Mathematics - 2023-02-15 (page 2 of 2)

Obliv

18:00

@Pm2Ring yeah, i didn't see anything mentioned about a sister on his wiki page. He had a younger brother alfred but no mention of a sister.

Prateek Mourya

Koro

and did you read about Cantor?

Prateek Mourya

please can anyone help me with the part 'the first term integrates to zero'

Koro

@Obliv I guess we'll never know what actually happened to Galois.

Ted Shifrin

@PrateekMourya Have you integrated by parts? That's what one always does in this subject, and then uses boundary values $0$ at $\infty$.

Oh, actually, in this case, I think integrating the first term by parts (and using those boundary values, of course) will cancel out the second term.

PM 2Ring

18:08

@Obliv I may have been wrong about the sister. But it definitely wasn't a woman that he had a strong romantic connection to. Actually, that Wikipedia article looks pretty good. I read about this stuff ~20 years ago, so my memory is a little bit fuzzy...

Obliv

I'm so rusty, how would I expand $\frac{x-25}{x^2+5x-24}$ I did $\frac{A}{x+8}+\frac{B}{x-3} \to \frac{A(x-3)+B(x+8)}{(x+8)(x-3)} = \frac{x-25}{(x+8)(x-3)}$

there is something called a heaviside cover up method but I don't even want to worry about shortcuts

the issue is if I set x=3 or x=-8 the denominator becomes 0 which is bad.

wait I think i figured it out

Sine of the Time

@Obliv if $A(x-3)+B(x+8)=x-25$ you must have $A+B=1$ and $-3A+8B=-25$

Ted Shifrin

Get rid of the denominators and consider the polynomials on the top ... They must be equal for all $x\ne -8,3$, therefore for all $x$.

Obliv

I think if I went from $\frac{A}{x+8} + \frac{B}{x-3} = \frac{x-25}{(x+8)(x-3)}$ and multiplied both sides by $x+8$ I'd have $A + \frac{B(x+8)}{x-3} = [\frac{x-25}{(x+8)(x-3)}](x-8)$ which should cancel the denominator right

so I'm left with $A+\frac{B(x+8)}{x-3}=\frac{x-25}{x-3}$

Sine of the Time

@Obliv yes now let $x=-8$ to get $A$

Ted Shifrin

18:20

Why are you not multiplying by $(x+8)(x-3)$ in one swell foop?

Obliv

@TedShifrin @SineOftheTime because $A(x-3) + B(x+8) = x-25$ seemed like a dead end

Ted Shifrin

It's totally not a dead end.

ペガサス Seiya

@TedShifrin wanna trade the Civic for a jellybean car?

Ted Shifrin

You have a polynomial $ax+b$ on the left and a polynomial $cx+d$ on the right. Since they are equal, $a=c$ and $b=d$.

D.C. the III

Obliv

18:22

Oh I get it

Ted Shifrin

@ペガサスSeiya What is a jellybean car?

D.C. the III

Had a question about what is happening here. I've done the question

ペガサス Seiya

@TedShifrin a car that's built !ike...a jellybean

Ted Shifrin

Namely?

DC, presumably this is done with change-of-basis?

ペガサス Seiya

@TedShifrin its called the Tata Nano I think

Obliv

18:25

Ooh so what I thought was going on was you were just setting the numerators equal to each other without considering the denominator. So when you set x=3 or -8 I thought it was giving you zeros in the denominator

Ted Shifrin

Oh, here we have the Smart car that looks like that. (Smart hails from Germany.)

D.C. the III

@TedShifrin But I'm trying to get a picture of what's occurring. So to do the question I projected my two standard basis vectors on to the subspace $W$ using the orthongonal projection expression $v_1 = \frac{\langle (1,0), (1,2) \rangle}{\|(1,2)\|^2} (1,2)$, same idea for the other basis vector. Then the matrix is obtained by writing my $v_i$ in terms of the standard basis vectors and using their coefficients.....The usual process.

Ted Shifrin

This is why I asked, DC. You did problems like that in my book in the first chapter. But now you should have a better way to do it (chapter 9 of my book).

What chapter of FIS is this?

Semiclassical

QM has me sorta programmed when it comes to projections. my brain goes immediately to the outer product

Ted Shifrin

That's yet another way to do it (chapter 5 of my book).

:D

Semiclassical

18:28

sounds right

Ted Shifrin

projection matrices, normal equations for inconsistent systems, etc.

ペガサス Seiya

@TedShifrin wait, why's it smart? (choked on a candy)

Semiclassical

QM do love $I=\sum_k v_k v_k^\top$ (assuming unit vectors)

Ted Shifrin

You could almost write half a final exam asking someone to do it all possible ways :D

And assuming orthonormal basis, in fact.

D.C. the III

@TedShifrin I now have my matrix and this would be the matrix for projecting a vector onto W. So say I take a vector and apply the matrix to it. I will be given a vector. Usually that vector is the coefficient vector. But here my vector space is only spanned by one vector so are the coeffcients just the vector itself projected onto the subspace? This is FIS - Sec 6.6

Semiclassical

18:30

yeah, i get so used to the assumptions that come with QM that i have to actively work against them outside of that

you get used to leaning on everything being finite-dimensional and self-adjoint

Ted Shifrin

No, you're writing a matrix for the linear map $V\to V$ that projects onto $W\subset V$. You should have a square matrix.

What is section 6.6? I do not own the book.

Sine of the Time

@Obliv this is a general method, in this case it's not a "dead end". But as you've previously said, you can use the cover up methos

Semiclassical

my own favorite use of this is how to exponentiate traceless matrices. the usual proof i see people do is with series, but i kinda roll my eyes at that now

Ted Shifrin

I never actually taught the cover-up tricks. (Of course, that shows us in complex analysis with residues.) They fail to work with repeated roots and/or irreducible quadratics, so I skip it.

Obliv

Yeah I'm not a fan of being taught tricks. I'd rather learn the general method and if I'm comfortable I'll use the trick

Koro

18:33

can anyone please explain to me splitting of cycles in A_n?

Sine of the Time

@Obliv I'm your opposite :)

Obliv

though to be fair I was taught the general method like 7 years ago so he probably taught us the trick because he assumed we knew it

Semiclassical

much prefer writing the spectral decompositions of $e^{iA}$ and powers of $A$, then use this to express the former in terms of the latter

ペガサス Seiya

An image is just a matrix and the pixels are its elements

Sine of the Time

@ペガサスSeiya didn't know you're a philosopher

D.C. the III

18:34

Yes I got the square matrix, I wasn't asking about the matrix, I was asking about the result of the matrix. So let's say $A$ is the matrix and $x$ a vector in $V$ then $Ax$ is the vector projected into $W$.............Oh I see what I was getting at. Even though I am projecting onto $W$ the vector is still a vector in $V$ so the vector is still of the form $[Ax]_{\beta}$ where $\beta = {e_1, e_2}$

ペガサス Seiya

@SineoftheTime I'm not. That's actually just how image processing works

D.C. the III

@TedShifrin Sec 6.6 is "Orthogonal Projections and THe SPectral Thm"

Sine of the Time

@ペガサスSeiya yes I know, I was just kidding

Semiclassical

i forget if that's how all image processing works tbh. i know it's true for bitmaps but i thought some of the formats were a bit weirder than that

Ted Shifrin

Yes, and you've already learned change of basis, @DCthe. Use it.

ペガサス Seiya

18:36

@SineoftheTime no wonder most computers find image processing difficult and demanding. Matrix calculations are difficult

Semiclassical

depends on the kind of calculations

ペガサス Seiya

@Semiclassical Edge detection, for example, is pretty difficult

Sine of the Time

@ペガサスSeiya yes, and depends also on the element of the matrix

Ted Shifrin

@DCthe I'm sure they have plenty of examples in the text, but if not look at Example 4 on p. 418 of my book.

Semiclassical

@ペガサスSeiya that i believe

ペガサス Seiya

18:38

@SineoftheTime they're usually not integers because colors are represented as something between 0 (black) and 1 (white)

Sine of the Time

and computers work with finite arithmetic

ペガサス Seiya

Now imagine a 1000x1000 matrix that has a lot of operations done on it, 60 times every second

Koro

what is the meaning of 'conjugacy class of an element in S_n splits'?

PM 2Ring

Yes, some image systems (mostly) don't work with matrices of pixels. They define shapes with vectors and curves, and only convert to a grid of pixels at the end, to generate the display. And in Ancient Times, there were displays & plotters that didn't use pixels at all.

Obliv

how come $\int \frac{-x}{x^2-9} dx$ is $-\frac{1}{2} \log(x^2-9)+C$ I did partial fractions instead of int by parts and got $\frac{-1}{2(x+3)} - \frac{1}{2(x-3)}$ as the expansion. integrating gives $\log (\frac{x+3}{x-3}^{-1/2}) + C$

Ted Shifrin

18:40

Normally one talks about splitting a polynomial. What's the rest of the sentence, @Koro?

ペガサス Seiya

@PM2Ring you're referring to how video games generates imagery

Ted Shifrin

@Obliv Do the integral by substitution from the beginning. But finish with partial fractions and you'll get the same thing.

Koro

38

A: Splitting of conjugacy class in alternating group

Note the following: (1) The conjugacy class in $S_n$ of an element $\sigma \in A_n$ splits, iff there is no element $\tau \in S_n\setminus A_n$ commuting with $\sigma$. For if there is one, for each $\tau' \in S_n \setminus A_n$ we have $$ \tau'\sigma{\tau'}^{-1} = \tau'\sigma\tau\tau^{-1}\tau'{}...

D.C. the III

@TedShifrin For whatever reason they showed no examples of this procedure in this part of the text and relied upon the previous idea of the vector $w = x + y$ where $x \in V$ and $y \in V^\perp$ and if I have an orthonormal basis $x$ can be written as $x = \sum \langle w, v_i\ rangle v_i$.

Obliv

why can't I start with partial fractions @TedShifrin

Ted Shifrin

18:41

@Koro Did you read the original question and the link for it to find out?

Koro

When I understand this, it'll help me write conjugacy classes of $A_n$.

Ted Shifrin

You can, @Obliv, but since $x$ is a multiple of the derivative of the denominator, substitution is way faster. You can in fact do it in your head, whereas partial fractions will take time.

Obliv

Wait what would I be substituting for u?

Ted Shifrin

One guess.

Obliv

ohh

I see

Ted Shifrin

18:43

Your partial fractions integral is wrong. It is not log of the quotient.

Watch out for signs.

Koro

I understand (a little) what the splitting would mean. Here's the idea: Some element $\sigma \in S_n$ may have some x elements in its conjugacy class. But if $\sigma$ is even (that is, in $A_n$) and if we want to find the conjugacy classes in $A_n$, then this $x$ should change in a certain way.

Obliv

I took the -1/2 out of the integral and subtraction of two logs is division

then raised to the -1/2 power

Koro

The way this x changes is what we call 'splitting'.

Ted Shifrin

The question is whether you still have one conjugacy class with only $A_n$ acting or two. Now splitting makes obvious sense.

Koro

Now, my question is: how does the linked answer prove the first statement in it? There is a proof in the answer but I don't understand how it proves.

Ted Shifrin

18:45

No, you can do better than your vague sentence.

@Obliv There is NOT subtraction of logs.

You have an arithmetic error in your partial fractions.

Koro

that is, the following part: (1) The conjugacy class in $S_n$ of an element $\sigma \in A_n$ splits, iff there is no element $\tau \in S_n\setminus A_n$ commuting with $\sigma$. For if there is one, for each $\tau' \in S_n \setminus A_n$ we have
$$ \tau'\sigma{\tau'}^{-1} = \tau'\sigma\tau\tau^{-1}\tau'{}^{-1}
= (\tau'\tau)\sigma(\tau'\tau)^{-1}
$$ and $\tau\tau' \in A_n$.
On the other hand, if $\tau\sigma\tau^{-1}$ and $\sigma$ with $\tau \in S_n\setminus A_n$ are conjugated in $A_n$, then for some $\tau' \in A_n$, we have $\tau\sigma\tau^{-1} = \tau'\sigma\tau'^{-1}$, giving

Ted Shifrin

Get rid of the $-x$ and make it $x$ from the beginning. You will die in all the minus signs.

Obliv

Oh dang you're right B is 1/2

equivalent expressions if I did it right :\

Koro

$\tau\tau'\in A_n$ as a conlcusion for $\Leftarrow$ implication.

But what does it achieve? How does it prove the implication?

D.C. the III

Ah, looking at your example as expected it clarifies things more. Because when you said use the change of basis matrix I didn't really kno wwhat you meant in this context. To do that I would have to get the basis vectors for $W^\perp$ in order to make the change of basis matrix. ALso they didn't refer to the matrix I was solving for as "standard matrix". I take it this is what it is usually called?

Koro

18:50

Ahh, I think I am starting to understand it now.

Ted Shifrin

So they're showing that if there is such a $\tau\in S_n\backslash A_n$, then the conjugacy class does not split. Because they produced an element of $A_n$ that conjugates to give the same result.

Koro

If there is an element $\tau\in S_n-A_n$ commuting with $\sigma$, then the conjugacy class of $\sigma\in A_n$ does not split.

Ted Shifrin

@DCthe I always call the matrix with respect to the standard basis the standard matrix (as in chapter 1).

In all my years, I have never before heard this terminology. It must be common among representation theorists, but I never have encountered it.

Koro

they are proving the contrapositive of $\Rightarrow$ implication.

Ted Shifrin

Yes.

D.C. the III

18:53

Ok. Thanks, back to work I go....

ペガサス Seiya

I remember when Ted was trying to explain community wiki posts to me and I just wasn't getting it.

He almost pulled his hair out back then

Obliv

$\frac{dy}{dx} = f(y)$ what is an equilibrium solution? I thought I knew but after re-reading I confused myself.

if $f(c) = 0$ is subbed in for $f(y)$ ?

Ted Shifrin

Zeroes of $f$.

Semiclassical

the phrase you also see is "steady state solution", mostly in the case where the independent variable is time

Obliv

that's what I thought, so $\frac{dy}{dx} = 0$ but then it says $f(c)$ is a constant solution/equilibrium solution to the d.e

Semiclassical

18:59

yes

Ted Shifrin

If $f(c)=0$, then $y=c$ is an equilibrium solution. Words matter.

Semiclassical

yeah, i missed that

point is that if $y=c$ initially, then $dy/dx=0$ as well. so if the system starts 'in equilibrium' it'll stay that way

Obliv

$y = x^2$ then $y(0) = 0$ would be the equilibrium solution?

since $y' = 2x$

Semiclassical

that's not of the form $dy/dx=f(y)$ tho

PM 2Ring

Like a ball sitting in the bottom of a bowl. It's not going to roll away from that position.

Semiclassical

19:02

you have to go a bit further and write $dy/dx=2\sqrt{y}$.

at that point, it's valid to note that $f(0)=0$

Obliv

Oh okay.

but as we know $y = x^2$ is just a particular solution the general would be $y = x^2 + C$

Semiclassical

right. in which case you can't really get this into the form $dy/dx=f(y)$

Obliv

the general solutions equilibrium solution would be $y(x^2 = -C)$

Oh right

Semiclassical

wait

$y=x^2$ isn't equilibrium

i think we got ourselves into trouble somwhere

Ted Shifrin

No. Obliv got it right.

The equilibrium solution is $y=0$.

Semiclassical

19:04

oh. right. it's just a singular solution

Obliv

Okay i will play around with some examples thank you.

Semiclassical

a pretty standard example is something like $dy/dx=y(1-y)$.

which has two equilibrium solutions, but of rather different nature

(in physics we'd call them stable vs unstable equilibria)

a manifold is a world in which Euclid can't be proven wrong unless you zoom out

Ted Shifrin

In math too.

@Semiclassical Not true. If you have a metric (or some connection on the tangent bundle) you can tell pointwise/infinitesimally.

You're talking about global topology, but that's actually not the issue.

Semiclassical

huh

Ted Shifrin

Whether a Riemannian/Lorentzian ... manifold is flat is an infinitesimal/local computation.

Semiclassical

19:09

ah true

Koro

sphere with hair is not a manifold

Semiclassical

does this come down to the fact that we're dealing with "locally homeomorphic to Euclidean space"?

rather than "locally Euclidean"

Sine of the Time

Is Module theory interesting?

Ted Shifrin

It depends on what structure you have on your manifolds. A metric is more structure than a plain old manifold. An affine structure is what one might mean by "locally Euclidean."

Yes @Sine

Sine of the Time

@ted I've to choose an exam and I was wondering what to do

this year I've chosen ODE

Ted Shifrin

19:15

Among what options?

Sine of the Time

Biomathematics, stochastic methods, and others about coding

ペガサス Seiya

Bio...what?

Sine of the Time

@ペガサスSeiya is basically an application of ODE to biology, chemistry...

I prefer theoretical exams tho

Ted Shifrin

ODE could be theoretical (dynamical systems), but is usually not.

Coding is interesting algebra/number theory.

ペガサス Seiya

@TedShifrin I love coding

Some of the best students in my computer science classes also happen to be mathematics students or students with majors that involve a lot of math classes. I wonder why

Ted Shifrin

19:25

I'm talking about cryptography (making and breaking codes), not programming.

Now I'm not sure what Sine meant, but I think it's mine.

Ah, chat is misbehaving ... again.

Every message appears twice.

ペガサス Seiya

@TedShifrin How many fingers am I holding? |||

Ted Shifrin

(I got a retry/delete message on three of those four.)

Koro

@SineoftheTime Biomathematics

I think that would also include modelling disease spread etc.

ペガサス Seiya

I always sucked at anything related to biology

Ted Shifrin

Yes, biomathematics is all about modeling. :)

Koro

19:39

@shintuku: Hi!!

which chapter are you at in the book?

shintuku

currently prepping for my semester so minimal maths atm :(

have not advanced beyond what we did in the reading group

Koro

ah okay, np. We can get the group de-frozen when you're ready.

shintuku

that will be in 4 and a half months heheh

but i will be continuing abstract algebra as soon as semester ends

Koro

np. My exams also start within next 2 weeks.

the exams will be in consecutive days (no leave in between)

shintuku

gl

Koro

19:46

thnks. gl to you too :).

Sine of the Time

@Koro you're right

Silly Goose

20:36

is the concept of taking a limit of a function related to taking a limit of a sequence?

I am still trying to figure out why the determinant being a continuous function implies what it does as in the following picture.

leslie townes

silly: depending on the specifics, sure. there are notions of limit that aren't captured well via sequences, but that probably isn't what you're asking about.

Silly Goose

Right now this is how I am reading it: we have a function $det(A) = 1$ for all $A \in SL(n ; \mathbb{R})$ by construction of the group. Since the det function is continuous...how do I think after this point?

leslie townes

silly: ok. so "det" here is a function from SL(n,R) to R.

D.C. the III

leslie townes

there's also a notion of what it means for a function on SL(n,R) to be continuous. note, this paragraph that you have quoted does not describe that notion at all. it is part of the background context in which this paragraph is taking place.

D.C. the III

20:40

Leslie quick question.

I'm asked to show that $I-T$ is an orthogonal projection on $W^\perp$ given that $T$ is an orthogonal projection on $W$. I have this theorem, COuld I use this and just do a few quick manipulations on $I-T$?

leslie townes

or more generally det is a function on GL(n,R), or even the set of all nxn matrices over R. there's a notion of continuity of a real-valued function on each one of these spaces.

and what this continuity captures is that if A_n is a sequence of matrices "converging" to some matrix A in any one of these spaces, then det(A_n) is a sequence of real numbers that converges to det(A).

Silly Goose

oh okay i see

leslie townes

i put "converging" in quotes because the notion of a sequence of matrices converging to another matrix is also not introduced or defined in the paragraph you quote. it's part of the background context.

Silly Goose

so the sequence of numbers det(A_n) = {1,1,1,1,1,1,1,1,1,1,...} when the domain is SL(n; R), which converges to $1$?

oh i see yes sorry that background context was indeed introduced in a previous part of the section

leslie townes

if (A_n) is a sequence in SL(n,R) then the sequence (det(A_n)) would be the constant sequence (1,1,1,1...) yes.

Silly Goose

20:45

so the determinant function's domain in this case is countable?

i think this is what i am also confused about because when I think of (naively) a continuous function i think of something, for example, with a domain over the real line (i.e. uncountably infinite domain)

and i have not encountered before a function defiend over a countable domain which is called continuous

but my calc is quite horrendous :P

leslie townes

here we're regarding the determinant as a function on a set of nxn matrices (the set of all nxn matrices, if you like).

then the paragraph says, if you have a sequence of things from that domain - such and such happens.

it would be like if i said let's consider the sequence sin(1/n) [which goes to 0 as n goes to infinity]. so we're then looking at the values of the sine function along a sequence of points in its domain. not the same thing as declaring the domain of the sine function to be countable.

Silly Goose

Hm and is that the usual notion of taking a limit

leslie townes

in my example yes, it is the usual notion of limit.

Silly Goose

you do not look at a general subset of points in the domain but a sequence?

leslie townes

not in that example.

you could also consider lim_{x to infinity} sin(1/x). if you look at how limits as x goes to infinity are defined in calculus books, the definition is probably not in terms of any sequential limit. but it happens to be the same limit as the limit in the example.

Silly Goose

20:53

Wait sorry I am confused. In the sin(1/n) example, i thought we are considering the values of the function along a sequence of points in the domain—is this different than what I said above

leslie townes

not really. it's different in that i'm giving an explicit sequence (1/n) of points in the domain of sine, and the paragraph does not consider any specific sequence of "points" (now matrices) in the domain of det. but not conceptually different.

Silly Goose

Or i guess is the more broad point that given this function sin(1/x), you can take the limit with respect to sequences of points of the domain, or you can take the limit as you mentioned is usually defined in calc lim_{x to infty} sin(1/x)

leslie townes

yeah. or another way of looking at the same example, you can think of lim_{t to 0} sin(t), or you can consider sequential limits lim_{n to infinity} sin(x_n), where x_n is a sequence that goes to 0 as n goes to infinity [the notion in which "x_n to 0 as n to infinity" is defined differently from the epsilon-delta sense of "sin(t) to 0 as t to 0"]

Silly Goose

oh okay i see this has helped a lot :D

and one last thing: so you are saying the two notions of limits do not necessarily have to agree?

shintuku

there's a proof somewhere that they are equivalent, if you mean the standard definition vs. arbitrary sequence one

proof at page 270 of Bruckner et al. Elementary Real Analysis

leslie townes

21:32

yeah, the fact that the real variable limit exists if and only if all appropriate sequential limits exist is included in a lot of texts.

i sometimes think of it as more useful in thinking than in practice. it's pretty unusual to have your hands on the set of all sequential limits that have to exist, for purposes of using one direction of this theorem.

Obliv

21:45

how does $\frac{dP}{dt} = P(a-bP)$ partial fractions apply here

$\frac{dP}{P(a-bP)} = dt$ ? then?

$\frac{A}{a-bP} + \frac{B}{P} = A(P) + B(a-bP) = dP$

leslie townes

@D.C.theIII do you have a characterization of "is an orthogonal projection" in terms of matrix algebra? (I-T)^t = I - T and (I - T)^2 = I - T - T + T^2 = I - T - T + T = I - T would show "is an orthogonal projection," if you did

Ted Shifrin

Or using the definition you were using in my book, it’s immediate.

leslie townes

and the "onto W^perp" could probably shake out of an identification of the range or kernel of T, if you have theorems about how subspaces relating to a matrix fit together

D.C. the III

@leslietownes Well this was in terms of the linear operators. The definition of orthogonal projection is (after al the other conditions) that $R(T)^\perp = N(T)$ and $R(T) = N(T)^\perp$. All of what you did is what I did, my only concern was that I haven't "shown" that $(I-T)^*$ is the adjoint of $(I-T)$

oh the other characterization is the theorem I posted earlier

but it depended on knowing $T$ has an adjoint (in terms of the original theorem)

Shaun

Hi :) Recently, my interest in topos theory has been renewed. See here for a question of mine on the legitimacy of topos theory & intuitionism. Rather than ask a very similar question, I thought I'd ask the following here: what are some reputable sources discussing the strengths & weakness of intuitionism?

Ted Shifrin

22:00

I’m not sure how many of our regulars know anything about any of this. I sure do not.

shintuku

alrighto, I wanna prove we can make any polynomial into a $w_0(z-w_n)\cdots(z-w_1)$ form. We use induction
The base case is obvious, now for the inductive step. Suppose it is true for any $n$ degree polynomial. Let $f$ be an $n+1$ degree polynomial. Then it has at least one root (this is given), say $w$, so it has a form $f(z) = (z-w)g(z)$ (this is also given), where $g$ is an $n$ degree polynomial. But by our inductive hypothesis, $g$ has the desired form, so $f$ will too. this concludes the proof

now this is all great, but where is my leading coefficient $w_0$ in $w_0(z-w_n)\cdots(z-w_1)$ coming from

Shaun

I thought that might happen, @TedShifrin. Perhaps there's a logic chat room I could try . . . Thanks anyway!

Ted Shifrin

@DCthe Note that $b-p\in V^\perp$ and $b-(b-p)=p\in V^{\perp\perp}=V$.

leslie townes

questions about the 'legitimacy' of an approach to math seem very early 20th century to me, at the latest, but this is definitely not my area. i'm guessing that for some folks, it's more about whether people think a particular approach is interesting, or worth investigating, than is 'legitimate.'

shintuku

maybe the foundation of mathematics is not about legitimacy, or about truth, maybe it was all along about the friends we made along the way

10

Ted Shifrin

22:03

Other than in the context of children, I’m not sure how to delineate legitimacy.

D.C. the III

Yea when I initially looked at the question I thought about your definition and did the proof in my head. I was just trying to work on "applying theorems" to my efforts...

leslie townes

dc3: oh. somewhere in the universe of life, it oughta be verified that * is a skew linear mapping on all linear operators (finite dimensional inner product space) or all bounded linear operators (infinite dimensional inner product space), so that you can just calculate without worrying about adjoints existing.

D.C. the III

not sure why I quoted it but ah well

leslie townes

the adjoint is a very subtle thing in the unbounded operator context.

Obliv

Anyone know where I went wrong in my partial fraction decomposition $\frac{dP}{P(a-bP)} = \frac{A}{a-bP} + \frac{B}{P} = A(P) + B(a-bP) = dP$ I'm confused if I put P = 0 I get B(a) = dP so B = dP/a ?

should reduce to $(\frac{1/a}{P} + \frac{b/a}{a-bP})dP$

Ted Shifrin

22:05

You should not write equals signs when things are not equal.

Shaun

I have asked to join the following chat room.

chat.stackexchange.com/rooms/info/44058/logic

D.C. the III

@leslietownes I'm not sure if what you said is the same as what was done in the past, but it was shown that all linear operators have and adjoint....

Ted Shifrin

If $P=0$, then $P=0$ throughout, Obliv.

D.C. the III

But I have a feeling that you mean much more....beyond what I have seen yet. What field would this be?

Shaun

@leslietownes That's what I have gathered about the Hilbert/Brouwer controversy, that it is out of date. My biggest fear in studying this stuff is to end up a crank or something . . .

Obliv

22:08

@TedShifrin oh so I set a = bP and then $A(a/b) = dP$

so A = bdP/a ?

nvm

Ted Shifrin

No.

What did I just finish saying?

Obliv

I think that's right, hence the factor of dP on the outside for the correct expression

OH

ペガサス Seiya

Someone please put out the sun

It isn't letting me sleep

Ted Shifrin

Just put out your eyes.

Obliv

Isn't it like midnight in japan or something

oh wait it's 12 hrs so around 5

ペガサス Seiya

22:11

@Obliv Its almost 7:15 AM

Obliv

So is that substitution correct at least @TedShifrin a = bP

I just have to take the derivative of a/b w.r.t. something..

Ted Shifrin

No.

Set $P=0$ every single time.

Obliv

but then $A(0) + B(a - b(0)) = dP$

D.C. the III

@ペガサスSeiya No being on your phone/ computer is not letting you sleep

Ted Shifrin

First of all, this is garbage from the beginning. What is $dP$ doing here?

Obliv

22:14

it's the logistic equation $\frac{dP}{dt} = P(a-bP)$

I did $\frac{dP}{P(a-bP)} = dt$

Ted Shifrin

The fraction is $1/ ….$.

Obliv

Oh..

in that case it's simple.

anyone have an example of a higher order diff eq with repeated complex roots I can solve

Ted Shifrin

Partial fractions occurs only when you separate variables. What do you mean higher order?

Obliv

I guess $\geq 2$ order

I feel cheap using Euler's formula as a tool to solve D.E.s without understanding it

Ted Shifrin

You mean $e^{ix} = \cos x + i\sin x$?

Obliv

22:24

yup, just using it for these $b^2 - 4ac < 0$ cases

Ted Shifrin

So what do you not understand? Euler’s formula comes from Taylor series.

Obliv

I haven't constructed it myself, so I just feel like I'm just borrowing it I guess. (Or if I have constructed it, I don't remember)

Also what does a root being of multiplicity k mean?

for example $\frac{d^4 y}{dx^4} + 2\frac{d^2 y}{dx^2} + y = 0$ gives us roots $m_1 = m_3 = i, m_2 = m_4 = -i$

in my textbook: In general, if $m_1 = \alpha + i\beta, \beta > 0$ is a complex root of multiplicity k of an auxiliary equation with real coefficients, then its conjugate $m_2 = \alpha - i\beta$ is also a root of multiplicity k.

Ted Shifrin

Factor the polynomial $z^4+2z^2+1$.

Obliv

$(z^2 + 1)^2$

Ted Shifrin

22:40

Keep going …

Obliv

Oh.. $(z^2 + 1)(z^2 + 1)$ ?

Ted Shifrin

Nah.

Obliv

OH $(z-1)^2$

Ted Shifrin

Um …

Obliv

$(z-1)^4$

mick

22:42

1

Q: About zero's of $f(s) = \sum_n p(n) n^{-s} = 1 + 2\cdot2^{-s}+ 3\cdot3^{-s}+ 5\cdot5^{-s}+\dots$

Let $f(s)$ be a somewhat zeta like function defined on the complex plane as : $$f(s) = \sum_n p(n) n^{-s}= 1 + 2\cdot2^{-s}+ 3\cdot3^{-s} + 5\cdot4^{-s} + 7\cdot5^{-s} + \dots$$ where the coefficients $p(n)$ are the noncomposite numbers ( $1$ and the primes ) Consider solving $f(s)=0$. It appears...

does anyone know how to solve it , or maybe restate it by an equivalent conjecture ?

Obliv

so multiplicity is the $4$

nvm says it's supposed to be 2 for that example

Well $e^{\alpha x}(\cos \beta x + x \cos \beta x + \sin \beta x + x \sin \beta x) = 0$ is the linear combination?

mick

when is sin(pi/p) not expressible by radicals NOR sin(pi/q_i) for q_i < p and p a prime ?

and is that a good question to post ?

Obliv

23:00

is $d/dx[e^{\frac{-1}{2}x}(-\cos(2x)+C_2 \sin(2x))]$ two product rules

Astyx

Factor $z^2+1$

Obliv

$(z+1)(z-1)$

Astyx

nope

Obliv

damn I am so tired lol I can't even factor

Astyx

try to find a root of that polynomial

Obliv

23:02

-1 and 1

wait

that's not factorable

I thought I was overlooking something

ted told me to keep going, what a dingleberry

i gotta take my test now, thank you for your help

Ted Shifrin

I had to star that!

kevinkayaks

(z+i)(z-i)

Ted Shifrin

Ayup.

mick

1

Q: When is $\sin(\frac{\pi}{p}) $ not expressible by radicals and $\sin(\frac{\pi}{q_i}) $?

We have the following identities: $\sin(\frac{\pi}{1})=0$ $\sin(\frac{\pi}{2})=1$ $\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{\sqrt{4}}$ $\sin(\frac{\pi}{4})=\frac{1}{2}$ $\sin(\frac{\pi}{5})=\frac{\sqrt{5-\sqrt{5}}}{\sqrt{8}}$ On the other hand it is known that for integer $n$ , $\sin(\frac{\pi}{n}) $ ...

abstract-algebra trigonometry galois-theory closed-form roots-of-unity

posted it anyway :)

im sick need rest. goodnight

Sine of the Time

@mick g'night

Transcript for

$Mathematics$ Mathematics