in the complex case there's like a conjugate or something

But if we restrict to real $x$, I think you can't automatically say that $\langle Ax,x\rangle = \langle Bx,x\rangle \implies A = B$, you'd need to already know that $A$ is self-adjoint or something.

Yeah in the complex case you get a conjugate

i never said we could conclude $A = p(A)$

I never claimed you did, I'm just wondering that since this doesn't guarantee equality, I'm not totally sure what information we can extract

i think maybe the most fruitful thing would be to study the $p^{2n}(A) = A$ thing

Perhaps

I mean this is a lemma to a more general problem which I'm starting to think is much easier than I'm making it out to be

It's that if a normal operator has all distinct eigenvalues, it's self-adjoint

My idea is that since it has all distinct eigenvalues, you can conclude that anything commuting with it is a polynomial of it. This now includes the adjoint since $A$ is normal, so that's where I've been trying to go

this is much easier than you make it

the challenge is more interesting than the normal operator thing lol

Okay I think I have an idea but I want to make sure it isn't stupid

So, assume $\langle Ax,Bx\rangle = 0$ for all $x$, then $A+B = 0$

The idea is that then $\langle Ax,Bx\rangle + \langle Bx,Ax\rangle = 0$ as well, so that $\langle (A+B)x,(A+B)x\rangle = \|(A+B)x\|^2 = 0$ for all $x$

Yeah that should do it if I'm not braindead

Because then $\langle Ax,x\rangle = \langle x,p(A)x\rangle$ for all $x$, so that $\langle (A-I)x,(I-p(A))x\rangle = 0$ for all $x$

But then this implies that $A - p(A) = 0$, so we're done

why isn't it the right answer

@AaronM 1) In both cases, the objects move in the positive x direction without turning around. That means that their displacement from the starting point is the same as their distance traveled.

2) How do you determine the velocity/speed from that graph?

I thought velocity was just delta x/delta t. And they both have the same current x and both started from t=o hence equal velocitys

And thanks for your explanation of choice #1

Delta x/ Delta t is the average velocity.

And that's certainly the same for both.

oh, so the final answer is choices 1,3,4?

But unless someone uses the word 'average' you should interpret it as the instantaneous speed.

Not quite. My point is that they're not asking whether they have the same average speed, but whether they've got the same instantaneous speed

What determines the instantaneous speed?

Slope, ant they don't have the same slope at A

Right.

Also, what's the difference between speed and velocity?

vector and scalar

Yeah, but mathematically, @AaronM

Other way around, actually.

Speed is the magnitude of the velocity vector.

correct ;)

Yes! Actually, you want to see something cool?

However: In 1D motion, by velocity one often just means "the x component of velocity"

in which case it's also a scalar, but a positive/negative one.

They talk about that in this video, and it's pretty funny.

You can't get out of a speeding ticket using calculus :) You can only get into a speeding ticket using calculus :P

It's a bit abusive terminology, but not that hard to understand.

lol

Whats the final answer to my question I won't understand the explanation until I Know what I did wrong or right

Let's go through the four options.

Which of the options is most obvious?

I am pretty positive 3 and 4 are coreect

Well, 3 is to my mind the most obvious one. The lines intersect at time t, so they must be at the same position then.

So 3 is correct.

Hi @ALannister

Hey @TedShifrin

The other one involving position is 1, and that has to do with distance traveled. But both particles are always moving in the positive x direction; they never turn around.

So is there a difference between the distance they've travelled and their displacement?

I agree with u on 1 and 4

I understand that if they don't turn around that displacement=distance

Right. So 1 is correct.

That leaves the bits re: speed and velocity.

Now, the important point here is that they could have asked for average velocity.

Had they done so, then the fact they've moved the same distance in the same time would indeed give the same average velocity.

However, they did not ask for average velocity.

Oh, so since they didn't it means instanenous and they don't have same slope

Right.

And since they traveled the same distance over the same time they must have the same speed?

Again, that'd be average speed.

Without that word, they mean instantaneous.

Oh ok, basically Im confusing average and instatntenous

Hmm, I'm just wondering (as I procrastinate on dynamics) what are some of your favorite problems/theorems, in general

Yes, but also I feel like this is a problem where they should say it explicitly.

glares at Demonark and tells him to stop procrastinating

Though, their phrasing does ask about what's going on "at point A."

Thank you for your help @Semiclassical , I will remember next time that if not mentioned, to assume instatntenous.

And there's no way to have an average just by specifying one point. You'd need two for that.

You have the origin as one point though

Well, you're assuming that the origin is the other point.

They didn't tell you to do so.

But one could also pick another point B and ask about the average velocity defined by points A,B.

Yes, Thanks again for your help

mmkay. Did the answer come out right now?

yes it did

Oh okay then yeah that's a good idea {\footnotesize I'll get back to it right now}...

glad to hear it. good luck with the rest of your problems.

Crap

Any ideas about finding isomorphism between rings of fractions?

Link to what I posted earlier: math.stackexchange.com/questions/2352954/…

@AaronM have you seen this?

@ALannister Best thing I can come up with is to stick with the case of $k=2$ for the moment.

Start with the simplest case and see if that gives any insight.

But @Semiclassical what should the mapping be?

Never said I had an answer. I've never had to work with rings of fractions myself.

All right.

But focusing on k=2 may provide some insight.

Possibly.

My question hasn't gotten a lot of attention, unfortunately. Only 12 views and I posted itseveral hours ago

Link please.

Well, "rings of fractions" is a bit obscure maybe? No easy fix to that.

But, like you said, look at $k = 2$.

@LasVegasRaiders the link is here, I put it on here 5 posts ago.

I'm not going to post it again, lest I be accused of spamming.

I see.

Ring of fractions isn't obscure. But, best I can understand, @ALannister, you're trying to look at inverting an arbitrary multiplicative system. This is usually written $S^{-1}R$ when $R$ is an integral domain and $S$ is a multiplicatively-closed subset (not containing $0$).

@TedShifrin correct.

When I looked at your link a few days ago I was confused by your terminology.

But, @Ted, surely anything I don't know about is obscure!! /s

In this case, yes.

Yes, @Semiclassic, of course you are always the universal arbiter.

About the not containing 0 part, that is. Sometimes, it does contain 0. But not in this case.

You can never invert 0. :)

Sooo tempted to star that :P

So that's not allowed (by reasonable mortals).

There are allowances made in that case, and we just say that $S^{-1}R$ is the zero ring.

I do not like that. But your teacher is in charge of you.

There's actually a very interesting explanation as to why that is, and it all has to do with equivalence classes.

But anyway...

I continue to say I do not like it.

So, I'm trying to figure out how I can show that those two things, $E^{-1}\mathbb{Z}$ and $D^{-1}\mathbb{Z}$ can be shown to be isomorphic.

So you're trying to invert less than everything nonzero.

If $E$ and $D$ are ... ?

Yes. And since $\mathbb{Z}$ has no zero divisors, we're in good shape.

$D = \{n_{1}^{t_{1}}, n_{2}^{t_{2}}, \cdots, n_{k}^{t_{k}}\}$

Ugh, the set of all products of such?

Yeah, $D$ is generated by the $n_{i}$

OK, and $E$?

then $E$ is generated by only one $n$

Whoa.

Let's try a concrete example.

Let's say $D$ is generated by $4$ and $6$.

lol, k=2 :)

Hehe, Semi ;)

glares

I'm just chuckling because that was literally the only suggestion of substance I was able to give earlier: Start with k=2.

And $E$ is generated only by 4

So I can invert $4$ and I can invert $6$.

OK, Semiclassic. You're so smart :P

No, if $E$ is all powers of $4$, inverting things in $E$ won't allow me to invert $6$.

But $E$ has to be generated by only one of the $n$'s, so what do we let it be generated by $6$ or something?

It needn't be one of the $n$'s, but I actually don't believe this will be correct.

So you mean, it can be any other integer n?

Not any other. But I'm not convinced it's right.

So if I can invert $6$, I can also invert $3$, since $\frac 26 = \frac 13$.

I can also invert $2$.

Oh, so I guess being able to invert $6$ will do it. Interesting. So $6$ works.

Can you sort that out?

I guess your homework is right, after all :)

;P

One must play with examples — my permanent moral of the story.

notes dead silence (not complaining about Semiclassic's mutedness for a change) :D

So, then a mapping between the two sets would be $\frac{l}{n_{1}\cdot n_{2}} \mapsto \frac{c}{n_{2}}$?

Pfffft.

That's an L in the numerator of the first fraction btw

Yeah, I got that.

You may interpret that as either me blowing a raspberry or passing gas, either one works in this context :P

No, we need to use the particular nature of the numbers for it to work.

smacks Semiclassic for rude behavior

Well, I need to establish some kind of isomorphism.

Yeah, yeah.

Well, slow down. Do you first see why inverting $6$ gives us everything in $D^{-1}\Bbb Z$?

Then, what is the general situation going to be?

How did I settle on $6$?

And, no, it won't be one of the $n_i$ in general.

@TedShifrin is it because 6 and 3 are not coprime, and 6 is not prime?

Well, I had $4$ and $6$.

Oh well, and 4 is not prime either, so there goes that idea. But, they're not coprime.

So how did I figure out that 6 is going to work?

Because if you can invert 6, you can invert 2, 3, and 4?

Right.

And you understand that?

Back later.

Bye, rude Semiclassic.

To be honest, only shakily right now. That might improve.

I'm totally serious, @ALannister ... and I gave this advice every time I taught algebra ... you really have to play with examples before you try to prove general theorems.

Well, $\dfrac 14 = \dfrac 9{6^2}$.

So, inverting just powers of $6$ allows you to invert any product of $4$s and $6$s.

$\frac{1}{2} = \frac{3}{6}$

Right, but we didn't really need to invert $2$. :P

Yeah, 2 is kind of like the annoying little cousin who wants to play, too.

So, if I had $n_1 = 8$, $n_2 = 9$, what would I use for $n$?

ROFL.

Thnking...

Hey Ted. Have you ever seen a geometry that treats two sides of a surface as different spaces to contain points and lines?

Hell no.

fair enough

:)

The surface consists of points. One doesn't see a side or not a side.

we were chatting earlier and thought you could say one way or the other

ah fair point

no pun intended

wasn't sure if any obscure varieties existed outside of toying around.

This case isnt oone where we can use one of the n's is it?

Nope, @ALannister.

A point has no size @Typhon

Granted, I can see where such a geometry in correspondence with a mobius might lead to... error.... cannot compute sort of issues.

@LasVegasRaiders I never claimed that?