Find the supremum of the set $A=\{\cos(10^n)\mid n\in\mathbb{N} \}$

Solution 1:

Either the first five decimal places of the supremum of $A$ are "$00000$" (if $\sup A = 1$) or "$99999$". No amount of computing elements of $A$ will resolve which is correct. (Proof: $10^k$ is rational and $2\pi$ is not, so no direct computation of a finite list of elements of $A$ will resolve whether $A$ misses a small open interval around $1$.)

For the latter, $\cos(10^{304}) = 0.999\,994\,33{\dots}$ and the supremum is at least as large as this element of $A$. (For positive powers of ten less than the $304^\text{th}$, the maximum value attained is $0.999\,94{\dots}$, so this is the first element of $A$ that resolves that the first five decimal digits of $\sup A$ are the two options in the first sentence.)

(If, bizarrely, you are somehow working with a version of cosine that takes degrees as argument, the sumpremum is the maximum, attained at the zeroeth power (footnote). In fact the degrees version is very boring: \begin{align*} \cos ((10^k)^\circ) = \begin{cases} 0.999\,847\,6{\dots} ,& k = 0 \\ 0.984\,807\,7{\dots} ,& k = 1 \\ -0.173\,648\,1{\dots} ,& k = 2 \\ 0.173\,648\,1{\dots} ,& k \geq 3 \\ \end{cases} \text{.} \end{align*} )

footnote: $0 \in \Bbb{N}$, as standardized in ISO 80000-2.

How did I find this?

For each $k$, find the (quotient, $q$, and) remainder, $r$ of $10^k / 2\pi$ to 1000 digits of precision. (This assumed I wouldn't need $k \geq 1000$, which turned out to be the case. To be confident in the computation, keep more than $k$ digits of precision.) $$ 10^k = 2\pi q + r, \quad 0\leq r < 2\pi $$ Then compute $\cos r$ to 10 digits of precision.

Mathematica code for this:

Module[{k, q, r, current, max},
  $MaxExtraPrecision = 1000;
  max = -Infinity;
  For[k = 0, k <= 1000, k++,
    {q, r} = QuotientRemainder[10^k, 2 Pi];
    current = N[Cos[r], 10];
    If[current > max,
      max = current;
      Print[{k, current}]
    ]
  ]
]

having output

{0,0.5403023059}
{2,0.8623188723}
{6,0.9367521275}
{13,0.9573637169}
{36,0.9766517640}
{67,0.9798253555}
{70,0.9889725210}
{83,0.9974446504}
{151,0.9999489800}
{304,0.9999943382}
{421,0.9999972157}
{901,0.9999988543}

It's not enough to narrow down the first six decimal digits. They're one of "$999998$", "$999999$", or "$000000$".

If we use sneakier code (and substantially more computation time), we can extend the above list.

{4428    , 0.99999986301}
{17540   , 0.9999998684877}
{24987   , 0.9999999739306}
{27797   , 0.999999999136465}
{120664  , 0.999999999996363727}
{1301493 , 0.99999999999901628196}
{4344039 , 0.999999999999609767808353}
{4910042 , 0.99999999999996379824433687}
{4911162 , 0.9999999999999946517749663489}
{74971140, 0.99999999999999997321787197875403}

and discover that the first sixteen decimal digits are either "$0000000000000000$" or "$9999999999999999$".

Solution 2:

Not a full answer: It has been proven$^1$ that the sequence $$a^n\bmod 2\pi$$ is equidistributed over $[0,2\pi[$ for Lebesgue-almost-every number $a>1$. If $a=10$ is one of the "almost every numbers" for which the above sequence is equi-distributed, then we are very happy, because then $10^n$ gets in particular arbitrarily close to a multiple of $2\pi$ and thus the supremum is $1$. However, if $a=10$ is not one of these "almost every numbers", then I don't know what to do.

$^1$ http://www.numdam.org/item/?id=CM_1935__2__250_0 EDIT: I noticed that they only talk about equidistribution modulo $1$, but I think the argument can be repeated for equidistribution modulo $2\pi$.

Solution 3:

Find $1/2\pi$ to many decimal places until you find a stream of three nines or three zeros. Then the appropriate power of ten is close to an integer multiple of $2\pi$.