no. for example, 5 is prime and not in the image of that map.
if you acknowledge the even prime, 2 is also not in the image of that map.
because it's one-to-one and not onto, there will be left inverses for it, but no unique left inverse. some arbitrary choices can be involved (e.g. if g is any given left inverse, you could redefine the value of g at 2, or at 5, and create other left inverses that are not the same as g)
i guess what im wondering is do all the primes exist in a sequence defined by the series above
0,1,1+1,1+2,3+3,6+4,10+5 etc
also is a direction field for dy/dx = f(x,y) just a graph of f(x,y) with x axis and y axis and each point is the slope
i remember seeing in class that some direction field presented to us had some arbitrary spacing between each slope point to demonstrate the solution curves pathing
ugh i guess i mean like is a direction field just y-axis is the f(x,y) function values for each x-axis value as the independent var
sorry, are you now asking if every prime appears as a prime factor of at least one of the numbers 1, 1+2, 1+2+3, ...? the answer is yes. any integer (whether prime or not) appears a divisor of infinitely many of those sums
I am quite confused (rightfully so due to my absence of any knowledge of this subject). Say we have Lie group SU(2) and its associated Lie algebra su(2). Then I know there is the exponential map $e: su(2) \rightarrow SU(2)$. In physics, it seems it is often written that this exponential itself acts on some Hilbert space. I am confused about explicitly what the mappings here look like
Say we have a Hilbert space $\mathcal{H}$. I assume we first define some sort of action (perhaps more particular some type of representation?) of $su(2)$ on $\mathcal{H}$ to make first contact with the Hilbert space?
and then okay, use the exponential map to go from su(2) to SU(2) and then how do we make contact from here with the hilbert space? in undergrad physics textbooks, we taylor expand the exponential, but is taylor expanding some sort of map from SU(2) back into su(2)?
Question: If you have a generating function for a sequence, does the behavior of that function tell you anything about the sequence? (Beyond the obvious fact that its derivatives generate the sequence, of course)
I've never quite had a primer on this topic. I've known of the existence of generating functions, but nothing more in-depth
@shintuku to see that one we just map the matrix $\begin{bmatrix} \cos\theta & -\sin \theta \\ \sin\theta & \cos\theta \end{bmatrix}$ to $(\cos \theta, \sin \theta)$.
but the SO(3) part is tricky (I don't understand it yet). The book gives a proof using quaternions.
@Thorgott ahh, I think I understand. I take (x,t) to xt.
I understand it intuitively though.
it's like take a cylinder with no bottom. Break the top of the cylinder and spread it into a disk, since there is no bottom of the cylinder-it will correspond to centre of the disk.
"cylinder with no bottom" sounds like a math inflected personal insult. hey, did you meet so-and-so's husband? cylinder with no bottom if i ever saw one.
did you read that "multivariable mathematics" book? no bottom to that cylinder.
My experience reading this book (Hyperbolic Knot Theory) has been: read a paragraph or two, close the book and look away, hallucinate vividly try to visualize
you can't hallucinate vividly if you microdose the book, akiva. you gotta do a few chapters all at once. make sure you have a friend and a few bottles of water around, and a controlled environment.
In any case, there's a certain type of topology book that I've been describing as "Being described an acid trip through text"
and I say this despite the fact that the book has pictures
Just, like, not enough lol
A Topological Picturebook also falls into that category even though it has tons and tons of pictures. I guess the pictures are incomprehensible without the text, and the text is describing stuff involving motion which is hard to do with pictures.
It's a shame The Geometry Center only ever made three videos
@AkivaWeinberger I just got to the part where they rotate around the center of a face in the honeycomb of ideal dodecahedra, and then the perspective made the space look flat. I think my brain exploded
@Rithaniel Yeah, that place is called a cusp. Notice that (a) every time it continues rotating and it looks like that again, they're in a different cusp each time
and (b) if the camera went further into the cusp, the parallel-looking lines would get closer and closer together
For any epsilon, there's only a compact set of points where the epsilon-ball around it doesn't intersect other copies of itself @Rithaniel. The cusps are "thin"
@Rithaniel Or, I mean the bit at 13:50
@Rithaniel When you're in a cusp looking out, you see what looks like a "plane" made of squares. That's a horosphere, a two-dimensional subset of hyperbolic space that has the same intrinsic geometry as Euclidean space.
Yeah, that makes sense. From the "honeycomb of hyperbolic space" pov, you're moving further and further into an infinitely-distant corner where the dodecahedra meet, and the lines are getting arbitrarily close together
And yeah, a horror-sphere is a plane where all points are equidistant from that ideal point, right?
In the half-space model, horospheres about the point at infinity correspond to horizontal planes, and horospheres about any other point on the ideal boundary correspond to spheres tangent to those points
The video shows "native" or "true" perspective (eg light following geodesics), but I think visualizing the half-space model is also a very important mental crutch as well
because you have coordinates so it's easier to reason about
I'm giving a short presentation on Bernoulli numbers and, during my prep, I was trying to see if there was any connection between hyperbolic space and the coth function (if there was, I could use that to link the Bernoulli numbers to hyperbolic space and make my talk about geometry)
@AkivaWeinberger Oh yeah, projections are immensely useful because you can then use some intuition with Euclidean geometry
So, the generating function for the Bernoulli numbers is $\frac{x}{e^x+1}$ and you can subtract out the $B_1x$ term to get that $\frac{x}{2}\coth(\frac{x}{2})$ generates the even Bernoulli numbers
I think the distance between two points in the hyperbolic ball/disk model is related to the hyperbolic tangent?
Probably if the first point is the center
@Rithaniel I'm taking a combinatorics class and let me tell you, generating functions are fantastic
Dunno how useful they are here though
@Rithaniel Oh another thing about this shot: there are infinitely many images of cusps, but because of the whole "multiple images of one object" thing you get here, there are actually only three cusps: one for each component of the Borromean rings
@AkivaWeinberger Yep, and there are some good identities, like $\cos(\prod(x))=\tanh(x)$ (where $\prod(x)$ is the angle of parallelism at distance $x$) but now you're talking about the inverse of tanh
If you cut off a bit from each cusp the result is a compact set (though it looks noncompact 'cause of the repeating)
@Rithaniel Incidentally, my combinatorics professor (following the textbook) doesn't really do segues. Like, we finish counting one sort of object, then it's "Let's look at this other thing to count"
We recently did alternating permutations (like 1324 or 1423 where it goes up, down, up etc)
and the two students next to me (out of like five total) were very amused to see the generating function: $\sec x+\tan x$
"Were you not expecting it to be so simple?" I asked. "We weren't expecting it to be that," they replied
(Exponential generating function, so you divide by $n!$ first)
Questions 3, 5d, and 6 were pretty cool I thought. I wrote about 7c earlier (I couldn't find an elegant way but someone else online figured it out). 9b was somewhat interesting.
For question 3, it's like 136245 - the only "descent" is 62
I didn't use generating functions in this p-set, though
Let $Z$ be the pushout of the diagram
$\require{AMScd}$
\begin{CD}
A @>{f}>>X\\
@V{g}VV @VVV\\
Y @>>> Z
\end{CD}
$Z$ is defined to be $Z:=(X\sqcup Y)/f(a)\sim g(a)\ \forall a\in A$
Let $W$ be any space. Then Prove that $Z\times W$ is pushout of the following diagram
$\require{AMScd}$
\begin{CD}
A...
I studied about categories a bit. $C$ is a category. Then we talk about $A\in $obj(C). I don't understand 1) why not say $A\in C$, 2) how is a class different from a set?
i'm not someone who ever cared professionally about this stuff, but i think the reason of distinguishing obj(C) and C is that a category is more than (i.e. not just determined by) its objects
and informally a class can be "bigger" than a set but sometimes, depending on the category, those collections are sets
e.g. the underlying set in a group structure can be essentially any set at all, so 'the collection of all groups' might contain within it 'the set of all singleton sets', 'the set of all two element sets,' etc and none of these are actually 'sets'
please don't take advice from me on what you can just say in category language, but if you insist, i would suggest avoiding it until it seems like you're wasting more time and space by avoiding it than you would be by using it
@leslietownes what is 'bigger' than a set? I mean there is no bound on the cardinality of a set. Given a cardinality a, we always have a larger cardinality $a^a$.
koro none of it will make any sense unless and until you study axiomatic set theory. it is necessarily part of a formal discussion, not an informal discussion. hence the use of quotes around 'bigger'
Proposition: For a set $X$ and its power set $P(X)$, any function $f\colon P(X)\to X$ has at least two sets $A\neq B\subseteq X$ such that $f(A)=f(B)$.
I can see how this would be true if $X$ is a finite set, since $|P(X)|\gt |X|$, so by the pigeonhole principle, at least two of the elements in $...
see the multitude of answers to that, with people bringing up all kinds of crazy crap about what you need to assume to get it or not and what it's consistent with.
Is Jacobson's book 'Basic algebra I, II' good? I'm mainly focusing on Module theory and representation of finite groups. I'm avoiding Dummit Foote and Lang's algebra textbooks. I felt it was more like an encyclopedia.
Suppose that D is disk, S is solid torus (that is the ends of a cylinder identified via identity map). I have shown that $S$ is homeomorphic to $D^2\times S^1$.
How do I show that the boundary of $D^2\times D^2$ is $\partial D^2\times D^2\cup D^2 \times \partial D^2$?
well, $D^2\subseteq\mathbb{R}^2$, so $D^2\times D^2\subseteq\mathbb{R}^4$ and you should interpret boundary that way
more generally, you can show that if $A\subseteq X$ and $B\subseteq Y$, the boundary of $A\times B\subseteq X\times Y$ is $\partial A\times B\cup A\times\partial B$ (where $\partial A$ is the boundary of $A$ in $X$ and $\partial B$ the boundary of $B$ in $Y$)
there's an intrinsic notion of boundary for manifold that is related to this one and yields the right result for $D^2\times D^2$, but you need not worry about this
because in general I'm getting: $\partial (A\times B)= (\partial A\times \bar B)\cup (\bar A\times \partial B)$.
Here is my working: $\partial (A\times B)= \frac{\bar A\times \bar B}{A^o\times B^o}$. Take any (x,y) in this. Then either $(x\in \bar A, x\notin A^o, y\in \bar B)$ or $(x\in \bar A, y\notin B^o, y\in \bar B)\implies (x,y)\in (\partial A\times \bar B)\cup (\bar A\times \partial B).$
the other direct can also be shown. This proves the equality.
Indeed, this answers my question as well.
Thanks a lot @Thorgott for suggesting the generalisation.
@MagnusAlexander put two copies of the fundamental square of a Möbius strip next to another and glue them, you will obtain the fundamental square of a cylinder
@Koro surely the relevant definitions are in every general topology textbook
the diagram just organizes the data
I guess there is a point at which, depending on context, you have to decide whether you wanna talk about "the pushout" or "a pushout"
Also, I think that your definition of gamification is too narrow. Gamification is about adding game-like incentives to a system in order to encourage people to engage and participate.
It is often used in education, but is broader than that.
E.g., the SE model, which uses game-ified incentives to get people to post content (reputation, badges, etc).
When you're right, you're right. I'm just still peeved that I googled Django gamification engine and all that came up were several GH projects with badges
@TedShifrin "I had a lot more fun teaching analysis in the context of Spivak" Would you recommend reading Spivak cover to cover for a math major and attempt all problems, or is there anything you would skip?
@ILikeMathematics If you really want to learn a topic, and you don't have an instructor or advisor to guide you, then I would recommend that you work through at least one text on a given topic in detail, but also keep a couple of other texts on hand to skim through or use as references. Work through as many exercises as you can, as this will build your "mathematical muscles". And don't get so hung up on which book you are reading. Just pick one, and get to work.
@ILikeMathematics Definitely there are things to skip. I would skip the appendices to chapter 4, chapter 15-17, 21, unless you're particularly interested. The course I taught stopped at chapter 24. There are plenty of exercises, but do selected "routine" ones — not all — and watch out for exercises with two and three stars.
Those are probably too hard, although most are quite interesting.
@leslietownes sorry im still on my phone cant find my last message: This doesnt seem trivial to me. As the sequence continues you have larger gaps between elements. It's like leapfrogging in the natural numbers if theres a chance you dodge a prime number forever how would you prove that
@Obliv there is a well known formula for the kth term of the sequence as an explicit polynomial function of k. i would prove the results we were discussing earlier using that formula.
although it is not unheard of for a student to say "i spent _ hours and still have no idea", where _ is, uh, somewhat larger than the time actually spent.
@Rithaniel When I read mathematical publications, I consider one page per hour to be a not entirely unreasonable pace. It often takes that long to actually work through and understand what is going on. I usually shoot for 2-3 pages per hour.
I don't see any reason to expect students to be able to go any faster.
