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12:26 AM
@Knight are you still unclear? (sorry, had to be said)
 
1:18 AM
Hi @robjohn
 
2
Q: Characterization of V-stages in the absence of foundation or replacement

user76284In the absence of foundation and replacement, we can say that $x$ is an ordinal iff $x$ is hereditarily well-founded (with respect to the relation $\in$) and hereditarily transitive, where $$ \text{$a$ is transitive} \longleftrightarrow \forall x \forall y (x \in y \in a \rightarrow x \in a) $$ ...

 
1:58 AM
@skullpatrol Hey there. Just back from getting a bit of food.
 
2:18 AM
@robjohn Hello
 
@Knight good evening (here, at least).
 
@robjohn Can we discuss my question?
Please ask which part was unclear
 
@Knight I was making a joke... you asked, "Please ask me if I’m still unclear," so I did.
 
@robjohn hahahahaha LOL
 
did you have a question about that still?
 
2:32 AM
No! But I’m always ready to learn everybody’s own part
 
Okay. I assumed you knew that there were many such $f$
 
@robjohn Matter is that $f$ can have just one property, our aim is to prove the existence of $d$ not $f$
Our aim is “we can find a $d$ for any $f$” (and $f$ have some property which I have described above)
I will call you “Robbie” or “Johnny” :-)
Which ever you like
 
2:51 AM
@Knight that's fine if you enjoy lightning bolts... robjohn too long?
 
 
2 hours later…
4:50 AM
@MikeMiller I suppose inverse limits of Banach spaces are naturally Frechet
Eg $C^\infty(\Bbb R)$ is an inverse limit of the Banach spaces $C^k(\Bbb R)$. The norm of the Banach spaces in each finite stage gives a seminorm and this family of seminorms generate the topology, making it Hausdorff and complete
Countable inverse limit rather
This stuff is so annoying man
 
@robjohn Robjohn is your name, I want to give you a friendly name
 
5:10 AM
@BalarkaSen i watched it last night
 
"it"?
 
5:37 AM
@BalarkaSen solaris
 
Oh
Did you like it
 
5:48 AM
@BalarkaSen the concept is interesting but it's a bit too slow and I didn't really understand the movie until I read the plot on wiki
I guess it's a bit 5subtle7me
 
fair. nah, everyone has different tastes
whats your favorite movies
like top 5 if i asked you to name
 
6:11 AM
life of pi
 
6:35 AM
Life of psi
 
interstellar
 
thats a good one
 
Can you point me to a proof of the existence of JNF that explains in detail how you would go about finding the form? :P
 
Artin does a good job but you can organize it better if you know a bit about minimal polynomials
Check out Artin I guess
 
 
2 hours later…
8:47 AM
@Knight Rob is mentioned in my profile
 
9:22 AM
@Thorgott Oh that's really neat! And when you say "anything that is improperly Riemann-integrable is also HK-integrable", does that mean there are no restrictions at all? So the function could be non-negative (I've seen this as a condition both in Tonelli's theorem and the corollary of Fubini's theorem for improper integrals)?
 
9:33 AM
@Knight your claim is true if a point $t$ where $f(t)=0$ is greater than $c$.
 
 
1 hour later…
10:49 AM
Anyone able to explain how modular arithmetic allowed this step here? math.stackexchange.com/questions/3610966/…
 
11:43 AM
@northerner $100000_2=1_2\pmod{2^5-1}$?
add $10000_2$ to both sides
 
$= 110,000_2 \mod{2^5−1} \\
= (10,000_2 + 1_2) \mod{2^5−1} \\$
how are these equal?
 
12:24 PM
Compute the number of ordered pairs (m, n) of positive integers such that (2^m − 1)(2^n − 1) | (2^10!) - 1.
 
@robjohn Wow! Really ? Please give me the proof
 
Just checking some things. If a polynomial $f(t) \in \mathbb C[t]$ annihilates $A:\mathbb C^n \to \mathbb C^n$, then every eigen value of $A$ is a root of $f(t)$ so we can factorize $f(t)=(t-\lambda_1)^{\alpha_1}\cdots (t-\lambda_r)^{\alpha_r}$ where the $\lambda_i$ are the unique eigenvalues.
So by Primary Decomposition we can write $V=\ker(A-\lambda_1 I)^{\alpha_1} \oplus \cdots \oplus \ker(A-\lambda_r I)^{\alpha_r}$. So if we consider $A_i$ the restriction of $A$ on $\ker(A-\lambda_i I)^{\alpha_i}$, its characteristic polynomial is $\chi_{A_i}(t)=(t-\lambda_i)^{n_i}$ where $n_i=\dim \ker(A-\lambda_i I)^{\alpha_i}$ so the characteristic polynomial of $A$ is $\chi_A=(t-\lambda_1)^{n_1}\cdots (t-\lambda_r)^{n_r}$, hence $n_i$ is multiplicity of $\lambda_i$ in $\chi_A$.
So $\ker(A-\lambda_i I)^{l}=\ker(A-\lambda_i I)^{m_i}$ if and only if $l \ge m_i$ where $m_i$ is the multiplicity of $\lambda_i$ in the minimal polynomial $\mu_A$.
Sounds right?
 
@FuzzyPixelz lol it's you again
 
You sound surprised
 
last time I used a lot of algebra words and you said you understood none of that :P
@FuzzyPixelz yeah sounds right
 
12:33 PM
Yeah module theory, never touched it
 
0
Q: Subbasis for a topology two problems

topologicalorientablesurfaceProblem 1: $\S$ $=$ $\{$ $(a,\infty)$ $:$ $a\in \mathbb{R}$ $\}$ $\cup$ $\{$ $(-a,\infty)$ $:$ $a\in \mathbb{R}$ $\}$ is a subbasis for $\mathbb{R}$ with the standard topology. Attempt: Clearly, $\S$ covers $\mathbb{R}$. I must show that $\tau$ (the topology generated by a subbasis) is the stan...

 
1:39 PM
Hey guys. I am trying to make mathematica do some equations to me. Can you check why it is not summing the variables in this situation)
How can I order it to sum the values? I am kind of a noob
 
1:54 PM
0
Q: Exponential inequality problem

maths studentHow to solve this inequality $e^{(x-2)}>\frac{1}{x-2}$ ? I tried to plot graph and able to get some bound as answer but not satisfy with the approach ? Anyhow we can convert this problem by substitution that $e^{x}>\frac{1}{x}$ Some help needed?

 
2:12 PM
The usual way is to have your numbers as a list of the form: {1.24, 1.25, 1.27, 1.28, 1.28, 1.29, 1.32, 1.34, 1.4, 1.54} and do operations on it.

But if you don't have that kind of input here is a more advanced way to sort the values and take partial sums.
(*start*)
nn = 10;
x1 := 1.54
x2 := 1.40
x3 := 1.34
x4 := 1.32
x5 := 1.28
x6 := 1.33
x6 := 1.25
x7 := 1.29
x8 := 1.28
x9 := 1.27
x10 := 1.24
a = ToString[Table[StringJoin["x", ToString[n]], {n, 1, 10}]]
b = Sort[ReleaseHold[MakeExpression[RowBox[{a}], StandardForm]]]
notice that you have double x6
 
What does nn represent
 
Just a variable. like n
N is a reserved letter therefore I wrote N
 
I see. Thanks.
Are you familiar with an online Course or a book for Begginers who have not had experience with programming
One guy here actually gave me once a quick guide. Was nice. But it wasn't for beginners.
 
@MadSpaces The Help documentation in mathematica gets you a long way
 
Alright. Well thanks!
 
2:17 PM
x = {1.24, 1.25, 1.27, 1.28, 1.28, 1.29, 1.32, 1.34, 1.4, 1.54}
Sum[x[[i]], {i, 1, Length[x]}]
@MadSpaces
 
@MatsGranvik Can I obtain something from this expression $$f^{-1} (x) = f \left( f(x) \right) $$
?
I’m thinking of getting $$f^{-1} (x) = f(x)$$, can I get that?
 
I don't know.
Inverse functions cannot always be found.
 
Please help me :)
 
Google surjective injective
And compositions are usually to complicated to solve in Mathematica.
 
Okay
 
2:43 PM
Reduce[Sqrt[x] == (x^2)^2, x]
https://www.wolframalpha.com/input/?i=Reduce%5BSqrt%5Bx%5D+%3D%3D+%28x%5E2%29%5E2%2C+x%5D
 
What is reduction ?
 
The same as Solve[]
 
Hi, does anyone here familiar with Haar measures?
 
Oh!
 
user131753
Please see the following meta post,
 
user131753
2:49 PM
6
Q: Autofilters for Hot Network Questions

Asaf KaragilaThe SE software allows us to request certain regular expressions be automatically excluded from the Hot Network Questions. I was asked by the CMs to make this post on the meta, and include the list. This way the community can express agreement or disagreement, as well as suggest improvements. L...

 
@MatsGranvik If I have something like this $$f^{-1} (x) = f \left( f(x) \right ) $$ and my claim is the only function that satisfy that property is the identity function. How should I go on for proving it?
After some mathematics we can obtain $$ f(x) = f^{-1} \left ( f^{-1} (x) \right) $$
:)
 
Knight: Let
$$
F(x)=\int_c^xf(t)\,\mathrm{d}t\tag1
$$
and $g(x)=e^{-x}F(x)$, so that $g(c)=0$ and
$$
g'(x)=e^{-x}(F'(x)-F(x))=e^{-x}(f(x)-F(x))\tag2
$$
Since $F(x)=e^xg(x)$
$$
f(x)=F'(x)=e^x(g(x)+g'(x))\tag3
$$
If $f(u)=0$, then $g(u)+g'(u)=0$, which means $g(u)$ and $g'(u)$ differ in sign.

If $g(c)=0$ and $g(u)$ and $g'(u)$ differ in sign where $u\gt c$, $g'(d)=0$ for some $d$ between $c$ and $u$. By $(2)$ that means $f(d)=F(d)$.
 
Here comes Robbie
Thank you so much Rob
I think the key step was taking $g(x) = e^{-x} F(x)$, ha?
 
3:44 PM
@Knight if $\omega^3=1$ then $f(x)=\omega x$ also satisfies the equation.
 
4:02 PM
@robjohn How you can come up with such a nice form for $f$ ?
 
@Knight you're looking for $f$ so that $f(f(f(x)))=x$, so...
 
@robjohn Hahahahaha, I have come up with a solution, you want to see it? Please validate it
Robbie please say you want to see my solution :-(
 
4:17 PM
Hello can someone tel me if the domain is correct: math.stackexchange.com/questions/3600804/…
 
4:42 PM
can someone help me with this: Prove that $p$ divides $ϕ(1+a^p)$, where $a≥2$ is a natural number, $p$ is a prime, and $ϕ$ is Euler's totient function?
 
@Knight why don't you just show it?
 
5:33 PM
@BalarkaSen What is the inverse limit of $\Bbb R^n$, where each map is projection onto the first n-1 coordinates
I guess you get $\Bbb R[[x]]$, but with what topology? The family of seminorms are 'norms of truncated power series'
So this is topologized as power series with convergence as convergence on first n terms
 
That sounds right to me
 
AKA, pointwise convergence
That sounds like a bizarre space
Ah no it's a standard example.
Let $X$ be a space with a compact exhaustion $\cup_n X_n$. Consider the seminorm on $C^0(X)$ given by $\|f\|_n = \|f\|_{C^0(X_n)}$ --- that is, measure distance on each compact set in the exhaustion
Then $C^0(X)$ with the pointwise convergence topology is still a Frechet space. Just not a Banach space, since functions may be globally unbounded
In the example above $X = \Bbb N$ and $X_n = [0, n]$
 
Makes sense, good example.
 
So I guess I believe your inverse limit comment
 
@TedShifrin Oddly, I just found out today that some functions don't have inverses... In all of my algebra and calc classes, I've never been given a problem where that was a possibility. It seems obvious now, though.
 
5:47 PM
You were definitely given such problems, but probably phrased badly
The map $f: \Bbb R \to \Bbb R$ given as $f(x) = x^2$ does not have an inverse, because $f(-1)=f(1)$
However the map $f: [0, \infty) \to [0, \infty)$ given as $f(x) = x^2$ has an inverse
We restricted the domain and the codomain so that $f$ is injective (restricted to non-negative input) and that $f$ is surjective (restricted to non-negative output)
 
Lame joke, forget I wrote that
 
I see what you're saying, it just was never defined that way in the classes I was in. Our "definition" was basically just telling us how to solve for the inverse functions. We didn't really pay so much attention to the restrictions and stuff.
 
I'm aware, I'm telling you that many of the examples where people talk about inverses don't actually have inverses
 
I'm not claiming the phrasing was helpful for you or rigorous and careful :)
 
5:55 PM
lol, ok
Well, I get it now. So whatevs I guess
 
6:18 PM
@robjohn the question is : Let $f: [0,1] \rightarrow [0,1]$ be a continuous function, such that $$ f \left ( f \left ( f(x) \right) \right) = x$$ is true for all $x \in [0,1]$. Prove that $f$ must be the indentity function
My solution:
It is given that $$ f \left( f
\left (f (x) \right) \right ) = x$$ for all $x \in [0,1]$ .That “for all $x \in [0,1]$” part is very important.
 
@Knight ah, it's a real function, so my counterexample doesn't apply.
 
Let’s assume that $f(x) \neq x$(that is $f(x)$ is not an identity function). Now, $f(x)$ is either greater than or less than $f(x) =x$ in some sub-interval of $[0,1]$. $$ f(x) \gt x \\
\textrm{let’s assume that f is an increasing function, well there is no possibility for it to be a constant function, it will either increase or decrease in some sub-interval (that same sub-interval in which it is greater/lesser than x} \\
f \left ( f(x) \right) \gt f(x) \gt x \implies f \left ( f(x) \right ) \gt x \\
Similarly, we can reach the contradiction if we assume $f(x) \lt x$ in some sub-interval, or if we assume $f$ to be decreasing, if it’s decreasing that inequality sign would flip.
@robjohn What was your counter-example? Please see if there is some mistake in my solution or something which needs more clarification
 
@Knight $f(x)=\omega x$ where $\omega^3=1$
 
Okay
Why you chose $\omega$ letter? Any particular significance?
@robjohn Please tell me how’s my solution
 
@Knight are you sure that $f$ maps an interval of increase to and interval of increase so that $f(f(x))\gt f(x)\gt x$?
 
6:30 PM
@robjohn Didn’t get you
 
"Now, $f(x)$ is either greater than or less than $f(x)=x$" makes no sense.
 
Inverse of $f$ is $f^{\circ 2}$, so $f$ must be a homeomorphism, therefore either strictly increasing or decreasing on all of $[0, 1]$. It will also have a fixed point by Brouwer fixed point theorem. Then the rest of Knight's argument plays out, I believe.
 
@Thorgott I know, I should have written $f$ instead of $f(x)$ in the first part, is that what you’re point at ?
 
No, simply the $f(x)=x$ part does not make sense, since you just assumed $f(x)\neq x$ in the previous sentence.
And, certainly, $f(x)$ is neither greater nor less than $f(x)$.
 
If it is not equal to $x$ then it must greater than or less than $x$ in some interval
 
6:36 PM
@Knight you assume that $f$ is increasing everywhere? if not, what happens if $f(f(x))$ is not increasing?
 
I think you mean the right thing, but you need to be more precise and provide an argument
 
I would say that the continuity is important, otherwise $$
f(x)=\left\{\begin{array}{}
x+\frac13&\text{if }x\in\left[0,\frac23\right)\\
x-\frac23&\text{if }x\in\left[\frac23,1\right)\\
1&\text{if }x=1
\end{array}\right.
$$
 
@robjohn No, I assumed $f$ is increasing in an interval where it is greater than $x$
@BalarkaSen Can we do it without the concepts of homeomorphism? Because I don’t know what it is :)
 
Yeah I was just ranting, you can write up some elementary argument
Don't pay heed to me
 
You don't need homeomorphy, just continuity + injectivity, but injectivity is crucial
 
6:40 PM
Hahahahaha
 
But continuity + injectivity is equivalent to homeomorphy in this context :p
 
WHAT!
Where my arguments made the use of (or just assumed) the injectivjty ?
 
You're assuming that $f$ is monotonic
 
I'm catatonic about my functions being monotonic
Just gets me everytime, you know
Full panic attack
 
@Thorgott in a sub-interval
 
6:45 PM
@Knight does your example work on my counter-example?
 
Still, it is not clear to me why that would be true
Generally, continuous functions need not be monotonic even in small intervals
 
@robjohn I couldn’t understand what that $\omega$ was
@Thorgott My assumption applies to any arbitrary small interval because that condition $f^{\circ 3}$ is true for every $x$. I assumed that $f$ will increase monotonously at least in a some small interval
 
@Knight in that counter-example, $\omega$ is a cube root of $1$, like $-\frac12+\frac{\sqrt3}2i$, but that requires a complex function, not one from $[0,1]\mapsto[0,1]$. I was actually talking about this one.
 
But why is your assumption true?
 
@robjohn Okay
@Thorgott Well I think every continuous function is monotonous at least in a interval (can be small), can you give a counter example as it would help me a lot
?
 
6:52 PM
@Knight There are continuous nowhere differentiable functions that are not monotonic on any interval.
For example, the Weierstrass function.
 
Okay! So, my assumption fails there, ha?
 
$f\colon\mathbb{R}\rightarrow\mathbb{R},x\mapsto\begin{cases}x\sin(1/x),&x\neq0,\\0,&x=0\end{cases}$
 
Can I prove that function of mine to be monotonic?
 
0
A: Is the derivative of the loss wrt a single scalar parameter proportional to the loss?

nbroThe derivative $f'(x)$ is correlated with $f(x)$ in a certain sense. In fact, $f'(x)$ is a function of $f$, so we could even say that there's a cause-effect relationship. The derivative at a specific point $c$ of the domain, i.e. $f'(c)$, can either be negative or positive. If $f'(c) > 0$, then ...

 
Yes, because it is injective
 
6:55 PM
is this loss?
 
@Thorgott How injectivty implies monotonicity?
 
Exercise: Take an interval $I$ and a continuous function $f\colon I\rightarrow\mathbb{R}$. Show that $f$ is injective if and only if it is monotonic.
 
I have a question for you guys, apart from checking that my answer to the following question above is correct
An injective function is a drug addict function?
 
yes
 
@Thorgott GOT YOU! It passes the horizontal line test, hence if it goes up it’s gonna continue to go upwards
 
6:59 PM
I'm not sure what that means
 
 
@Thorgott If the function is injective then it’s sure that it will cross any horizontal line just once, since it passes it once it cannot come down (else it will cut it once again) hence the only way left for the function is to go on
Upwards
 
I see, that's not wrong, but, of course, informal
 
You will be lost in formality
That should be the title of a film
 
@Thorgott How else can I prove monotonicity? (I’m going out of my problem) if that derivative condition is excluded.
 
7:06 PM
Wait, what derivative condition? None of this has had anything to do with derivatives.
 
@Thorgott If the derivative is greater than zero then the function is increasing
What are the other ways of proving the monotonicity of a function?
 
Proving it by definition
 
Please tell me
 
and if the integral of it is strictly increasing (with some other conditions) then the function is increasing @Knight
 
You can't argue by derivative since the function need not be differentiable
 
7:11 PM
@Thorgott Yes, that differentibilty shouldn’t stop us :)
 
If all you want is a solution, Google is your friend
 
No solution.
I want to know how to prove the monotonicity. If $$a \lt b \\ then ~~~~~~~ f(a) \lt f(b)$$
 
Can you prove monotonic => injective
That's the easier direction (and does not even require continuity)
 
Yes. I can do that
@robjohn Thank you so much for the friendly talk that we had today. Truly thank you for helping me.
 
Alright, for the other direction: Hint: IVT
 
7:18 PM
@Knight pleasure
 
@Thorgott Okay
I will do that, cya Thor gott
Thanks for helping me
 
Ricci curvature is trace of $T_p M \to T_p M$, $Z \mapsto R(X, Z)Y$ right
The second variation formula says if $\gamma$ is geodesic $dH(\gamma)(X, Y) = \int_\gamma g(X'' + R(X, \gamma')\gamma', Y)$
dimension of kernel of $dH$ measures the multiplicity of conjugacy of the endpoints of $\gamma$
So I wonder if I can swoosh something and get information about the diameter of $M$
Not sure what to swoosh
Take any vector field $X$ along $\gamma$, write it in orthonormal parallel fields as $X = a_i e^i$ and plug it in $dH(\gamma)(X, X)$?
Oh maybe I should take an orthonormal basis of Jacobi fields
$X = a_i J_i$, plugging and chugging gives $dH(\gamma)(X, X) =$ uhhh
$dH(\gamma)(X, X) = \int_\gamma g(a_i'' J_i + 2 a'_i J_i' + a_i J_i'' + a_i R(J_i, \gamma')\gamma', a_i J_i)$
Garbage in garbage out the last bit goes away by Jacobi equation
$\int_\gamma g(a_i'' J_i + 2 a_i' J_i', a_i J_i)$ now what the hell is this nonsense
No this cannot be right
I need curvature term to survive
Yeah I don't have a basis of Jacobi fields vanishing at both endpoints; I don't know the multiplicity
Parallel it is then
I guess I want my basis to be $\sin(\pi t) e^i$. $X = a_i \sin(\pi t) e^i$
$g(-\sin(\pi t) e^i, R(\sin(\pi t) e^i, \gamma') \gamma'), \sin(\pi t) e^i) = \sin^2(\pi t) (-1 + g(R(e^i, \gamma')\gamma', e^i))$
I should sum this crap? $\sum_i g(R(\gamma', e_i)\gamma', e_i)$ is $Ric(\gamma', \gamma')$
That should be $-\pi^2 + g(R(e^i, \gamma')\gamma', e^i)$ by the way
$\sum_{i = 1}^{n-1} dH(\sin(\pi t)e_i, \sin(\pi t)e_i) = 2 \int_0^1 \sin^2(\pi t) ((n-1) \pi^2 - Ricc(\gamma', \gamma'))$
Getting close. If the geodesic was length $c$ it'd be the same formula but $\sin(\pi t/c)$ and $(n-1)\pi^2/c^2$ accordingly. Then having $Ricc(\gamma', \gamma') \leq (n-1)\pi^2/c^2$ would make the integral positive
OK, if $Ricc(\gamma', \gamma') \geq (n-1)\pi^2/c^2$ then the integral is negative so some $dH(\sin(\pi t/c) e_i, \sin(\pi t/c) e_i)$ is negative so $dH$ is semidefinite so $\ker dH$ has stuff in it so $\gamma$ has conjugate points
If $Ricc \geq (n - 1)\pi^2/c^2$ then $diam(M) \leq c$. Bonnet-Myers boom
Cor: If $Ricc$ is bounded below by a positive constant $\pi_1 M$ is finite
Pf: Lift to universal covering where $Ricc$ is also bounded below by the same constant. Universal cover has finite fibers, so rip rip
 
8:05 PM
Oh
Pretty cool
 
8:26 PM
$dH$ is weird notation lol I meant $H$, the Hessian
 
Damn even after all this time
I log in and then immediately stumble upon some nerds
There's no escape truly
 
@AminIdelhaj get yike'd
 
you thought you'd see some people having a healthy homological algebra conversation didn't you
maybe talking about grothendieck riemann roch theorem
didn't you
 
Actually I'm probably an automorphic forms person now
 
8:34 PM
@AminIdelhaj makes sense, what with mathematics and its unusual applicability to the sciences
really, math is just a reflection of the universe
2
 
Hi people
 
@Alessandro it's time to drop it on Mike
do eet
 
I already showed him on facebook
 
oh lol
 
What's up Alessandro
 
8:36 PM
Still studying PDEs
Well I'm supposed to let's say
 
in what "directions" is binomial theorem generalized?
 
Towards the left, I would say
 
$$(x_1 + \cdots + x_n)^m = \sum_{\substack{a_1, \cdots, a_n \geq 0 \\ a_1 + \cdots + a_n = m}} \frac{m!}{a_1! \cdots a_n!} x_1^{a_1} \cdots x_n^{a_n}$$
 
Did you mean $m$ is non-integral, and factorial really means the fractional falling factorial?
 
what if $n$ is non-integral
4
 
8:41 PM
what would you suggest me to do, because in some elementary considerations i do, i got $$(\sum_{k=1}^{+ \infty}a_k)^w$$

i could do some limiting procedures
w is a natural number
 
If there is such a formula it would be given by my expression above, where i don't stop the $a_i$ at $a_n$. However, you probably need to work very hard to guarantee convergence.
I am not gonna think about convergence issues right now
 
ok, thanks
 
@MikeMiller Maybe $x_1 + \cdots + x_{p/q}$ will be an "object" in $k[x_i, i \in A]$, which, if you add $q$ other objects of the same "$p/q$" type to it, you get a genuine element of the polynomial ring
So maybe it's some kind of localization, you get my drift?
And then maybe you I-adically complete to get elements like $x_1 + \cdots + x_{\pi}$
Where $\pi = 22/7$
2
 
now you're thinking with quantum
 
But the real question is if binomial formula still holds in this adic complete local ring
 
8:49 PM
So we know the 0-categorical binomial formula
What's the 1-categorical binomial formula
 
@MikeMiller Shift?
 
Hi Demonark
 
Hey Ted, how's it going?
 
I am sure I can prove the binomial formula for $(x_1 + \cdots + x_n)^k$ if $n$ is a Liouville number, but in general it's really a diophantine approximations problem
 
Still here!
Re a @Balarka, @MikeM
 
8:54 PM
Hiya
 
Hi Ted
 
Hi Demonic
 
Hey @TedShifrin! Hope you are well!
I worked on the removable singularity problem and came to a solution. I wanted to ask you if you could take a look to double check if it makes sense what I came up with this time?
 
What've you been up to?
 
8:57 PM
Surely the categorification of the binomial formula is the construction of the tensor product complex
Why am I still doing this
I should sleep
 
@BalarkaSen I don't know if you should sleep, but you definitely shouldn't be doing this
 
$\Lambda^k V$ does categorify the binomial coefficient
but what's the binomial formula lol
The full exterior algebra categorifies $(1 + x)^n$ I guess
 
Well the 0-categorical binomial theorem (in proper context, of course) is an identity for elements of monoidal categories enriched in Ab
So presumably the 1-categorical binomial theorem is about iterated compositions of additive monoidal functors
 
Oh thats actually a good point
It's a formula for compositions of sum of two additive monoidal functors
perfect
So I think in light of Mike's comment the most general all-encompassing "motivic binomial formula" will be a spectral sequence of various compositions at different levels of $(\infty, n)$-functors
 
what are you smoking
 
9:10 PM
I did too much Riemannian geometry
does that explain it
 
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Q: What captures our intuitive notion of faces, edges, and vertices?

user76284This answer suggests that laypeople's intuitive notion of the meaning of these words is consistent with the following claims: A cube has 6 faces, 12 edges, 8 vertices. A cylinder has 3 faces, 2 edges, 0 vertices. A cone has 2 faces, 1 edge, 1 vertex. A sphere has 1 face, 0 edges, 0 vertices. ...

In case anyone is interested.
 
i cannot stand that CH is not yet resolved
 
Continuum Hypothesis?
 
yes
 
It is resolved, it's independent of ZFC, that's a resolution
 
9:14 PM
@eigenvalue Hi. Can you summarize what you found out about those $G_i$?
@BalarkaSen definitely no!
 
it is not resolved, that just means that ZFC is not enough to cope with it
 
right, yeah, the goalpost should be about 3 feet more to the right, thanks
 
@TedShifrin I started with the following Gaussian curvature term (and didn't look at the separate terms, per your advice): $K=-\frac{1}{2}\frac{\frac{\partial^{2}G}{\partial t^{2}}}{tG}+\frac{1}{4}\left(\frac{\frac{\partial G}{\partial t}}{t^{2}G}+\frac{(\frac{\partial G}{\partial t})^{2}}{tG^{2}}\right)=$.

Henceforth I write $G(0,x)$ just as $G_{0}$ in order to simplify the notation and $G^{(i)}$ for $\frac{\partial^{i}G(0,x)}{\partial t^{i}}$. Rearranging the coefficients of all the low powers of $t$ of the above Tylor series expression for $K$ yields
 
What's wrong with $G_i$??
 
For consistency, I just followed the notation I use in the rest my work
 
9:22 PM
OK, I worked it out as far as $G_1= G_2=0$.
Your formula for $K$ most things drop out, right?
 
for t=0, yes most terms drop out (as they are coefficients of a polynomial)
 
When proving an iff statement, I don't need to choose a new variable other than x when I start on the converse, right?
 
I mean the formula you typed with “I am left with”?
So I do not follow why that has to vanish.
 
yes, there is only one term left with a 3th partial derivative
To be precise: $-\frac{1}{8G_{0}^{3}}3G_{0}^{2}G_{0}^{(3)}$
 
With regard to your first question, have you taken complex analysis? Think about Laurent series.
 
9:28 PM
Nvrmnd, I figured it out.
 
no, no complex analysis. I had Taylor series during my analysis III course... That's why I am not really familiar with all of this
 
Why isn't $G_3/G_0$ perfectly fine at $t=0$?
 
@Ante And that's a solution, it completely determines what ZFC says about CH
 
But with you hinting at Laurent series, I might just check some text books to learn more about this topic
$G_3/G_0$ is indeed fine at $t=0$, but the Gaussian curvature might jump there? How do I know it is actually smooth?
 
@AlessandroCodenotti The independence just means this: "Hi Godel and Cohen, would you mind if I try to prove CH?" And they respond with: "You can try, but do not try inside of ZFC."
 
9:31 PM
Yeah, you need to take some complex analysis. The only singularity here comes from the $t^2$ in the denominator, unless $G_0$ vanishes (which you should analyze after you're done with all this). Once you cancel that out of the denominator, there is no more local issue near $t=0$.
 
@Ante Sure but as long as you want to use ZFC as your axiomatic system "CH is independent" settles it
 
Is the formula for $K$ valid for all $t$? Of course, for smoothness you need higher terms too ... are you assuming the metric is real analytic, I assume, so it is given by its Taylor expansion?
 
@AlessandroCodenotti It does not settle it, it shifts it to another domain.
 
Which domain?
 
There is no other natural domain
People have tried to come up with somewhat natural extensions of ZFC that settle CH, set theorists like the proper forcing axiom which implies $\mathfrak c=\aleph_2$ for example
 
9:36 PM
and Godel's first incompleteness theorem tells you that you will always have independent sentences
 
Yes, the formula is defined for all $t$. I have to assume the formula is real analytic to have a Taylor series. So I just made this assumption to make my life easier. Not sure if there is any other useful way to deal with the singularities.

You say " for smoothness you need higher terms too". Where can I read/lern more about the reasoning for this statement? Would that be covered in the "complex analysis" I still have to look into?
 
No, it's just coming from all the powers of $t$ in your computation. We don't know off-hand if it converges for all $t$, but you're interested only near $t=0$. Still not clear that the power series converges for $|t|<\epsilon$. YOu need to talk to your adviser about what he expects you to do.
 
@TedShifrin OK. Thank you for all your input and feedback! You have no idea how helpful you have been!
Once my advisor surfaces again, I will try to talk with him about further details. Will use the time being to look into the topics you mentioned.
 
Cool, and keep me posted. You can say thank you in a footnote to your thesis and email me a .pdf :)
 
9:52 PM
I certainly will do that :-)... I am still far from being finished, though. so you will have to be patient. And most likely I will get on your nerves again here in the chat sometime in the future (before I finish my thesis).
For now, please be safe and stay healthy!
 
I will soon be back to my computer and desk.
You Too!
@Captain: You certainly saw (perhaps did not absorb) in precalc and calc that $f(x)=x^2$ has no inverse until you limit its domain, and similarly with the trig functions.
 
@TedShifrin You're probably right. Pre-calc was a blur
 
And I always beat it to death in calculus, too.
 
pfft, my calc class was trash
I basically taught myself
 
10:12 PM
You should have taken my AoPS class :)
 
I wish I had
 
10:41 PM
which functions "satisfy" the IVT?
do they all have something in common?
 
thanks, but i do not have understanding of all the terms there yet, this Dbx_f seems too mystical a term
can you explain easily what it´s all about?
 
@TedShifrin dead inverses
 
@Thorgott ah, i found what i was looking for, thanks, i thought that i have found a new characterization of "IVT-property", but it is already known
 
10:57 PM
Heya @robjohn! Long time!
 
@TedShifrin how are things? keeping the room in line?
 
Nah. Been staying away from home for a month, but heading back in a few days. Risks everywhere ....
 
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