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Show that there are no even natural numbers between $\frac{\pi}{10^{-k}}$ and $\frac{\pi}{\arctan (10^{-k})}$ for integer $k \ge 2$. I encountered this while solving a physics problem and I am not even sure if it is true or not.

I tried to bound the difference of the two using the fact that $x > \arctan (x)$ for $x > 0$, but still there can be cases like $\frac{\pi}{10^{-k}} = n + 0.999999999$ for some integer $n$, regardless of how small the bound is.

Can anyone help?

EDIT : I have found Galperin's paper and it seems it is still left as a conjecture. All sources are in the comments of the first answer.

Vue
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  • Have you plugged in some values of $k$ to see what sort of values you end up getting? – CoffeeBean Nov 24 '22 at 01:14
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    Consider the Taylor expansion of $arctan(x) = x - x^3/3 + x^5/5$ .. As $k$ increases, $arctan(x) \approx x$ because higher-order terms decay faster. Therefore, your denominators are eventually indistinguishable. – algebroo Nov 24 '22 at 01:14
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    Is there any reason to imagine that this is solvable? Easy enough to check for large $k$, but for all $k$? Sure it seems likely to be true but I don't see how to get a handle on it. – lulu Nov 24 '22 at 01:18
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    @AryanDugar I have considered it but as I mentioned above there could be situations like $1.9999999 < 2 < 2.00000000001$. – Vue Nov 24 '22 at 01:21
  • @lulu The lower bound of $k$ can be increased since the rest can be checked by computer calculation. I have to check it for all $k$. – Vue Nov 24 '22 at 01:22
  • Hmm, would there be such situations? $\frac{\pi}{10^{-k}}$ is just taking the expansion of $\pi$ and moving the decimal point, so you get things like $3.1415..,31.415...,314.15...$. $\pi$ would never have an unending string of $9's$ in its decimals (it would be a rational number then), so I think what you're saying wouldn't arise. Having said that, this is all conjecture and I also don't have a proof, but perhaps this helps you. – algebroo Nov 24 '22 at 01:27
  • @AryanDugar I didn't mean unending $9$'s, not infinite but just enough to cause a problem since our bound cannot be zero. Also this seems to be unproved so take a look at Galperin's paper I linked below. – Vue Nov 24 '22 at 03:34

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We have $$ \frac{\pi}{\arctan (x)}-\frac{\pi}{x}=\frac{\pi}{x}\left(\frac{x}{\arctan(x)}-1\right)=\frac{\pi}{x}\left(\frac{x^2}{3}+O(x^4)\right)=\frac{\pi}{3}x + O(x^3) $$

Then the difference is around $\frac{\pi}{3}x \approx 10^{-k}$

To find an integer inbetween, then, roughly amounts to find, after $k$ digits in the decimal expansion of $\pi$, a sequence of at least $k$ consecutive nines, for some $k$.

This seems very unlikely. But also very difficult to prove. I don't think probabilistic approaches, assuming $\pi$ is normal, would lead anywhere.

Edit: as pointed out in the comments, the paper "Playing pool with π (the number π from a billiard point of view)" studies this (sec 10, question 1). It basically agrees with the previous paragraphs (the property "holds if and only if the string of $2N$ first decimal digits of $\pi$ contains $N − 1$ nines in its right half"). It concedes that "modern mathematics is powerless to answer it" but it also conjectures -with almost certainity- that it's not true. Some probabilistic analysis is provided.

leonbloy
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  • I read 3B1B's solution and the following was written there : "This difference between $\arctan(x)$ and $x$ could be problematic for our final count if, at some point when you're looking at the first $2n$ digits of $\pi$, the last $n$ of them are all $9$'s. It seems exceedingly unlikely that this should be true." – Vue Nov 24 '22 at 03:16
  • "For example, among the first $100$ million digits of $\pi$, the longest sequence of consecutive $9$'s is just $8$ of them in a row, whereas you'd need a sequence of $5$0 million for things to break our count!

    Nevertheless, this is quite difficult to prove, related to the question of whether or not $\pi$ is a "normal" number, roughly meaning that its digits behave like a random sequence. It was left as a conjecture in Galperin's paper on the topic. See sections 9 and 10 of that paper for more details."

     – Vue Nov 24 '22 at 03:16
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    https://www.3blue1brown.com/lessons/clacks is the explanation of the physics puzzle, and https://www.maths.tcd.ie/~lebed/Galperin.%20Playing%20pool%20with%20pi.pdf is Galperin's paper about the solution and his conjecture related to my question. – Vue Nov 24 '22 at 03:17
  • Please edit your post to put all this new information into the post itself, rather than burying it in a comment to an answer to your post, where few will see it. – Lee Mosher Nov 24 '22 at 15:35