Mathematics - 2017-01-06 (page 4 of 6)

Jessy Cat

18:00

Why did so many people not like this question.

Oh, I see now.

Semiclassical

That it's tagged as 'geometry' for no apparent reason probably doesn't help :/

Jessy Cat

Still, I feel bad :(

Yeah, that's weird.

Semiclassical

I don't see any reason to be rude to that person, but it's a question that should get closed.

Jessy Cat

0

Q: Wave equation on infinite line with piecewise $2\pi$-periodic i.c.

I need to solve the following p.d.e.: $\begin{align} u_{tt}(x,t) = a^{2}u_{xx}(x,t) & \,\,\,\, -\infty < x < \infty, \,\, t>0 \\ u(x,0)=0, \,\,\,\,\,\,\,& u_{t}(x,0) = g(x), \,\,\,\, -\infty < x < \infty \end{align}$ $\text{where}\, g(x) \, \text{denotes the}\, 2\pi-\text{periodic function ...

People love to be rude on MSE. It makes them feel slightly less not smart than they think they are but would never actually tell anyone.

Mary Star

Could you explain to me the meaning of a transitive group action?

Jessy Cat

18:03

Group Actions is in my next chapter :(

s.harp

@JessyCat instead of \, \, \, use \quad or \qquad, ie $a\quad b$, $a\qquad b$

Semiclassical

"People love to be rude on the internet"

Jessy Cat

@s.harp sorry.

I had forgotten about those tags

s.harp

no its just a comment, you will save yourself effort

Semiclassical

I think part of it is that there's so many bad questions that show up on main that it's easy to get jaded by it

Jessy Cat

18:05

It just got real now

Yeah, I probably shouldn't do that...

s.harp

@Semiclassical I think after a while most people just browse the titles of the main and only actually open questions if they are in the tags they follow or have an interesting titlte

Semiclassical

I try not to be an arse on main, but I don't necessarily spend time explaining why certain questions are a bad fit

It's a lot quicker to do a close vote than a comment

Not something that is necessarily good but eh.

s.harp

@MaryStar a group action (of $G$ on $X$) is transitive if starting from any point $x\in X$ you can "get to" any other point $y\in X$ via the action of $G$

Jessy Cat

ducks Please don't ban me.

Semiclassical

whu-oh.

Alessandro Codenotti

18:07

Sometimes you see a question that looks interesting but it's written in a way you can't really understand it

s.harp

ie for all $x,y\in X$ there exists a $g\in G$ with $g\cdot x = y$

Jessy Cat

There are some people I'd like to give that card to ;P

And they're not my aunt.

Balarka Sen

@AlessandroCodenotti That question sounds like nonsense to me

Semiclassical

@JessyCat Glancing at that question, separation of variables seems like a nice fit.

s.harp

The title sounds like its from a random generator

Jessy Cat

18:09

@Semiclassical even so, D'Alembert's is probably easier.

I just don't know how to make sense of that $g$

Mike Miller

It's probably not interesting.

Semiclassical

Possibly. I tend to fall back on tools I know how to use, though.

Alessandro Codenotti

to me too as it's written right now, but it's not from a new user so I guess there is a sensible question that was supposed to be asked there? @Balarka

Semiclassical

A good number of my edits nowadays tend to involve making the title be actually descriptive

Jessy Cat

Sometimes English is just so painfully obviously not their first language, and people write snarky comments and downvote because they don't like the grammar.

Mithlesh Upadhyay

18:10

This math.stackexchange.com/users/398545/vachoh?tab=profile id seems also attached with @KanwaljitSingh, It seems that he has many fake ID for voting purpose.

Jessy Cat

I actually saw an example of this the other day.

A---B

Which answer would you choose if you have 5 answers under your question all saying exactly same thing ? eg math.stackexchange.com/questions/2086428/…

Jessy Cat

Hello Roberto.

Semiclassical

@JessyCat I tend to flag such comments for being rude or not constructive.

Jessy Cat

I commented back and mentioned that the guy's first language is obviously not English and that they should give him a break.

So many Sheldon Coopers lurking...

Mary Star

18:12

Ah ok... In my notes there is the following:
$S_3=\{ id, (1 \ 2), (1 \ 3), (2 \ 3), (1 \ 2\ 3), (1\ 3\ 2)\}$.
The subgroups are: $id, S_3, \langle (1 \ 2)\rangle, \langle (1 \ 3)\rangle, \langle (2 \ 3)\rangle, A_3=\langle (1 \ 2 \ 3)\rangle=\{id, (1\ 2\ 3), (1\ 3\ 2)\}$.

Why are all the subgroups generated by one element?

Also why are only the $S_3$ and $A_3$ trasitive?

Balarka Sen

@Alessandro I guess

Semiclassical

I think that, if it's clear there's a language barrier, it's appropriate to edit the question for clarity.

Mike Miller

@SemiC Do you know any physical examples of things that aren't smooth? That are only like C^3 or something?

Jessy Cat

Not a bad idea.

Semiclassical

Step functions are an obvious one, though that's not even continuous.

Mithlesh Upadhyay

18:13

@robjohn, So, he math.stackexchange.com/users/401635/kanwaljit-singh is banned finally. Thanks SE.

s.harp

@MikeMiller solutions to the heat equation in presence of shock waves etc

Jessy Cat

Nice.

Mike Miller

@s.harp got it

Jessy Cat

@s.harp maybe you could take a look at my pde question. bats eyelashes

Semiclassical

My counsel would remain separation of variables.

s.harp

18:14

Just because I can talk a bit about where PDEs are used and how they look like doesn't mean I can do anything with them ;)

A---B

@MithleshUpadhyay No he just got a small suspension .

Jessy Cat

You probably know how to take care of a piecewise boundary condition that's given to be $2\pi$-periodic

Semiclassical

An integral transform technique might also work, though the two approaches are not incompatible.

Jessy Cat

actually, it's an initial condition

Pretty much every example in that section was done with D'Alembert's formula after it was derived.

Semiclassical

Hmm, fair enough.

Jessy Cat

18:15

But there are no examples anywhere with that kind of an initial condition.

Mithlesh Upadhyay

@A---B , However, I caught him. ;)

Semiclassical

I don't actually remember D'Alembert's formula so I probably can't say much.

A---B

@MithleshUpadhyay What he did ?

Alessandro Codenotti

the trajectory of a body with elastic collisions is continuous but not $C^1$

Jessy Cat

That wasn't supposed to happen

I meant to edit it not delete it!

Balarka Sen

18:17

@Alessandro Not even differentiable, is it?

Alessandro Codenotti

indeed

Balarka Sen

like path of a particle in a brownian motion

Semiclassical

Ah, Wikipedia has it: en.wikipedia.org/wiki/D%27Alembert%27s_formula

Jessy Cat

All I want to do is evaluate $\displaystyle \int_{x-at}^{x+at}g(z)dz$ when $\displaystyle g(x) = \begin{cases} -1 & -\pi < x < 0 \\ 1 & 0 < x < \pi \end{cases}$

Mithlesh Upadhyay

@A---B , He has multiple IDs to vote himself. You can check him. He was asking and answering on same post then voted himself(But, I don't know).

Jessy Cat

18:19

and $g(x)$ is $2\pi$-periodic

A---B

@MithleshUpadhyay I don't know what people gonna get by earning some rep.

Jessy Cat

How do I split that up?

Semiclassical

The most direct route I see is casework.

A---B

It certainly won't get them a degree in maths.

Jessy Cat

@Semiclassical explain

Semiclassical

18:20

For instance, if $x-at>0$ then $z>0$ on the domain of integration.

In which case, you're intergrating $1$ over an interval of length $2at$.

And similarly if $x+at<0$.

Mithlesh Upadhyay

@A---B , Now he commented this : math.stackexchange.com/questions/2086443/…

s.harp

@JessyCat the way I would do it is to take the fourier transform of $g$ (let it be $\mathfrak g$) then the equation becomes:
$\tilde u_{tt}(k,t)=-k^2\tilde u(k,t)$ (which is the harmonic oscillator) with boundary condition $\tilde u(k,0) = \mathfrak g(k)$

Jessy Cat

I wish i knew @Semiclassical there isn't an answer in the back of thebook for this one

Semiclassical

Where things become tricky is if $-at<x<at$.

But even there it's just a matter of being careful.

In that case, you're integrating $+1$ over some interval and $-1$ over another interval.

So it's just the difference in the two lengths of the intervals.

s.harp

$\tilde u_{tt}(k,t)=-a^2k^2\tilde u(k,t)$

i missed the constant

but thats really as far as my pde knowledge goes^

Semiclassical

18:23

I suspect for $-at<x<at$ it'll just be a linear interpolation between $f(x=at)$ and $f(x=-at)$.

Jessy Cat

Nevermind. Okay, so let $u(x,t) = U(\alpha,t)$ (I like my notation better). Then, $u_{tt}(x,t) = a^{2}u_{xx}(x,t)$ becomes $U^{\prime\prime}(\alpha,t) = -a^{2}\alpha^{2}U(\alpha,t)$

Then, the i.c. becomes $U^{\prime}(\alpha,t) = G(\alpha)$

A---B

@MithleshUpadhyay He is pretty good at making fake IDs.

Jessy Cat

But, that doesn't particularly help much.

Mithlesh Upadhyay

@A---B , Not good for his future.

Jessy Cat

I need to integrate $g(x)$

A---B

18:26

@MithleshUpadhyay I am not in this life going to pointed it out.

Jessy Cat

But, I guess I really don't understand the limits of integration. I know you tried to explain it to me @Semiclassical but I didn't understand

Akiva Weinberger

I have a $\Delta K<0$

Mithlesh Upadhyay

@A---B , :)

Akiva Weinberger

I mean, I have a cold :(

(I probably messed up the equation)

Jessy Cat

I'd say hahaha, but it's not funny to be sick

s.harp

18:28

$U(\alpha,t)=\sin(a\alpha t)\, G(\alpha)$, where $G(\alpha)=-\frac1{i\alpha}(1-e^{-i\alpha\pi})+\frac1{i\alpha}(e^{i\alpha\pi}-1)‌$ I think

Semiclassical

Not sure what there is to understand. For any given $x,t$ with $t>0$, you integrate $g(z)$ from $z=x-at$ to $x+at$.

Jessy Cat

Yes, but on each ofthe branches, what are $x-at$ and $x+at$?

Semiclassical

if z mod 2pi is between 0 and pi, g(z)=1. if not, it's -1.

What?

Jessy Cat

when I integrate say, -1, I'm integrating it from what to what?

Semiclassical

You can't say what x-at and x+at are. They'll be different for different points.

Balarka Sen

18:29

@AkivaWeinberger What's $K$

Jessy Cat

So, I just leave them in there like that?

Semiclassical

Yeah. However, I think I was misleading above: The periodicity of g(z) makes this trickier than I realized.

Jessy Cat

Shouldn't I at least plug something in for $x$?

Semiclassical

No. Your result will depend on x, t. (It had better, or this won't be a solution for all x,t.)

You may have to split it into some cases, but at the end of the day the solution must work for any x,t you plug in.

Jessy Cat

If I integrate $\int_{x-at}^{x+at}-1 dz$, I get $-2at$

For $1 dz$, it's $+2at$

Semiclassical

18:33

Yeah. So if the entire interval $[x-at,x+at]$ is between pi and 2pi mod 2pi, then that's what you'd get.

The problem is (and this is what I hadn't taken into account initially) that's by far the exception rather than the rule.

Jessy Cat

Why I said that about plugging in something for $x$ is that it's $2\pi$ periodic, so I thought I should be able to make the integral for $-1$ between $-\pi + at$ and $-\pi - at$

Semiclassical

I think you can do that for $x$, yes. But that still leaves $t$.

Jessy Cat

Anyway, what do I add the two integral results together now?

Mike Miller

what are we doing?

Semiclassical

A tedious integral which is harder than I realized. @MikeMiller

Jessy Cat

18:35

But then the solution is zero :(

Mike Miller

/peaces out

Semiclassical

You're not hearing me.

Jessy Cat

@MikeMiller math.stackexchange.com/questions/2086467/…

Semiclassical

Suppose you pick $x=\pi/2$. Then if $at<\pi/2$, you have $0<x-at<z<x+at<\pi$ for all $z$ in the domain of integration.

So in that case you'd have $g(z)=1$ and therefore the integral would just be $2at$.

Jessy Cat

Oh, so it's going to be a piecewise solution!

Semiclassical

18:37

However, if $at>\pi/2$, then the domain of integration extends outside of $[0,\pi]$

Akiva Weinberger

@BalarkaSen Wait, that should probably be $T$. (I was thinking Kelvin but that doesn't make sense)

Semiclassical

So therefore it starts picking up negative values and decreasing.

...but then again, if $at>\pi$, you'll again start picking up positive contributions. And so on.

Jessy Cat

So it's a piecewise, periodic solution?

Semiclassical

Yeah, in some manner. But I'm not sure what the best way to get it is.

I mean, the description I just gave was for $x=\pi/2$.

Jessy Cat

But that should be true for any $x$ on each of the intervals given.

Regardless of what $x$ is, it does not actually appear in the answers for either of the two integrals.

They cancel out.

Semiclassical

18:40

My point is largely that I'm not sure I know how to do it explicitly.

Not without some headaches.

Hence why this direct approach, while valid, may not be the simplest.

Jessy Cat

So, I'd think the answer would be $u(x,t) = \begin{cases} \frac{1}{2a} (-at) & \text{for}\, -\pi < x < 0 \\ \frac{1}{2a}at & \text{for} \, 0 < x < \pi \end{cases}$

It's dependent only on time, not on space, except in the sense of periodicity

Semiclassical

Could be right. I really don't know.

Jessy Cat

Hm... it seems to make sense.

Semiclassical

one thing I can say is that $u$ should vanish identically along the lines $x=n\pi$ for integer $n$.

Jessy Cat

It's not homework. I'm not getting graded on it. I'm just doing problems in the book to try to learn, so I'll sit on it for now and move on. If someone answers me in the next couple of days, great. If not, if I haven't had to either take my exam or figured it out on my own, I'll put out a bounty.

That doesn't seem to happen for my solution

Semiclassical

18:44

Yeah.

Jessy Cat

why should it vanish? It should definitely be constant along those lines...

Semiclassical

The reason I say that it should do that: First, the solution should be periodic in $x$. Additionally, I think the solution should be symmetric about the line $x=\pi/2$ since $g(x)$ itself is.

So I think I can focus without loss of generality on the case $x=0$.

In that case, we're doing $\int_{-at}^{at}g(z)\,dz$ with $g(z)$ being an odd function of $z$. But that integral is therefore zero by symmetry.

Jessy Cat

And in which case the whole darn $u$ is identically zero

because of the ic on $f$

Which would be the stupidest problem in history.

Semiclassical

I think that's a bridge too far.

I don't see how it being zero along $x=n\pi$ would lead to it vanishing everywhere.

GFauxPas

Is there a name for a plot where on the left you have contours and on the right you have images of those contours? Like showing how exp maps lines to circles or whatever

Jessy Cat

18:47

Using D'Alembert's formula, that's what it comes out to: $\displaystyle u(x,t) =\frac{1}{2} \int\left[ f(x+at) - f(x-at)\right] + \frac{1}{2a}\int_{x-at}^{x+at}g(z)dz$

oh right

Fixed now

Abcd

Does alternate segment theorem work if chord and tangent intersect externally?

Semiclassical

Just to check---$f=0$ identically here?

Jessy Cat

Crap. hold on

the formula's right except for the integral part in front of the f's here

Semiclassical

I figured. But $f(z)=0$ here since $u(x,0)=0$.

Jessy Cat

$\displaystyle u(x,t) = \frac{1}{2}\left[ f(x+at) - f(x-at)\right] + \frac{1}{2a}\int_{x-at}^{x+at}g(z)dz$

Yes, it is identically zero

So, the first part vanishes

So all I need to worry about is how to evaluate $\displaystyle \frac{1}{2a}\int_{x-at}^{x+at}g(z)dz$

Semiclassical

18:51

Right. My claim above was that, if $x=0$, then $u(0,t)=0$.

(for $t>0$, at least.)

To establish that, we need only note that $u(0,t)=\frac{1}{2a}\int_{-at}^{at}g(z)\,dz$ is the integral of an odd function over a symmetric interval.

Jessy Cat

So that doesn't affect the validity of my answer?

Semiclassical

Which answer?

I'm saying that that's a necessary condition for your solution to satisfy in order for the integral to have been done correctly.

Jessy Cat

13 mins ago, by Jessy Cat

So, I'd think the answer would be $u(x,t) = \begin{cases} \frac{1}{2a} (-at) & \text{for}\, -\pi < x < 0 \\ \frac{1}{2a}at & \text{for} \, 0 < x < \pi \end{cases}$

Semiclassical

Ah.

Jessy Cat

except the a's can cancel

Semiclassical

18:55

The problem is, suppose you consider $u(\pi,t)=\int_{\pi-at}^{\pi+at}g(z)\,dz$.

GFauxPas

do you need to label the angles in a polar plot?

Semiclassical

Suppose I make the substitution $w=\pi-z$.

That puts the integral into the form $\int_{-at}^{at}g(\pi-w)\,dw$. (there's a minus sign from the coefficient of z, but this cancels against a necessary flip of the limits of integration).

Jessy Cat

Okay...

Semiclassical

Thing is, what's the behavior of $g(\pi-w)$?

If w is between $0$ and $\pi$, then so is $\pi-w$. So $g(\pi-w)=g(w)=1$ on that interval.

Jessy Cat

Well, $g$ is $2\pi$ periodic, b ut since we don't know what $w$ is, we can't figure out how $g$ should behave there

Oh, okay

Then, if $w$ is between $-\pi$ and $0$...

Semiclassical

19:01

Well, a helpful observation in that case is that $g$ is odd, so $g(\pi-w)=-g(w-\pi)=-g(w+\pi)$ due to the 2pi-periodicity.

so if $-\pi<w<0$, we indeed have $0<\pi+w<\pi$ and therefore $-g(w+\pi)=-(1)$.

Jessy Cat

So, $g(\pi - w) = -1$ in that case

Semiclassical

Right.

So you've shown that $g(\pi-w)=g(w)$ for $-\pi < w< \pi$.

The overall thrust of this is that $g(w)=g(\pi-w)$ for all $w$.

This isn't so hard to believe if you think of this graphically.

Jessy Cat

So, what does that mean in terms of the whole problem?

Semiclassical

Anyways. That means that $u(\pi,t)=\int_{-at}^{at}g(\pi-w)\,dw=\int_{-at}^{at}g(w)\,dw$. That's the same integral we had earlier, and it once again vanishes.

Jessy Cat

So, we're no further along than we were before

Semiclassical

19:05

So $u(\pi,t)$ is also identically zero, just as $u(0,t)$.

Jessy Cat

Oh, so the solution is periodic

Semiclassical

In $x$, yes.

Jessy Cat

What's interesting is the problem in the book that follows it.

Semiclassical

It should also be continuous in $x$ for all $t>0$, since we're ultimately computing areas of functions.

Jessy Cat

Suppose $u(x,t)$ satisfies $u_{tt}(x,t) = a^{2}u_{xx}(x,t)$, $-\infty < x < \infty$, $t > 0$

Semiclassical

19:07

(that's a bit of a poorly argued statement on my part, but eh. details.)

Jessy Cat

$u(x,0) = f(x)$, $- \infty < x < \infty$, $u_{t}(x,0) = g(x)$ $-\infty < x < \infty$

Show that If $f(0) = f(L) = 0$ and $f(x)$ is $2L-$periodic and if $g(x) = 0$ for all $x$, then $u(0,t) = u(L,t) = 0$ $\forall t$

Semiclassical

Huh.

Jessy Cat

From what you just said that seems to be what's going on here

Semiclassical

Yeah, except the boundary conditions are different.

Jessy Cat

except $g(x)$ is not identically zero

right

Semiclassical

19:09

Huh.

Jessy Cat

$f$ is

Semiclassical

I think a crucial point is that, if you look at $g(0)$

Jessy Cat

Wait a minute

Mary Star

We have that $\sigma=k_1k_2\ldots $, where $k_1, k_2, \ldots$ are disjoint cycles of $S_p4, where $p$ a prime.
Then we have the following: $$p=ord(\sigma)=lcm \ (ord(k_i)), \ i=1, 2, \ldots$$ Why does this imply that $ord(k_1)=p$ ?

Jessy Cat

The next problem after that is when $f$ is...hold on . Ther

there's a solution

If $g(x)$ is odd and periodic with period $2L$, then $u(0,t) = u(L,t) = 0$.

Semiclassical

19:11

...which is what I was concluding. So I'm evidently not crazy.

Jessy Cat

No. I've got the monopoly on that, I'm afraid.

But I'm freaking adorable, so it all evens out.

Semiclassical

That obviously doesn't solve your original problem, but it does set some strong conditions on it.

Jessy Cat

right.

Semiclassical

My (educated?) guess is that you'll end up with $u(x,t)$ a sawtooth wave in $x$.

Jessy Cat

Holy crap

Semiclassical

19:13

piecewise linear and periodic in $x$

Jessy Cat

A bunch of people just got shot in fort lauderdale at the airport

Semiclassical

fuck

Jessy Cat

nytimes.com/2017/01/06/us/fort-lauderdale-airport.html?_r=0

Semiclassical

no word on how many victims yet, other than 'multiple'

Jessy Cat

It's not even clear of those how many have died

Semiclassical

19:15

Heck of a coincidence to have Bush's former press secretary on site. (I say that as a genuine coincidence, not in a conspiracy-theory way)

Jessy Cat

He was probably just snowbirding it.

Figured he'd go to Florida to avoid the cold. I guess I have no idea where that guy lives.

Semiclassical

Sounds like he was at the airport when it happened.

Jessy Cat

He was.

He tweeted about it

Anyway, thanks for all your help.

I'm going to transcript this and get back to work.

A---B

Sad $$\Huge \ddot \frown$$

Jessy Cat

Try to prepare as best I can :( AT least he gave us a list of topics.

One of them was to derive D'Alembert's

TheGreatDuck

19:17

@SimpleArt Yeah. Apparently I'm a rep hog.

Jessy Cat

Probably won't have to do any problems like this, but I don't like coming across a problem I can't do.

I tend to get bogged down in the details and stuck like molasses for hours.

Studentmath

@GFauxPas This is wonderful

GFauxPas

What is?

what did I do

Studentmath

What you linked, I think yesterday

GFauxPas

what did I link I dont remember

Studentmath

19:24

"Some Commonly Used Terms"

GFauxPas

oh lol that was someone else I was just sharing it

Studentmath

Still it's great

Simple Art

@TheGreatDuck Lol, ok then

Semiclassical

19:54

@JessyCat I just realized that there's a much simpler way to handle that integral, so much so that I feel silly.

Namely, while it's a pain in the butt to work out on what intervals $z\in[x-at,x+at]$ will give $g(z)=\pm 1$, it's not at all hard to write down the antiderivative of $g(z)$ is.

Vrouvrou

Who knows Harnack inequality ?

Semiclassical

For $0<z<\pi$, it'll be $G(z)=\int_0^z g(z')dz'=\int_0^z dz'=z$. For $\pi<z<2\pi$ it'll be $G(z)=G(\pi)+\int_\pi^z(-1)dz'=\pi-(z-\pi)=2\pi-z$. After that it just repeats with period $2\pi$.

And then the solution is just $u(x,t)=\frac{1}{2a}\left(G(x+at)-G(x-at)\right)$. Simplifying that is a pain, but easy enough to plot

TheGreatDuck

@SimpleArt nah i mean it's a pretty serious issue.

Semiclassical

For $a=1$, I get this picture:

x-axis is [-pi,pi], t-axis is [0,2pi]. (u-axis is marked as [-2,2] but should have been [-1,1]; I plotted the wrong thing.)

(I don't know why it has those annoying missing lines.)

Akiva Weinberger

20:14

Pretty

Semiclassical

Yeah. I just don't like the missing lines.

And it does that for contour plots as well, which is even less attractive.

Jessy Cat

@Semiclassical dafuq u talkin bout willis

Semiclassical

lol.

Jessy Cat

seriously though i ust downed a shot of buffalo grass vodka in the weee afternoon

but lets see what trhis antiderivativey thingy yo udid was

ooh that is a 70s colored nightmare

Semiclassical

I have a graphical way of deriving the same result as above, but it's more tedious.

Jessy Cat

20:17

start chatjax

Semiclassical

...Was that an instruction to me, or to yourself? :)

Jessy Cat

to myself

Semiclassical

heh, okay.

Jessy Cat

but let's take a look here. okay, but what is G(z)?

Null

how can I check if an equationsystem with 8 variables and 4 equations has exactly one solution?

^arbitary numbers

Semiclassical

20:20

$G(z)=\int_0^z g(z')\,dz'$. If $0<z<\pi$, that's just $G(z)=z$.

If $\pi<z<2\pi$, then that's $G(z)=G(\pi)+\int_\pi^z (-1)\,dz'=2\pi-z$.

And from there on it repeats period $2\pi$.

WDUK

20:31

im really confused with using the quadratic formula im getting a different result for solutions of x compared to simply factoring my quadratic normally in that the formula is giving me the negatives for x

Semiclassical

Example?

WDUK

i have 2x^2 -5x +2 i factor that to be (2x-1)(x-2) so x is 1/2 or 2

using the QF i get -1/2 and -2

Semiclassical

are you plugging b=-5 into (-b +- sqrt(b^2-4ac))/(2a)?

WDUK

i was doing (-5 +- sqrt(9)) / 4

Semiclassical

ah. yeah, it should be +5.

WDUK

20:34

ohhh

silly me !

forgot it was -b d'oh!

Semiclassical

happens to the best of us.

WDUK

lets hope it doesn't during my exam xD

thank you

Semiclassical

np

Balarka Sen

@MikeMiller What can one say about the group $[M(G, n), X]$ of (pointed) maps from the Moore space to $X$? The group structure comes from realizing $M(G, n) \simeq \Sigma M(G, n-1)$ and doing the same "pinching equator" construction like for spheres (which gives a map $M(G, n) \to M(G, n) \vee M(G, n)$)

Mahmoud

20:49

Hi.

Okay .. Dead chat.

Semiclassical

@Mahmoud Consider the second angle of that green right triangle.

on the one hand, it's complementary to the angle at the top (they belong to the same right triangle). but that same angle is also complementary to the angle at the bottom, since they're part of the same right angle.

Mahmoud

@Semiclassical Thank you $\phi=\frac \pi 2 - \theta$

Where $\phi$ is the second angle,

Semiclassical

So therefore both the angle at the top and the angle at the bottom are complementary to $\phi$. But if two angles are complementary to a third angle, they're equal.

or, more simply: if $\phi=\pi/2-\theta$ and also $\phi=\pi/2-\theta'$, then $\theta'=\theta$.

If you label the angle at the top as $\theta'$, that's exactly the situation you have above.

Mahmoud

@Semiclassical Thanks, I don't know how did I miss that.

Semiclassical

@AkivaWeinberger Finally got a nice contour plot:

Mahmoud

20:57

My trigonometry class is a nightmare.

Semiclassical

Mahmoud

All the identities ..

Semiclassical

It's easy to get lost in them, yeah.

To a certain extent it requires experience to see which ones are useful in a given problem.

For instance, in proving trig identities the sum-to-product identities tend to be more useful than the product-to-sum identities.

On the other hand, when doing certain integrals in Fourier analysis, the product-to-sum identities are quite handy.

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$Mathematics$ Mathematics