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00:01
i knew alan dershowitz in law school, he was actually very encouraging and not a madman. so picture my surprise when he mutated into whatever he is now.
@TedShifrin It is exactly this I'm asking you... Would you edit your own comment? Or just delet?
I will think about it, but probably not. I am tired of the way this site is abused, and I will stay on record for that.
3
ted may not realize this but i am looking to him for my own conduct. he can be the blueprint.
00:17
@TedShifrin ok
I love you, Ted.
So much love in this club.
Okay that is nice reading GH clears a lot of things.
Anyway. Thanks
good I will super commit to it + Andrea Gathman.
And I would like to apologize this community.
00:28
i can't speak for everyone, but apology accepted! i don't even know what it was for, but thanks.
Leslie, he made fun of you real badly
It was all over the front page, did you miss it?
possibly? i don't know. maybe? supposing i did.
Learn anything interesting recently, leslie?
i found math.stackexchange.com/questions/4067777/… to be pretty good today. i have a short term memory.
I like that Bill built off of Igor's answer. We need more of that on this site.
00:45
Two of the more — um — colorful personalities on MSE.
Leslie and Thorgott?
Is Thor colorful ?
He's a hoot.
Thor is always researching something interesting
To clarify, I respect Thor a great deal. My use of “colorful” was not intended to be complimentary.
01:01
william dubuque is a pill. i said it.
Hard to swallow?
Small and capsule shaped?
all of the above, and probably more.
a great number of his answers are high quality.
i'm weird. we're all weird. it isn't even criticism.
01:23
If R is a commutative ring with unity and $I_i$'s are ideals, then $rad(\bigcap_iI_i) = \bigcap_i(rad(I_i))$?
Here, intersection is arbitrary intersection
@leslietownes Lol, thanks
01:49
Quick calc qustion: if I have $\frac{d}{dx}[f(x)dx$ do the $dx$'s cancle?
Does that just evaluate to $f'(x)$ or $f'(x)dx$?
I guess the appropriate response might even be, read more...
where does this expression appear? it might not be nonsense. i just need more input.
02:27
the dxs not not cancel however.
02:39
@copper They actually do. $df = \frac{df}{dx}dx$ if we do differential forms, but the right answer for Andrew is yours.
03:28
Does polynomial function means the function in which all 3(+,-,x) algebraic operations are always present.
For example in x^2.
We have x X x(x multiplied by x) , X x X /1 but no subtraction.
So , it is not a polynomial f^n
Since there division by 1 present and no subtraction happened
No, @srijan. It is a polynomial, even though it is a monomial.
Where am I wrong @TedShifrin
Look up the correct definition.
03:46
Is there a difference between the matrix of a linear operator, $[T]_{\beta}$ and a square matrix $A$? I ask because in Insel he has been defining any concept for the matrix of the linear operator $T$ and a matrix $A$ separately. I've been noticing it again as I finish up this chapter on diagonalization.
Even though he defines diagonalization for a linear operator $T$ and a matrix $A$ separately, the only way the idea of diagonalization comes up for a linear operator is when you put that operator in matrix form. Is there something else going on behind why he would do this?
04:04
When you pick bases for the domain & codomain, then any linear operator has a corresponding matrix (in terms of those bases). If the dimensions of the domain & codomain are different the corresponding matrix will not be square.
Hey I found this problem on MSE
3
Q: Determine all the representations of the integer $2130797 = 17^2 \times 73 \times 101 $ as a sum of two squares.

user4204Determine all the representations of the integer $2130797 = 17^2 \times 73 \times 101 $ as a sum of two squares. attempt: Suppose $2130797$ is of the form $n = 2^kp_1^{a_1}....p_r^{a_r}q_1^{b_1}...q_s^{b_s}$ . Where $p_1,...,p_r$ are distinct primes congruent to $1 $ modulo $4$ and $q_1,....,q_s...

both divide 48 by 8 to get 6 representations, does that mean we could always divide by 8 the formula for the total representations as sums of squares? given $n=a^2+b^2$ then we could always take $\pm a, \pm b$ and interchange a^2 and b^2 thus 8 possible representations of the same sum, or am I wrong?
04:33
I agree @copper.hat, I guess I should've been more clear on how the terms are being used. Insel uses linear operator to describe a map going to the same vector space. i.e $T: V \to V$ so the bases will be the same and the matrices will be square ones
Then a matrix is just a matrix :-).
I mean the matrix of such an operator is indistinguishable from a square matrix.
Right, because they are the same type of object. I was curious as to why he would be so explicit in always stating them differently. ....maybe because eventually I'm going to get to situations where the linear maps won't be going to the same vector spaces, but there will be some sort of "diagonlization" idea to be discussed
04:50
Diagonalization makes sense really only with the same vector space both places (or working with an explicit isomorphism to identify them ).
Did someone say diagonalization?
:P how's it going everyone?
it's going good
0
Q: Extending $\sum_{n=0}^\infty s^{n^2}$ beyond its natural boundary

user76284Let $\mathbb{D} = \{s \in \mathbb{C} : |s| < 1\}$. Let $f : \mathbb{D} \rightarrow \mathbb{C}$ where $$ f(s) = \sum_{n=0}^\infty s^{n^2} $$ $f$ is analytic on $\mathbb{D}$. This is what it looks like: $\partial \mathbb{D}$ is a natural boundary, so ordinary analytic continuation cannot extend $f...

Any ideas?
05:14
What if there are too many questions on this site?
05:48
$-{1 \over 12}$.
06:07
hey guys good morning, I was wondering do you know about some reference treating the (approximately) heat equation with nonlinear source, something like $\partial_t u=\sigma(t)\partial_{xx} u +b(u)$ where $b$ is bounded and Lipschitz
In particular I would like to know if there's a Feynman.Kac representation for the solution
That's probably something you should ask on the main site.
@TedShifrin K. Got it now
Sorry, I probably should nt enter the chat room in the last minute of my lunch break.
I'm probably misunderstanding something pretty fundamental, but I'm trying to solve this integral $\int{x^2 e^{-\lambda (x-a)^2}}dx$.
I was playing around with substitution, but not having much luck so tried to use *integration by parts*, ie $\int{U}dv = UV - \int{V}du$. I had let $U = e{^\lambda (x-a)^2}dx$ and was computing $du$ as $\frac{d}{dx} [e{^\lambda (x-a)^2}dx]$. Which is where my original question came up.
Excuse me @TedShifrin, you said that to me? I don't know if you were discussing something above!
@AndrewMicallef that looks like the second moment of a Gaussian random variable with mean $a$ and variance $1/2\lambda$
@Chaos yeha this is an excercise for me to really focu all my misunderstanding of calculus it seams.
I probably should have mentioned that it is not a cosmic mystery, just a local one :P
But yeah ,y specific question is not, how should I evaluate this integral, but more, am I doing integration by parts right here (I'm not even asking if this is the best strategy to solve this problem)
Also I just noticed I made an error in the tex markup, that should be: $\frac{d}{dx} [e^{-\lambda (x-a)^2}dx]$
07:02
sidebar, if you reply to a message does the message owner get pinged? Is that poor form?
07:31
what can I research in electrical engineering using math
@AndrewMicallef only if you @username them. They will always get a notification if they are not on the chat page.
I use the links to make sure that the right person gets the comment, especially if the chat is busy.
I don't worry about the pings. If I am ever annoyed with pings, I turn down the volume or I turn off the pings in the "Sound Notifications" in the upper right of the page
0
A: When a subset $Y \subset \Bbb{Z}$ is periodic and closed under powering, what can we say about the size of a maximal interval contained in it?

StudySmarterNotHarderThe following code demonstrates that for all $n \geq 4$, there exists an interval $I = [p, q]$ such that $|I|$ is maximal with $I \subset Y$ and $p, q$ are both prime numbers. Therefore, the maximal size $|I|$ is (probably) bounded by the maximal prime gap in the interval $[0, m)$ where $m = p_...

Got a good one for you all. Yes, it's related to twin prime conjecture.
But I didn't tag it as such.
07:49
@robjohn thanks, I guess I should mention @leslietownes so they know I'm replying to their qestion earlier then :P
I think though I found the solution to my problem, I think it is just short hand used in the textbook examples, The actual definition of integration by parts is a bit longer than the mnemonic device: $\int{U(x)\frac{dV}{dx}}dx = U(x)V(x) - \int{V(x)\frac{dU}{dx}}dx$ I think if I use that longer definition, then nothing funny happens to the differentials
08:25
Here , for the graph of a modulus function. Why do we say the y axis as y = |x| Since the answer is the line towards right and the one to left. Second Q , why is it that x has to be one value >0 and other <0 . Since for few Q , I have got oth value of x as+ve also. For example ,2|2x-3| + 5= 27 gives x = +5 and + 7/3.
@SrijanM.T Those solutions do not look right.
Ohk. You mean second Q only right @robjohn
or the graph
the second question
Ohk. It is -7/3
Yes. Thank you.
I don't understand the first question @SrijanM.T can you rephrase it?
08:29
for the first, we don't say that the $y$-axis is $y=|x|$
I solved mg doubt for 2nd Q now.
The $y$-axis is $x=0$
Ohk.
So , there are two lines in the graph since the values of x have two values for a modulus of x
One negative and other +ve
@AndrewMicallef I got the answer . Don’t worry.
@robjohn Thanks
ok, no problem. Watcha working on there?
We say $y=|x|$ because the $y$-value is equal to $|x|$ for each $x$
08:34
is the line $y=|x|$ one dimensional?
does that even make sense to ask?
the x axis is a dimension, and y is a dimension, is the line a 2d object? Or is it 1d because there is only ever 1y for any x, so if you are on the line you are trapped in a single dimension?
@AndrewMicallef It's not a line; it's a curve. Yes it is one-dimensional.
Neither the $x$-axis nor the $y$-axis is a dimension. They measure the coordinate in a direction. Dimension is the number of independent directions.
It is not the same as the dimensions you measure when mailing a package
08:55
and it was at that moment that vitamnD walked in and I found myself on the wikipedia page for Hausdorff dimension
Depending on what you what the covering dimension might be more appropriate, Hausdorff dimension is not a topological invariant. You can draw curves in the plane (the Osgood curve) that have Hausdorff dimension 2
Hello!!
How many elements of order 35 has $Z_{350}$ ?
How do we check that?
Dowe check that with Euler function?
Or with Sylow theorem?
10:02
I haven't heard of weighted mean value theorem in my analysis course
but I think I can prove it with using intermediate value theorem
I just started writting out a question to main, and in the process of doing the typing, figured out the answer to my question, I feel I am making progress of a kind
This would make sense,
$$ <x^2> = \Big[ U(x) \cdot V(x) - \int_{-\infty}^{+\infty}{V(x)\cdot U'(x)}dx
\Big] \Big|_{-\infty}^{+\infty}
$$
The definite integral gets evaluated, to a number (incidently this case 0) and then I would evaluate the functions at the endpoints, and subtract like in evaluating any definate integral?
I deduced weighted mean value theorem for integral by using mean value theorem lol
suppose $h(x)=f(x)g(x)$ then $\int_a^bh(x)dx=h(c)(b-a)\implies \int_a^b f(x)g(x)dx=f(c)g(c)(b-a)=f(c)\int_a^bg(x)dx$
@AndrewMicallef what is $<>$?(just curious)
10:18
I think it is the notation for a "the expected value" of a probability density function
It's just funny notation in a book I'm working through
Which topic are you studying?
quantum mechanics
I went ahead and asked in main anyway, just in case I was wrong.
0
Q: Integration by parts: Am I doing this right?

Andrew MicallefSo I'm working through the problems in this book. While I'm not at school, I can totally see how this might be someones homework, so if I have made any egregious errors in anything outside of the specific question, be a sport and don't blurt it out. I should stress, I haven't taken a calculus cou...

and now must hit the hay
 
1 hour later…
11:44
Q is to find domain , range graph of f(x) = (x-1)^2. So , What I did is that let us say g(x) = x^2 . Then , x-1 to whole square in terms of g(x) is g(x-1). So , this is how I made the graph.
I wanted to know if there is any other way to solve it and whether why answer is correct or not.
jay
jay
12:40
@SrijanM.T the graph is $(x,f(x))$ ?
Can someone explain how weak convergence in $L^1(\mathbb{R}^d)$ doesnt imply strong convergence in $L^1(\mathbb{R}^d)$ ? If weak convergence of $f_n$ in $L^1$ means the convergence of the integral $\int f_n(x) g(x) dx $ for all $g\in L^\infty$ then cant I just take $g=1$ and this gives strong convergence of $f_n$?
how does that give strong convergence
jay
jay
12:56
it gives $ lim_n \int f_n = \int f $ ?
ah fair enough
i see
dw
if I have $\int (f_1(x) - f_2(x) )g(x) dx =0 $ for all compactly supported smooth functions $g$ can I say that $f_1=f_2?$
13:32
do you have something like continuity? f_1 and f_2 can disagree on a set of measure zero without affecting those integrals.
jay
jay
yes, I want $\|f_1-f_2\|_{L^1(\mathbb{R}^n)}=0$
14:28
hi all! Any idea about this question on pinch singularities?
2
Q: Question about pinch singularities

apt45I am reading the book "Analytic S-matrix" by Eden,Landshoff, Olive and Polkinghorne. In Sec.2, they discuss the different kinds of singularities of a function $f(z)$ defined as an integral of an analytic function $g(z,\omega)$ in the complex $\omega-$plane along some closed contour $\mathcal{C}$,...

15:18
i don't understand the question. it talks about singularities 'approaching' a contour. so i guess we've got a parametrized family of functions with singularities that move according to the parameter, but i'm not sure how. i might be able to evaluate the integral. is physics involved in this?
for the first time in my life, i'm not trying to be deliberately obtuse. just kinda wondering what went on in acts i and ii before i showed up in act iii.
If f(x) = x^2. Find the value for f(f(x)). So , here. I took the f(x) inside the function whose answer we have to find. So , it becomes $f(x^2) $= ($x^2)^2$. So , I don’t have any answer key with me. Please help in clarifying if I am right with my approach to the answer
looks good for me. i'm a little unsure of the question, "the value" suggests a particular value. but you have correctly computed a formula for f(f(x)).
Ohk. @leslietownes Thanks a lot. I’m sorry,it is supposed to be : if f(x) = x^2 , then find f(f(x)).
you have definitely done that. perhaps an annoying person would want you to 'simplify' the result to x^4. but i am not an annoying person.
well, i am, but not in that way.
if a function is not differentiable at a point, either the function is not defined there at all, or it is, but the limit that would serve as the definition of the derivative does not exist.
should i remove my answer too? :)
15:38
Hm. I realised that the lhd and rhd wouldn't match up in that case at the same time, sry
they may do worse than fail to match up, they may fail to exist.
i shaved my beard and my daughter keeps asking where my "fur" went and when it is going to come back.
she compares me unfavorably to the cat, who still has all of her fur.
@leslietownes yes but that happens only in case of vertical asymptote ( I wanted the function to be continous)
or is there another counter example?
you can have other stuff. think about e.g. x sin (1/x) at 0. or something like f(x) = 1 if x is rational and 0 if x isn't rational.
i forget if there was a continuity assumption in the original question. x sin(1/x) is a good example in that case.
hmm that makes sense. thanks
continuous functions can be very "wiggly." the sin example maybe obscures the fundamental point. imagine a sequence of straight line segments bouncing back and forth between two different lines. it's going to prevent the limit of the difference quotient from existing, even if it's continuous.
16:03
@leslietownes The singularity approaching the contour is the singularity of the integrated in the $\omega$-plane. So, you have two singularities, $\omega = z$ and $\omega = a$ which eventually coincide when $z=a$.
So, the singularity $\omega=z$ can approach the contour when $z$ lies on the integration path. However, I can always deform the path and I'll get the same answer for the integral, so this is not really an issue. The singularity for the integral arises when the contour is trapped between two singularities
what is known about the choice of contour? is it guaranteed to split the singularities? is it going around both of them? there's this feeling that stuff is potentially varying in time (or for lack of a better word, some parameter), but i don't understand how it's varying.
16:48
0
Q: Maximizing an integer variable such that modulo some prime $p \in \{p_1, p_2, \dots, p_N\}$ this complicated expression holds?

StudySmarterNotHarderConsider the union $U_N = \bigcup\limits_{i=1}^N (p_i\Bbb{Z} \pm 1)$, where $p_i$ is the $i$th prime number. Define a measurement $\mu(N)$ on $U_N$ namely the maximal interval of consecutive integers which it contains: $$ \mu(N) = \max \{ y - x : [x,y) \subset U_N\} $$ $\mu$ can also be defined e...

Any guesses?
$$
$$
$$
\sum_{V \subsetneq [x,y)} (-1)^{r - |V| +1} \prod_{v \in V} v^2 = ((x + (r - 1))\cdots (x + 2)(x+1))^2 \pmod p \iff \\
\sum_{V \subsetneq [x,y)} (-1)^{r - |V| + 1} \prod_{v \in V} v^2 = (r!)^2 \pmod p \tag{1}
$$
Maximize $r$ in the last expression such that the last expression holds modulo one of the first $N$ primes.
This is related to twin primes, if that helps :)
17:13
@StudySmarterNotHarder if you post a question on mse, you should give it some time (more than 30 mins) before dumping it here.
@copper.hat thanks, I'll remember from herein
And do not spam us with it, please.
Send Ted emails about it.
Suppose I have a sequence $(a_k)_{k \geq 1}$ of values in $(0, 1)$ so that $\sum_{k=1}^{\infty}a_k = \infty$. I guess $$\sum_{k=1}^{\infty}\dfrac{a_k}{(a_1 + a_2 + \cdots + a_k)^2} < \infty$$
Any first hint for showing this?
Did you set $s_k=a_1+\dots+a_k$?
17:17
Sure, you have $\sum_{k=1}^{\infty}\dfrac{a_k}{s_k^2}$
Oops, my statement is wrong
It should be $s_k \to \infty$ instead of $\sum_{k=1}^{\infty}a_k = \infty$.
Same statement.
That's what I suspected
Think.
It's dumb, but the first thing I thought was to try AM-GM
If you substitute, then substitute.
17:20
So I got $$\sum_{k=1}^{\infty}\dfrac{a_k}{n^2(a_1 \cdots a_k)^{2/n}}$$
as an upper bound for the above
Made a typo
But yeah, I tried that, thought maybe to do limit comparison with $1/n^2$, then got stuck
I guess the nicest way to write that is $\sum_{k=1}^\infty a_k (\sum_{j=1}^k a_j)^{-2}$
though I guess defining $b_k:=\sum_{j=1}^k a_j$ and $c_k:=\sum_{j=1}^k a_j/b_j^2$ isn't bad either
ugh, why can't i do indices today
So much for following my suggestion.
Substitution? Hmm.
17:26
oh
yeah
i guess i like doing a/b/c for labels in this case
what i find myself curious abuot is the exponent 2
If we assume $a_0 = 0$, we get
$$\sum_{k=1}^{\infty}\dfrac{s_{k} - s_{k-1}}{s_k^2}$$
oh, that's nice
(might have a typo there)
Keep working .
that looks awfully close to telescoping
17:31
Stop!
CEASE AT ONCE!
Ted and Semiclassical: what are your favourite matrix decompositions?
i usually work with either complex hermitian or real symmetric matrices, so typically all I need is eigendecomposition
but my answer should probably be SVD
(tho that's really a combination of factorizations)
SVD & Schur
I'm also fond of factorizations which take advantage of block structure
e.g. block-Gaussian elimination and the results you get from it
I am really warming up to Iwasawa lately.
Which is also a combination, I guess,.
17:35
I'll get back to this in a half hour or so. My guess is to work with the finite sum, see if that simplifies, and then take $n$ to $\infty$.
a lot of this just comes down to "what kinds of matrices are you dealing with"
like, if I was having to do stuff with $e^{i At}$ with non-symmetric $A$, then I'd have different opinions
@Semiclassical Can you give an example of what results here you are referring to?
but in QM you generally just deal with finite matrices with appropriate symmetry
sure. i mostly just have in mind Schur complement
This looks quite interesting. I have not seen it before. Does this come up in physics?
well
hmm
i'd say the place I encountered it was physically inspired but really more of optimization stuff
17:39
"The Schur complement is named after Issai Schur who used it to prove Schur's lemma" <-- though I have seen Schur's lemma, I have not seen a version which needed this for a proof.
Oh man sometimes I really hate the notation in GH
are you schur about that?
i'm schurly so
to sum up: $$\begin{pmatrix} I & 0 \\ -CA^{-1} & I \end{pmatrix} \begin{pmatrix} A & B \\ C & D \end{pmatrix}=\begin{pmatrix} A & B \\ 0 & D-CA^{-1}D\end{pmatrix}$$
there are also about twenty things called schur's lemma. doesn't help.
he was a sharp guy i guess
17:42
Hi all
fer schur
Salve
gauss has a similar problem. too many "gauss theorems" floating around.
you should count the instances of the use of "completeness"
one big reason the above is useful is to make it easier to check whether a matrix is positive definite
17:45
Gauss was a show-off, we get it: ur smart
If $C=B^\top$, for instance, the original block matrix is PD iff $A$ and the schur complement $D-B^\top A^{-1}B$ are as well
also, silly typo above: $D-CA^{-1}B$
Regarding my question, how do we deal with the fact that I have a bunch of constants with differing denominators? That's where I'm getting thrown off.
jay
jay
If I have a density $f$, and the push forward of $f$ by a map $T:\mathbb{R}^d\to\mathbb{R}^d$ denoted $T_{\#}f$, then does the change of measure formula say $$ \int_{\mathbb{R}^d} F(f) = \int_{\mathbb{R}^d} F(T_{#}f) $$ ?
where $F$ is a mapping $F: \mathbb{R} \to \mathbb{R}$.
18:10
Hi everyone! I'm happy to come back again. Today two guys won the Turing award of 2020, those are Jeffrey Ullman and Alfred Aho. There is a well-known quote by Alfred Aho who states the following: "Perhaps the most important principle for the good algorithm designer is to refuse to be content." Does he mean that the most important for an algorithmic designer is to think instead of being memorizing many algorithms. Is this what he said?
I bring this question here because algorithms is related to mathematics and I'd like to hear what do you think about this quote.
Okay I guess I knew what it mean. It just means an algorithmic designer should always not be satisfied. I think this works for any researcher in any field of study.
18:28
i didn't win the turing award, what do i know? :)
i first read that as "content" in the sense of fodder for online garbage, as opposed to "content" in the sense of the physical state of contentment. i think it's good if we all refuse to be content in the first sense.
what a different a 'd' makes
Regarding my prior question: any first hint on what to do once you have $$\sum_{k=1}^{\infty}\dfrac{s_k - s_{k-1}}{s_k^2}$$
I've gotta be making this too complicated. I tried using the fact that $1/s_k^2 \leq 1/s_1^2$, but then you end up getting a telescoping sum that diverges to $\infty$ if you look at the finite case.
19:19
@Clarinet: You can certainly be more clever than that. Come on.
19:30
Sure, we could also try $\sum_{k=1}^{\infty}\left(\dfrac{1}{s_k} - \dfrac{s_{k-1}}{s_k^2}\right)$
Other than that, my guess is there's something I must have to add and subtract to this.
@leslietownes I feel like the same could be said of laplace; I feel like I keep bumping into him all over the place in what seam to me unrelated places
What's frustrating to me is that it's gotta be something obvious, but my head is not seeing it at the moment.
It is. Ponder the numerator vis-à-vis the denominator and change something...
Side question: anyone know any good books about laplace that are part bio part math?
Is that even a thing?
I almost thought to multiply by the conjugate of the numerator, but that seems too ridiculous.
19:44
@Clarinetist think about the relationship between $s_k^2$ and $s_ks_{k-1}$.
Hello? I hope you are doing fine. Can you please help me as to why I got down votes on this question? I am trying to understand the matter and add as much details as I can, but I got down votes. Please help me if you can
-1
Q: Prove that one-to-one property is sufficient condition for a function to be countable

AvraFor a set $\mathcal{X}$ to be countable, it should be either finite or countably infinite. Now, to show that a set is countable, we can apply a function $f$ that will map it to another set, say $\mathcal{Y}$. If we show that this function is one-to-one, then it's sufficient to prove that $\mathca...

Math
Math is cool
@Avra i don't know why it was downvoted, but the question is loosely written. Just because a function is injective does not mean its domain is countable.
people downvote because they have indigestion. that's my theory.
@copper.hat Thank you very much. But some members should realize that other members are beginners and they put as much as they can to ask keeping in mind the guideline, but still some members always close my questions and condescending toward me
19:57
@Avra Maybe fix your question instead of getting annoyed?
Uh, guys, my book says: "Call $x\in\operatorname {End}V$ ($V$ fin. dim. vector space) semisimple if the roots of its minimal polynomial over $F$ are all distinct. Equivalently, ($F$ being algebraically closed), $x$ is semisimple if and only if $x$ is diagonalisable."
I'm confused; there exist diagonalisable matrices with repeated eigenvalues? (e.g., identity matrix)
there definitely do exist diagonalizable matrices with repeated eigenvalues.
Yea, I'm misinterpreting sth, but I don't know what
minimal polynomial =/= characteristic polynomial
@copper.hat Thank you.
19:59
ohhhh
thx
multiplicity (which comes from characteristic polynomial) > $1$ $\Rightarrow$ repeated eigenvalues
What is the reduction in acceleration down an inclined plane when rolling is substituted for sliding when both are frictionless?
I've contributed to the math zeitgeist....my work here is done.
@robjohn $s_ks_{k-1} = s_k(s_k - a_{k}) = s_k^2 - s_ka_{k} = s_k^2 - s_k(s_k - s_{k-1})$
20:11
@Clarinetist what about the relationship? (size perhaps)
Well, as $k \to \infty$, one would expect those products are pretty similar
Oh. I think I see what you're doing.
${a-b\over ab} = {1 \over b }- { 1\over a}$.
we were getting there
Yeah, I just hit that point. Telescoping sum.
that's it
20:18
Ugh, I feel like an idiot now having solved that. So all it boiled down to was the trick that $s_ks_{k-1} \sim s_k^2$, probably via limit comparison.
well, $s_{k-1}\le s_k$ so $\frac1{s_k^2}\le\frac1{s_ks_{k-1}}$
LOL
Thanks
@Clarinetist don't. add it to your bad of trick and move along :-)
or bag of tricks...
I've never seen that trick before today, but I will definitely be remembering it
I completely get it now, thanks for pointing it out
That's clever.
no, add it to your bad of trick
20:21
hindsight is sooo last year
only so many years i can use that line...
20:33
@robjohn Don't we want the reverse inequality?
@Astyx write it all out and see
I did
@Astyx $\frac{s_k-s_{k-1}}{s_k^2}\le\left(\frac1{s_{k-1}}-\frac1{s_k}\right)$
Oh ok mb
integration by parts
20:36
The name for sequences is "Abel transformation" IIRC
which is also called summation by parts
by any measure :-)
@robjohn why the angry egg?
@copper.hat what holiday might be approaching soon?
yeah, but why angry?
it's not angry; it's mean
20:39
average?
kinda has a flower power theme as well...
@copper.hat the eggs are usually decorated.
i had inferred ovo vegetarian...
21:06
Mean egg @robjohn
@TedShifrin that was the intended yolk
I saw it in the whites of your eyes.
hey chat
$\overline{\text{egg}}$?
sanity check: is $f\colon (0,1] \to \mathbb{R}$, given by $f(x) = \sin(1/x)$, sectionally continuous?
or $g$ given by $f$ everywhere at $(0,1]$ and $g(0) = 0$
both of these are sectionally continuous right?
21:18
what does sectionally continuous mean
you can find a partition of $(0,1]$ s.t. $f\big{|}_{(x_i, x_{i+1})}$ is continuous
Piecewise @Thor
@Lucas The first is continuous, period. Your definition is wrong, I believe. You need closed intervals.
$f$ is continuous everywhere on $(0,\infty)$. No need for sectional stuff.
@TedShifrin yeah... it's just weird. students are talking about this question because it looks like it's plaing wrong
we have to proof that $f$ is not sectionally continuous but it's integrable (using that the set of discontinuities has measure 0)
what???? it is continuous.
Lebesque or Riemann? presumably the latter?
21:29
yeah, riemann
@copper.hat ...yes. it is.
You cannot prove that is is not sectionally continuous.
that's it, we have no idea of what the professor meant
can't help you there...
that's my sanity check.
are you sure you posted the correct statement?
21:31
Second one. Second one.
It is not piecewise continuous, but it is integrable. No problem.
Exercise: If $f$ is bounded on $[0,1]$ and integrable on every $[\delta,1]$, then it's integrable on the whole interval.
time to read $\bar{\partial}$ Poincare lemma from Griffith and Harris
looks big haha
The $\partial\bar\partial$ lemma has a mistake.
@TedShifrin what is the mistake ?
Hypothesis too weak. I'd have to look it up.
21:41
Is it correct that if I have a basis for a Lie algebra, then exponentiating that basis is not necessarily a generating set for the Lie group?
For example, the Lie algebra of diagonal 2x2 matrices has a basis $e_{11}$ and $e_{22}$, but the exponential of these are diag(e,1) and diag(1,e) which does not generate the positive diagonal matrices.
@KarimMansour You need to assume $d$-exactness. $\partial$- or $\bar\partial$-exactness isn't good enough. I in fact gave a homework exercise in my complex geometry course to give a counterexample to it as stated.
oh very cool @TedShifrin
I have a note that the proof actually uses $d$-exactness because you need both $\partial\eta = 0$ and $\bar\partial\eta = 0$. You can play around.
P.S. 40+ years later, I don't remember my counterexample :D
I see this book altough it is great I have already found few errors
@TedShifrin I saw your name in front page btw
Oh, yeah, there are lots of errors. I never typed my list up, but I have a long list. (My list of errors in Guillemin & Pollack I did type up for my students years ago and it's still on my UGA webpage.)
21:57
that is great.
Holy cow, that book is $170 on Amazon
I will do the same thing as professor as well. It is important. I spent working on 3 hours reading some proof and it turns out it has error that I found.
@Clarinetist Yeah
My books are close to that. Math book prices are outrageous.
G/H has way more knowledge in it than all of my books put together. Just no exercises. Big flaw.
Unless they are Dover, then they are great prices.
there is not enough volume to get lower prices.
dover does volume.
used to be one could get cheap texts in India.
22:10
Dover does volume publishing wise, but does Dover actually do volume for a given textbook?
There is often a link between print volume and price, but Dover and James Stewart are counterexamples to the links (in opposite directions).
Yeah G/H has almost all I need for my research
What is G/H?
G is a group, H a subgroup and G/H the left coset space
f*ck
sniped
:-) as in book...
22:16
@copper.hat Griffith and Harris Principles of algebraic geometry.
@KarimMansour Much appreciated!
 
1 hour later…
23:29
there was a court case about textbook resale. kirtsaeng. lauded as pro-normal-folks but also legally destroyed getting things cheaper in india.
Except everyone still does the latter.
yes, at slightly higher prices. the legal regime is not determinative. physical access to the books and the cost of shipping remain obstacles.
gavin newsom has an outsized opinion of my ability to influence his recall campaign. out of my inbox, please.
23:45
@copper.hat yeah, unfortunately. the definition is also weird

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