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12:40 AM
"and nobody will know what that person does or why they are paid so much" hilarious
i wonder whether there's some sort of high-level view or diagram that makes sense of the bureacratic apparatus of a university
cool article reference
 
 
2 hours later…
2:25 AM
If anyone is able to comment on my last question in the chat above I would greatly appreciate it :)
 
3:03 AM
EE18: i'm not sure i understand the scrollback. there is the definition of multiplication in formal polynomial rings which is as you note a matter of definition. there is also just a formula for multiplying "polynomials in an element of an arbitrary ring." it sounds like you're looking to prove the latter and if so, yes, induction is one way to do that. lots of choices in how to structure the argument.
e.g. fix one of the polynomials and prove by induction on the degree of the other one, or induct on the total or maximum of the degrees of the pairs of factors, or something
if you get stuck you can probably google it. there will be a lot of results relating to infinite series but it wouldn't surprise me if it is also just an algebra book exercise.
 
3:27 AM
@LukasHeger how much are you looking to make for 1 hour of tutoring?
 
 
2 hours later…
5:40 AM
Yes I am looking for the latter @Jakobian :)
you would induct on the degree of one polynomial? I ask because I've been unsuccessful finding this online
 
there are all sorts of ways to structure the induction. i wouldn't let my recommendation be 'the way.'
you'll need to be clear about what your definitions are, though. you'll probably have, i dunno, a set of formulas for coefficients that is itself defined inductively. somewhere there is a definition having properties that are verified by induction.
i'd wonder slightly why you are bothering with this, if this is self study.
less important than the abstract "any definition that uses sigma notation can be thought of as an inductive definition" concept is being able to work with sigma notation, e.g. being able to re-index two different things written in sigma notation so that they can be clearly added together without the usual thing calculus books do, where they write out a few representative terms and do " . . . "
the best practice for this kind of thing might still be in calculus books, however, and stuff involving manipulation of series.
or maybe the right kind of discrete probability theory textbook.
look at the hash that the questioner and various answerers make of math.stackexchange.com/questions/1937543/… for example. although the green box (interpreted as suggested by the commenter) is a good example of something that it would make sense to base a definition around, and prove by induction. in that example the inner sum on k in the RHS is itself defined inductively.
math.stackexchange.com/questions/4778584/… is a cool question. someone asks how google sheets fills in 'smoothed' curves to fit a finite set of points, and someone posts various plausible guesses that come visually close but do not quite get there.
a good experimental question.
i'm assuming that google does not simply document how it does that somewhere, in which case it is a less interesting question.
 
6:33 AM
10
Q: Lower bounding $|1+z+\cdots + z^{n-1}|$ when $z\approx 1$

David AltizioI am trying to lower bound $|1+z+\cdots + z^{n-1}|$ when $z$ is a complex number close to $1$ (and $n$ is sufficiently large). My main concern occurs in the case $z = 1 + it$, where $t$ is small. In these cases, the terms of the geometric series spiral around the origin, but because $|z| > 1$ t...

also cool
 
6:45 AM
A Co-contributor of the site is an ntributor
 
7:20 AM
0
Q: Solve the system of equations $x_1+10x_2-x_3=3,2x_1+3x_2+20x_3=7,10x_1-x_2+2x_3=4$ using the Gauss-Elimination with partial pivoting.

Thomas FinleySolve the system of equations $$x_1+10x_2-x_3=3,$$$$2x_1+3x_2+20x_3=7,$$$$10x_1-x_2+2x_3=4$$ using the Gauss-Elimination with partial pivoting. I tried solving the problem as follows: We have the augmented matrix as $(A,b)=\begin{pmatrix} 1 && 10 && -1 && 3\\ 2 && 3 && 20 && 7\\ 10 && -1 && 2 && ...

Need some help with this.
 
 
2 hours later…
9:30 AM
integral from 0 to pi tanx dx is divergent right?
since integral tanxdx from 0 to pi/2 is divergent!
right?
 
Thank you :-)
 
9:56 AM
@MathCrackExchange 15$/h
 
 
2 hours later…
11:43 AM
I'm working with a non-zero sequence characterized by increasing by a factor of $2$, i.e. $$x_n\geq 2\cdot x_{n-1}.$$ Does this sequence diverge?
 
@LukasHeger do you do comm algebra/algebraic geo tutoring?
 
@sunny My educated guess is yes, since $x_n\geq 2x_{n-1}\geq 2^2x_{n-2}\geq\ldots\geq 2^{k-1}x_1$, and so as $k\to\infty$, the RHS of the inequality tends to $\infty$.
 
@shintuku yes. I'm more familiar with scheme theory than classical AG though
 
where could I contact you?
 
do you have Discord?
 
11:52 AM
yeah
 
 
1 hour later…
12:55 PM
@F.White I've made progress. Using (a) I can show that $\mathbb Z[1/2+1/2\sqrt{-3}]$ is a PID, since (a) tells us that taking the Euclidean norm is a norm function
so now what's left to do is argue (b) using that $\mathbb Z[1/2+1/2\sqrt{-3}]$ is a PID. yesterday I thought I had an argument, but it was faulty, so I'm revising it now
 
@EE18 you meant leslie?
 
can someone help me here?
1
Q: How can I show $\|XMY\|=\|M\|$?

user123234 Let $M\in \operatorname{Mat}_{d\times d}(\Bbb{C})$ and define $\|M\|:=\sqrt{\sum_{i,j} |M_{i,j}|^2}$. Let $X,Y\in \operatorname{GL_{d\times d}}(\Bbb{C})$ s.t. $X^HX=I, Y^HY=I$. I want to show that $\|XMY\|=\|M\|$ My Idea was the following. First we know that $\langle AX,Y\rangle=\langle X, A^H...

 
 
1 hour later…
2:13 PM
Ah yes I did mean Leslie, my apologies @Jakobian I got the icons mixed up!
 
2:37 PM
@LukasHeger does you offer still holds that I can ask alg numb thy questions? (I did some reading last year, but this year I'm actually taking the course)
 
@leslietownes To your point, I'm just trying to make rigorous the argument for reindexing the sum and what not but I don't know why I can't seem to find the argument online. Seems like none of the polynomial related wikipedia articles have it
 
2:57 PM
@ShaVuklia sure! I love to answer alg num thy questions :)
 
I need to find $M\in M_2(\Bbb{R})$ s.t. $e^M=\left(\begin{matrix}
-1 & 0 \\
0 & -1
\end{matrix}\right)$. They wrote that the eigenvalues should satisfy $e^\lambda=-1$ but I mean then I don't get a matrix with real coeff. Is there an error? @LukasHeger
 
I see how we have $c_k = \sum_{i,j,\, i+j = k} a_ib_j$ in polynomial multiplication, but how we get that to our classic $c_k = \sum_{i = 0}^k a_ib_{k-i}$ I can't quite see rigorously
 
3:22 PM
@user123234 do you know about how you can embed complex numbers into real 2x2-matrices?
you can use that
you know that $e^{\pi i}=-1$
now $\pi i$ corresponds to the matrix $\begin{pmatrix} 0 & \pi \\ -\pi & 0\end{pmatrix}$
@user123234 your mistake is that a real matrix can have complex eigenvalues
 
@LukasHeger great! I've also asked the question on the main site (with no response). shall I forward the link or shall I rephrase it here?
 
just linking is fine
 
1
Q: Fractional $\mathbb Z[\sqrt{-3}]$-ideals

Sha VukliaI want to show that each fractional ideal of $\mathbb Z[\sqrt{-3}]$ is of the form $\mathbb Z[\sqrt{-3}]a$ or $\mathbb Z[1/2+1/2\sqrt{-3}]a$ with $a\in\mathbb Q(\sqrt{-3})$. Initial considerations: Write $R=\mathbb Z[\sqrt{-3}]$ and $S=\mathbb Z[1/2+1/2\sqrt{-3}]$. Given a fractional ideal $I$ of...

 
it's enough to show that any $R$-fractional ideal is a $S$-fractional ideal, right?
 
but that's not true right?
 
3:28 PM
hmm, yeah
this would be easier if one can use the conductor of an order
 
unfortunately I'm just at the beginning of the course, so we have very little to work with
 
3:47 PM
My guess is that the non-invertible ideals are those of the form $S\cdot a$, since I also have to show that the Picard group is trivial
 
Let $I$ be a fractional ideal of $R$. Consider $\{x \in K:xI \subset I\}$. One can check that this is a subring of $K$
it contains $R$ and is contained in $S$ (because any element has to be integral)
There is no nontrivial intermediate ring between $R$ and $S$
so we have two possibilities: $\{x \in K:xI \subset I\} = S$, then $I$ is in fact a $S$-fractional ideal and hence of the form $aS$. Or $\{x \in K:xI \subset I\}=R$, then this implies that $I$ is invertible
 
ah right, our notes call that the multiplier ring $(I:I)$
I have to convince myself why it is contained in $S$
 
do you know different characterizations for when an element is integral?
and do you know that $S$ is the set of all integral elements over $\Bbb Z$ in $K$?
 
yes, $S$ is the ring of integers of $\mathbb Q(\sqrt{-3})$. I am aware of the characterizations, but I have to refresh them, so give me one moment
I suppose you're thinking of $s\in K$ is integral over $\mathbb Z$ iff $\mathbb Z[s]$ is a finite $\mathbb Z$-module?
 
4:02 PM
not quite
$s \in K$ is integral if there exists a faithful $\Bbb Z[s]$-module that is f.g. over $\Bbb Z$
 
hm, I might have seen that in Atiyah-Macdonald a long time ago
ah yes I have, Prop 5.1 (iv)
 
but it's useful here
 
ok, so my guess is that $I$ would be our fatihful $\mathbb Z[s]$ module
 
at this point, all that is left is to compute the Picard group of $R$, which you said you wanted to do anyway
@ShaVuklia correct
 
I guess because we have a domain
it is faithful
and it's a module by the above
 
4:09 PM
yup
 
ok, cool, let me parse the remainder of what you wrote
ok, I'm convinced there is no intermediate subring
@LukasHeger that is true, but the exercise I'm doing asks me to first prove that each fractional $R$-ideal is of the forms mentioned, and after that they ask to show that $R$ has trivial Picard group, so my intention was to use the description to show that the Picard group is trivial
my guess would be to show that an $R$-fractional ideal of the form $Sa$ is not invertible, from which we can deduce that the Picard group is trivial
 
4:34 PM
hm, and I also have to see why $(I:I)=R$ implies that $I$ is invertible
I've shown in a previous exercise that if an ideal is invertible, then it is proper (i.e., $(I:I)=R$)
so the logical way would be to argue that if $(I:I)=R$ then $I=Ra$, and then with the exercise just mentioned I find that the Picard group is trivial
 
5:19 PM
I've hit a wall on this one, so I'll leave it for today. I will probably ask my TA on Tuesday how that to show $(I:I)=R\implies Ra$. I'm really not comfortable with algnumb, so having solved the main part of the exercise (there were several subexercises) is already an achievement :')
 
5:35 PM
@LukasHeger let's make it $20 / hr
 
okay :)
 
So I can't start rn, but maybe later today
what time is it where ur at
 
19:36
 
Thx, let's connect over email
fruitful approach (one word) at the mother g-ship
Email me!
 
email sent
 
5:41 PM
K uploading the book now it's 80MB
K, so I'm confortable starting as far as convergent power series
We could use Discord, MS Teams, Google Meet, Zoom, MSE Chat rooms
 
H is a Hilbert space. e_n's are orthogonal basis vectors of H. T:H-->H is bounded linear operator. Is it possible that T(e_n) =e_1 for all n>=1?
 
Hi guys, if you have a function which is lipschitz and infinitely differentiable it is straight forward to show its derivative is bounded. Can anything be said about higher order derivatives?
 
Its solution says: clearly it is not possible.
 
@MathCrackExchange I've made an MSE chatroom
 
@Monty on a closed & bounded interval, yes.
 
5:48 PM
hmmm
so on the whole space there is no such condition that guarantees uniformly bounded derivatives
 
6:28 PM
@Monty Not if your whole space is not compact.
 
no yeh I meant $\mathbb{R}$.
 
For example $f(x)=x^2$ on $\mathbb{R}$.
 
ok thanks:)
 
@Koro Try $x_n = \sum_{k=1}^n \frac{1}{n}e_n$
$x_n\to \sum_{k=1}^\infty \frac{1}{n}e_n$, but $\|T(x_n)\|\to\infty$
 
6:53 PM
Is this proof correct? Problem: show that if $a_n>0$ and $a_n \to +\infty$ then $\frac{a_n}{e^{a_n}} \to 0$. Proof: since $a_n>0$, we have $\frac{a_n}{e^{a_n}}=e^{\log a_n -a_n}=e^{-a_n \left(1-\frac{\log a_n}{a_n}\right)}$. Since we proved that $\frac{\log x}{x} \to 0$ when $x \to +\infty$, from the theorem that binds limits of functions with limits of sequences I $\frac{\log a_n}{a_n} \to 0$ as well because by hypothesis $a_n \to +\infty$
Hence $-a_n \left(1-\frac{\log a_n}{a_n}\right) \to -\infty$ and finally $e^{-a_n \left(1-\frac{\log a_n}{a_n}\right)}\to 0$.
In our course, we proved both the theorem I quoted and the fact that $\lim_{x \to +\infty} \frac{\log x}{x}=0$; just to give context. :D
 
7:09 PM
Re-reading this, I could shorten this by using that $x/e^x \to 0$ as $x \to +\infty$.
 
@Monty Hardly likely without more info.
@Gwyn If you know that, yes.
 
Thanks for checking that, Ted! I suspect that a more elementary proof is required, I will think about that a little more. In the worst case scenario, I can always invoke this proof. :D
 
This is equivalent to that statement about limiting growth of $e^x$.
Which is equivalent to the statement about limiting growth of log.
 
Yes, indeed initially I was trying to use the ratio test or trying to find an inequality. I used $e^{a_n} \ge 1+a_n$ without success. I want to think about this more. Math is so fun.
 
7:24 PM
I would like to show that $e^X=\lim_{n\rightarrow \infty}\left(I +\frac{1}{n}X\right)^n$ where $X\in M_d(\Bbb{C})$. We should use the log-function as a hint. But I only know that $\log(e^X)=X$ for $\|X\|<\log(2)$ and this additional assumption is not made in the exercise. Can someone give me a hint?
 
@Gwyn Go one term more.
 
That's trueeee, adding one more term gives me the inequality with $\le$! This works! How could I not see that, I even knew that
Thanks again Ted, part of my graduation will be thanks to you :D
Maybe I need two more terms? Because I need $e^{a_n} \ge 1+a_n+a_n^2/2+a_n^3/6$.
 
8:03 PM
Nah, the quadratic does it.
 
8:16 PM
You're right, I don't know why I thought the inequalities obtained from Taylor's expansion alternate for the exponential as well, but that's not true. So the quadratic is indeed enough. Thanks!
 
8:31 PM
@user123234 Doesn’t it suffice to prove it for sufficiently small $X$?
 
@TedShifrin I'm not sure. I have now tried to show it for diagonalisable matrices and then used that the set of diagonalisable matrices dense is
but I have not used the logarithm for the diagonalisable part, I used the binomial theorem and the serie definition of $e^X$
 
8:46 PM
Right. I would not have tried to use log. But if it holds for $X$, doesn’t it follow that it holds for $cX$?
 
It should hold for $cX$ but I don't see where you want to go
 
Then we’re done if you prove it for all small $X$.
 
ah okey I see what you mean so I could then take the property that $\log(e^X)=X$ for $\|X\|<\log(2)$
 
9:04 PM
Right, if you see how to prove what they claim.
 
 
2 hours later…
11:24 PM
@user123234 Can you assume that $\lim\limits_{n\to\infty}\left(1+\frac1n\right)^n=e$?
 

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