Mathematics - 2017-05-02 (page 3 of 7)

Little Rookie

16:22

0

Q: Prove Absolute error for quadrature rule

Given the Legendre Polynomial $L_n(x)$ of degree $n$, and a quadracture rule for approximating $\int_{0}^{1} f(x) dx$ using $n$ points, Prove that the absolute error for applying the quadrature rule to $f(x)=\sin x$ is not larger than $\frac{1}{(2n+1)!}$ I totally have no idea how to deal with ...

Semiclassical

Probably you would start by writing out the quadrature rule for that integral. (Without that in your question, you've shown no work to the reader.) @LittleRookie

Astyx

Thansk @Steamy

Little Rookie

i wrote it out on paper, but doesnt give me direction on solving the problem

Astyx

And hi chat

Ted Shifrin

@Semiclassic: Answered.

Hi @Astyx, @Little

Goofing off well, @Astyx?

Abcd

16:27

How to do this question:

Little Rookie

Hello

Semiclassical

mmkay.

Little Rookie

Can you help me with the question please?

Astyx

Better than I should probably :p @Ted

Little Rookie

the term $\frac{1}{(2n+1)!}$ reminds me of taylor series

especially so since $x\in [0,1]$

Astyx

16:29

How are you ?

Ted Shifrin

Still a little sick, but bumbling along. About to disappear, it seems, as I have corrected pages to check out for an hour. Talk later!

Little Rookie

ahh chatjax is not working for me >.<

Abcd

Astyx

Catch ya later

Abcd

How do I do this question?

Astyx

16:31

@LittleRookie see Steamy's message on the starboard

This one

Semiclassical

@Abcd That seems pretty gross at first glance.

Abcd

@Semiclassical I didn't get you. Is it too easy?

Leaky Nun

@Abcd hint: $(1+\sqrt3)^2 = 4+2\sqrt3$

well y naught

@abcd: Can you solve for $y$ given $x$ and $xy?$

Semiclassical

No, I mean entirely the opposite.

That doesn't mean there isn't a nice solution, but that at first glance it looks unpleasant.

I mean, it's not hard to convince yourself that $y=2-\sqrt{3}$. (Multiply that by $x=2+\sqrt{3}$ to be sure.)

Abcd

16:34

@LeakyNun I tried that ... didn't work

Leaky Nun

@Abcd that should work

Semiclassical

I guess the other thing to do is to rationalize the fractions they give you.

Leaky Nun

$\sqrt x = \sqrt{2+\sqrt3} = \sqrt{\dfrac12(4+2\sqrt3)} = \dfrac1{\sqrt2}(1+\sqrt3)$

Abcd

@LeakyNun We have 2 + root 3 not 1 + root 3

SteamyRoot

Would it be easy to show that the expression is $<1 $ ?

Leaky Nun

16:34

@Abcd look at the right hand side

Abcd

@LeakyNun Is your LHS wrong?

Leaky Nun

@Abcd no it isn't

Anonymous

Find square root of $2 \pm \sqrt{3}$ @Abcd

Anonymous

You're done.

Abcd

@blue Thats impossible

Leaky Nun

16:35

@Abcd I just found it.

Little Rookie

Someone save me please :(
https://math.stackexchange.com/questions/2262385/prove-absolute-error-for-quadrature-rule

Anonymous

@Abcd -_-

Anonymous

Did you even try?

Leaky Nun

Why does everybody seem to be ignoring my solution?

Semiclassical

It's not impossible, even for a person.

Abcd

16:36

@blue I dont know how to do it.

Leaky Nun

@Abcd I just showed you.

3 mins ago, by Leaky Nun

@Abcd hint: $(1+\sqrt3)^2 = 4+2\sqrt3$

2 mins ago, by Leaky Nun

$\sqrt x = \sqrt{2+\sqrt3} = \sqrt{\dfrac12(4+2\sqrt3)} = \dfrac1{\sqrt2}(1+\sqrt3)$

which step do you not understand?

Semiclassical

Here's how I'd proceed.

Abcd

@LeakyNun How do I continue?

@LeakyNun I didnt understand why you took 1 + root 3 whole square

Semiclassical

First, it's a pain in the butt to have those sqrt()'s in the denominators. So we can rationalize by multiplying top/bottom with appropriate expressions.

Leaky Nun

@Abcd $(1+\sqrt3)^2 = 4+2\sqrt3 = 2x$

Semiclassical

16:38

If you don't have the patience to help people, then don't. @LeakyNun

well y naught

For real.

Leaky Nun

@Semiclassical sorry

Abcd

@LeakyNun still! x = 2 + root 3 ... I am sorry :(

Semiclassical

We were all algebra newbies at one point. :)

Leaky Nun

@Abcd what is 2x?

Abcd

16:39

@LeakyNun 4 + 2 root 3

Leaky Nun

@Abcd can you verify that $(1+\sqrt3)^2 = 4+2\sqrt3$?

Abcd

Yes.

Understood.

Semiclassical

That gives $$\frac{x}{\sqrt{2}-\sqrt{x}}+\frac{y}{\sqrt{2}-\sqrt{y}}=\frac{x(\sqrt{2}+\sqrt‌{x} )}{2-x}+\frac{y(\sqrt{2}+\sqrt{y})}{2-y}$$

Leaky Nun

@Abcd so can you find $\sqrt x$ now?

Abcd

:37112820 in LHS root 2 + root x

Semiclassical

16:40

What's nice in this expression is that $x=2+\sqrt{3}$ and (as I noted above) $y=2-\sqrt{3}$. So $2-x=-\sqrt{3}$ and $2-y=\sqrt{3}$.

Anonymous

Or one could solve $(x+y)^2=2+\sqrt{3}$ for $x$ and $y$ where $y$ is irrational and $x$ is rational

Abcd

@LeakyNun Yes.

Leaky Nun

@Abcd can you find $\sqrt y$ then?

Semiclassical

@blue if one recognizes that it should be of the form $a+b\sqrt{3}$, sure.

Rosie F

@blue Or even $(x+y\sqrt3)^2=2+\sqrt3$.

Anonymous

16:42

Yup

Mike Miller

@s.harp oh, those spaces disagree as coarse spaces, at least

Semiclassical

Bah, my expression earlier had a typo. Should've been sqrt(2)-sqrt(x) for the first term on the left-hand side and sqrt(2)+sqrt(x) on the right-hand side.

But I think writing $x,y$ in terms of 1,sqrt(3) is the smarter approach regardless. So I"ll leave it to the others.

Little Rookie

@Semiclassical can u help me with the question? :(

Abcd

I got confused because of too many solutions

Leaky Nun

@Abcd lmao

Semiclassical

16:44

Not really surprising, given how many of us were talking at once :P

I'm withdrawing my attempt, though. I don't think it really helps.

Abcd

yes.

Can someone help me with that question?

IP Address

quick question

Leaky Nun

@Abcd can you find $\sqrt y$?

Abcd

@LeakyNun is it 1/ root x

Leaky Nun

yes

but can you find its value?

Abcd

16:50

root 3 - 2?

IP Address

If an tanA is given and is said A is in Quadrant 1. Does it imply A is in quadrant 1 for sinA?

Abcd

y = 2 - root 3

Semiclassical

I don't know what "A is in quadrant 1 for sin A" could possibly mean.

Leaky Nun

@Abcd I'm talking about $\sqrt y$, not $y$

Can you rationalize $\dfrac1{\dfrac1{\sqrt2}(1+\sqrt3)}$?

Abcd

Just a minute

@LeakyNun Can you please resend your earlier steps.

Semiclassical

16:53

you can scroll up in the transcript.

Leaky Nun

20 mins ago, by Leaky Nun

@Abcd hint: $(1+\sqrt3)^2 = 4+2\sqrt3$

19 mins ago, by Leaky Nun

$\sqrt x = \sqrt{2+\sqrt3} = \sqrt{\dfrac12(4+2\sqrt3)} = \dfrac1{\sqrt2}(1+\sqrt3)$

Semiclassical

and the full transcript can be found here: chat.stackexchange.com/transcript/36

Astyx

Big brother is watching you

Abcd

@LeakyNun Is it: root3 root 2 - root 2 / 2

Leaky Nun

I don't think so

Little Rookie

16:57

anyone can spare some time to help me? https://math.stackexchange.com/questions/2262385/prove-absolute-error-for-quadrature-rule

please :(

Abcd

@LeakyNun Do you have a simpler approach to this question?

Semiclassical

Once you've gotten $\sqrt{x}=\frac{1+\sqrt{3}}{\sqrt{2}}$ the problem is not -so- bad.

But that's not something I'd easily be able to spot on an exam.

Leaky Nun

@Abcd $\dfrac1{\dfrac1{\sqrt2}(1+\sqrt3)} = \dfrac{\sqrt2}{1+\sqrt3} = \dfrac{\sqrt2(\sqrt3-1)}{(\sqrt3-1)(\sqrt3+1)} = \dfrac{\sqrt6-\sqrt2}2$

so you were close

IP Address

If A for tanA is in Quadrant 1. Can it be assumed A for sinA is in quadrant 1?

^ Quick answer pls

Leaky Nun

@IPAddress the question doesn't make sense to me

Semiclassical

17:00

As I said above, I have no idea what "A for sin A is in quadrant 1" means.

Leaky Nun

@IPAddress do you understand what a quadrant is?

Semiclassical

A is not "for sin A" or "for cos A." A is the angle, full stop.

IP Address

yes

Abcd

@LeakyNun Thats exactly what I had answered

Leaky Nun

@Abcd oh... you forgot a parenthesis

Abcd

17:01

Well, I only got confused whether root 2 root 3 is root 6 or not

Leaky Nun

(root3 root 2 - root 2) / 2

IP Address

y = tanA

Abcd

@LeakyNun I thought you would understand

IP Address

draw a graph

Leaky Nun

@Abcd sorry

Abcd

17:02

ohkay.

Next step?

IP Address

if I pick a value for A in quadrant 1

Semiclassical

y = x tan A, I think you mean.

IP Address

Nevermind! I've got the answer to my question anyway

Semiclassical

oook.

IP Address

Its really complicated to share problems on here... because u cant share photos...

crap...

just noticed the upload button

hahahahaha

Abcd

17:04

@IPAddress Use snipping tool

Semiclassical

I'm actually forgetting: How would one deduce $\sqrt{2+\sqrt{3}}=a\sqrt{2}+b\sqrt{6}$ if you didn't know it already?

Leaky Nun

@Semiclassical $\sqrt{2+\sqrt3} \in \Bbb Q[\sqrt2,\sqrt3]$ so your basis should have 4 elements

so it should be $a+b\sqrt2+c\sqrt3+d\sqrt6$

Semiclassical

Hmm. Sounds right.

Balarka Sen

How do you argue it's in Q(sqrt(2), sqrt(3)) though?

Semiclassical

hmm, yeah.

Leaky Nun

17:07

@BalarkaSen oh right...

Abcd

It's a very tough question. I surrender.

Ted Shifrin

Well, easy Galois theory will tell you when you can write $\sqrt{a+\sqrt b} = \sqrt c+\sqrt d$. See p. 472-473 of my algebra book. :P

Leaky Nun

@Abcd sorry I'm typing

Semiclassical

pfft.

Abcd

ohkay

Leaky Nun

17:08

@BalarkaSen one can find its minimal polynomial

Little Rookie

in need help here! >.< math.stackexchange.com/questions/2262385/…

Astyx

Doesn't fully expanding ${x\over\sqrt{2}+ \sqrt{x} }+{y\over\sqrt{2}- \sqrt{y} }$ work ?

Leaky Nun

@Abcd $$\frac{x}{\sqrt{2}-\sqrt{x}}+\frac{y}{\sqrt{2}-\sqrt{y}} = \frac{2+\sqrt3}{\sqrt{2}+\sqrt{0.5}+\sqrt{1.5}}+\frac{2-\sqrt3}{\sqrt{2}- \sqrt{1.5}+\sqrt{0.5}}$$

Semiclassical

Isn't the second denominator supposed to be sqrt(2)+sqrt(y) ?

Balarka Sen

Something like that. (x^2 - 2)^2 - 3 = 0. I think symmetry playing with this should tell it's in a degree 4 extension which has Galois group Z/2 x Z/2

Ted Shifrin

17:09

@Leaky: No. In general, the $c$ and the $d$ have nothing to do with the $a$ and the $b$. For example, $\sqrt{23+4\sqrt{15}} = \sqrt 3 + \sqrt{20}$.

@Balarka: Yeah, it's all about Galois group $\Bbb Z_2\times\Bbb Z_2$.

Leaky Nun

@TedShifrin in gerenal, $b$ are related to $c$ and $d$

Ted Shifrin

Well, of course, they're related. But, look at my example.

Semiclassical

@Ted Seems like that should have something to do with 15 = 3*5?

Leaky Nun

@TedShifrin $\sqrt3+2\sqrt5$

Ted Shifrin

The result is that you can do this precisely when $\sqrt{a^2-b}\in\Bbb Q$.

Leaky Nun

17:11

and $3 \times 5 = 15$

@TedShifrin intersting

Semiclassical

I guess you have 23+4sqrt(15) = c+d+2sqrt(cd)

Ted Shifrin

Things really depend on $\sqrt{a^2-b}$ for the answer.

Semiclassical

so you'd need 16*15=4cd -> cd=60 and c+d=23.

which has c=3, d=20 as a solution.

@ted this should boil down to the rational root theorem, shouldn't it?

Ted Shifrin

Of course, if we're going to write $\sqrt{a+\sqrt b} = \sqrt c+ \sqrt d$, then the Galois theory tells us that $\sqrt{a-\sqrt b} = \sqrt c - \sqrt d$ (with appropriate ordering), and then it's just high school algebra to solve.

Semiclassical

Fair.

But I stand by my initial reaction of "gross" to the multiple choice problem.

Ted Shifrin

17:16

I came in late, so I didn't see any such thing.

Semiclassical

45 mins ago, by Abcd

Ted Shifrin

Oh. Let me look.

Balarka Sen

@TedShifrin I can't do this explicitly, but sqrt(a+sqrt(b)) has min poly (x^2 - a)^2 - b = 0. This has Galois group Z/2 x Z/2 by working the symmetries out; the root belongs to a degree 4 extension of Q which contains sqrt(a), so it has to be in Q(sqrt(a), sqrt(k)) (because Galois group) for some k. That means sqrt(a + sqrt(b)) is a sum of a bunch of squareroots, but I can't reduce to 2 squareroots.

Semiclassical

I mean, it's easy enough to see that y=1/x=2-sqrt(3)

Ted Shifrin

Right.

@Balarka: I'll get back to you later.

Semiclassical

17:19

And you can rationalize the denominators that without much trouble, especially since 2-x=-sqrt(3) and 2-y=sqrt(3).

Balarka Sen

Ok, sure. I forgot a lot of the Galois theory I know

Ted Shifrin

The point is that $x,y$ are related by $\pm\sqrt3$, so I think we can do this by symmetry considerations.

Semiclassical

But I didn't see anything nice past that.

Yeah, should.

Ted Shifrin

This sorts of things (sqrt of sum with sqrt) show up in a lot of high school competitions, and the students are very fast at doing the $\sqrt c + \sqrt d$ thing (without knowing any of the fancy theory), so I'm used to seeing things like this in competition questions.

Semiclassical

That's fair.

This is the kind of thing I don't do very often so my skills are a lot weaker.

Ted Shifrin

17:23

(Note that, in terms of the theorem I announced earlier, in this case $\sqrt{a^2-b} = 1$, so things are pretty easy.)

Astyx

The algebra is strong with this one

3

Ted Shifrin

@Balarka: If it were a sum of more than two square roots, the Galois group would have more $\Bbb Z_2$ factors.

Anonymous

I learnt that whenever $a^2-b=r^2$ is a perfect square there is this handy shorcut formula $\sqrt{a \pm \sqrt{b}} = \sqrt{(a+r)/2} \pm \sqrt{(a-r)/2}$

Ted Shifrin

Right. That's the high school algebra to which I referred, @blue. I didn't realize you guys memorized such things. I sure never would.

But if you do 5 problems like this, you just know it.

Semiclassical

dang.

Ted Shifrin

17:26

If you guys insist, I'll type the whole proof here at some point.

Anonymous

@TedShifrin It helps a lot in time bound tests =P

Ted Shifrin

Sure, @blue. I agree that one shouldn't re-invent the wheel repeatedly in timed situations.

Balarka Sen

@TedShifrin Hmm? sqrt(2)+sqrt(3)+sqrt(6) is a pretty valid element of Q(sqrt(2), sqrt(3))

Semiclassical

The sqrt(a^2-b) thing seems to boil down to when $x^2-2a x-b=0$ has rational roots.

Ted Shifrin

@Balarka: But the original thing has $\Bbb Z_2\times \Bbb Z_2$ symmetry and this has more (two $\pm$'s). Interesting.

Semiclassical

17:28

which if you solve for $x=a\pm \sqrt{a^2-b}$ seems pretty straightforward.

Ted Shifrin

Oh, no, I see. $\sqrt6$ inherits its sign from those on $\sqrt2$ and $\sqrt3$.

Balarka Sen

sqrt(2)+sqrt(3)+sqrt(2)*sqrt(3) actually has +/- symmetry

Abcd

I am getting this in the numerator:

Balarka Sen

Correct.

Abcd

x root 2 - x/root x + root x/ x + root 2/ x

Balarka Sen

17:29

So I don't know how to argue it's sum of exactly 2 square roots.

Ted Shifrin

I guess the point is that if our element is $\sqrt c+\sqrt d$, then the Galois action has to be what I said earlier. I'm not arguing sufficiency.

Semiclassical

bah, x^2-2ax+b=0.

grumble

Leaky Nun

22 mins ago, by Leaky Nun

@Abcd $$\frac{x}{\sqrt{2}-\sqrt{x}}+\frac{y}{\sqrt{2}-\sqrt{y}} = \frac{2+\sqrt3}{\sqrt{2}+\sqrt{0.5}+\sqrt{1.5}}+\frac{2-\sqrt3}{\sqrt{2}- \sqrt{1.5}+\sqrt{0.5}}$$

Ted Shifrin

The key point is, as I said earlier, that if that holds, then $\sqrt{a^2-b}=\sqrt c-\sqrt d$ has to hold.

Leaky Nun

@Abcd did you see this?

Abcd

17:31

@LeakyNun yes. and I couldnt solve it further. It seems to be too complex :(

Balarka Sen

@TedShifrin I see. So you're not saying it's true $\sqrt{a+ \sqrt{b}}$ is always of the form $\sqrt{c} + \sqrt{d}$.

Ted Shifrin

Of course not. That's true $\iff$ \sqrt{a^2-b}\in\Bbb Q$.

Balarka Sen

Ah, that is believable.

Ted Shifrin

That's the theorem I announced when I came in ...

Balarka Sen

You didn't mention that here, but you said that in later.

Ted Shifrin

17:35

growls at @Balarka — time to un-sleep.

Anonymous

Ah! There's a whole wiki page dedicated to this en.wikipedia.org/wiki/Nested_radical

Anonymous

Nested radical

In algebra, a nested radical is a radical expression (one containing a square root sign, cube root sign, etc.) that contains (nests) another radical expression. Examples include 5 − 2 5 , {\displaystyle {\sqrt {5-2{\sqrt {5}}\ }},} which arises in discussing the regular pentagon, and more complicated ones such as 2 +...

Semiclassical

Which boils down to the fact that $x^2-2ax+b=(x-a)^2+b-a^2=0$ has rational roots only if the third coefficient is rational.

Ted Shifrin

There you go, @Semiclassic.

Semiclassical

Yeah, I had figured that much out just from the quadratic formula.

Ted Shifrin

17:36

And of course I've been assuming $\sqrt b\notin\Bbb Q$ all along.

Semiclassical

But I was trying to see if it was obvious from the rational root theorem directly

jeeze, those ramanujan identities on the wiki page

Ted Shifrin

My interest starts to wear out quickly on this stuff.

Alessandro Codenotti

Hi chat

Ted Shifrin

Hi @Alessandro

Semiclassical

agreed.

user193319

17:39

Hello. I am trying to determine the components and path components of $\mathbb{R}_\ell$, the real numbers endowed with the lower limit topology. I know that $\mathbb{R}_\ell$ is totally disconnected, so that if $C$ is a component of $\mathbb{R}_\ell$, then it must be connected and therefore a singleton. Now, if $C$ is a path component, then it must be path connected and therefore also connected, which implies it, too, is a singleton. Does this sound right?

Alessandro Codenotti

Yes

Ted Shifrin

Yeah, path components are no larger than connected components :)

Little Rookie

Can i get any hints for the following question https://math.stackexchange.com/questions/2262385/prove-absolute-error-for-quadrature-rule
please

user193319

Thank you both!

Ted Shifrin

@Little: We don't have numerical analysis experts here. I might be able to figure out what you're asking if I had a textbook in front of me, but I don't.

Alessandro Codenotti

17:42

You should have seen a (rather ugly) estimate of the error for Newton-Cotes formulas in class, right?

Do Green's identities have an intuitive explanation or should I just memorize them?

Little Rookie

No

We didnt cover error for newtwon cotes formula in lecture

Ted Shifrin

@Alessandro: Green's identities in vector calculus? To what do you refer?

Alessandro Codenotti

Oh, no, wait, you're not doing newton cotes here, just interpolation. Did you cover the error done with polynomial interpolation?

Little Rookie

yes

i tried to use the error form of hermite interpolation to solve the problem

but cant get to anywhere

Alessandro Codenotti

Those @Ted (in particular the first and the third, I don't think we ever used or talked about the second)

Ted Shifrin

17:45

Right. They're all just immediate from Stokes's or the divergence theorem. I personally don't memorize them. I just know to play the game. (They work nicely with differential forms, too, of course.)

Hi DogAteMy

Akiva Weinberger

$\sup$

Ted Shifrin

Sometimes you're so $\inf$uriating. :P

Mike Miller

$\infy$uriating

Secret

$\int$ ($\infty$ uriating) uriating

Akiva Weinberger

OK I get it

Secret

17:50

(Lol, I just like to explode any meme I saw)

Alessandro Codenotti

@TedShifrin hm, I guess I need to understand better the divergence and Stokes theorems then since we're using those identities everywhere in the PDE course

thanks

Ted Shifrin

Yes, they're very important in PDE. Let me know if I need to help. @Alessandro

Mike Miller

the point is that you can often integrate against a relevant function and get something positive plus a boundary term, say

Semiclassical

Shows up in physics a lot too, not surprisingly.

Alessandro Codenotti

@TedShifrin I'll almost surely need some help sooner or later, we saw a few results that feel like black magic to me. Right now I'm looking at some theorems about harmonic functions which are really neat

Semiclassical

17:56

Though with physics I guess I associate green's identities with reciprocity stuff.

Ted Shifrin

Yeah, it's all just $\int_M d\omega = \int_{\partial M}\omega$ :D

Alessandro Codenotti

@Semiclassical The PDE course is the continuation of the Lagrangian mechanics course, which is the most physics heavy course in the undergraduate curriculum here

Semiclassical

Sounds right.

Do you talk about the...oh, damn, why can't I remember the name of it.

It's the PDE formulation of Newton's laws. Something-Jacobi equation

Alessandro Codenotti

maybe? :P

Hamilton-Jacobi?

Semiclassical

Bah, yes.

Alessandro Codenotti

17:58

Those came up in Hamiltonian mechanics, we just had a quick introduction after the Lagrangian formulation

Semiclassical

Fair enough.

Hamilton-Jacobi doesn't get a lot of coverage in physics courses.

Transcript for

$Mathematics$ Mathematics