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1:01 PM
@user21820 Thank you for the suggestion. I have not intent in writing sudoku solver by itself. I am learning propositional logic and there is sudoku solver as an example and I am trying to understand it now.
 
Ah no wonder.
 
@LeakyNun why?
 
If you have 3 orthonormal vectors and from a3x3 matrix from them like $\begin{bmatrix}
e_{1}& e_{2} & e_{3}
\end{bmatrix}$
if you multiply this by its transpose it should return the identity matrix
 
@LeakyNun Just show that for any finite fragment of ZFC you can find a transitive model M, such as V[k] for sufficiently large k, and then use a Henkin construction to construct a countable elementary submodel M' of M, and then Mostowski-collapse M' to a (countable) transitive model of that fragment. I've read that one can even obtain a countable transitive submodel of M, but not sure why, or maybe I read wrongly.
 
i can see this from right multiplication but I can’t figure out how to get it from left multiplication
whenever I do it I end up with a scalar (3)
any help showing the left multiplication produces the identity matrix?
 
1:17 PM
@AlessandroCodenotti @user21820 then ZFC would be consistent by compactness theorem?
 
@LeakyNun No, because the finite fragment must be explicitly chosen.
 
Namely, in the meta-system within which you are studying ZFC...
 
I don't understand
 
ZFC cannot prove ( every finite fragment of ZFC has a model ).
By the compactness argument you alluded to.
 
1:19 PM
heh?
 
But for every finite fragment S of ZFC, ( ZFC proves S has a model ).
 
...
this is ridiculous
 
So if you explicitly write down finitely many axioms of ZFC, then you can prove within ZFC that that fragment has a model.
 
so ZFC |- (for every S, (ZFC |- S has a model))
 
Yes with "for every finite S ⊆ ZFC".
 
1:21 PM
this is ridiculous
 
In fact, Replacement is the biggest hurdle, so you can in fact throw in all the axioms except Replacement, and then V[ω+ω] already a model of ZFC−R.
And it turns out that the more unbounded quantifiers you allow in the defining formula in Replacement, the stronger it gets.
But I don't know the details of this.
Maybe @AlessandroCodenotti does. =P
 
Nope, sorry
 
well why don't we say Infinity is the biggest hurdle since V[ω] |= ZFC-I
 
Oh then we need to get Asaf to start using chat (besides the Mods Office). =D
@LeakyNun Because we believe PA is meaningful. At least I do.
 
ok so how does R fail in V[ω+ω]?
 
1:24 PM
Because V[ω+ω] already has the powerset of powerset of N and hence has uncountable well-orderings, and so by Replacement on one of those you can construct ω[1], which has uncountable rank, appearing only after V[ω[1]].
 
is replacement like substitution?
 
@user1732 Far from it.
 
:(
 
However, if you restrict Replacement to countable sets, then everything stops at V[ω[1]], because you cannot get out of it using countable sequences.
 
@user21820 so why don't we say V[ω+2]?
 
1:25 PM
@LeakyNun V[ω+2] fails powerset.
 
2 hours ago, by Alessandro Codenotti
The idea is that $V_\alpha$ already models most of the axioms (for $\alpha>\omega$ a limit), regular limit gives replacement and strongly inaccesible powerset
so what exactly does V[alpha] model?
57 mins ago, by Leaky Nun
extensionality, union, infinity, power set, foundation, comprehension, collection
 
Hi any idea how can I solve $\cos(x)^2=\sin(4x)$? I derived it so much but couldn't end up with something clear. As a comment, it can be transformed into $\cos(2x)=2\sin(4x)+1$.
 
@user1732: If you're really interested in set theory, and you already know first-order logic, then you can take a look at the wikipedia page on ZFC. Replacement is one of the axiom schemas in ZFC. If what I just said makes no sense, then it's probably best for you to ignore ZFC. =D
 
if M is a transitive set then what does M satisfy?
 
@LeakyNun Nothing much. Transitive is really a very weak condition.
 
1:28 PM
not even extensionality?
 
Um... a transitive set does not come equipped (so the algebraists like to say) with a relation on it.
 
membership
(M,∈)
which of the 7 axioms quoted above does (M,∈) model?
 
If you take the ∈ relation from the meta-system itself, which you choose to be ZFC, then every set satisfies extensionality...
Doesn't it?
And also, transitivity of M makes (M,∈) satisfy union.
 
not every set satisfies extensionality
 
Oh.
Oops.
 
1:33 PM
Extensionality, union, separation and foundation hold in every $V_\alpha$ (and ZF proves it)
 
@AlessandroCodenotti what is your list of axioms?
 
Looks like Pairing and Powerset and Infinity and Replacement are missing from the above, which seems right.
 
The usual, separation, union, extensionality, foundation, pairing, powerset, infinity, choice and replacement
 
Oh choice...
 
Is "the usual" an axiom?
 
1:37 PM
If $\alpha$ is limit then ZF proves that $V_\alpha$ models pairing and powerset, and if $\alpha>\omega$ infinity as well
Further than that you need ZFC and inacessible cardinals
 
separation: V[α]
union: any transitive set
extensionality: any transitive set
foundation: any transitive set
pairing: V[λ] for limit λ
powerset: V[λ] for limit λ
infinity: V[α] for α>ω
replacement: V[α] for regular limit α
is this right
 
The first 7 are provable in ZF, the replacement one needs ZFC, but they are right
 
2 hours ago, by Alessandro Codenotti
The idea is that $V_\alpha$ already models most of the axioms (for $\alpha>\omega$ a limit), regular limit gives replacement and strongly inaccesible powerset
then what did you mean by "strongly inaccessible [gives] powerset"?
 
Ah, no, wait, your list is wrong, you need strongly inaccessible for replacement
(Rather than powerset as I wrongly claimed earlier)
 
@LeakyNun and choice: V[k] for limit k (needs ZFC).
 
1:45 PM
Actually, if $\alpha$ is regular limit and $\lambda<\alpha\implies 2^{\lambda}<\alpha$ then $V_\alpha$ models replacement. If $\alpha$ is also bigger than $\omega$ it models the whole of ZFC
 
do you mean $2^\lambda$ when you say $2^\lambda$?
 
@LeakyNun What is the difference between the two?
 
I mean, "do you really mean it when you say $2^\lambda$"
 
Yes it's what guarantees things are 'stuck' below λ.
 
When I hear the word "cardinals" I can't help but thinking about cardiac arrest.
 
1:54 PM
@OskarTegby or high ranking priests
 
Sorry, typo... *below α, not λ.
 
or the bird
 
If $A, B \in L_\beta$, why $A \times B \in L_{\beta+2}$?
 
I'm off. Flap flap.
 
cya
\o @AméricoTavares
 
2:10 PM
If you have 3 orthonormal vectors and from a3x3 matrix from them like $\begin{bmatrix}
e_{1}& e_{2} & e_{3}
\end{bmatrix}$
Then how can you show multiplication by it’s transpose form the left is the identity matrix? It’s easy with the right side but I can’t figure out the left
@LeakyNun any help?
 
Is the $$\bigwedge_{i=1}^{9} \bigvee_{j=1}^{9} \bigwedge_{n=1}^{9}~p(i,j,n)$$ and $$\bigwedge_{i=1}^{9} \bigwedge_{n=1}^{9} \bigvee_{j=1}^{9}~p(i,j,n)$$ equal?
 
Why isn't the chat available in the app?
 
@TedShifrin by any chance?
 
I would prefer a favorite function in the chat
such that you can save certain history
 
2:42 PM
you can create a link to a group of messages by bookmarking a new conversation.
go to your chat profile and click on "conversations" on the right hand side
here is an example
 
@JakeRose Are you aware of the theorem and proof of ( if A B = I then B A = I )?
 
3:04 PM
[Random]
$💥(\beta)[0]=$Proper class for all $\beta < $ Inaccessibles
 
3:40 PM
Hey @TedShifrin :)
 
@JakeRose: That requires a theorem. You need to know that if $A$ is a square matrix and $BA = I$, then $A$ has to be invertible and $B=A^{-1}$, so $AB = I$ as well. It takes a bit of theory to get this. You can't get it by computation.
hi @Perturb
 
hello @TedShifrin
 
There’s actually a nice main site question about how to prove AB=I iff BA=I “by computation alone”, with David Speyer giving a nice answer
 
Let $A$ be a ring and let $A[[x]]$ be the ring of formal power series $f = \sum_{n=0}^{\infty}a_nx^n$ with coefficients in $A$. Show that $f$ is a unit in $A[[x]] \iff a_0$ is a unit in $A$.
I'm stuck on the reverse direction of this problem
 
Let me see if I can find it
 
3:45 PM
Just perform elementary operations and keep track of them.
 
I think the issue is how the number of elementary row ops involved grows with the matrix size
 
What I mean is to do it abstractly, not by hand.
Namely, given any matrix A, you can perform elementary operations to 'canonicalize' it to the form E R where E is a product of elementary matrices and R is a reduced row-echelon form.
From this alone you can observe a lot of things.
If A B = I then E R B = I and you can check that R must be identity.
So you would have shown that if a matrix has a right-inverse then it is a product of elementary matrices.
 
@Semiclassical How is that 'computation alone'? Lol..
 
"Then I started to wonder if there's any "brutal" proof that does not visit the "higher" domain of algebraic structures and just uses the simple componentwise algebraic operations to prove that the dot product of the i-th row of B and the j-th column of A equals to the Kronecker delta from the given condition"
that seems pretty close to 'by computation alone"
 
3:57 PM
I suppose the OP's question can be restated as: "How does one prove that a matrix over a commutative ring is a stably finite ring with algebra alone?" — Semiclassical Aug 5 '14 at 13:27
is that supposed to be "ring of matrices over a commutative ring"?
 
yeah, that's probably right. but that was four years ago, so I've no idea off the top of my head
 
@Perturbative let $g = \sum_{n=0}^\infty g_nx^n$ be an inverse, if it exists, for $f$. Try to compute $g_n$ in terms of $f_k$ and $g_{1},\ldots,g_{n-1}$.
 
@SohamChowdhury Well I have to construct an inverse $g$ for $f$ so not sure exactly what you mean
But regardless I found an answer to the question
 
@Semiclassical Well the proof I was sketching is literally elementary row operations alone, which I think is the easiest to explain. When you perform elementary operations E[1..n] on the left on A B = I, in that order, to reduce A to I, you will get A = E'[1] ... E'[n] and B = E[n] ... E[1], where E'[k] is the inverse of E[k]. It is then obvious that B A = I.
 
Let $f = \sum f_n x^n$. If an inverse $g$ exists, $fg = 1$, i.e. we must have $(\sum f_i x^i) (\sum g_j x^j) = 1$. Some computation gives $g_0 = 1/f_0$ and $$g_k = \frac1{f_0}\sum_{i=1}^k f_i g_{k-i}.$$ Hence if $g$ exists, $g_0 \in A$ so $f_0$ is invertible. Conversely, if $f_0$ is invertible, then with the given expression for the $g_k$, $g$ is an inverse for $f$.
 
4:07 PM
Thanks @SohamChowdhury that's the answer I found as well!
 
Just wanted to see that I remembered the nuts and bolts of the proof myself, you're welcome
 
reposting something I posted elsewhere: is there any sense in which $S^{-1}A$ "represents" the localisation functor $S^{-1}- : {\sf{Mod}}(A) \to {\sf{Mod}}(S^{-1}A)$?
(motivated by $S^{-1}M \simeq M \otimes_A S^{-1}A$ and the tensor-hom adjunction)
 
I think the issue there would be to show that any nonsingular matrix can be reduced to I using finitely many elementary operations
It can, but I'm not sure how obvious that is
 
If $x$ and $y$ are two bounded linear operators acting on some Hilbert space, how does one show there exists a unitary element $u$ such that $y = ux$?
 
4:12 PM
@Semiclassical Hey
 
I feel like my question might just be a ring theory problem, right? Isn't my question equivalent to showing that any pair of linear operators are associates?
 
sounds right
 
@Semiclassical Well you are right that details would have to be provided, but that is how Gaussian elimination is proven to work. There are a couple of variants, but all of them essentially proceed by iteratively increasing the number of pivots at each round, which takes at most n elementary row operations. In total then we would need at most n^2 elementary row operations.
 
I didn't see Hippalectryon for a long period of time.
 
What's the name of a ring in which all elements are associates? Whatever it is, I need to show that $B(\mathcal{H})$ has that property.
 
4:20 PM
Associates as in identical up to multiplication by units?
 
@SohamChowdhury Yes.
 
@Semiclassical didn't you get married yet?
 
Old & good times here are long gone.
What about the famous Jasper? Did he get married? Hmmm ... (hard to say)
@Semiclassical Also I wondered at the last election why someone nice like you wouldn't want to candidate as a moderator.
@Secret I didn't see you candidating at the last elections! Strange!
 
2
Q: Sufficient Condition for Positivity of Matrix with Operator-valued Entries

user193319Let $\mathcal{H}$ be some Hilbert space, let $B(\mathcal{H})$ denote the bounded linear operators acting on $\mathcal{H}$, and let $M_n(B(\mathcal{H}))$ denote the $n \times n$ matrices with operator-valued entries. Let $A = [a_{ij}]$ be one such matrix. My question is, If $\sum_{i,j=1}^n u_i...

 
4:30 PM
i'm not really active on the main site these days
 
No ring has that property, since 0 and 1 are not associate.
 
In the case where $0$ is excluded, what property is this, @Mike?
 
Can't you choose $u_i =1$ for all $i$?
@SohamChowdhury I think 'being a division ring'.
 
Hm, in the commutative case (which I just realised is almost certainly not the case!) I think DVRs satisfy this. All nonzero elements are associate means there's a unique ideal, so it's a ... local ring of dimension 1?
 
I'm concerned with the lack activity of Hippalectryon since I have something to tell him/her ... (and I'm sure it is of interest for him/her)
(actually, that's why I returned here for a few minutes)
Well, I might return back in a few weeks, months, maybe I'll find some activity.
 
4:37 PM
@user23571113 oh, a rather horrid integral I ran into the other day
 
A plane is missing and it is presumed that it was equally likely to have gone down
in any of three possible regions. Let $1 − \alpha_{i}$ denote the probability the plane will be found upon a search of the $i^{th}$ region when the plane is, in fact, in that region, $i = 1, 2, 3$. What is the probability that the plane is in the $i^{th}$ region, given that a search of region $1$ is unsuccessful, $i = 1, 2, 3$? My try : Given that search of $1$st region is unsuccesfull implies that the plane is in either region $2$nd or $3$rd with probability $0.5$, therefore probability that it will be found
 
@Semiclassical Let me know which one.
 
@MikeMiller Oh, $1$ is associate to everything too. I'm dumb.
 
$F(x,c)=\int_0^x e^{c(\cosh s-\cosh x)}\cosh s\,ds$
I don't think it can be done in closed form, alas
main neat thing about is that, if $c>0$, then the exponential factor is zero except in the vicinity of $s=x$ where it goes to $1$
 
@Semiclassical It looks interesting. Arising from physics?
@Semiclassical Well, you should probably look at a simpler form.
 
4:47 PM
@user23571113 yeah, from some of my own research that I've been playing around with
 
hi @TedShifrin
 
@Semiclassical $$H(x,c)=\int_0^x e^{c\cosh s}\,ds$$
 
the goal is really to infer how $x,c$ have to be related in order to get $F(x,c)=1$
 
@user23571113 I don't think my maths knowledge is solid enough to run for moderator. I expect a maths moderator to be relatively well versed in many fields of maths
 
@user21820 @Semiclassic: I wasn't suggesting the theoretical arguments were anything deep or difficult. I just meant to say that you can't in the case of @JakeS's explicit question see it in a direct computational fashion.
hi @Shobhit.
 
4:48 PM
can you help me with the problem above?
 
@user23571113 in truth, that's probably a good approximation anyways since $\cosh s\approx \cosh x$ for $s$ in the vicinity of $s=x$
 
@Shobhit: Just because the search of region 1 was unsuccessful, you cannot conclude that the plane is not in region 1 !!
 
oooh
 
@Semiclassic: rather horrid? Ugh.
 
true
 
4:50 PM
lol
I'll probably just have to go with a saddle-point approximation and be happy with that
 
You could try differentiating with respect to the parameter $c$, @Semiclassic, but that looks equally ugh.
 
I don't expect indefinite intergrals of the form $$\int a^{a^{p(x)}} dx, a > 0, p \in P(\Bbb{R})$$ has closed forms as special functions
 
The only theoretical tool that might be handy is the Jacobi-Anger expansion of $e^{z \cosh \tau}$
 
the above integral basically have two $e^{e^x}$ terms in it and people don't play attention to tetration functions alot to give them special functions
hmm... [random]
 
4:56 PM
i.e. $e^{z\cosh\tau} = I_0(z)+2\sum_{n=1}^\infty I_n(z)\cosh n\tau$ where $I_n(z)$ is the nth modified Bessel function of the first kind
 
Let $y=e^x$. Then
$I = \int e^{e^x} dx = \int e^y dx$
$I' = e^y$
$I'' = y'e^y=ye^y$
$W(I'')=y$???
 
@Semiclassical Did you check the table of integrals? Not sure if you can get something neat as expected.
 
which I guess gives $$\int_0^x e^{c \cosh s}\,ds= I_0(c)x+\sum_{n=1}^\infty \frac{2}{n}I_n(c)\sinh n x $$
@user23571113 not yet. can't say I'm really expecting anything nice
 
$$\int_0^x e^{\frac{c}{2}(e^s - e^{-s})} ds$$ hmm...
that's the wrong form to do gaussian stuff
hmmm...
Let $u = e^s - e^{-s} \implies du = e^s+e^{-s} ds$
 
@Semiclassical Kept it in mind and return later if I get anything useful.
 
5:03 PM
kk
this is coming out of me trying to do an old problem in a novel way, so there's a good chance I'm doing something nonsensical :p
 
$$\int_0^x e^{c \cosh s} ds = \int_{-\infty}^{stuff} \frac{e^{\frac{cu}{2}}}{(e^s + e^{-s})_{s \to u}} du$$
$$ = \int_{-\infty}^{\ln \bigg(\frac{1}{2} (\sqrt{x^2-4} + x)\bigg)} \frac{e^{\frac{cu}{2}}}{\ln \bigg(\frac{1}{2} (\sqrt{u^2-4} + u)\bigg)}du$$
The denominator does not seemed able to be converted into elliptic integral stuff
I wonder if this integral has symmetry of the form $e^{\int f(x) dx} = \int g(f(x)) dx$ where $g$ is known...
 
how to do counter integral of a elliptic curve.
 
What does that even mean, @Adarsh?
@LeakyNun We have enough cardinal sins on earth these days, Leaky.
 
lol
 
5:35 PM
Can I bump my question here in chat?
-2
Q: Are the row sums of the lower triangular part of matrix $T_{2}$ equal to the row sums of the lower triangular part of matrix $T_{4}$?

Mats GranvikThe starting point for this justification is the Dirichlet series associated with the formula for the von Mangoldt function: $$\Lambda(m)=\lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|m} \frac{\mu(d)}{d^{(s-1)}} \tag{1}$$ which is a infinite symmetric matrix where the $n$-th row has peri...

Latex is so bad. It is like art. You never really know if it is true or not. Whereas Mathematica code never lies. Or at least it gives you an indication if you are on the right track. I like Mathematica. I wish everybody did.
 
Ehhhh
I usually like Mathematica
but it has its own obscure bits
(Curse you NDSolve)
 
Mathematica is not perfect. It has a rather serious bug in the sum rule of integration.
among other known bugs
 
I think there's also some error messages which haven't been properly updated to account for version changes
But this is again mostly to do with NDSolve and its obscure magic
 
5:54 PM
How do I find the $(n+1)th$ derivative of $\ln(1-x^2)$ ?
 
@Tanuj: The easiest thing is to write down the Taylor polynomial for $\ln(1-x)$ and substitute $x^2$ in it. I have no idea what you know (rigorously or not).
 
@TedShifrin okay
@TedShifrin What after that ? Do I differentiate every term individually ?
 
Yes, but you should realize that the first $n/2$ terms (roughly) will disappear.
Why are you doing this?
 
Let $G$ be a group and $\phi:G\to G$ be a homomorphism. If $G/\text{ker} \phi$ is isomorphic to $G$, then is it necessary that $\ker \phi=\{0\}$?
 
I mean, I assumed you mean the $(n+1)$st derivative at $0$. You want it at general $x$?
 
6:00 PM
Actually I'm finding the remainder term for the taylor polynomial at x=0
$$R_n (x)=\frac{x^{n+1}}{(n+1)!}\cdot f^{n+1}(x)$$
 
So what you want to do is find the remainder term for $\ln(1-x)$ and then think about what happens when you compose with $x^2$, @Tanuj.
 
I get something of this sort
 
@user170039: What do you know about $G$? Is it finite?
 
$$\frac{(n+1)!}{(1-x)^{n+1}}$$
 
@TedShifrin $G$ is an arbitrary group.
 
6:03 PM
Your formula is wrong, @Tanuj. The derivative is evaluated at some $\xi\in (0,x)$, not at $x$.
 
@TedShifrin I get it , in the question , it is given to be between 0 and 0.5
 
If it's totally arbitrary (not even finitely generated), I'll bet the answer is NO, @user170039.
 
@TedShifrin But I can't find a counterexample.
 
@user170039: Try playing with an infinite direct product or infinite direct sum.
 
howdy y'all
 
6:04 PM
howdy, @EricSilva.
You healthy again now that you didn't give your wonderful talk?
 
im healthier certainly
hopefully will have time to squeeze in the talk early next week before i depart from chi-town
if i dont though, i wont be too upset over it
no use crying over spilled milk
 
I'm sure the kidlets are despondent.
 
@TedShifrin If you have time ,
0
Q: Evaluation $(n+1)th$ derivative of $\ln(1-x^2)$

TanujThe function $\ln(1-x^2)$ is approximated about $x=0$ by an nth degree Taylor's Polynomial. Find n such that $|Error|<0.1$ on $0\leq x\leq0.5$. My Try: So I know to evaluate the remainder term of the Taylor polynomial, I can use the following expression $$R_n (x)=\frac{x^{n+1}}{(n+1)!}\cdo...

 
a few of them are hopeful and that'
certainly encourages me
 
Other than when I was in the hospital for major surgery, I never missed a class. I did teach pretty sick a few times. Probably shouldn't have.
 
6:06 PM
yeah i had a throat issue so i couldnt vocalize
 
Yeah, I've been there :(
@Tanuj: Fix the remainder formula and put in $\xi$.
 
poorly coincided w a flare up of my many chronic pain issues
 
Then realize that the T.P. of degree $n$ for $\ln(1-x)$ will give you the T.P. of degree $2n$ for $\ln(1-x^2)$.
 
@TedShifrin done
 
ugh, @EricSilva. You need better stress management in your life ... cuz grad school is gonna be stressful at times.
 
6:08 PM
@TedShifrin yeah
 
So you need to find the error term for which polynomial for $\ln(1-x^2)$? Which degree are we using? Don't tell me $n$.
 
@TedShifrin yeah :( i really do
and just better genes would be nice too
 
things like yoga and meditation have been recommended to me, but I'm not a very stressed person, @EricSilva.
LOL, well, I doubt you can put in a work order for different genes.
 
sad
 
@user170039 No, but I won't spoil a counterexample
 
6:10 PM
yeah, @EricSilva, I know ... as I suffer with my neck degenerating and a few major heart surgeries ... my genes aren't so wonderful, either.
 
ive been better in recent years than when i was a teenager at least
so im a little optimistic
 
me too (for you) :)
hi demonic @Alessandro
 
@AlessandroCodenotti Where can I find such a counterexample? Can you point me to a source?
 
i had to do MRIs like once a week back then :(
 
6:12 PM
OMG
 
I know such a counterexample, but I want you to think about it
 
that does a number on my stress about my own body being against me lol
 
yup @EricSilva. I had MRIs regularly before/after my cancer surgery on my arm. Being jammed in that noisy constricted space for an hour or two ain't fun.
Any fun math to talk about instead? :)
 
@user170039 look at @Ted's hint
 
First of all you need to find a group $G$ that has a proper subgroup isomorphic to $G$, obviously this can't happen for finite $G$ but it'd be even better if $G$ weren't finitely generated, just to be sure
 
6:13 PM
Ted already was more explicit than that.
 
@TedShifrin n+1 ?
 
But Ted didn't give a complete answer, of course, Ted being Ted.
4
 
Oh, I didn't see
 
@Tanuj. Let me start this way. Take the $n$th degree poly for $\ln(1-x)$. When you plug in $x^2$ what degree T.P. is this for $\ln(1-x^2)$?
 
@TedShifrin i helped a kiddo w a hyperbolic geometry lecture that im about to watch
i anticipate it will be good, she's a v good student
 
6:15 PM
Oh that's cool. Oh, what is she talking about explicitly?
 
@SohamChowdhury Ok. I will think about it.
 
deriving the isometry group and getting the geodesics
 
Oh for $\Bbb H^2$, I presume?
 
2n
 
indeed
 
6:16 PM
Hmm, what about this easier problem: Can you find a vector space $V$ and a linear map $L\colon V\to V$ such that $V/\ker(L)\simeq V$ but $\ker(L)\neq\{0\}$? If you can then by saying "$\Bbb Z$-module" instead of vector space you'll have the counterexample you're looking for
 
OK, @Tanuj. It could even be $2n+1$. Why? So which remainder term will give you a better result?
 
she is also the one who lectured about conformal maps earlier on when we were doing complex analysis so it was some nice continuity
 
@EricSilva: I assume they're ignoring my notes this summer? :P
 
@TedShifrin 2n+1 I suppose
 
we gave her an amalgam of sources to prep from on this one including your notes
 
6:18 PM
Too many sources is very difficult for newbies. :)
 
yeah but she can handle it i think
 
Neat, @EricSilva. If there's a video, let me see it :)
@Tanuj: OK. Regardless, what is the error term for $P_n$ for $\ln(1-x)$, precisely?
 
there unfortunately is not but ill tell you how it goes
 
Okey dokey.
 
@TedShifrin $$\frac{x^{n+1}}{(n+1)!}\cdot \frac{(n+1)!}{(1-x)^{n+1}}$$ ?
 
6:22 PM
Rehi @TedShifrin
 
Where's the $\xi$? @Tanuj
rehi @Perturb
 
this is some advanced level of dark sorcery
 
Oh, yeah, I was reading that earlier, @Alessandro. Really cool result I hadn't known.
 
@TedShifrin x=0.5 to make this whole value max ?
 
No, no, no. I'm not worrying about intervals yet. Get the formula for the error correct.
 
6:26 PM
@TedShifrin What's your views on olympiad/contest math?
 
I'm not particularly good at it, @Perturb. Some of the stuff interests me, some doesn't.
 
Neither am I, I'm writing one tomorrow and I don't really wanna prepare for it, I'd rather spend time doing some algebraic topology
 
IT's sorta hard to prepare for it, although there are some bags of tricks people who do that stuff are good at.
 
@TedShifrin I don't know how to do it .
 
Yeah you're right @Ted
 
6:30 PM
@Tanuj. I made you put the $\xi$ in the formula so that the remainder formula would be correct. And yet then you didn't do it in this example.
 
@TedShifrin Okay , so $\xi$ will just replace $x$ in the expression I wrote but only in the second part of the expression ?
 
So the remainder term will be $\dfrac{x^{n+1}}{(n+1)!}\dfrac{(n+1)!}{(1-\xi)^{n+1}}$ for some $0<\xi<x$. Right.
Now ask yourself, "Self. What's the largest that second term can be?"
(Note: In a minute we're going to be looking ONLY at positive $x$, hence only positive $\xi$.)
hi @Oskar
 
@TedShifrin when I put $\xi=0.5$
 
Well, no. Hold on.
But putting $\xi=x$ will give us the maximum (supremum).
Next we're going to substitute $x^2$ for $x$ everywhere to get our function $\ln(1-x^2)$. THEN we ask what interval we're on.
 
@TedShifrin okay
 
6:38 PM
Hi, @TedShifrin!
 
Did you sort out that uniform convergence question (plus the extra exercise I gave you)?
 
I'm going to look at it now. I worked on an other exercise today. :)
 
@Tanuj: So, for the T.P. of degree $2n$ or $2n+1$, what is our error estimate?
 
What are your thoughts on the effectivity on passive learning such as lectures, @TedShifrin? Research shows that it's not the most effective way to do it, but everyone everywhere still does it.
 
@TedShifrin replace n by 2n everywhere ?
 
6:46 PM
I still like non-passive lectures, @Oskar. I tried to engage students as much as I could. Plus then working with them individually or in groups. If I hadn't retired, I might have tried having students watch my videos of the lectures and then having class be a bit more just problem-solving. I'm not sure.
No, @Tanuj. You replace $x$ with $x^2$. But the $\xi$ term doesn't change ... except that now $\xi$ lives where?
 
@TedShifrin oh of course ! 0 to $x^2$ ?
 
Okay. I've noticed that I learn much more by doing projects than in class. I guess that everyone's different. Maybe that's just the best way to do it for me.
 
right, @Tanuj.
 
@TedShifrin okay something right at last. Oof
 
@Oskar: A good lecture can give you lots more intuition than just working in a corner by yourself can ...
So now you see that you have $x^{2n+2}$ (so that's matching being the error formula for a polynomial of degree $2n+1$) @Tanuj.
@Oskar: The really interesting question is how to learn how to read research papers and do research. That's a huge step from being in classes.
 
6:50 PM
@TedShifrin yup
 
OK, so now, if $x\in [0,0.5]$, then $x^2\in ?$ and so $\xi$ lives where?
Now you should be able to finish up, @Tanuj.
 
@TedShifrin $x^2$ should lie from 0 to 0.25.
$\xi$ lives between 0 and 0.25 ?
 
Right, so now you can figure out the maximum value of that second coefficient.
 
@TedShifrin I surely can. Thanks.
 
Now you just have a bit of calculator work to finish up. :)
Nicely, the factorials cancel out.
 
6:55 PM
yeah
 
Cool. You understand what we've done now?
BTW. The fact that the error term we just found has $x^{2n+2}$ proves that the polynomial must be the T.P. of degree $2n+1$ for our function. :)
 
@TedShifrin got it .
 
I think that working on a problem and talking with someone who's giving you clues is a really good way of doing it. I've read a few research articles, and it's definitely an acquired taste.
 
@Oskar: Yes, and I was fortunate enough to be able to do that a lot with students, but it is hard to work that way effectively with 50 or more students (or even 25).
 
@Ted quel est votre histoire avec mathematique?
 
7:17 PM
Yeah! Here at Uppsala University I heard that for every class, which is two hours here, the professor gets eight hours of preparation time. That's a lot of time to sit around in small groups and talk about questions that have popped up.
 
7:30 PM
Hey guys, I'm not sure what I'm doing wrong with my calculations here, I'd appreciate help.

We know that given some number $x$ in the range $a$ to $b$ we can work out what percentage of $x$ that is in between $a$ and $b$ with $\text{Percentage}(x,a,b)=\frac{x-a}{b-a}$ as explained in this post https://math.stackexchange.com/q/754156.

Now, given the following function $f(x) = x^2 *10$. The values for the functions are as such
https://gyazo.com/197f5e950772a5844ae509868f0c5ce8

Let us arbitrarily choose $a=10$, $b=40$, and $x=25$ which is $50%$. We can confirm $25$ is $50%$ using the equation
 
7:49 PM
When I said $x=25$ is $50$, I meant $50$%
 

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