Mathematics - 2021-04-24 (page 2 of 3)

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8:11 AM

user image

uh, does $D$ here mean derivative of?

from: proofwiki.org/wiki/Derivative_of_Inverse_Function

yes

yes

thanks!

@shintuku I would assume so.

$\forall x\in I^o:Df(x)\gt0$

try using MathJax instead of pasting images. It is easier for others to use.

alright!

8:20 AM

you do have ChatJax working on your browser?

yes, thanks a lot!

managed to set up the whole thing

very smooth

great. I'd hate to have posted that for you only to see \forall x\in I^o:Df(x)\gt0

kinda ruins the effect

9:05 AM

Signal processing question associated with numerical analysis - help!

1

Q: Numerically finding impulse response from a wave equation

I'm developing a song synthesizer (like Yamaha's Vocaloid). I decided to mimic the acoustics of human speech. I modeled the space between articulators as a solid of revolution, and came up with the following PDE: $$ (PA)_{tt} = c^2 (P_x A)_x $$ where $P$ is air pressure (difference to the atmosph...

impulse-response numerical-algorithms

epsilon-emperor

9:15 AM

@robjohn Yes, thank you!

9:40 AM

@epsilon-emperor So you see how to motivate that answer?

. o O ( talking to ghosts )

epsilon-emperor

10:00 AM

Yep, I do

10:43 AM

Sorry for this silly question, but I want to approximate $2^{73}$. The exact value that I got was 9 444 732 965 739 290 427 392. Is it $9.45 \times 10^{19}$?

or $9.45\times 10^{21}$?

2 hours later…

1:09 PM

Hi @Ted. I have two (complex) 1-submanifolds of $\mathbb{CP}^1\times\mathbb{CP}^1$, whose Poincare duals are represented by $(p_1, q_1)$ and $(p_2, q_2)$ in $\mathbb{Z}\oplus\mathbb{Z}$. Suppose they are transverse. I want to count their intersection.

I know its given by the cup product $H^2(\mathbb{CP}^1 \times\mathbb{CP}^1) \times H^2(\mathbb{CP}^1 \times\mathbb{CP}^1) \to H^4(\mathbb{CP}^1\times\mathbb{CP}^1)$. But I don't what map it induces on $(\mathbb{Z}\oplus\mathbb{Z})\times(\mathbb{Z}\oplus\mathbb{Z}) \to \mathbb{Z}$.

@soupless actually, $9.44\times10^{21}$

@robjohn Thank you so much.

Wait, it's not 9.45, but 9.44?

it's closer to $9.44$ than $9.45$

Oh, sorry. I thought it was 947, but I was wrong. It was 9447.

1:31 PM

wanting to approximate 2^73 is a very peculiar need.

@leslietownes maybe he has a chess board with an extra row and wants to compute how much rice he needs...

he'll get one grain too many

as mitch hedberg once memorably said, rice is great if you're hungry and want to eat a thousand of something.

@leslietownes It is for our project, by the way. We need to model the number of bacteria in a youtube video

maybe not that memorably. internet says two thousand of something.

numbers do get very big when you start counting bacteria.

not if you dump a cup of bleach in with em

1:37 PM

It is this one.

WCYDWT: Bacteria

not a video for a germophobe.

give it another day and there will be more bacteria than atoms in the universe.

It looked as if there started out to be 2

reminds me of teaching exponential decay out of a book where all of the exercises were poorly written. they'd ask, how much americium or whatever is left after X time of decay. the answer was less than the mass of a proton. and all of the exercises were like that.

i began to think maybe it was on purpose.

i remember pointing it out to the class. you literally cannot have this much americium. i don't know why it's written out to three significant figures. blank stares.

Oh, yes. Those blank stares. You know you're making contact, for sure.

epsilon-emperor

Hey, I need some help

1:51 PM

calculus for business and social sciences they called it. i don't know why anybody bothered.

epsilon-emperor

Arzela Ascoli theorem says that if X is a compact metric space, and F is a subset of C(X), then: F is compact if and only if F is closed, uniformly bounded and equicontinuous.

but, lots of teaching hours. they usually let me teach 125% time.

@leslietownes I taught several classes of that. 200+ students each.

not 125% of full time but 125% of half time. i needed money.

epsilon, that sounds about right.

epsilon-emperor

From this, how do we prove this corollary:
Let X be a compact metric space. If (f_n) is a uniformly bounded, equicontinuous sequence in C(X), then some subsequence of (f_n) converges uniformly on X

1:54 PM

how can a subset be equicontinuous?

epsilon-emperor

what else can be equicontinuous?

equicontinuity is defined for families of functions right

sequences of functions

epsilon-emperor

C(X) is the set of all continuous functions on X

take some subset

@robjohn no actually sequence is a specific case

@epsilon-emperor oh, I missed that F was a subset of C(X)

epsilon-emperor

equicontinuity is defined for sets of functions

@robjohn Yes!

does the question make sense now?

2:00 PM

The question made sense, it was the description of AA that I was wondering about

it's been a while. see if you can prove that the closure of {f_n} is compact.

then it's just the jazz about a sequence in a compact metric space having a convergent subsequence.

Now that we have the set of functions is compact, we can treat it like a compact set and cover each function with a ball of radius $1/n$

sometimes your formulation is taken to be the AA theorem itself.

your problem, i mean.

There is a finite subcover, and one of those must contain an infinite subsequence

this way we can construct a cauchy subsequence.

2:18 PM

it's 7:20 and somehow my daughter is still not awake. it's a miracle.

two fighter jets flew overhead at midnight last night, waking my daughter up and terrifying her. i am sure there was a good reason for that.

2:55 PM

Can anybody please help me on how to evaluate infinite products? The definition of the product that I am using is similar to definition of series : we say f(1).f(2).f(3)... f(k)f(k+1)... converges to a finite value L (non zero value) if the limit f(1).f(2)... f(n) converges to L as n tends to infty and then we write f(1).f(2).f(3)... f(k)f(k+1)... =L

In particular, I want to find the infinite product for f(n) = (1+x^(2n-1))

as n tends to infty.

The necessary condition for convergence of the product is clearly $f(n)\to 1$ as n tends to infty

mm is that 1/(1-x)?

no. It is like: (1+x) (1+x^3)(1+x^5)....

@leslietownes

2n-1 or 2^(n-1)?

2n -1 (odd powers)

n is natural number

mm, i don't see any simplification. is there more context?

3:00 PM

And this is to be solved using limit definition. It is not expected to use continuity, differentiability etc. But I would like to see that as well. It can also further be deduced that $|x| $ should be less than 1

yes, that seems necessary for the convergence.

if you formally expand it you get various x^n's where n is expressed as a sum of odd parts. i don't see that simplifying.

It's an exercise problem. Using this "Viete" formula is to be deduced, which gives value of \pi/2 in terms of 1/2, sqrt (1/2) etc

hrm, i must be missing something.

Expansion into a power series will complicate the matter

that's just how i think. but i'm also dumb, so i should point that out.

3:04 PM

though then summing it up will be another challenge. I believe there must be some way apart from that by taking log etc. I tried that but got stuck

user image

I am unable to understand after the construction.

Should I post the weblink instead of the image?

i see the image. can you be more specific about where you are uncertain? this proof does assume a number of preliminaries.

i'm still kicking that infinite product over in my head.

@leslietowneshttps://math.stackexchange.com/questions/4114714/how-to-find-the-product-pi-n-1-infty-1x2n-1

I posted my question a while ago

ha, there's also math.stackexchange.com/questions/344571/… with a vintage answer from this chat's very own robjohn.

@leslietownes "Now equal angles stand on equal circumferences when they are at the centers, therefore the circumference $BK$ equals the circumference $EF$. by I.26" which is side and two angles congruency.

3:09 PM

or comment, i should say.

@leslietownes: Thanks a lot :) I think that's fair. :)

translating that into normal reasoning is tough. i think the idea is that if you have two congruent circles with centers O, P, and an angle at O and and angle at P that are congruent, then they cut off equal circumferences. the fact that the circles have the same radius seems important to this.

side and two angles, hrm.

Shouldn't it be SAS(side-angle-side)?

3:27 PM

that's what i was wondering.

also wondering how to relate triangles to circumferences. this stuff is not like riding a bicycle, you stop doing it, you forget all of it.

@leslietownes I also found another proof that relates segment and circumference :-/

that looks like a scan from the dover edition. i had all of those books and the postal service lost them.

3:56 PM

brb, filing a complaint in the court of federal claims.

4:13 PM

@feynhat You can see this with cohomology by pulling back Kähler forms from the factors, or you can see it geometrically. If $L_1 = \Bbb P^1\times \{q\}$ and $L_2 = \{p\}\times \Bbb P^1$, then you can see intersections geometrically: $L_1\cdot L_2 = 1$ and $L_1\cdot L_1 = L_2 \cdot L_2 = 0$ (just "wiggle" the submanifold by taking a different fiber, and then they are clearly disjoint).

4:24 PM

Why would the submanifolds be lines?

They could have non-trivial projections onto the factors, no?

4:39 PM

cohomology and intersections!!

These are the generators, @feynhat.

Now use bilinearity.

4:55 PM

Aah, I see. So, $p_1q_2 + p_2q_1$?

Right.

is there a nice way to express the determinant of a matrix whose columns are shifts of one another?

Sounds like a circulant matrix.

Circulant matrix

In linear algebra, a circulant matrix is a square matrix in which each row vector is rotated one element to the right relative to the preceding row vector. It is a particular kind of Toeplitz matrix. In numerical analysis, circulant matrices are important because they are diagonalized by a discrete Fourier transform, and hence linear equations that contain them may be quickly solved using a fast Fourier transform. They can be interpreted analytically as the integral kernel of a convolution operator on the cyclic group C n...

You can find eigenvectors and eigenvalues :)

4:58 PM

ah, exactly, I knew there was something, thanks

epsilon-emperor

@robjohn I'm not sure I understand this proof

Could you give some more details?

I've also created a post on MSE

5:30 PM

i love toeplitz matrices.

Kind of a silly question, but suppose you have two ODEs, y' = f(x) and y' = g(x). Fix a starting point, and flow time t along the flow of the first, then time t along the second, then time t along the first, and so forth, alternating. As $t\to 0$, does the resulting "flow" converge to something?

not sure how the alternating 'flow' is intended to interact with t going to zero. in general i would not expect solutions or solution families to different differential equations to have much to do with one another.

Yeah, I didn't think so. I had one person show me demos where they said it converged to the average of the flows.

But like I would have thought something like that would be easy to see if it were the case, but I can't put my finger on how to even formalize this alternating flow concept as t goes to 0.

that seems goofy to me.

we are on the same page.

Is there a way to easily show it is goofy to my friend?

I tried to think about how flows don't necessarily commute, but I don't know if this comes into play anywhere here.

5:37 PM

it's not even clear that if you grab a solution to y' = f(x) at 0 and use it to guide you from x = 0 to x = t that y' = g(x) will even be well posed at x = t.

Well I think we can for the sake of it, assume that the flow is globally defined on the plane or something like that.

Slow down. This is in the plane and prime is wrt to $t$?

Prime is with respect to x.

t is the time flowed along any given flow before alternating back to the other.

dy/dx = f(x) and dy/dx = g(x) being the DEs

So the vector field is $(0, \cdotj$ in the plane.

You're just doing plain integrals.

Flow is a fancy word for this.

Yeah, the flows are just the integrals of f and g, with a suitable +C.

And so what my friend suggests is that the "limit" of this process (however one would formalize this) is the average of the integrals.

5:41 PM

so you're doing int 0...t f(x) dx up to t = t_1. then int 0...t_1 f(x) dx + int t_1...t g(x) dx, up to t = t_2, and so forth?

this limiting process seems like it needs more definition.

Right. The $C$ involves the first integral for the second one. Etc.

All $t_i=t$?

you have weird friends.

If I understand what Leslie wrote, all the t_i would be the same t

err

Convergence is far from obvious.

they've got to alternate but somehow also something subsuming this alternation is going to zero. i'm not sure how to analyze further.

5:45 PM

I don't see how the limit as $t\to 0$ can be a flow, even if exists.

@leslietownes Maybe I don't understand what you were saying. It's imagined like a dynamical system: pick a point x_0, flow for a fixed time t along the flow (integral curve) of f(x) containing x_0 to a point x_1. Then flow x_1 along the flow (integral curve) of g(x) containing x_1 to a point x_2. Then just continue doing this, alternating to get a sequence of points x_0, x_1, x_2, ... .

yeah, and then somehow x_0, x_1, x_2, ... , x_n are confined to an interval [0,t] and we push t to 0? like we're partitioning an increasingly smaller interval into the same number of pieces?

@TedShifrin I agree, I don't see how it is obvious what this actually ends up being. They are basically running little programs to plot these points x_0, x_1, x_2, ... and then if they make t small enough, it appears as a curve which they claim is lining up with the integral curve of y' = (f + g)/2 containing x_0.

So we're plotting this in the plane as the flows of $(1,f)$, etc.

Yes.

5:53 PM

So they iterate $n$ times and want a limit as $t\to 0$ and $n\to\infty$, whatever the hell this means.

Yes, indeed, whatever that means.

weird, weird friends.

I felt it was ill-posed. They were not doing the flow even in their experiment. They were just doing linear approximations of the flows (along the tangent lines).

People who draw pictures on computers but don't make math sense.

Is there a nice way of conveying that this is silly (assuming it is?).

5:57 PM

possibly. i'm not a "nice way" kind of guy. you might as well ask me to speak ancient greek.

I need one clarification in definition of algebraic structures

i guess, i would ask for more precision about what limiting process one has in mind. sure, if your limiting process amounts to averaging f and g, then you might get an average of f and g. i don't think that there's enough detail in what is envisioned to even figure that out.

As I understand the definition states: A set with binary operations defined on it forms an algebraic structure

I wonder what becomes of it if the number of binary operations become infinite

σὲ φιλῶ, leslie.

@leslietownes so basically, how one defines the limit formally?

Write down this linearized thing for step $h$ arbitrarily small. I don't remotely see averages. I see products ....

6:00 PM

In case of finite binary operations, we may have groupoids, semigroups, monoids, groups, rings

koro often one fixes a handful of operations at a time. it's rare to consider, say, a multitude of ring structures (going to infinity) on the same underlying set.

@TedShifrin products of what exactly?

Until you show us an interesting example, @koro, I don't think this is interesting. Who cares?

The functions, @anakhro. I just did it in my head. Who knows.

@TedShifrin: Fair point. I read about algebraic structures today so thought about it. I’ll think about the examples:)

@anakhro are you just integrating? when you write $y'=f(x)$ i interpret this as the integral of $f$?

6:03 PM

@Koro you do have things like en.wikipedia.org/wiki/A%E2%88%9E-operad :P

@anakhro: Dear Lord! What’s that :D

@copper.hat Yes. The solution to the ODE is just the family of integrals $\{F + C \mid C\in\mathbb R\}$ for a primitive $F$ of $f$.

@Koro sadly I can't tell you much more. But it's kind of cheating your idea, from what little I do understand. It's like having infinitely many algebraic objects (and hence infinitely many operations).

@copper The picture comes from the ode system on the plane. I dunno.

so you are integrating some 'switched' version of $f$ and $g$?

@copper.hat I am not sure I know what you mean by that. The geometric idea is to view it as a dynamical system where you develop a sequence of points on the plane x_n by starting with an x_0, then flowing along F for a fixed time t to a point x_1. Then flow along G for the same fixed time t to x_2, and then alternate back and forth.

6:09 PM

i an trying to understand why that is not just $\int_Af+ \int_B g$ where $A,B$ are a partition of the 'time ' interval?

i am severely caffeine deprived, at least that is my excuse...

Ah. I was being stupid, forgetting that only $x$ shows up in the functions. I think copper's view is right.

@copper.hat what do you mean about $A,B$ being a partition of the time interval?

your function looks like int 0...t_1 f + int t_1..t_2 g + int t_2..t_3 f + int t_3...t g

[0,t_1], [t_1, t_2], [t_2, t_3], etc being a partition of the time interval. its length and how many subintervals are in it and how it goes to zero, kind of unknown.

not to stuff words in the mouth of an unruly irishman who is severely deprived of his 'caffeine' (we all know what that means). but that's my best guess.

i'm totally ruly.

truly ruly?

6:23 PM

yes, that is what i meant.

so @anakhro, for 'nice' functions it seems reasonable that some sort of average could be obtained with suitable conditions on the 'time switching'.

there is a vaguely related issue in control systems called relaxed controls. the idea being that a sort of relaxed/limiting system is obtained by wildly (my term) switching around the controls.

here you would have a system like $y'=u \cdot f+ (1-u) \cdot g$ where $u$ 'switches between $0$ and $1$.

Maybe I misunderstand, but A and B being a disjoint union of the intervals on which you are flowing along respectively the flow of f and of g, and the constant of integration being chosen suitably on that interval?

see what the lawyer said above...

yes. when you do perform integrals you're implicitly choosing a constant of integration, chosen by the endpoints of the intervals you are integrating over.

Well I just ask whether I understood it correctly. :P

can't say i was ever comfortable with indefinite integrals other than as a precursor to definite

6:30 PM

i guess in the formula i wrote out, i'm implicitly choosing a solution F with F(0) = 0.

the constant of integration is sort of irrelevant here surely?

but all the other +C's are fixed by the way the formula would work. you make things continuous and you proceed via integration.

yes, i think so.

to me it is an averaging system

@copper.hat so you'd think that just defaulting to $t = 1/n$, and then taking the limit as $n\to \infty$ doesn't necessarily get you the average? That you'd have to be more thoughtful than that?

if $f,g$ are reasonable then yes, why not?

6:32 PM

i think at a minimum, you need to specify a length of the interval that we are finding this solution for, and then some parittion o the interval. keeping track of both how t goes to 0 and n goes to infinity.

Well since you are dealing with a family of curves which you are flowing along, surely the constant by which the curves differ by is important to the set up.

it does seem you could make an analogue apparatus to perform this in real time with someone frantically flipping a switch.

if $f$ is slowly varying and you integrate over $[0, {1 \over n}] \cup [{2 \over n}, {3 \over n}] \cup ...$ then I would expect to get ${1 \over 2} \int f$.

anak something like int 0..t_1 f(t) dt + int t_1...t_2 g(t) dt + int t_2...t_3 f(t) dt would automatically glue these things together with the right +C's.

@copper.hat I think I see what you mean now. Thanks, this helps a lot.

@leslietownes I see, it's implicit in your definition.

6:34 PM

kind of sucks to write out the general formula for that, let alone analyze it as some upper bound of something goes to 0 and n goes to infinity, but we can't have everything.

it's the doctrine of original sin.

Heh.

Leslie, do you like to read? You sound like you read a lot of fiction.

for some reason yesterday my brain hit its formal limit and the engineer took over. so formal analysis is out for today :-).

leslie reads a lot of fRiction

these days i read mostly nonfiction, but yes i love to read.

this deleted problem sent me to be in a bad mood last night math.stackexchange.com/questions/4114425/…

i have been reading the dulles brothers (steven kinzer) for 5+ years. it is too depressing to take in large quantities.

the pleasures of counting is an antidote, along with the quantum computing for everyine

i sometimes quote poetry or shakespeare, but i don't actively read it. i just internalized it at a young age.

6:37 PM

lots of discussions about bras, makes me uncomfortable and blush

koerner's naive decision making is also pretty good. he teaches how to bet on the horses.

i like to read about the lives of poets and writers more than their work

behan is obviously an influence...

esp about sex.

that's a joke (just in case)

ahh, behan.

have i posted the one were he was invited to oxford to give a talk about the distinction between prose & poetry. probably needs some uk/irish vocab to get the immediate impact plus the degree to which he stood on local sensibilities :-)

i like him, he has no respect but gives lots of dignity

wow, i really do degenerate the tone of any conversation i take part in...

died so young. if i go another few months i will have outlived him.

i have not made similar contributions to literature.

6:41 PM

had a few facetime eyerolls from my daughter

my poor son wonders what he did to deserve such a useless lump of dad

i sent her a picture from my uncle's suitcase (he was a parish priest in livingston outside edinburgh): Medical Fact: Flies spread DISEASE. Button yours.

no wonder the victorian prudes at mse were emailing me

about underage maths.

corrupting the youth.

(not to in any way lighten the awful things i am implicitly referring to.)

somewhere i have a bunch of manuals my grandfather was given when he landed in north africa in WW2. they are chock full of offensive stereotypes but also contain useful information about social disease.

it is so amazing how backward we are as a people

to be fair, i have heard such generalisations about people from across the river growing up.

not to mention by name, but the people from cobh are...

irredeemable.

6:46 PM

last landfall of the titanic (queenstown then)

technically not a landfall in cobh, but gimme a little leeway so to speak

when words appear in brain all mathematical ability, rare as it is, disappears

as an engineer, i was always a bit appalled by some of the modeling performed. odes are reasonable, but pdes involve all sorts of infinities and a depth of knowledge that any practitioner (as in solving, finite elements, etc) cannot possible know.

i need some exercise :-)

not sure where the above came from.

i do like the context of someone left to flip a switch between f and g and let's see where everything ends up.

i would like to wager on the outcome.

presumably you could reach the convex hull of the separate solutions. i love convex stuff.

been a while, but there used to be control systems that were loosely based on averaging of some sort. More so you could replace a linear actuator (which cost $$$) by a simple on off switch.

there were dithering systems as well.

ha, dithering. i love that word. it sounds like boring conversation.

i am an expert

7:04 PM

Guys, if you have random variables $X,Y$ such that $\mathbb P(X<Y|X>1)\neq\mathbb P(X<Y)$, does it follow then that $X$ And $Y$ are independent?

completely off the cuff, my guess is no. i do not have examples.

My guess is no too, but a quarter of my students claimed this in a homework assignment, and I don't want to embarrass myself by deducting points if this is true after all x') Also, a fellow TA of mine (who did a master in stochastics) gave full points for this argument...

oh shit.

I might ask him then

oops. my oops.

7:08 PM

my wife has a masters degree in statistics but she mainly knows, in considerable depth. the specific models that she uses and not what people would regard as general theory. probability theory in general is just a nightmare world.

i would be interested in the answer to that question.

I'll let you know!

I vote no, @Sha.

ted, we need to stop agreeing.

Lol

I shall try not to agree.

7:10 PM

Well, even if it turned out to be true, they are losing points, since it's not clear why it's true

so I'm in the clear

The object of a proof is to convince your audience. If a student's argument fails to convince you, it's not a crime to say "this is not clear to me", and deduct points. If you wanted to be open to an explanation, you can also offer "if you have an elementary argument for this, drop by my office hours and show me".

that's a good point. being right is never enough.

It seems ludicrous.

I am still going to ask that TA why he gave the students full points tho...

how about $Y=X+1$ and the domain $[0,1]$ (for $X$) with uniform prob.

7:11 PM

that's something that seems missing, any idea of how Y would relate to 1.

when i taught analysis i would give people a fairly generous set of axioms. i said, if you can't prove it in terms of these axioms, with each line citing something on the formula sheet, i'm certainly willing to listen, but don't be surprised if i take points off. it's not enough to be right. spell it out for me. pretend i'm dumb. (i'm dumb!)

@copper.hat But $\mathbb P(X>1)=0$

but maybe something along those lines, let me think

oh wait, that still proves the point

hmm

Let's say that $Y=0$ whenever $X>1$ and $Y=X$ whenever $X\le 1$. These are not independent. If $X$ is uniformly distributed on $[0,2]$, then $P(X<Y) = 1/2$, but $P(X<Y|X>1) = 0$. Now how do we decide independence?

@ShaVuklia i always look for extremes :-)

@TedShifrin Nice one, thanks. Maybbeeee I should have put a bit more thought into this myself, oops

you just need one counterexample (or an elaboration of context).

7:16 PM

i guess it's ok if ted and i agree on objectively true things. we cannot agree on anything subjective.

i suppose you both agree on that?

Hell no.

there we go.

Oh wait.

:-). i'm outta here. i found a cycling buddy

7:17 PM

Bye, copper!

enjoy your ride and subsequent physical fitness. i am eating leftover french fries.

Leslie is high on life.

fried potatoes can get you most of the way there.

I like Lebanese style potatoes.

we used to do a dish like that. my wife learned she had a potato sensitivity, so now it's just my daughter and i who eat that dish. we use a recipe from one of my high school classmates who immigrated from lebanon.

7:21 PM

Oh, ooopss....

what I meant was: if they are not equal, can we say then that they are dependent

anyways

stop saying oops! i feel like i'm falling down a well.

Oh x'D okay xD

as an attorney, i say things like: on further review of relevant information, we have deduced an alternative outcome.

there's a landscape of possibilities. we are wandering around the landscape.

7:38 PM

does an integral transform of f(x) give one a new way to view f(x)

sure? it depends on the thing inside the integral. 'kernel' is what they sometimes call it although it might not be that.

lots of interesting results on the integral of K(x,y) f(x) d something d something.

some of them can even be analyzed via the withcraft i know as functional analysis.

if the integral only converges for a very small number of functions does this mean it's a "bad" transform?

there are no bad transforms. only good questions.

@leslietownes does your wife have a sensitivity to nightshades? Or just potatoes? And any potato, or just non-yam ones?

you don't see people publishing papers on something where the kernel is ghastly and only makes sense for some apparently randomly selected subset of the function space.

it seems to be nightshades in general although she can be OK with them when limited in quantity. it's not an on-or-off thing as much as not eating too much of any one thing in a week.

i have a similar thing with cheese. i can eat about half of a pizza's worth of cheese, in about a week's time, but if i eat more than that i break out in hives. so i have to monitor the situation.

7:46 PM

@leslietownes let's say you take the Mellin Transform (kernel is $x^{s-1}$) of $f(x).$ What do you do with the Mellin Transform of $f(x)?$ Or does it really depend on context

why not just use $f(x)$ to begin with

i only studied the mellin transform in connection with louis de branges's largely failed attempts to prove the riemann hypothesis. he proved some deep results and also glossed over really superficial details.

please use f(x). if you can avoid an integral transform, do so.

okay

i'm an idiot. i need to bring that up. integral transforms do reveal unperceived truths.

but they are not a panacea.

@epsilon-emperor I've answered it, too. (Some epsilons in there for you, too)

my daughter is doing a very entertaining performance of the alphabet song.

7:50 PM

can you give me a simple example of an f(x) that when transformed gives a better function to work with

my daughter is saying: "i want a bunny to watch me." i have no idea what she's talking about.

living with a little psycho is what you agree to when you have a child.

@copper.hat lots of things seem to make you uncomfortable and blush. Unfortunately, there is no Hairy Bra Theorem.

lil psycho

that will be her rap name.

@leslietownes Her spirit guide is a bunny

Psycho Bunny, qu'est-ce que c'est

7:53 PM

Psycho Bunny

that's it

she's a nutcase. her hair is redder than mine ever was. she never had a chance.

@leslietownes A Deeper Shade of Irish

the demon rabbit will helpfully advise her, and give us advice on mathematical questions.

like is there any useful use for integral transforms

convolutions are good examples of integral transforms.

7:55 PM

Fibonacci the demon rabbit

facts

bad bunny

when they test orchestral halls for acoustic stuff, they do so by firing a starting pistol or its equivalent and recording the response. the response to the impulse tells you how the room is shaped. it gives you the kernel.

convolution of f with itself is a special case of general convolution

in general a vector fields integral curves in (0,1)^3 will not project to integral curves on (0,1)^2

in general a vector fields integral curves in (0,1)^3 will not project simultaneously to the faces of [0,1]^3 yielding integral curves

I wonder can it happen for only constant integral curves

is everybody taking a nap?

8:16 PM

@geocalc33 . o O ( zzz... )

@geocalc33 why don't they project? Is there a non-vanishing derivative requirement?

@robjohn somebody said that I don't know

huh, I'll have to look into that.

@robjohn they project yes

but whether they project onto integral curves I guess is what they are saying

I would think so, but it's been a while since I've worked with them.

Looking at the definitions, I don't see why they wouldn't. I'd like to understand why, if they don't.

@robjohn yeah I was questioning this claim as well, but the person is not answering anymore so

8:38 PM

Each point is hit by uncountably many points when you project. You just get mud.

@TedShifrin Oh, I see. The field does not project, the same point is the image of many points whose vectors point in many different directions.

I was thinking of integral curves on a surface projecting onto a plane. Not very general

Vector Mud®

8:54 PM

@robjohn i do like procol harum :-)

@copper.hat Good ol music

in reality few things make me blush, i have zero modesty. changing on irish beaches after a freezing swim rendered such things irrelevant :-)

making a mathematical mistake (which is frequent) might...

Usually, if I make a mathematical mistake, it's because I was careless. It's just a wake up call. I treat is as such.

Be more thoughtful

then i would say nothing...

Is it our charge now to make copper blush so we can see it?

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