Mathematics - 2012-06-19 (page 1 of 2)

Peter Tamaroff

00:11

@Eugene I have one from Burton here

Little Fermat Again

$$(a,35)=1 \Rightarrow a^{12} \equiv 1 \mod 35$$

I have this so far

$$(a,35)=1 \Rightarrow (a,7)=1, (a,5)=1$$

By the LFT

$$a^6 \equiv 1 \mod 7$$
$$a^4 \equiv 1 \mod 5$$

Now I guess I can use this one

$$(m,n)=1, a \equiv b \mod n, a\equiv b \mod m$$ then $$a \equiv b \mod mn$$

...somehow.

Eugene

00:34

this doesn't work. you need to consider $a^{12}$ instead but you're close.

Peter Tamaroff

@Eugene OK.

Oh, I got it.

$$\eqalign{
& {a^{12}} \equiv {1^2} \equiv 1\bmod 7 \cr
& {a^{12}} \equiv {1^3} \equiv 1\bmod 5 \cr} $$

Done.

Eugene

yup

Peter Tamaroff

@Eugene Yay!

Eugene

@PeterTamaroff 9K user shouldn't be asking questions like these!

lol

Peter Tamaroff

@Eugene Hahaha I will get better with practice!

@Eugene Now I'm solving

$(a,42)=1 \Rightarrow a^6 \equiv 1 \mod 168$

Eugene

00:43

@PeterTamaroff i'm not worthy!

Peter Tamaroff

@Eugene Commmon! Modular forms guy! Strap up!

Eugene

@PeterTamaroff lol.

sorry i'm watching batman and doing my homework right now so i'm a little occupied

Peter Tamaroff

@Eugene Hahha its OK dude, I'm kidding.

@Bill Hello, Bill.

Eugene

but the idea is the same as the last problem

Peter Tamaroff

@Eugene I pressume so

Eugene

00:49

$168 = 2^3 \cdot 3 \cdot 7$

Bill Dubuque

@Peter Just poked in for a minute due to your query.

Peter Tamaroff

@BillDubuque Yes, thank you..

I wrote this, well, actually copied something from you:

$$\rm f_n=f_{n-m}+k f_m \rightarrow (f_m,f_n)=f_{(m,n)}$$

So define

$$\rm \widetilde n = \frac{{{x^n} - 1}}{{x - 1}}$$

Then since $$\rm \widetilde n = {x^r}\widetilde m + \widetilde {m - n}$$

One has

$$\rm \left( {\widetilde n,\widetilde m} \right) = \widetilde {\left( {m,n} \right)}$$

That is correct right?

Bill Dubuque

@PeterTamaroff Yes, that's what's in my proof here.

Peter Tamaroff

@BillDubuque OK. And then $a^{(m,n)}-1=(a^m-1,a^n-1)$ is just another simple case of that.

Bill Dubuque

@PeterTamaroff Yes, note that I added the more general answer later.

Peter Tamaroff

00:57

@BillDubuque Yes! I enjoy generalizations quite a lot.

@BillDubuque That's why I really enjoyed reading the first easy parts of Truesdell's Essay.

@BillDubuque Do you know the book Elementary Number Theory by Burton? I'm doing some excersices from it, from the Chapter on the Fermat Theorem (Euler's Theorem for primes)

What is the important use of $a^{p-1} \equiv 1 \mod p$?

The one that seems tough is $p^{q-1}+q^{p-1}\equiv 1 \mod pq$, but I guess you have a slick result to prove it :P

Bill Dubuque

@PeterTamaroff I've never used Burton's book, so I can't comment on it. The Euler-Fermat theorem may be viewed as a special case of Lagrange's theorem. That's a prior question that I answered.

Peter Tamaroff

@BillDubuque Oh, I see.

One considers $1+x+x^2+\cdots+x^n \equiv 0 \mod p$?

What polynomial does the job?

Bill Dubuque

@Peter Think about it a bit using CRT. I'm off now, good luck.

Peter Tamaroff

@BillDubuque Oh, OK. Same to you. Thanks, again.

Bill Dubuque

@PeterTamaroff See my answer here.

user19161

01:27

It seems many of the reference requests are duplicates.

user19161

For example there is a complex analysis book request now and the answers are just a rehash of the ones on the older question.

Peter Tamaroff

01:52

@JasperLoy Yep, seen that, been there.

Eugene

free points though.

Steven Li

02:14

hi

Peter Tamaroff

@Eugene Is it normal to see numbers in colours?

Or relate them to colours, more precisely.

leo

@PeterTamaroff it sounds like this

Peter Tamaroff

@leo I don't know if it happens to you, but I relate the letters like this $\color{blue}{n}$,$\color{yellow}{a}$,$\color{orange}{b}$,$\color{purple}{m}$

$\color{red}{r}$

leo

it don't happen to me

Peter Tamaroff

@leo It only happens with those, though. Not that is something common in general.

Bill Dubuque

02:24

For some mathematical synesthesia see John Conway's The Sensual (Quadratic) Form

Steven Li

hello?

leo

@StevenLi HI??!!

Steven Li

hi

leo

@PeterTamaroff I see. What a rare thing!

Peter Tamaroff

@BillDubuque Nice. "Sinestesia" sometimes helps a lot with problems, like to not mess up letters between each other.

Steven Li

02:26

can i ask math questions here?

leo

I used to confuse the $5$ with the $7$

Peter Tamaroff

@StevenLi It depends on how long they are. But shoot.

leo

but that's other thing

Peter Tamaroff

@leo How so?

Eugene

@PeterTamaroff well different people have different ways of remembering

leo

02:28

@PeterTamaroff don't know. And I hate when I have to write something that involves $5$ or $6$. I try to avoid them

Peter Tamaroff

@Eugene What's yours, Euge?

Eugene

@PeterTamaroff i don't really know either.

i just do

Peter Tamaroff

Please don't tag, I have my earbuds on.

@Eugene [Puts on glasses.] Walks away from explosion scene, unscathed.

Bill Dubuque

I'm glad that I cannot smell or taste many of the proofs on MSE!

Eugene

@PeterTamaroff oh sorry i didn't know

Peter Tamaroff

02:29

@BillDubuque Hahahahaha.

@Eugene Really, dude come on!!!

Eugene

ok ok

Steven Li

is the intersection of all maximal ideal of a ring the ring itself

Eugene

no

trivial example

let R be a field

Eugene

03:03

@PeterTamaroff have a go at this

Benjamin Lim

@Eugene Next sem:

multivariable analysis

Eugene

i don't like analysis multivariable or not.

Benjamin Lim

amazon.com/…

based on this

Eugene

yuck

Peter Tamaroff

03:18

@Eugene It is a very messy integral, but I'm writing something.

Eugene

@PeterTamaroff it has had a lot of answers since then. maybe not worth it any longer.

Peter Tamaroff

@Eugene Yeah.

Benjamin Lim

@Eugene what's wrong?

Eugene

anything that isn't rudin is ugly

Benjamin Lim

that's such a biased view

Peter Tamaroff

03:23

@Eugene You'll convince me one day of this, its either Rudin or Apostol. (And Rey Pastor)

Eugene

@BenjaminLim it's not

@PeterTamaroff screw apostol. the only good book he wrote is on modular forms IMO

i posted a question on the main site just now.

Peter Tamaroff

@Eugene I won't bother to read it <|:^) <|:^o

Eugene

@PeterTamaroff why not?

Peter Tamaroff

@Eugene I don't get a word of it.

Eugene

huh

anon

03:31

I just realized something and derived a question from it. Even when $U\le G$ is not normal, the coset space $G/U$ is a $G$-set, and given $U\le V\le G$ there is a well-defined canonical quotient $G$-map $G/V\to G/U$ (mapping a coset of $V$ to the coset of $U$ containing it). Thus the quotients of $G$ form an inverse system in the category of $G$-sets. Furthermore all $G$-sets are disjoint unions of orbits, orbits are transitive $G$-sets, and all transitive $G$-sets are isomorphic

(in our category) to a quotient $G/U$ of $G$. This begs the question: what is the multiplicity of a given $G/U$ in the decomposition of the inverse limit?

I guess we might want to instead say the quotients $G/U$ over all conjugacy classes of subgroups $U\le G$. Or if we wanted to be really fancy we could go over the subroups of $n$th powers $\{g^n:g\in G\}$...

Peter Tamaroff

@Eugene Vinogradov showed that if $A(x)$ is the number of even integers $n < x$ that are not the sum of two primes, then $lim A(x)/x = 0 $ This allows us to say that "almost all" even integers satisfy the conjecture. As Edmund Landau so aptly put it, "The Goldbach conjecture is false for at most $0%$ of all even integers; this at most $0%$ does not exclude, of course, the possibility that there are infinitely many exceptions."

BTW What's the difference between Holland and an "$i$"?

The $i$ has a point.

anon

I could have swore Eugene asked Dubuque about the motivation of 3-descent in chat before.

Eugene

@anon his was a "maybe it's this..." answer though.

anon

03:47

Ah good, I was right. Just wanted to check my memory. I don't actually know what 3-descent or the Sha group are...

Eugene

@PeterTamaroff just stick to doing anal and skip the jokes.

@anon don't worry nobody really knows either

Peter Tamaroff

@Eugene Hahahaha, big Holland fan? What's the deal here?

Eugene

@PeterTamaroff yes i am

Peter Tamaroff

@Eugene Don't succumb to passions.

Eugene

@anon why don't you ask on the main site?

anon

03:49

I am thinking about asking it on the mainsite.

I want to finish the previous question first though if possible, that Jyrki partially answered.

Eugene

i see

anon

I need to see that (1) GL(n,F) acts transitively on rank n-1 matrices and (2) rank <n matrices decompose as a product of rank n-1 matrices.

robjohn

05:54

@PeterTamaroff: I am sorry if I come across as heavy handed on your answer. I am just trying to point out a few errors. Even if you fix up that $F(\varphi)\color{red}{-}F''(\varphi)=0$, and the sign of $F''(\varphi)$, I really haven't seen any argument that leads me to think that $\mathrm{PV}\int_{-\infty}^\infty\cos(\varphi x)\,\mathrm{d}x=0$.

anon

seems like it could work out to $\delta(\varphi)$ in a distributional sense. though I have no idea what problem/answer you're talking about.

Eugene

mattE to the rescue again

i'm beginning to wonder if i can ask a question he CAN'T answer.

robjohn

@anon this answer

Dylan Moreland

@Eugene I haven't found such a thing yet.

Eugene

@DylanMoreland neither have i! =)

robjohn

06:04

@anon I just used contour integration :-)

Eugene

@robjohn do you like baseball? ever watched moneyball?

Gustavo Bandeira

06:19

58. What number is expressed in octal by m consecutive ones?

What method can I use to answer such a question?

anon

The method of understanding what base-8 means.

In decimal it represents (1)8^8+(1)8^7+(1)8^6+....+(1)8^1+(1)8^0.

derp, ignore deleted comment. geometric sum formula says this is 8^9-1 divided by 8-1.

(I was mixing it up with binary.)

Gigili

07:08

Willie Wong

@henningmakholm : Henning, do you mind taking a look at this Meta discussion and tell us your thoughts? meta.math.stackexchange.com/a/4000/1543 Thanks.

(I mean, especially in regards to the Recursion tag)

David Wallace

08:19

@RagibZaman Since you didn't reply to my earlier question - I find this remark very offensive. It's insulting to everyone who has ever been genuinely cyber-bullied, as well as insulting to Gigili. I know it's a few days back now, and flagging it isn't going to achieve anything; but maybe next time, you could think more carefully before you go making ridiculous allegations such as this.

experimentX

09:30

HI .. anyone?? here

I was wondering if Laplacian operates both on vector and scalar

anon

There is a Laplacian operator that acts on scalar fields, as well as a vector Laplacian that acts on vector fields.

experimentX

hi anon

thank you for confirming!!

$user image$

is this different from $ \nabla \cdot \nabla $ ??

anon

you mean is $\nabla^2$ different? well, the definitions get more general when working with tensors, but they're they are still equivalent

experimentX

09:46

okay ... i'll ask this after i'm done with tensors

thank you ...

Zhen Lin

Well, actually, there's a significant complication when you use non-cartesian coordinates...

Gigili

10:48

@DavidWallace I didn't see those remarks until yesterday or something. How can I repay you?

Even though I know There is no charge for awesomeness.

robjohn

11:15

@Eugene I don't really watch sports, and no, I haven't seen Moneyball.

Ragib Zaman

11:26

@DavidWallace I don't understand why my remark is so offensive to anyone other than perhaps Gigili. Whether it occurred in reality or not, I had and still have good reason to believe there was a significant chance that it did. I decided that mildly insulting Gigili if I was incorrect was worth voicing my suspicion, due to the severity of the action if it indeed occurred. You asked what caused me to make such a serious allegation, so I will tell you why I thought there was a good chance such an event occurred.

Rajesh D

Hello

Ragib Zaman

11:38

@DavidWallace Further, if you take offense on Gigili's behalf for my suggestion that it was possible he was intentionally misleading Jordan, perhaps you could also take offense on my behalf that my mathematical/teaching ability was blatantly insulted. I must be quite an idiot, because my "mistake" warranted no explanation or discussion, all that was required was to be told I was wrong. He even warns Jordan "follow his step-by-step guide and it'll be solved by tomorrow morning".

anon

(she)

Ragib Zaman

My mistake then, she*.

David Wallace

You asked me to explain why your accusation of cyber-bullying was offensive. Well, it's about the effects that cyber-bullying has had, on all sorts of people. Teenagers have committed suicide. Children have been afraid to use technology, and afraid to discuss the matter with their parents or teachers. Lots of people have become stressed, withdrawn, depressed and so on. The fact that it relates to modern technology means that parents and other concerned parties haven't really worked out how to deal with it.

Rajesh D

Hey @anon

anon

hey

you've caught me at about my sleeptime

Rajesh D

11:52

ok good sleep then

Henning Makholm

@DavidWallace The missing assumption in that argument is "when he/she clearly wasn't". I wasn't around at the time, but is seems clear to me that Ragib was actually perceiving Gigili's behavior as part of a larger tendency towards bullying of Jordan. That may or may not be how Gigili intended it, but if you want to argue against Ragib, the point you need to support by argument is not "cyber-bullying is serious business" but "this is not part of a pattern of cyber bullying".

(And how do you make those mammoth posts? The chat server rejects everything I say if it is more than about four lines).

David Wallace

@HenningMakholm I see what you mean. However, this particular incident is the first I've heard of anyone bullying Jordan. Can you (or anyone else) perhaps point me at any other incidents that you think might be part of this pattern?

Benjamin Lim

@anon Hey

Do you think it is a problem if say you learn one definition for a manifold in $\Bbb{R}^n$

that's different from the standard definition given?

David Wallace

Do you know whether the definitions are equivalent?

Benjamin Lim

I think they all are

for example the one I have is about graphs of functions

David Wallace

12:01

So if you can prove it, it's not a problem.

Benjamin Lim

I believe the standard one is every point having a neighbourhood homeomorphic to an open ball in $\Bbb{R}^n$

David Wallace

If you can't prove that they're equivalent, it might be worthwhile asking for help from your professor.

Benjamin Lim

@DavidWallace but don't you think that may cause some confusion?

like when looking for references on something, etc

David Wallace

Actually, I think the opposite.

Can I give you an unrelated example?

Benjamin Lim

yes

Ragib Zaman

12:03

@DavidWallace I disagree with your statement that "cyber-bullying" would be not an appropriate descriptor had what I suspected been correct. You describe some of the extremes of cyber-bullying, but that doesn't mean anything less is no longer considered such. This is analogous to how saying "you are a stupid idiot" goes under same the heading of "abuse" that severe physical beatings do.

Had Gigili been intentionally leading down Jordan on the wrong path, some potential effects could have been regressing Jordan's knowledge in how to do related problems correctly, or making Jordan feel even more unwelcome than present, knowing that people are now giving him false solutions. I'm sure the feeling Jordan experiences of being disliked is not easy on him.

@DavidWallace, If you read the meta thread that I linked to above, you will see discussion of some instances of bullying towards Jordan.

David Wallace

At the university where I used to work, the first year students would learn that an "equivalence relation" was a relation that's reflexive, symmetric and transitive; then learn how to prove that a relation is an equivalence relation iff it induces a partition. Some of them would get horribly confused by this, because they couldn't picture what reflexivity, symmetry and transitivity meant.

So if they came to me for help, I would throw away that definition, and teach them that an equivalence relation is a relation that induces a partition; and get them comfortable with that concept. Then I…

@RagibZaman OK, thanks, I'll go and read that.

David Wallace

12:20

@HenningMakholm OK, I understand your argument. I guess, where I come from, people are innocent until proven guilty. I find it completely incongruous that the Gigili that I know and love would take part in any such campaign against an individual. In that light, Ragib's comments just seem like an attack out of the blue, likely motivated by Gigili's slurs against his own teaching.

Much as I feel sorry for Jordan if such a campaign is in fact taking place, I hope I never witness this kind of accusation here again.

Anyway, it's really late here. Good night everyone.

Ilya

12:37

TL;DR

user19161

@Ilya I wanted to say that too!

user19161

@skullpatrol And you should have the first word!

skullpatrol

@JasperLoy My words don't mean anything here.

user19161

@skullpatrol Don't worry! You are too anxious!

user19161

I did not know there is so much drama in this room as well...

skullpatrol

12:49

@JasperLoy What did you mean by "And you should have the first word!" ???

user19161

@skullpatrol I just meant that we should hear what you have to say first!

skullpatrol

@JasperLoy Like I said " My words don't mean anything here."

user19161

@skullpatrol Don't read too much into mine either.

N3buchadnezzar

@skullpatrol "interpreting this this as Skullpatrol never utters anything useful ever"

Its my thoughts. It was complicated making bubbles around my text.

$\boxed{\text{hi}}$

Heh, I can make squares. although I doubt circles.

skullpatrol

Try :D

user19161

12:58

So @skull your bounty ends tomorrow.

Ilya

@Jonas: the cool guy is right here

skullpatrol

@JasperLoy Yes, and the question has been down voted 7 times :(

user19161

@Ilya Oh, pics or it did not happen.

Ilya

@JasperLoy pics?

user19161

@Ilya Pictures. It's just a common expression, what I said.

Ilya

13:01

@JasperLoy that I understood. I didn't get the sense of your sentence

user19161

@Ilya I just meant pictures of the cool guy or he is not there.

Ilya

@JasperLoy I think, Jonas called me a cool guy

so I sent him a message that I'm here

user19161

@Ilya Oh, sorry. You are so cool...

Ilya

I also could say 'they cool guy is in da haus' but 00s have passed already

skullpatrol

The down voters are very active on this web site.

skullpatrol

13:20

Frank Science

I have a problem in limits.

Ilya

@FrankScience shoot it

Frank Science

Yeah

For example, $\lim_{x\to x_0}f(x)=y_0$ and $\lim_{y\to y_0}g(y)=z_0$, find a sufficient and necessary condition that $\lim_{x\to x_0}g(f(x))=z_0$.

Ilya

@FrankScience well, continuity of $g$ at $y_0$ is sufficient, right?

Frank Science

I'm not sure whether it is a theorem in most calculus books.

Yes

Ilya

13:26

is it necessary?

Frank Science

No

Ilya

example

Frank Science

For example, there exists some $\delta$ such that $f(x)\neq y_0$ whenever $0<|x-x_0|<\delta$.

Ilya

I am rendering your formulas. that wasn't addressed to you, sorry

Frank Science

?

By the way, how to edit my post in chat?

Ilya

13:31

just to the left of each post there is an arrow for a drop-down menu

or you can just press up button on the keyboard

Frank Science

I see

Ilya

you have 2 minutes to edit/delete posts

Frank Science

Incidentally, you can use this to display $\LaTeX$ immediately.

Rajesh D

Hey @Ilya

Ilya

@RajeshD hi

Rajesh D

13:35

So watz up?

Ilya

@Frank: that's what I am using. Still, if you write a new formula, it doesn't render for me until I write a message

I think, it's a common problem for Chrome users

Frank Science

I think that, the sufficient and necessary condition is that $g(y)$ is continuity at $y=y_0$ or $x_0$ is NOT the limit point of $\{\;x\;|\;f(x)=y_0\;\}$

Rajesh D

render

Frank Science

I don't know whether it is a famous theorem.

Rajesh D

$x$

render

Old John

13:39

Hi folks

Rajesh D

Hi @OldJohn

Frank Science

First, it's sufficient. Supposing that $x_n$ is any series which converges to $x_0$, thus $f(x_n)$ converges to $y_0$, and from some term, it would not be $y_0$, so $g(f(x_n))$ converges to $z_0$.

Rajesh D

render

Frank Science

Next, it's necessary. We suppose that $g(y)$ is NOT continuity at $y=y_0$ and $x_0$ is the limit point of $\{\;x\;\vert\;f(x)=y_0\;\}$.

Now we can find a series $x_n$ which converges to $x_0$ and $f(x_n)=y_0$ for all positive integer $n$, thus $g(f(x_n))$ is always $g(y_0)$. For $g(y)$ is NOT continuity at $y=y_0$, we have $g(y_0)\neq z_0$.

I wonder whether it is a famous theorem in calculus book.

Any idea?

Frank Science

14:11

I've found a ref

Here

Gustavo Bandeira

I understand what base 8 means but when trying to answer, I first calculated these values and then I got:
1 - 1
11 - 9
111 - 73
1111 - 585
11111 - 4681
111111 - 37449

I've made a division from the 37449 and then I've made the oposite process, and I got this:

1
8 + 1
9*8 + 1
(9*8 + 1)*8 + 1
((9*8 + 1)*8 + 1)*8 + 1
(((9*8 + 1)*8 + 1)*8 + 1)*8 + 1

From here, I don't know how to answer it.

I can see the pattern of repetition but I can't figure out a way of expressing it mathematicaly.

Eugene

16:22

@robjohn I got my first email from a crank journal! it invited me to publish in their journal at $25 per page...

Peter Tamaroff

@Eugene Hahahahaha what?!?!?!

robjohn

@Eugene Cool! Which journal?

Eugene

@robjohn Journal of Mathematical Sciences: Advances and Applications

they also called me a professor...

robjohn

@Eugene You gotta frame that :-D

Professor Eugene

Peter Tamaroff

@robjohn The ODE qoq suggests has a "parabolic cylinder" solution. Yikes.

Eugene

16:23

@robjohn and the most surprising thing is that people actually seem to publish there! that's so weird!

@robjohn indeed! they should have at least done their research

and they also accept papers typed in microsoft word...

Peter Tamaroff

lol the page

@Eugene See the page. It makes sense.

Eugene

@PeterTamaroff it's crazy!

anyway i have to go teach calc now. bye all!

Peter Tamaroff

Subscription Rates for 2008 Price in USD for USA and Canada USD 300.00

@Eugene Teach em well, professor!

robjohn

@PeterTamaroff I agree, that is not the way to go :-)

Peter Tamaroff

@robjohn My solution has to have an explanation. It works out.

I'm not saying it is the most rigorous solution.

robjohn

16:29

@PeterTamaroff Just because you get the right answer, doesn't mean you've done everything right :-)

Peter Tamaroff

@robjohn Yes I know, but the intuition is right. =)

robjohn

I will fight against $\int_{-\infty}^\infty\cos(\varphi x)\,\mathrm{d}x=0$

Peter Tamaroff

@robjohn Yes, that's the big deal.

@robjohn But it depends where you split the integral.

robjohn

@PeterTamaroff No it does not.

Peter Tamaroff

@robjohn It's like the Grandi series for example.

robjohn

16:32

There is no singularity, the only problem is at $\infty$ and things are bad there.

Peter Tamaroff

@robjohn That's why.

robjohn

@PeterTamaroff It is exactly like that series

Peter Tamaroff

If you choose to expand from the origin, you get the non convergence. But why is not accepted to expand from $\pi 2$ for example?

@robjohn Yes, that's why.

I mean

You get

Splitting in $$\int_{k{\pi /2}} ^{(k+1){\pi /2}}$$ gives the Grandi series, if I'm not wrong.

robjohn

But that series does not converge.

and you want $(2k-1)\pi/2$ to $(2k+1)\pi/2$

We get that $\displaystyle\int_{-\infty}^\infty\frac{\cos(\varphi x)}{1+a^2x^2}\mathrm{d}x=\frac{\pi}{a^2}e^{-|\varphi|/a^2}$

We can let $a\to0$, and the integral goes to $0$

Scratch that...

$\displaystyle\int_{-\infty}^\infty\frac{\cos(\varphi x)}{1+a^2x^2}\mathrm{d}x=\frac{\pi}{a}e^{-|\varphi/a|}$

Same thing as $a\to0$

So in this summation the integral is $0$

Peter Tamaroff

16:53

@robjohn But we have a two variables function right?

@robjohn Yes yes.

N3buchadnezzar

-Mein gut.

Planetes is an awesome manga, great story and characters.

robjohn

17:39

@PeterTamaroff The idea here is the same as with Cesàro summation

Peter Tamaroff

@robjohn OK.

robjohn

However, using the sums computed that way is very hard, and often they can't be used.

Peter Tamaroff

@robjohn rob, see this

I'm writing this out, but I think the OP wants another thing.

The area of one triangle of the $n$ possible ones will be

$$\alpha_n= \sin \frac \pi n \cos \frac \pi n$$

Since we have $n$, we get

$$A_n=n\alpha_n= n\sin \frac \pi n \cos \frac \pi n=\frac{n}{2}\sin \frac{2\pi}{n}$$

Thus, we have that $$\lim_{n \to \infty}\frac{n}{2}\sin \frac{2\pi}{n}$$

$$\pi\lim_{n \to \infty} \frac{n}{2\pi}\sin \frac{2\pi}{n}=\pi$$

With some special calculuations one can get Vieta's formula for $\pi $, but that's another deal.

@robjohn Do you think using $$\lim\limits_{\lambda \to \infty}\int_0^{\lambda}\int_0^xf(y) dy $$ should work? in our PV dilemma?

experimentX

18:10

HI ... can I say $$ \oint \vec F \cdot (\hat i dx + \hat j dy) + \oint \vec F \cdot (\hat i dx + \hat j dy) + \oint \vec F \cdot (\hat i dx + \hat j dy) = \oint \vec F \cdot dr $$

Peter Tamaroff

Is $r$ the position vector?

experimentX

yes

Though i think it is not ... it's the last part of stokes theorem ... I did it myself ( i don't know what i did though)

Peter Tamaroff

@experimentX Hm, no idea.

experimentX

still have a look at my question

0

Q: How to arrive at Strokes's theorem from Green's theorem?

I would like to verify the identity $$ \oint \vec F \cdot (\hat i dx + \hat j dy) + \oint \vec F \cdot (\hat i dx + \hat j dy) + \oint \vec F \cdot (\hat i dx + \hat j dy) = \oint \vec F \cdot (\hat i dx + \hat j dy + \hat k dz) $$ If it is incorrect then what would be the correct identity....

Jordan

18:26

Well I am an idiot, got a D on my calc test

experimentX

I'm gonna flunk tomorrow

David Wheeler

I really wish I could help you, Jordan, I honestly do. But your troubles with calculus go deeper than just the class you are in at the moment.

what you need to do, is re-visit analytic geometry and trig, and really get comfortable with it. and that's going to take some time.

see, right now, you could learn all the formulas in the world, but if you have no intuition about "which one to use", it's not much help

from what i have seen, you're not an idiot. you just have "gaps" in your knowledge, that need to be "filled in", so the other stuff that BUILDS on that, won't be so mystifying

Américo Tavares

@Jordan Was it very different from your questions here?

Jordan

@AméricoTavares Not really, I just need to learn to study I guess

David Wheeler

and the sad thing is, colleges and professors...they don't care (i'm not trying to slam professors, here), your individual fate is only a small part of their future

Jordan

18:38

@DavidWheeler I have taken college algebra 3 times, calculus 1 twice. I can't keept doing that

David Wheeler

i agree. you need a real, live person, to work with you, for a lot of time

someone has to get to know you a bit, so they can see "how" you run into your road-blocks. re-taking classes is going to be more of the "one size fits all"

Américo Tavares

@Jordan I agree with David Wheeler's words: "from what i have seen, you're not an idiot. you just have "gaps" in your knowledge, that need to be "filled in", so the other stuff that BUILDS on that, won't be so mystifying"

Jordan

I can't really afford to take all that math over again, I am just so incredibly sick of school. Sick of everything about it

If I delay that much I will need 8 years of school for a 4 year degree

David Wheeler

well, you can augment your "school-learning" with self-study. nowadays there are all sorts of free-ebooks, and video lectures on the web.

Jordan

I can't stand the people in my calss either, 2 people infront of me just talk about how easy the class is and they spaz out over every little thing in the class about how amazing and fun it all is. I really wish someone would jump those people

Gigili

18:47

@DavidWheeler Why don't you? You both live in the US, I presume.

Américo Tavares

@Jordan Was the time you had to do the test not enough for you? Or did you fail in the computations or didn't you know how to start solving the questions.

David Wheeler

it's easier to like something when you're good at it. when i was a child (and in fact to this day) i was short. the only sport in alaska (where i grew up) going was basketball. i grew up not liking sports.

Jordan

@AméricoTavares I didn't have enough time but I don't think more time would have helped me I just forgot how to do a lot of things, I got some processes mixed up and wasn't sure which way to do things

David Wheeler

@Gigili i would be happy to help Jordan as much as i can.

but realize, the US is a "big" place. it's like 2 europes put together.

Gigili

Oh? TIL.

Jordan

18:49

And those people infront of me answer every single question right away and then they ask another question completely just for their satisfaction. They dominate the class basically

Gigili

@Jordan Perhaps your book doesn't have enough exercise or it's not well written?

David Wheeler

i had a guy like that in my abstract algebra class. his name was stewart. always showing off. i was like: dude, i'm here to learn this stuff, not win an award.

Jordan

@Gigili It is the most popular calc book I think

stewart early transcendentls

David Wheeler

@Jordan...you may find this interesting to read. it's rather "old-fashioned" but charming in its way: djm.cc/library/Calculus_Made_Easy_Thompson.pdf

robjohn

@PeterTamaroff I don't see a way to use it. Do you?

Jordan

18:55

Honestly I have never hated anyone as much as those two guys in my class

David Wheeler

yes, stewart is very widely-used as a textbook.

Jordan

I think I can still pass the class if I do really well on everything else, like at least a B and atleast a C on the next two tests

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