How and where could I compute the decimal reminder of a product of square roots times ten:
$$Dr\left( \prod_{x=1}^{k}x^\frac{1}{2} \right) \times 10$$
Where $k$ is a power of $10$.
I would like to be able to generate a table of values as $k$ approaches $\infty$.
I tryed using wolframalpha but it gets stuck when computing $k>5$.
This is interesting for me since I noticed that for $k=1,2,3,4,5$ we have: $$\approx9,4$$ $$\approx6,6$$ $$\approx3,4$$ $$\approx3,3$$ $$\approx3,14$$
Which seems to approach $\pi$?
Or is this just a coincidence, and the values have a different pattern when considering more values of $k$?
Anyhow, I would want to be able to compute a table of those values somehow.
The $k$ is a power of $10$ since if it was not a power of $10$, then the final result could vary and seems to not follow a pattern, for example if $k$ was $10^5\pm1$ then the result is $\approx 6,8$ .
But since $k$ is a power of ten, it seems to be stable in a sense of tending to $\pi$ but only for the first $5$ values so far, since I couldn't find a way to compute the next one.
As far as I know, all of this is most likely a coincidence?
Anyway, I need a more efficient way to compute this final value since I want to explore its properties. (And with the help of that then later likely explore similar expressions in a sense of taking out the decimal part of the number in some way.)
Basically you want the leading-order digits, starting from the second one, of $\sqrt{n!}$, where $n$ is a power of ten. Stirling's approximation will give you what you need. Specifically, use $$ n! \sim n^n e^{-n}\sqrt{2\pi n}\left(1 + \frac{1}{12n}\right). $$ If $n=10^k$, then $n^n$ is a large even power of $10$ and its square root doesn't affect the decimal digits. So the leading-order digits you want are the same as the ones in $$ e^{-n/2}\sqrt[4]{2\pi n}. $$ According to WolframAlpha, taking $n=10^2,10^3,\ldots$ gives leading-order digits of $9.66$, $6.343$, $5.335$, $5.314$, $2.875$, $3.468$, $\ldots$. So the apparent pattern of convergence to $\pi$ from $k=3,4,5$ does not continue to $k=6$.