Mathematics - 2016-02-24 (page 1 of 3)

PVAL

00:06

I seem to have given a rather simple proof for a really nice sounding statement.

I wonder what I did wrong.

Ted Shifrin

LOL, I've been there lots of times, @PVAL.

PVAL

If this is correct this must be folklore.

I am almost sure it isn't in the literature and I'm sure someone has come up with before.

Ted Shifrin

If it were folklore, wouldn't you have known it?

PVAL

I think some people (for instance my advisor) write down a very small percentage of the things they have proven.

Ted Shifrin

There have been some mathematicians who're notorious for that, yes.

You should ask him in this instance, I guess. The trouble with posting a question on MO is that you can end up scooped if in fact it isn't written down anywhere or even folklore.

OK, I'm out of here for now. Let me know what happens, @PVAL.

PVAL

00:14

I like Arnold's philosophy on this. Either tell no one what you are doing or tell every single person you talk to everything you've done so no one can dispute it (he said he abided by the latter.)

Jeff

00:57

Hi folks. May I ask what technique to use to integrate $\int \sqrt{2+2 \sin t} dt$?

Or should I ask what trig identity do you use?

Akiva Weinberger

Perhaps let $x=\sin t$? @Jeff

Then ${\rm d}x$ would be $\cos t~{\rm d}t$, so we'd have ${\rm d}t=\frac1{\cos t}~{\rm d}x=\frac1{\sqrt{1-x^2}}~{\rm d}x$.

MickLH

Ooh that one is surprisingly nasty, it breaks my program! Thanks for the new test-case

Jeff

01:13

@Akiva Working on it now. But I'm having trouble getting it into a form I can integrate.

@MickLH What program is that?

MickLH

It's an analysis engine

Jeff

@AkivaWeinberger I keep getting zero. Because (I didn't mention) the bounds go from $t=0$ to $t=2\pi$. Then $x(0)=0$ and $x(2\pi)=0$ and the integral becomes zero (but my TI-84 says it's 8).

MickLH

I'm gonna go with the TI-84's answer there :P

Intuitively, I can find at least one point where your integrand is greater than zero, and so assuming you're using the positive square root it can't be zero.

Jeff

Well, that's what I keep getting. Using Akiva's subst. and factoring inside the radical I get $$\int \sqrt{\frac{2}{1-x}} dx.$$ Then do another substitution $u=1-x$ and I get the answer $-\sqrt{8} \sqrt{1-\sin t}$. Evaluate that from $t=0$ to $t=2\pi$ and I get 0. What'd I do wrong?

MickLH

What happens when $t = \frac{3\pi}{2}$ ?

Jeff

01:29

the answer evaluates to 0

i don't see your point

MickLH

I don't have a magical solution, I'm just going off my gut with where I'd start looking for weirdness

Jeff

Oh. Well in the graph of the integrand, there is sharp point at x=4.7124. But I'm not sure why that would affect the integration.

Here's what the integrand looks like from 0 to 2pi: desmos.com/calculator/el1vfmlaux

MickLH

So clearly, not zero :P

Ted Shifrin

01:55

@Jeff: If I wrote the problem as $\int \sqrt{1+\cos 2\theta}d\theta$, would you have an idea? Indeed.

Jeff

I could use a trig identity on that.

@Ted

Ted Shifrin

So remember you can turn sin into cos and vice versa by using complementary angles.

Jeff

But there are not trig identities I can find for 1+sin(t)

Ted Shifrin

Aha ^^^

It is a sneaky problem, but it shows up in real life, believe it or not.

Jeff

well, of course it shows up in real life! how do you think they discovered the homework problems in the texts? :D.... so you're saying use $\sqrt{2+2\cos\left(\frac{\pi}{2}-t\right)}$?

Ted Shifrin

01:58

um, I meant more real than that :P

Yes, I'm saying once you figure out the cos one, you do a $\pi/2-\theta$ substitution :)

Jeff

and then use the cosine sum identity?

Ted Shifrin

Yes, @Jeff ... you're on it.

PVAL

@TedShifrin My advisor actually has a construcion that implies the "simple" thing I proved was in some sense optimal.

Ted Shifrin

And the 2's in there will work out so that you don't have any $\sqrt2$ to deal with.

Oh, cool, @PVAL. So is it folklore or not?

BTW, I hope you say hi to him for me sometime :)

Jeff

After using $\pi/2-\theta$, use the power reduction formula? that seems a bit unwieldy (not that there's anything wrong with that)

PVAL

02:00

I don't know I haven't talked to him yet, I just figured out that something that he did proved that you couldn't do any better than what I got. (His construction had very different motivations).

Ted Shifrin

no, no, $1+\cos 2\theta = 2 \cos^2\theta$, @Jeff

ohhhh, cool, @PVAL

Jeff

@Ted Yeah, isn't that a power reduction formula (in reverse)? where $2\theta = \pi/2-t$.

Ted Shifrin

no, it's a double-angle formula. Power reduction is all with the same $\theta$, usually.

Jeff

@Ted OK, I got the names wrong (usually I get more than that wrong :D). But in this case $\theta$ is rather cumbersome, $\theta = \pi/4-t/2$

Ted Shifrin

Do the problem with $\cos$ in there, and then convert at the end (with the appropriate chain rule correction).

BTW, adorable kitties, @Jeff.

Jeff

02:06

@Ted Thanks. I lost all my pics of them, and the brown one died recently. That's all I have left of her :D

Ted Shifrin

Aww, I've lost a few, myself. My heart goes out to you.

BTW, @Jeff, with regard to the $0$ discussion up there ... Be very careful when you integrate square roots of squares. :) True/false: $\sqrt{x^2} = x$.

Jeff

@Ted false

Ted Shifrin

So, if you're going to do an integral, you need to write ... $\sqrt{x^2} = $?

Jeff

$|x|$

Ted Shifrin

Righto. The $0$ mistake comes from omitting the absolute values.

Jeff

02:10

Of course, I never lost the square root symbol the way I was trying it.

PVAL

My students always had trouble writing $| |$ as a piecewise function.

Jeff

so that shouldn't have been a concern

Ted Shifrin

well, now you know the right way to do it, @Jeff :)

Jeff

@PVAL mine, too. even when I show them. they are like "huh"

Ted Shifrin

yup, @PVAL ...

but a lot of your students also use $\sqrt{a^2+b^2}=a+b$. :P

Jeff

02:12

@Ted Well, I see how to do it. I'm not actually finishing it. It's too much effort for the time I have available. Thanks!

Ted Shifrin

LOL, sure.

It's really not hard, but ok.

Jeff

@TedShifrin @Ted So do a lot of mine. I even get some that thing (a+b)^2=a^2+b^2.

@Ted OK. I will try it. But over the weekend.

Ted Shifrin

I didn't get that so often, interestingly.

Akiva Weinberger

It's cause they think that the distributive property, $a(b+c)=ab+ac$, is a property of parentheses rather than of multiplication. I think.

Ted Shifrin

OK, @Jeff. I'm not here to give orders (except to @Balarka).

Jeff

02:14

I had several students use the product rule on the derivative of $\tan(7x^2)$.

Akiva Weinberger

'Cause the multiplication is hidden.

Ted Shifrin

Whoa. What? DogAteMy

Very interesting comment, DogAteMy ...

Akiva Weinberger

Which is why they'd write $(a+b)^2$ as $a^2+b^2$…

Ted Shifrin

How do you connect those?

Akiva Weinberger

'Cause "can't you always do that with parentheses?"

Jeff

02:15

And then I have students who think the derivative of $\ln(\sin x)$ is $1/x (\cos x)$

Akiva Weinberger

@TedShifrin I didn't come up with this. Did you know there's a Mathematics Educators Stack Exchange? :P

Ted Shifrin

Well, let's not spend the rest of the night talking about how students in high school and college don't understand what a function is.

CommonerG

Where do you guys teach?

Ted Shifrin

Yes, DogAteMy ... I've been on there, but I don't go there frequently. I was once a math educator, you know? :D

PVAL

indoors

CommonerG

02:16

:)

Ted Shifrin

LOL @Pval

Jeff

I teach at several community colleges in NJ. (they're desperate for cheap, part-time math teachers)

Ted Shifrin

@Commoner: I've just retired from 34 years at UGA.

CommonerG

Ted, I'm from Georgia, did undergrad at GT

Akiva Weinberger

Oh, I thought you were still teaching?

CommonerG

02:17

Congratulations on retirement.

(If it is appropriate)

Ted Shifrin

nope, DogAteMy ... only unofficially (teaching a 2-person topology class long distance) and volunteering in high schools ... and here.

Sure, thanks, @Commoner.

Akiva Weinberger

So… I can call you Dr. Shifrin but not Prof. Shifrin?

Heh, is Balarka in that "topology long-distance class"?

PVAL

@CommonerG Did you ever take a course from Etynre?

CommonerG

I was an economics major.

PVAL

Oh i see

CommonerG

02:20

Anyway, looks like he started after I graduated.

MickLH

@Jeff I took a minute to attack your problem and I just want to encourage you. Following @TedShifrin's suggestion led me to a nice way to show that the area is 8.

Ted Shifrin

no, no, not Balarka. He's more advanced in topology than I've ever been (except for certain aspects). Two former UGA students.

Jeff

@MickLH Yeah, I'm on it now. But I have to type it up. I get too messy and lose my thought writing anything too long.

Ted Shifrin

DogAteMy, I'm fine with Ted. Even most of the high school kids I'm working with call me that. ... I've never insisted on fancy titles.

@PVAL: One of my former students finished her Ph.D. with John.

MickLH

@Jeff If it's any help to you, I did have to treat that point at $t = \frac{3\pi}{2}$ with care while evaluating the anti-derivative I obtained

CommonerG

02:23

He (Etnyre) has an impressive CV.

Ted Shifrin

@Jeff: I'm always in favor of writing stuff on paper and then typing in LaTeX. Unless you get really good (as I did after several textbooks authored).

@Commoner: Yeah, he's good!

Jeff

i'm pretty good at typing and latexing. but when i write, it gets too messy. of course, when i'm helping students i have to write. but then i (usually) know the problem in advance

MickLH

I got two different values when evaluating the limit from each side, which was a crucial part of my strategy. Without treating that point specially, I found that my anti-derivative was periodic!

Jeff

@Ted I just looked you up. You're a geometry specialist?

At georgia?

Ted Shifrin

I used to be, @Jeff. I just retired.

Jeff

02:25

Oh. congrats.

Ted Shifrin

I still love teaching (or I wouldn't hang around here), but I decided it was time to move on to a different life.

Jeff

i have a tutoring student who liked geometry and hates precalc. he asked if i could relate it to geometry. i failed. i couldn't.

what do you do now? what books did you author?

Ted Shifrin

I try to draw pictures for everything, including abstract algebra. Certain things in precalc can be related, for sure, but not others (exponentials?). Books on abstract algebra, linear algebra, multivariable calculus and linear algebra mixed (more proof-oriented), and differential geometry.

Jeff

@Ted @MickLH Well, I got $-4 \sin \left(\frac{\pi}{4}-\frac t2\right)$. But that doesn't seem right. Since it's derivative is not anything as complicated as the question. But I will point out I violated the $\sqrt{x^2}=|x|$ rule.

@Ted Oh. I'm talking about a HS student.

Ted Shifrin

I'm currently volunteering with high school precalculus classes at a charter school, @Jeff. If he means proofs in geometry, not so much, but we can bring lots of geometric problems (similar triangles, trig) into precalculus, for sure.

MickLH

02:30

What did you "get" that for?

Jeff

For the integral I was trying: $$\int \sqrt{2 + 2 \sin t} dt = -4 \sin \left(\frac{\pi}{4}-\frac t2\right)$$

I didn't plug in the bounds yet. I like to do indefinite integrals before I do the definite one (call me strange)

Ted Shifrin

No, you're right, @Jeff, but be careful with the absolute value issues.

Jeff

@Ted I'm actually not sure how to address that. Do I just add the absolute values in this step: $$-2 \int \sqrt{4\cos^2 d\theta} d \theta = -2 \int 2\cos \theta d\theta$$

PVAL

@Jeff There's actually many definite integrals cannot be done this way.

Jeff

Then do the integral piecewise?

Ted Shifrin

02:34

Yes, $\sqrt{\cos^2\theta} = |\cos\theta|$.

Or, use symmetry, and reduce the problem to some multiple of what happens where things stay positive.

OK, I have dinner guests coming over. I need to skedaddle. Talk to y'all later!

Jeff

I'm still not getting quite the right answer. It must be the absolute value. Am I at least getting the right answer (not counting the ab val)?

MickLH

@Jeff I have obtained a much different result

(I needed a radical to write my anti-derivative)

Jeff

@mick Did you start with this step $$\int \sqrt{2 + 2 \sin t} dt =\int \sqrt{2 + 2 \cos \left(\frac{\pi}{2}-t\right)} dt$$

MickLH

Very similar

Jeff

Then next is let $2\theta = \frac{\pi}{2}-t$, so you get $$\int \sqrt{2 + 2 \cos \left(\frac{\pi}{2}-t\right)} dt = -2 \int \sqrt{2 + 2 \cos 2\theta} d\theta$$

Then factor out the 2 to get $$-2 \int \sqrt{2\left(1 + \cos 2\theta\right)} d\theta = -2 \int \sqrt{2\left(2\cos^2 \theta\right)} d\theta = -4 \int \cos \theta d\theta$$ (not counting the absolute value issue which I'll come back to later)

@PVAL What way?

PVAL

02:42

@Jeff By first finding the indefinite integral.

There are many definite integrals with a closed form for which the corresponding indefinite integral has none.

Jeff

Really? I like to do that to check myself. After I get the answer, I can check on WolframAlpha.

Such as? @PVAL

PVAL

$\int_{-\infty}^{\infty}e^{-x^2}dx$

Jeff

That's the normal distribution (almost)

OK. I'll come back to this tomorrow maybe. I have to get to work writing lesson plans now. Thanks, all.

MickLH

@Jeff I'm skeptical of this step

Jeff

@MickLH which one?

MickLH

02:48

The one I aimed at, -4 cos ...

Well I guess you said "not counting the absolute value issue"

But I think you're torturing yourself by leaving that factor of $\sqrt{2}$ in there

This is the very first thing I did: $$\int_{0}^{t}{\sqrt{2\,\sin x+2}\;dx}=\sqrt{2}\,\int_{-{{\pi}\over{2
}}}^{t-{{\pi}\over{2}}}{\sqrt{\cos x+1}\;dx}$$

God, that machine-generated latex is ugly

Jeff

@Mick i have to think about that. I've not seen this technique of putting variables in the bounds.

@MickLH Yeah, you can't even see negative signs on some of them.

MickLH

I wrote it that way to show the thought process I used, secretly I just did indefinite integration

Jeff

@MickLH I don't get that step at all. I don't even see the reason for writing the integral as $\int_0^t \sqrt{2\sin x + 2} dx$

I'm starting from the top again. I'm starting with $$\int \sqrt{2 + 2 \sin t} dt$$

OK. I see now (sorry for so many messages). You switched the angle, then changed to cosine.

MickLH

Yeah, it's just $\sin(t) = \cos(t - \frac{\pi}{2})$

Jeff

That's an unusual way of doing it (well, for me it is).

MickLH

02:57

also I used $\sqrt{a\cdot b} = \sqrt{a}\cdot\sqrt{b}$

Jeff

That part I figured out :D

MickLH

You didn't remove the $\sqrt{2}$ you copied it instead

$2+2\sin(t) = 2\,(1+\sin(t))$

Jeff

You're right. But the part I'm worried about is the bounds. $$\int \sqrt{2 + 2 \sin t} dt = \sqrt 2 \int_0^t \sqrt{1 + \sin x} dx. $$ That gives you $\sqrt{1 + \sin t} - \sqrt 2$.

MickLH

I don't see it, how do you mean?

Jeff

When you integrate the right side, you get some function $F(x)$. Then you evaluate it at $t$ and subtract from that evaluate it at 0, or $F(t)-F(0)$.

It doesn't equal the function on the left hand side.

MickLH

03:06

Ah ha! Guess what?

Jeff

what?

MickLH

The issue you've illustrated became apparent in the next step, the way I did it.

Jeff

oh.

MickLH

lets simplify that just to draw some conclusions so far

with $f(t) = 1+\sin(t)$ we can more easily see that $\sqrt{2\cdot f(t)} = \sqrt{2}\cdot\sqrt{f(t)}$

Jeff

OK. Let's try it without that fancy change of variables technique that I haven't seen before (and am now dubious of). We agree on this (obvious) first step $$\int \sqrt{2 + 2 \sin t} dt = \sqrt 2 \int \sqrt{1 + \sin t} dt.$$

@MickLH Agreed. Same thing I just typed, too.

MickLH

03:09

Also we know that $\sin(t) = \cos(t - \frac{\pi}{2})$, so we can substitute that in

Jeff

So we get $$= \sqrt 2 \int \sqrt{1 + \cos \left(\frac{\pi}{2}-t\right)} dt$$

MickLH

At this point, I used a known result: $$\int {\sqrt{\cos x+1}}{\;dx}={{2\,\sin x}\over{\sqrt{\cos x+1}}}$$

Jeff

How do we get that? by rationlizing the numerator?

MickLH

Hmm lets put deriving / proving that off until later

I checked that result and it holds

Jeff

OK. But I think it might be easier to let $2\theta = \frac{\pi}{2}-t$ in the prev. step and apply the trig id we were discussing earlier.

@MickLH But let's continue this way first

MickLH

03:14

It very well may be... I didn't find any optimal route, just one of them

Ok so we substituted $x = t - \frac{\pi}{2}$ earlier, lets come back to that

Jeff

OK, so this is what I have now $$= \sqrt 2 \int \frac{2\sin \left(\frac{\pi}{2}-t\right)}{\sqrt{1 + \cos \left(\frac{\pi}{2}-t\right)}} dt$$

MickLH

too much! notice the result above has only integration on one side

Jeff

OK, but instead, let $x = \frac{\pi}{2}-t$.

Oh.

OK, so now we have $$= \sqrt 2 \frac{2\sin \left(\frac{\pi}{2}-t\right)}{\sqrt{1 + \cos \left(\frac{\pi}{2}-t\right)}}$$

MickLH

Ok, so now the roadblock issue is visible

Jeff

oh?

I think we should have substitute $x=\frac{\pi}{2}-t$ before using that result. no?

MickLH

03:17

We already did

You have a rational function there, which is not defined in some circumstances

Jeff

@MickLH Oh, so you have the result directly as the one i am pointing to now

MickLH

Yes, sorry if I wasn't clear

That's just a result I've obtained at another time

Jeff

@MickLH You're going to fast for me. Also, I think you have the subtraction backwards in the complementary angle formula.

(Sorry)

MickLH

It doesn't matter

$\cos(x) = \cos(-x)$

Jeff

Ok, right

MickLH

03:23

@Jeff Is this where we are at? $${{\sqrt{2}\,2\sin \left(t-{{\pi}\over{2}}\right)}\over{\sqrt{\cos \left(t-{{\pi}\over{2}}\right)+1}}}$$

Jeff

Yes, that's where i'm at, too. OK, so the problem is where the denominator is 0.

MickLH

exactly

Jeff

or where cosine is -1, which is where $t-\pi/2 = -\pi$

plus or minus $2\pi$.

MickLH

Right, so we encounter this undefined value once every $2\pi$

Which means that without even checking, since we are integrating over at least $2\pi$ units of $x$, we already know we will hit it

Jeff

I think we hit it at $t=3\pi/2$.

MickLH

03:30

That is what I got also: $\sqrt{\cos \left({{{\pi}}\over{2}}-{{3\,\pi}\over{2}}\right)+1} = 0$

So, this screws over our approach of obtaining the area by straightforward evaluation of the indefinite integral

Jeff

actually wait. i think we get zero when $t-\pi/2=-\pi+k\pi$, right?

MickLH

Well I'm sure there are at least countably infinite zeros, but we only really care how many are in our bounds of integration

Jeff

so that $t = -\pi/2+k\pi = \pi/2, 3\pi/2$ within our range

if i did that right isn't it both $\pi/2$ and $3\pi/2$?

but $\pi/2$ does not give zero in bottom.

MickLH

Let me investigate before I say anything

$$\sqrt{\cos \left(x-{{\pi}\over{2}}\right)+1}=0$$
$$\cos \left(x-{{\pi}\over{2}}\right)+1=0$$
$$\cos \left(x-{{\pi}\over{2}}\right)=1$$
$$\sin x=1$$
$$\sin \left(2\,\pi\mathbb{Z}+{{\pi}\over{2}}\right)=1$$
@Jeff

Jeff

negative 1 on RHS

MickLH

03:40

oops lol, well mirror the relevant parts

This is a tangent, I'm getting claustrophobic in this little tangent here

Jeff

BTW, I did mention that this is from a polar coordinates problem. Maybe that makes it easier.

MickLH

the points of interest are generated by $t = 2\pi\mathbb{Z} - \frac{\pi}{2}$

Jeff

I think you overcomplicated this part. the points are when $\cos(t-\pi/2)=-1$ (just looking at the bottom. We know that cosine is -1 when the inside is $-\pi$ plus multiples of $2\pi$ (aha! there's my mistake -- it's *two*$\pi$....

therefore it's when $t-\pi/2=-pi$ which means $t=-\pi/2$, then the only point in our interval is $3\pi/2$, which is the answer we knew we should have gotten. OK.

MickLH

So now you see why my gut wrenched about $\frac{3\pi}{2}$ initially?

Jeff

If I'm correct, that means we can evaluate the integral results $\frac{\sqrt 2 2 \sin\left(t-\pi/2\right)}{\sqrt{\cos\left(t-\pi/2\right)+1}}$ from 0 to $3\pi/2$ and then again from $3\pi/2$ to $2\pi$. Is that right?

MickLH

03:50

Yes that's where I'm headed with this, though we have to be very careful doing the evaluation

Jeff

What else do we have to be careful of?

MickLH

$\mathrm{F}(\frac{3\pi}{2})$ is not defined, and the two-sided limit does not exist either

Jeff

Limit? we already did the integral (it was your result). what limit is left?

MickLH

We need to use a limit because we need to actually reach the undefined point, to have correct integration

Jeff

oh yeah. in improper integreal (duh)

MickLH

03:54

since $\mathrm{F}(\frac{3\pi}{2})$ is not defined, we have to evaluate both $\lim_{t\uparrow {{3\,\pi}\over{2}}}{F\left(t\right)}$ and $\lim_{t\downarrow {{3\,\pi}\over{2}}}{F\left(t\right)}$

Jeff

sigh sadly I don't have time to finish this. sorry. but i am still paying attention while you do (I realize that may not be too exciting to you). i just have a ton of work to get done

MickLH

@Jeff Well I've already evaluated them which is why I'm sure that your TI-84 has the correct answer

Jeff

OK (I never really doubted it). You basically took the limit of $F(t)$ as $t \to 3\pi/2^-$ minus $F(0)$ and added $F(2\pi)$ minus the limit as $t \to 3\pi/2^+$. Right?

how did you do those limits? i'm seeing an infinity at first glance.

MickLH

Yes that's what I did, if I'm reading the + - notation correctly

Jeff

that's debatable, since it doesn't print well. the notation $t \to 3\pi/2^-$ means $t$ approaches $3\pi/2$ from the negative side (or left side).

and $t \to 3\pi/2^+$ means approach the value from the positive side.

MickLH

04:02

yes that's what I did

Jeff

@MickLH I've never seen your notation before. I like it better.

How did you do that limit? because I'm seeing 0/0 at first glance. did you rationalize the denominator?

MickLH

My notation probably makes someone out there cringe lol

Jeff

i like yours better. it's easier to read and understand

MickLH

I cheated on the limit and used another known result lol

Jeff

how would you do that limit had you not known that?

MickLH

04:14

Well it's not comprehensive, but I re-wrote the point of interest as $$\sqrt{8}\,\left(1-{{\sin \omega}\over{\sqrt{1-\cos \omega}}}\right)$$

MickLH

04:25

Bleh, I don't want to derive it. Computer algebra confirms the result, and I'm not sure you're even here anymore anyways @Jeff $$\lim_{\omega\rightarrow 0}{\frac{\sin \omega}{\sqrt{1-\cos \omega}}} = \pm\sqrt{2}$$

Jeff

I'm paying attention. but i understand if your done for the night. thanks very much for your help.

@Mick

MickLH

Well I'll need to get a glass of water but I'll be around for at least a couple more hours

Jeff

@mick Thanks for helping. :D

MickLH

np! hope it gets you somewhere

Jeff

i learned something from it. mostly an integration technique. but also a reminder of improper integrals (which I'm probably going to have to teach this summer, anyway).

MickLH

04:57

@Jeff I couldn't resist poking at it again... the limit can be evaluated just using the fact that $\cos(2x) = 2\cos^2(x)-1$

Jeff

how did you use that?

MickLH

I massaged the expression in a way that I ended up with the part that goes to zero on the top and bottom of the ratio

Then I just let them cancel each other, and evaluated what was left as usual

MickLH

05:15

Forgive the latex bombing please:
$$\mathrm{limit}={{\sin {\omega}}\over{\sqrt{1-\cos {\omega}}}}$$
$$\mathrm{limit}^2={{\cos \left(2\,{\omega}\right)-1}\over{2\,\left(\cos {\omega}-1\right)}}$$
$$\mathrm{limit}^2={{\cos \left(2\,{\omega}\right)-1}\over{2\,\left(2\,\cos ^2\left({{{\omega}}\over{2}}\right)-2\right)}}$$
$$\mathrm{limit}^2={{\cos \left(2\,{\omega}\right)-1}\over{4\,\left(\cos \left({{{\omega}}\over{2}}\right)-1\right)\,\left(\cos \left({{{\omega}}\over{2}}\right)+1\right)}}$$
$$\mathrm{limit}^2={{2\,\left(2\,\cos ^2\left({{{\omega}}\over{2}}\right)-1\right)^2-2}\over{4\,\left(…

GGG

Define $f: \mathbb{Z} \to \mathbb{Z}$ via $f(n) = n^2$ for all $n \in \mathbb{Z}$. Why does $f^{-1}(\left\{0,1,2\right\}) = \left\{0,-1,1\right\}$? The definition I'm using is $f^{-1}{(T)} = \left\{a \in A: f(a) \in T \right\}$ so we have $f^{-1}({ \left\{0,1,2\right\} }) = \left\{n \in \mathbb{Z}: n^2 \in \left\{0,1,2\right\} \right\}$. So I could think about this as all the integers whose squares is the set $ \left\{0,1,2\right\}$ but nothing I square would give me $2$.

L33ter

06:00

how can I determine that there must be 30 distinct polynomial over F_2 of degree 8.

Eric Stucky

irreducible, presumably?

L33ter

yeah

@EricStucky

I got there can be no more than 2^6 = 64

Eric Stucky

I think there is a general formula which you might try to prove

let me see if I can find it

Yeah here it is: it's $\frac1n\sum_{d|n} \mu(d) q^{n/d}$ for irreducibles of degree $n$ over a field with $2$ elements

If you specialize to $n=8$ then the sum has only two terms: $d=1$ and $d=2$, so it's probably easier to do that than the general case, actually.

Rajesh Dachiraju

Can you comment on this answer?

0

A: A problem on real valued functions in $\mathbb{R}^2$ with least variation

I roughly sketch a solution without a proof, and also a possible minimum possible value for total variation of $f$. Let $f$ assume values of $J$ on the boundary curve $\alpha$. We construct infinite number of scaled down versions of the curve $\alpha$ with scale factor ranging from $1$ and conve...

I want to know if its correct or any mistakes

L33ter

I have an idea how many ways can I choose 4 things from element of 8 ? well, it is (8,4)

Because irreducible polynomials must be of form x^8 + a_7x^7 + ... + a_1x + a0

polynomial is irreducible when it has a linear root

reducible

I meant

since we are in binary those x^n doesn't do anything

hm 1 sec

L33ter

06:55

@EricStucky here?

MichaelMitchell

07:18

hey

im having a little trouble understanding why in Z[x]

well

how can you show that the ideal (2, x) cannot be generated by one element?

Tobias Kildetoft

07:55

@MichaelMitchell That element would have to divide both $2$ and $x$

So I noticed a paper today on arXiv that looked interesting. When I was about to put it in the correct folder based on the authors I realized that one author is called Clinton Boys. That has got to cause some problems with google searches and such.

MichaelMitchell

@TobiasKildetoft Thanks!

@TobiasKildetoft So if $d | 2$ and $d | x$, would that mean $d$ is either 1 or 2?

Tobias Kildetoft

@MichaelMitchell if it divides $2$ then it is $\pm 1$ or $\pm 2$.

But $2$ does not divide $x$.

MichaelMitchell

why can we say that?

Tobias Kildetoft

@MichaelMitchell well, we know this is true for integers, so we just need to rule out non-constant polynomials. But there it is clear by considering the degree

(if $f$ divides $g$ then $f$ has at most as large degree as $g$).

MichaelMitchell

I don't understand, in this case isn't 2 degree 0 and $x$ degree 1? So 2 should divide $x$?

Tobias Kildetoft

08:07

@MichaelMitchell that was a necessary, but not sufficient criterion

MichaelMitchell

:(

sharaf zaman

check this link math.stackexchange.com/questions/1598719/… and see the last comment of answer

anyone has seen??

Tobias Kildetoft

@sharafzaman seen what?

sharaf zaman

@TobiasKildetoft the link i sent see the last comment under answer

Tobias Kildetoft

@sharafzaman what about it?

sharaf zaman

08:16

@TobiasKildetoft limits

Tobias Kildetoft

@sharafzaman Please ask a full question

sharaf zaman

@TobiasKildetoft Assuming you have a finite sum, we can have sum of limits equal to limit of sums if each of the limits in the expression converges.

then what happens for infinite sum

@TobiasKildetoft ??

Tobias Kildetoft

@sharafzaman Pretty much anything could happen, depending on the sum

sharaf zaman

@TobiasKildetoft did you see the answer and question?

Tobias Kildetoft

@sharafzaman briefly. Why?

sharaf zaman

08:30

in the answer it is given"Because the number of terms goes up exactly as the size of each term goes down." does this mean if we sum up all the small $\frac{1}{n}$ this will tend to 1

Tobias Kildetoft

@sharafzaman Depends on how many you sum. Please just ask whatever question you have precisely.

Josué Molina

What topological property do metric spaces have that makes sequential compactness and compactness equivalent?

Is it that they are Hausdorff?

sharaf zaman

@TobiasKildetoft i am having exactly the same question as given in the link math.stackexchange.com/questions/1598719/…

Tobias Kildetoft

@sharafzaman and in what way does the answer not help?

sharaf zaman

@TobiasKildetoft i didn't understand answer you please explain that!

Tobias Kildetoft

08:34

@sharafzaman which part of the answer?

sharaf zaman

@TobiasKildetoft you explain the whole answer!

Tobias Kildetoft

@sharafzaman No, the answer looks fine to me, so you need to explain what part you are having trouble with

sharaf zaman

@TobiasKildetoft "Because the number of terms goes up exactly as the size of each term goes down" this part

Tobias Kildetoft

@sharafzaman Just ignore that part. It is specified what he means on the line just below it

sharaf zaman

@TobiasKildetoft the next line he has just given the general representation that doesn't explain my answer

Tobias Kildetoft

08:39

@sharafzaman The next line is quite clear to me, so again you need to explain your problem with it

sharaf zaman

@TobiasKildetoft i think i got it by reading the comments again and again

GGG

09:02

Hello!

Suppose we have a pair of functions $f: A \to B$ and $g: B \to A$ such that $g \circ f = 1_{A}$. I've read that $f$ has a left inverse if and only if $f$ is injective. I've proven the $ \Rightarrow$ direction. How do I prove the $\Leftarrow$ direction?

Tobias Kildetoft

@GGG To make a left inverse, you just need to pick what each element in the image is sent to and then send the rest wherever

Huy

@DanielFischer: "if eigenfunctions (of the Laplacian) are our aim, we first need to show compactness of the inverse". here, is compactness of the inverse likely meant to be that the image of any bounded set under $\Delta^{-1}$ is relatively compact? or rather $(\Delta - \lambda)^{-1}$ in this context?

Daniel Fischer

09:20

@Huy I don't know.

GGG

09:36

@TobiasKildetoft how would you write that? Suppose $f(a) = f(a') \implies a = a'$ for all $a' \in A$. Then by definition of identity mapping, we have the map $1_{A}: A \to A$. Also if we have a function defined by $f: A \to B$ then $f \circ 1_{A} = f = 1_{B} \circ f$. Therefore there exists $h: A \to B$ such that $h \circ f = I_{A}$.

For all $a,a' \in A$ I meant.

Tobias Kildetoft

@GGG You have not actually written what I said. You need to construct that left inverse

GGG

@TobiasKildetoft Oh, I see what you mean now I think. I'll try to construct it.

Huy

10:04

@DanielFischer: no problem. something else I'm wondering: the Sobolev space $H^1$ is often defined as the completion of $C_c^\infty$ with respect to the Sobolev norm. what happens if I instead consider the completion of $C^\infty$ with respect to the same norm? will I get something bigger or end up with the same space?

Daniel Fischer

10:19

@Huy The Sobolev norm isn't finite on $C^{\infty}$ in general. If $\Omega$ is a bounded region, then one can look at the completions of $C_c^{\infty}(\Omega)$ and of $C^{\infty}(\overline{\Omega})$ with respect to that norm. The first is often denoted by $H_0^1(\Omega)$ and the second by $H^1(\Omega)$. These are different spaces, $H_0^1 \subsetneqq H^1$.

Regardless of what $\Omega$ is, the completion of $C_c^{\infty}(\Omega)$ with respect to the Sobolev norm is often called $H_0^1(\Omega)$, and $H^1(\Omega)$ is often defined in a different way, as the space of square-integrable functions whose first distributional derivatives are given by square-integrable functions, for example. Then typically you have $H^1_0(\Omega) \subsetneqq H^1(\Omega)$, but $H^1_0(\mathbb{R}^n) = H^1(\mathbb{R}^n)$.

But notations for Sobolev spaces are a terrible mess, everybody uses different conventions.

(and different definitions)

GGG

@TobiasKildetoft Suppose $f$ is injective. Define $g(a) = a'$ if there exists $a' \in A$ such that $f(a') = a$, in which case we have $g \circ f(a) = g(f(a)) = g(a) = a'= 1_{A}.$ If not, suppose that $\alpha \in A$. We say that $g(a) = \alpha$, in which case $g \circ f = \alpha \circ f$. However, I can't get this equal to the identity.

Tobias Kildetoft

@GGG when considering $g\circ f$ it doesn't matter what happened to those things no in the image of $f$ (as whatever you evaluate the composite function on, it will involve evaluating $g$ at something in the image of $f$)

GGG

10:36

Ah, I understand. Many thanks @TobiasKildetoft

Jessy Cat

0

Q: Find and sketch the image of the straight line $z = (1+ia)t+ib$ under the map $w=e^{z}$

I need to find and sketch the image of the straight line $z = (1+ia)t +aib$, where $-\infty < t < + \infty$, $a,b\in \mathbb{R}$, and $a \neq 0$, under the map $w = e^{z}$. In order to accomplish this thus far, I have substituted the expression for $z$ into the expresison for $w$ to yield $...

I am so very, very confused :(

Please take a look at my question and see if you can help!

Jessy Cat

10:52

Maybe it's a bad time of morning to post a question on Stack Exchange.

So far, that question has only had 3 views :(

GGG

@JessyCat I've noticed that when it's late in North America there's less activity in the site.

Jessy Cat

@GGG, yeah, it's awful.

Hopefully it gets some attention.

Anyway, I've got to get ready to go to school.

Huy

11:47

@DanielFischer Thanks for your answer, and I agree about the terrible mess.

GGG

12:21

Below I'm trying to prove that $f$ has a right inverse if and only if $f$ is surjective.

Suppose $f$ has a right inverse $g$. To show that $f$ is surjective, we need to show that for all $y \in B$ there is $x \in A$ such that $f(x) = y$. If $y \in B$ we have $y = f(g(y)) $ since $f \circ g = 1_{A}$. Therefore $y$ is of the form $f(x)$ for $x = g(y).$ Hence $f$ is surjective.

To prove the converse, suppose that $f$ is surjective. Let $\alpha \in B$. Since $f$ is surjective, there exists at least one element $x$ such that $f(x) = \alpha$. Choose the minimum of these and call it $x_0$. Define $g(\alpha) = x_0$. Then we have $f \circ g = f(g(\alpha)) = f(x_0) = \alpha = 1_{A}.$

Tobias Kildetoft

@GGG What do you mean by the minimum? Have you put some well-order on the set?

GGG

@TobiasKildetoft You're right. I should have said "choose one of these".

Tobias Kildetoft

@GGG It amounts to the same thing (applying axiom of choice in some form).

GGG

@TobiasKildetoft How do I fix this?

Tobias Kildetoft

@GGG You can't (i.e. you will need to invoke choice to do this)

GGG

12:32

That's surprising to me!

Tobias Kildetoft

@GGG Well, you might be able to get away with a bit less (though as far as I recall, it is still open how much less, if at all)

Agawa001

13:15

@robjohn how long hav u not seen chris so far ?

skull petrol

aka the artist formerly known as...

robjohn

@Agawa001 I haven't seen her here for over a week.

@Agawa001 But she's been on the main site an hour ago.

skull petrol

How about anon?

Mithlesh Upadhayay

13:31

Hello everyone, would you like to check this shortest path question.

0

Q: Computing shortest path including specific edge

Consider the weighted undirected graph with $4$ vertices, where the weight of edge $\{i, j\}$ is given by the entry $W_{i, j}$ in the matrix $W$. $$W = \begin{bmatrix} 0&2&8&5\\ 2&0&5&8\\ 8&5&0&x\\ 5&8&x&0\\ \end{bmatrix} $$ The largest possible integer value of $x$, for which at least one short...

algebra-precalculus discrete-mathematics graph-theory arithmetic trees

robjohn

@skullpetrol Not since Dec 29.

GGG

Problem: Prove that if $g \circ f$ is injective then $f$ is injective. Why is the proof below wrong?

Proof: Let $g(x) = b$. Then $g(f(x)) = b$ and by injectivity $f(x) = b$ which implies $x = b$.

robjohn

13:47

@GGG why does $g(x)=b$ imply that $g(f(x))=b$?

what you want to show is that if $f(x)=f(y)$, then $x=y$.

GGG

13:58

@robjohn what I haven't quite grasped is how to infer $x = y$ from $g(f(x)) = g(f(y)) \implies f(x) = f(y)$

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