{2n}}=4\sqrt{\pi n}{{\left( -1 \right)}^{n-1}}{{\left( \frac{n}{\pi e} \right)}^{2n}}\left( 1+O\left( \frac{1}{n} \right) \right)$ and $\Gamma \left( z \right)=\sqrt{2\pi }{{e}^{-z}}{{z}^{z-\tfrac{1}{2}}}+O\left( {{\left| z \right|}^{-1}} \right)$

so

$$\frac{\Gamma \left( n-\tfrac{1}{2} \right)}{\Gamma \left( n \right)\sqrt{\pi }}-\frac{{{\left( -1 \right)}^{n-1}}{{2}^{2n}}\left( {{2}^{2n}}-1 \right){{B}_{2n}}}{\left( 2n \right)!} \tilde{\ }\frac{1}{\sqrt{n\pi }}-\frac{2/e}{{{\left( \pi /2 \right)}^{2n}}}$$

So at the very least, asymptotically, we see the coefficients are always positive. It should be fairly easy to extend it back to show this is true for all n.

mathstackuser12
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  • An interesting approach. Now, what you can show with this is that our function ( $\tan(x) - \tan(\sin x) $ ) composed on the right with $\arcsin $, has a positive Taylor series. That should be a consequence of the hypothesis, since also $\arcsin$ has positive coefficients. But the converse would not follow right away, you would need to compose your function back with $\sin x$, which has an alternating series. So, indeed, what you are claiming should also hold, but would not solve the problem, would be just a confirmation of it, quite useful in fact. Nice so far! – orangeskid Aug 14 '17 at 12:11
  • yes you're right. hmmm. more thinking required. – mathstackuser12 Aug 14 '17 at 12:21