Mathematics - 2013-05-03 (page 2 of 5)

@AlexanderJones My personal preference lies with the latter two. They come in useful as background information. A warning needs to be in place, though: Once you know analysis, you may get frustrated with the lack of precision in the other subjects.

Oh indeed, as @JasperLoy says, a bit of linear algebra comes in handy with multivariate calculus.

Alexander Jones

What do I need to know before I start linear algebra? Right now I know high school algebra, calculus and analysis

Lord_Farin

@AlexanderJones High school algebra should be enough.

Dominic Michaelis

a good advice is that everything you learned in school is wrong ;)

Alexander Jones

2:07 PM

I agree! High school only teaches you to prepare for the exam, doesn't teach real maths...

Which textbooks for analysis and linear algebra?

I have Appstol for analysis and "Introduction to Linear Algebra" by Gilbert Strang

Lord_Farin

@Vrouvrou Well the map $r_1: M^b \to M^a$ restricts to the identity on $M^a$, and it is continuous. Hence it is a retraction.

@Alexander Perhaps you could look on MSE for previous questions on this; I'm sure they exist.

Alexander Jones

Ok, thanks everyone!

robjohn

@AlexanderJones Of course it was in the 70s, but I learned some real math in high school. Our geometry was enough that I was able to develop most of spherical trig.

user19161

@robjohn I think I will retire at 1k, lol.

Dominic Michaelis

@JasperLoy I DARE YOU !

robjohn

2:23 PM

@JasperLoy cul8r

Charlie

@JasperLoy if you do, I'll get a flight to singapore and obligate you to make another account!!!

robjohn

@AlexanderJones I wrote most of this paper during high school

user19161

@robjohn WOW!!!

robjohn

@JasperLoy LaTeX wasn't around then, so I had to put it into LaTeX later :-)

Charlie

Hi @peter

user19161

2:30 PM

Maybe I should delete my latest answer, since it might mislead the asker, lol.

Dominic Michaelis

@robjohn the figure 1 is a bitmap isn't it ?

Charlie

Don't use this word again

Dominic Michaelis

@charlie i know bitmap is bad but when it is bitmap what else should i say :(

Charlie

@DominicMichaelis I.was talking to @jasper sorryyyyy!

robjohn

@DominicMichaelis you mean "Central Plane of a Unit Sphere Containing the Side $\alpha$"?

Kasper

2:37 PM

Why are all those moderators suddenly resigning ??

Dominic Michaelis

@robjohn yeah or is it only the dropbox pdf viewer sucking again ?

robjohn

@DominicMichaelis When I view it at high magnification, it is still sharp, so it must be a pdf

@DominicMichaelis oh, I don't know about that. I am using Adobe Reader

@DominicMichaelis download and use Reader

@JeannePindar: welcome. It has been a while (85 days)

Jeanne Pindar

Hi

Dominic Michaelis

@robjohn oh you are right they look like my tikz graphics :D

robjohn

@DominicMichaelis yeah, I try not to use bitmap graphics if I can help it.

Dominic Michaelis

2:41 PM

bitmaps + word + comic sans :D

Peter Tamaroff

@Charlie Hi, Charles.

Chris's sister and pals

It's a lovely question when trying to solve it without using Taylor series, Euler sums$$\int_0^1 \log(1-x) \log(1+x) \ dx $$

Peter Tamaroff

@Chris'ssisterandpals That is like soooo trivial.

Chris's sister and pals

@PeterTamaroff maybe you can evaluate it with double integrals as you previously did with other similar integrals :P

Dominic Michaelis

isn't it $$ 2 \cdot \int_0^\frac{1}{2} \log(1-x) \log(1+x)\, \mathrm{d}x$$

Peter Tamaroff

2:48 PM

@DominicMichaelis Nah.

Dominic Michaelis

oh wait that

is shit

Peter Tamaroff

You want $$\frac 1 2 \int_{-1}^1$$

Dominic Michaelis

$1\neq \frac{1}{2}$ :D

Chris's sister and pals

:D

user19161

Whoa, I got 2 downvotes today...

Peter Tamaroff

2:50 PM

@Chris'ssisterandpals Didn't you try to IVP?

user19161

Haha, the downvoters are really vicious today...

Chris's sister and pals

@PeterTamaroff I think it works by triple integrals. (I check that right now)

Peter Tamaroff

@Chris'ssisterandpals You're making it too complicated

Just integrate by parts.

Chris's sister and pals

@PeterTamaroff It doesn't work that way.

Peter Tamaroff

Take $u=(x-1)\log(1-x)+(x-1)$ and $v=\log(1+x)$

Then $du=\log(1-x)$ and $v=\log(1+x)$

MphLee

2:53 PM

Hi, someone can give me a tip, I think there is something wrong in how I ask my questions (1st the horrible english xD ) but someone can say me how to improve my questions? or maybe are not enough intresting?

Charlie

@peter how are you doing?

Peter Tamaroff

Your integral is $$-\int_0^1 \frac{x-1}{x+1}\log(1-x)dx-\int_0^1\frac{x-1}{x+1}dx$$

Charlie

@anon ANON!

Peter Tamaroff

@Chris'ssisterandpals Did you work it out?

Chris's sister and pals

By other means, not going this way.

Peter Tamaroff

2:58 PM

@Chris'ssisterandpals What?

Chris's sister and pals

Double series.

Peter Tamaroff

-\int_0^1 \frac{x-1}{x+1}\log(1-x)dx+\int_0^1\frac{x-1}{x+1}dx

It is like this.

@Chris'ssisterandpals OK. What is the value?

Note that IVP really pays off here.

Chris's sister and pals

$$2-\pi^2/6+(\log 2)^2-2\log 2$$

Charlie

@Lord_F do you have any email you could use for MSE purposes?

Lord_Farin

@Charlie What do you mean with "MSE purposes"?

Charlie

3:01 PM

@Lord_Farin would you give me your email adress so I can email you?

Peter Tamaroff

@Chris'ssisterandpals Should check, but I'm too lazy now.

Chris's sister and pals

@PeterTamaroff ok. I'm here working on more ways. Let me know if you see other ways.

Peter Tamaroff

@Lord_Farin Ruuuuuuuuuuuuuuuuuuuuun!

@Chris'ssisterandpals But I have found one. Why would I want any other? =)

Chris's sister and pals

@PeterTamaroff lol , ok. :D

Lord_Farin

@PeterTamaroff :)

@Charlie I suppose you can make an educated guess from my profile :).

Charlie

3:04 PM

@Lord_Farin some people here don't like to give their emails, so how you barely use your real name, I suppose you may have a secretive email

Lord_Farin

@Charlie I have email addresses dedicated to my online activities :).

Charlie

Okay, forget what I asked...

user19161

(removed)

Lord_Farin

@Charlie Did you catch that?

J. M.

Why hello, dark Bieber, I see you've reconstituted.

Charlie

3:07 PM

@Lord_Farin yes

user19161

@J.M. I am the real JL, not an imposter...

J. M.

As you say. :)

Vrouvrou

@Lord_Farin thank you, you tell me something about the use of $\varphi_{t(a-f(q)}$

user19161

You can see from my answers: short and misleading. =)

MphLee

Someone knows the name of this operator? I can't find it, even under another name/notation :S
http://math.stackexchange.com/questions/380139/has-this-operator-a-name-alpha-gf-g-ominus-f

Peter Tamaroff

3:09 PM

Who knows how to count here?

MphLee

me maybe... but only form zero to ten

xd

J. M.

@PeterTamaroff How many fingers would be required for this job?

Lord_Farin

@MphLee Well it seems to be conjugation in the group of bijections on a set.

Peter Tamaroff

@J.M. Well, as many as $10!$.

Lord_Farin

@Vrouvrou Well it sends $M^{f(q)}$ to $M^{f(q)+t(a-f(q))} = M^{a t + f(q) (1-t)}$.

Charlie

3:12 PM

@lord_F I didn't find it....

nonagon

Hello

I asked a problem here earlier regarding basic newton's laws

can anyone help with me with it ?

This was the link

Lord_Farin

So as $t$ varies from $0$ to $1$, $\phi_{t(a-f(q))}$ will send $f(q)$ on a path from $M^{f(q)}$ to $M^a$.

nonagon

-2

Q: Work Done by Tension

At an instant the velocity in horizontal direction is $0$, gravity acts vertically downwards and tension acts horizontally. Then it is said tension will do no work. Now, at the very next instant on time, There is a finite velocity now in the horizontal direction. Gravity couldn't have prov...

J. M.

@Lord_Farin, prolly too quick for Charlie to see. Maybe wait until she gives some confirmation before redacting?

nonagon

And more justification I added was ::             Because tension is there in horizontal direction , and it acts for a however small but finite amount of time , hence , since a=F/m , and for a small time it acts , hence it must increase velocity by a x t. where t is the time for which it acts and F here is tension and m is mass of the ball/stone which is rotating in circular path
Kindly see the edit too
And if only tension is there , then acc. to Work Kinetic Energy theorem , the Kinetic energy must not change , hence forth , speed must not change , but it is changing here according to newt…

Peter Tamaroff

3:15 PM

Dude, this is math.SE. Are the physicists giving you a hard time?

Lord_Farin

@J.M. You're probably right.

J. M.

@PeterTamaroff Apparently they do. They're even less tolerant of HW than here.

nonagon

No one had replied and since the problem is regarding Newton's laws , I was hoping mathematicians will know

Peter Tamaroff

@J.M. Yes, I have noticed.

MphLee

@Lord_Farin conjucation in the group.. grop theory?

@Lord_Farin srry is than i never haerd that before (I'm a bit newbye xD)

nonagon

3:17 PM

@PeterTamaroff so ?

Peter Tamaroff

@nonagon So what?

nonagon

I mean , can you please answer ?

Peter Tamaroff

@nonagon No, sorry. I don't know physics! =)

nonagon

Alright

Lord_Farin

@MphLee Conjugation in a group $(G, \circ)$ is the operation $c_b(a) := b\circ a\circ b^{-1}$. Alternatively, just read some wikipedia; this and links there should provide you with some information.

user19161

3:20 PM

Hey @jayesh was that your new email?

Jayesh Badwaik

@JasperLoy Yes, that was my new mail.

user19161

@JayeshBadwaik Aha! I got 2 downvotes today, lol...

Lord_Farin

I'll have to go (dinner). Will remain here, afk.

Jayesh Badwaik

@JasperLoy Hehe.

MphLee

@Lord_Farin Bingo! seems me what i was looking for, bye and thanks

N3buchadnezzar

3:24 PM

Anyone remember seein this integral before?

$$ \int \frac{\mathrm{d}u}{(1+\sqrt{u})(u+1)^2} $$

I know I have, but I can not seem to find it again...

robjohn

$$
\begin{align}
\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{kn(k+n+1)}
&=\sum_{m=2}^\infty\sum_{k=1}^{m-1}\frac1{k(m-k)(m+1)}\\
&=\sum_{k=1}^\infty\sum_{m=k+1}^\infty\frac1{k(m-k)(m+1)}\\
&=\sum_{k=1}^\infty\sum_{m=k+1}^\infty\frac1k\frac1{k+1}\left(\frac1{m-k}-\frac1{m+1}\right)\\
&=\sum_{k=1}^\infty\sum_{m=1}^\infty\frac1k\frac1{k+1}\left(\frac1m-\frac1{m+k+1}\right)\\
&=\sum_{k=1}^\infty\frac1k\frac1{k+1}H_{k+1}\\
&=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+1}\right)H_{k+1}\\
&=\lim_{n\to\infty}\sum_{k=1}^n\left(\frac1k-\frac1{k+1}\right)H_{k+1}\\

user19161

@JayeshBadwaik You look very different from your earlier pic.

Heber Sarmiento

Hello folks , anyone could give me an idea on how is true that : $$1+\int_{1}^{N-1}\frac{dx}{x}>\sum_{j=1}^{N-1}\frac{1}{j}$$

I don't know how to go about that

Chris's sister and pals

@robjohn beautiful!

user19161

@Chris'ssisterandpals Oh I am glad you did not quit the site, lol.

Chris's sister and pals

3:27 PM

@JasperLoy I'm an active user anymore.

Jayesh Badwaik

@JasperLoy Yes.

robjohn

@HeberSarmiento $\displaystyle\int_{j-1}^j\frac{\mathrm{d}x}{x}\gt\frac1j$

add 'em up

Chris's sister and pals

@robjohn I used the $3$rd identity from Double infinite summation and I got more or less the same thing functions.wolfram.com/GeneralIdentities/12. I kept the initial variables.

robjohn

@Chris'ssisterandpals That just looks like change of order of summation

Chris's sister and pals

Yeah.

robjohn

3:38 PM

another user removed! bah!

Chris's sister and pals

@robjohn actually, you applied the variable change but after some work things are relatively the same. Anyway, I like your way!

user19161

@robjohn They kept my votes. I hope you know.

robjohn

@PeterTamaroff Yoda quotes?

Hayaku

if i want to know the probability of something taking on values in a range, what is the general set up?

Charlie

Oh noes

robjohn

3:40 PM

@JasperLoy I know. I would have lost a lot if they didn't

@Hayaku we'd need a bit more information than that...

Chris's sister and pals

@JasperLoy where are your points? :-)

Jayesh Badwaik

@Chris'ssisterandpals Argon.

Hayaku

like if i am rolling dice until the sum exceeds L

Chris's sister and pals

@JayeshBadwaik how's that?

Hayaku

i want to know the probability that the final value rolled is a 1, or a 2, or between a 3 and 4, or whatever

robjohn

3:42 PM

@Chris'ssisterandpals he lost them all when he abdicated.

Jayesh Badwaik

@Chris'ssisterandpals Like all good chemical jokes argon.

Chris's sister and pals

hehe

Hayaku

i don't know how to set up the proper integrals or summations

Jayesh Badwaik

;-)

Chris's sister and pals

I'm ready to upvote him!

:D

Heber Sarmiento

3:43 PM

@robjohn $$\sum_{j=2}^{N-1}\frac{ln(j)-ln(j-1)}{j}$$ you mean like this?

user19161

@Chris'ssisterandpals I have 4 answers now...

Chris's sister and pals

@JasperLoy it's a beginning ... :-)

robjohn

no... If you sum what I wrote for $j$ from $2$ to $N-1$, you get
$$
\int_1^{N-1}\frac{\mathrm{d}x}{x}\gt\sum_{j=2}^{N-1}\frac1j
$$

@HeberSarmiento then add $1$ to both sides

and get $$
1+\int_1^{N-1}\frac{\mathrm{d}x}{x}\gt\sum_{j=1}^{N-1}\frac1j
$$

Jayesh Badwaik

@GitGud Or a zero tolerance to trolls. :-)

Chat Manners or Etiquettes | $\LaTeX$ Rendering in Chat

12

@robjohn Please pin. :-)

Thanks. :-)

CBenni

If the fundamental group $|\pi_1(X)|<\infty$, doesnt that imply $\pi_1(X)=\{1\}$?

because the order of any representative $\omega: [\varepsilon]\neq[\omega]\in\pi_1(X)$ is infinite, isnt it?

Tobias Kildetoft

3:53 PM

@CBenni why?

robjohn

can elements of the fundamental group handle snakes?

user19161

Why don't I get bonus 100 now that I have 100 rep?

Will Jagy

@Hayaku, I notice you deleted your question on sum of uniform variables above 1. Have you had multivariable calculus?

Dominic Michaelis

@jasper you need 200 to get the bonus

CBenni

@TobiasKildetoft what I said above. Additionally, I cant think of a space that has a finite, but not trivial fundamental group.

user19161

3:55 PM

@DominicMichaelis Ah, but I got 100 bonus on Eng yesterday because the system linked me back to my old network and chat ID, lol.

robjohn

@JasperLoy Are those accounts deleted?

Hayaku

I understand the basics of calculus fine, but my problem is figuring out how to set the problems up

i feel like that much, at least, is woefully undertaught (or maybe I just suck, hah)

robjohn

@Hayaku are you rolling standard 6-sided dice?

Hayaku

yes

user19161

@robjohn I created all brand new accounts but using an email I used before...

Will Jagy

3:57 PM

@Hayaku, calculus in more than one variable

Tobias Kildetoft

@CBenni I just seem to recall that one can make a space where going around twice gives something homotopic to a point

Hayaku

@WillJagy In terms of a formal class no

robjohn

@Hayaku Then rolling a $1$ has a $\frac16$ chance, rolling a $2$ has a $\frac16$ chance and rolling a $3$ or $4$ has a $\frac13$ chance.

Hayaku

how do i render the latex you just posted

it isn't rendering automatically like it does on the site

Will Jagy

@Hayaku then how did you happen on a problem in sum of n variables? It is, officially, a multiple integral. Since it is the integral of 1, it is actaully an n-volume, area for n=2, volume for n=3, etc.

CBenni

3:59 PM

@TobiasKildetoft might be somethig like a moebius strip or the $\mathbb{R}P^1$

Tobias Kildetoft

@CBenni It seems if you take the homogeneous space $G/H$ where $H$ is a finite subgroup of the Lie group $G$, then that has fundamental group $H$

mobius strip has same as the circle (just looked that up)

Hayaku

@WillJagy Probability problem, not a math problem, if that matters. Multivar may be a strong prereq but isn't required for most of the problems I'm looking at. This is just a problem that looked really interesting and I want to learn it

brb

Will Jagy

@Hayaku, are you studying out of a particular book? Is there a professor involved?

Tobias Kildetoft

@CBenni yeah, it seems if you take $\mathbb{R}P^n$ with $n\gt 1$ then that has fundamental group of order $2$

CBenni

ok, thanks for clarifying.

Hayaku

4:04 PM

@WillJagy No, problem the prof put up on the board as "extra learning" for the curious. technically his version involved uniform variables but i changed it to discrete because i wanted to ask about it here without getting spoiled. I figure if I learn the discrete version with dice I can maybe figure out the continuous one

i don't want to be told how to do it, but i do want to understand how to set it up

user19161

Debian 7 should be released this weekend.

Charlie

Hmm

Will Jagy

@Hayaku I recommend you ask your professor about this in office hours. Lacking multivariable calculus, you have a long way to go before doing this yourself.

Charlie

@WillJagy hi Jagy

Hayaku

I don't think so; I somewhat understand the answer given in the other thread. My problem is just setting it up because I'm never sure when to integrate what or condition on what

Will Jagy

4:19 PM

@robjohn, this is about math.stackexchange.com/questions/379949/… I am trying to be, um, kind, as per your request.

@Charlie, any luck with email and Somaye?

Charlie

@WillJagy I talked once, now that she has my email it may be easier for her

Will Jagy

@Charlie, I prefer Will or O Almighty one

Charlie

@WillJagy ah, okay, Will

Will Jagy

@Charlie, if you have a title that emphasizes your exalted spiritual position we can use that.

Heber Sarmiento

@robjohn Thank you, I got it. I don't know how you could see that (the trick with the integral from j-1 to j), I guess a lot of practice huh?Thank you very much!

Hayaku

4:23 PM

"Trying to be kind"? That's rather condescending

Charlie

Hahaha @will

robjohn

@Hayaku beleve me, there have been other respondents in the same position who have not been "kind"

Hayaku

@robjohn I don't really care, that's no excuse. Nothing said warrants anything "unkind"

I will try elsewhere

goodbye

robjohn

@Hayaku Will is not being condescending to you, he is talking about the FAQ saying that we should be nice and not condescending to others

sigh

Hayaku

@robjohn The point is that he has to apparently "restrain" himself from doing so

Being kind just because you're following some external rule is like being good to others because you're just following the Bible or something, instead of actually wanting to be nice to others intrinsically

leaving now

skullpatrol

4:28 PM

@Hayaku Bye

Charlie

@skullpatrol bye, skull, how are you?

skullpatrol

@Charlie Fine thanks. How are you?

robjohn

If you want help, you can stay and get help. Pampering is not in the job description.

skullpatrol

>8(

Charlie

@skullpatrol fine, fine

robjohn

4:30 PM

@skullpatrol I am mean, am I not?

skullpatrol

@robjohn You are not mean.

Babak S.

Hi Friends. Is there anybody who knows a non euclidean geo?

robjohn

@BabakS. I know one :-)

Charlie

He's standard deviation

skullpatrol

@robjohn You are the square of mean ;-)

robjohn

4:32 PM

@Charlie I have one standard deviation and one non-standard deviation

Will Jagy

I had a deviated septum fixed

robjohn

but I shouldn't go into that here :-)

skullpatrol

Please do :D

robjohn

@WillJagy only the nose knows

Babak S.

@robjohn: I don't get your time. Just a point. Can I use the following picture for the student who want to know it for the first time?

robjohn

4:38 PM

@BabakS. I like it :-)

Babak S.

And you think they also like it?

skullpatrol

@robjohn "(with one standard deviation and several unusual ones)"

robjohn

@BabakS. they may not understand it if they don't know the idea of non-Euclidean geometry.

Babak S.

Thanks @robjohn. :-)

robjohn

@skullpatrol I just added that :-)

skullpatrol

4:40 PM

Orly?

robjohn

@skullpatrol as a result of the foregoing conversation

skullpatrol

Yes, that foregoing conversation was an unusual deviation.

Rush

what does "f(x) is supported in [-A, A]" mean?

Charlie

@skullpatrol it was an outlier

robjohn

@Rush outside of $[-A,A]$, $f(x)$ is $0$

Peter Tamaroff

4:45 PM

@Rush Google "support of a function". You'll get lot's of info

Rush

thanks very much

robjohn

Perhaps we need a banner... This is the Math chatroom; spoonfeeding is down the hall

Babak S.

@robjohn: I see. I will talk to them about the parallel and the 5th axiom first.

robjohn

@BabakS. that would probably make the comic more entertaining

@Rush that was not aimed at you, but at an earlier customer :-)

Vrouvrou

@Lord_Farin , i don't understand ,first how to prove that if $a\leq f(q)\leq b$ for $q\in M^b$, $r_t(q) \in M^a$ , how to prove that $f(r_t(q))\leq a$ ? please

Rush

4:52 PM

@robiohn I have understood :-)

Vrouvrou

@Lord_Farin that's ok $r_t$ is well defined

Peter Tamaroff

@robjohn

robjohn

@PeterTamaroff

there, that is better

Peter Tamaroff

@robjohn Can I ask for some sanity check in counting?

robjohn

@PeterTamaroff not above 3

Will Jagy

4:58 PM

@robjohn, I was aware that Hayaku would see my use of the word "kind" and might take offense. However, your not yesterday or the day before about leaving questions fort others to solve, if the OP went too far, left me wondering what room I had left. Much of the time my immediate response is "Why can't you do this?" What is your background?

Peter Tamaroff

@robjohn Above 3 what?

skullpatrol

@Charlie Have you seen this movie?

robjohn

@PeterTamaroff I know my limits. Counting to 3 is it :-)

Peter Tamaroff

@robjohn Ah. OK.

Let $A=\{1,3,4,5,7,8\}$.

Tobias Kildetoft

@WillJagy I frequently see questions that causes the same question in me (the type of question where no trick is involved and one just needs to apply the definitions)

Transcript for

$Mathematics$ Mathematics