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12:08 AM
The symmetric tensor algebra? @Astyx
 
12:24 AM
hat must be it thanks
 
12:38 AM
Thought you were asleep!
 
1:37 AM
In complex analysis, if two closed paths $\alpha,\beta$ are homologous in an open set $U\subset\Bbb C$, then for holomorphic function $f:U\to\Bbb C$, $\int_alpha f = \int_\beta f$. Is this related to some kind of homology on $\Bbb C$ or $\Bbb C^*$?
 
2:18 AM
Is there a place on Math Stack Exchange Chat where it would be appropriate to ask for feedback on one's mathematical writing, say on a blog?
 
this is the mse chat.
 
Hi folks
 
@subrosar this is the place. or at least a place. for exactly that.
we also discuss other topics but math is permissible.
 
@leslietownes Thanks, good to know.
 
i mostly talk about ducks and poorly tarmacked drives but there's all sorts of stuff going on in here.
people have already clicked over to your profile, read the blog, and will be providing unsolicited feedback shortly.
math blogging was considered risky in my professional day. then overnight it became risky not to do it.
 
2:52 AM
Anyone know if it's possible to write the Pythagoras theorem in terms of the gamma function?
 
I mean... yes?
Question is what form of the expression you're looking for.
 
How would you express it?
actually i'm not sure, i'm just curious to know if it's possible
 
$x = \frac{\Gamma(x+1)}{\Gamma(x)}$. The rest should be pretty straightforward for $a^2 + b^2 = c^2$.
 
this is where $\Gamma(x+1) = x!$ right?
 
Gamma function is the Gamma function, not factorial.
 
2:59 AM
But am I not right in saying that $\Gamma(x+1) = x!$ because $\Gamma(x) = (x-1)!$ and making the substitution would lead to $\Gamma(x+1) = ((x+1)-1)! = x!$?
 
$\Gamma(x+1) = x!$ only holds for integers.
That's what you were missing
 
I don't quite follow, could you elaborate more please.
 
Finally finished my homework. Once again, I'm not sure if I might actually be estupido ;-;
 
Deep down we're all a little stupid
 
3:05 AM
Oh, I was confused because I thought the gamma function only held for complex numbers
 
Man, you probably right.
 
@mohan10216 The entire page on mathworld explains the domain and codomain of Gamma.
 
Yeah just read through, feel a little stupid now.
 
@UnderMathUate Case in point
 
My confidence just boosted by 2%.
frikin 2
 
3:07 AM
Nice. If you grind some more you can get to confidence 100
 
probably adults right?
How old r u guys?
 
mathworld is like mathematics if Wikipedia were actually good.
 
I'm adulthood
incarnate
 
I'm 21
Anyways...
 
3:08 AM
I'm 20
 
It doesn't have the smol brain policy of requiring extraneous sources.
 
wait, why'd u ask that
usually people under 18 ask that
 
^
 
I'm 15
 
I knew it
 
3:11 AM
What are your vocational interests? Do you have any idea of what you want for a profession yet?
If you don't mind me asking.
 
Me or Mohan?
 
You can if you want, but the question is directed at Mohan
 
I'm not quite sure but I definitely want to do math as a major, I might do economics as a minor.
 
Oh, lol.
Ye, do math. Return to mathematician.
 
Do any of you guys take topology?
Is it hard?
 
3:14 AM
I haven't taken topology. The general consensus is that it is a grad-level course. This doesn't necessarily mean it will be "hard" if you have the right background.
I'm going to do an independent study with a prof in a semester or two.
 
I find it really hard, I just can't wrap my head around it.
 
What math have you studied before? Do you have experience with analysis and abstract algebra?
I don't think these are hard pre-requisites, just the ones I've seen most people suggest.
 
I have experience with abstract algebra and a little real analysis
 
Sehr gut. You want to be a mathematician. Math and anything that exercises the intellect is based. Anyone who tells you otherwise is a fool.
 
I have studied lot's of things like graph theory, set theory, algebra-precalc, trig 1 and 2 etc
 
3:18 AM
Yeah, so it's probably just down to experience. Once you mess around with the definitions/theorems and ones related to it, it will suddenly make more sense. Usually.
And by mess around, I mean pictures and examples. Lots and lots of them
 
Unfortunately I cannot say that pre-college courses are very good for teaching maths because they have cring philosophies that deny things like substance and accidents.
 
Are pre-college courses even considered math at this point?
2
 
At least here in America. Not sure about others.
@UnderMathUate FACTS
There's also this silly thing called PEMDAS.
 
I'm so sad I didn't know PEMDAS wasn't real until senior year of high school.
 
I use BIDMAS and BODMAS
What does the I and O mean in these?
 
3:21 AM
Bruh, just learn laws, not a general order of operations. Far more powerful, but I guess we can't learn those because those are universals or what I suppose in math is called axioms.
Parentheses? Distributivity.
Exponents? Distributivity.
Multiplication? Distributivity.
You get the point.
 
ok thanks
 
Division? $\frac{x}{y} = x\cdot \frac{1}{y}$.
 
Yoo, I was just explaining this to someone today.
 
kek
 
3:22 AM
It was over text though, so idk if they were amazed or not
 
Well what will blow some minds is that there's a way to multiply by adding and dividing.
 
Out of curiosity, is there a way to determine if a function is transcendental or not?
 
it's difficult. depends on what you mean by transcendental. liouville's theorem is helpful but maybe not the last word on this.
 
Sorry, if i'm asking too many questions, i'm just really curious
 
Every yes or no question is associated with a boolean decision algorithm. The only issue is finding the algorithm which, in my humble experience, is pain.
You can never ask too many questions; you can always ask too few.
 
3:25 AM
Words to live bye.
 
Leaving D:
 
Dang, I must be getting tired. I really thought that sentence was correct.
 
Sorry. The 1 iq jokes just can't be stopped.
 
i'm very confused as too what is going on now...
 
It's ok, reading them might help me finally make the jump from 0.5iq :D
 
3:26 AM
The real question is why do they call it an intelligence quotient if it's a constant? :thinking:
 
AMDG, always asking the real questiosn
 
that stuff has a pretty ugly history.
 
Listen. I have like half a working brain cell at any given time yet I still somehow manage to find things that, according to academia, I shouldn't be able to find. Like an infinite sum for computing quotients.
I forget where I was going with that...
 
sometiems one rain cell is all u need ;-;
wait u said half, i dont believe u now
 
Don't believe what?
 
3:29 AM
see ya guys, coming back in in like 10 mins
 
shrug
 
o/
 
kk
Me too. I'll be back in 15 - 20 min.
 
Aight. Later.
 
ye, you too.
 
3:31 AM
Oh yes, right. So my point in stating that was uh... people who have more than half a brain cell don't have much excuse in my opinion.
 
Nvm I’m back now
Took less time than expected
 
Can someone who knows Riemannian geometry check me on something
So suppose we have a conformal metric meaning $g$ is a multiple of the identity matrix everywhere
Visually, that means that the Tissot ellipses (circles of size $\epsilon$ in the metric, which are distorted in the chart) are all circles (of varying size)
If the metric is flat, meaning it's (locally) isometric to a subset of the plane, then - we can square the metric (square the size of all the Tissot circles) everywhere and it's still flat
 
Hi all, I'm trying to understand the math behind the pythagorean harmonics. If anyone can help that would be great.
The idea is that Pythagoras took a string, plucked it and heard a frequency he found "appealing".
Then he held the string down at the halfway point, plucked it again and heard twice the frequency (again, a nice sound).
And so on and so forth. My questions are these: How exactly does this result in the circle of fifths? How are 12 tones produced? Forgive my lack of musical vocabulary (I'm very new to the subject) And what does this have to do with harmonics?
 
The idea was about ratios of frequencies I thought - that he played two strings at the same time and compared their frequencies
Specifically, a 2:1 ratio is always an octave
and a 3:2 ratio is a fifth
 
@AkivaWeinberger Ah, so he had two strings?
 
3:39 AM
If you stack two 3:2 ratios on each other, you get 9:4
Two notes an octave apart are generally considered to be equivalent (this is called "octave equivalence")
 
Hm, I'm afraid I don't know what an octave is. Google says it is "an interval whose higher note has a sound-wave frequency of vibration twice that of its lower note." Does that mean any pair of notes that are the result of strings that are twice the size of each other are an octave apart? So if I have a string of length $x$ and one of length $2x$, the notes they would produce would be an octave apart, right?
 
so if we start from (say) a C, then the note 9:4 above it (two fifths gives us a high D) is equivalent to the note 9:8 above it (low D, a "whole step" above the C)
Yeah
 
Got it.
 
I really recommend listening to some examples
Do you have an instrument nearby
 
Actually, no, but I can listen online.
 
3:43 AM
In any case, we keep on stacking fifths, going down octaves to keep our ratio less than 2
 
Hm, I see
 
Eventually we get to (3/2)^12 / 2^7
which is 531441:524288 ≈ 1.01364:1
which is really close to 1
 
Looks a bit like the golden ratio ...
But perhaps I'm wrong; Spoiler: I'm wrong. Golden Ratio is 1.6
 
So we call it close enough and say that twelve fifths is the same as seven octaves
This difference is called the Pythagorean comma
In musical tuning, the Pythagorean comma (or ditonic comma), named after the ancient mathematician and philosopher Pythagoras, is the small interval (or comma) existing in Pythagorean tuning between two enharmonically equivalent notes such as C and B♯ (Play ), or D♭ and C♯. It is equal to the frequency ratio (1.5)12⁄27 = 531441⁄524288 ≈ 1.01364, or about 23.46 cents, roughly a quarter of a semitone (in between 75:74 and 74:73). The comma that musical temperaments often refer to tempering is the Pythagorean comma.The Pythagorean comma can be also defined as the difference between a Pythagorean apotome...
So OK, close isn't exact. How do we "fix" it?
 
Hm
Multiply by some constant factor?
 
3:46 AM
Solution a: just don't, just live with the fact that we can't get back to where we started (up to octaves) by going up by fifths
Solution b: fudge it a little bit
 
@AkivaWeinberger Interesting, I'm going to read up a bit more on this. Thank you for the help and references! In particular, I find the pythagorean comma intriguing.
 
Instead of using 3:2 = 1.5:1 for our fifths, there's a system called "equal temperament" which uses 2^(7/12):1 = 1.4983:1
So we sacrifice the "purity" of our fifth intervals, but now going up by a fifth twelve times gives us seven octaves exactly
In "12-tone equal temperament" (the full name) there are 12 notes per octave. If the starting tone has a frequency of 1, then they're 2^(1/12), 2^(2/12), 2^(3/12), etc
Why 12? Because doing this happens to contain a note that's "close enough" to that pure 3:2 ratio called a fifth
 
lol 1.0 is way off from Phi
 
Another reference to look at:
In music, just intonation or pure intonation is the attempt to tune all musical intervals as whole number ratios (such as 3:2 or 4:3) of frequencies. An interval tuned in this way is said to be pure, and may be called a just interval; when it is sounded, no beating is heard. Just intervals (and chords created by combining them) consist of members of a single harmonic series of an implied fundamental. For example, in the diagram, the notes G3 and C4 (labeled 3 and 4) may be tuned as members of the harmonic series of the lowest C, in which case their frequencies will be 3 and 4 times, respectively...
A really good composer who experiments with dividing the octave into numbers other than 12 is Sevish @rb3652
Incidentally, notice that these are all ratios. That's because humans perceive pitch "logarithmically"
That means that we perceive notes with frequencies x and 2x to be the same "distance" as notes with frequencies 2x and 4x
(The same is true with volume and lots of other things)
(There's a connection with the Riemann zeta function but I don't remember details)
Something about looking at $|\zeta(\frac12+2\pi i x/\ln2)|$
Some variation somehow gives that chart, which is like a "quality measure" of x-tone equal temperament (dividing the octave into x equal pieces) but I don't understand the details
 
4:15 AM
i'm back fr now
Hi @AMDG @UnderMathUate
 
 
1 hour later…
5:26 AM
@mohan10216 Hey, I was away longer than intended. I gotta go to bed now, so talk to you later. And @AMDG too.
 
 
2 hours later…
7:45 AM
quick question guys:
Is the Cauchy product of two polynomials simply just expanding the brackets and adding like terms together?
 
 
4 hours later…
11:50 AM
yes
 
 
1 hour later…
12:56 PM
@love_sodam yes. It means that integration over a closed cycle is only dependent on its homology class, and thus integration induces a bilinear pairing $\omega,[\gamma]\mapsto \int_{[\gamma]}\omega$ (where $\omega$ is a $1$-form and $[\gamma]$ is the homology class of a loop $\gamma$)
 
 
2 hours later…
2:43 PM
Let $(X, \mu)$ be a probability measure space, let $\theta$ be a probability measure preserving automorphism of $(X,\mu)$, and let $f$ be an essentially bounded function on $X$. I'm pretty certain the following is true, but I'm not sure from which first principles I can deduce it: $\int_{\theta (E)} f = \int_{E} f \circ \theta$, where $E$ is any measurable(?) (borel?) subset of $X$.
Intuitively, it seems right.
 
3:27 PM
I guess what I am looking for is some sort of change of variables formula, and $\theta$ doesn't necessarily need to satisfy the strong properties that it does; $\theta$ can simply be a measurable function.
 
do you think i can find this theorem somewhere in the literature >
 
You can probably find it in any book on calculus or real analysis.
In fact, the page you linked cites Rudin's Principles of Mathematical Analysis.
 
it cites it but is only doing so to make use of a theorem in it
definition*
 
I bet that it contains a proof of the fact or all the ingredients needed to prove it.
 
3:59 PM
@homemath1homemath1 here is a proof for $\limsup$, you should be able to adapt that since $\liminf(a_n)=-\limsup(-a_n)$
or were you looking for literature to cite?
 
@robjohn I was trying to use the result quoted in this post jyotirmoy.net/posts/2013-12-17-liminf-products.html but am a bit worried since ``my $A$'' might be negative ( it is definitely finite )...
 
4:12 PM
@robjohn thanks for the link though. But I cant spot the error in the proof jyotirmoy.net/posts/2013-12-17-liminf-products.html if $A$ is negative ?
 
5:02 PM
@homemath1homemath1 why must there be an error?
 
5:29 PM
@robjohn I cant see why it is true when they say `` direct consequence of the usual theorems on limits of products and quotients ''
 
@homemath1homemath1 That is very hand wavy. I wouldn't use that proof, especially for getting an understanding of the situation.
 
what if for such a sub-sequence $\lim_k a_{n_k}$ didnt converge, but $\lim_k a_{n_k}b_{n_k}$ did
I want to do : Consider two sequences $\{a_n\}_{n\in \mathbb{N}}\subset \mathbb{R},\{b_n\}_{n\in\mathbb{N}}\subset \mathbb{R}_+$.

We assume that :

- $(*)$ $\liminf_{n} a_n \geq A \in \mathbb{R}$

- $(**)$ $\lim b_n=B \in \mathbb{R}_+$

$\textbf{Claim :}$ $$ \liminf_n a_nb_n \geq AB. $$
I want to prove that Claim
(im not sure if it is true)
 
@homemath1homemath1 since $B\gt0$, for the tail of the sequence $a_{n_k}=a_{n_k}b_{n_k}\cdot\frac1{b_{n_k}}$
apply the other direction
since we know that $\frac1{b_n}$ converges to $\frac1B$
 
ah so this proves that $a_{n_k}$ converges
 
yes
 
5:35 PM
Thank you <3
 
5:52 PM
I think that the answer here math.stackexchange.com/questions/1654510/… is wrong.
Clearly, the limits for $r$ should be from $\sec\theta \tan\theta$ to $\sec \theta$.
 
@robjohn Now that you're an accomplished CD LaTeXer, I can't wait for better.
 
wow, he has amazing hair.
 
Hi Ted, the answer I linked seems to be wrong. The region of integration in r theta doesn't look right.
 
There are lots of wrong things on this site.
 
6:05 PM
Why I ask is because the answer is upvoted and may mislead somone.
 
Ross is generally very reliable. Yeah, he went too fast and messed up. $r\sin\theta = r^2\cos^2\theta$. Not rocket science.
Not sure why you specified $[0,\pi/4)$ rather than $[0,\pi/4]$.
 
no major reason behind leaving the interval open at $\frac \pi 4$. I was thinking that since that segment of ray, is of content zero, we can remove it so I removed that.
 
Why bother and why make the distinction? The original integral goes from $x=0$ to $x=1$, including endpoints. Your region in $r\theta$-space should likewise be compact.
 
Right, no need to bother about that. I'm new to the stuff so I was experimenting :).
 
Well, for purposes of posting to correct an incorrect solution, I see no point in muddying the waters.
 
6:24 PM
But the statement in my comment is not wrong. I just didn't say anything about what happens at $\theta=\frac \pi 4$. So it's not really muddying the waters?
Should i delete the comment if it's causing some confusion?
 
I'm telling you that when you go out of your way to omit $\pi/4$ it makes people wonder why. The rest of the comment is fine, and I added my own comment.
 
I have taken note of your advice. Let people wonder for that one :). And thanks for adding the comment :).
 
7:15 PM
@Koro No, sorry, you're right.
 
8:06 PM
o/
 
Hello if $f\circ f\equiv 0 $ then always $f\equiv 0$ ?
If I suppose that $\exists x_0\in\mathbb{R}, f(x_0)\neq0$
Then
$f\circ f(x_0)\neq f(0)$ ??
We don't have that f is linear
 
8:22 PM
vrou you could have something like f(x) = x^2 if x is negative, 0 if x is not negative. f(f(x)) will be 0.
the condition only requires f to be 0 on the range of f.
 
Ah so it is not always true
Thank you
 
$f(x)=|x|-x$
 
8:46 PM
Thank you for this example
 
Anyone know of a linear $2$nd-or-higher-order ODE with non-constant coefficients and a constant but nonzero forcing function that plays nicely with the initial conditions $y(0) = 0, y'(0) = 0, ...$? The Euler ODEs (with descending power coefficients) breaks under these initial conditions because the entire part of the solution from the associated homogeneous ODE gets zeroed out.
 
9:05 PM
I want to prove that $\lim_{t\to1} \frac{2x+1}{x-2}=-3$ using the epsilon delta definition
When I get $5\frac{|x-1|}{|x-2|}<\varepsilon $
How to delete x-2
To get $\delta$
 
9:23 PM
Ana suggestions please ?
 
@Vrouvrou how does the value of $t$ affect the value of $x$?
 
X not t sorry
$\lim_{x\to1} \frac{2x+1}{x-2}=-3$
 
Hi
Well, you know that $0<|x-1|<\delta$.
 
What is ${2x+1\over x-2}+3 =$ ?
Presumably, if $|x-1|<1$ then you can show that ${1 \over x-2} \le 1$
 
9:43 PM
@Vrouvrou the idea behind that is to use the hypotheses $0<|x-1|<\delta$ into $|f(x)-L|<\varepsilon$. WLOG you can start with $\delta:=1$ and use that in $|x-1|\frac{1}{|x-2|}<\frac{\varepsilon}{5}$.
 
And what is delta ?
 
I'm trying to draw a diagram using LaTeX. But all the reference seems very difficult. Does anyone know a nice manual for drawing with LaTeX? Maybe is there a program that miraculously converts a drawing to LaTeX?
 
If $|x-1|\le\delta$, then $1=\overbrace{|(x-1)-(x-2)|\le|x-1|+|x-2|}^\text{triangle inequality}\implies |x-2|\ge1-|x-1|\ge1-\delta$
So now we have $5\frac{|x-1|}{|x-2|}\le5\frac{\delta}{1-\delta}$ and we want $5\frac{\delta}{1-\delta}\le\epsilon$ Now you should be able to solve for $\delta$
 
9:59 PM
some OPs are very entitled. just got annoyed in a comment.
 
@copper.hat sort of like here?
or in the comment to your answer?
 
@robjohn i'm sensitive today :-) the comment to my deleted answer annoyed me.
i was in the middle of adding more detail when the comment came in so i canceled the edit and deleted the question. entitlement bugs me.
cut off my nose to spite my face
 
:-( maybe a little music can help you on this day or maybe some exercise?
whenever someone adds a "!" looks like he's screaming or something like
 
10:16 PM
i did my little 5k detox this morning already :-)
 
How long will it take you to run 5 km? When I was 18 years old, I participated in speed races, 1200 flat meters. However, an injury prevented me from advancing professionally. Now I'm in terrible physical condition, I don't exercise anymore.
unless chess is considered an exercise to lower calories :-)
 
i am 61 and have a hip/knee that needs replacement, so if i can manage 30mins now i am happy. i could manage under 20mins a few years ago.
by few i mean many decades
on the flats
my preference is trail or x-country
i miss being able to run distances
 
That's a good time. I like to see the speed competitions in the Olympics, the training is very hard.
 
11:15 PM
@Ted: I added an answer to the question to which you and Koro commented earlier.
 
the OP is likely long gone but i think we can all appreciate the merits of a corrected record.
 
hello, i'm having a hard time proving the following: a module $M$ over a commutative ring $A$ is flat if and only if $I\otimes_AM\rightarrow IM$ given by $i\otimes m\mapsto im$ is an isomorphism. the definition of flatness i am working with is that $-\otimes_{A}M$ is exact, and i can show this implies the ideal condition, but not the other way around
 
I didn't think it needed an answer after 15 years.
But, yeah, region under, not over.
So I just tried to draw a figure for a silly question using Inkscape. I was so good at Illustrator and Inkscape is balky and impossible to figure out. And then it crashed and it can't reopen the document. Agh.
I'm downloading the latest version of Inkscape. Clearly I'm going to have to spend days learning how to use it. Frustrating.
 
@TedShifrin Possibly, but I thought since the existing accepted answer is wrong, perhaps a bit more emphasis was needed. Comments can get overlooked or even deleted.
@TedShifrin I have never used Inkscape.
 
^oops, i mean $I\otimes_AM\rightarrow IM$ is an isomorphism for every ideal $I$ of $A$
 
11:24 PM
@leslietownes yes, comments are ephemeral.
@copper.hat read the first few comments on this answer (you have to look at all the comments as the comment that annoyed me is hidden by more upvoted comments). So I can understand your annoyance.
 
those comments sure aren't ephemeral. at least, they lasted longer than mayflies do.
 
@Ted: I wrote a contour integration answer the other day for which the integral on the encircling contour does not vanish absolutely, as they usually do. The upper part cancels the lower part, but the integral of the absolute value on the encircling contour gets big.
 
@robjohn At least they apologized! Your answers are so pretty that they deserve up voting regardless of correctness :-)
 
@leslietownes yes, there are many comments that have survived a long time.
@copper.hat Yes, I was just about to mention that they did apologize.
 
some of my comments have lasted a long time too, which i am ok with. before i knew of their ephemeral nature, i put more effort into them than now. although certainly i do not take the view that "well, it's just a comment" and put out the first thing that comes to mind.
although some people do essentially that. with posts that have obvious problems even first-impression comments can be very helpful.
 
11:36 PM
@leslietownes If they provide good content, they should survive.
@leslietownes I agree with this answer (I even say so in a very old comment :-)
Willie Wong was a moderator here when I joined
 
11:55 PM
as long as my comments advertising lesliecoin are not deleted, i will be a happy customer.
 
deletes leslie in its entirety
@robjohn Quite unusual.
 

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