runway44

3b1b's explanation of quaternions and visualizing 3D/4D rotations was basically something that I had idly imagined making a youtube video on

@PseudoLooped I haven't actually seen that many of them. There are so many topics it makes more sense to find what to watch based on what sounds interesting, not what is "better" on some objective scale. Also of note is the #SoMe2 tag - these are videos from other youtubers submitted to Summer Of Mathematical Exposition started by 3b1b

Is that a cinnamon stick in a pumpkin-based thing?

heh

I assume Parler was failing and losing money. If Ye can't revitalize it, it's a lot of money lost. Maybe that's justice for his offensive ignorance. But OTOH, I feel like Owens and her husband can cause more injustice with dozens of mill than Ye could have. Or maybe her and hubby are just grifters and'll hoard it, idunno.

@amWhy candace owens' husband just made billions, no? doesn't sound good to me.

For future reference, you are supposed to show your work in the question itself, not in the comment section below the question. If you have more to add to the question, use the edit button to include it into the question. And it is considered bad form to spam the comment section with a dozen one-line comments - try to collect and condense your thoughts into just one or as few comments as possible for brevity and readability. It is not like texting or chatrooms.

$\exp x=1+O(x)$ means $\exp x-1$ is $O(x)$ which means $|\exp x-1|\le K|x|$ for some $K$ in a nbhd of $x=0$

(from $\exp x=1+O(x)$ as $x\to0$)

@Gwyn $\exp O(e^{-2u})=1+O(e^{-2u})$ as $u\to\infty$, and $e^{-2u}O(e^{-2u})=O(e^{-4u})$

anybody see a copy of $S_4\times S_4$ in $A_8$, and/or a copy of ${\rm Aff}_2\Bbb F_2\times{\rm Aff}_2\Bbb F$ in ${\rm Aff}_4\Bbb F_2$?

True, it's an injection and its range is the canonical basis

This user has been suspended for voting irregularities. Are mods aware of the dozen other accounts they've used? (If you put the suspended user's ID into this feature, it will list all dozen-and-a-half other users that favorited their one question (including me). Thirteen of those accounts were created in the last two weeks, like the OP's was.

If you use one-line notation, you can list all elements of $S_n$ systematically using lexicographic ordering. For example, $123,132,213,231,312,321$ lists the elements of $S_3$ based on the ordering of the corresponding $3$-digit numbers. If you want cycle notation - yes, you'll have to find all integer partitions of $n$, which reduces the problem to listing all cycles of all lengths. All $k$-cycles in $S_n$ can be listed as you would expect - they are the lists of $k$ out of $n$ elements, up to cyclic shifting of those elements.

Think about what you're saying. Are you suggesting you think if two elements are not generated by a common element, they must be the same element? That's absurd. Like, in $\Bbb R^2$, we know $(1,0)$ and $(0,1)$ are not multiples of a common vector, but why would we think that implies $(1,0)$ and $(0,1)$ are the same vector? Plus, $(123)$ and $(132)$ are generated by a common element - they generate each other. If anything, we would suspect this would imply they are not distinct (but this implication is also false).

Well $(abc)(acb)=e$ automatically implies $(abc)=(acb)^{-1}$. That's like going from $2x=1$ to $x=1/2$ (with the caveat that left and right inverses aren't always equal and don't imply each other's existence in the most general circumstances, but you should work out that they do in group theory). You use the fact $(acb)^3=e$ to get $(acb)^{-1}=(acb)^2$.

Nobody said $S_3$ is generated by one element. By the way, you're looking for the word distinct - $(123)$ and $(213)$ are distinct elements of $S_3$, not disjoint permutations. What even is your question?

Let us continue this discussion in chat.

No. First, $s(t)$ doesn't need to a be a bump function. And second, even if $s(t)$ is a bump function supported on $(0,P)$, the summand $s(t-0\cdot P)$ will not be the nonzero one if $t$ is outside of $(0,P)$. In general, if $nP$ is the closest integer multiple of $P$ to the left of $t$, then $s(t-nP)$ will be the unique nonzero summand.

If $s(t)$ is supported on $(0,P)$, then $s_P(t)=s(t\bmod P)$ if you interpret $t\bmod P$ as the unique value in $[0,P)$ that is an integer multiple of $P$ away from $t$. In other words, in the sum $\sum_n s(t-nP)$, only one of the summands will be nonzero (I guess that's what you mean by "fire"?) and all the other summands will be $0$. Was this not all obvious? Note that if $t$ is outside $(0,P)$, then the summand with $n=0$ will not be the nonzero summand, contrary to what you seem to be claiming.

Huh? What does that mean?

If $s(t)$ is supported on $(0,P)$ then you don't "read out" a nonzero value for $s$ when $t$ is outside of $(0,P)$. By definition, in this case, $s(t)$ is $0$ when $t$ is not in $(0,P)$. How is this a problem?

What's the issue? Just because $s(t)$ is only supported on $(0,P)$ doesn't mean it's only defined on $(0,P)$. By definition, if $s(t)$ is a bump function supported on $(0,P)$, that means $s(t)=0$ for $t\not\in(0,P)$. And this isn't even an issue if $s(t)$ isn't a bump function anyway. For example, if $s(t)=e^{-x^2}$, look at the graphs of $s(t)$ and $s(t+\pi)$, then look at the graph of $s(t)+s(t+\pi)$. It combines the two bumps into one graph!

No, I don't see your problem. Please tell. (Also, please say "infinite number of peaks" or "infinitely many peaks" - when you say "infinite peaks" it makes it sound like each peak is a singularity, which is not true.)

If $s(t)$ is a bump function supported on $(0,P)$ then the periodic summation will indeed have zeros at the integer multiples of $P$, but that's just a discrete set of valleys. In this case, you would say $s_P(nP)=0$ for all integers $n$, not $s_P(t+nP)=0$ for all $n>0$ (which would be incorrect). That doesn't necessarily happen in general, and I don't see why you think this is an issue when it does happen anyway.

The formula does not convey that idea clearly? I mean, if the graph of $s(t)$ is a bump, then the graph of $s(t-nP)$ is a translation of that bump by $nP$, and the sum of all these translations adds all the graphs together which gives a repeating pattern of bumps. I don't see what's not clear about this.

Why would you say $s(t+nP)=0$ for $n>0$? Nothing in the article indicates that. Writing functions as sums of other functions is pretty ubiquitous in math. We write polynomials as sums of powers, and lots of things as sums of trig functions or complex exponentials, etc. Why you would do it depends on the context at hand, but the broad idea is that symmetry allows for stuff to work out nicely (e.g. cancellation). But if you look at the graph of a periodic function and see a bunch of bumps, it seems pretty natural to think of it as the periodic summation of a single bump, just intuitively, no?

No. First, $s(t)$ doesn't need to a be a bump function. And second, even if $s(t)$ is a bump function supported on $(0,P)$, the summand $s(t-0\cdot P)$ will not be the nonzero one if $t$ is outside of $(0,P)$. In general, if $nP$ is the closest integer multiple of $P$ to the left of $t$, then $s(t-nP)$ will be the unique nonzero summand.

If $s(t)$ is supported on $(0,P)$, then $s_P(t)=s(t\bmod P)$ if you interpret $t\bmod P$ as the unique value in $[0,P)$ that is an integer multiple of $P$ away from $t$. In other words, in the sum $\sum_n s(t-nP)$, only one of the summands will be nonzero (I guess that's what you mean by "fire"?) and all the other summands will be $0$. Was this not all obvious? Note that if $t$ is outside $(0,P)$, then the summand with $n=0$ will not be the nonzero summand, contrary to what you seem to be claiming.

If $s(t)$ is supported on $(0,P)$ then you don't "read out" a nonzero value for $s$ when $t$ is outside of $(0,P)$. By definition, in this case, $s(t)$ is $0$ when $t$ is not in $(0,P)$. How is this a problem?

Huh? What does that mean?

What's the issue? Just because $s(t)$ is only supported on $(0,P)$ doesn't mean it's only defined on $(0,P)$. By definition, if $s(t)$ is a bump function supported on $(0,P)$, that means $s(t)=0$ for $t\not\in(0,P)$. And this isn't even an issue if $s(t)$ isn't a bump function anyway. For example, if $s(t)=e^{-x^2}$, look at the graphs of $s(t)$ and $s(t+\pi)$, then look at the graph of $s(t)+s(t+\pi)$. It combines the two bumps into one graph!

If $s(t)$ is a bump function supported on $(0,P)$ then the periodic summation will indeed have zeros at the integer multiples of $P$, but that's just a discrete set of valleys. In this case, you would say $s_P(nP)=0$ for all integers $n$, not $s_P(t+nP)=0$ for all $n>0$ (which would be incorrect). That doesn't necessarily happen in general, and I don't see why you think this is an issue when it does happen anyway.

No, I don't see your problem. Please tell. (Also, please say "infinite number of peaks" or "infinitely many peaks" - when you say "infinite peaks" it makes it sound like each peak is a singularity, which is not true.)

The formula does not convey that idea clearly? I mean, if the graph of $s(t)$ is a bump, then the graph of $s(t-nP)$ is a translation of that bump by $nP$, and the sum of all these translations adds all the graphs together which gives a repeating pattern of bumps. I don't see what's not clear about this.

Why would you say $s(t+nP)=0$ for $n>0$? Nothing in the article indicates that. Writing functions as sums of other functions is pretty ubiquitous in math. We write polynomials as sums of powers, and lots of things as sums of trig functions or complex exponentials, etc. Why you would do it depends on the context at hand, but the broad idea is that symmetry allows for stuff to work out nicely (e.g. cancellation). But if you look at the graph of a periodic function and see a bunch of bumps, it seems pretty natural to think of it as the periodic summation of a single bump, just intuitively, no?

(from slide 3 here: math.miami.edu/~armstrong/Talks/The_McKay_Correspondence.pdf)

any ideas what package can make this?

the before and after picture for the vector wrt the reference frame will look like it rotated clockwise, yes

$d\alpha$ is supposed to be a small angle

it's called change-of-coordinates

You can rotate the coordinate system (aka reference frame) while keeping the vector the same, yes.

bro, it's rotating the vector and keeping the reference frame the same

seems like much ado about nothing

What are you confused about? Rotating the coordinate system while fixing the vector is different from rotating the vector and fixing the coordinate system - they have, in a sense, opposite effects on how the vector is represented with respect to the coordinates.

different formulas for the same thing should give equivalent results. different formulas for different things generally do not.

I know now that's what you're doing. But that appears not to be what the uncited source with the formula is talking about.

(0,0,1)x(0,1,0)=(-1,0,0) you mean, yes

What's up?