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12:00 AM
@0celo7 Yeah I mean only math people would potentially deny that math is science but that's the standard thing
 
i used the $ x^2 + y^2 = (13^2) $ and took the derivative of the entire expression with respect to time
does that make sense?
 
@BalarkaSen Who thinks math is science?
 
Akinanator asks pretty generic questions. It's not so amusing anymore
@0celo7 Katy Perry?
 
@0celo7 One thing that does take getting used to about the pilot wave story is how bound states behave
 
@Balarka i got grothendieck
 
12:02 AM
@Eric oh damn how
 
german speaking, french, researcher, been dead for less than 20 years probably narrowed it a lot
 
suppose I have a particle in a box. if you've got an energy eigenstate, then by definition it's got a definite expected energy; since the particle experiences zero potential inside the well, it's specifically got positive expected kinetic energy
 
it just asked dead but not less than 20 years for me. interesting
 
So if you were going to think in terms of trajectories, you might expect that the particle is bouncing back in forth inside the well
 
but yeah he's doing the generic narrowing thing
the first few of my tries were so common that he got in a few tries and asking apparently random questions
i can see the algorithm now
 
12:04 AM
im shocked by the fact that it got to grothendieck so quick for me, strikes me as a niche character
 
but the velocity field in pilot wave theory is just current / density, and the probability current of an bound energy eigenstate is zero
 
@EricSilva Did you get the insane question?
 
so the particle trajectories for any bound energy eigenstate just consist of "the particle doesn't move"
 
what's the insane question
 
@EricSilva "Is your character insane?"
 
12:05 AM
oh lmao
 
for Grothendieck, idk what the answer is
he wasn't all there
 
i would say yes
 
me too
 
okay, i'll bite. wtf is this
 
@Semiclassical yeah that's how I think of the particle in a box tbh
@Semiclassical yeah that's how I feel like I'm supposed to think about it
 
12:07 AM
yeah
 
@Semiclassical like we learn that an electron has no velocity, but then atomic physicists treat it as a billiard ball
 
i don't have books on this stuff, so presumably that'd be explained in such sources
 
I don't really know what's going on
 
I did come across this sentence in one place: " Pilot wave trajectories are not well suited to bound state problems, which are really steady state phenomena."
which would sorta explain it
I suspect it comes down to what it really means, experimentally speaking, to have a system described as the bound state of some potential
 
@Semiclassical 20 questions
 
12:11 AM
and if that's the case, that to me is actually a vote in favor of the pilot-wave mindset: it forces one to think more carefully about what the simple textbook calculations actually correspond to
gotcha
 
i wonder if arkinator can get the composer kid of cartan
 
Hi, I think I need to clear up some definitions. The question is: The set X contains 6 vectors (1,1,0,0) (1,0,1,0) (1,0,0,1) (0,1,1,0) (0,1,0,1) (0,0,1,1). Find two different subsets Y of X whose member are linearly independent, *each of which yields a linearly dependent subset of X whenever any element x in X with x not in Y is adjoined to Y. *
The sentence within * * is what I don't understand
 
so Y is linearly independent, but if you put one more vector from X into Y, then it becomes linearly dependent
 
wow that clears everything up
 
howdy fellaz
 
12:39 AM
hi @Meow
 
hi ted
wanna see some cool code
too bad youre gonna see it
sending email right now
 
1:38 AM
hey how do I derive the lengths of the sides of the special triangles for example the $30$ , $60$, $90$ triangle from scratch?
 
@CausingUnderflowsEverywhere reflect along one edge to form an equilateral triangle
 
Can someone explain how I do this...?
I can't figure out the f(x,y) form
or how i factor in the (2,-3,9) into it.
 
@10Replies what have you tried?
have you tried drawing a diagram?
 
The equation for a plane is n.(x - p) = 0 where n is a vector normal to the plane and p is a point on the plane.
Because the vector from a point on the plane (p) to any other point on the plane (x) must be perpendicular to the normal of the plane (n).
What are p and n?
 
1:54 AM
n is the slope, P is the offset?
 
n is a vector normal to the plane, p is a point on the plane
n is the direction the plane is 'facing'
 
oh that "." is a dot
 
Yes, the dot product
Two vectors a and b are perpendicular when a.b = 0
 
What is x?
 
Like I said, any other point on the plane
So any other point on the plane is going to satisfy that equation (when substituted into x)
 
1:58 AM
Ok, so how do I get from that to z= f(x,y) = ....
 
Thanks @LeakyNun so I set each side to a variable $x$ then split the equilateral triangle being left with two 60 angle and two 30 angles, so the long sides remain $x$, the shorter sides of the base where it has been split are each $(1/2)x$ and to solve for the remaining side I can use pythagorean theorem. and if I have anymore questions to go more in depth I'll have to find out how pyth theorem is derived. Is that all correct?
 
Solve for x. I guess I should call it q to avoid confusion with the x coordinate. So n.(q - p) = 0.
 
@CausingUnderflowsEverywhere yes
 
You know n and p. What do you get for q?
 
30(x-2)+22(y+3)+26(z-9)
=0?
 
2:02 AM
I got 10 replies??? sweet!
 
@10Replies You got it
 
@PhysicsGuy yes they should rename stackexchange to everything stackexchange because you can find the answer to everything on it
 
@user76284 Noice.
 
2:23 AM
hey @MeowMix I want to see some cool code. what language is it in?
 
can a sequence be represented by a real number? I originally had this inequality $||(\alpha d_1+\beta d_2)-x||<1/n$ and then I wanted to change the real numbers $\alpha,\betha$ by the sequences $||(\alpha_{1/n} d_1+\beta_{1\n} d_2)-x||<1/n$. Can I do that? Is it wrong?
 
2:41 AM
moprning
@TedShifrin yeah i figured that out had a friend of a friend ask if i could tutor calc I though oh sure should be fine had no problems with all the other questions but that one i got to that differentiation and was like this looks way too hard i eventually figured out cos and it was really easy.
Anyway will show him the easy way when we meet up on wednsday
@BalarkaSen that quote has the best since the semester started in chat. @TedShifrin
 
 
1 hour later…
4:17 AM
@Faust: Don't forget basics with product and quotient rules!! :) ... You mean the middle name one? :)
 
@TedShifrin yeah its hard to tell with implicit differentiation wether i need to use the product rule if i dont think about wether its dependent
but i figured it out
 
Totally irrelephant. You must always use product rule.
 
@TedShifrin yeah the middle name thing made me laught so hard
let x be dependent on time
 
Also, be careful about parentheses so that multiplying by a negative doesn't look like subtracting. You remember what I'm talking about?
 
let y=4
 
4:19 AM
No. You substitute values only after diff is done, unless y is literally constant throughout.
 
whats the derivative with respect to time of xy
no im defined y to be 4
always non dependent on time
yeah
lol\
 
Then don't even write y. Write 4.
 
there is a diffrence though
i just didnt think bout it enough
 
But try to teach your tutee to think about how to approach it to make it least complicated, not most.
You have to be on top of things to do that, but you need to try.
 
i asked him to send me the questions before we met but he didnt for our first lesson
 
4:20 AM
Excellent idea.
 
i havent done calc I in 9 years
i was impressed i was able to answer all his questions without looking anything up O.o
 
That's probably part of your problem with diff geo ... you need to be good at Calc 1 as well as 3 !!
 
im super awesome at 3
 
You're too old, man. Almost as old as I am! :D
 
not reall im still in my 20's
i can practically speak in trig
well most of 3
 
4:22 AM
We did law of sines and cosines in class yesterday. Next week is addition formulas :)
 
eveyrhting after vector stuff
 
Calc 3 has nothing to do with trig. I'm confuzled.
 
you have to remeber ted i was working full time so i never actually attended a single calc I,II or III class
i just did the assined hw and random stuff from my textbook
i took days off to write midterms and that was it
 
So a lot of understanding (as opposed to just cranking problems) you really need to work on learning at this point. I learned so much in classes ... compared to the books.
 
i have flled in alot of the gaps lately
 
4:24 AM
OK, well, ask some of us if you need some help explaining some of the stuff you need to explain to your tutee.
 
this summer i re did ALOT of calc and algerba
Eventually i wanna get to the point where i can take a real whack at Diff geo again
 
I had a few students in my diff geo who couldn't do Calc 1 correctly, even though I assigned stuff for them to work on to learn it again.
 
also in physics multivariable
you do get like x is dependent on z and t
y dependent on z and x
but z independent of time
so when u take the derivative of z with respect to time
its 0
 
sure
You could have $y=y(x(z,t),z,t)$.
Just to make it more realistic. :)
 
exactly
yeah
lol
 
4:27 AM
It's the best way to do it and it's a hell of a lot less writing. Best linear approx of composition is composition of best linear approxs.
 
i'll confess, I never learned how to do the multivariable chain rule in terms of the Jacobian
 
Weird.
Out of place. Oh, I realize why.
 
anyway my mistake he says he gunna send me the questions in advance this time so i can make sure i dont use $\tan \theta =y x^{-1} $ lol
 
so I'd just write $\frac{dy}{dt}=\frac{\partial y}{\partial x}\frac{dx}{dt}+\frac{\partial y}{\partial t}$
assuming I haven't forgotten something obvious...
 
lol i think i do the same thing
anyway i should sleep thanks @TedShifrin
 
4:31 AM
Night, @Faust.
I truly hate that notation, Semiclassic, because the $y$ on both sides of the equation represents different functions completely. Just like your $f(x,y)=x^2+y^2$ what's $f(r,\theta)$ issue? Or was that someone else's?
 
yeah, that's me
i'm fond of that one
i come down on $f(r,\theta)=r^2$ for that :P
 
But it makes my point why that standard notation is completely horrible.
You have to use a different letter than $f$.
 
I tell you. This confuses students no end.
 
in my first PDE class we made a big deal out of it
when going to polar coordinates, etc. one needs to keep these things straight
 
4:33 AM
Yup.
 
I mostly dislike it because it makes little sense physically to distinguish between, say, the electric potential written in polar coordinates versus Cartesian coordinates
 
I always wrote $F(r,\theta) = f(r\cos\theta,r\sin\theta)$, etc. Most of my students did as I did.
They're different functions, dammit.
The output is the same if you put in corresponding inputs, but truly this is horrendously crap in calculus books.
 
i'll agree to the following: $f(x,y)=x^2+y^2$, $f(r,\theta)=r^2$ makes no sense if you think of it as a literal function of $r,\theta$
 
It's not going to be productive to discuss this.
You're spewing crap.
I'm no happier with $f(x)=\cos x$ and $f(t)=\cos(t^2)$.
Equal crap.
 
One place where I do tend to put my foot down is with Fourier transform pairs
Going from $f(x)\to f(k)$ is just a bad idea
 
4:37 AM
I'm done with this. Seriously.
 
in other news, class groupmes are so strange
I never have the issues other people do on the homework
 
for Fourier transform pairs I typically do $f(x)\leftrightarrow \tilde{f}(k)$
 
instead I spend hours on some other thing no one else had an issue with
 
why doesn't text get converted to latex in chat
 
4:46 AM
read the upper right corner...
 
 
3 hours later…
7:30 AM
can someone tell me how can we simply just forget the set C when we are taking the pullback of g along f (f: A --> C, g:B-->C)? It seems like we should take first the set AxBxC and then take a subset of it?
 
Well you look at the subset of A x B consisting of pairs (a, b) such that f(a) = g(b).
 
I know that's the answer and i know that inverse limit is unique, so we are done
but see page 77, it seems like we should have first take AxBxC and then do something
because C was an object in our free diagram and taking the pullback of this diagram, we must have a subset of AxBxC
Ahh, I think I get it: They are actually isomorphic, one takes the C component as f(a) (or equivalently g(b)), so there is no new information on the C component, every tuple is (a,b,c), where f(a)=g(b)=c, so c is fixed
Am I right?
 
Hello!!

I want to calculate a triple integral over the space $D=\{(x,y,z)\mid |x|\leq 1, |y|\leq 1, z\geq 0, x^2+y^2+z^2\leq 1\}$.

We have the following:
- $|x|\leq \Rightarrow -1\leq x\leq 1$
- $|y|\leq \Rightarrow -1\leq y\leq 1$
- $x^2+y^2+z^2\leq 1 \Rightarrow z^2\leq 1-x^2-y^2\Rightarrow -\sqrt{1-x^2-y^2}\leq z\leq \sqrt{1-x^2-y^2}$, since $z\geq 0$ we get $0 \leq z\leq \sqrt{1-x^2-y^2}$.

So that the square root is defined it must hold that $1-x^2-y^2 \geq 0 \Rightarrow x^2+y^2\leq 1 \Rightarrow x^2\leq 1-y^2 \Rightarrow -\sqrt{1-y^2}\leq x\leq \sqrt{1-y^2}$, right?
 
If you define your relation in $\mathbb{R}$ as $xRy$, if $x-y \in \mathbb{Q}$. Why every equivalence class has an element in $[0,1]$?
 
Think about the decimal expansion
 
7:43 AM
@Kirill Because $\mathbb{Q}$ is dense in reals.
 
if I knew what that is :) @TastyRomeo
 
$\pi = 3.1415....$
It's clearly equivalent to $0.1415...$, no?
 
sure
 
The same holds for any real number...
 
I just cannot find any mathematical proof for this statement, even it seems clear.
The statement was given just as it is in the first lecture about measures.
 
7:46 AM
Just, like, write down the idea of what I said?
 
@KonformistLiberal (Sorry, my internet momentarily disconnected) I don't see where in pg 77 you want to look, but no, that's not how pullbacks should work. Think about a fiber bundle $p : E \to X$ and a continuous map $f : Y \to X$. Pullback of $p$ along $f$ gives rise to a new fiber bundle $f^* p : E' \to Y$ and the fibers of the total space $E'$ over $y$ should send to the fibers of $E$ over $f(y)$, hence the relation $f(y) = p(e)$.
 
ok, I will :) thank you, @TastyRomeo
 
@Balarka Sen Prop 6.16
look at the embedding of the inverse limit
(first part of the proposition)
I am actually right now sure about my thoughts, it also works for pushout
 
@KonformistLiberal wow, the theorem comes right now in the lecture
 
@Kirill I am also taking a really similar course, and probably we began a week earlier :)
 
7:58 AM
@KonformistLiberal I aam reading about the reals the third time now, it wonders me that they are introduced differentyl each tme :)
 
@Kirill Haha :) . Check amazon.de/Surreal-Numbers-Donald-Knuth/dp/0201038129 this out, for a (really) different and fun construction
 
@KajHansen Hey!
 
Hi @kaj long time no see
 
Hey Balarka. Have you started in on IUTeich yet? ;)
 
8:08 AM
Soon, soon
 
Hey there Alessandro
I'm trying to see if Newton ever calculated the polynomial expression for x^6 + y^6 + z^6 in terms of elementary symmmetrics
because he did an enormous number
 
@BalarkaSen What do you think about my question?
 
I didn't know that Newton proved that $k[x_1, \cdots, x_n]^{S_n}$ is generated by the symmetric polynomials.
 
He actually did
 
@Konformist So you wanted to look at the subset of A x B x C consisting of triplets (a, b, c) such that f(a) = g(b) = c?
 
8:12 AM
Newton made incredible strides in the study of algebra
most people don't know of it
 
Because that's how the inverse limit of the diagram embeds in A x B x C, according to nlab I guess
 
nlab <3
 
lmao
@Kaj Oh I know the Newton's identities
But that's just for symmetric polynomials of the form x_1^d + ... + x_n^d isn't it
 
8:14 AM
$p_k(x_1,\ldots,x_n) = (-1)^{k-1}ke_k(x_1,\ldots,x_n)+\sum_{i=1}^{k-1}(-1)^{k-1+i} e_{k - i} (x_1, \ldots, x_n) p_i(x_1, \ldots, x_n)$
huh, mathjax rendering doesn't work for me anymore
p_k here is $\sum_k x_1^k
well, \sum_i x_i^k
 
You should use the new link by robjohn. The older version of mathjax expired
 
ahhh, there we go
 
@KonformistLiberal You're probably right that A x_C B and your thing are isomorphic. I mean it's (A x_C B) x_C C where the pullback is taken with the projection F : A x_C B --> C and the identity map i : C --> C
Because, you know, it's the tuple ((a, b), c) such that F(a, b) (= f(a) = g(b)) = i(c) = c.
But that should just be isomorphic to (A x_C B) because you took pullback with the identity map.
@KajHansen I think the most general theorem along the lines is that for any finite group $G$ the invariant subring $k[x_1, \cdots, x_n]^G$ is finitely generated.
Given an action of $G$ on $k[x_1, \cdots, x_n]$, I meant. Sorry.
 
It can probably be generalized further :P
It always amazes me just how far generalizations can be taken
 
Maybe it can be generalized to higher toposes.
 
8:25 AM
Let's query nlab
 
nlab you're slacking
where are the infinity-topoi?
where is the n-point of view?
 
hahahaha "n-point of view"
the $\aleph_0$ point of view
 
that's literally the slogan of the site
 
haha, I know. It makes me crack up every time
 
true
I was trolling Ted with nlab a few days back
Sep 15 at 22:27, by Balarka Sen
"One way to exhibit this statement nicely is:

A differential n-form on X is a smooth n-functor $P_n(X) \to \mathbf{B}^n \mathbb{R}$ from the path n-groupoid of X to the n n-fold delooping of the additive Lie group of real numbers."
 
8:29 AM
So the vitali set $V$ is not measurable, i.e. there is a set $A$ such that $\mu^*(A) \ne \mu^*(A \cap V) + \mu^*(A \cap V’)$?
 
8:45 AM
@BalarkaSen, this probably has a generalization to rational functions instead of just polynomials
 
@KajHansen Hmm
Other than the obvious and boring generalization to denominator/numerator being symmetric, I don't see how
 
9:36 AM
[Random]
The function you hate to encounter in social context: $\text{sgn}(-x)$
 
?w?
 
The graph of the function is like a step, positive chemistry with other people are y values above the x axis and negative values below the x axis
 
chemistry ?
 
So $\text{sgn}(-x)$ is a fancy way of saying abrupt social connections turned sour
 
time ?
 
9:42 AM
yeah
Btw, chemistry:
In the context of relationships, chemistry is a simple "emotion" that two people get when they share a special connection. It is not necessarily sexual. It is the impulse making one think "I need to see this [other] person again" - that feeling of "we click". It is very early in one's relationship that they can intuitively work out whether they have positive or negative chemistry. == Definition == While the actual definition of chemistry, its components, and its manifestations are fairly vague, this is a well documented concept. Some people describe chemistry in metaphorical terms, such as "like...
 
hides away
 
I am those group of people that do't understood social interactions well. Besides learning from others, I make various plots of my chemistry with others and also analyse group discussions
What I found interesting is that the topics covered by my social groups are more or less independent of how emotionally close I am to other people
therefore, I suspect the notion of closiness in my social group may be more to do with memories and how long I knew the person
There are still a lot of open questions about that, and I hope some social science literature may help to shed some light on it
I won't say mathematics can fully model something as complex as emotions, but there are other ways to learn more about it
 
 
2 hours later…
user84215
11:53 AM
@mixedmath What will (or may) happen If I insist on rolling back?
 
@MathematicsAminPhysics If you roll it back, then I will remove it again and kick you from the room
4
 
user84215
@mixedmath Can you delete your messages from those rooms?
 
@MathematicsAminPhysics I believe you can delete them as room creator, right? Feel free to do so
 
user84215
@mixedmath I can not delete them; I can only move them to the trash room. But I want them to be completely deleted.
 
12:09 PM
[Random]
One way for me to discover my implicit bias is to do the same test again after one year, and noticing how I got the exact same questions wrong, be it maths or human ethics
It does make me wonder, whether what people called destiny is basically implciit bias not made aware , causing you to retrace history and do the same thing again and again
However, I have a deeper problem: I am kinda aware of what my implicit bias are, but I still retracing history like crazy
(This is the first [random] that does not involve maths, perhaps next time I should just state it instead of confusing the curation with the [random] tag)
 
12:24 PM
Quote: "Recall that $\Bbb{R}_K$ denotes the real line in the $K$-topology. Let $Y$ be the quotient space obtained from $\Bbb{R}_K$ by collapsing $K = \{1/n ~|~ n \in \Bbb{N} \}$; let $p : \Bbb{R}_K \to Y$ be the quotient map".
Okay. That seems pretty nondescript...What does the rest of $Y$ look like? It isn't even clear what this "collapsing" amounts to. What exactly is meant?
 
what's the $K$-topology ?
 
The $K$-topology consists of all the standard topology plus sets of the form $(a,b)-K$.
 
Can anyone explain how you Z-transform 3^-n?
 
I prefer the W-transform
is $A-B = \{a-b \mid a \in A, b \in B \}$ ?
 
It's supposed to be solved using the Z-transform
 
12:35 PM
well "solve" would mean there is a question
or a problem
 
Well, we should get the same results as mathematica. That is: 3z/(-1 + 3 z)
I know how to solve the regular a^n. Now the exponent is negativ and I'm not sure how to solve that.
 
maybe use that $3^{-n}$ is $(1/3)^n$ ?
 
I've tried that without success. Do one need to z-transform the 1 aswell or is it enough to z-transform 3^n?
 
I have no idea what you're talking about though
 
If i Z-transform both the denominator and voter. I get this: (z/(z-1))/(z/(z-3))
 
12:42 PM
what the hell is a Z-transform
 
In mathematics and signal processing, the Z-transform converts a discrete-time signal, which is a sequence of real or complex numbers, into a complex frequency domain representation. It can be considered as a discrete-time equivalent of the Laplace transform. This similarity is explored in the theory of time scale calculus. == History == The basic idea now known as the Z-transform was known to Laplace, and it was re-introduced in 1947 by W. Hurewicz and others as a way to treat sampled-data control systems used with radar. It gives a tractable way to solve linear, constant-coefficient difference...
some kind of fourier transform like operation
 
can't he just replace $a$ with $1/3$ and call it a day ?
 
Never mind
solved it
 
so the Z in the name is just because you want a fancy letter ?
like in "K-mean clustering" ?
 
mercio
that's what i did. I totally missed that
 
12:49 PM
wait so the problem is simply to compute the power series $\sum 3^{-n} z^n$ in terms of $z$ ?
 
1:03 PM
Man I have a terrible misreading problem: Consider the following:
Oct 14 at 8:25, by Alessandro Codenotti
In such a model the irrationals will have countable subsets, but also a subset $X\subseteq\Bbb R\setminus\Bbb Q$ that has no countable subsets
This is what I read:
 
@mercio it’s pretty much just a generating function
 
In such a model the irrationals will have countable subsets, of which but also a subset $X \subseteq \Bbb{R \setminus Q}$ that has no countable subsets
 
The main difference is that you consider Laurent series rather than power series
So X(z) will be analytic in some annulus in the complex plane
 
hola
 
and the terrible thing is, until someone point that out for me, I never aware I have misread. It's as if I am living in another universe seeing different things and hallucinating without knowing
 
1:06 PM
tbh that sentence is a bit hard to parse @Secret
am i a bad guy if I really don't like the name "Z-transform" ?
also in his case, the thing doesn't converge for any $z$ ?
if it's two-sided
 
Again, think Laurent series
 
are we talking about $f(z) = \sum_{n \in \Bbb Z} 3^{-n}z^n$ ?
 
Hmm
Pretty sure that shouldn’t make sense anywhere in the complex plane
I mean, it would seemingly have to satisfy z*f(z)=3f(z)
 
$f(z) = 0/(z-3)$ o..o
 
Which only can be true for all z if f(z)=0
 
1:12 PM
this reminds me of my favorite thing about $p$-adics numbers
$...1111.1111... = 0$, and so $...1111 = - 0.1111... = -1/9$
 
Yep
I actually have seen stuff like that show up in physics, though in a decidedly handwavey way
 
1=0.9999999....
0=...111.111...?
 
nods
I think I wrote an answer about that for fun once
 
hmm, so -1/9 in p adics has a very "large" representation
 
blah, this is going to bother me until i track down what I'm remembering
is that for $p$-adics in general? i'd have expected it's only true for a specific $p$
 
1:20 PM
well i kinda used $p=10$ up there, don't eat me
 
lol
i figured
 
zero divisors break fields
so p adics has to be based on prime bases
(plus other reasons I forgot)
 
I like how in $10$-adics, there are $4$ solutions to $x^2 = x$
0
A: Relating the base-p periodic expansion of a rational to its p-adic representation

mercioif $r$ is rational whose denominator is coprime with $p$ then it can be written as $r = a/(p^n-1)$ for some integers $a$ and $n$ and it gives you a relationship between $r$ and $p^nr$. If you write it as $r = ap^{-n}+rp^{-n}$, you get a way to write $r$ as an infinite sum that converges in the u...

apparently i didn't read the question completely
 
Knock knock
 
1:36 PM
Guys, my book says the following: Let $R$ be a commutative ring, $\alpha\in R$, $f\in R[X]$. There there exists a $q\in R[X]$ with $f=q(X-\alpha)+f(\alpha)$. We know that $f=q(X-\alpha)+r$, with $\operatorname{degree}(f)\leq 0$. Thus $r$ is a constant polynomial; $r\in R$.
I don't see where they use the fact that $R$ is commutative
However, that does seem to be a requirement later on, because $X^2+1\in\mathbb H[X]$ has infinitely many roots according to my book, because $\mathbb H$ is not commutative.
or maybe they use it in the theorem that builds upon this one
Let $R$ be a domain, let $f\in R[X]$, and let $\alpha_1,\dots,\alpha_n\in R$ be distinct roots of $f$. Then there exists $q\in R[X]$ with $f=q(X-\alpha_1)\cdots(X-\alpha_n)$.

In the proof I only see them use the fact that $R$ has no zero divisors, but I don’t see where they use commutativity. If I have to write out the proof, let me know.
ah, never mind, I asked someone.
 
1:55 PM
@Leaky you have time?
oh never mind:P
 
Please take a look for this question here: Let $\cal A$ be subset of power set $X$, and union of elements $\cal A$ need not be equal to $X$. So, $\cal {A}$$\cup \{X\}$ forms subbasis for $X$. Prove that topology generated by $\cal {A}$$\cup \{X\}$ equals the intersection of all topologies on $X$ that contain $\cal {A}$.
 
@Martin well I got it now. The devision thing doesn't require commutativity, but the point is that when you fill in a value in a polynomial $f=gh$, then you don't need to have $f(\alpha)=g(\alpha)h(\alpha)$
so polynomials as functions of a ring kind of have a multiplication as if the ring is commutative (at least when you fill in values), so the ring better be commutative when you start talking about roots and the like:P
 
2:30 PM
I am trying to learn partial fraction expansions by experimenting with wolfram alpha. :-) Can anyone explain why $\frac{z}{(z^2+1)}$ cannot be wtitten as $\frac{A}{z-i}+\frac{B}{z+i}$ ?
 
@Adam who said it cannot?
 
wolfram alpha...
 
$$\frac z {z^2+1} = \frac1{2i} \left[ \frac 1 {z-i} - \frac 1 {z+i} \right]$$
@Adam just because wolfram alpha says something doesn't mean something is true
 
I see.
 
2:43 PM
Hi
Any cool stuff ?
 
@Astyx Come up with a function $f: \Bbb R \to \Bbb R$ such that $\forall x \in \Bbb R: f(x+1)=-f(x)$ and $f$ has no period.
 
Wolfram Alpha only does partial fractions into real rational parts
So it’ll do z/(z^2-1) but not z/(z^2+1)
I’m not sure how well it deals with irreducible denominators either
 
in a parallel universe, Wolfram Alpha only does partial fractions in quaternions
 
@LeakyNun You know much about PDEs?
Or does anybody know much about PDEs?
 

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