3 ms·

I'm also not getting your proof. What does it mean every time you hit a power of 10? The sequence n! does not hit powers of 10 since all terms of the sequence a

I'm also not getting your proof. What does it mean every time you hit a power of 10? The sequence n! does not hit powers of 10 since all terms of the sequence after n=3 are at least divisible by 3. I agree though, density in (0,1) of the fracPart(log10(n!)) is a good thing to try to prove, since 10^fracPart(log10(n!)) should have the same digits as 10^intPart(log10(n!))*10^fracPart(log10(n!))=10^log10(n!)=n!

SamReidHughes 7y ago
Every time n hits a power of 10, not n!. E.g. https://www.wolframalpha.com/input/?i=Table%5B%5Bfrac%28log10%28%2810%5E6+%2B+n%29%21%29%29%2C+%2810%5E6+%2B+n%29%21%5D%2C%7Bn%2C0.0%2C20.0%2C1.0%7D%5D https://www.wolframalpha.com/input/?i=Table%5B%5Bfrac%28log1... In general for b << 10^k, we have fracPart[log10((10^k + b)!)] ≈ fracPart[log10((10^k)!) + b(b+1)/(2 log(10) 10^k)]. So it takes sqrt(2 log(10) 10^k) steps for these factorial values to work their way around the unit interval. Increase k and you fill it arbitrarily finely. Edit: ^^^ To be clear, the approximation there is formed using the linear approximation log(1 + b/10^k) ≈ b/10^k, and basic log multiplication/addition/division formulas.
- markisus 7y ago
  The approximation holds only for b << 10^k. Then you don't get to take O(10^k) steps to work you way around the unit interval, do you? After the approximation breaks down, you don't know that the spacing is approximately uniform anymore.
  - SamReidHughes 7y ago
    You only need sqrt(2 log(10) 10^k) steps. The spacing is not uniform, it's just bounded by O(1/10^(k/2)). The spacing is, at its greatest, about 1/sqrt(2 log(10) 10^k).
    - markisus 7y ago
      I'm not sure you only need that many steps. As you start out, each step gives you a lot of distance. But log gets flatter and flatter the more steps you take. If you knew the number of steps, I think you could only make a lower bound on spacing, not an upper bound. Since at worst all the steps could be clustered really close, except for the last one. But now I have to think more because it seems like it should actually help the argument that the spacing gets finer and finer as you step, because that is what we want anyway. And is the number of steps actually important to the proof?
      - SamReidHughes 7y ago
        The distance you cover with step b is about b/(log(10) 10^k). The exact distance is log10((10^k + b)/10^k). That's a funny way of saying fracPart(log10(10^k + b)). The number of steps tells you how big the last step is.
        markisus 7y ago
        Okay I think I buy it. Great proof, thanks!

SamReidHughes 7y ago

Every time n hits a power of 10, not n!. E.g. https://www.wolframalpha.com/input/?i=Table%5B%5Bfrac%28log10%28%2810%5E6+%2B+n%29%21%29%29%2C+%2810%5E6+%2B+n%29%21%5D%2C%7Bn%2C0.0%2C20.0%2C1.0%7D%5D https://www.wolframalpha.com/input/?i=Table%5B%5Bfrac%28log1... In general for b << 10^k, we have fracPart[log10((10^k + b)!)] ≈ fracPart[log10((10^k)!) + b(b+1)/(2 log(10) 10^k)]. So it takes sqrt(2 log(10) 10^k) steps for these factorial values to work their way around the unit interval. Increase k and you fill it arbitrarily finely. Edit: ^^^ To be clear, the approximation there is formed using the linear approximation log(1 + b/10^k) ≈ b/10^k, and basic log multiplication/addition/division formulas.

markisus 7y ago
The approximation holds only for b << 10^k. Then you don't get to take O(10^k) steps to work you way around the unit interval, do you? After the approximation breaks down, you don't know that the spacing is approximately uniform anymore.
- SamReidHughes 7y ago
  You only need sqrt(2 log(10) 10^k) steps. The spacing is not uniform, it's just bounded by O(1/10^(k/2)). The spacing is, at its greatest, about 1/sqrt(2 log(10) 10^k).
  - markisus 7y ago
    I'm not sure you only need that many steps. As you start out, each step gives you a lot of distance. But log gets flatter and flatter the more steps you take. If you knew the number of steps, I think you could only make a lower bound on spacing, not an upper bound. Since at worst all the steps could be clustered really close, except for the last one. But now I have to think more because it seems like it should actually help the argument that the spacing gets finer and finer as you step, because that is what we want anyway. And is the number of steps actually important to the proof?
    - SamReidHughes 7y ago
      The distance you cover with step b is about b/(log(10) 10^k). The exact distance is log10((10^k + b)/10^k). That's a funny way of saying fracPart(log10(10^k + b)). The number of steps tells you how big the last step is.
      - markisus 7y ago
        Okay I think I buy it. Great proof, thanks!