Show $\tan(n) < n^q$, conjectured $q < 1.1$

Question

Show $\tan(n) < n^q$, $n \in \mathbb{N}$, $n > 1$. The argument of the $\tan$-function is in radians. It is conjectured that $q < 1.1$.

In fact, search for the maximum of $q$ in $n\in [2, 10^9]$ gives $\tan(260515)= 383610.707744 = 260515^{1.031031}$ and then the next higher $q$'s only at $\tan(122925461)= 326900723.479835 = 122925461^{1.052508}$, and further $\tan(534483448)= 1914547468.536829 = 534483448^{1.063489}$.

It is known that $\tan(n)$ is unbounded (see math.stackexchange.com/questions/1056119). So it is clear that with increasing $n$, ever larger $\tan(n)$ will eventually be found. While we know that $n\ne(k+1/2)\pi$, that doesn't mean that we know how close $n$ comes to some $(k+1/2)\pi$ with growing $n$.

In the range of $n$ above, it appears that no higher values of $q$ are found, and the conjecture is that such high $n$ are required for higher values of $\tan(n)$, that no higher $q$ will be attained. Possibly the limit for $q$ will have to be made more loose than $q < 1.1$, where derivations of such looser bounds would certainly be appreciated.

For possible relations to the irrationality measure of $\pi$, which is unknown, see discussions in here: math.stackexchange.com/questions/2977461 .

Answer 1

This is not a complete answer, but I suppose it might be useful (btw I don't know how to answer this question without using the irrationality measure of $\pi$).

Let $\mu$ be a positive real number such that there are infinitely many rational numbers such that $$ \left\vert\pi - \frac{a}{b}\right\vert < \frac{1}{b^\mu} $$ if $\mu$ is the largest such number then we say that $\mu$ is the irrationality measure of $\pi$. It is known (see https://en.wikipedia.org/wiki/Liouville_number#Irrationality_measure) that $$ 2 \le \mu \le 7.60630853 $$

Let as make the assumption that for the true value $\mu$ of $\pi$ we can find infinitely many $a/b$ such that $a$ is even and $b$ is odd and in addition $\pi > a/b$, (this seems very reasonable but I don't have any idea of how it can be proved) then

$$ 0 < \pi - \frac{2n}{2m+1} < \frac{1}{(2m+1)^\mu} $$

and so $$ \pi/2 + m\pi - \frac{1}{2(2m+1)^{\mu-1}} < n < \pi/2 + m\pi $$

observing that $m \approx n/\pi $ we can finally write $$ n = \pi/2 + m\pi - \frac{\alpha}{n^{\mu-1}} $$ for some bounded $\alpha >0$.

Now we can use the expression for $n$ in the expansion of $\tan x$ about $\frac \pi2$: $$ \tan x = -\frac{1}{x-\pi/2} + \frac{1}{3}\left(x - \frac{\pi}{2}\right) + \frac{1}{45}\left(x - \frac \pi2\right)^3 + \cdots$$ and we get: $$ \tan n = \alpha n^{\mu-1} + O(1) $$

You can see that your conjecture would be true (at least for large enough $n$) if $\mu \le 2.1$ and the assumption above is true.

remarks: If you consider the continued fraction expansion of $\pi$: $$ \pi = 3 + \frac 1{7 + \frac 1{15+\frac 1{1 + \frac 1{292+\dots}}}} $$ Then the candidates to the fractions $a/b$ above are those obtained from stopping the continued fraction at any point, or the same adding 1 to the last partial quotient. If we keep those fractions that verify the assumption above we get the following candidates for $n$:

4                344 0.929204 *
8             260515 1.031030 * 
12           4846147 0.986072 
15         122925461 1.052508 *
17         534483448 1.063489 *
19        3083975227 1.067087 *
22      902209779836 1.026923
26    74357078147863 1.018592 
27   214112296674652 1.087606 *
30 18190586279576483 1.020496
 .....

I have looked for the first 10000 convergents and there isn't any one which gives a larger value than tan(214112296674652). As the number increases it seems to approach 1. For example for the convergents found with 1000 partial quotients or more the largest has exponent 1.0033.

Edit: the code I have used to obtain the above data is the following Pari-GP program:

\\ change to 100 digits precission
\p 100
A = contfrac(Pi);
M=0;  
h=[A[1],1];
k=[1,   0];
{
    for(r=1,length(A)-2,
        u=h;
        h=u*A[r+1]+k;
        k=u;
        if(h[1]%2==0 && Pi> h[1]/h[2], 
            ex=log(tan(h[1]/2))/log(h[1]/2);
            printf("%3d %20d %.8f %d\n", r+1, h[1]/2, log(tan(h[1]/2))/log(h[1]/2),if(ex>M, M=ex;1,0))
        );
        u = u+h;
        if(A[r+2] != 1 && u[1]%2==0 && Pi> u[1]/u[2],
            ex=log(tan(u[1]/2))/log(u[1]/2);
            printf("%3d %20d %.8f %d\n", r+1, u[1]/2, log(tan(u[1]/2))/log(u[1]/2),if(ex>M, M=ex;1, 0))
        )
    );
}

Answer 2

Suppose that the irrationality measure of $\pi$ is $\mu$. By definition, for all real $\epsilon,C>0$ and all integers $p,q$ with $q$ large enough: $$\left | \pi - \frac{p}{q} \right | > \frac{C}{q^{\mu+\epsilon}}.~~~~~(1)$$

Now, let $n$ be any positive integer and write $n=m\pi+c$, where $-\pi/2<c<\pi/2$. If $c<0$, then $\tan n<0<n^q$ for any $q$. Thus, suppose $c>0$. Since $\tan x (\pi/2-x) <1$ for all $x\in (0,\pi/2)$, we have $$\tan(n)=\tan(c)< \frac{1}{\pi/2-c}\leq \frac{2/(2m+1)}{\pi-2n/(2m+1)}\leq \frac{2}{C}(2m+1)^{\mu+\epsilon-1}\leq \frac{2}{C}(2/\pi)^{\mu+\epsilon-1}n^{\mu+\epsilon-1}.$$ If we let $C=2(2/\pi)^{\mu+\epsilon-1}$, we have shown that: for every $\epsilon>0$, the inequality $$\tan n <n^{\mu+\epsilon-1},$$ holds for $n$ large enough.

It is currently known that $\mu < 7.11$. So we have the following:

Theorem. For $n$ large enough: $\tan n<n^{6.11}$.