Activity for Grove‭

Type	On...	Excerpt	Status	Date
Comment	Post #290318	So as I suspected my claim was (technically) wrong, but because most numbers are normal, not (as I thought) because most numbers aren't. (more)	—	5 months ago
Comment	Post #286961	I don't think negative values for $k$ or $j$ produces any functions that are really different from what we get if we choose $y$ properly, ($-\sin(x)=\sin(-x)=\sin(x+\pi)$), so having a lower bound of $0$ on them makes sense. The upper bound on $k$ is to make sure $f$ stays below $1$, the upper bound ... (more)	—	over 1 year ago
Comment	Post #286961	Oops. That was just a typo here (I made it consistently though), the hand-written notes I had on this, actually said $[-1,1]$. When I fixed that, I realised that $k$ can also range down to $-1$, and $j$ can also be negative, but to what degree do I get new functions, and to what degree is it just fun... (more)	—	over 1 year ago
Comment	Post #282642	Technically $\Leftrightarrow$ just means that the statements on either side have the same truth value. As both are true, the statement in your title is true. But as "$2=2$" and $\text{FLT}$ (Fermat's last theorem) are both true, we can also write $2=2 \Leftrightarrow \text{FLT}$, but that doesn't rea... (more)	—	almost 3 years ago
Comment	Post #278332	No. $B_1$ does not have area $1/1$, $n$ is fixed throughout the procedure so all the $B_k$'s have area $1/n$., and I thought that was conveyed by using $k$ as index on the $B$'s when defining those. (And of course it's a mistake that I forgot an index on that $S$, I will add it right after posting th... (more)	—	over 3 years ago