12:04 AM
I totally forgot that work is area under a curve

12:15 AM
Every single integral is area under a curve. So?

12:51 AM
I don't understand linear stability analysis in dynamical systems.

3 hours later…
3:32 AM
@Thorgott I revised the proof. I wonder if this proof should be this long.

3:57 AM
@TedShifrin yes professor Ted, the extension lemma has been proven in the book. I understand your argument now. I was afraid to consider a basis of null $T_1$ at the start as I thought V could be infinite dimensional (as in the question V wasn't given finite dimensional) but you resolve this problem by transfinite induction. I'll try to learn more about transfinite induction though as I have never heard of it before. Thanks a lot professor Ted. :)

4:39 AM

i'm guessing 0 gets cast into different types depending on what its being compared to?
either that or we live in a cruel world

it is just javascript's demonic behavior
though it works fine if we use === (false in all cases)

5:09 AM
@Koro Well, at least the version of Axler I have does assume only that $W$ is finite-dimensional. Since he's not assuming $V$ is as well, my proof is not the preferred proof. Including that would have clarified things. I assumed everything was finite-dimensional at this point in his book. I was wrong.

5:27 AM
i thought that axler was purely finite dimensional at that point too. i wonder if that's changed in a new edition. i had the first ed.
this evening my daughter found a pen that our cat had knocked onto the floor and batted around. "i'm playing like livvy!" she announced as she crawled around on the floor swatting it from place to place. "i have paws too!"
this was a little before 6:45.
she remained in human form, however.

5:58 AM
@TedShifrin Yes professed Ted, only W is given to be finite dimensional in the exercise problem.

6:10 AM
@leslietownes I am still waiting for the transmogrification.

@leslietownes Just after that exercise, there is an another exercise (in 3rd edition) that says: Suppose V is finite dimensional and $T_1,T_2\in L (V, W)$. Prove that range $T_1\subset$ range $T_2$ iff there exists $S\in L(V,V)$ such that $T_1=T_2S$. Here, V is given to be finite and nothing has been said about dimension of W. I managed to solve this exercise problem.
@robjohn How's your health now professor Rob? Are you well now?

i know he eventually introduces infinite dimensional spaces, the inner product spaces section deals with C[0,pi] or something with an L^2 norm. but i don't recall him proving the basic stuff in the infinite dimensional setting, unless he just says 'trust me on this.' but i was working with the earliest edition.

@Koro still getting better, but still not great. Appetite is still not great.

Get well soon.

2 hours later…
8:19 AM
could I have a hint for this? Prove $$\sum_{n=2}^{\infty} \log ( 1 + \frac{(-1)^n}{\sqrt{n}}) = -\infty$$
the divergence is very slow , im guessing that there may be some improper riemann integral that obviously diverges which I can compare this series to, but i dont have a clear idea of what it would look like

8:40 AM
In mathematical analysis, the alternating series test is the method used to prove that an alternating series with terms that decrease in absolute value is a convergent series. The test was used by Gottfried Leibniz and is sometimes known as Leibniz's test, Leibniz's rule, or the Leibniz criterion. == Formulation == A series of the form ∑ n = 0 ∞ ( − 1 ) n a...

how does that apply in this situation?
I don't see why the absolute value decreases monotonically

well, then you should try to combine terms for $\log\left(1+\frac1{\sqrt{2n}}\right)$ and $\log\left(1-\frac1{\sqrt{2n+1}}\right)$

ah yes, that is a good idea

Those are the two approaches

we will get something like$log (1 - \frac{1}{2n} + \epsilon(n))$ where hopefully $\epsilon(n) = O(\frac{1}{n^k})$ for $k > 1$, and then we need only prove the divergence of $\sum_{n=1}^{\infty} log(1 + \frac{1}{2n-1})$ which should follow from the divergence of $\sum_{n=1}^{\infty} \frac{1}{2n-1}$
thanks @robjohn

8:59 AM
$\log\left(1+\frac1{\sqrt{2n}}\right)+\log\left(1-\frac1{\sqrt{2n+1}}\right)=\log\left(1+\frac1{\sqrt{2n}}-\frac1{\sqrt{2n+1}}-\frac1{\sqrt{2n(2n+1)}}\right)=\log\left(1-\frac{1+\sqrt{2n}-\sqrt{2n+1}}{\sqrt{2n(2n+1)}}\right)=\log\left(1-\frac{1-\frac1{\sqrt{2n}+\sqrt{2n+1}}}{\sqrt{2n(2n+1)}}\right)$

9:13 AM
$\log\left(1-\frac{1-\frac1{\sqrt{2n}+\sqrt{2n+1}}}{\sqrt{2n(2n+1)}}\right)\le\log\left(1-\frac{1-\frac1{2\sqrt{2n}}}{2n+1}\right)\le\log\left(1-\frac{1-\frac1{2\sqrt2}}{2n+1}\right)$
@porridgemathematics: and that should do it.

2 hours later…
11:01 AM
I didn't notice that my account is almost an year old
I am still worthless

11:36 AM
Does $\not\equiv$ means "not always equivalent" or "always not equivalent"?

1 hour later…
12:40 PM
@love_sodam This is detailed and correct. You could combine the proof that $S$ and its complement are open into one if you want to shorten it, there isn't really a need for the repetition

1:29 PM
@Thorgott I also did the converse which is much shorter. There is a post math.stackexchange.com/q/3133406/668308 which also proves the converse but there is some further 'closure' assumption during the proof which I didn't assume. Could you check this one too?

1:53 PM
Trying to figure out how to say something simple. Suppose I have the word AABBB. Then there’s two ways to permute the As without changing the word, and 4! ways to permute the Bs
So there should be 4! * 2! permutations which leave the word unchanged. Intuitively that makes sense, but I feel like there’s something I’m not saying
I guess the fact that the permutations are disjoint?
(I’m trying to say this for an N-letter word with letter frequencies n1, n2,etc, so saying it simply would be nice)

2 hours later…
3:43 PM
@Semiclassic Presumably you meant 4 Bs and not 3. There are $4!2!$ permutations of the six letters that fix the original configuration. This means that for every configuration of the letters, there are $4!2!$ permutations that spell the same word.
You can mumble subgroup and cosets if you want ...

4:01 PM
@TedShifrin hey
are there any manifolds of dimension $\ge 4$ with constant positive sectional curvature which have complex structures?
or is it a well known conjecture that none of them do, perhaps

Hmm ... Well, certainly spaces of constant curvature are classified. There's Joe Wolf's book, for example.

yeah sure haha
hence why I'd expect someone to know instantly if this is possibly true or false

I've never thought about this question. But I assume there aren't any, modulo the $S^6$ mess.

someone recently posted an S^6 paper
which was wrong
but then they followed up with a paper claiming this
I am wondering if it's even true on the nose

Doesn't sound like a trustworthy source.

4:07 PM
thats why i asked someone more trustworthy

I didn't think about these sorts of things. Robert Bryant would know the answer in a negative millisecond.

@love_sodam calling the element in $M$ $\mu_B^{\prime}$ and the one in $M^{\prime}$ $\mu_B$ seems like bad notation. anyway, the point is that you want $H_n(M^{\prime},M^{\prime}-B)\rightarrow H_n(M,M-B)$ to be an iso. check the hypotheses of the excision theorem to see what you want to assume about $B$.

If I still had Wolf's book, I'd look at it, but with retirement it disappeared.

connected (WLOG) manifolds of constant positive sectional curvature are universally covered by $S^n$ with the round metric
if $\pi\colon\tilde{M}\rightarrow M$ is a covering of Riemannian manifolds, then we have an isomorphism of Euclidean bundles $T\tilde{M}\cong\pi^{\ast}TM$, so a complex structure on $TM$ induces on $T\tilde{M}$, no?

@Thor Did my eyes deceive me? Pictures and everything.
Yeah, that's right. So it does just boil down to the $S^6$ morass.

4:13 PM
ok yes fair enough
lol

@Ted lo and behold, how far this algebraist has come
I'm currently writing my thesis, I have a bunch of similar arguments in there

What'll be next? :D

4:40 PM
@TedShifrin woops
but yeah
mostly i'm trying to figure an short but convincing way to say "it's number of permutations that fix the As, times the number of permutations that fix the Bs"
I guess that point is how to justify saying that every such permutation may be written as $\sigma_A \sigma_B$ where each permutation acts only on $A$'s or $B$'s.
i can sorta see how this is a subgroup thing but not quite explicitly
$S_2\times S_4$ being a subgroup $S_6$ i guess, but i feel like there's more i should be saying there
(like, $S_2 \times S_2$ would also be a subgroup but not the 'largest'? i'm reaching)

4:56 PM
This is just standard stuff. $S_2\times S_4$ stabilizes the original configuration.
More precisely, it is the stabilizer subgroup.

stabilizer subgroup definitely sounds like what i'm forgetting, yes
so would the difference between $S_2\times S_4$ vs. $S_2\times S_2$ is that the former is a stabilizer subgroup of $S_6$ whereas the latter isn't?
hmm. i guess saying that doesn't make a lot of sense. a stabilizer subgroup only makes sense in the context of a group action on some set
i dunno. if i ever learned this stuff proper it's been a decade

Right, you are acting on the set of words you can spell with the letters. Orbit of a word is group mod stabilizer subgroup of the word.

Hello,

Can you please show why the capacity of cut (capacity of a cut is the total capacity of forward edges) is $C(\mathscr{X}_1) = 12$for the left cut below? (We have 2 cuts)
Also total flow of a cut is total flow of forward edges - total flow of backward edges, where forward go from source $s$ to sink $t$ and backward goes in the opposite direction? Backward edges going in the opposite direction but I see no backward edges above :|?

5:29 PM
Got it
It's straightforward application of definitions of forward and backward edges of a flow
I did not notice that backward is a vertex that starts in V_t and ends in V_s and not opposite in direction
:/
Sorry

6 hours later…
11:17 PM
2

For example, $\forall x \in X, P(x)$ can be viewed as a map $P : X \to \text{TrueProp}$ the collection of true propositions. Why do we make the distinction in type theory, which seems to want to reduce all constructions down to a few building blocks? In type theory $\forall$ is a handled using w...

Who's more "right" in the above argument betwee Malice Valdrine & I?
I think Pi-types are just specializations of map, Malice Valdrine has another persepective.