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12:00 AM
Alright, well, it is rather possible to reach that and stay at that level essentially
 
xrightarrow
works better than overset
 
Though in order to maintain it, it might be better to make a smaller change that you'll feel less, and then slowly let it set in
 
I like fruit so I've been eating fruit.
And not buying snacks at school.
 
That's a good way of going about it
 
And when I do eat junk, make sure I don't go above like ~1300 calories for that day.
@Dami So, what classes are you taking right now?
 
12:10 AM
This coming quarter, I'm taking 4 classes
First is the third quarter of Classics of Social and Political Thought, one of the core social science sequences
Second is Text and Performance, which I'm also using to satisfy a core arts requirement
Third is Intro to Differentiable Manifolds and Integration on Manifolds, which is going to use Guillemin and Pollack's Differential Topology
 
Oh, that's cool.
 
@Daminark One of these things is not like the others...
 
Fourth is the third quarter of analysis, which doesn't use a book, but last year covered a pretty wide range of topics
A lot of it is measure theory to start with (including some topics that are off the beaten path, like Banach-Tarski and topological games), then some functional analysis, and a bit of Fourier at the end
 
Sounds awesome. I'll be taking... sighs, Algebra 2.
 
LOL, poor, poor Zach.
 
12:15 AM
Though there's not much of a set curriculum for the class, they kinda want us over the course of the year to cover Rudin, measure theory, and functional analysis, and then whatever the professor feels like doing
Is algebra 2 where you learn rings?
 
@MeowMix Oh man, Dummit and Foote?
lol
 
No, it's where you learn how to do the sines and the cosines
 
That's not Algebra.
 
I mean in school.
And complex numbers
 
I know. But that course used to be called Trigonometry or Math Analysis.
 
12:16 AM
We don't do a lot of trig in geometry, the class I'm in for now.
All we learned was SOH-CAH-TOA, Algebra 2 covers unit circle stuff and what not
 
No, geometry shouldn't have trig in it.
Boy, they've really messed up the secondary curriculum.
But I knew that.
 
Well, we learned how to calculate angles and stuff
 
Oh @Fargle now I know why a lot of people don't like that book
At least for group theory
 
Meanwhile, the fifth and sixth graders I tutor have to do basic addition and multiplication on their fingers and usually mess it up.
 
It drags out f o r e v e r
 
12:17 AM
I can't put my finger on it, but I don't particularly care for it.
 
I haven't taught out of Dummit/Foote, but I like it for the exercises.
 
It's just a little obtuse to me.
 
But it is a huge tome.
It is more advanced than Artin, however.
 
Like, Laci said to go through either it or Herstein and to make sure I know the equivalent definitions of solvable groups, nilpotence of p-groups, and Jordan-Holder
 
@TedShifrin Most of what I learned about trig in high school came through advanced algebra and trig, which I had to take because graduation requirements in TN specifically for my year were messed up.
 
12:18 AM
I totally dislike Herstein.
 
Now, Herstein is beautifully compact, what I used my first time around (which is why Laci recommended it), but it
 
It makes algebra about manipulating symbols, and it divorces linear algebra from the rest of algebra. Integrating it beautifully is one of Artin's strengths.
Artin was a broad mathematician. Herstein was not.
 
's not merely about being concise, it loses some content
 
@Daminark Isn't every math book compact as long as you're not reading it? They're real, closed, and bounded.
 
smacks Fargle with a yard-long ruler
 
12:20 AM
@TedShifrin Wow okay, like you've never made a pun. :P
 
I have standards, @Fargle, standards.
 
Wait if we're not using the metric system then how can we even talk about Fargle being a yard away?
 
@Dami: yards are imperial.
 
Demonark: Whatcha talkin' about, boy?
 
We don't have a metric!
 
12:21 AM
Ahahahah
 
braces for smack
 
It's a "ruler"
It's "imperial"
 
lets the children play in their punbox
 
@Ted is running a math daycare in here.
 
When I was first learning Rudin topology I kept making so many puns and it was driving everyone crazy
 
12:24 AM
Earplugs are a wonderful thing.
 
A surefire way to protect yourself is Eyeplugs
 
"I hate it when my internet is the union of 2 disjoint open sets"
 
@Adeek A \xrightarrow{f} B becomes $A\xrightarrow fB$
 
@AkivaWeinberger yeah I got it
thanks
I used xrightarrow as well
 
Interesting ... I always used overset and underset.
 
12:29 AM
@TedShifrin I use white noise on my headphones when I am alway commuting so I can do math peacefully
 
The nice thing is that if the arrows get really big, you can swap those out with overproperclass
 
Karim: Peace is overrated.
 
@Adeek Wait aren't you always commuting in your algebra class right now?
OK sorry I'm gonna tone it down a bit
 
haha
 
@Daminark :D
 
12:31 AM
Can someone help me understand this answer what is $\overline{2}\in \mathbb{Z}_4$ in 1×1 matrix?
 
If anyone would like to check my question math.stackexchange.com/questions/2195887/…
 
It's not a matrix, @Ramanujan. Some people use $[x]$ to represent the equivalence class.
 
And why 2×2=40?
 
I fear that if I take a complex analysis course I won't be able to stop myself from humming the Polish anthem every time poles are brought up
 
I just want to show that hom is right exact functor with respect to left exact sequence sequences
 
12:32 AM
I don't know why you'd have an overbar and the brackets, though. Can you give me exact context?
DogAteMy: You'd also make sure your dog is well housebroken.
 
@TedShifrin your asking me ?
 
@TedShifrin ?
 
Don't want her leaving residues where they don't belong, DogAteMy.
Huh? Karim?
 
> If $A$ is invertible then from $AB=0$ you get $B=0$ by multiplying by $A^{-1}$ on the left on both sides.

Otherwise, consider the following example: $\overline{2}\in \mathbb{Z}_4$ is non-zero and it is a $1\times 1$ matrix. However
$$\overline{2}\cdot \overline{2}=\overline{4}=\overline{0}.$$
 
Ugh. :D
 
12:33 AM
sorry @TedShifrin mind going overdrive. Anyway I am gonna go I will do some work. brb
 
Complex analysis leads to a continual stream of puns
Eh that didn't work out perfectly well
 
artic it would be awesome to check my question when you come back..
 
Oh, I see. They actually did mean the matrix, @Ramanujan. But the elements of $\Bbb Z_4$ are written $\bar 0, \bar 1, \bar 2, \bar 3$.
@Ramanujan: They could also give $2\times 2$ matrix examples where $A$ and $B$ are nonzero and yet $AB=0$.
You should figure that out.
 
$\Bbb Z_4$ means matrix in 4-dimensional coordinates?
 
$2\times2$ real matrices rather than $\Bbb Z_4$ matrices?
 
12:35 AM
No, it means $\Bbb Z$ mod 4.
Right, DogAteMy.
 
@Ramanujan Do you know modular arithmetic? Are you familiar with the notation $\Bbb Z/4\Bbb Z$?
This is that
 
@AkivaWeinberger No,I will prefer next answer.
 
I'd usually see matrices as $M_n(R)$, where R is the set that entries come from
If a matrix is singular, it has no inverse
What's your definition of singular?
 
.__. inverse doesn't exist.
 
12:40 AM
That was right, @Ramanujan. If $A$ is not singular, $A$ is invertible, and $A^{-1}$ is of course invertible, too. What is its inverse?
I thought you said not singular.
 
It was edited. But yeah the reason I ask is that Laci defined it to be non-invertible, but I've also seen it defined as having a non-trivial kernel, in which case proving invertibility would require use of rank-nullity
 
That link went nowhere. I need to leave now. But you can ask me tomorrow.
 
Bye
 
@Adeek k
 
12:43 AM
See you @Ted!
 
I'll be out for a while, see you guys around!
 
1:01 AM
@AkivaWeinberger no, this link shows history of my deleted message.
 
1:12 AM
Hi
 
Hello frenz
 
@Daminark Do you want to consider a problem with me? I don't know a solution, maybe your insight can help.
 
Sure!
 
So, it's about finding a circle in $\Bbb R^2$ such that exactly $n$ points lie on it
And proving that for any $n$, such a circle exists
I've already proved it for $8 | n+4$... do you see how?
 
1:29 AM
Not quite
 
So, here's how I did it
Consider a circle with center at 0
OOPS! I meant "$n$ integer points"
Anyways, so with radius 1, you see how it's $4$, right?
We have $(0,1), (1,0), (0,-1), (-1,0)$
 
If radius is large,say $\to \infty$ then number of integers also $\to \infty$ (?)
 
Yeah
 
@Daminark Well, let's say we have radius $k$
The integers $(a,b)$ satisfying $a^2 + b^2 = k^2$ will be the only points, right?
 
That's right
 
1:36 AM
So, our 4 trivial points are $(0,k), (k,0), (0,-k), (-k, 0)$
The other integer points are all pythagorean triples whose $c$ is the same as $k$
And so each pythagorean triple with both non-zero integers with $c = k$ will actually add 8 additional points
 
@Ramanujan Doesn't work for me
Maybe it's only visible to the person who posted it
 
Because we can reflect over the x- or y-axes, giving us 4 distinct points, and we can also switch the coordinates, which will be the same because of addition's commutativity
So, if we prove that for every $n$, there exists a $k$ such that there are exactly $n$ non-zero pythagorean triples whose $c$ value is $k$, then we can QED, right?
 
Seems like it
 
QED means?
 
Quod Erat Demonstrandum
 
1:41 AM
It's what you say after a proof is done.
 
I used to think it was "Demonstratum" but I was wrong..
 
(Here, he's using it in a slightly strange manner, turning it into a verb.)
 
And when to use "hence proved"?
 
(As Calvin once said, "Verbing weirds language")
 
@Akiva likes being critical of me. :P
 
1:43 AM
@Ramanujan They're the same
@MeowMix I'm not being critical. I just felt like it might be confusing to a non-native English speaker without an explanation.
I fully believe that anything's a verb if you verb it.
 
I was joking
 
That's cool!
 
What circle has only three points on its circumference?
 
Take a set with 4 points and the discrete metric
 
Oh, you don't know the context
I meant three lattice points
 
1:49 AM
I know I know
I was trolling
My immediate inclination would be to play some weird game with partitions of unity
 
@MeowMix You never finished this
 
Hello
Does anyone have recommendations for basic to intermediate number theory books?
 
Basic Number Theory by Weil
snicker
JK you'll want number theory for beginners by Weil
 
Thanks!
 
2:06 AM
I am troubled by the expectation that everything complex should by typed out in mathjax
It's almost unreadable when a subscript falls within an exponent
 
Even real analysis questions should be in mathjax as well :P
Jk
But really like, I've found it alright to do subscripts in exponents
 
must be a font thing
 
And @Freeman Basic Number Theory scared me when I first opened it, starts with some madness on locally compact fields
$e^{f_n}$
Alright I can see how this is a bit iffy, but I dunno
 
Maybe you can find a better form for my infinite sums here math.stackexchange.com/a/2195569/72719
 
2:25 AM
@AkivaWeinberger my phone died
 
Take $n$ pythagorean triples whose $c$ values are prime and distinct. multiply each of them together to get a value , and then multiply each triple so that the c equals this value
 
why?
 
By this, we'll have exactly $n$ unique triples which have the same $c$ values and no more
 
um.... ok? Thanks...
 
2:35 AM
@TheGreatDuck He was continuing a conversation that he and I were in earlier
@MeowMix Are you sure that works? Take $(3,4,5)$ and $(5,12,13)$. You want $c=5\cdot13=65$, giving us $(39,52,65)$ and $(25,60,65)$
But how do we know that these are the only triples with $c$-value $65$?
In fact, $(16,63,65)$ is another triple.
@MeowMix So that doesn't work.
 
@AkivaWeinberger I thought he was talking to me.
 
I invoke Hahn-Banach and extend this question to: "Can Richard funktional?"
 
Suppose that for $n \geq 1$, $X_n$ is uniformly distributed
on {1, 2, ..., n}. how to show
$\lim\limits_{n\to\infty}P(\frac{X_n}{n}\leq y)= y$ for $y\in(0, 1)$.
 
3:21 AM
@AkivaWeinberger so what is answer? Are there "n" integer points on circle or infinite?
 
@Fawad What? A circle can never have infinitely many lattice points. The question was if, for every $n$, there's a circle that has $n$ points on its circumference.
@Daminark Also
 
anyone still here?
 
3:40 AM
@TheGreatDuck yes
 
@Fawad do you have any inkling what might be wrong with this answer math.stackexchange.com/questions/2104702/…
?
i mean, there are minor arithmetic errors, but it seems like someone is claiming there to be a fundamental flaw?
 
I'm back guys!
@Akiva l m a o
 
@Daminark hi. How's it going?
 
Everything's aight, how about you?
 
3:57 AM
All my codes are currently working, thus the task should be done in a few days.
@TheGreatDuck what maths you are currently investigating?
 
4:13 AM
@Secret why do you presume I am "investigating" any Maths? :p
 
well, that's mostly why we all hang around here in this chat room, : D ?
 
no...
we hang around the chat room to chit chat
besides... we cannot possibly be all engaging in research.
 
\begin{align*}
\varphi_{X}(t)&=\int_{-\infty}^{\infty}\frac{e^{ixt}e^{-(x-\mu)^2/(2\sigma^2)}}{\sqrt{2\pi}\sigma}\,dx\\
&=\int_{-\infty}^{\infty}\frac{e^{i(u+\mu)t}e^{-u^2/(2\sigma^2)}}{\sqrt{2\pi}\sigma}\,du,\;\;\;\,u=x-\mu\\
&=\frac{e^{i\mu t}}{\sqrt{2\pi}\sigma}\int_{-\infty}^{\infty}e^{iut-u^2/(2\sigma^2)}\,du\\
&=\frac{e^{i\mu t}}{\sqrt{2\pi}\sigma}\int_{-\infty}^{\infty}e^{(2\sigma^2iut-u^2+(\sigma it)^2-(\sigma it)^2)/(2\sigma^2)}\,du\\
&=\frac{e^{i\mu t}e^{-(\sigma it)^2/(2\sigma^2)}}{\sqrt{2\pi}\sigma}\int_{-\infty}^{\infty}e^{-(u-\sigma it)^2/(2\sigma^2)}\,du\\
 
In the immortal words of Theodore Roosevelt. imgur.com/a/fCT3T
 
there's no image...
 
4:15 AM
well, I also count in personal research (those that has no ties with the research institute's goals), but yeah you had a point there
 
my point still holds
i was not referring to formal research
 
ok
 
are you engaging in any research? :)
 
Well for starters:
Doing quantum chemistry calculations in my PhD

As for the maths side, it is mostly personal research. Currently thinking about topology, symmetries of integrands and infinite sets
 
"symmetries of integrands"?
you've perked my interest.
 
4:22 AM
I am pretty sure formally it wil be studied as part of algebraic and complex varieties (alien terms that I don't really know what they mean). However I have used basic calculus to play around with integrals and in turn, learnt a bit about how the integrand controls some integral's results
 
i don't know what you mean by symmetries.
but integrand perks my interest
@jmac hi
 
For starters, $\int e^{nx} dx = \frac{1}{n}e^{nx}$. If you think of $e^x \in \mathcal{C}^{\infty}(\Bbb{R})$ as a vector, then the integral operator is like a functional. Thus for this example you can read that $e^{nx}$ is an eigenfunction of integration with eigenvalue $\frac{1}{n}$
 
whoa
mind = blown
 
Generalise further, you can derive quite general looking formulae like these:
Mar 10 at 13:24, by Secret
Jan 21 at 16:06, by Secret
[Integral symmetries] Cauchy Riemann theorem like analogues in reals:

$$\frac{d^n}{dx^n}f=\lambda\int^{(m)}fdx$$

(where $f=\sin, \cos,e^x$ (possibly more?),$\lambda \in \mathbb{R}$)

For example

$$\int x^n \sin x e^x dx$$
Consider

$(uv)''=u''v+2u'v'+uv''$

Plug $u=\sin x$, $v=e^x$. Then

$(uv)''=-uv+2u'v+uv$

Note how the antisymmetry of $\sin x$ wrt the functional $()''$ results in cancellation. Hence

$(uv)''=2u'v$

Now it is easy to check that $\cos x$ is also antisymmetric wrt $()''$ thus we could have start with $u' =\sin x\implies u = -\cos x$. Hence
 
@Danu @SteamyRoot How does TikZ compare to PSTricks
 
4:25 AM
So that explains why some integrals are easy, because they "cycle" around via differentiation
 
Also, I found this PSTricks thing that draws the Fox-Artin wild arc (though each loop is kinda lopsided)
 
That is, just like in complex numbers, for some integrands, integration is like differentiation to them
 
@Secret Never thought of it that way.
That's cool.
 
@Secret "just like in complex numbers"?
i've never done calculus with complex numbers
 
@TheGreatDuck Well in complex numbers, for any holomorphic functions, you have the cauchy integral theorem
 
4:28 AM
uuuh
ok
sure
 
My latest result in the investigation (though so far only checked for a(x)=polynomials, sin nx, cos nx) is the folllowing:
yesterday, by Secret
Akiva: Actually, not all is lost, if $n>1$, $a(x) \in \mathcal{C}^{\infty}(\Bbb{C}or \Bbb{R})$ and $\lim_{R\to\infty}a^{R+1}(x) < \infty$ (These conditions kill off the $R+1$th term), then

$\int a(x)e^{-nx}dx=e^{-nx}\sum_{k=0}^{\infty}-\frac{a^{(k)}(x)}{n^{k+1}}$

So in a sense one can pull out the exponential from the integral, converting it into an infinite series of derivatives of $a(x)$ in the process
That explains quite a lot of integrals involving simple exponentials in the table of integrals (though the trigonometic a(x) cases are a bit shaky cause they somehow can converge even if their geometric series should not, and I am tyring to understand why)
 
well I don't have anything that advanced. I suppose I have something from a while ago.
$y^{\to} = \{x | \lim_{h \to 0^+} \frac {y(x+h) - y(h)}{h} = x \lor \lim_{h \to 0^-} \frac {y(x+h) - y(h)}{h} = x\}$
now
 
y tends to x iff its left and right limit agree which each other?
 
let $f(x,y,y^{\to},y^{\to\to},\dots) = 0$ and $f(x,y,y',y'',\cdot) = 0$ be two differential equations. It is a conjecture of mine that the solution set of the first is a super-set of the second and that all continuous solutions of the first are solutions of the latter.
@Secret no. That's a solution set of a different type of derivative of y at x.
rather than be the intersection of the solution sets of the left and right hand limits, it is the union of the solution sets of the left and right hand limits
try putting a step function into it and see what you get. You'll get $h^{\to} = 0$ whenever h is a step function.
@Secret you still here?
 
I am, I am digesting
 
4:41 AM
alright
 
Ah I see, so your conjecture relates the solution sets of different types of differential algebras
 
it should be noted that solutions to the equation $y^{\to} = f(x)$ vary by step functions rather than constants. This is why I said "continuous solutions"
@Secret precisely. And the truthfulness of it would trivialize differential equations involving step functions. At that point, most of them would be solved as if they were constants. The only challenge would be seeking out the continuous solutions.
 
how is "continuous" defined in your investigation, is it still the usual no jumps, $\lim_{x\to a} f(x)=f(a)$ or something different such as they can differ by step functions (i.e. jumps are not considered as discontinuity)?
 
no. the definition of continuity never changes. I'm not bending the rules. I'm just relating two different types of operator equations.
 
Ok I see
 
4:45 AM
the reason I say continuous solutions is because all solutions to differential equations are continuous.
and even if it were only true, for lets say, linear differential equations, that would still make most of equations people use much easier to solve (assuming they were solvable beforehand).
 
yup, I can also see that it will be quite useful to solve differential equations with discontinous forcing terms because in the other differential algebra, it will act like a constant term
 
Maybe at some point I'll ask TeX Stack Exchange to turn my hand-drawn image into a computer generated thingy
 
@Secret exactly. One would just use the method of undetermined coefficients. Far easier than a laplace transform.
 
There's something mezmerising about the way it tails off in those 4 directions
but I don't know how to describe that
 
this is a view of the real number line sitting above a reflective pool...
 

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