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1:00 AM
Same goes to @Semiclassical. Thanks, I think that clears it up.
 
@Semiclassical and then it's just some numerical linear algebra calculations
 
right.
the nontrivial part of that, now that i think of it, is establishing that that system has full rank
 
Huh, that can be shown in general?
 
i think so, though i'm having to juggle some algebra in my head
 
I would think you could do induction on $d$
 
1:02 AM
that's probably a lot easier, yes
 
Sure, I can buy that.
 
If you're working strictly over the reals, you need to show you can integrate $\int \dfrac{dx}{ax^2 + bx + c}$ and $\int \dfrac{x\ dx}{ax^2 + bx + c}$
because you might have a quadratic factor with no real solutions
 
Right. that can be done by completing the square, IIRC
Speaking of partial fractions, do they have any use other than the integration of rational functions? That's the only place I see them taught
 
But yes, these types of integrals are usually established "before-hand"
 
well, they come up in complex analysis big time
mostly because complex integration of stuff like $1/x$ is a lot more interesting than real integration
 
1:08 AM
@Semiclassical Oh, right, with the residue theorem and all. I can see that
 
right
and while it's not something I know a lot about, i know that in spectral theory they care a lot about objects of the form $(z I-A)^{-1}$ where $A$ is some matrix and $I$ an appropriate identity matrix.
in that case you're no longer working with polynomials, but you still get useful results regarding how to decompose that matrix inverse into a sum of other inverses.
 
It occurs to me they might also come into play in perturbation theory, where you might want to reduce a 2nd-order approximation in a quotient to the sum of two first-order approximations, which might "blow up" slower
 
aye, or make it obvious which of those fractions actually gives a large contribution
one can probably link that with the resolvent formalism i just alluded to, but eh
 
Interesting. So there's also a matrix analogue of partial fractions?
 
yeah. let me see if i can find a relevant link
okay, check out the statement of this question to get a bit of the flavor of it
1
Q: Decomposition of resolvent in projections

AdamI am reading the book Perturbation theory for linear operators from Kato. He defines (ยง5 Section 3) for an operator $T : X\to X$ on a finite Banach Space the resolvent as $$ R(x) = (T- x)^{-1}.$$ He then consideres the Laurent Series of the resolven on a singularity $\lambda$ of $R$ and find...

 
ah, nice
 
Neat! There's something I've learned today
 
Moral of the story: if you find a "trick" that makes stuff easier-abuse the living daylights out of it
Much of higher mathematics boils down to taking the "sliced bread" theorem, and extending it to the "make me a sandwich" case.
8
 
theorem 32 of this link gives a good statement of what i was referring to, though i can't find the preceeding part of the notes: people.math.osu.edu/costin.10/603/Resolvent.pdf
@DavidWheeler and if you're a prof, you can further extend that to the "make someone else make me a sandwich" case
 
Thanks. But you needn't find more, this goes far enough over my head already.
 
1:22 AM
that holomorphic functional calculus stuff is something i really wish i knew better, but am too lazy to get good at
 
The idea is basically: complicated stuff in denominator = BAD, simple stuff in denominator = GOOD, even if we have a lot more of them
 
I suppose that's the same idea behind rationalizing by multiplying by the surd conjugate
 
In the complex plane, for example, it's simpler to go around holes, "one at a time", then have to worry about a region "peppered" with them. One hole is easy to miss.
 
aye, especially since once you do that you're in a position to decompose the resulting denominator into partial fractions
now i find myself wondering what the widest generalization of partial-fractions decomposition would be, lol
 
1:26 AM
Say, did the two of you happen to see the question I posted earlier on dual spaces?
 
Well, I think you'd want to work over a UFD, at least
 
saw it in passing, but i'm not remembering the statement
 
Basically I was asking for some intuition on what dual spaces "look like."
I can imagine the dual space of R^n easily enough -- it's the space of "row vectors" of length n, so to speak.
But is there a simple characterization of the dual space of R^\omega (space of sequences of reals) or R^R (space of functions: reals -> reals)?
 
for the former, maybe you could characterize the sequence as being analogous to an infinite column vector, and so the dual would be the space of infinite row vectors
 
Those two are just examples, of course. In general I somehow don't understand the duals of infinite-dimensional spaces, which seem to be much more complicated than their finite-dimensional cousins
But aren't there some kind of convergence issues?
 
1:30 AM
here is where i point out that i'm a physicist, aka "rigour, what rigour?"
 
@DavidZhang Here is a simple example, take $\Bbb R^{\omega}$
 
i.e. i don't actually know the technical details well enough to talk about that clearly
 
Fair enough.
 
A "vector" is just an infinite sequence of real numbers
 
Sure. so an element of the dual space is a linear function mapping infinite sequences of reals to reals
 
1:33 AM
BUT...in order to ensure linearity, we have to require that all but finitely many "coordinates" are 0
It can be seen that $\{e_i = (0,\dots,0,1,0,\dots,0,\dots)\}$ form an (infinite) basis
 
That makes sense. So the dual of R^\omega is the set of real sequences which are eventually zero?
 
HI I'm working with Conway's analysis book and in one question I really appreciate a clarification. This says: Find an open connected set G in the complex plane and two continuous functions f and g defined on G. s.t. f^2= g^2 = 1-z^2 for all z\in G. Can you make G maximal? Are f and g analytic?
My approach is the following, since when we taking a branch for the square root we cut some ray emanating from the origin we can choose for \sqrt (1-z) and \sqrt (1+z) two rays say (-\infty, -1] and [1,\infty) and define G as the complex plane minus the two rays.
 
The dual is all linear functionals $\Bbb R^{\omega} \to \Bbb R$
 
Wait, so are some of these linear functionals not of that form?
I know I'm sort of abusing notation here by identifying a functional with its sequence of "coordinates"
 
and in G the square function is well-defined and we're done, but for the other part I'm not completely sure
 
1:38 AM
Now, here's the thing: ANY infinite sequence (even one with only a FINITE number of 0's) yields such a functional.
 
Oh. Then what happened to the convergence issues from before?
 
For $x = (a_0,a_1,a_2,\dots)$, for $v = \sum_{i = 1}^n v_ie_i$ we define $x(v) = \sum_{i = 1}^n (a_iv_i)e_i$
This works, because only a finite number of the $a_i$ survive.
 
Finally I think I solved another.
 
is anyone here knowledgeable about probability to reply y/n to my relatively simple and unattractive questions? math.stackexchange.com/questions/1184834/…
 
Another spawn of satan eliminated.
3
 
1:43 AM
Here's the thing: $(\Bbb R^{\omega})^{\ast}$ is WAY bigger than $\Bbb R^{\omega}$
 
How so?
 
It's the same difference between "power series" and "polynomials"
 
Wait, now you've got me confused. I understand that x(v) converges if v has a finite number of nonzero components
But are there not elements of R^\omega which have an infinite number of nonzero components, and hence cannot be written as sum(v_i e_i, i = 1...n)?
 
It always does, that's how we define $\Bbb R^{\omega}$: FINITE linear combinations of the $e_i$
 
Oh, my apologies. I was unaware that was the standard definition
In that case, what do we call the vector space of possibly infinite real sequences?
 
1:47 AM
that is (isomorphic to) the DUAL of the first space.
 
Oh, I see. That makes sense, then.
 
When we "hit" the finite L.C. with an infinite one, we still get a finite sequence.
 
So in that case, what is the dual of the dual of R^\omega?
 
which we can sum to get a real number.
 
as can be seen from the fact that the following is a well-defined mapping to the reals: "What is the sum of all the terms of my sequence (including all the ones that are zero)?"
 
1:49 AM
Is it R^\omega again?
 
Since there's no assumptions regarding the size of the entries of that sequence, i'd think it'd have to be to ensure convergence
 
The dual is often written $\Bbb R^{\Bbb N}$ -not $\Bbb R$ to the $\Bbb N$-th power, but the set of all FUNCTIONS from the naturals to the reals
Given any such function, to get something that goes from $\Bbb R^{\omega} \to \Bbb R$, we take the "dot-product" (of sorts)
 
i.e. the only way to make sure you get a finite result from an arbitrary sequence in $\mathbb{R}^\mathbb{N}$ is to take only finitely many of its terms
 
I see. There's just one thing bothering me at this point
 
Sure, go ahead
 
1:52 AM
Isn't there a theorem that states something like "the dual of an infinite-dimensional vector space has strictly larger dimension"?
 
oh, i re-edited the chebyshev-polynomial question i referenced a bit earlier
1
Q: Extending a Chebyshev-polynomial determinant identity

SemiclassicalThe following $n\times n$ determinant identity appears as eq. 19 on Mathworld's entry for the Chebyshev polynomials of the second kind: $$U_n(x)=\begin{vmatrix}2 x& 1 & 0 &\cdots &0\\ 1 & 2x &1 &\cdots &0 \\ 0 & 1 & 2x &\cdots &0\\0 & 0 & 1 & \ddots & \vdots \\ \vdots & \ddots & \ddots &\ddots &

 
@DavidZhang It's pretty much the same argument
 
right, but R^N and R^\omega seem to both have countable dimension
*countably infinite dimension
 
No, $\Bbb R^{\Bbb N}$ does not have a countable basis.
 
wouldn't $\mathbb{R}^\mathbb{N}$ include the decimal expansion for $\sqrt{2}$?
if so, then it'd better not be countable!
 
1:54 AM
Oh, right, since we can only take finite linear combinations of basis elements.
So the basis of R^N is uncountable? That's pretty intuitively strange to me.
I'm having a hard time imagining what such a basis would look like
 
.
 
actually, since each entry is itself a real number, i'm not sure what the cardinality of the two would be
 
right-that's the key-L.C. are by definition, finite (the same restriction is made on "formal linear combinations" in other algebraic entities)
 
Is this one of those bases which can't be written down explicitly (like the basis of R over Q)?
 
i mean, i'd still say that the dimensions are strictly different. but i don't know what their cardinalities would actually be
 
1:56 AM
@Semiclassical The cardinality is $|\Bbb R|^{|\Bbb N|}$
 
hmm. i suppose that really is all that can or need be said
 
Does this look like an easy integral? $\int arctan(\frac{1}{x})dx$
 
Interesting. What does that leave for the dual of R^R, then?
Is it, say, the set of functions R -> R which are nonzero at a finite number of points?
 
It's "big".....I don't want to think about it.
 
at that point you're beyond my horizon, alas
 
2:00 AM
@DavidZhang Think about this: what makes $\Bbb R^{\Bbb R}$ a vector space?
 
@DavidWheeler What do you mean? The ability to add and multiply by scalars?
 
if i had to take a leap of faith, i'd guess the relation of R^R to its dual would be rather like that of Hamel bases to Schauder bases (reference
 
Huh. Can we simplify the situation at all by restricting to a "nice" subset of R^R (e.g. continuous functions, square-integrable functions, etc.)
 
@DavidZhang my point is, we want the elements of the dual to be linear functionals, not just functionals. That's going to limit our choices, a bit.
So just choosing "any old function" $\Bbb R^{\Bbb R} \to \Bbb R$, and then trying to limit the domain isn't the way to go.
 
Right, I see that not just any functional will work. My thinking behind it being the space of functions nonzero at a finite number of points was this:
One simple example of a linear functional is the "evaluate at x" functional.
(where x is a fixed real number.)
since that's a linear functional, (finite) linear combinations of "evaluate at x_i" functionals will also be linear functionals
and those can be conveniently encoded as functions R -> R which are nonzero at a finite number of points.
 
2:08 AM
i think that amounts to working with a finite sum of delta functions, with the inner product now being integration
 
But of course, the set of functions nonzero at a finite number of points is a strict subset of R^R, so there must be other classes of linear functionals that I've missed
 
Yeah, the problem is we can "embed" the original vector space isomorphically in the dual, but it's not going to be exhaustive.
 
And I can't even imagine what those other functionals would look like.
 
There's going to be a lot that "look like" integrals-except we'll need "work-arounds" for non-integrable functions
 
i almost think it's enough to do distributions with compact support
 
2:12 AM
My feeling is (I'm not an expert) that we'll have things that act a lot like "distributions" (a kind of generalization of functions)
 
though i'm probably being too simple
 
Hmm, will the space of distributions with compact support have dimension strictly larger than that of R^R?
Speaking of which, what even is the dimension of R^R?
 
well, the delta function isn't a function from reals to reals
 
It's "uncountable", which is as high as I go
 
but, hmm, not every function from R to R has compact support either
 
2:14 AM
Yeah, we don't have a strict inclusion either way
 
right
 
The space $\Bbb R^{\Bbb R}$ is ridiculously big. There are functions out there NO ONE HAS EVER SEEN
 
hmm. from wikipedia's page on schauder bases: "In mathematics, a Schauder basis or countable basis is similar to the usual (Hamel) basis of a vector space; the difference is that Hamel bases use linear combinations that are finite sums, while for Schauder bases they may be infinite sums. This makes Schauder bases more suitable for the analysis of infinite-dimensional topological vector spaces including Banach spaces."
 
It sounds like your original conjecture was right
 
that sounds a lot like our discussion of R->omega v. R->N earlier
 
2:18 AM
Then I guess that theorem works only with Hamel bases, not Schauder bases.
 
The point being-with dual spaces, things can get ugly very fast.
 
No wonder I had such trouble building intuition for them.
 
once the phrase 'banach space' becomes a standard phrase in those articles, i'm decidedly out of my knowledge base
 
With a Hilbert space (a "normal" kind of infinite-dimensional vector space), if we restrict our attention to continuous linear functionals, things "get nice" again.
 
right. at that point we're back into stuff i know, since hilbert spaces are mathematical underpinning of a lot of quantum mechanics
hence phrases like "the wavefunction is a ray in hilbert space!"
 
2:25 AM
If, as most people like to do, we assume our "original functions" are at least twice-continuously differentiable, it's a happy place.
 
hey everyone! I have an interesting question (from a calc course). Basically I'm dealing with an "infinite trumpet" (i.e. $1/x$ revolved around the x-axis where $x \ge 1$). The question has asked me to find volume and surface area. I did this. Then it poses the question: Suppose we fill up the trumpet with the finite amount of paint (equal to the volume), it would seem that we have coated the infinite interior SA with a finite amount of paint. How is this possible?
 
Oh, that's a classic. The problem basically boils down to the fact that you're using a mathematical idealization of "paint"
 
In the real world, paint has "thickness" but a surface doesn't have any.
 
I'm not sure how to integrate $\int 2^x$
 
Pretty much. You can't spread real paint infinitely thin, which you would need to do to coat the inside of the horn
@Owatch You would agree that 2 = e^ln(2), right?
 
2:29 AM
use $2^x = e^{x\log(2)}$
 
Then 2^x = (e^ln(2))^x = e^(x ln 2)
 
Uh.
 
Was this too blatant? :-)
 
Surely you've seen how to integrate a function of the form e^(cx), where c is a constant?
 
That's of the form $\int e^{ax}\ dx$ with $a = \ln(2)$
 
2:31 AM
$2^x$ = $e^{\ln(2)x}$?
Oh no it messed it up
 
No, you've got that right.
 
You can fix it You need $\{\}$ and the $x$ needs to be out of one exponent and into the other
 
You can then further simplify that to e^(x ln 2)
Wait, I take that back. you've got the parentheses in the wrong place
2^x = (e^ln(2))^x.
 
$e^{\ln(2)x} =( e^{\ln(2)})^x = 2^x$
 
Is it right now?
 
2:33 AM
@robjohn well linked :-)
 
@Owatch I fixed it for you :-)
 
ok
 
Moderator abuse! Oh, wait...
 
@DavidWheeler That's clever.
 
So now you just need a "fudge factor"
 
2:35 AM
Alright, so now that I've transformed it from $2^x$
What should I do?
Can I move everything down in front of e?
 
See, $(e^{ax})' = e^{ax}\cdot a$, by the CHAIN RULE
 
Wait, but I'm integrating, what will the chain rule do for me at this point?
 
You already have the "$e^{ax}$" part inside the integral, so now....
 
Yes, I have $\int e^{ln(2)x}$
Oh.
Wait, I have a question
 
what do you have to "put inside" to get it of the form $\int e^u\ du$, with $u = \ln(2)x$
 
2:39 AM
Can $e^{ln(2)x}$ be changed into $e^{ln(2)} * e^{x}$ ?
No it can't/
 
Because that would be adding exponents
Nevermind. Foiled
 
look at what I wrote: what is $du$?
 
Well if U = ln(2)x
 
Yes, it's a function...can you differentiate it?
 
2:41 AM
= 1/2?
 
Suppose $f(x) = ax$, what is $f'(x)$?
 
f'(x) = $(a * n) x^{n-1}$
 
I don't see any $n$ in what I wrote
 
every injective K-linear transformation
is surjective and so invertible.
 
Well, there is always a power there.
 
2:43 AM
how do you prove it?
 
What power of $x$ is $x$?
 
1
 
and so....
 
= a?
f(x) = ax, where x power is 1, then f'(x) = a.
 
yes. now $a$ could be any constant, and $\ln(2)$ is certainly a constant.
 
2:44 AM
It is a constant. Should I not have made it 1/2, but left as ln(2)?
I think that was a big mistake.
 
so if $u(x) = \ln(2)x$, what is $u'(x)$?
 
ln(2)
 
correct
 
so if $u = \ln(2)x$, what is $du = u'(x)\ dx$?
@EnjoysMath Only true if the domain and co-domain have the same dimension
 
2:46 AM
du = ln(2)dx
 
you're missing the "dx" part, but yeah
 
I add, no problem.
 
so how do we get a constant factor of $\ln(2)$ inside the integral, without changing it?
 
You sub du for dx?
And remove ln(2)?
I don't know.
 
Hint: $\int f(x)\ dx = \int \frac{a}{a}\cdot f(x)\ dx = \frac{1}{a}\int a\cdot f(x)\ dx$
 
2:50 AM
I cannot understand, I am sorry.
I see what you did there, you moved 1/a out
 
$\frac{a}{a} = 1$, yes?
 
Yes
 
multiplying something by one doesn't change it
 
Yes.
 
and we can move constants out, so I took the denominator out, and left in the numerator. Clear?
 
2:52 AM
Yes, that makes sense. But I've never seen that done before. Or at least I don't think so.
 
So: $$\int 2^x\ dx = \int e^{\ln(2)x} = \dfrac{1}{\ln(2)}\int \ln(2)e^{\ln(2)x}\ dx = \dfrac{1}{\ln(2)}\int (e^{\ln(2)x})'\ dx = $$?
 
Okay, I see what you did, but how does that make an integral of a derivative at the end?
 
Because you didn't think the chain rule mattered...
Now, maybe you can see it does
 
I cannot. I see it is a derivative. I do not know why it is.
I'm too tired for this.
 
OK, you could do this right: $\int e^x\ dx$?
 
2:58 AM
Sure.
It's just e^x
 
What are we doing here?
 
Oh.
 
Now, the point of integrating $f(x)$ is to "find a function whose derivative is $f(x)$"
 
Yes.
 
If we had $(e^{\ln(2)x})'$ to start with, it'd be easy, the integral would be $e^{\ln(2)x}$
But we don't, so we put the factor in we need, and its reciprocal outside the integral
 
3:01 AM
Aren't you missing an x in the first?
 
Yes, I was
 
phew
 
Good catch!
 
So you are 'removing' ln(2) by placing it's reciprocal outside?
 
So, after all this, we get: $$\int 2^x\ dx = \dfrac{1}{\ln(2)}e^{\ln(2)x}$$
But...we're not quite done, we can simplify this
$$\int 2^x\ dx = \dfrac{2^x}{\ln(2)}$$
 
3:03 AM
@DavidWheeler Could you explain this part again?
 
Which part?
 
Click the arrow next to the @David
But we don't so we . . .
Putting the factor in we need, and reciprocal outside?
 
Let's say we know $f'(x)$. Can we calculate $(f(ax))'$?
By the chain rule, this is $(f \circ g)'$, where $g(x) = ax$.
 
Yes.
 
And $(f \circ g)'(x) = f'(g(x))\cdot g'(x)$
 
3:07 AM
Yes.
 
But $g'(x)$ is really easy-$g'(x) = a$ (a constant)
So $(f(ax))' = f'(ax)\cdot a$
 
g'(x) = a?
or (a * x)?
No!
Just A
It's g'
Go on
 
This is exactly our situation with $e^{\ln(2)x}$
It's $e^{(-)} \circ a\cdot(-)$
 
Ay, it would be multiplied by ln(2)
As 'a'
 
I wrote it in the form $e^u$, whose derivative is $e^u\ du$
If $u = \ln(2)x$, then $du = \ln(2)\ dx$.
 
3:12 AM
Isn't that a U substitution?
 
Now we have $\int e^{\ln(2)x}\ dx$, but what we WANT is $\int e^{\ln(2)x}\ln(2)\ dx$
 
So because that is what is the result of a chain rule
You want to put* 1/ln(2) in front, so you can keep original meaning, but integrate it?
 
so we write: $$\int e^{\ln2x}\ dx = \int e^{\ln(2)x} \dfrac{\ln(2)}{\ln(2)}\ dx = \dfrac{1}{\ln(2)}\int e^{\ln(2)x}\ln(2)\ dx$$
now we can make the $u$-substitution
So, with $u = \ln(2)x$, we get: $$=\dfrac{1}{\ln(2)}\int e^u\ du$$
 
Nice.
 
And that $$=\dfrac{1}{\ln(2)}e^u + C$$
 
3:17 AM
any topology people here?
 
now we "substitute back"
 
approximately
 
hey mike!
 
$$= \dfrac{1}{\ln(2)}e^{\ln(2)x} + C = \dfrac{2^x}{\ln(2)} + C$$
 
@MikeMiller Was it you that was giving me advice on $A$ an $m\times m$ matrix, with $A^n = I \implies A$ is diagonal?
 
3:18 AM
mhm and hey
 
@Committingtoachallenge not true
 
can you look at my answer here, specifically my argument about continuity
 
On $\Bbb F = \Bbb C$
Or any algebraically closed I mean
 
@Committingtoachallenge still not true
 
A is diagonalisable***
 
3:19 AM
@Committingtoachallenge still not true
 
What? It is meant to be?
 
I wish I was a computer.
 
Oh shoot, I thought you were still talking about milpotent, nevermind
 
And not a stupid ghost.
 
@Owatch So there is "the complete answer"
 
3:20 AM
@DavidWheeler Oh okay haha, was worried for a min
 
@DavidWheeler Thanks so much.
 
Is mhm = oui @Mike?
 
Qui parle francais?
 
@Committingtoachallenge But your original assertion that such an $A$ is diagonal is untrue.
 
je fais
Yeah I meant diagonalisable
$A$ is a matrix such that $A^n = I$ implies that $A$ is diagonalisable on an algebraically closed field
Je don't actually speak french btw @Owatch
 
3:23 AM
Oh.
Yes, it was a bit weird
You said "I make"
 
Well, it might not work in a field of characteristic 2 @Committingtoachallenge
 
I know a few words
 
When I asked who spoke french.
 
I thought that was I do lol
 
3:24 AM
I thought it was some sort of new slang.
 
Jajajaja
Wait is it spanish that laughs in j's?
@DavidWheeler Field of characteristic $2$?
 
Russian use xa I believe?
 
Are tu french?
Je suis bad at this lol
 
My mother is French, I was born there, and lived there a while.
 
In any ring (and fields are rings) the chracteristic of a ring is the smallest positive integer $n$ such that $1+1+\cdots+1$ ($n$ summands) $= 0$
 
3:26 AM
So somewhat, yes.
 
@DavidWheeler Ahhh so in $\Bbb F_2~=\Bbb Z / 2 \Bbb Z$?
 
@Committingtoachallenge there is no $\Bbb F_1$ (though some people would beg to differ)
Yes, in $\Bbb Z_2$ we have $1 + 1 = 0$.
 
Okay well I will just put it in $\Bbb C$
 
And my thinking on this, is that we can have $x^2 = 1$ only having one root.
 
3:29 AM
I'm going to sleep now. bye.
 
@Owatch Bonne nuit
 
A demain.
 
Has tomorrow is how I read that before google
 
In that field $\begin{bmatrix}1&1\\0&1 \end{bmatrix}\begin{bmatrix}1&1\\0&1 \end{bmatrix} = \begin{bmatrix}1&0\\0&1 \end{bmatrix}$
 
I am confused why the question says $A^n = I$, without specifying $n$, is there an obvious(not to me) reason for this?
Is it saying any $n$ probably?
 
3:32 AM
In the complex field (which has characteristic 0), this doesn't happen, and we get $n$ distinct roots over $\Bbb C$
 
$A^n=I\implies (A-I)(A^{n-1}+\dots+A+I)=0$
 
i'm trying to remember a certain notation: given a sequence $\{t_n\}$, how do I write this sequence with the $k$-th entry dropped?
 
I might write $\{ t_{n \ne k} \}$.
 
@Semiclassical I've seen $\{\widehat{t_k}\}$
 
that's the one
 
3:43 AM
Huh, I've never seen that one before. Where is it commonly used?
 
Bib
I'm new to abelianization (and commutator subgroups in general). Does Z have a finite index subgroup with infinite abelianization?
 
@DavidWheeler for context, i'm starting with a $n$-variable function $f_n(t_1,t_2,\ldots,t_n)$ and wanting to express $f_{k-1}(t_1,\ldots,t_{k-1})$ and $f_{n-k}(t_{k+1},\ldots,t_{n})$ as simply as possible
 
Oh my god yall fuckin' haters need to S T O P STOP! I'm tryin to get smart here and yall fuckin turbonerds stoppin me from achievin' my dreams! WTF? — xXxSkIttLEzDeStRoYeRxXx 2 mins ago
-5
Q: Circumference of a Heart?

xXxSkIttLEzDeStRoYeRxXxI need yo help. Look see, mah girl was like "You don't love me Romeo" and I'm like "Baby I love you" and shes like "No you font you made me lobster and then didn't turn on the water chef boy-r-dumb" and I'm like "how can i prove that I love you?" and shes like "Get a brain." How do i do some of ...

 
...a simple "troll post" would've sufficed
 
Oh I meant to copy the link the first time
Nooooooooooooo I have class in 6 min
I did it again
bye
 
Bib
3:55 AM
Wow at "et tu, brute?" Well played...
 
@Semiclassical I think what you've got there is already quite concise and understandable.
 
With regards to responding to trolls: meta.math.stackexchange.com/a/19330/137524
@DavidZhang: yeah, i think i'll just go with that
 
@Semiclassical Good call.
 
and it's gone. RIP, troll post, may you not return (but you'll probably try)
 
trollers gotta troll :-)
 
4:05 AM
yep, hence don't feed em
 
If he comes here it looks like a job for skullpatroll!
 
4:17 AM
@Semiclassical there was an interesting discussion by Stan Shunpike about the idea of "distance" in the physics room where he quotes what people said to him in here.
in The h Bar, 3 hours ago, by Stan Shunpike
@DavidZ I was talking to some dudes in the math chat and they said something to the effect that...distance is a useful way to get spatial information just not necessarily information that can be described with real numbers. But thats math. Agree or disagree when it comes to physics?
 
4:38 AM
heh, now this is a fun formula which i don't know how to proceed with further:
for a particular n-variable function $F_n$, I found $$dF_n(x_1,x_2,\ldots,x_n)=\sum_{k=1}^n F_{k-1}(x_1,\ldots x_{k-1}) F_{n-k}(x_{k+1},\ldots x_{n-k})\,dx_k$$
 
5:05 AM
https://upload.wikimedia.org/wikipedia/commons/9/98/End_of_universe.jpg

I see this image just about every time I see someone talk about the geometry of space. I was generally following the lecturer's analogies and descriptions in my mind pretty well, and liked the idea of using a the sum of angles in a triangle to bring the point home for different values of omega. Actually looking at the triangle when omega < 1 always threw me off though. I see another triangle with angles > 180 degrees.

Any tips on what I'm missing?
 
5:28 AM
Have you tried using an actual globe of the world to model the >1 case @Louis?
 
5:49 AM
 

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