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

Question

I have just finished learning in class that every non-empty bounded above subset of $\mathbb{R}$ has a least upper bound, but my professor then showed us the following set: $$A=\{\cos(10^n)\mid n\in\mathbb{N} \} $$ and asked us to compute the first $5$ decimal places of the supremum of $A$. At first glance, I want the supremum of this set to be $1$, but that is not possible for integer values of $n$, as $10^n$ is never an integer multiple of $2\pi$. There seem to be no clear patterns concerning the periodicity of the function. My hypothesis is that there is no method to find the supremum of this set, but all we know is that it exists.

I started out by noticing that $10^n$ must equal some integer multiple of $2\pi$, so I got the equation $10^n=2\pi m$ where $m\in\mathbb{Z}$. This leads to $n=\log(2\pi m)$ which has no integer solutions, but the question is what value of $m$ gets $\log(2\pi m)$ the closest to an integer, which I have no idea how to begin showing.

So my question is, is there a way to find the supremum of this set, and if so, how can I compute the first $5$ decimal places?

Answer 1

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 $2\pi$.

Answer 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 $2\pi$.

Answer 3

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.

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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 "$9999999999999999$".