12:46 AM
Is there a better way to compute $\varphi$ (Euler function) than brute-force searching for the number's prime decomposition and then using the formula which comes from that:
This is like a "just-curious" question not a "I need help"
including an approach via seiving
Cool, thanks for the link Semiclassical!
So basically my question is equivalent to RH ;)
@EE18 searching for prime decompositions not using brute force?
@EE18 that's about the limit of what i understand too
What do you mean Jakobian?
12:56 AM
Like primes less than $\sqrt{n}$
There is also some other ways to slightly improve this
yeah, i'm not sure what the best known algorithm for Euler phi is
best i'm seeing via google is using seiving to get O(n ln ln n) time complexity
Sorry the improvements I've been thinking of are for checking if a number is prime. My bad
(all primes larger than 3 are of the form 6k+1 or 6k-1)
Though I suppose this might be useful if n is really large...?
1:12 AM
well, thats seiving with 2 and 3. so not surprising that its better to seive more
beware: 3n+1 approaching
1:54 AM
<joke>
2:13 AM
<_<
>_>

2 hours later…
4:41 AM
the room has lost its edge
5:13 AM
Yup, I think Ted was part of that edge.
🔪
⚔️
5:44 AM
🥱
😴

1 hour later…
6:50 AM
Let $f:\mathbb R^n\times\mathbb R^m\to\mathbb R^m$ be $C^1$ in an open set containing $(a,b)$ and $f(a,b)=0$. Under some additional conditions, the implicit function theorem gurantees there is a differentiable function $g$ such that $f(x,g(x))=0$ for $x$ in an open set containing $a$.
In Spivak's book, he then differentiates a component of $f$ in the equation $f^i(x,g(x))=0$, that is, he takes $D_j$ on both sides and gets $$0=D_jf^i\left(x,g\left(x\right)\right)+\sum _{\alpha =1}^mD_{n+\alpha }f^i\left(x,g\left(x\right)\right)\cdot D_j g^{\alpha}\left(x\right),\quad i,j=1,\ldots, m.$$
I just don't understand where the first term is coming from on the RHS and why $j$ ranges as it does. Suppose $n=1$ and $m=2$, and let $j=2$. Then $D_2 f^i(x,g^1(x),g^2(x)),$, but does $D_2g^2(x)$ makes sense?

2 hours later…
8:30 AM
i would have to see what he's using that result for to be sure, but absent any surrounding context it would certainly make more sense if i ranged from 1 to m and j ranged from 1 to n. i expect that that is a typo.
in your example g is a function from R^1 to R^2 so you have g(x) = (g^1(x), g^2(x)) where g^1 and g^2 are functions from R^1 to R^1 and f is a function from R^3 to R^2 so it has two scalar valued components of three variables. forgetting spivak for a minute it should make sense for a scalar valued function h of 3 variables that d/dt h(t, g_1(t), g_2(t)) = (D_1 h)(t, g_1(t), g_2(t)) + (D_2 h)(t, g_1(t), g_2(t)) g_1'(t) + (D_3 h)(t, g_1(t), g_2(t)) g_2'(t).
where D_1 h and D_2 h and D_3 h are just the partial derivatives of h with respect to its first, second, and third variable.
ok, I've elaborated a bit here, using other notation, maybe it makes it clearer
so you'd get that for h being either of the components of f. just one variable (for j = 1 only, ssuming it's a typo, there would be no "j = 2" if 1 <= j <= n = 1)
and that rewritten in spivak is what spivak's formula says, so it makes sense to me.
fixing the component i of f, he's looking at the function x mapsto f^i(x, g(x)), which is a function from R^n to R, and evaluating each of its partial derivatives (there are n of them, which is why it would make sense when he considers the evaluating the jth one for j only to range up to n)
ok, so it makes sense to you that it is a typo (or not a typo)? :)
8:51 AM
yes. assuming the j range is a typo, it should make sense where the first term is coming from on the RHS. these f^i's are functions from R^(m+n) to R^1, so formally D_j of x -> f^i (x, g(x)) could at least be expected to be a sum over all m+n partial derivatives of f^i, but from the chain rule, the first n terms work out pretty simply: because of what the partial derivative of x^k is with respect to x_j, only the term involving jth partial derivative of f^i will contribute to the first n terms
and the remaining m terms look the way they do because of the chain rule again.
I feel the same, that it is a typo
i just went looking to see if anyone had compiled a list of errata for 'calculus on manifolds,' since it seems like the type of book where students would bother to find and report errata, and found nothing, but someone on MSE asked your question a few years ago, and reached the same conclusion.
4

Did I find an error on page 42 of my edition of Spivak's Calculus on Manifolds regarding computing the partial derivatives of the implicit function defined by the Implicit Function Theorem? I don't find this error listed in any errata online, but can't make sense of what is written. For convenien...

Thanks a bunch, I was googling like crazy :)

4 hours later…
12:57 PM
I've been studying category theory, and I find it a little odd that while there is so much emphasis on "universal properties", there does not appear to be a formal definition of what a "universal property" is. Is it fair to say though that in most (all?) cases where this phrase is used, it is meant in the following sense: the object $A$ is universal with respect to a property $P$ if, in some appropriate category, $A$ is the initial/final object of that category.
@Joe I understand it like you do
(of course, initial/final are dual definitions so technically one could stick with just initial or just final, but not something that matters)
> Technically, a universal property is defined in terms of categories and functors by means of a universal morphism (see § Formal definition, below). Universal morphisms can also be thought more abstractly as initial or terminal objects of a comma category (see § Connection with comma categories, below).
@Jakobian: So you mean that technically we could always just stick with, say initial objects, and then if something is final in a category, then we could just consider the opposite category?
yeah
I need to learn about comma categories but they look a little ... abstract
they're not that abstract, think of category of functions as objects $f:X\to A$ where $X$ is some fixed set for example
this is what I think about when talking about universal property of tensor products for example
I know that it looks intimidating when said out loud but its not that abstract if you can trust me on this
1:08 PM
@Joe There's a number of equivalent ways of characterizing what a universal property "is". One perspective is indeed that a universal property describes something as an initial resp. terminal object in some associated category.
Another perspective that is internal to our category is that a universal property describes an object as representing a certain (co-)representable functor. This is a very useful perspective with the Yoneda lemma in mind cause it explains in what sense the "universal morphisms" are universal.
1:57 PM
@Leslietownes I've kinda circled back, I no longer wish to transcribe everything into symbols. I think logic is embedded in natural language because it's an inherent quality in the way we think so as long as we can communicate, idc if it's symbols, natural language, braille, gestures etc
computers are just extensions of that quality, we're literally manifesting it irl by encoding logic into silicon chip cpus
@user85795 I'm still watching it, that is a good video though thank you.
I like the quote "mathematics is different for different people"
u can just insert anything for mathematics though I suppose :P
lol
2:04 PM
It's also important not to be overly concerned with deducing true statements all the time. Sometimes positing statements that aren't formal in nature helps to mix up the soup in the mind (need good soup)
I think it belongs in this room also.

1 hour later…
3:23 PM
@Obliv a soup that very little is known about
re: philosophy of the mind
3:47 PM
Man, I answered a question a couple of days ago. It is a question for which I am probably one of maybe a dozen or two people in the world who really can answer the question (like, it's pretty niche---it is about a homogeneity condition which is used by some authors when defining the Assouad dimension). I took time to write a good answer. And silence from the asker. :(
(I am not linking the question, and please don't go looking for it---I don't like pity upvotes or artificial engagement---it is just kind of frustrating to have something right in my wheelhouse pop up, and to get nothing from the question author).
@user85795 Oh, I know. I'm just complaining.
:-)
@Jakobian Hi Jakobian, why is the following true: Assume $M$ is a compact manifold such that $M= U\cup U'$, where $U,U'$ are open sets and $V,V'$ are open in $U$, $U'$. Given homeomorphisms $f:U\times \mathbb{R} \rightarrow N$, where $N$ is another manifold and $f':U\times \mathbb{R}\rightarrow N$, then observe that $dist(f'((U'\cap V')\times 0),N-f(U\times \mathbb{R}))>0$ and so for $t$ small, $f'((U'\cap V')\times (-t,t))\subseteq f(U\times \mathbb{R})$
4:03 PM
@monoidaltransform I can't right now. Walking
no problem whenever you can :)
Hi
I believe the following is true: if $f_n$ is a sequence of continuous functions converging uniformly on $A$, then it converges uniformly on $\text{cl }A$

1 hour later…
5:10 PM
@monoidaltransform why would it be true
even the first inequality doesn't have any reason to be true
5:26 PM
by the way the above remark about sequences of continuous functions follows from $\sup_{x\in A} h(x) = \sup_{x\in \overline{A}} h(x)$ for continuous function $h$
idea is to take $h = f_n-f_m$
@Jakobian What if the first inequality is true, then will it be?
I don't think so
the neighbourhoods $U, V$ and whatever else probably might be chosen so that it all holds, but this doesn't automatically hold, is what I suspect
6:11 PM
I am working on this:
What I have done:
By Theorem 6.5 it is enough to show that $(s_n)$ is Cauchy. We want to figure out how large we need to take $N$ given a fixed $\epsilon > 0$. Let's first consider $n,m$ with $n>m$ WLOG.
$$||s_n - s_m|| = ||\sum_{m+1}^n \frac{x_k}{k^2}|| \leq \sum_{m+1}^n \frac{||x_k||}{k^2} = \sqrt{\alpha^2 + \beta^2}\sum_{m+1}^n \frac{1}{k^2}$$
I don't know where to go from here though. I "know" that the sum I have at the end can be taken less than any $\epsilon$ because the infinite series $\sum k^{-2}$ famously converges (and so is Cauchy)
But I don't know that yet. I am basically not sure how to pick $N$ large enough that this sum is small enough
then what do you know
I think I need to get down into the weeds basically. I need to demonstrate an $N$ such that $\sum_{m+1}^n \frac{1}{k^2} < \epsilon/\sqrt{\alpha^2 + \beta^2}$
For $m,n \geq N$
one way or the other you want to prove $\sum_{n=1}^\infty \frac{1}{k^2}$ converges
so - what do you know?
Well my question is sort of twofold. Yes, proving the Basel problem there is sufficient for me here. But is it necessary? i.e. I imagine there's an easier way to solve my problem here than solving a problem which I think comes a few chapters ahead?
So yes I don't know that the series converges truly. I "know" it but don't know it
6:26 PM
its equivalent to it
isn't it quite obvious
there is no "solving it by different method" because the question is about convergence of $\sum_{k=1}^\infty \frac{1}{k^2}$ either way
Ah OK. I will think on it/refer to proofs of basel problem then. Thanks Jakobian (I wasn't sure if it was indeed equivalent)
its not the Basel problem
EE18: if you expressly evaluate the "telescoping" partial sums of something like 1/k - 1/(k+1) you can see that sum 1/(k^2 + k) converges. if you have/prove the limit comparison test (or toy around with numerical coefficients in a similar example like that and have/prove the comparison test) you get sum 1/k^2 converges
by the way
EE18: mere convergence (not evaluating) can be done in a very low tech way although in a calculus class it is often done via an integral test. search MSE for short proofs
6:30 PM
Basel problem asks for value of $\sum \frac{1}{k^2}$, here you only determine convergence
That's true Jakobian
Appreciate you pointing that out
EE18: note that the limit comparison test and comparison test have very short proofs from the definitions (shorter than e.g. evaluating most 'complicated' definite integrals in a calculus class)
they're even often included in calculus books, for some reason
yeah, sorry for not saying that earlier, I didn't fully remember what Basel problem is until right now
@leslietownes Just thinking on this right now. Not quite sure what you're aiming at but will spend some time thinking before asking again
maybe that reason
EE18: the vibe is "you can prove that something easily comparable to sum 1/k^2 is convergent by expressly evaluating a formula for that something's nth partial sum and taking a very elementary limit"
6:34 PM
anyway, when I say "equivalent to" above I mean equivalent to convergence of $\sum k^{-2}$ and not equivalent to Basel problem
@leslietownes Understood now. Working on that now
How to expand $r^{d-2}/(r^d-m)$ about $r = m^{1/d}$? Obviously it should be a Laurent series, whose coeffs I can in principle find doing the contour integrals as in the "definition" of Laurent series. But sometimes, taking some factor out, one can use standard binomial expansions to write down the Laurent series. I was wondering whether that sort of thing can be done here
@leslietownes You are saying to use that $$\sum \left(\frac{1}{k} - \frac{1}{k+1} \right) = \sum \left(\frac{k+1 - k}{k(k+1)}\right) = \sum \left(\frac{1}{k^2+k}\right).$$ Now this sequence converges because it's equal (given that it telescopes) to $x_k = 1/k - 1/2$ which converges to $1/2$ easily.
@Sanjana this can be done by writing $r^d-m$ as a product $(r-e_1m)...(r-e_dm)$ where $e_1, ..., e_d$ are roots of unity, decomposing the fraction and using geometric series
@EE18 wrong logic
x_k is not constant here
6:42 PM
I need another sequence to bound my $\sum \frac{1}{k^2}$ from above in the comparison test, is that what you mean Soumik?
@Jakobian That's what I would do if I had to find out the residue. I have done integrations like $\int \frac{dx}{1+x^n}$ for some small $n$, but I don't know how to deal with the general $n$ case
That's what I'm thinking on now if so
@Jakobian Namely, how do I decompose the fraction???
Oops what I wrote above converges to 1
My error
@EE18 you mean to bound \sum 1/(k^2+k)?
6:44 PM
Then for the other I guess I consider $\sum_2 \left(\frac{1}{k-1} - \frac{1}{k} \right) = \sum \left(\frac{1}{k^2-k} \right)$ which also converges to 1
@Sanjana for non-zero $m$, $\frac{1}{r^d-m} = \frac{A_1}{r-e_1m}+...+\frac{A_d}{r-e_dm}$ implies $A_k = \frac{1}{m^{d-1} \prod_{i\neq k} (e_k-e_i)}$
Hmm I don't think I'm quite right because of my starting at 2 in that second series. Can't apply comparison test as naively as I thought
@EE18 you can bound $\sum 1/k^2$ by $\sum 1/k(k+1)$, its okay
@Jakobian Umm. Is that obvious? I tried to derive this but couldn't
That gives me a bound from below right Jakobian? What about from above?
6:51 PM
@Sanjana multiply both sides by $r-e_km$ and take $r = e_km$
@EE18 no it gives you both
I just resigned in a drawn position:"(
@Jakobian Where does the $m^{d-1}$ come from?
@Sanjana write it down on paper
Oh got it
@Jakobian But I don't see the geometric series, now
@Sanjana $\frac{1}{r-a} = -\frac{1}{a}\cdot \frac{1}{1-r/a} = -\frac{1}{a}\sum_{k=0}^\infty (r/a)^k$
7:05 PM
@Jakobian What? I have to do this for each term in the series $\sum \frac{A_i}{r-e_i m}$?
@Sanjana what do you mean series
thats just a finite sum
but yes, for each fraction $\frac{1}{r-e_im}$ you do that
and then you sum it up, bring in the terms $r^k$ together
and you obtain the desired series
7:21 PM
hi. erdosproblems.com/297 I have a problem with this... the number of possibly subsets/fraction sums is 2^N isn't it? if so then 2^0.91N-2^0.93N would mean "almost all" such fraction sums equals 1?
@chx That depends on what you mean by "almost all". Generally speaking, "almost all" is used in mathematics when talking about measures---some property holds for "almost all" elements of a set if the measure of the set of things which have that property is equal to the measure of the whole set.
So, for example, almost every number in $[0,1]$ is irrationals, as the measure of the set of irrational numbers in that interval is $1$.
Personally, I would not use "almost all" as you have proposed.
@Jakobian But each coefficient is itself a sum, right. The answer I am expecting is simple. E.g. For $d=3$, the answer is $$\frac{1}{3 \sqrt[3]{m} \left(r-\sqrt[3]{m}\right)}$$
@Sanjana that doesn't sound right
But maybe it is. Not because what I am doing is wrong, rather the case d = 3 probably allows some symmetry which makes certain terms disappear
7:37 PM
@Jakobian What you are doing is right, I think. But similar results hold for all integral $d$
@Sanjana and what is O(...) term?
@Jakobian Its just the terms of the order of $(r-m^{1/3})^1$ and higher
@Sanjana actually I am positive this is incorrect
Ah okay I get it
But as I showed above, Mathematica gave out this result, and additionally integrating this result w.r.t. $r$ I get a log term which matches with the result of a paper
Sorry my bad I know what you're doing and what I am not getting above
7:41 PM
@XanderHenderson ok, yes, fine, but I mean, say there are only 2^0.07N fractions which are not 1 and 2^0.93N -- that's really counterintuitive
Also works for $d=5$
All the terms 1/(r-e_km^(1/d)) for e_k different than 1 will just be standard holomorphic functions (around the point)
So they can be disregarded
Oh sorry I forgot the root
@Jakobian I didn't get you. Why is this true?
59 mins ago, by Jakobian
@Sanjana for non-zero $m$, $\frac{1}{r^d-m} = \frac{A_1}{r-e_1m}+...+\frac{A_d}{r-e_dm}$ implies $A_k = \frac{1}{m^{d-1} \prod_{i\neq k} (e_k-e_i)}$
Here after equality the m should have a dth root
@Sanjana because the only singularity of them is $e_km^{1/d}$
So the relevant part of $\frac{1}{r^d-m}$ is $\frac{A_1}{r-m^{1/d}}$ where $A_1 = \frac{1}{m^{1-1/d}\prod_{i > 1} (1-e_i)}$
sorry for misunderstanding you earlier
and $A_1$ can be simplified because $e_1 = 1$ and we know what $\frac{z^d-1}{z-1}$ is
its just $z^{d-1}+z^{d-2}+...+1$ so the whole product will actually be just value of this at $z = 1$
so it will be $d$
so $A_1 = \frac{1}{m^{1-1/d}\cdot d}$
(assume the roots of unity $e_1, ..., e_d$ are chosen so that $e_1 = 1$)
7:55 PM
There's an extra factof of $m^{1/d}$ somewhere. But it solves the problem
Thank you so so so so so much
yeah, sorry for misunderstanding you earlier, I was thinking of taylor series at $r = 0$ for some reason, and you obviously wanted Laurent series for integration
(so only the term near $(r-m^{1/d})^{-1}$ matters)
@Jakobian It is completely fine. Thanks for the help.
@Jakobian But sorry for the trivial doubt. How do you know that the product is this geometric series
@Sanjana you have this equality for partial fraction decomposition above, and you multiply both sides by $r-m^{1/d}$
so then you have, up to terms which disappear when taking $r = m^{1/d}$, $A_1 = \frac{1}{\frac{r^d-m}{r-m^{1/d}}} + ...$
the fraction is $r^{d-1}+m^{1/d}r^{d-2}+...+(m^{1/d})^{d-1}$
and then you plug in $r = m^{1/d}$ you get $A_1 = \frac{1}{dm^{1-1/d}}$
this is even more direct for the problem at hand
oh sorry we need to include the $r^{d-2}$ term
$$\frac{r^{d-2}}{r^d-m} = \frac{A_1}{r-m^{1/d}}+...$$
okay this is more appropriate and then $A_1$ will be $\frac{r^{d-2}}{\frac{r^d-m}{r-m^{1/d}}}+...$ at $r = m^{1/d}$
okay sorry for this being so messy
but yeah you get $A_1 = \frac{m^{1-2/d}}{dm^{1-1/d}} = dm^{-1/d}$
sorry $A_1 = \frac{1}{dm^{1/d}}$
yeah it will be for the best if you were to write this on your notebook or something on your own, I made a lot of mess with my writing here
36 mins ago, by Jakobian
59 mins ago, by Jakobian
@Sanjana for non-zero $m$, $\frac{1}{r^d-m} = \frac{A_1}{r-e_1m}+...+\frac{A_d}{r-e_dm}$ implies $A_k = \frac{1}{m^{d-1} \prod_{i\neq k} (e_k-e_i)}$
in here you want not only the right side to be $m^{1/d}$ but the left side should have $r^{d-2}$ in nominator and formula for $A_k$ is wrong then as well
8:31 PM
@Jakobian Yeah. The idea was important. Thanks for that. I could work it out
Although it seems like magic.
I am not sure whether the geometric series converges.
But it's alright even if it doesn't. This method works atleast at the level of rigour of math used in physics
@Sanjana which geometric series?
You mean $\frac{1}{r-e_km}$ for $k\neq 1$?
@Jakobian This one. Does it converge for $r=m^{1/d}$?
@Sanjana what do you mean? Its a polynomial
@Sanjana no no, this is fully mathematically rigorous
@Jakobian I mean using the series directly obviously it converges but using the formula for the sum in which it is replaced $\frac{r^d-m}{r-m^{1/d}}$, it seems indeterminate
series?
so you just mean the replacement of $\frac{r^d-m}{r-m^{1/d}}$ by this polynomial
treat it like a formula for $r\neq e_1m^{1/d}, ..., e_dm^{1/d}$ for example
and you obtain $A_1$ not by plugging in $r = m^{1/d}$, but by taking a limit $r\to m^{1/d}$
is this convincing enough for you
and this is the same as plugging in $r = m^{1/d}$ in a formula where $\frac{r^d-m}{r-m^{1/d}}$ was replaced by polynomial $r^{d-1}+...+(m^{1/d})^{d-1}$
8:42 PM
@Jakobian Yes. The polynomial gives a finite answer, but the expression $\frac{r^d-m}{r-m^{1/d}}$ doesn't. Gosh, this happens for every finite geometric series. I haven't noticed this before!
@Jakobian Wait lemme go through these messages...
8:53 PM
Got it, thanks.
9:30 PM
@Jakobian Possible to explain how?
How can one sequence be above and below the other?
9:55 PM
@EE18 thats not what I'm saying
and I could explain it to you, but I'd rather you provide an argument on your own
its not hard - and I really mean this
don't count on me explaining you something, but rather try to figure out how to use this series to bound the other, on your own
I can do it, I know I can, but its about your abilities, you've got to explain it
I know this is not beyond you
try to work with what someone is saying, not exactly what they are saying, this is not about verifying if something is true, its about you forming an argument
if you feel like you can't do it right now, say, you're tired, do it when you don't feel tired
I will work on it, thank you for the hint
10:30 PM
In Jaconbson's Basic Algebra I, in the context of monoids $M$, is defined in $M$ a binary operation $p$. The author says that usually $p$ is denoted as $p(a,b)=ab$. The question is: the associativity of $p$ is the equality $p(ab,c)=p(a,bc)$ for each $a,b,c \in M$?
yes
b associating with a is the same as b associating with c
@ZaWarudo this is usually written in this way: $(ab)c = a(bc)$
the left side is understood as $p(p(a, b), c)$ and the right side as $p(a, p(b, c))$
10:46 PM
or (a+b)+c=a+(b+c)