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12:00 AM
(In intuitionistic logic, $\lnot\lnot p$ isn't equivalent to $p$.)
 
1=0 is true in some systems... such as the trivial ring
 
($\lnot\lnot\lnot p$ is equivalent to $\lnot p$, though.)
 
@AkivaWeinberger typo
 
Thanks, fixed
 
@Secret in that case FLT is false lol
 
12:04 AM
anyone want to help me with yet another contour integration question?
 
I kinda like the basic idea of institutionistic logic, that you need to present proof before a statement's truth value is determined
 
@GFauxPas have a look at my longest proof xd
 
The aim of a probabilistic logic (also probability logic and probabilistic reasoning) is to combine the capacity of probability theory to handle uncertainty with the capacity of deductive logic to exploit structure of formal argument. The result is a richer and more expressive formalism with a broad range of possible application areas. Probabilistic logics attempt to find a natural extension of traditional logic truth tables: the results they define are derived through probabilistic expressions instead. A difficulty with probabilistic logics is that they tend to multiply the computational c...
 
maybe later! right now I'm confused about a contour!
My question is, is that curve a contour? Because it crosses the negative real axis
 
12:06 AM
well contour is any curve on the complex plane
 
I thought a contour is continuous by definition
anyway does the curve make sense from beginning to end
there's no branch cut that completely avoids the curve
so I'm not sure the curve makes sense, especially if I want to integrate over it
 
Give a counterexample to "if (x_n) approaches L then floor(x_n) approaches floor(L)"
x_n is a real sequence
 
The expression strongly reminds of a gamma function after suitable rearrangement, I am not sure however if that will help in the integration
 
is this question-worthy
yes Secret!
that is why I know I can integrate that expression
because it is so closely related to the gamma function
but I'm not sure if the graphical interpretation is correct
 
remind me again what happens when a contour passes through a branch cut, does it mean in order to do maths with that region you need to use the principal value?
 
12:10 AM
@DHMO i could, do you?
 
that's pretty much what I'm asking Secret ;D
i don't know what happens when a contour passes through a branch cut
 
$1-\frac{1}{n}$ approaches 1, but the floorfunction approaches 0.
 
if it helps any it appears to only go through a branch cut a finite number of times
whenever I google "contour integration through a branch cut" I find sites doing exactly the opposite
i.e., how to avoid contours going through branch cuts\
 
coffee white without sugar is the only thing I learned today :)
 
I'd ask a question about it but I'm not sure what precisely the question is
 
12:13 AM
How is it not just 0
if you can split in into two regions that are homotopic to a point
 
Akiva can you help me form a question to ask
 
did you see my contour
@akiva I think it isn't homotopic to a point because it's not allowed to go through the removed point
 
@GFauxPas This doesn't look like it crosses the axis
@GFauxPas But you can split it into two things that are
See the image with the red and green in the article
 
Akiva it does, look at the axis labels
it crosses it once
 
12:19 AM
Oh, I see
 
possibly more if I zoom in more
 
What are you integrating
 
it's on the graph
as t goes from epsilon->0 to T->inf
 
Oh, I thought that was the equation of the curve
 
it is
 
12:20 AM
So what are you integrating then?
 
the integral $\int_\epsilon^T \text{that} \, \mathrm dt$ is the integral
but if you plot the integrand, it gives you the curve in the picture
 
it's parameterized
 
Should be fine; choose a branch for one of the points and define it on the rest of the curve so that it's continuous. I think the only issue is when you have a loop, because going around the loop might end up giving you a different branch value
 
but I don't see any place to put a branch that wouldn't go through the contour somewhere
 
12:24 AM
Doesn't need to be a straight line.
 
hmmm, let me see if it loops
 
hi @TedShifrin
 
Hi Karim
 
I fucked up completely in my analysis today only did like 5 and half questions out of 7
and not all of them are correct :S
 
I was going to say that 5 1/2 out of 7 isn't so bad
 
12:26 AM
 
It sucks I did well in my algebra exam yesterday
 
so I'd need to have a branch that's ... spiral shaped?
 
What in the world are you doing this time, @GFauxPas?
 
oh wait, when you zoom in that much you
're away from the axis
 
yeah I mean I could have done better @TedShifrin. Last month I only concentrated on algebra.
 
12:27 AM
I'm playing with contours, Ted
 
I have bad time management skills ...
 
let me post the original question
 
I assumed as much, @GFauxPas :)
 
tomorrow is my last exam which is geometry
 
12:27 AM
the usual branch cut for this type of curve is the negative real axis
 
then I am done for this semester.
 
but this guy crosses the negative real axis.
 
I think what I should for next semester I need to properly manage my time properly.
 
@TedShifrin The contour can go through the usual branch cut (negative reals), right? Because you can draw a curvy branch cut
 
well, it's learning styles, too, probably Karim ... But good luck tomorrow.
 
12:28 AM
to avoid it
 
@Adeek how are geometry excercises typically?
 
@Null differential geometry.
@Null for example any compact connected lie group is a complex torus
 
so in esscence, nothing where a ruler helps :d
 
this result is cool any compact lie group is a complex torus
 
@Adeek You are missing an adjective.
 
12:30 AM
That's blatantly false, Karim :P
 
What is ":d"? Trying to lick your nose? @Null
 
@PVAL-inactive two
 
You're missing "complex Lie group"
 
@AkivaWeinberger if I can!
 
oh, fine
 
12:31 AM
:d
I can almost do it
 
complex lie group * @TedShifrin
yeah I forgot complex part.
 
@TedShifrin @AkivaWeinberger I zoomed in and it appears to only cross the negative real axis 1 to 3 times
before it starts behaving
 
I guess we needed the maximum modulus principle in part of the proof.
 
and it's simple
according to my graphing technology at least
so can I have a curvy branch cut that walks along holding my contour's hand and making sure it stays safe?
 
This sounds pretty crazy to me, @GFauxPas, but I've resigned myself to your obsession.
 
12:34 AM
Well okay, what would you do with that contour?
 
I have shown that for $A,B\in \mathbb{R}^{2\times 2}$ the $U=\{X\in \mathbb{R}^{2\times 2}\mid AX=XB\}$ is a vector subspace of $\mathbb{R}^{2\times 2}$.

Then when $A=\begin{pmatrix}a_1 & 0 \\ a_3 & a_4\end{pmatrix}$ and $B=\begin{pmatrix}b_1 & b_2 \\ 0 & b_4\end{pmatrix}$ I want to show that $$U=\{0\} \iff \{a_1, a_4\}\cap \{b_1, b_4\}=\emptyset$$

How can we show the direction $\Rightarrow$ ?
When we have the zero matrix, doesn't it hold for every $A$ and $B$ ?
 
I can't think of how else to make sense of the contour
 
What is the precise question?
 
can I integrate $\int_\epsilon^T t^{1/2 - i}e^{-(5-12i)t} \, \mathrm dt$ for $\epsilon > 0$ arbitarily small and $T > 0$ arbitarily large?
k I'm leaving the computer, if you want to talk to me make sure you @ me so I 'll see it when I come back
take care
 
I don't see why this is such a difficult problem, @GFauxPas. Just choose a usual branch of log.
@MaryStar: Did you write down the system of linear equations for $X$? When does it have only the trivial solution. (Think of $X\in\Bbb R^4$.)
 
12:45 AM
@TedShifrin It goes through the branch cut (negative reals), and perhaps all straight lines from the origin
 
What's it?
All I need to do is define $t^{1/2 - i} = e^{(1/2-i)\log t}$.
 
$$AX=XB \Rightarrow \begin{pmatrix}a_1x_1 & a_1x_2 \\ a_3x_1+a_4x_3 & a_3x_2+a_4x_4\end{pmatrix}=\begin{pmatrix}x_1b_1 & x_1b_2+x_2b_4 \\ x_3b_1 & x_3b_2+x_4b_4\end{pmatrix} \\ \Rightarrow \left\{\begin{matrix}
a_1x_1=x_1b_1 \\
a_1x_2=x_1b_2+x_2b_4 \\
a_3x_1+a_4x_3=x_3b_1 \\
a_3x_2+a_4x_4=x_3b_2+x_4b_4
\end{matrix}\right.\Rightarrow \left\{\begin{matrix}
(a_1-b_1)x_1=0 \\
(a_1-b_4)x_2=x_1b_2 \\
a_3x_1=x_3(b_1-a_4) \\
a_3x_2=x_3b_2+x_4(b_4-a_4)
\end{matrix}\right.$$

For the direction $\Leftarrow$ we have that $\{a_1, a_4\}\cap \{b_1, b_4\}=\emptyset$, so we have the following:
 
The curve is also $t^{1/2-i}e^{-(5-12i)t}$ (from $T=0$ to $\infty$) is my understanding @TedShifrin
 
Write a matrix for that linear system in 4 variables and see about the rank of the matrix, @MaryStar.
 
@TedShifrin what do you think of my proof. First I will prove that any compact connected lie group is commutative.

Suppose G is as given. Fix $g \in G$. Consider $\phi_g : G \rightarrow G$ given by conjugation by g. That is $y \mapsto gyg^{-1}$. Then this is automorphism of G. So if this mean that $d(\phi_g)_e : T_e(G) \rightarrow T_e(G)$ is automorphism of $T_e(G)$. Consider the map given by $\psi : g \mapsto d(\phi_g)_e$, the map $(x,y) \mapsto yxy^{-1}$ is holomorphic in both variables. This means that $\psi$ must be holomorphic map from $G$ to $GL(T_e(G))$. Since G is connected and com
 
12:47 AM
I think the function he's integrating is also the contour
 
No, that's nonsense.
The integral is along an interval on the real axis. @GFauxPas
 
Ping him then
 
@TedShifrin You mean a matrix X, or not?
 
No, I mean that system of equations in variables $x_1,\dots,x_4$.
 
@TedShifrin are we only interested in the reals roots, because of the way $\sqrt{ }$ is defined?
 
12:48 AM
Write it in the way we always do homogeneous systems of linear equations.
@Null: Yes, your context is real functions.
 
@TedShifrin that's the problem, we have to state a working domain. It is not given.
 
If a function is single-valued (no branch cuts) but possibly has poles, is the Pochhammer contour zero?
 
I'm quite confident you're doing the real cube root function, @Null, or else it's not a function.
 
I feel like it should be because it can be broken into two nullhomotopic contours
 
@TedShifrin for {0} as the domain, and {0} as the codomain, it still will be a function, just a boring one
(i agree basicly, just asking)
 
12:51 AM
DogAteMy: I've never seen that in all my years of topology and complex analysis. I just looked quickly (too much going on); I guess the point is that curve is not null-homotopic but is null-homologous.
Ordinarily, @Null, when domain is not specified, it's considered to be the "natural domain," i.e., the largest set on which you have a well-defined function.
 
0
A: Prove $\forall\epsilon>0:\exists m\in\mathbb{N}:\forall n\ge m:b_n-a_n<\epsilon$ for nested intervals

Jorge Fernández HidalgoWe have $a_n<b_n$ by the harmonic-arithmetic mean inequality. Therefore $a_n<a_{n+1}$. It follows $b_{n+1}-a_{n+1} < \frac{a_n+b_n}{2}-a_n=\frac{b_n-a_n}{2}$. The proof follows.

why does the proof follow? Simply because it gradually becomes smaller or what?
 
@TedShifrin ah, didn't knew that convention ;)
@NaCl long time no see, hope you are fine
 
@Null ye kind of
 
@NaCl: Positive numbers can get smaller and smaller without going to 0.
But if $0<c_{n+1}<c_n/2$, then the sequence must go to 0.
@Null: No, that sequence still goes to 0 !!
 
If $0<c_{n+1}<\frac{c_n}{2}$ then $c_n$ goes to zero as $n$ becomes larger and larger
 
1:04 AM
Can you prove that?
 
I didn't try yet
 
@TedShifrin so $1+\frac{1}{n}$? (gets smaller, but >0)
 
Sigh
 
Right, @Null. That's a good example :)
 
polishes @usukidoll nails
 
1:06 AM
Thanks I guess :/
 
@Akiva: DogAteMy, note that I was correct. That curve has net winding number 0 around each singularity, so the integral will be 0. (And it is nullhomologous in $\Bbb C-\{P,Q\}$.)
 
Imagine how social life would be, if we could take statements back, or deleting them.
could be either a nice utopia or dystopia
 
Dumb question, but is there a formal way to prove that P(x) <=> {x | P(x)}?
 
That doesn't make sense to me, @Alias.
 
@TedShifrin, I'm an idiot
One sec
 
1:08 AM
($\{x|P(x)\}$ is no statement)
 
P(X) <=> x \in {y | P(y)}
Thank you -- oops :)
:%s/X/x
 
ah, that's cool, @Alias :P
 
haha, thanks :)
 
@AliasUser have you tried something already?
 
You're saying that $P(x)$ holds if and only if $x$ is in the set of all $y$ for which $P(y)$ holds.
 
1:09 AM
@TedShifrin exactly
 
Basically you're saying $x\in A\iff x\in A$. So you're cool :P
 
@Null Intuitively, it's obvious that if P(x) holds, then x is in the set of all y such that P(y) holds, and vice versa, which seems trivial
I was just wondering if there was a formal rule that would prove such a statement
@TedShifrin - awesome, thanks! Is there a rule in first order logic or set theory that states as such?
 
those statements should have some name. with those I mean all of the kind $a\iff a$
 
woo
booked cambridge trip
 
I don't know the official logic rules, and I survived math for 50 years or so :P
 
1:11 AM
@TedShifrin hahaha :)
 
Which Cambridge, @MikeM?
 
Ah, you jet-setter, you.
 
not looking forward to the flight tbh
 
Didn't you just do a longer one?
Oh, maybe not.
But close.
 
1:13 AM
How difficult is the gre?
 
Difficult.
 
Iceland is closer than the UK! Not by much, but it is.
 
Are you serious?
 
Yes. I'm always serious.
You talking about the advanced math exam?
 
X. X
Yeah the one in April
 
1:13 AM
mmh, i think I analyize $f(x)=x^2$ when $x\geq 0$, $f(x)=-x^2$ when $x\leq 0$.
 
I intend to sleep through it, anyway. I've got some California-grown sleep aids for the flight there.
 
The general exam is stoopid for math, but the verbal is quite challenging.
No comment @MikeM.
 
Huh?
 
For what @Null?
 
@Ted: I'm serious, though. It's an effective medicinal use. Sometimes it's helpful when my restless leg gets really bad.
 
1:14 AM
So should I focus on Praxis instead?
 
Oh, I believe that completely, @MikeM.
 
@TedShifrin I assume, just by looking at it's graph, that the function could be one function instead of piecewise
 
Praxis is a test for teachers to be certified. I'm confused, @usukidoll.
Sure, @Null, if you use the signum function ...
 
Duh that's one of the requirements to enter the masters in education for teaching program
 
But it's defined by cases, because of 0.
 
1:16 AM
@TedShifrin So when can it be nonzero? When it's a multifunction?
 
Duh yourself, @usukidoll.
 
Don't be mean
:(
 
You had no call to say "duh" to me.
DogAteMy: So the usual Cauchy and residue theorems don't apply to multifunctions, so then we're back to undefined territory.
 
@TedShifrin okay, I can't seem to get it proven, I somehow need to get the damn $\epsilon$!
 
"Undefined"?
 
1:17 AM
Praxis Math content knowledge is specific for secondary math teachers but at the last minute the professors who wrote the recommendation letters for me wanted to try enter for the MA in math for the upcoming spring. A. K. A late admission or something
 
Multifunctions are not well-defined. So how do you even define what you mean by the integral, DogAteMy?
 
I'm like wtf k let's try it. X. X
 
That's a good point. So what's going on in the Wiki page, then?
 
@NaCl: But your size is controlled by halving, isn't it?
DogAteMy: I try not to read wiki pages. I find mistakes :P
 
I think the discussion of multifunctions is often unhelpful, pedagogically, since we only work with the sections of the multifunction that are continuous. Secretly we've been doing sheaves all along.
 
1:19 AM
That doesn't help, @MikeM :D
But, even with local continuity you don't have global continuity :D
 
It might for Akiva. For the average class, just never ever ever talk about set-valued functions.
@TedShifrin Of course not. That's why you want the sheaf picture.
 
Has DogAteMy learned sheaf theory already? Geez.
 
No idea.
 
@MikeMiller You think I know what sheaves are??
 
@TedShifrin so I have $I_{n+1}\le \left(\frac{1}{2}\right)^{n}\cdot I_{n}$
 
1:20 AM
Ask your parents.
 
I mean, I've heard of them
 
No exponent of $n$ there, @NaCl, unless you put $I_1$.
 
No idea what they are
Something category-theory-y
 
no, no, no category.
 
Oh, yeah
 
1:20 AM
Or else I wouldn't have spent my career using them.
 
Sheaves are about as complicated as spectral sequences.
 
No, way easier.
 
Not something category theory-y then
 
Confusing until you understand them, and then very simple.
 
Like always, you can couch things in category terms, but it's crap to think you need to.
 
1:21 AM
Quick question I hope one of you can help me with: Does there exists a separable Hilbert space with a continuous dual space consisting of purely non-injective mappings?
 
Something algebra-y, then?
 
Sheaf theory is natural and important because its cohomology allows you to phrase patching (gluing) questions very naturally, DogAteMy.
I think about it more analytically.
 
huh?? Functionals are never injective!
 
Dim 1, @MikeM? :D
 
I knew someone was going to say that. I didn't expect it would be you.
 
1:23 AM
Of course it would be I.
 
Today's xkcd is relevant here
 
@MikeMiller was that a response to my question?
 
yes, @Martin
 
Tired and hungry x. X
Should I heat up some lunch or sleep?
 
1:25 AM
feeds @usukidoll some sweet snacks, and reads a goodnight story
 
Well consider the identity mapping from R to R, that is surely a injective bounded linear map in the dual of R right?
 
Probably, food and then sleep.
 
But, failing that, sleep is better than doing nothing for half an hour
 
(@Martin: Note that I already gave the dimension 1 objection. But other than that, what he said is totally right.)
 
I heard that eating before sleeping makes fat tho
 
1:26 AM
Eat, sleep, elliptical
Not all at once tho that would be hard
 
@AkivaWeinberger for a dolphin only the third^^
 
Dolphins eat while sleeping?
 
they sleep with one hemisphere, as far as I remember
so eating would be possible, I dunno
 
The elliptical would be hard, though, yes
 
@TedShifrin Ahh i see, yeah i can see that if the hilbert space in question has more that two elements in its basis, then no dual element is injective. Thanks.
 
1:29 AM
actually, what would change if human would sleep too with one hemisphere constantly o_O
 
@Martin: Don't thank me. Thank yourself (and maybe Mike).
 
@te
@MikeMiller Thanks mate :)
 
So according to my textbook, Pres. Jackson's informal assistants were called his "Kitchen Cabinet"
 
What should we name Trump's cabinet? I can't think of anything not obscene.
 
The obscenities would rightly go to Jackson's cabinet, too.
 
1:33 AM
is $\neg a\land a\iff \neg b\land b$ a sensical statement?
 
Sure. I think it's true, too.
 
Sure, @Null.
 
$\rm F\iff F$
 
Any statement that is always false is equivalent to any other (I guess).
 
Would someone mind checking this proof for me?
I'm reading through Velleman, and I wanted to try to formally prove one of the exercises they use Venn diagrams for
 
1:36 AM
$\forall\epsilon>0:\exists N\in\mathbb{N}:\forall n\geq N:b_n-a_n<\epsilon$

Let $\epsilon>0$, then there exists $N=1$, now for all $n\geq N$ the following holds:
$$b_n-a_n \le \frac{1}{2}\cdot\left(b_{n-1}-a_{n-1}\right) \le \frac{1}{2}\cdot\left(b_{n-2}-a_{n-2}\right)\le\cdots\le\epsilon$$

But that is decreasing $n$, not increasing
 
You need to give a recipe for $N$ in terms of $\epsilon$, @NaCl.
(And you lost powers in your 1/2.)
 
Strange way of saying "function"
 
@NaCl: $N=1$ is hardly ever going to work.
 
In terms of $\epsilon$? So I need the archimedian property?
 
Well, yes.
 
1:38 AM
humm....
We can find $m$ such that $\frac{1}{m}\le\epsilon$
 
Nope.
 
By the archimedian property
 
Well, of course, but not relevant.
 
ok, I'm lost
 
@TedShifrin is $\leq\varepsilon$ even ok? i only saw strictly less than it.
 
1:40 AM
So, what you should have written is that $b_n-a_n\le \frac1{2^n}(b_0-a_0)$. Can you make that less than $\epsilon$?
Strictly less is great.
 
But not mandatory? I thought $\epsilon$ could be even the same as $\frac{1}{m}$
 
If you don't have $a_0,b_0$, then you need $\frac1{2^{n-1}}(b_1-a_1)$.
NO, if $\epsilon = \pi/12$, you're not going to have it equal to $1/m$.
 
Sure, but it can be any value
Since there is $\forall$
 
But I give you $\epsilon$. You have to find an $m$.
 
oh, ok
then obviously $<$
 
1:43 AM
$<$ versus $\le$ isn't really the point.
 
I know $b_1=x$ and $a_1=1$.
 
Irrelephant.
 
@TedShifrin I thought it was, because of this
 
Oh, sorry. All that matters is that $b_1-a_1$ is some positive number; call it $c$.
 
@TedShifrin Another irrelephant fact: When Hannibal crossed the Alps, he was relying on the elephant of surprise
 
1:45 AM
I know that $b_n-a_n$ is positive
 
Thanks, DogAteMy. Let me know when I should go back to giving you problems to hush you up :P
 
Theoretically I'm reading my history textbook
 
so $b_1-a_1$ is positive as well
 
Jackson vs. the Bank
 
But give it a name, @NaCl, as I suggested. You'll need that quantity to say how large your $N$ or $m$ or whatever it is has to be.
 
1:46 AM
And how do I find $N$??
So $N$ must be dependent on my interval
 
@AkivaWeinberger If you cared, apparently FLT is provable in PA
 
Yes, it is.
 
and not on $\epsilon$?
 
It depends on both.
Mostly on $\epsilon$.
 
How do I make it depend on both?
 
1:47 AM
Use high school algebra and solve.
 
@AkivaWeinberger theoretically I am reading my theory book.
 
I don't have a connection from one to another
 
I'll give you a hint, @NaCl. You need to use logarithms.
 
wtf why
 
because you have $1/2^n$ appearing.
 
1:49 AM
en.wikipedia.org/wiki/Bayes%27_theorem i.e. probability of intersections are conserved under partitioning
 
@MikeMiller O_O whoa
Also, the bank is dead
Jackson wins
 
@AkivaWeinberger i never played a game where the bank doesn't plays
 
$$\frac{1}{2^{n-1}}(b_1-a_1)\le\frac{1}{2^{n}}(b_2-a_2)\le\frac{1}{2^{n+1}}(b_3-‌​a_3)\le\cdots\text{and the potentially for whatever reason }\le\epsilon$$?
 
oooh, I misunderstood. ok
 
1:53 AM
@AkivaWeinberger I think it was a certain fascist politician who said that. Nevertheless, I find this quote quite epic. (tho I can't find the exact wording >:| )
 
$$\frac{1}{2^{1-1}}(b_1-a_1) \le \frac{1}{2^{2-1}}(b_2-a_2)\le\cdots$$
 
@Null I think you may have made a typo
 
good evening
 
Quick question: I have a PDF for the loss Y, $f(y) =.02*(1-.01*y), 0<y<100$. Company Z pays 80% of loss that exceeds the deductible of 10. I got $E(Z) = \int_{0}^{100} .8*(y-10)*.02*(1-.01*y) dy = 18.667$ I attempted to find $V(Z) = E(Z^2) - E(Z)^2 = \int_{0}^{100} (.8*(y-10))^2*.02*(1-.01*y) dy - 18.67^2 = 355.28$, but that is drastically different from the listed answer of $100416$. I think I'm making an obvious, dumb mistake that one of you will be able to point out immediately.
 
@AkivaWeinberger "to which [...] replied that throughout his life he had always played vabanque." - wikipedia.
 
1:59 AM
hi @Null
 
@ZachHauk hi^^
 

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