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00:05
Is there anyone out there who is able to help on a question about approximations of trig ratios over x, specifically the limit as x -> 0 (for homework).
00:20
Aha, I had my calculator in degrees mode when it should have been in radians mode. That was giving me unexpected results, but everything makes sense now.
00:37
Well, it looks like there really is something very, very wrong with me.
Because apparently, I just had an "attitude" with someone.
Nothing wrong with that
you're human (hopefully), so you're allowed to make mistakes
But the thing is, I don't mean to! I never mean to! And yet, I get called out by people for it on MSE all the time!
I'm just being myself, and people think I'm being rude.
Some people are just rude
00:39
So, I'm just rude?
lol, I meant the other people
Har har.
@ALannister Would you like to play the large number game?
I have a feeling they're all old, too.
What's the large number game?
(har har har)
00:40
Isn't that from a movie or something?
Whoever says a larger number wins, so you go first, you say a number. Then, I just say a number larger than that, and then I win?
Given a sheet of paper, you are challenged to make the largest number you can. Do take it serious and use whatever mathematics you know or can explain on that sheet of paper, and no, this is not from a movie
Meh, I made it a single player game, since it wouldn't be fair if I played it against you
Simply this sounds like an interesting challenge I wonder what the number theorists of this channel might come up with
can you give me an example of your beautiful art? @SimplyBeautifulArt
@ZION You may play the game as well if you wish to try
@ForeverMozart Depends what your art is, but here is one of my simple ones: math.stackexchange.com/questions/2118082/…
Simply i'm not sure if i'd stand a chance i'm an analst by chance
00:43
@ZION :P You can still try. There is no losing, only learning
all right but it may take a while for me I like going slow
:-) Of course, you are meant to spend a few months before you feel finished with the game, if you ever do
it is an unsatisfying game of "what's the next step?"
To anyone playing the game, you may find me in this chat room.
how would one approach the problem of finding the exact value of sin^-1(1/2)?
@micsthepick Inverse sine?
@SimplyBeautifulArt yes
00:49
Special right triangles
@micsthepick Got it?
@SimplyBeautifulArt yes, once again my calculator was in the wrong mode (this time radians instead of degrees), so I missed that simple answer (30˚=pi/6)
Haha, happens all the time
Can you edit your question, I think you are forgetting "-" minus sign in the Fourier coefficients. You are using circular coordinate $\theta$ and the linear coordinate $x$ in a confusing way. Your question is not clear either. — Aymane Fihadi 2 mins ago
Can you edit your question, I think you are forgetting "-" minus sign in the Fourier coefficients. You are using circular coordinate $\theta$ and the linear coordinate $x$ in a confusing way. Your question is not clear either. — Aymane Fihadi 3 mins ago
1
Q: Verifying the Indictor Function:$X_{[a,b]}$ can be expressed as a Fourier Series?

ZIONIn the text "Fourier Analysis: An Introduction" by Elias M. Stein and Rami Shakarchi. I'm having trouble attempting to verify the that $f(x)$ can be written as a Fourier Series in $Propostion \, \, (1.2).$ Note my initial approach to verify $Proposition \, (1.2)$ can be seen within, $Lemma \, \,...

^ Is my question not clear
@ZION Man, I hardly ever write formal proofs
I mean in questions and on MSE
Simply I did get a comment that my question was not clear, I clearly defined everything
I used
there was only tiny latex erros nothing big through
01:04
always understandable
did I use my variables in a confusing way the person who commentated on my question seemed to get it
Perhaps you should ask them
Simply I see the initial error I made ops misapplied the definition simple fix
@AkivaWeinberger Hey!
01:13
Hi Akiva
@BalarkaSen Prove or find a counterexample: For any two disjoint compact sets in $\Bbb R^3$ such that the complement of each is simply connected, the complement of their union is also simply connected.
@Akiva Do you happen to know of the fast growing hierarchy?
I believe I found a counterexample (it's weird), but I'm not sure it works.
@SimplyBeautifulArt Vaguely
You understand limit ordinals and fundamental sequences?
That's the one where if the subscript is a limit ordinal you diagonalize, right?
And if it's not, you repeat the function with subscript the predecessor ordinal
Yup.
I tried to make a faster growing function. Mind joining my realm for some tea?
Well it looks like Balarka is finally off the starred list. It was ironic while it lasted.
I know it is possible to use a right angled triangle diagram in finding the value of an expression such as sin(cos^-1(x)). Is there no such trick for finding the value of an expression where the inverse function is on the outside (e.g. cos^1(sin(pi/3)) )?
you meant the inverse cosine at the end @micsthepick ?
Indeed, just draw a triangle for that one too and you will see the answer
01:28
@SimplyBeautifulArt of course. arccos(...) if you prefer
@micsthepick Well you have cos^1, which is a bit odd
@SimplyBeautifulArt oops... forgot the negative sign
:P Whatever now
oh, wait
drawing triangle does not work for these
oh wait, I take that back. Drawing triangles is the way to go
@SimplyBeautifulArt So I have cos(α) = sin(π/3) where α is the value I am looking for, do I draw a right triangle with one angle equal to π/3?
01:35
@SimplyBeautifulArt ... I see now, I would proceed by finding the complimentary angle, right?
Mouth is all cottony
By the transitive property then, my mouth is all kingy.
Not too long ago, the FB page "Mathematical theorems you had no idea existed, 'cause they're false" posted the following false theorem:
> Every order-preserving bijection of the rationals is piece-wise linear.
Someone posted a ridiculously long counterexample in the comments
01:46
Beautiful
not simple
not really artsy either
so it get 1/3 from me
but $\begin{cases}\frac52-\frac1x,&\frac52<x<2\\x,&\rm else\end{cases}~$ works
They forgot $\frac1x$ is locally bijective in the rationals, apparently
Whoops, typo.
$\begin{cases}2.5-\frac1x,&0.5<x<2\\x,&\rm else\end{cases}$
I'm back
@AkivaWeinberger Want to do QQ questions?
If you're still online
QQ questions?
Qualifying Quiz
01:59
I'm actually about to go to bed, sorry
OK, good night.
is it possible to solve tan^-1(3*tan(7*π/6)) by drawing a right angled triangle?
'tis a horrible looking question
@micsthepick First let's look at the innermost
$7*\pi/6$ do you see any way to make this easier ?
Well, it's $6\pi/6 + 1\pi/6 = \pi + 1\pi/6$
@micsthepick You could use a circle, but in this case it is unnecessary
02:09
But tangent has a period of $\pi$
So tan of that will just equal $\pi/6$
So, draw a triangle with angle $\pi/6$
That's just a 30-60-90, huh?
Now, the tangent is just the slope of our hypotenuse, correct?
@MeowMix Hisssss
He said the unspeakable words
he used degrees
It burnsss ussss
@MeowMix yes
@SoumyoB I need more of your advice!
Well, to be frank, $\pi/6$-$\pi/3$-$\pi/2$ is a lot less catchy.
@micsthepick Multiply it by three and you should see another familiar value for a tangent.
02:12
and good night!
@micsthepick So we're looking at the slope of our 30-60-90 triangle
And multiplying it by 3
Well, what's the slope of a 30-60-90 triangle? (What I'm really asking here is, what's $\tan(\pi/6)$
@SimplyBeautifulArt g'night
If you know your sine and cosine values, it isn't too hard :P
What's $\sin(\pi/6)$?
@MeowMix 3^-0.5
yes, 1 / root 3
Wait, which one are you responding to
02:16
@MeowMix The first, just hover over the reply.
@MeowMix simply 0.5 (for the second)
It's sqrt(3)
$\tan(\pi/6)$ is sqrt(3)
@MeowMix My Casio calculator appears to disagree.
@MeowMix $\pi / 3$
$tan(\pi / 6)$ is $\frac{1}{\sqrt{3}}$
well... mine is disagreeing
one second
not sure why its doing this
yes, you're correct
Ignored :(
02:19
its 1 / sqrt(3)
I read your message, @Tim
:D
Senpai noticed me
@micsthepick Anyways, when we multiply by 3, we get $3/\sqrt{3}$
Do you know how to simplify that?
(hopefully you do)
@MeowMix its simmilar to taking a half from one, you get sqrt(3)
correct
now, do you know what angle has a tangent of sqrt(3)?
Hint- you can use the same triangle
which shall not be named
except in radians
02:22
@MeowMix the complimentary
There are some people who call me...
Yes!
Do you understand why?
@ALannister Ah, a noble traveller. I have been waiting.
clicks my coconut shells together
That is, why $\arctan(x) + \arctan(1/x) = \pi/2$?
02:24
You like those before you, seek the holy grail.
@Tim I fixed the angle for you
@MeowMix feelsgoodbro
@MeowMix You swap the hight with the width, since the gradient has been inverted
Yes.
Anyways, the complimentary of our angle is.... $5\pi/6$
So, we have completed the problem! And all we need to do was simplify, innermost first
@MeowMix isnt that the supplement?
02:26
Supplement would be $\pi$
OH
yeah, $\pi/3$ sorry
The complement is $\pi /3$
@MeowMix Therefore tan^-1(3*tan(7*π/6) = π/6
I knew the names I was forced to memorize in school would be useful someday.
It's pi/3 I made a mistake
@micsthepick Principal branch my friend.
02:28
I'm going to look for Akiva's contour again
And see if I can solve it
Actually @TimTheEnchanter I seek the Iron Throne.
@ALannister Ah, but then it is a different castle and monster you seek. You've gone too far, your quest lies with the dragon you met in the cartoon sequence.
Now that sounds like a Farscape reference.
Prove that this contour has an integral 0, even though it's not nullhomotopic
(Red and Green points are the only singularities)
02:31
You know, those curves look naughty.
Oh, I see it :D
@Akiva I have a solution to your contour! Break it at the intersection near the bottom left between the purple and black parts. Then, both contain only the green singularity, but are in reverse orientation, thus they cancel out
02:57
Hi, is there a divergence theorem for the integral $\int S(x):\nabla g(x) dV$. Here $S: R^3 \to$ Linear Transformation, is a 3 by 3 matrix and $g : R^3 \to R^3$ is a vector. I want to get an area integral which has $g(x)$ simply instead of $\nabla g(x)$.
Here : in the integral is the Frobenius inner product between 3 by 3 matrices.
 
3 hours later…
05:42
any1
if every subgroup of a finite group G is normal, then G is solvable.
18
Q: Groups with all subgroups normal

Kevin VentulloIs there any sort of classification of (say finite) groups with the property that every subgroup is normal? Of course, any abelian group has this property, but the quaternions show commutativity isn't necessary. More generally, the group $(redacted)$ will have this property.(See answer below)....

I don't remember (nor how to prove quickly) whether all dedekind groups and hamiltonian groups are solvable
06:20
hello
if a function is continuous and surjective can we deduce that is injective ?
$f(x) = x^{2}$
?
it is continuous
@Secret in that answer it never mentioned anything about solvable ?
where $f:\mathbb{R} \rightarrow \mathbb{R}^{+}$
@Vrouvrou
@Vrouvrou $f(x)=x^{even}+x^{odd}$ this is continuous and surjective but not injective)
yes thank you
06:25
is my answer wrong there ?
@BAYMAX I am not sure, if that's the case, it means all dedekind and hamiltonian groups (since by definition they are groups where all subgroups are normal) are solvable, which I don't know how to prove
Take a look at the function$x^{2}$
@Secret
$x^{2}$ is surjective right?
@BAYMAX That's not surjective unless you restrict it to $f : \Bbb{R}^+\to \Bbb{R}^+$
chat.stackexchange.com/transcript/message/36134526#36134526
$f : R \rightarrow R^{+}$ ?
@Secret
@BAYMAX O wait, sorry yes, the preimage of $\Bbb{R}^+$ is nonempty (alternately the image coincide with the codomain), thus it is a valid counterexample
06:30
I was just wandering whether my solution was correct or not it seems that it is correct,right?
ok
07:23
[Random rambles]
Time value of money goes beyond the fact that money can be accumulated or lost via interests
Money, like time, impose an ordering on its targets.
While time imposes an ordering on events, money (and more generally, price) impose a priority on how resources are distributed
This is because of the reality that (as far we know) resources are finite
If anyone can even found one example of physical infinity that is usable, it will be groundbreaking because time and money will no longer be needed
physical infinity of joy ,happiness
08:03
@AkivaWeinberger I am thinking of Antoine's necklace. That's constructed by starting with a solid torus, and at each step replacing each solid torus by a chain of smaller solid torii with another chain of solid torii interlocking with each other.
Call that the "A-chain" and the "B-chain"
Now piece it up into two sets: 1) At each step delete the A-chains. So you just replace the solid torii in each step with a chain of disjoint solid torii filling it.
2) Similarly, each step delete the B-chains.
In this way I think you decompose Antoine's necklace into two Cantor subsets which have simply connected complements. But Antoine's necklace doesn't have that.
I wonder if there is a reason why counterexamples tend to look like fractals
(The reason those pieces are simply connected is because the only potential nontrivial loops are ones which "wind around" some solid torus at each step. But the solid torus itself decomposes into disjoint union of smaller, standardly embedded unlinked chain of solid torii - so just nulllhomotope the loop through the "gaps")
@Secret Well, the whole point is that it's a Cantor set. So it's more or less obvious that you need to come up with something fractal-like.
No I mean in a more general context. A lot of counterexample and pathological examples of some given proposition in analysis, topology and geometry tend to have "Fractal looking" structures
Hi, @MikeMiller.
08:17
e.g. Weistrass function is the counterexample that all contnuous functions must be differentiable at least somewhere. It has a fractal like structure
I think I just gave a counterexample to a question Akiva asked. I hope it's right.
what was the question?
The counterexample that forever mozart proved, while cannot be drawn, is also a fractal
If two compact, disjoint subsets of R^3 have simply connected complement each, need their union be simply connected?
I think the "horizontal" and "vertical" Cantor subsets (ones which are unlinked) in the Antoine's necklace gives such a decomposition.
O wait a minute, I think I found a counterexample to my claim about counterexamples:
In mathematics, the Fabius function is an example of an infinitely differentiable function that is nowhere analytic, found by Jaap Fabius (1966). The Fabius function is defined on the unit interval, and is given by the probability distribution of ∑ n = 1 ∞ 2 − n ξ n , {\displaystyle \sum _{n=1}^{\infty ...
This one does not look like a fractal
(as in if you zoom in, you don't get bumpy behaviour all over the place)
08:23
Your observation is at best meta-philosophical. It's easy to justify fractal-like counterexamples by saying you want stuff to be complicated enough, and self-similarity (spaces modeled on itself) can get really complicated.
Hmm I see, so self similarity can get pretty complicated which is why some of the common counterexamples we learnt in text tend to be fractal like, despite the "Set of all complicated counterexamples" does not necessary have to have all members to have a property of being self similar
, and yeah I sometimes accidentally drifted into philosophy when I thought I am seeing patterns among a bunch of mathematical objects
I guess I freely generalise things too easily
is $\displaystyle\lim_{x\to\infty}\sum_{n=0}^x f(x,n)=\sum_{n=0}^\infty \lim_{x\to\infty}f(x,n)$ always true?
for example there's proof that $e=\sum_n \frac 1{n!}$ by taking the limit of $\left(1+\frac 1 n\right)^n$ and using a particular case
assuming all the limits exist, of course
20
Q: Under what condition we can interchange order of a limit and a summation?

zzzhhhSuppose f(m,n) is a double sequence in $\mathbb R$. Under what condition do we have $\lim\limits_{n\to\infty}\sum\limits_{m=1}^\infty f(m,n)=\sum\limits_{m=1}^\infty \lim\limits_{n\to\infty} f(m,n)$? Thanks!

08:42
[Random] Bizarre function idea: Let $f : \Bbb{R}\to \Bbb{R}$ and $n \in \Bbb{N}$. Then $f^{n}(x)=x,n < c$ and $f^{c}(x)=x+1$ Solve for $f$
Actually, thinking about it, if one treats $f^n(x)=g(n,x)$ then it is just a piecemeal function

$$g(n,x)=\left\{\begin{matrix}x, n < c \\ x+1, n \geq c\end{matrix}\right.$$
09:34
Hello Everyone, I am confused about the transformation used in method of characteristics. To have change of coordinates, we use the slope dy/dx (=y/1 say) and we get eta=xy. But for the second one, in most cases, it is assume that zeta=x out of the blue. My question is, how we arrived at zeta=x? any thoughts?
10:10
how do I prove that $\lim_{x\to\infty}e^{\beta x}\int_x^\infty e^{-y^2}dy=0$ for any real constant $\beta$?
You can first turn that integral into erfc(x), then proceed from there
note $\lim_{x\to \infty} erfc(x)=1$ thus regardless of what happened to $\beta$ the integral will be bounded, thus whether the limit exists depends solely on $e^{\beta x}$
10:35
@BalarkaSen I'm not sure I understand what the $A$ and $B$ chains are
Call the big torus a level-0 torus, the smaller tori level-1 tori, the smaller ones level-2 tori, etc. Are the $A$ chains every other level-1 torus, and the $B$ chains every other level-1 torus? But once you finish the construction, each level-1 torus becomes a smaller copy of the necklace, which has nonsimply-connected complement.
On the other hand: if you make the $A$ chains every other level-1 torus, only made of every other level-2 torus, only made of every other level-3 torus, etc., and similarly for the $B$, then their union doesn't have the desired property. None of the level-1 tori will be made of all level-2 tori, so its complement will be simply connected.
@AkivaWeinberger No, I mean, you do that at each level. From the big torus you get to a picture that looks like this. The "horizontal" torii are A-torii and "vertical" torii are B-torii.
Right, but what happens when you divide those tori further? @BalarkaSen
Consider the whole construction where you remove the horizontal torii after each step, and iterate.
@Akiva After division you get more vertical torii from the torii. Remove them.
So the subsets are "all horizontal torii" and "all vertical torii"
(The things in that picture are what I'm calling level-1 tori) I don't think their union is the entire necklace. Which set has a horizontal level-2 torus contained in a vertical level-1 torus?
And if their union isn't the entire necklace, what guarantees that their union's complement is simply connected?
@AkivaWeinberger Oh, I guess that's a good point.
10:46
I think you're trying to find dense clopen subsets of the Cantor set
Can you even get dense clopen subsets of any closed set, actually?
Can you, like, take the two sets to be consisting of all horizontal tori ever, regardless of whether it comes from a vertical torus or not?
I am not sure
@AkivaWeinberger Are these subsets that I want dense? I don't think so.
@BalarkaSen If either one contains an open ball (intersected with the necklace), then that one contains some miniature version of the whole necklace, 'cause it's a fractal
So neither one can contain an open ball, which means each open ball centered at a point in one of the sets must intersect a point in one of the other sets.
(And they need to be closed because the original question asked for compact sets)
@BalarkaSen What do you mean? Like, a horizontal level-2 torus contained in a vertical level-1 torus should be in the "all horizontals" set? I don't think that makes sense. What about the point that's in a level-1 horizontal torus, a level-2 vertical torus, a level-3 horizontal torus, a level-4 vertical torus, etc., alternating the whole way down
I don't think that point can be assigned to either set easily
Yeah, so that won't make them disjoint.
Annoying.
Ok this is not getting anywhere, it just explodes

\begin{align}
I_1(0,x,n)=\int a^{(0)}(x)e^{-nx}dx & = a(x)\left(\int e^{-nx} dx\right)-a'(x)\left(\iint e^{-nx} d^2x\right)+a''(x)\left(\iiint e^{-nx} d^3x\right)-+a'''(x)\left(\iiiint e^{-nx} d^4x\right)+\cdots +(-1)^{R+1}\int a^{(R+1)}(x)\left(e^{-nx}\right)^{(-(R+2))}dx\\
& =\sum_{k=0}^{R}(-1)^ka^{(k)}(x)\left(e^{-nx}\right)^{(-(k+1))} +(-1)^{R+1}\int a^{(R+1)}(x)\left(-\frac{1}{n}\right)^{R+2}e^{-nx}dx\\
& =\sum_{k=0}^{R}(-1)^ka^{(k)}(x)\left(-\frac{1}{n}\right)^{k+1}e^{-nx} -\int a^{(R+1)}(x)\left(\frac{1}{n^{R+2}}\right)e^{-nx}dx\\
So I'm out of ideas. What's the answer, @Akiva?
10:54
@Secret What are you even doing
I should also note that I haven't tried to prove it, so it might as well be true.
Maybe by some M-V argument. Shrug.
@AkivaWeinberger That grew out of trying to get a continuous version of the method of snake oil, which previously, a overlooked careless mistake in the calculation result in it to look like a Larent series, which prompt my further curiosity. and then today recalculating it proved that nope, there is nothing nice in that sum
@BalarkaSen It's based on the Fox-Arron wild arc
Wikipedia's version is two-sided, so it has a nonsimply-connected complement:
Consider what happens if you take Wikipedia's version and skip a loop:
(Luckily, Google images has these images for me)
I think the latter is also simply-connected.
I actually thought about Fox-Artin a little but I couldn't come up with anything useful. How do you break it into two simply connected compact disjoint subsets?
what's the question you're talking about?

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