12:24 AM
I had a teacher who used $e_n$ for the function $x\mapsto e^{2\pi inx}$

i think i prefer the zeta.

12:37 AM
How many subsets of $\{1\mathbin{..}17\,000\}$ have a sum that is divisible by $17$?
The answer is $\frac1{17}(2^{17\,000}+16\cdot2^{1\,000})$. Why?
(and why does this generalize to all odd primes?)

One thing we can do is say, there's no reason why we consider the set $\{1\mathbin{..}17\,000\}$ rather than the multiset $\{1\mathbin{..}17\}\cdot1\,000$
eg we can think of it as just a thousand copies of each residue class mod 17
Not sure where to go from there

find some kinda group action and get it out of group theory? i dunno.
Z_p always wants to act on stuff.
and yes i write that because i don't believe in p-adic numbers

If we wanted "divisible by 2" (instead of 17) we could just add in or take out the element $1$ and get that it's $\frac122^{17\,000}$
Maybe something about doubling each element?
Where I'm thinking of 1000 copies of each residue class... that permutes the residue classes
I know the multiplicative group of nonzero numbers mod 17 is cyclic (this is true for all primes, though finding a generator is not easy in general)
Not sure how to use that tbh

12:59 AM
Is there even a nice way to find the number of subsets that add up to a given sum?

if i had more of an attention span i would code this up and look for patterns in small values

1:12 AM
Munchkin can deal with your gnathood.

@AkivaWeinberger I usually see $\omega$ or less often $\omega_n$ for a primitive $n^\text{th}$ root of unity.
@AkivaWeinberger integers, or positive integers?
what is the context? does order matter?

the context is akiva fulfilling his established role as the resident quizmaster and collector of esoterica.

well, I'm fulfilling my role as occasional guest. Leaving to get food. BBL

mm, food. i wonder how many saltines my daughter will want this evening.
i've established a precedent of 3.

1:33 AM
The SCOTUS has utterly destroyed the notion of precedent.

i'll tell her that previous judicial decisions are not binding on this Court.
i can't find any mention of any number of saltines in the constitution.

Most definitely.

i'll shut up, i don't want my own blood to boil.

Saltines postdate the Constitution. They should be banned.

submitting a draft act to criminalize possession of saltines to the local state legislature.

1:37 AM
Not to mention nutella. Yuck. Way to ruin two of my favorites.

my daughter's favorite saltine pairing is hummus.

Wheat crackers would be better.

i had some pita crackers a while ago that went over pretty well.

As well they should.
And leeks agridolce.

my cat has spent the last 3 hours nesting in a pile of clean laundry i was planning on filing away this evening. i think my house has developed something of a culture of entitlement.

1:41 AM
Encouraged by whom, I wonder …

i did fold it and make it look kind of like a cushion, so maybe it's my fault.

Warm napping place. Duh.

at least they're white shirts and she's a black cat.

wrt to math.stackexchange.com/questions/4452993/…, I realize it's going to be closed soon; but can I answer it here to get some feed back on whether I got it right?

1:58 AM
This is a homework question, either from Apostol or from Spivak.

I assumed that much.

You can post any proof you want.

@TedShifrin Thanks.
$(f(x))^2 = \int_0^x{f(t)\frac{sin(t)}{3 + cos(t)}$, using FTC, right hand side = $f(x)\frac{sin(x)}{3 + cos(x)} - f(0)\frac{sin(0)}{3 + cos(0)} = f(x)\frac{sin(x)}{3 + cos(x)}$
oh.. formatting doesn't work here..

@ewong: differentiate both sides of the equation in the posting you are interested. Form there a simple equation for $f'$ follows.

You have to argue $f$ is differentiable.

2:09 AM
@OliverDíaz right. so I find out that f(x) = -1/2 ln|3 + cos(x)| am I right?

@robjohn Thanks. Yes, it makes sense. My question: was bringing $\log (\epsilon)^\frac 1{a+1}$ in $\lambda=1+\log (\epsilon)^\frac 1{a+1}$ obvious?
I understand that choosing $\lambda$ this way works but was there any way of choosing it (the term $\log \epsilon^{\frac 1{a+1}})$?
@OliverDíaz: Thanks for the answer and especially for the comment at the bottom of the post. I also tried with Bernoulli's inequality yesterday but everything for the lower bound seemed to converge to $0$ or to something negative.
I think I got how that term was selected: $\lambda$ should be $\sim \epsilon^{\frac 1{a+1}}$ (I have no problem with this). In the limiting situation $\lim_{a\to \infty}, \epsilon^{\frac 1{a+1}}$ becomes $1$. $\epsilon^{\frac 1{a+1}}=e^{\frac {\log \epsilon}{1+a}}\ge 1+\frac {\log \epsilon}{1+a}$
so RHS here was taken as $\lambda$.

2:38 AM
@Koro My first attempt was to use Bernoulli's inequality within the interval $[0,1/a]$. Passing to the limit gave $1/2$. I did not tried the interval $[0,1/\log a]$ *which would include a negative part. That may also do the trick.

@TedShifrin Nutella between cream crackers is heaven.

@D.C.theIII unrelated to all this, masala milk with dry fruits embedded in it is also very yummy.

I'll have to try this one. I'm not sure they would sell it at any Indian restaurants though would they?
There is a very large Indian community here in Toronto so it just might be available

@Koro: actually, Bernoulli inequality seems to be hopeless to get the right limit.

@D.C.theIII i hate nutella. Appallingly sweet. But hazelnuts are my favorite.

2:46 AM
@D.C.theIII I doubt if it’ll be available there. A new shop opened nearby where I live and I also tried it the first time very recently.

I could see why you would say it is too sweet.... That's where the cream cracker sandwich comes into play. To mute the excessive sweetness
I don't think I've ever tried hazelnuts on their own outsize of nutella

@OliverDíaz :’(

One of my favorite desserts to make is a frangelico chocolate hazelnut torte.
Oh, toasted hazelnuts are to die for!

I know this is a cake because of "torte"

Amazing.

2:52 AM
Lol....I could see the sarcasm oozing from your words.......

Don’t say “tort” to @leslie. He’ll go crazy.

i'll get tortious.

@Koro that is the idea.

nutella is like 5% hazelnuts, 95% coconut oil and sugar.
i like it though, in the right context.

@robjohn thanks a lot, I understand now. :-).

3:04 AM
for some reason, I thought I saw the inequality in the other direction, but your statement is correct

@leslietownes I went to a coffee shop and a staff there asked me if I would like that to be topped up with hazelnut something. I asked if he knew how hazelnut looked like, he didn’t know.
Similarly with vanilla.

So they'd never seen a hazelnut or a vanilla bean?

Neither have I. I have tasted these only in ice-creams etc.
😬

3:26 AM
@leslietownes Some % chocolate
Hazelnuts are worth making the acquaintance of.

vanilla beans, by contrast, are disappointing when you meet them. they are best in tiny shavings. or in extract form.
the bean should be abstracted away so only the vanilla is left.

@robjohn screen might have rotated.
:P

@robjohn Context was trying to figure out why the number of subsets of $\{1\mathbin{..}17\,000\}$ whose sum is divisible by $17$ is $\frac1{17}(2^{17\,000}+16\cdot2^{1\,000})$
(see the earlier messages)

4:17 AM
akiva: is that number an integer?
i'm looking for as many simpler ways to understand it. before counting.
and any hint as to where that formula originated would be welcome
i'm getting weird stuff with alternating sums of binomial coefficients and not huge powers of 2.

@leslietownes $2^8\equiv1\pmod{17}$, so yes.

well, i mean something bijective. modular arithmetic doesn't tell me anything that it counts. anything simple that it counts would be great, even if it's not the main thing.

It tells you that it's an integer, which is what I was answering.

yes. my question was imprecise. it was meant as a lead to: is there something simpler than the task before us, that the sum can be said to represent. where like, yes it's an integer because it's a number of ways of doing something, anything combinatorial, not necessarily the thing we want.
it's just a weird kind of sum.
like i can write any integer as a sum of binomial coefficients in all sorts of ways due to the strong law of small numbers, but what is that formula trying to tell us in a way we can grasp.
combinatorics is all smoke and mirrors to me. witchcraft more than math.

i got a spam call today and i still do not know whether it was a human or a bot.

4:30 AM
turing, we have a winner!

i think human, but a tough call.

skynet and the singularity coming soon

it was some republican fund raiser.
a bit scary that imitating a human is so easy

well, imitating a republican fundraiser? i think i could do that with a chatterbot.
low taxes, limited government, give give give
plus or minus MAGA and some racist dog whistles

:-). unfortunately what used to be republican is no longer

4:32 AM
i sometimes get calls because i donated to the campaign of a republican i personally know. i tell them, "i only donate to candidates i personally know."

i only know the pelosis
but i'm not giving up my hard lost cash

and i suspect the pelosis can afford not to have your contribution.

no ship
any of my friends can afford to not have my contribution

one time in 2004 i was stopped on the street to donate to john kerry. i said, what kind of donations are you talking about and the guy opened with something like $300. i said, i think his wife has that kind of money but i don't. look at my shoes. i might have had$5 for john kerry.

a gazillion years ago when folks thought i might amount to something the irish consular mission used to include me in their circle. one of the things that kept coming up was some fancy dinner at $500 a head. as a student i had to decline 4:43 AM and that's how you ended up here, among the roustabouts upon a little internet searching i see just how naive i was. the chap who used to get me invited was Ireland's First Honorary Consul in Los Angeles. i had absolutely no idea. i am certain in my wine induced merriment that i have communicated many irreverent perspectives with the poor gentleman presumably not all upon the subject of convexity i did get slapped by the head consular officer in sf a long time ago I don't know how to expand pentagonal number theorem summination formula.$n=-\infty$doesn't makes sense to me. presumably$|x|<1$4:52 AM I don't know if i should write it as$\lim_{k\to\infty}k=-\infty$one Or may be I split sum in two... 0 to infty and -infty to 0 Is this da way? If that so then n/2(3n+1) should be used for the power of x for n=0,1,2,..... and n/2(3n-1) for n=0 -1,... may be not I am completely brainwashed by pentagonal number Are you asking how to get the formal expansion? Yes I don't know how to expand those infinity. @copper.hat Thanks for help. 5:29 AM Ah so I was right at first few statements and wrong at last few 2 hours later… 7:38 AM Dumb question: When I type in an equation on my calculator, and it returns the answer “All”… what does that mean? What can I do with my answer? 3 hours later… 10:56 AM A friend from undergrad asked me a basic question about conditional probability and I had to google each and every basic thing. I don't think I've ever so clearly seen how much I've forgotten Probably an excuse to go through my old notes. Anyone got tips for remembering stuff from a past life? 12:00 PM Don't play too much video game. JK I usually use one note to remember Hard stuff. Others are trivial to me. Can anyone help me why S(x)=sin(x)-sin(2x)/2+sin(3x)/3-... is not continuous I didn't wrote note about this :( Just an intutive understanding not rigorous analysis one OkI understand it now 12:44 PM I always forget the conditions under which the limit of sequence of differentiable functions is differentiable. :( @robjohn It's not ick! I agree it would be bad if Numpy claimed that it's mathematically valid to add matrices of different sizes, but it's not doing that. Numpy is a Python library for operating on arrays, and many of those operations run much faster than plain Python code. Yes, it supports various matrix & tensor operations, but it's not like you're manipulating matrix or tensor objects, you're just manipulating arrays. The thing that micsthepick alluded to is called broadcasting. It makes array code easier to write & read, and more efficient in its use of RAM. 1:03 PM @Koro don’t know a N&S condition but uniform convergence of the derivative should do (C^1 is Banach) hi. is it true no matter what set of things (maybe simple set of numbers, maybe some exotic computer science data structures) I have and a binary operation on those things and it follows (a) closure, (b) associativity, (c) commutativity, (d) distributive, (e) it has some element such that the operation doesn't change the thing and (f) every element has it's pair which when binary operated with it's pair, it becomes the previous element. Then I can do all the normal algebra on it. @CalvinKhor I've not yet studied Banach's. I was referring to a theorem on this in Rudin's PMA. I think that you have stated a strong condition. Convergence of the sequence of functions at only one point+ convergence (uniform) of sequence of derivatives implies: 1) The sequence of the function converges uniformly to say f. 2) The sequence of the derivatives converges uniformly to$f'$(derivative of f in 1)). 1:37 PM @Koro yes, i would think that the condition i stated is not very good. But it is easy to remember 1 hour later… 2:50 PM @Koro in case you do not know, I thought I would share: it is possible for a sequence of differentiable functions to have a pointwise limit which is differentiable, but the sequence of derivatives does not converge to the derivative of the limit. E.g. sin(nx)/n yes :). @Koro did you come across distributions (generalised functions) in your engineering study? 3:23 PM @CalvinKhor I came across Dirac delta. in a course called Random Vibrations. we followed this text for the course: D. Newland's an introduction to random vibrations, spectral and wavelet analysis :) cool. I wanted to say, the sequence of derivatives of sin(nx)/n does converge to zero, but only in a suitably weak sense (like the sense of distributions) Is there a difference between symmetric transitive closure and transitive symmetric closure? I searched about it but couldn't find a verified source on it. 3:41 PM @CalvinKhor I don’t think that the sequence of derivatives converges if x is not an integral multiple of$2 \pi$. For, the derivative is cos nx. Or, am I missing something? @Koro uniform convergence is stronger than pointwise convergence, right? There's something weaker than pointwise convergence. And it does converge in the weaker sense 4:06 PM @CalvinKhor I see. I'm not yet familiar with convergence that is even weaker than pt.wise. 4:22 PM Hey guys, im wondering does a non pure atomicand a non difusive measure exist? 4:57 PM @bumblebee I don’t see where they explain what the display answer for “ALL” is… 5:12 PM @AlekMurt what's a diffusive measure?$A_{\mu}=\{x \in X, \mu(\{x\})>0\}$a diffusive measure is the one such that$A_{\mu} = \emptyset$OK, so you want a measure that is neither atomic nor diffusive? lebesgue + dirac at 0? Ohh ya it works, ty 5:44 PM Np! 6:07 PM @CalvinKhor are you talking about the sequence$\cos(nx)$, or something else? 3 hours later… 8:57 PM trying to find a proof or counterexamplel to the following: Let$f$be a$2\pi$-periodic square-integrable function. If$f(0)=f(\pi)=0$, must$f(x)=-f(-x)=-f(x+\pi)$? The converse is true but the statement itself seems dubious. ($f$is the solution of a second-order ODE with periodic coefficients in the example of interest, which seems ilke it'd be a strong enough condition to make the statement true. but i'm curious if that's actually needed) Beyond dubious. Insane. So you’re saying the Fourier series has only sin terms. But leave the world of continuous functions. it's a lilttle stronger than that, no? should just be$\sin(m x)$with even$m$Take a continuous even function and change its values at$0$and$\pi$. ah, hmm. Why even$m$? 9:11 PM$f(x)=f(x+\pi)$oh$f(\pi)=0$...huh you're right ah, i see what i'm doing wrong. I meant$f(0)=f(\pi/2)=0$Oh. Then you’re right, but my outrage stands. yeah, i can imagine that being periodic and square-integrable is insufficient Periodic and everywhere continuous is still unclear to me. but if you've got a sum of such sines, then isn't that already giving you$f(x)=-f(-x)=f(x+\pi)$? But who says it’s a sum of sines? 9:16 PM hmm. I guess its Fourier series being a sum of such sines isn't quite the same thing What about$\cos(4x)-1$? that'd do it I was thinking random functions that are$0$as required and extend periodically. it seemed unlikely but having a counterexample is reassuring I’ll take a notarized thank you :) 9:21 PM lol the relevant ode is$f''(x)+(a-2q\cos (2x))f(x)=0$with$2\pi\$-periodic boundary conditions. i'm not seeing it asserted but presumably they're assumed to be smooth

2 hours later…
11:38 PM
Can we prove that for a>1 it is a^n->infinity as n->infinity using the binomial theorem the way it follows? "Since a>1, it is a^n=(1+h)^n for some h>0 and so (1+h)^n=bin(n,0)h^n+bin(n,1)h^(n-1)+...+bin(n,n-1)h+bin(n,n)h^0>nh->infinity as n->infinity."

yes. that is a standard way of proving it. you don't even need the full binomial theorem, if you don't have it.
sometimes if a book hasn't proved the binomial theorem, they prove some version of en.wikipedia.org/wiki/Bernoulli%27s_inequality first, e.g. by induction (as is done on that page)