« first day (2879 days earlier)      last day (2144 days later) » 

2:00 PM
hmm, sure
 
but you truncate past the power you're interested in
 
Mine is usually to get it in the form of 1/(1+stuff)=1-stuff+stuff^2+...
which isn't always so nice
 
yes that's what we did in class but I'm likely to make sign errors that way
 
makes sense
 
and then you have to prove |stuff| < 1
whereas long division is "formal"
for what it's worth
 
2:01 PM
tru
 
which can be a bad thing I suppose
 
another is to recast the problem as a recurrence relation
 
well if you want a formal manipulation you can just pretend |stuff|<1 I suppose
I'm listening
let's take $1/e^z$ as an example
 
Are you guys doing series expansions?
 
pretending we don't know it's equal to $e^{-z}$
 
2:02 PM
o..o
 
e.g. $F(x)=1/P(x)\implies P(x) F(x) =1$
 
right
 
I have in mind something more like $F(x)=1/(1-x-x^2)$
 
sure
so $(a + bx + cx^2 + O(x^3))(1-x-x^2) = 1$
so solve for one coefficient at a time?
 
then $$1=(1-x-x^2)F(x) = (1-x-x^2)\sum_{k=0}^\infty a_k x^k =a_0+(a_1-a_0)x+\sum_{k=2}^\infty (a_k-a_{k-1}-a_{k-2})x^k$$
 
2:05 PM
why
 
where $F(x)=\sum_{k=0}^\infty a_k x^k$
 
why are the coefficients those
$a_0, a_1 - a_0, a_2 - a_1 - a_0$?
 
That's what you get when you distribute the product.
You're doing $(1-x-x^2)(a_0+a_1 x+a_2x^2+\cdots)$
 
ah
 
I'm skipping some manipulation of the summation for $k\geq 2$
but that's just tedious
so you identify term-by-term and get $a_0=a_1=1$ and $a_k=a_{k-1}+a_{k-2}$ for $k\geq 2$
 
2:08 PM
well when the coefficients are just $\pm1$ then $k = 2$ isn't so bad
 
but generality etc
can you please walk through one with other coefficients with me
 
It's not really too bad in any case, though. $$A(x)B(x)=\sum_{i=0}^\infty A_i x^i \sum_{j=0}^\infty B_j x^j =\sum_{k=0}^\infty \left(\sum_{l=0}^k A_{k-l}B_l\right) x^k$$
 
say $1/(1-x^3/3)$
oh the Cauchy product
 
yup
Which is handy here when only one of the two series is infinite
Anyways. What I expect to see is $a_k=a_{k-3}/3$ for $k\geq 3$
 
2:11 PM
sure
$-$?
 
no
with $a_0=1,$ $a_1=0$, $a_2=0$
 
oh I see
okay that's a good guess
 
If $A(x)B(x)=C(x)$ with $B(x),C(x)$ polynomial, then for large enough powers we must have that the Cauchy product contains a specific number of terms and that this Cauchy product vanishes
This Cauchy product equaling zero will then just be the recurrence relation
 
neato
 
Yeah
Doesn't help with something like $1/e^x$, though, since that's not polynomial on the bottom
I'll also point out that one usually does the reverse for $1/(1-x-x^2)$ i.e. start from the recurrence and work to get the power series
that's because the recurrence in that case is just the Fibonacci sequence :)
 
2:17 PM
well let's say I want the residue of $1/\cos z$ at $z = \pi$
can't I just consider the first 2 terms of $\cos z$?
 
what i'd do is first work out the order of the pole as the multiplicity of $z=\pi$ as a zero of $\cos z$
 
that would be $1$ because $\sin \pi \ne 0$
 
no?
 
trig functions man
 
2:18 PM
maybe I don't remember the defn for order of a pole
oh I'm thinking of the order of a zero lel
 
it's more that you're not remembering the values of your trig functions :)
$\cos \pi =$?
 
$0$
 
No...
 
i'm losing my mind
oooh I meant $\pi/2$
damn it
 
lol, happens
 
2:20 PM
I don't know why I said $\pi$
 
anyways. then yeah, $f(z)=\cos z\implies f(\pi/2)=0$ but $f'(\pi/2)=-\sin(\pi/2)=-1\neq 0$
so $z=\pi/2$ is a zero of multiplicity one of $f(z)$ and therefore $1/f(z)$ has a simple pole at $z=\pi/2$
 
right, and $\pi/2 \ne \pi$ in general
 
yes, arithmetic is generally assumed to make sense :>
 
sure. okay, right
so we want the coefficient of $1/x$ on $(1 - x^2/2 + O(x^4))^{-1}$
 
No. That's the expansion around $x=0$.
 
2:22 PM
what does that have to do with $x= \pi/2$ ?
 
oh right
$f(\pi/2) = 0, f'(\pi/2) = -1$
 
if you're just doing a simple pole, it's easiest imo to just do $\lim_{z\to z_0}(z-z_0)/f(z)$
 
yes I know but I'm using this as an example to understand Cauchy Product method
 
Fair enough
 
and $f''(\pi/2) = 0$
 
2:25 PM
So you'd have $$f(z)\cos z=\sum_{i=0}^\infty a_i (z-\pi/2)^i\sum_{j=0}^\infty b_j (z-\pi/2)^j$$
where $b_0=0$, $b_1=-1$, $b_2=0$ and so forth
 
right
 
And all you're really interested in is the leading term
 
right
 
But that'll just be the lowest-order terms on both, so $a_0\cdot b_1(z-\pi/2)^1=-a_0(z-\pi/2)$
hmm, silly mistake on my part
It's a Laurent series for $f(z)$, so the series should start at $i=-1$
 
oops
 
2:29 PM
kids understand arithmetic w/o group cohomology because they have a TEACHER which is holding dominion over the classroom. Creating a GROUP.
 
hence the lowest-order term is really $a_{-1}(z-\pi/2)^{-1}\cdot b_1 (z-\pi/2)^1=-a_{-1}$
and since we have $f(z)\cos z=1$, we must have $a_{-1}=-1$
 
bam
thanks Semi :)
 
at which point we realize this is overkill since all we've really done is obtain $f(z)=1/\cos z = 1/[-(z-\pi/2)+O[(z-\pi/2)^3] = -1/(z-\pi/2)+O(1)$ :)
 
shhhh
 
Hey guys
 
2:31 PM
we're pretending we don't know that :P
hi geo
 
@GFauxPas
Is a class of functions less rich than a group
 
when you talk about a group on a set, you have to decide what your multiplication will be
it might be composition if the domains are equal to the codomains of all, for example, or it might be pointwise addition
wait the first is not a group
you'd need invertible functions, for that example
but the point is you can't define a group unless you define the operation on the group
 
By composition you mean multiplication?
 
$f \circ g: x \mapsto f(g(x))$
$(f \circ g )(x) = f(g(x))$
 
2:44 PM
But I'm sure classes of functions are useful objects to study
 
for sure
 
Are there any really important classes of functions in mathematics
But I'm sure classes of functions are useful objects to study
 
yup
normal families are very important in complex analysis
they are functions that have uniformly convergent subsequences on the same compact sets
rather
they are a collection of sequences
errr let me try again
they are functions such that
every sequence of functions in that family
has a subsequence that converges uniformly on compact sets
a very important one in analysis is
the set of all functions $\mathbb R \to \mathbb C$ which are square-integrable over $\mathbb R$
that one is denote $L^2(\mathbb R)$ and it's important because it's a Hilbert space with inner product $\displaystyle \int_{\mathbb R} f g^*$
lots of important families of functions
set of functions with compact support, set of continuously differentiable functions, set of functions that are equal to their power series
etc
 
3:42 PM
hey does anyone know how to compute this:
 
You may need to use that, not entirely sure though
 
@NicholasRoberts I'm not sure but I doubt it based on the material so far
I think that it should first simplify to 2 * int[(a+b)/2, ->b]
 
o.
.o
how did i mess this up
 
so that's how one root is eliminated because x is never =a
 
yeah it looks like it's symmetric
so you can say that
but
then the integrand gets more complicated
there is a change of variable that can turn it in to a rational fraction
 
3:53 PM
what about partial fraction decomposition? @jcora
 
idk I think I'll give up
this is bothering me too much
 
Do you kow partial fraction decomp? It seems like thats the obvious method for this.
 
well, yeah, I do, but I thought it was likely some trick to get that polynomial-looking thing centered at 0, 0 or something like that and then it would seem easy..
I'll try that
 
you could maybe see it geometrically
then invoke a well know fact about the area of certain shapes
 
@NicholasRoberts but wait it's under the square root what good will decomposition do?
 
4:07 PM
Wow, i did not see the root there. My bad!
 
@GFauxPas, is compactness important or can you have open intervals just as easily
 
@jcora $$ \begin{align} \int_a^b\frac{\mathrm{d}x}{\sqrt{(x-a)(b-x)}} &=\int_0^1\frac{\mathrm{d}x}{\sqrt{x(1-x)}}\\ &=\frac{\Gamma\!\left(\frac12\right) \Gamma\!\left(\frac12\right)}{\Gamma(1)}\\ &=\pi \end{align} $$
No Principal Value needed
However, I did use the Beta function to simplify things.
Though we can use a trig substitution instead
$$ \begin{align} \int_a^b\frac{\mathrm{d}x}{\sqrt{(x-a)(b-x)}} &=\int_0^1\frac{\mathrm{d}x}{\sqrt{x(1-x)}}\\ &=\int_0^1\frac{\mathrm{d}x} {\sqrt{\frac14-\left(x-\frac12\right)^2}}\\ &=\int_{-1}^1\frac{\mathrm{d}x}{\sqrt{1-x^2}}\\ &=\left[\sin^{-1}(x)\right]_{-1}^1\\ &=\pi \end{align} $$
 
4:32 PM
Sigh I'm an idiot
 
@jcora Give up 12 minutes after asking? No!
 
somehow i misread the integrand
 
4:48 PM
Anyone know what $\Subset$ means?
 
@Kari hairpin turn?
 
I have no idea what that means, @robjohn
For context, it's in a proof of the existence of a partition of unity
 
It looks as if it might mean "proper subset" in that context
It could be that $U\Subset V$ means that there is a closed $K$ so that $U\subset K\subset V$, but I'm not positive. You might need to see if the book says anything more.
 
@robjohn Ah right, I usually see that as $\subset\subset$
Sadly it's page 3 of this pdf and there isn't a section on notation
Thanks for the help!
 
5:03 PM
@Kari is that the one with the closed set that you see as $\subset\subset$?
 
@Kari maybe it means subset of a subset . it probably doesn't, but maybe that's what the notation should mean
oh rob beat me to it
 
so my professor once denoted a map from a set to itself as an arrow doubling back around
like $f: A \supset$ kinda, but with an arrow
i can't find an equivalent in latex :(
$f:A \curvearrowleft$
not quite
 
@robjohn Yep, that's the one. I often see $U \subset\subset V$ referred to as $U$ being compactly contained in $V$.
(In $\mathbb{R}^{n}$ anyway)
@GFauxPas Yea, you'd need to rotate that symbol
I don't think it's standard notation. In fact, some professors I speak to advise against that notation as it's non-standard. I like it though
 
5:22 PM
he just did it like once or twice
I think he was feeling whimsical
 
5:48 PM
Hi all, if p and 2*p+1 are prime, can p-2 be prime as well for some p?
 
p=5
 
p>=7 *
I forgot to mention the condition p>=29 @mercio
But for all p, p=2,3,5 are only examples I found
 
6:06 PM
Can it be shown that if p, 2p+1 are prime and p>5, then p-2 is divisible by 3?
 
yeah you can look at them modulo 3
 
I am trying to prove the following
 
@Maximus just filter on the union
 
as a proof technique for the direction (--->)
do you assume existence and uniqueness as well for C and prove x \in A and x \notin B?
@LeakyNun not familiar with filters, what about the technique I just wrote
 
not filters
 
6:16 PM
I am more concerned with the proof technique when you have show that there exists a uinique C and then the condition
 
@robjohn I haven't been looking, but thank you so much!
 
$\exists!C[\forall x[x \in C \iff (x \in A \oplus x \in B)]]$ is what you want to prove
 
when you go for the (--->) direction do you also assume the eistence and uniqueness
 
so I think you're misinterpreting it
 
so in that logical statement, when you have to prove it what can you assume
 
6:19 PM
@Maximus: Draw a picture. What is the set $C$?
Hint: Do you know about set difference?
 
I'll leave it to Ted, I have to go now
 
no worries cheers
@TedShifrin It's the union, I am more asking what is the proof technique
 
@TedShifrin setual dimorphism?
 
No, union of WHAT?
 
when it comes to proving the logical statement written Leaky Nun wrote
 
6:21 PM
@robjohn!!!!!
Wow. Nice to see you back ... Now if only DanielF and Pedro would show up :P
I don't like sentences that are all symbols. Let's use words, @Maximus. But you haven't told me what $C$ is. That's important.
 
@robjohn hey can you just explain the justification for the first equality? like I see that if the integral is constant, you could plug in any value for a and b, but I sort of though it'd depend on them, so how can you do that?
 
hi @Mathein
 
Hi @Ted
 
x in A and x not in b is the set difference
 
How do you notate that with sets?
 
6:23 PM
A\B
 
OK. So what's $C$?
 
so C is A in the first case
sorry
 
Huh? NO.
 
C is A\B in the first case
 
No. What is $C$? There are no cases.
 
6:24 PM
doesnt the or in the if and only if imply that there are cases?
 
I'm asking you to define the set $C$ that will make the sentence correct.
 
@jcora substitute $x\mapsto\frac{a(1-x)+b(1+x)}2$
 
How do you use sets to say one of the cases happens?
 
disjoint union?
of A and B>
 
not disjoint ... just union ... in math "either or" isn't exclusive.
NO!!!!!!!!!
I don't know what disjoint union of two sets means.
If you have a definition of that, let's use it.
 
6:26 PM
A\B union B\A
 
OK, finally.
 
but I don't it's just the standard
I am doing a set theory book they havent defined disjoint union
 
So we want to show that $x\in A\setminus B \cup B\setminus A \iff$ their second thing.
Precisely, they haven't. So you have to use set difference as I suggested.
 
disjoint union of sets does have a definition, but it's not htat one: mathworld.wolfram.com/DisjointUnion.html
 
Shaddup, Semiclassic.
I don't like the term the way Maximus (and maybe Leaky) was using it.
 
6:28 PM
I am aware of that definition, creating isomorphic copies @Semiclassical
 
well, my main reason to bring it up is that that definition is definitely not "A\B union B\A"
 
Alright Ted so we expressed the second part of the iff in terms of set
 
@Maximus: The point is that now that you've defined $C$ correctly, it's a tautology that the iff holds.
 
ah I see
so we can jsut define that set that way
we constructed it
 
Right. So now you just have to use definitions of checking "or" and "in" $A$ but "notin" $B$.
Right.
 
6:29 PM
while I was trying to prove it by logic
with 4 cases
 
It's unique because we have given exactly what it is.
But you can't hope to succeed if you don't say what $C$ has to be.
 
@robjohn thanks but I'm slow today how would a go to 0 then?
 
yes correct, btw I'm working on this book on my own so I can learn foundations. amazon.ca/Introduction-Revised-Expanded-Applied-Mathematics/dp/…
 
Oh, Jech was a well-known text for decades. Although I don't know it. I presume somebody revised it. Who knows if it's better or worse.
 
@TedShifrin ah I see, so how did you know to do that? It's obviously some kind of technique. When I have to prove existence, most of the time I'd have to construct it or do some kind of contradiction, correct? (as a proof technique to keep in mind)
 
6:31 PM
I've never learned foundations other than what's in Munkres's point set topology book. I survived fine.
 
oh wow
 
oh damn I'm likely looking in the wrong direction for changing the integral bounds
 
Right, to prove existence, you either need an existence theorem (like the Intermediate Value Theorem) or you have to produce the thing.
 
so then my approach, which is a tunnel vision approach was wrong
 
@jcora: I haven't seen your question, but if robjohn gave you a substitution $u=u(x)$, then you need the $u$-limits in your new integral.
@Maximus: I'm not good at formalism in mathematics. If there's a formalism for doing this, I don't know it.
 
6:33 PM
I generally just look at the logic and follow the steps without this overhead view that you have. Any tips on developing that? Not to mention that you don't look at logical symbols? Then how do you prove things in general? I was taught to look at the logic and when I prove things I generally have the logical statement in my mind.
 
Indeed, Munkres (and other professors in college) taught me NOT to write symbolic sentences, that mathematicians (other than logicians) hate those. So I write sentences. I'm allowed to use $\in$, $\implies$, $\cup$, $\cap$, etc., but not $\forall$ and $\exists$ and so on.
You have to understand how to negate quantifiers. That's the most important skill. And to negate an implication.
But you don't need a fancy logic set theory book to learn these.
 
yeah I am already familiar with those
 
Leaky will no doubt differ with me, because he loves foundations.
 
math.stackexchange.com/q/2828583/98602 This is kind of fun: "figure out all actions of $S^3$ on $S^4$"
 
What math do you want to learn?
 
6:36 PM
presumably they are all linear, but that needs to be proved
 
I saw that, @MikeM. Except when I saw it is was asking for some nontrivial action.
 
@TedShifrin was that question directed to me?
 
Yes, @Maximus. I hate pinging in a conversation.
 
Yeah, that's what he's asking, but that's not the fun question
 
@MikeM: Is the suspension one linear?
 
6:37 PM
Yeah
It's the same as the action on the unit sphere of $V \oplus \Bbb R$.
 
Oh, just like suspending a rotation to be a rotation.
Right, I see.
Your claim seems a lot harder. :)
 
No worries, Ted I am a 4th year undergrad student at the moment, and I was just self studying foundations due to a lot of people stressing how important it is. Hopefully to learn topology later and more algebra and analysis.
including multivariable analysis which I saw you have a book written for!
 
sup chat
 
Maximus, if you want to study formal logic (set theory and model theory), then I think it's reasonable. But to do analysis, algebra, topology, what I've said is really all you need. (Or the first chapter of Munkres.)
 
Ted I'd argue that it's useful to be able to identify whether variables are bound or free, whether or not you write $\forall$ explicitly
 
6:38 PM
Yup, I did, @Maximus. You can see lectures on it on YouTube.
 
@jcora That substitution gets us to the integral from $-1$ to $1$.
 
is foundations important? lol news to me
 
Sure, @GFauxPas, but that's part of thinking like a mathematician, not formal symbol crap.
 
Ted, yes I actually checked them out
 
sure, agreed
 
6:39 PM
So you're agreeing with me again, @EricSilva. Shock shock :P ... Hey, when do I need to think about Bryant?
 
@TedShifrin Yeah, which is why it's the fun one.
 
im finally done w the summer school in GR and going back to chicago today
 
With $\dim S^3 = \dim S^4 - 1$ I figure these actions must be very rigid.
 
Agreed, Mike. All sorts of topology questions are like that ...
 
@TedShifrin pinging is the only way I can figure out to which conversation various comments belong.
 
6:40 PM
so gonna crack open a cold can of bryant this sunday most likely
 
@robjohn: Well, until Mike came in, Maximus and I were the only ones talking, so I don't bother.
 
What about Cardinals and ordinals? aren't those important with math and constructing the real numbers? isn't real analysis based on that?
 
@TedShifrin jcora and I were, but we are commenting much more slowly.
 
Oh, I didn't realize you'd gone away, Eric. Just warn me what I should look at if we're gonna talk. I'm here until July 11, when I'm gone for a week. Gone for 9 days in August, too.
Oh, yeah, @robjohn, my apologies. And I did interrupt with something for jcora.
 
ah mmk
yeah i was in boston
 
6:41 PM
Did you visit my alma mater, Eric?
 
where are you going in August? I will be in Mammoth the second week
 
indeed i was there
 
I'm writing a blog post on representation theory and I want to include some geometric example. Is this okay? "Let $M$ be a smooth connected manifold and let $\pi:E \to M$ be a vector bundle with a flat connection. Let $G=\pi_1(M)$ and $K=\mathbb{R}$. Let $x \in M$ be a base point.
If we take a smooth loop $ \gamma: S^1 \to \pi_1(M)$ based at $x$, parallel transport along that loop defines an automorphism of $V=T_xM$. The flatness condition gives us that this automorphism only depends on the homotopy class of $\gamma$ and by smooth approximation, every homotopy class of continuous loops may
 
@Maximus maybe this is a hot take but i think constructing the reals doesnt matter lol
 
Driving up to the Bay Area, as is my annual tradition (until I get more feeble), @robjohn.
 
6:42 PM
Oh, interesting
 
what matters is that someone did it, and that that person is not me
 
hahahaha
 
@TedShifrin That takes about what 8 hours? It takes us about 6 hours to get to Mammoth.
 
@robjohn ok so when I use that substitution, I apply the expression to the integral bounds, so a -> (a(1-a) + b(1+a))/2, however that doesn't get me to -1?
 
but im probably way more bored w foundations than others so i mean, study it if you think it's cool
 
6:43 PM
Depends how long it takes me to cross LA, @robjohn. I now tend to go north on 101, so I stop in Santa Barbara for lunch and then head to Palo Alto. I come back down on 5. [Insert "the"s as you wish.]
@Mathein: Of course the base point is irrelephant.
 
@jcora no, $x$ goes from $-1$ to $1$. Maybe you need to put the substitution in the other direction
 
Unless you base $\pi_1$ at $x$.
 
@TedShifrin Ah, we go up the middle of CA on the 395
 
@TedShifrin ah, thanks I meant to include the notation, but forgot
also a loop is not a map from $S^1$ to the $\pi_1$ lol
 
But the geometry is right, @Mathein.
Oops. I wasn't paying attention to that.
 
6:45 PM
@TedShifrin thanks
 
Oh, you said $\pi_1$ based at $x$. Just write $\pi_1(M,x)$.
 
$(x-a)(b-x)\mapsto\left(\frac{a(1-x)+b(1+x)}2-a\right)\left(b-\frac{a(1-x)+b(1+x‌​)}2\right)=\frac{b-a}2(1+x)\frac{b-a}2(1-x)$
 
@robjohn: UGH. (Just to remove all ambiguity of whom I might be addressing.)
 
@robjohn ahhh right, I'm used to writing x = f(t) or t = f(x) explicitly, I mixed up what's the case here
 
I tend to teach substitution with naming the function explicitly.
@jcora: I think that's wise. You just need to decide which direction you're going :)
 
6:47 PM
@TedShifrin you mean like what I wrote here in the last message?
yeah writing it like that makes the direction explicit because of the different variable names and separation
 
Yes, @jcora. Well, the arrow notation is explicit enough. But you want to name the function $t$, say, so that you can then write $dt$ and write your new integral in terms of $t$ and $dt$ (with new limits).
Using $x$'s on both integrals is bound to lead to horrid confusions.
 
holy christ my screen is full of [Math Processing Error]
what a sight
 
Oh oh, Eric.
 
holy knuth*
 
@robjohn: I don't think I've quite done 395. I think that's a road I could have taken from the east side of Yosemite, but my Yosemite visit was essentially late enough for snow issues and road closures, and I didn't want to risk anything.
 
6:51 PM
@TedShifrin Since you are going to the bay area, PCH makes sense.
395 is for eastern sierras etc
 
Right, I know.
Well, I'm certainly not doing the 1. It's still closed from the mud slide in Big Sur. I came back that way a few years ago. Maybe I'll try that again another year when I have more time.
 
@TedShifrin my wife and I enjoy the drive along the 395. It is a bit more arid than the 1, but it's an acquired taste.
 
I've yet to do any desert stuff or go to Las Vegas or the Grand Canyon. I better hurry up before I'm too old and feeble.
 
and we are usually driving in 100+ temps
 
Yeah, of course I'll have those coming down the 5 too.
 
6:54 PM
how do people live in hot places
 
I dunno, Eric. I couldn't.
Oh, I lied, @robjohn. I want to Anza-Borrego once. I need to go back next spring.
 
@TedShifrin yeah the main thing that confused me was reusing the x in the substitution now it's all clear
 
i already spent too many years in a place that's too hot
 
I'm on your side on that, @jcora :P
 
@TedShifrin I went to Borrego Springs a couple of years in a row for an astronomy conference there.
 
6:56 PM
Very cool, @robjohn. I wouldn't want to spend a week there, but I will probably go for overnight next time.
 
@TedShifrin if you don't reuse variables, you run out of letters in more complicated proofs. Sorry, I carry that over to places where one could use different letters
 
I come from the Chern school of having a table of ranges of indices. It's never been as bad as my thesis with various flag spaces and roman and Greek indices. So bah.
 
@TedShifrin My conference was over a 3-day weekend, so it was not a whole week, thankfully.
 
LOL ...
Anyhow, @robjohn, you've taught enough calculus courses to know that it's pedagogically important to be as "simple" as possible :P
 

« first day (2879 days earlier)      last day (2144 days later) »