Mathematics - 2021-09-12 (page 1 of 2)

robjohn

00:28

@copper.hat yeah, we sold my uncle's house last year and the appraisal was way higher than the sale price and so we tried to get some of the loss off on taxes, but our preparer is not comfortable claiming a loss since she feels that everything should have gone up last year, if anything. So we will not be getting a refund on that probably.

Ted Shifrin

00:45

@robjohn: Integrity is over-valued.

copper.hat

@robjohn Bummer! I'm drowning in taxes and related stuff. I have to file an Irish tax return but at least do not owe anything, and need to sort out my daughter as her taxes became complicated this year and are likely to continue so.

robjohn

@copper.hat yeah, our taxes are quite complicated and thanks to an ex-President, are a nightmare now.

@TedShifrin certainly is. No value to it at all according to our tax bill...

Semiclassical

imprecise question i'm thinking about over at h Bar too. Hilbert spaces are also Banach spaces, and closed subspaces of Banach spaces are also Banach. Do there always exist closed subspaces of Banach spaces which are also Hilbert spaces? And if so, how does one classify/identify them?

Ted Shifrin

Closed? Well, $L^2$ won't be a closed subspace of $L^1$, will it?

Semiclassical

hmm

Ted Shifrin

00:50

You can find finite dimensional subspaces, of course.

Semiclassical

maybe i'd better drop that, then

Ted Shifrin

I'm not sure it's a good question, irregardless.

Semiclassical

@TedShifrin yeah, true.

shintuku

Prove that if a matrix $Y$ is obtained from $X$ by performing one or more row operations, then their row spaces are identical.
Proof: Given the row space $R_X$ of $X$, a row operation consists of substituting one of the rows of $R_X$ by a linear combination of vectors in $R_X$. Therefore, any $R_B$ obtained through row operations on $R_X$ is nothing else than a set of vectors spanned by (produced by a linear combination of $R_X$).

Semiclassical

I started with this as "C* algebra" instead of Banach spaces, but maybe changing it was a mistake

shintuku

00:52

does anyone believe me or do i have to start talking about bases

Ted Shifrin

One or more operations? Care to be specific?

You do not need bases at all.

shintuku

whoops, i meant row operations

Ted Shifrin

Just argue inclusion both ways.

Semiclassical

the logic was: If I start from a Hilbert space $H$, I have the C* algebra $B(H)$ of bounded linear operators on $H$, and within that algebra I can find the Hilbert-Schmidt operators $B_2(H)\subset B(H)$.

Ted Shifrin

Well, if you're talking $C^*$ algebras, then I withdraw and page @leslie.

shintuku

00:54

i was thinking doing double inclusion, but i thought that maybe the above was more intuitive and faster, if it worked, even if it is slightly informal

Ted Shifrin

To show identical, you can't just show subset.

What if row operations allowed multiplying one row by $0$?

shintuku

hm, but would that still not be a linear combination of vectors in $R_X$?

setting coefficients to 0

Ted Shifrin

You drop row space when you do this.

So you will get inclusion one way, but not the other.

That's the whole point.

shintuku

oh, right. if I have a set $V$ where $v \in V$ is a linear combination of vectors from $W$, that doesn't mean $W$ is a set of linear combinations of vectors from $V$

robjohn

@TedShifrin Your statement started me on a journey that ended here

Ted Shifrin

01:00

ROFL

robjohn

read the whole thing. It is interesting

Ted Shifrin

One of my best friends and I use it humorously, and most of our friends have quit correcting us.

Semiclassical

@TedShifrin fair

the question on the C* algebra side was essentially "how to classify C* subalgebras which are also Hilbert spaces"

Ted Shifrin

That seems a better question than your original, but maybe an analyst would explain why I'm wrong.

Semiclassical

possibly

Ted Shifrin

01:03

@robjohn It seems like my usage fit within her parameters.

@shintuku That truly is the whole point.

Semiclassical

I know that the difference between Banach and Hilbert spaces is whether the norm on a Banach space is an inner product norm (satisfies the parallelogram law and thus determines an inner product)

shintuku

great, thanks!

Semiclassical

so i was wondering when you can find subspaces for which the norm is an inner product norm but not on the entire space.

(and how to classify those)

Ted Shifrin

Oh, well, using the $L^1$ norm on $L^2$ doesn't help.

Checking the parallelogram law occurs on two-dimensional subspaces, so I guess if it fails, it's going to fail in finite dimensional subspaces as well.

Semiclassical

finite being larger than one, at least

Ted Shifrin

01:07

Could it happen to work on some two-dimensional subspace. Probably?

Semiclassical

yeah, that's what i wonder

of course, just being Banach may be too general. so restricting to the case of C* algebras seems a healthy simplification

Ted Shifrin

@Leslie would know this in a millisecond.

Semiclassical

I sorta know an answer, though it feels like it's more circuitous than it should be. If I have a C*-algebra $A$, I can pick a state $\omega$ and use the GNS construction to get $\mathcal{A}=B(H_\omega)$ as a representation on the Hilbert space $H_\omega$. From there I can obtain $B_2(H_\omega)\subset B(H_\omega)$ which is naturally a Hilbert space.

SmokenSieEinBitteChebaHitBits

What you need is an abstract spacecraft with which you can view this space

Semiclassical

that would seemingly push the problem back to representation theory of C* algebras but that's a lot better known.

copper.hat

01:15

@Semiclassical what are you trying to do?

Semiclassical

ultimately, figure out what math i should know for certain QM stuff

leslie townes

@Semiclassical math.stackexchange.com/questions/112844/…

Semiclassical

that's addressing the case when the entire C*-algebra is a Hilbert space. I'm aware that doesn't happen much

I'm interested in subalgebras

leslie townes

norm-closed subalgebras?

copper.hat

as opposed to domalgebras

Semiclassical

01:17

that sounds plausible

copper.hat

sorry

leslie townes

note that the hilbert space operators are a hilbert space, but with a different norm than the operator norm. do you care about renorming?

renorming changes the game

Ted Shifrin

Well, if we can renorm, then $L^2$ is a subspace of $L^1$.

Semiclassical

hmm

Ted Shifrin

That was my first comment.

leslie townes

01:17

well there are gradations of renorming other than arbitrary renorming.

Ted Shifrin

That wasn't arbitrary, dammit.

leslie townes

i mean, you can quantify 'renorming' in such a way that would rule out renorming L^2 with L^1 but might not rule out other norms.

hanche-olsen's comment about the geometry of the unit ball is worth looking into. that is often how easy embedding results are proved or disproved

because of the nice relationships between subspaces and subspaces of duals and double-duals.

Semiclassical

sounds like i've got stuff to read, then

Ted Shifrin

The final comment on his answer seems important.

leslie townes

the "modulus of convexity" might also be a worthwhile jumping off point. it has been a long time since i studied whether you can use it to rule out potential embeddings. but it sometimes gives more information than less quantitative aspects of the geometry of the unit ball.

Avra

01:21

Can you please give me a source for one-to-one correspondence with examples on graphs?

Semiclassical

looking at those answer, should I be looking specifically for one-dimensional subalgebras? That'd be disappointing.

Ted Shifrin

What are you even talking about, @avra?

leslie townes

did i just say the g-word? forgive me.

Ted Shifrin

Yes, @leslie, you sinned for the last time.

Semiclassical

are you asking "what can we say about the graphs of bijective functions?"

Avra

01:24

@Semiclassical. Yes please

@TedShifrin. :(

bijection relation = one-to-one + onto

Semiclassical

"horizontal line" and "vertical line" tests

Avra

I am just looking for a source with examples please if you have one at hand

Semiclassical

well, wikipedia has articles on each: en.wikipedia.org/wiki/Horizontal_line_test, en.wikipedia.org/wiki/Vertical_line_test

Ted Shifrin

Are we talking only functions $\Bbb R\to\Bbb R$? What is the context?

Avra

I mean for graphs of vertices and edges

Semiclassical

01:27

so you mean like the picture here?

Avra

Yes!

This is bipartite graph

this is one-to-one

I am looking for source with examples

Or systematic way to prove this for graphs

Semiclassical

what do you, 'examples? they all look the same

Avra

most what I found is either basic or too complex

Ted Shifrin

This is why I said I had no idea what you were talking about. I rest my case.

Semiclassical

^

only difference between the various graphs is which vertices are connected and how many there are

Ted Shifrin

01:29

It sounds to me like you need to sit down and do some examples for yourself and decide what the issue is.

leslie townes

semiclassical, this reminds me a bit of the right group action thing. the arrows going left to right. if you use a diagram like that for f: X -> Y and "compose" with a diagram of g: Y -> Z placed to its right [by connecting the arrows and pulling the strings taut], you get a representation of $g \circ f$, although the diagram for $f$ appears to the left of that of $g$ :)

Avra

@Semiclassical. Some graphs get very complex, so doing it manually seems crazy.

I am looking for simple explanation how to do it in a systematic way

Ted Shifrin

I still have no idea what you're talking about.

leslie townes

avra, people tend to prove general statements about graphs without reference to a pictorial representation. the visual representation is for working examples and identifying trouble spots.

Semiclassical

i have no idea what you mean. for each vertex in X you pick exactly one vertex in Y and draw a line between them. what's complex about that?

Ted Shifrin

01:31

Where are the edges coming into play?

What is the function?

Avra

I meant isomorphism

Thorgott

what does a graph have to do with being one-to-one

Semiclassical

@Thorgott graph of a relation more generally, presumably

Avra

So is isomorphism same as bijection please

Thorgott

define isomorphism

Semiclassical

01:32

that's entirely too vague to answer. in what context?

Avra

two graphs are isomorphic if there is a one-to-one correspondence

Semiclassical

now you're talking about graph isomorphism.

that's entirely different than "bijection"

Avra

:/ sorry

Semiclassical

if you're wondering how to tell if two arbitrary graphs are isomorphic, in general it's not easy

copper.hat

i guess the vaccination was a bijection?

shintuku

01:34

pls someone ban copper

Avra

So this is different from bijection. So proving that a relation on a graph is bijective does not prove that it's isomorphic

copper.hat

:-) at last

Ted Shifrin

You are throwing words around like crazy. What is a relation on a graph? Are you talking about a relation on the vertices only?

Semiclassical

if the bijection is acting on both the edges and vertices, then it's a graph isomorphism. if it's only on the vertices, it's in general not

Avra

@TedShifrin. I see that I am tooo broad. Pardon me. Yes on vertices

not edges

Ted Shifrin

01:35

I need to start smacking @copper.

Semiclassical

a graph isomorphism is a relabeling of vertices in such a way that the edges are preserved

Ted Shifrin

So the graph is totally irrelevant.

shakes head

Thorgott

yeah, the terminology seems all over the place right now

Avra

hahaha :(

@Semiclassical. tHANKS. Okay so I am mixing both isomorphism and bijection

Semiclassical

@leslietownes to make sure I'm understanding that answer from earlier: The only C* algebra that is also a Hilbert space is $\mathbb{C}$. So if we want C* subalgebras which are Hilbert spaces, we need subalgebras which are isomorphic to $\mathbb{C}$.

Thorgott

01:39

@Semiclassical indeed, as far as I know, the algorithmic complexity of determining whether two graphs are isomorphic still isn't even known

Avra

@Semiclassical. I found this statement that confused me, "there is a one-to-one correspondence between R and G=<V,E>, E for edges and V for vertices

R is a relation on set A that is a subset of $A \times A$

Ted Shifrin

Relation on what?

And the vertices are in $A$?

So we get an edge any times two vertices are related.

Avra

Yes!

Ted Shifrin

That's how they're defining the correspondence.

Semiclassical

@Thorgott yeah

Avra

01:40

So as you can see above, we are trying to prove a bijection between R and graph G

Semiclassical

if you've got small graphs i imagine there's brute-force algorithms

Ted Shifrin

With what I said to explain how.

Avra

@Thorgott. Wow!

@Semiclassical. Wow!

leslie townes

semi, sounds right. the examples given in that solution show that in a (norm closed) C* subalgebra with dimension >= 2, a counterexample to the parallelogram inequality can be found within that subalgebra.

Avra

" the algorithmic complexity of determining whether two graphs are isomorphic still isn't even known" :0 -- Thorgott

Ted Shifrin

01:42

@leslie Did you ever try that recipe I sent you? Now I've forgotten what it was.

Semiclassical

which rules out the two-dimensional possibility floated earlier

leslie townes

ted: the leek one? not yet. the last time i tried to go to the store they didn't have good leeks, which defeated the purpose. maybe this week.

Ted Shifrin

Ah, right, that's what it was. Too bad.

Yeah, I usually find fine ones at Sprouts, but occasionally they are just too old.

Avra

@Semiclassical. So, what is meant here please by, "there is a one-to-one correspondence between R and G=<V,E>, E for edges and V for vertices R is a relation on set A that is a subset of A×A., where A is vertices of G"

SmokenSieEinBitteChebaHitBits

@Avra a lot can be done with brute force on small graphs, so as long as your application breaks everythign down in into relatively small graphs, you're okay for complexity concerns

Ted Shifrin

01:45

I answered that for you, @Avra. Did you read what I said?

Avra

"So we get an edge any times two vertices are related.
"

leslie townes

when i lived in berkeley i lived next to a small market that for some reason always had perfect leeks. i was in heaven.

Avra

Yes. Thanks

Ted Shifrin

That's the answer.

Semiclassical

What confuses me about that, is if I take the story I gave earlier (start from algebra $\mathcal{A}$, pick a state $\omega\in\mathcal{A}$, use the GNS construction to get $\mathcal{A}$ as $B(H_\omega)$, and thereby obtain $B_2(H_\omega)$)

leslie townes

01:45

and for cheap.

Ted Shifrin

Monterey Market or something else, @leslie?

leslie townes

similar spirit but smaller and less selection. yasai market on college/claremont.

Ted Shifrin

Oh, after my time.

Semiclassical

then it seems like I'm getting Hilbert-space subalgebras, indexed by the choice of $\omega$.

Ted Shifrin

I lived on the south side, but shopped on the north side.

Semiclassical

01:47

so does that mean that $\mathbb{C}$-subalgebras of $\mathcal{A}$ are indexed by elements of $\mathcal{A}$?

leslie townes

the hilbert schmidt norm is not the same norm as the norm of A (or B(H_omega))

unless i'm missing something

Semiclassical

yeah, fair

this is something that's confusing me, the more i think of it

leslie townes

i think in QM applications it's "natural" to consider the hilbert schmidt norm whenever you have a representation on a hilbert space, but it doesn't interact very well with abstract C* algebra structure. i don't know that it has a universal property or something without reference to a choice of representation, for example

Semiclassical

i'm fine with it being with reference to a choice of representation

SmokenSieEinBitteChebaHitBits

Semiclassical

01:51

i was trying to find a way to "get at" the possible ways to have $B_2(H)\subset B(H)=\mathcal{A}$, without focusing on H itself

leslie townes

it's been a long time since i thought about this. you might be interested in the theory of 'operator spaces.' they can be defined abstractly but somewhat concretely are banach spaces paired with a choice of an embedding in some B(H). the pairing matters.

Thorgott

objects without a universal property, what horror

leslie townes

i don't know that many QM people work with them, but maybe they ought to

SmokenSieEinBitteChebaHitBits

@leslietownes that's how far the editor is

Semiclassical

the motivation is roughly: hey, operators on a Hilbert space are themselves a Hilbert space w/r/t H-S. i wonder how this latter Hilbert space relates to the C* algebra on H

and whether there's a way to understand that without constructing an H explicitly

leslie townes

01:53

i always thought of them as different things. never really thought of connecting them up in any way (not to say that it can't be done)

Semiclassical

it's entirely possible it's an unnatural thing to do, of course

I do need to get a better handle on GNS representations, though

leslie townes

02:40

it's good to know the basics. they come up a lot in operator theory.

there are a bunch of theorems that generalize the GNS construction in varying directions. i don't know how important any of that is to QM though.

Semiclassical

02:56

the person i should probably talk to about this is Paul Garrett, the guy who wrote that bit on "Hilbert space tensor product does not exist"

if only because he's at the UMN and i know where his office is :P

(i have met him a few times, though that was a while ago)

shintuku

i'm in single variable analysis. the limit definition gives us $g'(a) = \lim \limits_{x \to a} \frac{g(x) - g(a)}{x-a}$ and without altering the definition, since $f(a)$ is a constant, we can say:
$g'(f(a)) = \lim \limits_{x \to f(a)}\frac{g(x) - g(f(a))}{x-f(a)}$

am I allowed to say: "since for all $x$ approaching $f(a)$ there exists an $x^*$ such that $x=f(x^*)$, and for any such $f(x^*)$, we have a well defined $g(f(x^*))$ , we have:
$g'(f(a)) = \lim \limits_{f(x) \to f(a)}\frac{g(f(x)) - g(f(a))}{f(x)-f(a)}$"

leslie townes

let f(x) = x^2, a = 0, and consider approaching f(a) = 0 via negative x. are there x*'s for which x = f(x*) in this case?

note also that there's nothing thus far that rules out the possibility that f(x) = f(a) for infinitely many x near a, making the interpretation of the denominator of your limit a problem

stick with a textbook proof of the chain rule

calculations of this type motivate the result but don't prove it in generality

shintuku

03:13

argh both are good counterexamples

leslie townes

some calculus books don't bother to include a complete proof of the chain rule, but rely on arguments about that difference quotient in the cases that it makes sense

shintuku

i'm trying to figure out rudin's proof. he tells us that the definition of the derivative gives us $g(s) - g(y) = (s-y)[g'(y) + v(s)]$, where $v(s) \to 0$ as $s \to y$, which makes perfect sense, but then he sets $s = f(x)$

Semiclassical

oh hey, this question on MO: mathoverflow.net/questions/35207/…

leslie townes

doesn't seem like an easy problem to me. the commenters are no slouches at C* algebras. :)

Semiclassical

yeah

the problem i have is that i have the expertise to come up with kernels of questions, but not to properly formulate the full question

leslie townes

03:22

life is the search for the full question

Semiclassical

yes

shintuku

well, thinking of it, at no point would a composition $(x^2)^2$ have $x^2$ negative, so maybe only the justification is wrong but there is a way to properly define the limit

leslie townes

certainly, and it might be worth studying how rudin's proof deals with that (or rather, gets around having to explicitly address that case)

shintuku

sound advice sound advice

leslie townes

this is a real drawback of the rudin approach. everything is arranged exactly so that the tricky details you might run into if you tried it any other way don't even appear

it's also what people like about the rudin approach

shintuku

03:31

alright, we can account for the second argument by noticing that the limit definition requires the approaching number not to equal the constant

since the condition is $0 < |x-c| < \delta$

...right?

leslie townes

in the original limit you only have a restriction on the values of x approaching a. you don't have any control over how f(x) approaches f(a). there may be no values of x for which f(x) is unequal to f(a)

as far as i can tell, "lim f(x) to f(a)" has no independent meaning

at least rudin doesn't separately define what that might mean

i'd focus on figuring out why rudin is right and not why some other approach isn't wrong

even if the other approach isn't wrong (note i don't see how you get f'(a) appearing in your limit formula, although maybe that's in a next step)

it feels like reinventing the wheel to me

others may differ in opinion

shintuku

oh! i get it. $g(s) - g(y) = (s -y)[g'(y) + v(s)]$ is indeed from the derivative definition, and if that equality holds, then it must also hold with $s = f(x)$

no need to think up an alternative derivative definition or funky limits

Koro

are you proving Chain Rule @shin?

shintuku

yessir

Koro

03:47

while you're at it, you may look up Caratheodory theorem, and then you'll see how Chain Rule follows in 5 to 6 lines :)

(spelling of Caratheodory may be wrong)

copper.hat

imagine if rudin was a lawyer...

Koro

imagine if Rudin was a doctor :)

leslie townes

rudin seems more like the type who would go from law school to clerkship to judge without ever getting his hands dirty

Ted Shifrin

Right to the worthless Supreme Court. Super qualified.

leslie townes

yep

Ted Shifrin

03:51

Yes, I’m in a bitchy mood.

shintuku

@Koro you're on convex optimization?

on an unrelated note, i've completed the proof. it's not very insightful, but it is so tremendously short i'm not sure i could forget it

Ted Shifrin

The right proof, which works in multivariable, is best linear approximation.

But no calculus books teach this.

shintuku

the $f(x) - f(a) = (x-a)[f'(a) + \varepsilon(x)]$ form of the derivative sort of hints at this but at no point does he mention linear approximation

Ted Shifrin

Well, that’s what that equation is, once you quantify the error.

Linear approx of composition is composition of linear approx.

PM 2Ring

04:07

The Mona Lisa Twins (originally from Austria) love 60s music, and have been doing great Beatles covers since their early teens. Their latest original song is a hilarious sarcastic piece with a music hall flavour. I Bought Myself A Politician.

shintuku

@TedShifrin oh nice way to put it

PM 2Ring

It's nice to see that there are some young songwriters who aren't afraid to get political.

copper.hat

05:36

our supreme court is just providing job opportunities for poor bounty hunters.

have we lost our way entirely?

BannedUser

You guys make want to read Rudin again >:(

shintuku

i'm eyeing that four line fundamental theorem of calculus proof

leslie townes

i put an upholstered chair in my office today and the cat's spent all evening making it her own. sleeping on the seat. sleeping on the armrests. sleeping on the backrest.

at least cats have their priorities in order

copper.hat

there needs to be a condensed rudin

cats hate me

leslie townes

cliff's notes: rudin principles of mathematical analysis

could be worse. one of my friends is hated by birds, and there are a lot more wild birds than there are wild cats.

Ted Shifrin

05:54

@copper.hat Bluntly, you and I haven’t. But yes!

copper.hat

:-(

Ted Shifrin

Hell, it’s already telegraphic. Condensed?

copper.hat

:-)

An ostrich beaked me once, but it was deserved (and it hurt).

Once a crow buzzed me on Key Route Boulevard while I was running.

leslie townes

i thought my friend just had a phobia of birds, but then i was out with her once and she got divebombed by a crow. it wasn't even nesting season so they wouldn't normally be doing that.

apparently something like that happens once or twice a year.

copper.hat

Does her hair look like a nest?

leslie townes

05:57

you didn't hear it from me, but sort of, yes.

copper.hat

The Bodega Birds movie spooked me slightly.

leslie townes

but most birds don't orient toward nests outside of nesting season. it's baffling to me.

i love that movie.

i love all of hitchcock's north bay work. shadow of a doubt is his best.

copper.hat

the stuff of conspiracy theories :-)

last week's economist had an article on said topic

leslie townes

my daughter is a bird whisperer. she can walk into a crowd of canada geese and just hang out with them.

copper.hat

hissterics?

leslie townes

05:59

sometimes she says "hi, geese"

Ted Shifrin

Amazing.

leslie townes

geese make the worst sounding noises at us, but tolerate her.

copper.hat

one of my homeworks for my kids was to estimate how much goose poop was deposited in Ocean View field per hour.

leslie townes

good exercise in the use of scientific notation.

copper.hat

:-). at least one of them found it amusing. perhaps this is why my son only talks to me for 5 mins per day

leslie townes

06:02

i'm really not looking forward to my daughter's teenage years. as annoying as she is now, i have a feeling i'm going to miss it.

copper.hat

oh yessss.

leslie townes

she spent half of dinner telling the cat and my wife that she was going to put both of them in the garbage truck and they would 'take you away'

copper.hat

whoa.

that would have got us whipped.

leslie townes

she was giggling hysterically. nonsense threats are one of her favorite genres to work in.

one time she kept saying to my wife, in a threatening way, "you're gonna eat a burger."

followed by laughter

copper.hat

i was just going to say that i dodged the bullet, as headstrong females seems to be the norm in our gene pool.

just as well i am such a great peacemaker

leslie townes

06:07

my daughter takes after me. i used to refuse home cooked food by accusing my mother of having put "cat food" in it

copper.hat

we were marketing guinea pigs for the meat factory my dad was associated with. we had to work our way though many boxes of tinned stewed beef which bore a remarkable resemblance to our dog's canned food, except our tins had no labels.

ok. i think i am done for the day. still smarting after my 1040-ES.

did your daughter get a cast?

leslie townes

06:25

we weren't able to get an appointment until monday. we hope that a cast is all she needs. the person in urgent care gave us the impression she'd need something more than the splint we're using now.

of course this stuff has to happen around a weekend. just like the water heater that broke on a thursday night, the air conditioner that has to break friday morning, ...

shintuku

06:39

is there a function that isn't trigonometric and intersects some $f(x) = c$, with $c$ constant, at least an arbitrary number of times, and nowhere intersects $c$ more than once at a time?

i.e. all intersections are points and not intervals

leslie townes

you could just draw a path made up of line segments /\/\/\/\/\/\/\/\/ that keep crossing the line y = c

you choose how high they go and how spaced apart each peak or valley is from the next

shintuku

hm

i'd like to add the condition that it has to be possible to encase the function in a formula, but i have no idea how to say that precisely

leslie townes

you can write formulas for line segments

convolve with a smooth bump function to get smooth examples

convolution is a formula

the distinction you're looking to make is unlikely to be a useful one

shintuku

oh, i was also looking for smooth

i'll look into "convolution with a smooth bump function"

thanks!

leslie townes

Mollifier

In mathematics, mollifiers (also known as approximations to the identity) are smooth functions with special properties, used for example in distribution theory to create sequences of smooth functions approximating nonsmooth (generalized) functions, via convolution. Intuitively, given a function which is rather irregular, by convolving it with a mollifier the function gets "mollified", that is, its sharp features are smoothed, while still remaining close to the original nonsmooth (generalized) function.They are also known as Friedrichs mollifiers after Kurt Otto Friedrichs, who introduced them....

shintuku

06:49

very neat, thanks!

btw, do you know if there is a precise statement for, "has a formula"?

because I can image drawing some lines and curves on the 2d plane in a way that it is impossible to give an analytic formulation for them

but maybe that's actually some sort of theorem

leslie townes

there are the 'elementary functions' you run into in formalizing the idea that the antiderivative of e^(x^2) [for example] can't be written in terms of finite comibnations of the usual precalculus functions

i would like to emphasize that this is an analytically useless distinction. it's maybe interesting if you are interested in algebra

from an analysis point of view it's quicksand

shintuku

i'll start with elementary functions, thanks

robjohn

07:35

A bit more about mollification

shintuku

i've gotten lost in the wikipedia article and its referenced articles

under what heading is this stuff usually studied?

robjohn

convolution, but some is under smoothing operators

shintuku

a nice intro in your post, thumbs up

Mohcine

08:52

hi,

In DTMC, I would like to show that the equivalence class for i Cl(i)={j in S / i <--> j }={i}, what must I show to proof that Cl(i)={i} Thanks

Adil Mohammed

09:28

6*x^3*y^12 is a polynomial in 2 variables. What is its degree? is it 15 or 12?

Mohcine

09:39

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Q: Chapter 2 Exercise 7 Question (a) Page 85 Linda J. S. Allen 2010

Chapter $2$ Exercise $7$ Question (a) Page $85$ Linda J. S. Allen $2010$ $7$. Verify the following statement. Assume the period of state $i$ in a DTMC model is $\rm{d}(i)=0$. Then the set $\{i\}$ is a communication class in the Markov chain. for all $i \in S$, $d(i)=\rm{gcd}\{n/ p_{ii}^{(n)}>0...

Koro