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12:01 AM
Well, I think I learned something here. And the main takeaway is stop "optimising". Cheers!
 
we humans are doomed to optimize the slowest parts of the system.
 
So I'm going to have to read up on Fisher's out of mathematics ideologies then huh? Reminds me of a BBC article about the most detailed atlas of the human body and how it was done in Nazi Germany, but the detail it presents is so good that it is still used today, but there is a large "moral quandry" about the whole thing
 
i think a lot of people in the 'first world' ~ 100 years ago were informally into eugenics and other stuff. which isn't to say they were raving nazis.
 
well didin't have to read far to find out about Fisher's beliefs.....second paragraph of his wikipedia entry.....
 
i do hope that if i'm ever written up on wikipedia, my bio doesn't include references to dr. mengele
we should not think of people as machines, and maybe the brightest of us should not imagine themselves as super-men or super-women or encourage others to believe in anything that places people in hierarchical boxes. that's my hippie chant of the day.
i'm actually in a good mood today. this is rare.
 
12:10 AM
well put......must mean you won a case or got a good settlement arranged
 
we did have a good result the other day. a judge made the other side pay for our fees. and didn't question my time entries either although he did imply that someone else on our team may have been padding the bill.
 
It's criminal that CORDIC isn't well-known. I passed along misinformation for decades until I got educated about 20 years ago.
 
that's about when i learned of it.
i'm also happy because a relative who was incapacitated several weeks ago made a lucid post to facebook for the first time since 2017. almost anything mental can be managed but step 1 is acknowledging that there might be a mental thing.
i sometimes blend CORDIC and CRANDIC. CRANDIC meaning Cedar Rapids and Iowa City, a railroad thing. a park near our rented house had this name and was a great place to watch bald eagles in the winter.
they'd fly over a bend in the river and pull fish out.
 
What was the misinformation Ted?....that trig functions are calculated by power series?.....I ask because a PhD student that was teaching the first real analysis course I took answered my question about "how exactly are trig values calculated" by saying exactly that
It is only here I found out that CORDIC is what is used
 
the idea that anyone uses power series is a lie, i might even call it the Big Lie, propounded by those with a stake in selling math books.
i can't speak for ted's experience but that was my own.
 
12:19 AM
Right. I stupidly assumed (having been told by someone ignorant) that it was Taylor series, plus knowing some basic facts.
 
another one of the beliefs I have coming crashing down.....I thought power series were the end game for all of these estimates
 
Are the only closed sets on the topology {∅, X, {a}, {b, c}} are ∅ and X?
 
i might need to know the complements of the things in that list. what's X?
sierpinski would like a portion of his licensing fee donated to me.
 
X is {a,b,c}
 
the complement of {a} appears to be open.
 
12:21 AM
Oh right, so {b,c}, and {a} are both closed and open
 
sounds reasonable. this topology does not provide a mechanism for separating b from c. but does provide a mechanism separating a from b and or c.
 
What would {b,c} be?
 
its complement is open..
it is also in the topology.
 
Oh sorry I meant what is {a,b}
or {c,a}
 
these sets appear to be neither open nor closed.
 
12:25 AM
@BalarkaSen That is a surjective system isn't it?
 
Thank you @les
@leslietownes
 
i like "les" as a shortening. i may go by that from now on.
 
less
 
12:46 AM
Les is more?
 
Les is a Moor?
 
LSS
1:01 AM
Guys, quick question. The set of polynomials of degree greater than 6 isn't a subvector space of Pn, right? The book i am doing says it is, but i don't see why:

x^6+x^4+(-x^6) = x^4
greater than or equal, over real field
 
lower bounds on degree will not generally be preserved by vector space operations.
x^n - x^n = 0 seems troublesome. the book might need editing.
 
that set doesn't contain the 0 polynomial
if you include the 0 polynomial it is though
 
you can certainly consider a span of a set of high degree polynomials as a vector space, and this space might not include certain lower degree polynomials, or all polynomials of lower degree.
 
oh I was thinking of the one that only had non-zero coefficients of degree > 6
so the one spanned by $x^7,x^8 \dots $
 
as you noted, a vector space's need to include a zero vector is troublesome here.
 
1:17 AM
@Onir wrong.
 
yeah I was thinking of the one that only had non-zero coefficients of degree > 6
 
LSS makes a point more important than $0$
 
well, when a book says something I tend to believe it
that's my initial response
but yeah, in general in cancelative structures I think closed things cant have closed complements
 
please buy my book, Invest in LeslieCoin by Leslie Townes. it instructs all readers to put any assets they have into something called "LeslieCoin."
 
or something like that
I can't do that but I can shout you out on my tiktok
I only ask 10k If LeslieCoin gains 20k marketcap 24 hours after my post
 
1:30 AM
I got an irate email from someone telling me that an exercise in Fraleigh's algebra book contradicted a claim (with proof) in my algebra book. Clearly Fraleigh had to be right and I had to be wrong. Oops.
 
that's funny.
fraleigh was my first introduction to abstract algebra. in hindsight, there are sections of it that i like and sections that i don't.
 
I don't like his spoonfeeding style, but I liked the eclectic contents in the early editions.
 
there's stuff in my edition of the book that seems far too dumb. this might be his spoonfeeding style. i liked how he pointed out that counting arguments can be powerful. he failed to leverage this and other proof strategies in his own book.
 
Yeah, after my book, I choose Artin :)
Very different audiences.
 
he also puts groups before rings and maybe that is the abstract mathy ordering instead of the teach people stuff ordering.
 
1:35 AM
That's what 80% of texts do.
You're voting for me, I see.
 
i hate how much we agree these days.
 
Oops.
 
prof. wu is also in our camp on this. rings before groups.
or he was when i last talked to him about this.
 
I think he may have used my book one time. Hard with Berkeley's course syllabus.
 
they discontinued 'transition to upper division mathematics' after i taught it twice. i said, just make everybody take discrete math. they'll get proof techniques in there.
they ended up doing that.
a lot of undergrads at berkeley are only interested in teaching high school, which is a laudable goal, but also makes a fraction of these students highly hostile to anything that isn't directed at teaching high school.
 
1:42 AM
From Jordan Ellenberg: You can’t maintain a dislike for any part of mathematics because all of its branches touch each other.
That’s to you!
 
that point is well taken. i will stubbornly maintain a dislike for anything i want.
 
Just like I hate most lawyers.
 
my daughter is a comedian. "now i'm playing with livvy." livvy is the cat. livvy did not want to play with her. there were annoyed meows.
 
My kitten would gouge — bite and scratch deeply — your comedian.
 
lawyers are generally garbage.
myself included in that.
 
1:47 AM
Granted.
 
one time we went to a party and later my wife said 'XX was super weird, where do you know him from?" and i had to say, i'm sorry, mathematicians are weird, someone decided to award that guy a seven figure cash prize for something. which is wrong because clearly it should have been awarded to me.
my cat bit my daughter.
my daughter said "i was trying to play with her." she wasn't trying to play with you.
learning when people are trying to play is about 90% of becoming an adult.
 
My kitten has a long way to go on learning restraint.
 
hopefully her tetanus shots are current...
i'm lurking.
 
Copper's daughter's? Probably not at this age .
 
:-) I was thinking Leslie's since the cat bit her.
 
1:53 AM
Oops. Leslie's daughter.
My bad.
 
My daughter is current.
Helicopter Joe.
 
LOL
 
I know some lovely lawyers. One is delightfully named Lou Liberty.
 
2:05 AM
my name is 1-900-buy-lesliecoin.
we have washed and hopefully disinfected the bite location.
when i enrolled in law school i had no proof of my shots being current because my childhood doctor had quit his job 20+ years previously and then sold his practice to someone who quit 15+ years previously.
so they refused to give me an ID card until i had the whole run of vaccinations administered by the campus health center.
"you're wasting your time," i said. "i actually had chicken pox." they didn't care.
kids these days don't know the first thing about chicken pox.
when the blood test eventually came back it was through the roof for chicken pox antibodies. because i'd had it. and now i've also had the vaccine.
 
3:09 AM
@leslietownes You won't get shingles.
which might be bad for a lawyer... wait, that's A shingle.
 
i want exactly one shingle.
 
I'll have a shingle, my dear - S. Connery
 
3:30 AM
I need to get my tetanus shot updated every few years. Dogs seem to like me.
 
i stepped on a nail once. it was very unpleasant. the idea i will share with everyone, so you can not remove it from your minds, is that when i reflexively lifted up my leg, the board i had stepped on also lifted up. the nail stayed stuck until i manually removed it.
 
Is there a numerically-stable way to compute $\theta(x, y) = \arccos\left( \frac{x \cdot y}{|x| |y|} \right)$, i.e. the angle between two vectors $x$ and $y$?
The main issue I'm having is that $\nabla_u \arccos(u) \to -\infty$ as $u \to \pm 1$.
 
I probably mentioned this before, I kicked a pitchfork that was hidden underneath some hay. It went through the shoe, my foot and out the other side. Was a bit awkward to remove. Took a few moments before the pain hit thankfully.
@user76284 Why will computing the normalised inner product not suffice?
 
i remember being more confused than in pain.
 
@copper.hat Good question. Suppose $x = y$. Then $\theta(x, y) = 0$ since the argument inside the arccos is 1.
But the gradient of arccos at 1 is infinite.
 
3:44 AM
Yes, I understand that, but why do you want an angle as opposed to the value of $\cos$?
 
(I'm working in the context of backpropagation/automatic differentiation, so I need to propagate gradients through the computation.)
 
i could imagine treating the function differently, perhaps with a different algorithm, if the thing you're plugging in is near 1 or -1.
 
@copper.hat Geodesic distance. It's related to the problem I mentioned yesterday :)
@leslietownes Yes, I was thinking maybe some kind of "transformation" of the problem could avoid the issue of the infinite gradient.
Because the infinite gradient does not really "exist" when you take the normalization into account.
i.e. there is no value of $x$ or $y$ for which the gradient of $\theta(x, y)$ explodes (I think).
 
i'm thinking, maybe, find out where the line will be where the function goes nuts at your level of numerical precision, and cut off slightly close to that and just put in arccos(1) or -1 at those values.
 
If you have an infinite slope you cannot avoid it. You want to check the values and do something different instead.
There is always another way. But you may have to give up something.
 
3:47 AM
i've offered a stupid truncation as a first approximation to something smart that you might actually want to do.
 
Yeah I tried clipping the argument of arccos by some small epsilon, i.e. (-1+epsilon, 1-epsilon), but was wondering if there's something more elegant and numerically stable.
 
well we agree on that. you wouldn't want to trust your arithmetic to be good very close to a point where the function you're attempting to compute has a vertical tangent line.
 
@copper.hat This is what I'm referring to (assume the second vector is just $(1,0)$).
This is what I mean by the singularity being "fictitious".
 
I might be missing something here, but you are asking how to compute $\arccos$ or the slope?
 
I guess a way to state my problem is how to compute $\theta(x,y)$ without using $\arccos$ directly, which is where the numerical instability comes in.
Note that $\theta(x, y)$ is well-defined and well-behaved everywhere (except the origin), unlike $\arccos$. That's really the heart of the problem.
 
3:53 AM
this seems similar to atan2.
 
@leslietownes Ah, something like this
7
A: Numerically stable method for angle between 3D vectors

D0SBootsShort answer: Method 2 is better for small angles. Use this slight rearrangement: $$ \theta = 2~atan2(||~||v||u - ||u||v~||,~||~||v||u + ||u||v~||) $$ This formula comes from W. Kahan's advice in his paper "How Futile are Mindless Assessments of Roundoff in Floating-Point Computation?" (https://...

 
Certainly atan2 if not using it already, but I have rarely found a need for an explicit angle.
i would listen to Kahan.
 
kahan knows quite a lot. do what he says.
 
He was on one of my committees if I recall correctly.
Certainly talks.
 
i only rode elevators with him. i was afraid of him.
 
3:56 AM
Himsself & my advisors are of the same mind when it comes to tormenting advisees.
 
i found the notes from my qualifying exam earlier today. traumatizing. i passed but only barely. i told my committee to use a version of the krein milman theorem that was obviously inapplicable to the context i wanted to apply it in (state spaces of C* algebras).
 
This is a neat answer!
 
He was very good at punching through bs.
 
I'm surprised this isn't more well known.
 
as a fan of convexity you might enjoy the agony of that.
 
3:58 AM
Everyone knows the arccos formula, but few people seem to know the better formula.
 
Is it true that in any topology space $(X,\tau)$, $X$ and $\varnothing$ are clopen? Because they are in $\tau$ by definition, and their complement is each other so they are closed and open?
 
what's tau?
 
Any tau
 
tau seems to be of a different sort of thing than X. tau seems to be a family of subsets of X, while X is among other things a subset of X.
 
Ooops sorry I meant $\varnothing$ instead of $\tau$!!
 
4:01 AM
the empty set and the whole space are clopen. i think i agree with that
 
Thanks you @les !
 
i think my committee figured that this was a good way of getting rid of me.
Kahan was well known for saying "no student before its time"
(as in wine)
 
Is that an Orson Welles reference?
 
i used to use harry lime and holly martins as internet pseudonyms, before i adopted my present moniker.
one time i was in an elevator and someone who was old and esteemed and a former student of kolmogorov turned to me and said, with significance, "your system is incomplete."
i puzzled on the meaning of that for some time. my officemate observed that i'd failed to button the top button of the shirt i was wearing that day.
which i think is what he meant but i'll never know.
 
cute :-)
 
4:32 AM
@leslietownes the topology
 
privet, or alternatively, privyet.
 
a simple link will do.
 
Never gonna give you up
Never gonna let you down
 
5:08 AM
does anyone want to try some analysis? i'm struggling with a small part of a proof lol
 
it its easy ill have a go :-)
 
cool! i've made a post because it'd be too much to type on chat. rudin uses it in theorem 3.14's proof, but you don't really need any context to prove it: math.stackexchange.com/questions/4153707/…
 
just to be clear, the only issue is $\sup$ vs $\operatorname{ess sup}$ ?
 
yes correct
we need to show that the two are equal for simple measurable functions
 
are you comfortable showing $\sup \ge \|s\|_\infty$?
 
5:16 AM
hmm not really
 
How about this (the other side): Suppose $\alpha$ is such that $ \mu(|s|^{-1}((\alpha,\infty])) = 0$. Then you must have $|s(x)| \le \alpha$. Hence $\sup |s| \le \alpha$ and so $\sup |s| \le \|s\|_\infty$.
I am supposing the sets $A_k$ are disjoint and non null.
 
It makes sense, except one step
The fact that the measure is zero, only tells us that $|s(x)| \le \alpha$ a.e.
a.e. doesn't tell us anything about all $x\in X$?
 
If $|s(x)| > \alpha$ means that the entire $A_k$ will lie above $\alpha$ too.
Do you see the latter point?
 
I understand that, yes
 
So then you follow this direction comfortably?
(I don't want to revisit, so would rather nail one direction first.)
 
5:25 AM
$|s(x)| \le \alpha$ a.e. means $|s(x)| < \alpha$ on a set of measure zero. What did you do next?
 
Do you see that if $|s(x)| > \alpha$ then $\mu(|s|^{-1}((\alpha,\infty])) \ge \mu A_k > 0$ ?
Where $|s(x)| = |c_k|$.
 
I don't :(
Does this have to do with the disjoint, non-null assumption?
 
Nothing yet.
 
Wait, I do see it. It follows from monotonicity alone
 
If $|s(x)| = |c_k| > \alpha$ then $|s(x)| > \alpha $ for all $x \in A_k$.
Since $|c_k| \in (\alpha, \infty)$ then $\mu(|s|^{-1}((\alpha,\infty])) \ge \mu A_k > 0$.
 
5:30 AM
Yes, agreed
So if $|s(x)| > \alpha$ on a set of measure zero, then $\mu A_k = 0$
which doesn't make sense, because A_k are non-null
 
I'm confused now. The question says $\mu A_k >0$
 
I completely follow this direction
 
Excellent. Now suppose $\sup |s| < \|s\|_\infty$.
 
@copper.hat $\mu(A_k) = 0$ is a contradiction to the assumption that $|s(x)| > \alpha$ on a non-empty set of measure zero. Hence, that set is empty.
@copper.hat Go on
 
Can you guess what I am going to do next?
Pick some $\beta$ such that $\sup |s| < \beta < \|s\|_\infty$.
 
5:35 AM
Hmm, I see
 
What is $\mu(|s|^{-1}((\beta,\infty]))$ ?
 
One interesting thing, the essential supremum lies inside the set of which it is the infimum, if that set is non-empty. Does that help?
 
Actually, what is $|s|^{-1}((\beta,\infty])$ ?
No, don't wander.
If $\sup |s| <\beta$ then how many values of $x$ satisfy $|s(x)| \ge \beta$?
 
and the measure of the empty set is?
 
5:38 AM
0
 
which contradicts the definition of $\|s\|_\infty$.
Hence $\sup |s| \ge \|s\|_\infty$ and so we are finished.
 
Yes, makes sense! This is since $\beta$ gets inside that set, so the infimum is at most $\beta$, right?
Nice proof. Thanks a lot! @copper.hat
 
This is a fairly standard proof, not mine at all :-)
I prefer the following characterisation of the essential supremum:
$\inf_{N \text{ null}} \sup_{x \in X \setminus N} |s(x)|$
 
Ah okay, cool!
 
An exercise for another day to prove that :-)
 
5:42 AM
Yep! I'll do it
 
Authors are a bit (in my opinion) sloppy with essential vs. $\sup$ just as they are with distinguishing functions from equivalence classes in, say, $L^p$ spaces.
 
Right, I see.
Would you be adding this as an answer to the post? If not, should I?
 
When you are comfortable with it, it is easy to work with, but very confusing until you get to that point of comfort.
I won't be adding it, I am pretty zonked for today :-). By all means add it yourself.
 
Alright, I'll do that. Thanks again!
 
You are welcome, good luck!
 
6:02 AM
hi all
0
Q: searching for $f(z) \ne -1 $

mickIm looking for a real-entire function* $f(z)$ such that for any (finite) complex $z$ we have $$ \lim_{t = +\infty} |f(z+t)| = \infty $$ $$ \lim_{t = +\infty} f(z-t) = 0 $$ $$f(z) \ne -1$$ (* real-entire means a function is entire and maps the reals to a subset of the reals. In other words an enti...

urgently looking for solutions :)
 
what's so urgent? slooow down :P
 
life is short
and the slower you get some answers , the slower you will be able to conclude others
besides i need to keep my muscles intact too
 
 
4 hours later…
10:20 AM
Hey
For $f \in C_c(\mathbb R^n)$, $$\sup_{x\in\mathbb R^n} |f(x)| = \|f\|_\infty$$
I'm trying to prove this, and wondering if math.stackexchange.com/questions/4153707/… would be useful in any way?
 
uh, what exactly are you trying to prove?
that's the definition of $\lVert f\rVert_{\infty}$ as far as I'm concerned
 
same here
 
oh no no
the infinity norm is the essential supremum
so i'm trying to prove that supremum = essential supremum for all functions of that class
(please check the linked MSE post for definitions if needed)
 
Let $E=\mathcal{C}([0,1],\mathbb{R})$ and $T$ a map defined on $E$ by
$$\forall x\in[0,1],\qquad T(f)(x)=\int_0^x f\big(4(t-t^2)\big)dt. how to show that $T$ is not surjective
 
hey @Thorgott remember I asked you yesterday about change of variables, basically how to justify integrating over a ball (of radius R) is the same as integrating over a sphere of radius r and integrating all radius $<r<R$ i.e how to rigorously justify $$\int_{B_{R}(0)}\frac{1}{\|x\|^s}dx= \int_0^R \int_{\partial B_r(0)}\frac{1}{\|x\|^s} dSdr $$
 
10:33 AM
ah, you wanna do $L^{\infty}$, ok
for continuous functions, the notions of essential upper bound and upper bound agree (why?), so the notion of supremum and essential supremum do too
 
idk what essential upper bound is
 
@ioch It's a special case of en.wikipedia.org/wiki/Coarea_formula. You can probably also derive this special case explicitly by using spherical coordinates, but it seems like a messy computation (I haven't done it).
@epsilon-emperor an essential upper bound is a $t$ such that $f^{-1}(t,\infty)$ has measure $0$
so the essential supremum is the infimum of all essential upper bounds by definition
 
okay makes sense
let me think
 
10:48 AM
an idea please ?
 
@Thorgott thanks for that, yeah sorry I should of spent longer looking myself this explains it math.stackexchange.com/questions/4150476/…
as you said its coarea
@Thorgott sorry I meant this math.stackexchange.com/questions/130341/…
:) cheers !
 
@Thorgott if t is an essential upper bound, why is it an upper bound? the reverse direction is clear
 
what can you say about $f^{-1}(t,\infty)$?
keep in mind that $f$ is continuous
 
11:04 AM
@Thorgott it's an open set
oh cool so non-empty open sets have positive measure
so that set must be empty. that's the argument isn't it
then since upper bounds and essential upper bounds are equal, we see that the supremum and ess sup are equal. hope this is correct
 
yeah
 
@PeterJohn No, multiplication by $x$ takes $\Bbb C[x]/(x^n)$ to the ideal $(x)$. This does not contain constant polynomials.
 
11:30 AM
@BalarkaSen Ah, it's different to quotient map right?
 
11:52 AM
does anyone know what a "Lipschitz singularity" is?
 
Is it possible to cover 8*8 chess board with 13 many 4\times 1 tetrominoes and three 2\times 2 tetrominoes?
 
12:30 PM
Hi, I am stuck in a problem
I want to prove that $\frac{\mathbb{C}[x,y]}{\langle x^2+y^2-1 \rangle}$ is a dedekind domain. I was stuck and refered to this link
2
Q: Showing $F[x,y]/(ax^2+by^2-1)$ is a Dedekind domain

Sam Birns Let $R = F[x,y]/(ax^2+by^2-1)$, where $F$ is a field with $\text{char}(F) \neq 2$ and $a, b \neq 0$. Then $R$ is Dedekind. My attempt: $R$ is Dedekind if and only if it is Noetherian, integrally closed, and if every prime ideal is maximal, so it suffices to show these conditions. Noether...

The answer by Vincenzo Zaccaro says that $F[x] \to R$ is integral. Is it true?
 
 
2 hours later…
2:31 PM
What is the motivation behind the quadratic gauss sum? I know how it could be used to prove the quadratic reciprocity but why consider such sum?
 
 
1 hour later…
3:48 PM
If $R$ is a complete ring w.r.t. $I$-adic topology then $R^\times\to (R/I)^\times$ is surjective?
I've seen this question for local ring case but not in complete ring case
 
Please shed some light on understanding above passage
 
@PeterJohn I don't understand what you mean. Multiplication by $x$, as an endomorphism of $\Bbb C[x]/(x^n)$, is not surjective.
 
4:23 PM
The pages related to topic from the book.
 
glS
4:48 PM
hi all. If I may ask a naive question, which however probably doesn't deserve to be a question in the main site: the wikipedia page for irreducible polynomials states in the examples that $9x^2-3=3(x\sqrt3-1)(x\sqrt3+1)$ is a reducible polynomial over the integers. Is that a typo? I don't understand how this is reducible when $x\sqrt3\pm1\notin \mathbb Z[x]$... or am I missing something?
(on second thought, they probably are using a different definition here for irreducibility over $\mathbb N$, which is not a field; never mind, that's probably the reason)
 
@BalarkaSen You mean multiplication by $x$ followed by a quotient map? The binding maps should be a map $C[x]/(x^n)\to C[x]/(x^{n-1})$ if I understood correctly.
 
@glS As the Wikipedia article notes, the third one is reducible because the factor 3 is not invertible in the integers. The polynomial is reducible because $9x^2-3=3(3x^2-1)$ is a factorization over the integers in which neither factor is a unit.
The factorization $9x^2-3=3(x\sqrt{3}-1)(x\sqrt{3}+1)$ is not a factorization over the integers, but we simply do not care about it.
 
glS
@Thorgott yea I was confused because I was thinking of reducible polynomials as simply those that can be written as product of linear factors. I didn't realise that if operating over a non-field, a constant polynomial can also be irreducible and thus count as part of a decomposition
 
@PeterJohn No. All the objects are $\Bbb C[x]/(x^n)$.
The inverse system is literally $\Bbb C[x]/(x^n) \stackrel{\times x}{\to} \Bbb C[x]/(x^n) \stackrel{\times x}{\to} \cdots$
None of the binding maps are surjective, but it's clearly a Mittag-Leffler system
 
5:11 PM
@Balarka If I have two inner product spaces $(V,\langle-,-\rangle_V),(W,\langle-,-\rangle_W)$, do you agree that the right choice of inner product on $V\times W$ is given by $\langle(v,w),(v',w')\rangle_{V\times W}=\langle v,v'\rangle_V+\langle w,w'\rangle_W$?
 
@BalarkaSen For $n = 2$ then $\operatorname{Im}(\Bbb C[x]/(x^2)\to\Bbb C[x]/(x^2)) = (x)$ but $\operatorname{Im}(\Bbb C[x]/(x^2)\to\Bbb C[x]/(x^2)\to\Bbb C[x]/(x^2)) = 0$.
 
@PeterJohn So?
@Thorgott Yes.
 
Oh after composing 2 maps it's just 0
 
Ok, now tell me why
 
@PeterJohn Exactly.
So everything stabilizes after 2 steps
That's why it's Mittag-Leffler
 
5:13 PM
Thanks very helpful
 
@Thorgott No.
 
no fun
 
Ok, I'll play but don't flame me if I'm wrong. It's the unique inner product in which $V \times \{0\}$ and $\{0\} \times W$ are orthogonally embedded isometric copies in $V \times W$?
Sounds like diagonalization.
That should work
 
I agree that's a characterization of this inner product, but I'm somewhat confused by why this is the right condition. The thing is that "being isometrically embedded" is a property of maps into $V\times W$ (demanding the canonical injections to preserve the norm), yet this is not a coproduct in the category of inner product spaces and norm-preserving maps (there are no coproducts). Meanwhile, "being orthogonal" seems more like a "map out of $V\times W$" condition, e.g. it would be forced by demanding the projections to be norm-preserving, but that's not what this norm does either (and ther
 
5:29 PM
If you have just orthogonal then you get $c_1 \langle -,-\rangle_V + c_2 \langle -, -\rangle_W$
I don't know formalism so I have nothing more to contribute :)
 
$-,-$
 
5:46 PM
-_-
 
:-
i love ambiguity
 
one noteworthy thing is that an inner product induces an iso $V\rightarrow V^{\ast}$ and this is the only choice of inner product on $V\times W$ such that the iso $V\times W\rightarrow (V\times W)^{\ast}\cong V^{\ast}\times W^{\ast}$ is the direct product of the isos $V\rightarrow V^{\ast}$ and $W\rightarrow W^{\ast}$ induced by the respective maps
 
Ah yeah this is a good perspective
 
is there a better characterization of the isos $V\rightarrow V^{\ast}$ that come from inner products?
 
6:18 PM
Hi, I'm trying to prove (from scratch) that Cauchy sequences in C_0(X) converge (no fancy properties of Banach spaces, etc.)
Can I get some help, please?
I can point to places where I'm stuck
 
What is $X$?
 
and what is C_0
 
so many questions look like the following to me: if $\phi \in L$ and $\varphi \subset J_0$ with $\Pi_\rho(L)\xrightarrow{\#} (J_0)^p$, why is $\Pi_\rho(L)\sim_{\phi,\varphi}J_0$ regular?
 
@copper.hat X is a locally compact Hausdorff space
@Thorgott C_0(X) is the space of all complex continuous functions on X which vanish at infinity
 
6:33 PM
What does infinity mean here?
 
@copper.hat Good question! A complex function $f$ on a locally compact Hausdorff space is said to vanish at infinity if for every $\epsilon > 0$ there exists a compact set $K\subset X$, such that $|f(x)| < \epsilon$ for every $x\notin K$.
By the way, I have written out my attempt in detail here: https://math.stackexchange.com/questions/4154220/cauchy-implies-uniform-convergence-in-c-0x-w-r-t-the-sup-metric
I have started with a Cauchy sequence, and proposed a limit function. I need to show that the proposed function is actually the limit, and also within C_0(X). I seem to have done the former (though not sure), but stuck with the latter.
 
6:46 PM
note that you can bound $|f(x)-f_n(x)|$ uniformly in $x$, this should resolve most of your issues
 
@Thorgott How?
 
use your hypothesis!
 
You have some distance on $C_0(X)$ presumably?
 
yeah, sup metric
 
@Thorgott I don't understand what you mean by "bound uniformly"
 
6:57 PM
Have you never learned about uniform convergence, @epsilon?
You want a bound on $|f(x)-f_n(x)|$ that holds for all $x$.
 
@epsilon-emperor I would suggest proving it in $l_\infty$ first.
 
Greetings @copper and @Thor.
 
@TedShifrin Afternoon Ted!
 
hello, let $E=\mathcal{C}([0,1],\mathbb{R})$ and $T$ a map defined on $E$ by
$\forall x\in[0,1],\qquad T(f)(x)=\int_0^x f\big(4(t-t^2)\big)dt$. how to show that $T$ is not onto
thank you
 
Please fix your MathJax.
 
7:11 PM
at the very least fix your tex if you're just gonna repost your question
also hi Ted
 
@TedShifrin Right, trying this. Are you proposing I use Cauchy-ness?
 
@TedShifrin sorry : $\forall x\in[0,1],\qquad T(f)(x)=\int_0^x f\big(4(t-t^2)\big)dt$
 
@epsilon-emperor I was just answering your question. But, yes, the sup metric will give you uniform convergence.
 
Convergence alone is enough. I know that convergence and uniform convergence are equivalent in the sup metric.
 
So how do you relate $d(f_n,f_m)$ and $d(f_n,f)$?
@Vrouvrou: Your question is very easy. What have you done other than asking us to do it for you?
 
7:21 PM
@TedShifrin $d(f_n,f) \le d(f_n,f_m) + d(f_m,f)$
 
That isn't what I had in mind. Also, have you established that $f$ is continuous?
 
once you have the uniform bound on $|f_n-f|$, that follows easily
 
@TedShifrin I see that T is derivable and it is continuous so Im T is include in C^1 but I don't know if it is sufficient ?
is it possible to define a function which has antecedent ?
 
Think about basic facts about integrals, @Vrouvrou.
 
7:42 PM
Is there a special name for a (possibly nonsquare) matrix with unit rows? Or the space of such matrices?
 
Not that I've ever seen.
 
i have similarly not seen this pointed out in discussions of matrix theory.
 
@TedShifrin Haven't yet
@TedShifrin limit as m\to\infty d(f_n,f_m) = d(f_n,f). Is that what you had in mind?
 
@love_sodam No, it's impossible. You can show that with a mod 4 numbering of the chessboard squares.
Like this:
$$\begin{array}{c} 0 & 1 & 2 & 3 & 0 & 1 & 2 & 3\\ 1 & 2 & 3 & 0 & 1 & 2 & 3 & 0\\ 2 & 3 & 0 & 1 & 2 & 3 & 0 & 1\\ 3 & 0 & 1 & 2 & 3 & 0 & 1 & 2\\ 0 & 1 & 2 & 3 & 0 & 1 & 2 & 3\\ 1 & 2 & 3 & 0 & 1 & 2 & 3 & 0\\ 2 & 3 & 0 & 1 & 2 & 3 & 0 & 1\\ 3 & 0 & 1 & 2 & 3 & 0 & 1 & 2\\ \end{array}$$
 
7:59 PM
@epsilon-emperor Yes.
 

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