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 |
| Edit | Post #290318 | Initial revision | — | 5 months ago |
| Question | — |
Is it known whether most numbers are normal or not In a comment over at mathematics.stackexchange.com, I just (well, now stuff has happened and it's 8 hours ago) claimed that for most numbers it is not known whether they are normal. And then I got to think whether that is actually true? Is there some set of numbers that (in some way, whether my me... (more) |
— | 5 months ago |
| Edit | Post #286961 |
Post edited: Chnaged formatting |
— | over 1 year ago |
| Edit | Post #286961 |
Post edited: Described that you can add a constant to the functions solving the orgonal problem |
— | over 1 year 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 |
| Edit | Post #286961 |
Post edited: |
— | over 1 year ago |
| Edit | Post #286961 | Initial revision | — | over 1 year ago |
| Question | — |
Is $f(x)=\sin(x)$ the unique function satisfying $f'(0)=1$ and $f^{(n)}(\Bbb R)\subset [-1,1]$ for all $n=0,1,\ldots$? > Question. Is there a function $f:\Bbb R \to \Bbb R$ with $f'(0)=1$ and $f^{(n)}(x)\in [-1,1]$ for all $n=0,1,\ldots$ and $x\in \Bbb R$, other than $f(x)=\sin(x)$? Going through some old notes, I stumbled upon this question, I don't remember where it take from/what inspired me to look at it. If t... (more) |
— | over 1 year ago |
| Edit | Post #283591 | Initial revision | — | over 2 years ago |
| Answer | — |
A: What is "continuous" in Math? There are more possible cardinalities above the the Cardinality of the continuum which is $2^{\aleph0}$ (e.g. $2^{2^{\aleph0}}$ - that wikipedia page even mentions this, have you actually read that), sets of those sizes generally don't contain discrete things, so your distinction is simply wrong, but... (more) |
— | over 2 years 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 |
| Edit | Post #278332 |
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— | over 3 years ago |
| Edit | Post #280373 |
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— | over 3 years ago |
| Edit | Post #280373 |
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— | over 3 years ago |
| Edit | Post #280373 | Initial revision | — | over 3 years ago |
| Answer | — |
A: Are 3 10% chances better than one 30% chance (when penalized by a variable for failures)? As you seem to know (the $10\times 10\\% \neq 100\\%$-stuff - but continue doing anyway) and celtschk's answer shows you can't add the probabilities like that. (The events aren't independent) The expected number of upgrades are the same, but what you really need to weigh against each other is the ... (more) |
— | over 3 years ago |
| Edit | Post #280348 | Initial revision | — | over 3 years ago |
| Question | — |
Normal in one base and not in another? Examples of numbers that are normal in some bases are known (Champernowne's $0.12345678910111213\ldots$ is probably the most well-known example of a number that is normal in base 10). To my knowledge (but please correct me if I'm wrong) no numbers that are normal (in all bases) are known. But a... (more) |
— | over 3 years ago |
| Edit | Post #278332 |
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— | over 3 years ago |
| Edit | Post #278332 |
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— | over 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 |
| Edit | Post #278332 |
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— | over 3 years ago |
| Edit | Post #278332 |
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— | over 3 years ago |
| Edit | Post #278332 | Initial revision | — | over 3 years ago |
| Question | — |
Cutting the square I don't know how simple/diffcult this is, but it has been on my mind for some time, and I haven't managed to do anything about it, so now I'll try here to see if anyone has any input. Let a natural number $n\in\mathbb N$ be given. Start with a square $S0$ with sidelength $1$. For $k$ in $\\{... (more) |
— | over 3 years ago |