Mathematics - 2022-06-08 (page 1 of 2)

D.C. the III

00:11

@TedShifrin having trouble framing your non overlapping rectangles with sides parallel to coordinate axes question....How to capture the triangle and rectangles in a unified set of variables?

Ted Shifrin

This is a standard, standard, standard 2-variable problem.

Doing all the compactness/max. value theorem checks is less standard for a calculus student, but the set-up should be easy.

What are you using as your variables?

D.C. the III

I did the simpler version of trying to maximize area of one rectangle.

Ted Shifrin

That's a one-variable calculus problem.

But if you know how to do that one, you shouldn't have trouble with the problem.

D.C. the III

That one I got a result of length and height both being $1/2$.

Ted Shifrin

Well, so I return to my question. What are you going to use as the independent variables for the actual problem?

D.C. the III

00:21

hmm...one thing I had thought of was perhaps it may be easier to minimize the area left over from the two rectangles. in that case one variable would be $1 - (x_a + x_b)$ where $x_i$ is the length of the bottom of the rectangle

but I'm still trapped with more than two variables....so writing that out woun't work

Ted Shifrin

The sum of the two areas is easy enough. So why not use those two $x$ coordinates?

D.C. the III

but I would also need their $y$ coordinate analogues as well and that wold be four variables

Ted Shifrin

No. Not independent of these two. Think!

It goes without saying that a simple drawing clarifies everything. Even leslie would concede that.

D.C. the III

first thing i did was a drawing

leslie townes

conceded

D.C. the III

00:40

Ted Shifrin

Well, no wonder. Your picture does not comply with the legal terms of the problem.

Clearly, a picture such as yours can’t result in maximal area, btw.

D.C. the III

I see....let me go rework the picture then.

leslie townes

hilroy, eh? only the finest of papers for you. big spender.

Ted Shifrin

Huh?

leslie townes

just noticed the branding at the bottom of the piece of paper. i don't think it's actually fancy paper.

D.C. the III

00:45

I triggered leslie back to his days of scribbling notes on lower tier paper and the rough dragging of his pen over it

Ted Shifrin

Does the problem not specifically specify inscribed rectangles?

leslie townes

i don't even know the problem statement, but you wanna nudge the rectangle on the left up and to the right, and the rectangle on the right to the left if you wanna maximize area.

at least it'll get you into a smaller parameter space.

Ted Shifrin

no student of mine has ever had problems drawing the right picture.

D.C. the III

wait...does inscribed mean that they MUST touch the edges?

Ted Shifrin

duh.

D.C. the III

00:49

@TedShifrin 😭😭

Ted Shifrin

when did you last inscribe a triangle in a circle by having it well inside the interior?

D.C. the III

true...

leslie townes

people don't inscribe triangles in circles, ted

Ted Shifrin

I appoint leslie to persecute and prosecute.

leslie townes

that's something they only do on the old parchments you read

maybe euclid did that, but he walked so we could run

Ted Shifrin

00:52

As a thought exercise you should convince yourself that to get maximum area the rectangles need to be not only inscribed, but also contiguous.

D.C. the III

Ted Shifrin

Much better.

D.C. the III

just idea scribbling, not formal by any means

inscribing was the game changer....

Ted Shifrin

01:17

No, there’s a significant error.

D.C. the III

I was working things out and I see an error in what I did as well because I may have calculated the "length" that maximizes the individual rectangles, but those won't give the right coordinates

based on my picture, the coordinate for $x_b$ would have to be the sum of the length of the base of rectangle A and base of rectangle B

Ted Shifrin

Right.

D.C. the III

and what I have above as my sum of areas ended up giving me $1/2$ for each of $x_b$ and clearly if I sum those I get 1 which isn't right because that would be a flat rectangle.

Some playing around with relationships ensues then.

Akiva Weinberger

01:36

Heyo what's going

Koro

Let A be an n by n , $n\ge 2$ matrix whose characteristic poly. is $x^{n-2}(x^2-1)$. Then rank of A is

dimension of eigenspace of 0 is n-2, which is same as nullity A.

So rank A= n-nullity A= n-(n-2)=2.

But this is apparently not correct.

the given correct answer is: rank A is at least 2.

Can anyone please let me know what went wrong in the above solution?

Akiva Weinberger

@Koro What's the rank of {{0,1},{0,0}}?

$\begin{bmatrix}0&1\\0&0\end{bmatrix}$

Koro

1

Akiva Weinberger

And what's its characteristic polynomial?

Koro

$x^2$

Akiva Weinberger

01:45

What's the eigenspace of 0?

Koro

all scalar multiples of (1,0).

Akiva Weinberger

And that has dimension 1

even though the characteristic polynomial is a multiple of $x^2$

Koro

agreed.

yes.

Akiva Weinberger

What about something like, a 3x3 matrix with 0s on the diagonal and everything below, 1 all above

$\begin{bmatrix}0&1&1\\0&0&1\\0&0&0\end{bmatrix}$

(Not even sure that top-right 1 is necessary)

Koro

Oh, I understood my mistake now.

I thought earlier that power of x in the poly. is actually eigenspace of 0.

which is not true at all!

Akiva Weinberger

01:48

I think $\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}$ also works for what I wanted

I don't know much of the theory here beyond the examples I've just given, but I think this is probably relevant:

Jordan normal form

In linear algebra, a Jordan normal form, also known as a Jordan canonical form or JCF, is an upper triangular matrix of a particular form called a Jordan matrix representing a linear operator on a finite-dimensional vector space with respect to some basis. Such a matrix has each non-zero off-diagonal entry equal to 1, immediately above the main diagonal (on the superdiagonal), and with identical diagonal entries to the left and below them. Let V be a vector space over a field K. Then a basis with respect to which the matrix has the required form exists if and only if all eigenvalues of the matrix...

Koro

yes, it does. char. poly. is $x^3$ but eigenspace of 0 is of 1 dimension.

The point here is that: dim of eigenspace of 0= GM of 0 $\le n-2$

So rank A = n- nullity A $\ge n-(n-2)=2$

Akiva Weinberger

Incidentally, you may find this pdf interesting: maa.org/sites/default/files/pdf/awards/Axler-Ford-1996.pdf

Koro

thanks a lot @AkivaWeinberger

Akiva Weinberger

It goes through characteristic polynomials and eigenvalues without using determinants

"Down With Determinants!" by Sheldon Axler

Ted Shifrin

Your formula above is wrong.

Akiva Weinberger

01:56

There were a handful of formulas

D.C. the III

$(x_A, y_A) = (1/3, 2/3), (x_B, y_B) = (2/3, 1/3)$

leslie townes

koro is a big fan of the book that developed from that article :)

D.C. the III

@leslietownes THought he was persona non grata around here...

Ted Shifrin

Not you, DogAteMy. sorry.

Akiva Weinberger

02:12

Koro, then?

Or DC3?

Koro

@AkivaWeinberger :-)

D.C. the III

It's interesting how at least to me how some things in math are used absent mindedly........I most likely have in the past, but I just noticed doing the maximization question how I would just treat the cartesian plane as is and not as a proper tool.....just my mind wandering...

@AkivaWeinberger isn't the characteristic polynomial defined in terms of a determinant? well at least up to now what I've read in my Lin-Alg book it points to that

Akiva Weinberger

Well, if you think that, then do I have the pdf link for you

(I think it does it in terms of the eigenvalues)

D.C. the III

@AkivaWeinberger yeah that's what I'm talking about specifically....

I've glanced at characteristic polynomials in other spaces, but since I'm not that advacned yet haven't delved into them

Akiva Weinberger

The definition's on page 7

but the interesting bit is the definition of "generalized eigenvalues" which occurs a bit earlier

(page 3)

D.C. the III

02:18

the characteristic polynopmial how he defines it is already split.

Akiva Weinberger

I will say that I don't share his enthusiasm

(Axler's enthusiasm against determinants)

D.C. the III

haven't played enough with them to have a position to take. beyond the basic linear algebra stuff

Ted Shifrin

I definitely disagree with him. Stooopid analyst.

D.C. the III

is there a reason he takes that position that is at my level of comprehension at this point?

Akiva Weinberger

He thinks they're non-intuitive

Koro

02:27

@leslietownes 😊😊

🫠

D.C. the III

@AkivaWeinberger as a descriptive object? or how to compute them?

Akiva Weinberger

Literally just read the article

maa.org/sites/default/files/pdf/awards/Axler-Ford-1996.pdf

I'm not answering myself because I'm finding it hard to empathize with his position

He thinks there's basically no time you're ever gonna need them

and also in practice they're hard to compute and give you no real information about the matrix

(I suppose the sign does, but (according to him) not the magnitude)

I personally think it's useful to know that the "volume" of a linear transformation is a polynomial in terms of the entries

and not too unintuitive to think that $\bigwedge^n\Bbb R^n$ is isomorphic to $\Bbb R$ (given one knows what $\Lambda$ means)

I think it's easier to leverage intuition already built up about polynomials and simple algebraic rules like $\det(AB)=\det A\det B$ then build up a new intuition about how to reason about generalized eigenvalues

but I'm not Axler

robjohn

@TedShifrin who's a stupid analyst?

Akiva Weinberger

@robjohn Sheldon Axler

Koro

Nooooo

leslie townes

02:39

akiva i think his position is more of a pedagogical one. lots of great reasons to study determinants, some of the russian texts in particular often start with them and do a great job. but they can be very badly taught. and maybe were more in his day than now.

Koro

I read somewhere that working with operators is more natural than working with matrices as the former doesn’t depend upon basis.

leslie townes

i've met people who knew that det L was the dual of the nth wedge power of L and could not calculate one. that's its own kind of pedagogical problem.

anyway when you're coming out with a textbook, you need branding. axler was building his brand. these days he would build it by beefing with other mathematicians on tiktok.

like how ted sells copies of multivariable mathematics by sitting in his yacht and spitting champagne at you, the viewer, talking about how bad all the other books are, while nas 'hate me now' plays in the background

4

Ted Shifrin

03:03

Yup. I agree only to the extent that describing a linear map as a $\theta$ rotation about an axis is far more meaningful than a matrix representation in the standard basis. But I think analysts think more indinite-dimensionally and not necessarily in terms to teach undergraduates.

Maybe I should specify operator theory types, not geometric analysts. So robjohn will not take it personally. :P

@robjohn I was very careful to specify stoopid.

robjohn

@Ted what was your function that was injective on a dense subset? Or should we not discuss that yet?

I don't know if the underlying question is still at large.

one potato two potato

03:19

Let $G$ be a finite group and $K<G$ be a subgroup. If any $g\in G\setminus K$ has order $2$ then $K$ is abelian. How can I show this?

Calvin Khor

@robjohn was it a function $[0,1]\to\mathbb R$?

D.C. the III

@leslietownes 🤣🤣🤣....I could picture this vividly in a red robe with a large diamond encrusted chain with a "math god" pendant dangling from it....

robjohn

@CalvinKhor mine was different

Ted Shifrin

@robjohn I took $x$ on the rationals from $0$ to $1/2$ and $1-x$ on the irrationals to $1$.

robjohn

I took $\sin(x)$ on $\mathbb{Q}$.

Before I saw that, I took something like yours, however.

Ted Shifrin

03:25

Interesting. A little harder :) i don’t need $\pi$ irrational :)

robjohn

$f(x)=|x|$ on $[-1,1]$ with the dense set $(\mathbb{Q}\cap[-1,0])\cup((\mathbb{Q}+\sqrt2)\cap[0,1])$

That was my first one

Ted Shifrin

Yeah, isomorphic to mine.

Calvin Khor

i thought you guys were talking about continuous function on X, injective on dense subset of X, non-injective on X

Ted Shifrin

We are.

robjohn

Indeed

Calvin Khor

03:26

oh wait lol let me reread then

oh i see haha

nice

leslie townes

one: by "any" do you mean "every" there? you must.

Calvin Khor

given the target space wasn't defined i had the much dumber example (sin t, cos t) on [0,2pi] with [0,2pi) dense

Ted Shifrin

Related, injections aren’t a stable family of functions unless you also have immersions.

Well, the original question was about maps $X\to X$.

Calvin Khor

sorry, didnt see that... :)

just saw you guys discussing it midway

Ted Shifrin

He wanted two functions that were inverses in a dense subset to be globally inverses.

Calvin Khor

03:30

hmm lol

Ted Shifrin

Tempting, but …

Akiva Weinberger

@onepotatotwopotato You want "every" rather than "any"

(otherwise $S_3<S_5$ is a counterexample)

one potato two potato

Ah right.

Akiva Weinberger

So OK, say $a,b\in K$ and $g\in G\setminus K$. We know $ag,bg,abg\in G\setminus K$ also

('cause if for example $ag$ and $a$ are both in the subset then so is $g$)

one potato two potato

yes

Akiva Weinberger

03:32

so we know $g^2=agag=bgbg=abgabg=e$

one potato two potato

yes

Akiva Weinberger

I'm gonna try random stuff 'til it works

If $agag=e$ then $gag=a^{-1}$, similarly $gbg=b^{-1}$

so we want to show $aba^{-1}b^{-1}=e$ (to show $a$ and $b$ commute), let's substitute in those formulas for $a^{-1}$ and $b^{-1}$

so $aba^{-1}b^{-1}=ab(gag)(gbg)=abgag^2bg=abgabg$

${}=(abg)^2=e$

one potato two potato

oh right

Akiva Weinberger

That's a really neat fact, though, I don't think I knew it before

I wonder if there's a more direct way

I guess a kind of prototypical example here would be $C_n<D_n$

(where $C_n$, sometimes written $Z_n$ or $\Bbb Z/n\Bbb Z$, is the cyclic group of size $n$, and $D_n$ is the dihedral group of size $2n$)

one potato two potato

If $K$ is such a case then $K$ is always normal subgroup of $G$?

Ted Shifrin

03:37

What’s the neat fact?

Akiva Weinberger

If you have a subgroup $K<G$ with the property that every element of $G\setminus K$ is order $2$, then $K$ is abelian

Oh here's a more direct way to see it

Because $(ga)^2=e$ and $g^2=e$, this means that $ga=a^{-1}g$

thus $gab=a^{-1}gb=a^{-1}b^{-1}g$

but at the same time $gab=(ab)^{-1}g$

thus $a^{-1}b^{-1}=(ab)^{-1}$ which means $K$ is abelian

Ted Shifrin

Hmm, never encountered that. So $G/K$ is a direct sum of $\Bbb Z_2$s.

Akiva Weinberger

$G\setminus K$, not $G/K$

Set complement

Ted Shifrin

I understand.

Akiva Weinberger

The example I gave is $C_n<D_n$ (cyclic group in dihedral group)

Abstract Space Crack

03:42

I just thought of the world's greatest number theory game.

Akiva Weinberger

Is it "I think of a semiprime and you find its nontrivial factors"?

leslie townes

i found this in rose. under these hypotheses, if k^2 isn't e for some k in K, then G is the dihedral group on K

Ted Shifrin

Consider the quotient, nevertheless.

one potato two potato

We don't know K is normal

Akiva Weinberger

Do we know that $K$ is a normal subset?

Oh, I suppose we do actually

Ted Shifrin

03:43

When you write < that’s what I assume. I don’t use that for subset.

Akiva Weinberger

'cause $gag^{-1}=a^{-1}$

@TedShifrin $K\lhd G$ is normal subgroup

Ted Shifrin

Yeah, I never use those notations. Good reason …

Subgroup, not subset

Akiva Weinberger

Right, yeah

Ted Shifrin

@leslietownes Interesting

Abstract Space Crack

It's based on Bertrand's postulate: I'll put own the first part of the sequence, then you must continue the sequence:
$1 \leq p_1 - 1 \leq p_1 + 1 \leq p_2\leq 2p_1 - 1\leq \dots$ When they're $=$ you can put them in any order you'd like. The goal is to list only numbers that are $= \pm 1 \pmod {p_i}$ for some prime number $i=1..n$. So by the end you have to have come up with a formula in $n$ from experience (hopefully).

That's the game, go. You can use only Bertrand's basically when you must come up with a formula with $n$ in it

Akiva Weinberger

03:46

$K\subseteq G$ means $K$ is a subset (not necessarily a group), $K\le G$ means subgroup, $K\unlhd G$ means normal subgroup. That's how I learned it

It is actually true in this case that it's normal, though

which you can kinda check directly

Koro

@onepotatotwopotato I think you should write G/K explicitly.

Akiva Weinberger

04:01

Without determinants, we could not have this puzzle^

(I have not yet started thinking about it)

Calvin Khor

05:53

78. (GCD matrix) Let $D=(d_{ij})_{n\times n}$ where $d_{ij} = \operatorname{gcd}(i,j)$. Then,
$$
\det D = \varphi(1)\varphi(2)\dots \varphi(n)
$$where $\varphi$ is Euler's $\varphi$ function: $\varphi(m) = \lvert \{ k \mid 1 \le k \le m , \ \operatorname{gcd}(k,m) = 1 \} \rvert $

leslie townes

from the odd spacing in the image it looks like they might've used \mid outside the { } too

Wave

is it correct that the only prime ideal in $\Bbb{Z}/9\Bbb{Z}$ is $[3]$?

leslie townes

sounds right to me

Wave

perfect thanks!

Calvin Khor

@leslietownes all the more reason to re-type it correctly

leslie townes

05:57

no \textbf{GCD matrix} for you

Calvin Khor

Yeah I gave up on why bold didn't work. Maybe its the brackets. test. (test). (test).

leslie townes

my aesthetic would be for \cdots and not \dots there

Calvin Khor

Oh, now it works. fuck you too mathjax

leslie townes

the important thing is that the conversational ball is rolling

Calvin Khor

@leslietownes what about $a_1\cdot a_2 \cdot \dots \cdot a_n$

leslie townes

06:00

yeah, no

automorp15m

how do you interpret an RMSE of 100000+ ?

Calvin Khor

@automorp15m root mean square error? in what units, microns?

automorp15m

leslie townes

size matters

automorp15m

we're dealing with millions (money)

so we get 100000+ as RMSE

thought this was reliable, it has 18 recos

leslie townes

06:03

stat education is full of rules of thumb like that that have no basis in anything objective

they can become objective if you toss a lot of models and theory on top of it, but then you're not just staring at the size of one number

automorp15m

Did Hamid actually mean RMSPE

Calvin Khor

what on earth is rmspe

automorp15m

6

Q: What is the correct definition of the root mean square percentage error (RMSPE)?

Göçken et al. define the root mean square percentage error (RMSPE) as \begin{equation} \text{RMSPE} = \sqrt{\frac{100\%}{n} \cdot \sum_{i=1}^n \Delta X^2_{\text{rel},i}} \end{equation} with \begin{equation} \Delta X_{\text{rel},i}=\frac{X_i}{T_i}-1, \end{equation} where $T_i$ is the desired val...

leslie townes

even if you normalize in some way it's still a rule of thumb and not something objective

all of this stuff is pretty context specific, in terms of what matters and what a good model is, it's always a judgment call but it's more subtle than looking at one number

or at least oughta be

if they're teaching your class throw the model out and start again until you get something less than 0.5

it's called learning

Calvin Khor

@automorp15m this post wondering where to multiply by 1 is amusing

automorp15m

06:12

@CalvinKhor yeah! noticed that too lol

researchgate.net/post/…

Calvin Khor

i wouldn't put too much stock into an online comment

automorp15m

i'm just frustrated can't find any other (yet)

and it's researchgate,
it has 18 recos from other researchers

trust issues 📈📈

Calvin Khor

theres some silly people on research gate too

with enough money anyone can "publish"

must you have some sort of verified publication in order to reco? Or is it a similar issue with twitter botting? I would assume the latter

automorp15m

06:28

not that familiar with it
thought it's vetted through a rigorous process

assumed*

copper.hat

06:50

personally i look for an RMSWTF of $\aleph_0$

robjohn

08:58

@copper.hat is that the aleph-null test?

Wave

Could maybe someone help me here?math.stackexchange.com/questions/4468027/…

Monty

09:20

can anyone help me understand why a flow is invertible

think im missing something very obvious

Im looking at these notes on the continuity equation and the corresponding flow/ODE

perso.crans.org/moussa/dipernalions.pdf

why is the flow $x$ of $b$ in theorem III.1 invertible?

ignore me- i see it comes from the property of being a flow.

Calvin Khor

you mean (53) right? @Monty

and those aren't notes, that's the paper of DiPerna-Lions

Wave

09:44

are there some easy examples of integrally closed domains which are not fields?

I mean sure PID and UFD are integrally closed domains but are there further ones?

Monty

10:21

yup sorry calvin

porridgemathematics

10:35

@Wave since PID -> UFD, anything that isnt a UFD but a domain and not a field would work as an example

so for instance, something non-noetherian that is a domain will work, e.g. R[x_1,x_2,...] where R is some ID

also usually integrally closed just follows from something being a domain

or rather all the time, as far as rings are concerned, domain = integral domain = integrally closed ring

oh whoops, what i just wrote is a UFD

my bad

it isnt noetherian though, but that doesnt mean you cant factorize

you can try something like formal $\mathbb{C}$-linear combinations of nonnegative rational powers of $t$, this isnt a UFD but it is a domain

VLC

11:03

Can somebody help me with a parametrisation problem: We have points A(0,0,0), B(1,0,0) and C(1,1,1). I need to parametrise in the orientation: $A \rightarrow B \rightarrow C \rightarrow A$

To get a closed curve

Wave

11:14

@porridgemathematics perfect thanks!

Koro

Suppose that A is a 2 by 2 matrix whose trace is 5 and determinant is 6. Now I define $T: M_2(R)\to M_2(R)$ as $T(B)=AB$.

Is T diagonalizable?

So here it can be observed here that A is diagonalizable.

porridgemathematics

11:32

@Koro if $A$ is diagble, then $T$ is too, let $v_1,...,v_n$ by an eigenbasis of $A$, and consider $(v_j | 0)$, $(0 | v_k)$, for $j,k=1...n$, these form an eigenbasis of $M_2(R)$ with respect to $T$

Koro

Eigenvalues of A are infact 3 and 2. Let u and v be corresponding eigenvectors.

porridgemathematics

uh i mean of $M_n(R)$, i.e. what I said should work for any dimension

Koro

(u, 0),(0,u),(0,v),(v,0) are linearly independent eigenvalues of T.

porridgemathematics

the eigenvalues of $T$ are the same as that of $A$, but $T$ can still be diagble with eigenvalues that have arithmetic multiplicity greater than one, i.e. it does not need to have distinct eigenvalues to be diagble

that is sufficient but it isnt an equivalent condition

Koro

dim M_2(R)=4 so T has a basis consisting of eigenvectors.

u and v are LI.

porridgemathematics

11:42

eigenvectors

T's eigenvalues are $2$ and $3$

Koro

:-)

Jakobian

12:05

So, this might be a dumb question

Let's say that a set of vectors $(e_\mu)$ of a Hilbert space is an orthonormal basis if it's a maximal orthonormal set

How do I show that $(e_\mu)$ spans the whole space

Oh, I guess, if $H$ is our Hilbert space and $F = \overline{\text{span}}\{e_\mu\}$ then $H = F\oplus F^\perp$ so that if $F\neq H$ then $F^\perp \neq \{0\}$.

researchgate.net/publication/…

Interesting, so they study Schauder bases in Hilbert spaces as well

porridgemathematics

12:37

but this means the span is dense in the whole space, not that the ON basis really spans the whole space

Daniel Adams

what am I missing here : where is it used that the map $\phi$ is invertible?.

$\mathcal{H}$ is relative entropy.

Jakobian

14:13

@porridgemathematics semantics

Jakobian

14:24

$\varphi\#\mu(A) = \mu(\varphi^{-1}(A)) = \int_{\varphi^{-1}(A)} \frac{d\mu}{d\nu}(\omega)d\nu(\omega)$, since $\varphi$ is invertible we have $\varphi\#\mu(A) = \int_A \frac{d\mu}{d\nu}(\varphi^{-1}(\omega))d\varphi\#\nu(\omega)$

@DanielAdams I believe this is where invertibility is coming from

Without it, we wouldn't be able to use the formula $$\int g d(f\# \mu) = \int g\circ f d\mu$$

Wave

14:37

Hello do someone know abut forgetful functors?

one potato two potato

forget structure?

Wave

Yes right I am looking at forgetful functur from R-modules to abelian groups for a ring R

So I look at a prove where they show that this functor is exact.

@onepotatotwopotato I mean it is clear that we need to take a short exact sequence of R-modules 0->M'->M->M''->0 and show that if we apply the functor we still get a short exact sequence right?

one potato two potato

don't ask me I've never studied category theory seriously

Wave

Me neither I'm taking commutative algebra

Daniel Adams

@Jakobian yes I got it, to get the relation between the radon-Nikodym derivative and the push forward you need the invertibility.

one potato two potato

14:50

I took CA last year and it's like nightmare to me. I remember I couldn't solve almost half of the problems in final exam. After that I don't take any advanced algebra course.

Wave

Hahah I'm at the same point now... it's horrible and my exam is in 4 weeks... I don't see why we should use this and how...

one potato two potato

Funny thing is that even if I tried to avoid algebra, after all, the only thing that mathematician can do (or human can do) is essentially just algebra or analysis

Wave

Sometimes it's like mystery with a lot of symbols and handwaving.

Jakobian

15:26

@Wave yes

Wave

@Jakobian ah so can I ask you about my question

Jakobian

Okay

Wave

@Jakobian Perfect thank you. So as I said I need to take this exact sequence of R-modules. Then applying the functor I get 0->F(M)->F(M')->F(M'')->0. Let me denote f:M'->M and g:M->M''. Then the corresponding maps would be F(f):F(M)->F(M') and F(g):F(M')->F(M''). So then to say if it is exact I need to check that ker(F(g))=im(F(f)) right?

Jakobian

Well the maps don't change, kernel and images are the same too

@Wave You also want F(f) to be injective and F(g) to be surjective, but yes

Wave

Ahhhhhhhh so the maps don't change only one operation gets lost @Jakobian

And therefore also injectivity of F(f) and surjectivity of F(g) don't get lost?

Jakobian

15:35

No, they don't change

Wave

sorry my english is not so well do you mean with no, that I said something wrong or do you mean no nothing change and what I said is correct

Jakobian

The maps don't change

Wave

perfect thanks!

VLC

15:50

Does anybody know how jacobian in spherical coordinates changes if:

$x = r/3 * cos(phi) cos(theta)$

And others stay like $y = r sin(phi ) cos(theta)$ and $z = rsin(theta)$

Hasini

16:25

Hi!

When considering $1^{\frac{5}{2}}$, can I say it is equal to 1?

I mean if we think of it as ${{-1}^{2}}^{\frac{5}{2}}$, then it is equal to $-1$ so then should I say both $1$ and $-1$ are solutions?

Ted Shifrin

NO.

$[(-1)^2]^{1/2} = 1$. You're being sloppy with "rules."

Hasini

Hi, @TedShifrin

Is it by thinking that we should simplify inside the brackets first?

Ted Shifrin

Yes. Only for positive numbers $a$ do you know that $(a^p)^q = a^(pq)$.

Semiclassical

you always want to do the sanity check of plugging in roots back into the original equation

it won't tell you if you've missed roots, but it will tell you if the ones you came up with are right

Ted Shifrin

Well, in this case, it's an expression, not an equation.

Hasini

16:35

Hmm, okay. Then when taking square root of 1 = + or -1

Xander Henderson

No.

Ted Shifrin

Square root means positive square root, always ...

Xander Henderson

The square root of $1$ is $1$.

Semiclassical

for stuff like 1^(p/q) you really need to do complex algebra

if you want values other than 1

Xander Henderson

The equation $x^2 = 1$ has two solutions, which are $\pm 1$. But that doesn't mean that the square root of $1$ is either $-1$ or $+1$.

Hasini

16:36

Ohh, okay

Now I get it :)

Semiclassical

eh, it depends on how you use the phrase "square root". for instance, Wikipedia's initial definition is "In mathematics, a square root of a number x is a number y such that y^2 = x."

reserving the positive square root for "principal square root"

Xander Henderson

Unless otherwise noted, the notation $\sqrt{x}$ denotes the (rolls die... 1, which is odd) principle square root of $x$. If $x>0$, then this principal square root is well-defined, and is positive.

(the second die roll was a 4)

Hasini

So when you just take the square root like $\sqrt{9}$ it is always equal to +3

Semiclassical

i mean, that is certainly one definition. but i would hardly be shocked if certain texts don't define it like that

Xander Henderson

@Semiclassical Right, which is why I emphasized the definite article, "the".

Ted Shifrin

16:39

This is why we write $\pm\sqrt{b^2-4ac}$ in the quadratic formula.

Hasini

@TedShifrin To specifically say that the - one also should be considered?

Ted Shifrin

Right, because $\sqrt{\text{blah}}$ is always $\ge 0$.

Hasini

Okay, cool

Ted Shifrin

@Xander I wish you had bolded the correct principal.

Hasini

Thank you very much @TedShifrin @Semiclassical and @XanderHenderson

Ted Shifrin

16:41

You're welcome.

Xander Henderson

@TedShifrin Yeah. But, at this point, it is more fun not to.

Ted Shifrin

You just want to drive me maximally nuts.

Semiclassical

there's also the annoying point of what exactly $\sqrt{}$ means if you don't have real roots

Xander Henderson

But, just to make you happy, I have bolded it now. :D

Ted Shifrin

Gee, thanks.

Xander Henderson

16:42

No problem!

Semiclassical

but doing $\pm$ washes over that too

Ted Shifrin

Well, in the complex case, both values are represented by $\sqrt{}$, and so you don't need the $\pm$ :D

Semiclassical

lol

Xander Henderson

@Semiclassical Sure, but, again, unless otherwise noted, the default assumption is that it means the (4) principal square root, which is not unambiguously defined, but which should be contextually well-defined.

Ted Shifrin

I disagree. What if your number lands on the "principal branch cut"?

Semiclassical

16:45

1^(p/2) at least has the generosity to only have at most two possible values

Ted Shifrin

To me, $\sqrt z$ is a bifunction when $z\in\Bbb C$.

Semiclassical

ze^z=w, by contrast...yikes

Ted Shifrin

Unless specifically specified.

Xander Henderson

@TedShifrin Which is why I note that the (2) principal square root is not unambiguously defined, but it should be contextually well-defined.

Ted Shifrin

With logarithm, we can make you write Log. But I know no analogous notation for $\sqrt{}$.

@Xander I think I disagree, having worked in $\Bbb C$ almost all my mathematical life.

Semiclassical

16:47

with the Lambert W function you've got the subscript to denote which branch you're on

tho the way the branching works in that case is tricky

Xander Henderson

@TedShifrin We work in different areas, but I am not sure that I have ever seen the notation $\sqrt{\cdot}$ used in a place where this becomes problematic---most of the texts I have prefer to write everything in terms of exponentials and logarithms.

But the work I do only touches on complex analysis---I don't live there on a daily basis.

Ted Shifrin

In the context of Riemann surfaces (where it's more algebraic) one is not going to write exponentials and logs :)

Anyhow, this is silly.

Xander Henderson

@TedShifrin Ah, that makes sense.

Ted Shifrin

Anyhow, I don't think high school algebra leads to a sufficient understanding of the exponent rules, and the $a^{pq} = (a^p)^q$ conundrum is a common confusion.

Xander Henderson

@TedShifrin Sure. Two or three "I've shown that $-1 = 1$ by abusing exponents!" questions get posted to the main page every month.

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$Mathematics$ Mathematics