Someone help poor, little old me.

Is there a way of rewriting $(O+U) \setminus (A+U)$.

$O+U \setminus A + U$

@RandomVariable I would tend to agree with Ted, but I will have to think of a counterexample. That will have to wait a while as we are getting food and I am going to walk dogs.

I tried searching this problem in google but nothing came up :(

$\int_0^\infty\frac{a}{(1+ax)^2}\,\mathrm{d}x=1$ so that convergence of the integral is uniform but the limit of the function is $0$ for all $x\gt0$. This works whether $a\to0$ or $a\to\infty$.

is it possible that a good encryption used multiple times like encrypt(encrypt(something important) produces the same output like any of the previous steps? Such that doing it 1 Googol times is the same as doing it x times where x is significantly smaller.

For linear and quadratic hash function that probes, e.g., the former search for bucket to hash value to in linear manner by probing sequentially. The second one does that by squaring probe step from 2, 4, 16, etc. Why both linear and quadratic probe hash functions can not generate more than $N^2$ probing sequences please?

@SAJW. It depends on your encryption

@SAJW. If you have a good encryption algorithm, it should be independent on any pattern found in the text

@robjohn On what interval are you claiming $J(a) = \int_{0}^{\infty} \frac{a}{(1+ax)^{2}} \, \mathrm dx$ converges uniformly?

@RandomVariable. Have you tried taking ratios?

@Avra What kind of ratios?

why you don't substitute $\infty$ with $t$ and take $t$ to infinity?

if you got result $\infty$ then the function diverges

otherwise converges

@RandomVariable. What do you think please?

@Avra I'm not trying to determine if the integral converges or not. I'm trying to determine on what interval the integral converges uniformly, a concept that's not talked about much.

@RandomVariable. Wow! This sounds amazing!

This is area of research in number theory or real analysis I guess?

Define precisely what you mean by that phrase.

@TedShifrin. Me Prof Ted?

No, @RandomVariable

@TedShifrin What phrase?

“The integral converges uniformly on $I$”

Don’t ping me. Just post your question for general attention.

I’m leaving soon to cook dinner.

Way too long for me now.

@BahSoh. It's almost 10 pm here!

We’re all in different time zones.

you silly geese, it's not much after 7.

There are 3 time zones in my house alone.

my zone is 2 hours ahead of my wife's zone which in turn is 2 hours ahead if my son's zone.

@RandomVariable you asked whether if, for certain range of $a$, the convergence was uniform, and for all $a$ the convergence is the same. It didn't seem as if you were asking anything about the function being integrated. However, that function does uniformly vanish as $a\to0$.

my daughter is claiming to have seen a dragon. "in the dark."

@leslietownes I saw many things in the dark when I was young.

one time while very feverish and young i heard moaning voices. it was terrifying.

that would be

because of constant ear infections at about age 2, i have a constant ringing in one ear and a tone that sometimes varies in pitch in the other. when it drops i know i'm about to be sick.

in normal environments background noise overcomes them, but if i put in earplugs it's like my own concert.

@robjohn this is why I asked for a precise definition of the utterance

@TedShifrin as $a\to0$, the function does uniformly go to $0$, but not monotonically. I was also in a rush to leave, but I thought of that function, so I threw it out there.

hello.

hi

I understand, robjohn, but I’m still waiting for a precise definition.

Hi EM4

@robjohn I think we might be talking about different things. I'm referring to this.

how you guys doing?

@RandomVariable Okay, so the improper integrals converge uniformly. There are many meanings for uniform convergence and it was hard to infer which was meant.

So if the sequence of functions converge uniformly and their improper integrals converge uniformly (as given in your citation), then the sequence of the integrals of the functions also converges. Pick an $\epsilon\gt0$. Uniform convergence of $f_n$ allows us to pick an $n$ so that $|f_n(x)-f(x)|\le\frac{\epsilon}{3K(\epsilon/3)}$. Then

@robjohn I got used to saying "the uniform convergence of an improper integral" because that's how some textbooks refer to it, but we're not really dealing with a single integral.

@RandomVariable "uniform" usually does indicate a family of functions or sequences, so that is the proper term here. However, there are several types of uniform convergence.

I don't think one would talk about a single integral converging uniformly.

@robjohn Agreed 1000%.

in other words the term uniform is not used uniformly.

@robjohn I find it strange that most analysis textbooks don't even bother to cover the topic anymore. It sometimes comes in handy when the dominated convergence theorem is not applicable.

the coverage of uniformity is not uniform either.

the whole system is not uniform. attica, attica.

rant: why are programmers so imprecise with their definitions :/

because they are programmers not mathematicians

If the intervals [s,s+Dt] and [t,t+Dt] are nonoverlapping, s+Dt<=t. Please, can someone explain to me what it means interval is nonoverlapping?

I believe it to be false, I can always takek 2 abelian group and their multipication won't be the same

does anyone else have a proof why group of order 1225 must be abelian?

@Mohcine it means their intersection is empty

if $A$ and $B$ are nonoverlapping intervals, then $x \in A$ if and only if $x \notin B$ and vice versa

@shintuku Thank you so much

it just like when we say A and B are two disjoint sets

yep

@SAJW Of course. The output of the encryption function has a fixed size, so by the pigeon-hole principle it must eventually repeat. Eg, if we iteratively encrypt a block of 128 bits, there are only $2^{128}$ different blocks, so we must get at least 1 repeated block after $2^{128}$ iterations. Note that 1 googol = $10^{100}\approx 2^{333}$

i have a matrix $M: S \to V$ and i don't want to have problems with figuring out from what bases it takes inputs and to which bases it outputs. what do i need to suppose? is there something like a standard matrix?

hm, an abstract vector space doesn't necessarily have a standard basis no?

that question does not make sense to me

hm, thanks i'll need to review stuff

@thecorrectanswer $1225=5^2\cdot7^2$ is what they are trying

but then they need to show something about products of primes

this should be useful in finishing the proof

oh, in order to properly defined a matrix between two abstract spaces I need to designate its corresponding linear map and specify it is a matrix with respect to which bases

if both $i, j$ are integers such that $i<j \le \frac{M}{2}$, where $M$. Then if $(j-i)$ is divisible by $M$, why $(j-i)$ please being less than $M$ is a contradiction?

Is it because of the fact that if $(j-i)$ is divisible by $M$, then there is integer $k$ such that $kM= (j-i)$ please?

@Avra because if $M$ is greater than $j-i$ the division does not result in an integer

if $j-i = 1$ and $M = 2$, we have that $1$ is divisible by $2$, which is false. $1/2$ is not an integer

@shintuku. :(

Sorry I did not notice that

Last question please. If $M$ is a prime, then it must be the case that either $(j-i)$ or $(j+i)$ equal $M$. I see that here we can not have both $(j-i) = M ~and ~(j+i)=M$ because if $i<j$, then $j-i$ and $j+i$ would definetly results in two different primes and thus at most one is equal to $M$ .

Is this correct conclusion please?

if $j$ is prime and $i= 0$, then both $j-i, j+i$ are prime

edited: whoops, can't have negative primes, but yeah, if $j$ is prime and $i=0$, this contradicts your statement

But we have in the second question $i>0$!

So, based on this added info (sorry to forgot to add), my cocnlusion is correct please?

yes, but note that this isn't a property of prime numbers but of all numbers (if you don't allow 0 for i)

@shintuku. Thanks apprecaited!

it is also faster to say: $j-i \neq j+i$, the condition $i<j$ is not necessary. but if you're talking about primes, you also have to add that $j+i, j-i$ are positive

i guess this is covered if you say $i>0$

@shintuku. Indeed! Thanks

shortest path j−i≠j+i,

@TedShifrin , $u$ is the integration variable and it is unbounded. Why doesn't it?

could somebody please answer this question?: math.stackexchange.com/questions/4116801/…

I have no familiarity with this text, but just from skimming it, it seems as if you would at the very least also need a broad background in homotopy theory and K-theory

@schn You just answered your own question.

my daughter spent the morning screaming. we're thinking of renaming her Screech.

How original of you!

@Oxide I would expect you need to be familiar with vector bundles and the basics of handlebodies and surgery theory.

do they do this for a specific reason or are they totally unpredictable

do they have a system

if you're asking about my daughter, it's usually that there's something she can't do immediately. if she gets a new puzzle or is asked to perform a task, she can usually do it instantly, and when she can't, complete meltdown.

adjusting to this will be her introduction to life.

i can't count the number of puzzles we've been given as gifts where she takes the pieces out of the box and immediately solves it, and it's just like, OK, i guess we're done with that. she is keeping the puzzle industry in business.

smart one!

she's terrifying.

she isn't even 3 and can hold fairly coherent conversations with my friends over zoom.

clearly to be technologically fluent at 3

i complimented her on her shirt, which has an egyptian-styled falcon on it, this morning. she said "thanks. it cost a lot."

hahahahhaha

she was riffing on something that happened earlier. she had been destroying bandages for a scratch on her knee, and my wife informed her of the cost.

people are sometimes hesitant to give their child to day care for most of most days because of attachment. or they cannot afford to do so. but i think it has been very good for her to be around a ton of talking people and not just mr. and mrs. townes.

@robjohn can you please show me? this isn't clear to me

@Thorgott @TedShifrin thanks.

@leslietownes You will owe me weekly royalties.

too pricey. we'll stick with her great-great-grandmothers name. it's already in all of the paperwork.

I figured you would chicken out.

it would be hilarious to have people refer to my daughter as Screech, but would require effort and payment of royalties to do so.

A small price to pay for all the notoriety.

there's also that dumb character from a 90s sitcom, i am only dimly aware of it but it's not the best association.

Yes, that's where everyone assumes I stole the name. Saved by the Bell. I thought of that only after the name was picked.

it's a pretty narrow demographic who are familiar with that. people my age plus or minus a few years.

and i missed it entirely because i was a nerd who didn't watch TV.

I think I may have seen a few in reruns years later.

the actor had a very troubled post-show history, perhaps because of how he was such a punchline on the show, and maybe also because he was not that good of a person. he died of lung cancer.

happy saturday, everybody.

This Screech has taken to climbing in to the (closed) upper drawers of my desk (from the little opening on the inner sides). Oy.

olivia does that in my wife's dresser. you thought you had clean clothes. oops, they're all covered in black cat hair.

LOL, it's like the two of them went to sneaky cat school

humanity's companionship with cats is difficult to parse. if i had a store of grain in my yard that i wanted to keep free of rodents, i could understand it. there's a lot of olivia that i don't understand.

probably nobility house pets

olivia is asleep in a pool of sunlight right now, completely stretched out. she's the only nobility in this house.

my cat can fit in any place

She is a fluid

my cat does the thing where she seeks out any box or otherwise cat-sized delimited location and lies in it. shoebox, blanket on the floor, anything.

Yup, Screech has various boxes, but particularly loves the smallest. I can just barely close him up in it, although usually I get scratched deeply if I try.

@Oxide Homotopy theory, mainly. Hatcher Ch 4 would be a good place to start.

When I started reading the first few chapters off it I had a decent idea why one would try to localize or complete spaces.

the really bad rip in my leg from olivia's latest massive attack has begun to heal. no more bruising, no infection.

That's good. I always do thorough hydrogen peroxide.

my not-even-three-year-old daughter can identify all of the ways a cat can harm a human. "that's a livvy scratch." "that's a livvy bite." she could be a forensic examiner.

my wife's cousin is considering getting a cat. she keeps asking us questions where the answers are mixed. "well, it might attack you. you're in its territory."

we still recommend cat ownership. my wife's cousin has had an unpleasant post-covid professional life and needs companionship. we adopted our cat for similar reasons.

hola

@leslie Maybe your wife's cousin would like to take over Screech :P

No idea, @BahSoh. What is it?

@TedShifrin she maybe could. she's searching the LA and OC based shelters for something that isn't a kitten.

diagrams like that are often automatically generated. the placement of vertices is not that material to the data.

i made considerable use of graphviz when i was a mathematician. cool product.

if you feed it the adjacency data for the united states, you get a map that actually resembles the united states.

Yeah, an older cat would be wiser, @leslie.

It's not geometry, @BahSoh. It's a graph.

it's tough to give advice on pet ownership to someone who has no experience, at least with cats. "there's going to be this thing in your life that you'll have to care for and a lot of the time it's going to hate you."

get a dog if you want positive feedback.

yeah, cats play mind tricks.

My cats in GA were mostly dog-like. I'll have to see what happens when Screech works his way into adulthood. He certainly hangs around me and follows me around all the time.

livvy does that. the minute a center of attention materializes, she appears. she also greets any member of the family with rubs and meows upon arrival home.

she didn't get the memo that cats don't do that.

I am wondering if someone could help with this question: math.stackexchange.com/questions/4235224/…

Huh?

What does "For" mean? Is $h_1$ relatively prime to $N$, for starters?

One second Professor please. Sorry I will add to that

We know that if $(h_{1}(k_{1}) \equiv h_{1}(k_2) \mod N)$ and $(h_{2}(k_{1}) \equiv h_{2}(k_2) \mod N)$, then we cab sum both as $h_{1}(k_{1}) + h_{2}(k_{1}) \equiv (h_{1}(k_2) + h_{2}(k_2)) \bmod{N}$

OK, not sure why you're writing such complicated expressions.

Let's just make this simple. If $a=b$ and $c\ne d$, can we conclude that $a+c\ne b+d$?

When you have so many letters, you just get lost.

Thanks Prof.

Yep same logic. So does this mean that h1(k1)+h2(k1)≢(h1(k2)+h2(k2))mod N?

Given that if $a \equiv b \bmod N$ and $c \not \equiv d \bmod N$

So can we conclude that $a+c \not\equiv (b+d) \bmod{N}$

Working with equivalence mod $N$ is like doing any other algebra with equalities.

Just think about the contrapositive if you want a proof.

Thanks Prof. I know that if a≡bmodN and c≡dmodN, then a+c≡(b+d)modN

But I have never saw the one above

Write the contrapositive of your statement.

Assume $a=b$ and $a+c=b+d$. What do you conclude?

Take $a=b$ as an assumption. Then the statement is $c\ne d\implies a+c\ne b+d$. Take the contrapositive of that.

if a+c = b+d, then c=d

There you go.

is true

original is true then

since contrapositive proof shows the original is true

Thanks Prof.

@leslie You might want to weigh in on this one.

i'll think about it. i love the signal boost to axler. he is sometimes on math.SE.

Well, since I am not fond of his approach, it's only fair to plug the book in a case like this.

a lot of my work was in the axler vein but additionally failed to associate polynomials or indeed analytic functions with operators entirely.

I don't know how to get around algebraic closure, though, because he's going to base it all on eigenvalues and multiplicities.

yeah, i have the same uncertainty.

Polar decomposition?

Could someone give me a (non-trivial) example of a Stein space as an exercise, so that I can prove it is a Stein space?

:-)

Are you talking about complex manifolds or something else?

Yes.

These exercises are mostly deep theorems. Even showing a non-compact Riemann surface is Stein takes work.

That's unfortunate.