« first day (3333 days earlier)      last day (28 days later) » 

12:01 AM
@Rithaniel: It's really taking the coordinate vector of a vector in your new basis back to the coordinate vector in the original basis, etc.
@s.harp I think you're missing a factor of $i$ here.
Of course, we can get this easily by Green's (Stokes's) Theorem ... to generalize.
 
Thanks, guys. Like I said earlier, this particular professor doesn't make things entirely clear during lecture, so you might see me asking questions about linear algebra stuff in here through the semester.
 
Hey @Ted
 
It’s always fun when I manage to get an interesting question by thinking about a poor one on the main site
 
hi @Mathein
 
This question, for instance, is essentially inscrutable: math.stackexchange.com/questions/3361542/find-arctan-cos3x-4
 
12:07 AM
As I said, @Rithaniel, you might find some of my lectures helpful (if not interesting), although I was assuming your course was more advanced.
This question seemed interesting to me, @Semiclassic, @Akiva, although I believe the OP cannot have a correct proof as he claimed.
 
But out of curiosity I plotted arctan(cos(x)), and found that it’s remarkably well-approximated by pi/4*cos(x) for all x
So that’s a bit more interesting
 
reversing the trig and inverse trig would be more standard.
 
Yeah, exactly
 
Some probably trivial diff geo question: how is a flat connection on a vector bundle a PDE on the underlying manifold?
 
@Mathein: The PDE is the statement that the curvature is identically 0.
 
12:11 AM
The other question in this vein which comes to mind is what the Fourier coefficients of arctan(cos(x)) would be
 
oh so a flat connection is a solution
 
hlo
@Ted do you know of a source in which someone actually works out the geodesics for the triaxial ellipsoid
 
That’d quantify what “well-approximated” actually means here
 
Yes, you can think of it that way, @Mathein. Although I wasn't. The space of connections is an affine vector space.
Hell no, @Eric.
 
i found the computation in some book from the 19th century
 
12:12 AM
Yeah, it's meant to be a graduate course, but what is "advanced" is subjective, I've found.
 
@ÉricoMeloSilva i want to say Arnold has something on that in his classical mech book
 
Works out in what sense, @Eric?
 
ik but he doesn’t do the computation
 
I mean, can we work out the geodesics on a torus? I describe them using the Clairaut relation. I guess that's an implicit ODE.
 
Gotcha
 
12:13 AM
@TedShifrin jacobi characterizes all the geodesics
 
Like, most math courses here are applied math, statistics, and analysis. Everyone thinks even basic group theory is "advanced"
 
Yes, but you need a global coordinate system and then an "explicit" solution, @Eric.
 
Yeah, I remember that Arnold references Jacobi
 
cause the triaxial ellipsoid is cool cause it’s sharp in that it has exactly 3 embedded closed geodesics
 
@Rithaniel: Good grief. Applied mathematicians, more than anyone besides differential geometers and representation theorists, need to know a lot of linear algebra.
 
12:14 AM
and every S^2 has at least 3 yadda yadda
 
@TedShifrin yeeeep
 
I didn't know that, @Eric.
The ellipsoid, I mean.
 
i wanted to see if it was like feasible to get my hands on this and work it out but it might be too much of a hassle
 
So you'd need to work in ellipsoidal (stretched spherical) coordinates?
 
yeah the only ones are the three obvious ones
idk how it’s done at all
 
12:16 AM
Someone I help sometimes posted on main an exercise from his homework to find the umbilics on a (specific) tri-axial ellipsoid. I told him good luck :P
My intuition is that there are none, but I haven't thought about it.
 
yeah no it’s nightmarish to do this kind of stuff it was just bothering me that literally the only source i could find was from 1840
@TedShifrin it’s a conjecture that every closed convex surface has at least 1 umbilic i think
or maybe it’s at least 2
 
I once knew this sort of thing, but no longer. I remember hearing a problem about umbilics on ellipsoids ...
I supposed we could google and find out that A'rnold wrote a paper on it. That happened to me several times years ago :P
 
actually i think this conjecture has been proven for things this smooth
so the triaxial ellipsoid definitely has at least 2 umbilics
but that’s hard
 
What if we take an ellipsoid that's a surface of revolution?
 
wouldn’t the two extreme points along the axis of revolution be umbilics lol
 
12:22 AM
Yes. I was wondering if there were obviously more.
 
uhh might be but it’s not obvious to me
 
For this there's some hope to see geodesics, using Clairaut.
 
yes this is much easier to work out than for the bad ellipsoid
 
Hello. I asked a question about scalars on the exchange and wanted a bit of clarification. Anyone willing to discuss?
 
Have you found good places to get Brazilian food in NYC yet? :)
Scalars, @Kieran?
 
12:24 AM
no but i’m part of a whats app group to find brazilian food
 
aha
 
@TedShifrin Yes, scalars.
 
Philadelphia has some fabulous restaurants, too.
 
In what context, @Kieran?
 
12:27 AM
Well, I think that I understand the definitions of vector spaces and fields, and the desire to keep things "content free" in a mathematical formulation, but I am used to the idea of building up more complex things from simpler ones - e.g. groups first, etc. But the answers I am getting imply that a scalar is a higher level concept than a field because I cannot define what I mean by scalar without first defining a field.
 
No, no, no.
Oh, I see what you're thinking. You do definitely need to know what a field is, and then you're just talking about elements of that field. More generally, scalars can be thought of as elements of a ring (when you discuss modules rather than vector spaces). But fields are very basic. :)
Most math students, I would say, learn fields LONG before they learn groups, rings, etc.
 
Ok, that helps a bit.
 
I mean, we take $\Bbb R$ and $\Bbb C$ as "known" structures, pretty much, and do linear algebra over $\Bbb R$ and $\Bbb C$ without making a big deal about it. Later on, after you've learned groups, rings, etc., you might discuss vector spaces over more general fields.
 
I would like to be able to say that a scalar is any element in a set that I form that has addition and multiplication analogs that allow you to form a field.
I would like to be able to say that a cabbage is not a scalar. :-)
Is there any context in which a say... a set of matrices are scalars?
 
if you allow for a non-commutative basering (which some people definitely do), yes
 
12:32 AM
Oops, i"n which to say."
 
The set of diagonal matrices with the same entry down the diagonal (in other words, constant multiple of the identity) can be interpreted as "scalars" when the vector space is the set of (square) matrices.
My honest reaction is that you're making too big a deal out of something, but I can't tell what you're really after.
 
square matrices form a ring, you have addition and multiplication satisfying all the necessary axioms (but no inverses and no commutativity), so you can do modules over that
in that case a matrix would be a scalar
 
(I understand the diagonal matrix point, but that's sort of an edge case). But your earlier comment is helpful. I had thought that there was no context in which matrices could be called scalars.
 
I think he's sticking to fields/vector spaces at this point, @Mathein.
Although I alluded to modules above.
 
for example, nxm matrices are a module over nxn matrices
 
12:35 AM
We could take a cabbage to be a scalar if we could take cabbages mod 3, for example. Or allow real numbers of cabbages.
 
another example where you have a field of matrices is if you embed $\Bbb C$ as a subring of $\mathrm{M}_{2\times 2}(\Bbb R)$
 
And I suppose we could multiply cabbages by vectors of cabbages in the obvious way ... or maybe even some other vectors (cabbage times parsnip is another parsnip). :P
 
you could also look at the field of formal Laurent series in a variable called "cabbage"
 
I am teaching a course of Math Methods in Physics. (I know, never let a physicist teach math). One approach to defining scalars vs. vectors vs. tensors is how they behave under coordinate transformations. But then I started to look at how a mathematician actually defines a "scalar" and it doesn't connect well to this approach.
@TedShifrin You are confusing a cabbage with the representation of a cabbage :-)
 
I can confuse things if I want. That's what isomorphism is for.
Yes, physicists muck up tensors badly, although I do agree that the cross product of two vectors should be a pseudo-vector and not a vector (because we suppress the exterior product, which is really what's going on).
You're just going to say that the scalars are unaffected by changes of coordinates to the underlying vector space. Why is that different?
 
12:40 AM
Thank you for the discussion. If a matrix can be a scalar in some mathematical contexts then we are using the words very differently. (I think) (If it is a Cartesian matrix,)
 
to this day I don't understand this "a vector is something that transforms like a vector"-thing
 
Sure you do, @Mathein. Change of basis formula for vectors, matrices, tensors. That's all it is.
 
I get that, but how is that supposed to be a definition of what a vector is?
 
And @Kieran: You should teach your students differential forms and not to be scared of them.
 
We say that because in computer science a vector can be two column of social security numbers and grades. That set of numbers can't rotate or transform from one element into another.
So we don't want to call them vectors.
 
12:43 AM
Well, a vector is an element of a vector space, and we think of the vector space as one copy of itself, as opposed to multiple copies of itself and its dual.
 
I wouldn't call the computer science vectors vectors either
but not because I can't rotate them
 
There is an underlying structure (from our point of view) that when your make a transformation the elements must obey certain rules.
 
Well, they are inside the space $\Bbb R^2$, but the units have no meaning.
Anyhow, I'm going to cook dinner. Have fun, guys!
 
Thank you very much for your patience.
 
Enjoy @Ted
 
12:45 AM
Good night.
 
BTW, @Kieran, you might try the math pedagogy pages, although I haven't been there in ages.
 
@KieranMullen from my pov, a vector is something that can be scaled and added to other things from the same vector space. The fact that there are transformations of vector spaces is not integral (to me) for what a vector is. How do you even define a transformation of a vector space without knowing what a vector is?
 
1:39 AM
@LeakyNun yes.
note that if $(x_1, \dots, x_n) \in \Bbb Q_p^n$ is a solution to a homogenous equation, then we can rescale it to a solution $\Bbb Z_p^n$
so we find a solution in $\Bbb R \times \widehat{\Bbb Z} \subset \Bbb{A}_{\Bbb Q}$
one that is non-trivial at each place even
if we would only require non-trivial at a single place, then the answer is even easier
 
1:55 AM
cool
 
2:21 AM
Multitopology:
Let t1, t2 be two different topologies on the set {A,B,C}
For any net beginning on A or B, the convergent nets are those in t1, and for any net beginning on C, the convergent nets are those in t2
It remains to be proved whether the resulting construct is always a topology
 
2:56 AM
So, using the Sierpiński space centered at 0 or 1 respectively as an example, taking the union of them will give the trivial topology, while their intersection give the discrete topology
However, the multitopology constructed by having all sequences starting at 1 to take the Sierpiński space center at 0 and rise versa for all sequences starting at 0 has a set of convergent sequences intermediate between those of the respective Sierpiński spaces and the discrete topology
Thus the multitopology is something stronger than a topology, of which it forms a tree with the other topologies
For those who want the real deal (which has nothing to do with the stuff I made above), read here: pdfs.semanticscholar.org/c89b/…
 
 
1 hour later…
4:22 AM
ok on second thought, the above is still a topology
We just have open sets defined in terms of what features its convergent sequences have
It is thus an example of a topology which cannot be visualised easily despite the underlying set is only of cardinality 2
Let $\tau =\{X,A,\varnothing\}$
where $A$ are open sets such that given any $\phi : X \to X$ and $x \in I$, $\lim \phi_x = p$ for each $p$, and $\phi_0 (X) = q$ where $q\neq p$
We can shorthand this by defining a relation $R$ thus the above means satisfying $R(\phi,p,q)$
(I will figure out how to prove the general case later)
 
2
A: Choosing squares from a grid so that no two chosen squares are in the same row or column

Gerry MyersonHere's the hard way to do the problem: inclusion-exclusion. There are $25\choose3$ ways to choose 3 squares from the 25. Now you have to subtract the ways that have two squares in the same row or column. There are 10 ways to choose the row/column, $5\choose2$ ways to choose the two squares in...

 
Let $X = \{0,1\}$. Then we have the convergent nets satisfying:
 
In this problem, what if the chosen squares are distinguishable?
Would we just multiply by n! where n is the number of squares we are selecting
 
4:39 AM
You will do the same thing except using permutations instead to take account of the ordering in picking the squares. The relationships between permutation and combination is the factor $r!$ which here it should be $r=3$
 
$R(\phi, 0,1)$ or $R (\phi,1,0)$
Thus $A$ is open if $R(\phi,0,1)$ or $R(\phi,1,0)$
Taking unions gives all the convergent nets. Since the tail of any convergent sequence has to end somewhere, we have this union being $X$. Meanwhile its intersection are all the sequences that converges simultaneously to 0 and 1. This is possible because $\tau$ is not required to be $T_1$
Hence unions, and intersections are both closed and $\tau$ defines a topology
In particular, this topology is unusual because the constant nets are not convergent
 
@Secret How is that different from what I said?
 
ok sorry, I used to see $r$ as the selecting element, yes what you said is the same as mine
Distinguishable squares is basically the same as labelling the whole array with 1 to 25, and then select 3 of them, so it is a permutation as expected
the formula for the exclusion will probably be a lot more complicated as it need to account for which number in which column/row
 
4:58 AM
Right makes sense
But, doesn't that get more complicated if we only have certain pieces that are indistinguishable from one another but distinguishable from others?
I.e. what if we have 3 pawns, 2 rooks, 2 bishops and we are putting them on an 8x8 board such that no two pieces of the same type are in the same row or column.
So having 2 pawns on a row is not allowed but having a pawn, rook, and bishop all in one row is allowed
 
If that is the case, then you will need to model all the choices with the Permutations of multisets formula
which obviously becomes more complicated
 
I was thinking of inclusion exclusion and just handling every case that wouldnt satisfy this
But im not sure if that works
Or is doable
 
Another way I can think of modelling this is to use coordinate pairs (x,y)
Then whether pieces share the same column or row reduces to the problem of how many possible ways you can pick (x1,y1), (x2,y2) such that x1=x2 or y1=y2
which maybe is a lot more doable
 
in this case, what are x1, x2, y2, y1 representing
 
x= rows, y= columns
you have a 5x5 grid so x and y ranges from 1 to 5
Using this, the problem reduces to how many 4 digit numbers such that the odd positions and even positions do not agree
 
5:12 AM
Yea but whats the significance of making it such that x1 = x2
4 digit numbers? do the individual digits represent x1y1x2y2?
 
You do not want the squares selected to be on the same row, right? That is the same as you do not want the coordinates of the squares to share the same x values
 
Yea but they can be the same row if they are distinct
i.e. you can have a rook and a pawn in the same row
But you cant have 2 rooks in the same row
or 2 pawns in the same column
 
Also since you pick squares one by one, you have three coordinate pairs (x1,y1),(x2,y2),(x3,y3).

You can also add an extra parameter to denote what piece you are handling, e.g. 1=rook, 2=pawn etc.
Thus for each piece you have a tuple (t,xi,yi)
where t is the type of the piece, and (xi,yi) is the position of that piece
Then assuming you have 3 pieces, you thus basically have the tuple (t1,t2,t3,x1,x2,x3,y1,y2,y3)
Thus your requirements translates as follows:
"you cant have 2 rooks in the same row" => (t1=t2=2 and x1=/=x2) or (t1=t3=2 and x1=/=x3) or (t2=t3=2 and x2=/=x3)
"you can have a rook and a pawn in the same row" => (t1,t2 in {1,2} t1=/=t2 and x1=x2) or (t1,t3 in {1,2} t1=/=t3 and x1=x3) or (t2,t3 in {1,2} t2=t3 and x2=x3)
 
Why would we have three coordinate pairs?
 
the first entry denote the type of piece you are using, the other two denote its position on the board
you have 3 tuples because one for each piece and its position
Thus my board can look something like this:
(1,1,1) => rook at (1,1)
(2,5,4) => pawn at (5,4)
(1,3,2) => rook at (3,2)
Thus the above configuration of my broad summarises into:
(1,1,1,2,5,4,1,3,2) assuming I place the pieces in that order
 
5:23 AM
I get the logic but I still don't really follow how computing this by hand would really work. How would you work this out with say a 4x4 grid and 2 rooks and 2 pawns?
looking at (t,x,y)
 
Using your example, we knew all possible board configurations will have something like this:
(1,x1,y1,1,x2,y2,2,x3,y3,2,x4,y4)
we knew that no two rooks can occupy the same row, thus we have:
x1=/=x2
This means, for any pawn configurations, the rook has 3-4 possible places to choose given x1,x2
 
Is this correct: if we just look at the rook (lets assume t = 2) then each rook position is denoted by (2, x , y) so with the first rook, you have 4 slots for x, 4 slots for y, then for the second rook you either have 4 slots for x and 3 slots for y or 3 slots for x and 4 slots for y. So you have (4x4)[(4x3) + (3x4)] = 384
No wait, you would always have 3 slots for the x value for the second rook, right?
 
The first rook will have 4 slots for x and 4 slots of y, yes. The second rook will only have 3 slots for x (because no two rooks share the same row) and 4 slots for y
but yes the logic proceeds in that fashion
 
So after getting the total number of orderings for the rooks and pawns individually, do I just multiply those? Does product rule apply in this case?
 
Presumably the ordering of placing the chess pieces should not affect the conclusion, but it is a bit tedious for me to check that by hand, otherwise, you have to compute all possible permutations for each rook and pawn order. That's 4! times of the above reasoning, which the details may differ if the order matters
actually, the order will matter, placing pawns first the first rook can end up with less than 4 slots for rows and columns because pawns already occupied the board
so you have 4!=24 different terms in your sum possibly governed by some symmetry such as the column and row rules of the rooks and pawns
Product rule will probably only apply for some of the terms had the rules are something like, "no two rooks nor two pawns should occupy the same row" then you have a symmetry between rooks and pawns, and some of the 24 cases will be degenerate
 
5:45 AM
Thats tedious...hmm
There must be a simpler way to generalize it to bigger boards
 
5:57 AM
The eight queens puzzle is the problem of placing eight chess queens on an 8×8 chessboard so that no two queens threaten each other; thus, a solution requires that no two queens share the same row, column, or diagonal. The eight queens puzzle is an example of the more general n queens problem of placing n non-attacking queens on an n×n chessboard, for which solutions exist for all natural numbers n with the exception of n = 2 and n = 3. == History == Chess composer Max Bezzel published the eight queens puzzle in 1848. Franz Nauck published the first solutions in 1850. Nauck also extended the puzzle...
The 8 queen problem is basically a subset of the problems we tried to model above, so yes, expect it to be very hard if there is little symmetry
 
1
Q: 2 white and 2 black rooks on chessboard

AlexI realize this is a fairly basic combinatorial problem, but I'm still confused. In how many was can 2 black and 2 white rooks be placed on a standard chessboard such that white and black don't attack each other. The way I saw it first, there are 2 cases: Case 1. Both white are on the same ro...

It may be closer to this problem
 
6:11 AM
I think the problem becomes a lot simpler if we suppose no two pieces can be on the same row or column, regardless of their type. @Secret
But in this scenario, would we just have to consider the permutations among the distinct pieces (rook vs pawn)
 
 
6 hours later…
12:17 PM
It seems there is a variant of the Kelvin-Stokes theorem stating something along the lines of $\int \mathrm d\mathbf r\times\mathbf F=\iint (\mathrm d\mathbf S\times\nabla)\times\mathbf F$. I hope it has a tidy derivation. approach0 doesn't seem to help, but I found a couple MSE posts basically deriving individual components of it. I'd like to see a reference/standard-derivation for this form. Do you think it's worth posting a reference-request question for?
 
1:11 PM
I don't know what the notation means but if that's true it is going to just be a rephrasing of the general form of Stokes' theorem with differential forms to the special case of objects in R^3. This is a very common phenomenon.
 
1:31 PM
I just realized addition in an elliptic curve agrees with addition of points in the Picard group of the elliptic curve. That's so surprisingly natural.
(In fact, Pic E = E x Z by some pure homological algebra arguments. The Z copy comes from the degree homomorphism Pic E -> Z)
 
Indeed
 
Is it somehow clearer from thinking in terms of line bundles instead of divisors?
 
What's the slickest proof you can come up with of that?
I know the fact but I do not see an easy argument right now
 
If P, Q, R are distinct points of E, then P + Q + R = 0 means there's a line through P, Q, R in CP^2. Restrict equation of that line to E to get a meromorphic function. This is the easy direction.
 
how does E sit in CP^2?
 
1:36 PM
Conversely, if (P) + (Q) + (R) = (f) as divisors, f = 0 gives a curve in CP^2 intersecting E at P, Q, R with multiplicity 1 each
By Bezout, f has to be degree 1
That's a line passing through P, Q, R
@MikeMiller E is y^2 = x^3 + ax + b in C^2. When you compactify, you get y^2z = x^3 + axz^2 + bz^3, so you just add a single point (0 : 1 : 0).
It intersects the line at infinity CP^1 = {z = 0} at (0 : 1 : 0) with multiplicity 3, the so-called inflection point
That's "the" point at infinity of the elliptic curve.
Dinnertime for me, but I'll come back
 
2:26 PM
Does Halmos conclude B is not a set (last page of Section 2 and Section 2 in Naive Set Theory ) ? Book Link from google: piazza.com/class_profile/get_resource/is25bi6c6o1oh/…
Does Halmos conclude B is not a set (last page of Section **2** and last page of Section **3** in Naive Set Theory ) ?
Book Link from google: https://piazza.com/class_profile/get_resource/is25bi6c6o1oh/isv14pknyni2z3
 
2:54 PM
@NewStudent Is there a phrase in section 2 that you're focusing on where it sounds like that is being said? If you were already familiar with Russell's paradox or thinking about section 3, it's important to note that Halmos's B in section 2 is not the same problematic set as in alluded to in section 3.
@MikeMiller yes, but 1. That rephrasing is tidy for people not very familiar with the general form of Stokes' Theorem and 2. Whether it uses differential forms or not, I'd like to see a fairly direct derivation (maybe you could write one). I think I probably will post a question once I can gather my thoughts/references.
 
If I have a bimodule over some ring, is there a convenient space into which the left-linear and the right-linear maps embed into?
 
I'm suggesting that if you want to see a proof you should try to derive it from the general theorem and then rephrasing. I am not interested in writing such down.
I am certainly aware that the differential forms phrasing is not familiar to most Calculus students (and hence physicists, engineers, etc).
 
3:25 PM
Can someone explain why the integers cannot be decomposed into two cyclic subgroups. I'm trying to prove it and I can't seem to get a grip on it.
 
because there are no cyclic elements in the integers? (else $1$ or $-1$ would have to be cyclic)
or what does cyclic subgroup mean
 
A cyclic group is a group with a single generator.
 
but $\Bbb Z$ is generated by $2\Bbb Z + 3\Bbb Z$ (these are not disjoint, so I guess you want that)
if you are looking at generators $p, q$ then $p\Bbb Z \cap q\Bbb Z$ contains $p\cdot q$
 
3:43 PM
@MikeMiller thanks for your attention/response. I think I wasn't very clear that I've already proved it through those sorts of methods, I just don't have an elegant derivation.
 
 
1 hour later…
4:56 PM
@krauser126 it certainly will, because in this new case, the only difference are the rook and pawn pieces
 
5:25 PM
@MarkS. Even with forms, I see no elegant approach. You have to do it component-by-component, I do believe.
 
6:16 PM
This was really involved to write on chat, can someone have a look at this question I posted:
https://math.stackexchange.com/questions/3362151/generalizing-chains-of-connected-sets-to-arbitrary-indexing-sets
 
6:59 PM
@TedShifrin I am at a loss on the non-vanishing section thing now.
I thought it was something like Sard's theorem applied to the map $S^1\times S^2\to\mathbb R\colon(\theta,p)\mapsto\|K'(\theta)\times p\|$.
But I don't see it if it is.
 
7:34 PM
it seems there should be a general argument that if multiplication is continuous the product of two closed sets is closed
 
 
1 hour later…
8:55 PM
@s.harp product being $A\cdot B$? Or $A\times B$?
And where are A,B located?
And closed in the topological sense?
 
$A\cdot B$ on some normed algebra or topological group etc
 
9:56 PM
@MarkS. the only intelligent remark that comes to mind for me is that, since your formula is vectorial, you can compute components by dotting with a constant vector $\vec{c}$
At which point the LHS becomes $$\int \vec{c}\cdot(d\vec{r}\times \vec{F})=\int d\vec{r}\cdot(\vec{F}\times \vec{c})$$
In which case the usual form of Stokes-Kelvin applies
What isn’t so obvious is the surface integral
 
10:35 PM
@anakhro I suggested finding a point that was on no tangent line to $K$.
Its existence is immediate by "silly Sard."
 
I am working on a heuristic algorithm that tries to find a representation in radicals for polynomial roots of degree >4, in part based on mathematica.stackexchange.com/a/134551/7288, mathematica.stackexchange.com/q/34011/7288 and mathematica.stackexchange.com/q/105933/7288.
It appears that many roots "naturally occurring" in analysis are solvable in radicals. E.g., see results in math.stackexchange.com/a/3345906/19661, math.stackexchange.com/a/3353410/19661 and math.stackexchange.com/a/3353303/19661. Can anybody please provide more polynomials of higher degrees that might be solvable in radicals, which I could use to validate and improve my algorithm?
 
11:18 PM
does anybody know anything about quasi-nilpotent operators? ive been looking through google for any useful general result about them, they seem to be hell
 
@TedShifrin I thought that would involve finding a particular function whose regular values I can guarantee.
 
11:41 PM
@s.harp they're almost nilpotent
 
@anakhro Hmm ... can't use my ipad to chat any more. Hint: Consider an appropriate map $TK \to \Bbb R^3$.
 

« first day (3333 days earlier)      last day (28 days later) »