Mathematics - 2012-04-22 (page 1 of 2)

@Alex I might have asked you this question, but again, how much time do you spend on blogposts, please? How do you write them up? Would love your telling me something about it.

Alex Youcis

Hey

Hmm, I probably spend (on an intense day)

six hours blogging

What do you mean "how do I write them"?

Kannappan Sampath

Write them into TeX straight away or write it out first and then....TeX them?

@AlexYoucis 6 hours... impressive.

(So,I assume that is on a daily basis?)

Jordan

I don't understand this question, I am suppose to find the integral of $2|x|$ from -1 to 2

Alex Youcis

2:57 AM

@Jordan Try posting this on the main forum

Kannappan Sampath

@AlexYoucis Nice group theory answer. Knew it but was waiting for the OP to show some work. With the advent of home work, posting, I answer genuine questions alone...these days.

Not to say, you should not answer, but telling you,..., +1 btw.

Alex Youcis

@KannappanSampath Yeah, I knew he was too. But for a beginner I could see that being a semi-difficult question

Kannappan Sampath

Did you read my previous question @AlexYoucis

Alex Youcis

@KannappanSampath Which one?

Kannappan Sampath

12 mins ago, by Kannappan Sampath

Write them into TeX straight away or write it out first and then....TeX them?

@AlexYoucis ^ :-)

Hi @Mariano

The Chaz

3:15 AM

@Jordan, you on?

Jordan

yes

Kannappan Sampath

@Jordan Found where you went wrong?

The Chaz

Want to talk about the absolute value integral?

Jordan

sure

I thought the absolute value of -1 was 1

The Chaz

That's true, but not what the question is asking.

What is the integral of x - 2x?

If you decided not to write x - 2x as "-x", you might write
x^2/x - x^2

for the antiderivative

That is also what you said the antiderivative of "x - 2|x|" is.

Jordan

3:19 AM

I dont understand why that isn't right, any x will always be positive

The Chaz

Do you know what the graph of |x| looks like?

Kannappan Sampath

@Alex What edit are you talking about in your group theory answer? I find it completely right. @AlexYoucis

Jordan

sort of like x^2

The Chaz

Sort of... it's more of a "V"

Kannappan Sampath

\/

Alex Youcis

3:20 AM

@KannappanSampath See my edit now

The Chaz

Sharp point, right? Well wouldn't
int(x = -5 to x = +5 of |x|) be zero by your reasoning?

Jordan

yes

The Chaz

How can the area under a curve be zero if that curve is *always above the x-axis?!?!

Kannappan Sampath

@AlexYoucis Argh! Sleep Deprivation can get you to this.

The Chaz

int(x = -5 to x = +5 of |x|) is actually 25

(not zero!)

Alex Youcis

3:23 AM

@KannappanSampath Haha, it's no big deal! It happens to us all.

Kannappan Sampath

Anyway, breakfast time. Then, I should be off. See you all later, people. Bye.

Jordan

isn't it 10?

actually I don't know, I got zero

Alex Youcis

Bye

Jordan

from 0 to 5 the value is 5 and from -5 to 0 the value is 5 and I find the difference and it is 0

Alex Youcis

@MarianoSuárezAlvarez Do you do any sort of complex geometry?

The Chaz

3:26 AM

@Jordan "The value" does not mean what you think that it means.

The AREA under |x| from 0 to 5 is 12.5

Eric Gregor

is this integrable: $\int_0^\infty \frac{1-e^{-4\pi i \xi}}{\xi}\ d\xi$?

The Chaz

The area under |x| from -5 to 0 is also 12.5

Jordan

oh I forgot to get antiderivative

Eric Gregor

somehow i can't find a nice expression for that...

The Chaz

It happens.

But you cannot get an antiderivative of |x| without using the definition

Jordan

3:27 AM

Isnt it -12.5?

Alex Youcis

@EricGregor Yes. You should ask it on the forum though

Jordan

then I do 12.5 - (-12.5)? for 25

Eric Gregor

i kind of did

0

A: Help with an inverse Fourier transform calculation

There's a mistake in the penultimate line of the equation array – the $|\xi|$ is missing in the exponents. If you correct that, you'll find that the whole calculation is one big detour and you could have directly obtained the last line by using $$\sin(2\pi t|\xi|)=\frac1{2\mathrm i}\left(\...

The Chaz

Yes, when I said "area", I meant without a sign

So there you have it.

But now let's think about x -2|x| from -1 to 2

Eric Gregor

i haven't left a followup comment, but the offered solution seems to imply that this has a nice expression

@Alex are you considering giving another response to the question? you seem to imply that you know how to integrate that guy

The Chaz

3:31 AM

It's a little trickier to visualize the graph, so we use the definition of |x| like in one of the answers

The interval [-1, 2] crosses x = 0, which is significant in the definition of |x|

When x \geq 0, |x| = (just) x

Alex Youcis

Well, I'm not so sure now. Wolfram seems unable to integrate it. The basic idea in these cases is to rewrite things like this as a double integral. Namely, I think the integral from 0 to 1 of e^(-i*pi*x*y) dy should give you (1-e^(-i*pi*x))/x

@EricGregor Yes?

The Chaz

So when x >= 0, we can leave the absolute value symbols off

When x >= 0, the integral (x = 0 to 2) of x - 2|x| is
integral(x = 0 to 2) of x - 2x (= -x)

So we just integrate -x from 0 to 2

Eric Gregor

hmm

The Chaz

When x is negative (like in the interval [-1, 0)), we can drop the absolute value signs and add a minus sign (since x < 0 --> |x| = -x)

Jordan

so I have to seperate the negative numbers, I can't ever write one expression to do this problem?

The Chaz

3:34 AM

So the integrand x - 2|x| = x - 2(-x) when x is negative.
And x - 2(-x) = x + 2x = 3x, which has antiderivative 1.5x^2

YES!
But here's the catch: It might not always be positive/negative. Let me explain.

Int(x = 0 to 6) of |x - 4|

|x - 4| = x - 4 when x - 4 is positive or zero. This happens when x >= 4

So you don't split the integral into two @ zero, you split it into two at x = 4.

Integral(0 to 6) = Integral(0 to 4) + Integral(4 to 6)

Jordan

so y = |x-4| y can never be negative but x can

The Chaz

True

(So the area under y will always be positive)

Jordan

but that is true for any function, when x is <0 then x is negative isn't it?

well lots of functions, like y = 3x

The Chaz

x <0 means "x is negative"

Jordan

then why dont I split up the integral of that?

The Chaz

3:39 AM

If y = 3x, if x is negative then
x < 0
y =3x < 3*0 = 0
So y < 0

The absolute value function is a piecewise function. It is two pieces!

Jordan

so it is split when y = 0?

The Chaz

|x| changes sign when x = 0
|x - a| changes sign when x = a

Worth repeating:
**The absolute value function is a piecewise function. It is two pieces!**

Jordan

x^2 has two pieces too though doesnt it

The Chaz

No.

Mariano Suárez-Alvarez

$x$ and ${}^2$

The Chaz

3:41 AM

@MarianoSuárezAlvarez cute

:)

Jordan

well I mean the values of the function are always mirrored around x right?

The Chaz

There is y-axis symmetry

Jordan

or 0 I mean

The Chaz

That's not what I mean by piecewise function.

Jordan

I guess I just don't quite get the absolute value stuff

Mariano Suárez-Alvarez

3:42 AM

@TheChaz, Thank you.
I'll be here all week

Jordan

if I have -4 it is 4, if I have 4 it is 4

The Chaz

I'm gonna go work out some more. It sounds like @Mariano wants to take over.

Jordan

thanks

Mariano Suárez-Alvarez

nonononono

The Chaz

@Jordan: Don't get discouraged about the absolute value for the moment. It throws of EVERYONE at some point.

@MarianoSuárezAlvarez What happened to all that "here all week" talk ??!?! :D

Mariano Suárez-Alvarez

3:44 AM

at the moment I am reading flags...

Eric Gregor

evening, @Mariano

The Chaz

@MarianoSuárezAlvarez Oh that's right - you have responsibilities :)

Jordan

a derivative can tell me the rate of change of a function and an integral can tell me the area under the function (the y values?) so why does it work out that the antiderivative of a function will tell me the area and the derivative of a function will tell me the rate of change. Why does it work out like that?

I mean, these are like inverse operations, why does it just happen that they are so related

Mariano Suárez-Alvarez

make pictures with linear functions, and you'll see the relation

the reason is very apparent when you see the proof of the fact that the two operations are "inverse" to each other

anon

ah, the fundamental theorems

Jordan

3:59 AM

stewart never really connects them

I mean if I want to find how fast a car was travelling I find the derivative of the function for location?

So that tells me at x time they were travelling x mph, but that is just an estimate isn't it?

anon

the definition of the derivative is basically designed to be the "rate of change." the fact that integration is the reverse of the antiderivative can be seen geometrically. let me pull up the mspaint.

Jordan

I have pictures in my book, I get the geometric proofs I guess

Mariano Suárez-Alvarez

stewart does not prove the fundamental theorem of calculus?

Jordan

he does but he never really ties it all togther

there is a part 1 and part 2

It is in a chapter on integration and is just used to teach how to do 2 types of integration

he just says that they are connected, and never expands on that really

Rajesh D

Hi

Limitless

4:13 AM

Hi, @RajeshD.

How are you?

Rajesh D

fine thanks, how are you doin

Limitless

Fine. Just thinking about the next coming days.

Rajesh D

Why Whats up?

Limitless

Generally, I like to plan my life in my head.

As they say around here, "Getting your ducks in a row."

Mariano Suárez-Alvarez

isn't that for when you want to shoot them?

Limitless

4:21 AM

@MarianoSuárezAlvarez... Maybe I'm a little on the morbid side with my particular life mottos. :P

Raven Dreamer

6:00 AM

As per this question

1

Q: What Parametric Equations are required to move along a circle while moving left?

I'm working on a program where I can set objects along arbitrary parametric paths. Moving left is easy: X = x - dT(V) Y = y Moving in a circle is easy: X = x+ Cos(dt*Pi) Y = y+ Sin(dt*Pi) So I tried to combine them to move left while also moving in a circle, likeso: X = x- dT(V) + Cos(dt...

parametric

How do I read this equation:

Which, from this page, looks like the answer to my question?

Specifically, what is Phi? (Is that a Phi?)

anon

looks like phi is just some initial angle

your question isn't very clear to me

I'm not this person

hi

anon

Raven: you talk about a point moving left and around a circle, so I suppose the circle itself is moving left, in which case the cycloid comment looks like what you need (though the reparametrization so that the speed of the circle's center leftwards is constant and the rotation around the circle is also constant might be unwieldy)

hello Imnotthisperson

hmm, this seems to be incorrect: "it only defines an inner product on V when $\phi$ is"; looks to me like injectivity is necessary and sufficient.

Raven Dreamer

@anon I think I'm either missing a trig function or an angle in my equation. At the moment, I appear to be rolling an oval, rather than the circle (the cycloid) that I was aiming for.

anon

Also, I should mention the second line does not follow from the first line. The differential equation given just represents a parabola. || You say you appear to be rolling an oval, but since $r_1$ is the same in both components x and y it should be a circle. Is this appearance just your subjective impression or ... ?

(Note that we can use $\LaTeX$ in the chatroom using a simple bookmarklet the current room owner robjohn made, here, if you're into that sort of thing.)

Raven Dreamer

6:19 AM

I know it's an oval, because when I change the radius, the x values fluctuate more than the Y. (I have two objects following the same path, supposedly differing only in the radius.)

So there's definitely a bug in my equation.

anon

Err, I assumed the first line was a DE, but I guess it might represent the circle center.

@anon : )...

Hey @MattN

Hey anon! : )

Oops. Missing variable. That'd do it.

anon

6:22 AM

Indeed x2. :-)

Mike

Since this is a math chat room, I figure I might as well ask. Anyone else here work on Project Euler problems?

anon

Some of the problems end up on MSE. I'm not a coder so I'm not personally familiar.

Raven Dreamer

Well, it looks like I had the answer originally, it just a) didn't look like I was expecting (which implies the path I'm looking to recreate is not quite circular + linear movement), and b) had to futz with the initial parameters a bit.

Mike

Been working on problem 149 there. Got a program up and running, answer's not being accepted. Troubleshooting's been a pain. I've manipulated the program to work on a smaller size I can verify by hand and I can't find anything wrong. Was just curious if I could find someone who's done the problem that I could pm my answer to. Knowing whether I'm too high or too low would help immensely. I've posted in that problem's topic on their message board, currently awaiting any responses.

anon

lol, "knowing whether I'm too high." I like how the rest of that sentence is on the next line.

bah, that doesn't look like a fun problem

Mike

6:34 AM

I came up with a method I thought would work. I've verified by hand most of the values on the 10x10 upper left corner. I'd do a larger size, but I can only fit about 10 numbers in a row on a page printed in landscape. It's been a lot of work though.

Best case scenario is my answer's too high. Then I can take a closer look at the value I've obtained and verify or disprove whether that sum is possible.

Kannappan Sampath

Hi @anon

anon

hey kanna

Kannappan Sampath

I'd like a gravatar...I am in the process of finding a replacement which will still keep the gravatar unique and different.

anon

I just fired up the mspaint and made curvy lines. My usual MO is to rip something off of DeviantArt though.

Kannappan Sampath

MO?

anon

6:41 AM

modus operandi

Kannappan Sampath

Oh!!! That is also abbreviated these days, sigh.

Raven Dreamer

You hear it all the time on Crime Dramas.

Kannappan Sampath

hear what?

Raven Dreamer

MO.

12 mins ago, by anon

modus operandi

Kannappan Sampath

Oh, OK.....

Benjamin Lim

7:05 AM

@anon What did you say about me having exactly 4000 rep?

anon

At one point you had exactly 4k rep.

Benjamin Lim

ah ok

hahaha

anon

On the dot.

Benjamin Lim

that was like

4000 exactly yeah

anon

woops, I meant to remove a comment on the mainsite, not here

Kannappan Sampath

7:09 AM

!!!!!!!111

anon

apparently the wolfgang guy in henning's linked comment is actually a professor or something

lol, your old grav is still in your profile

Matt N.

Hey if you look at this guy's questions and answers they're all more or less high school maths. And yet, there he is, commenting on a general topology question. I wonder whether he can actually answer the question he's commenting on.

Kannappan Sampath

@MattN The Chaz AFAIK is a graduate student.

Benjamin Lim

@KannappanSampath The Chaz is a graduate student?

Matt N.

@KannappanSampath Grad of what?

Kannappan Sampath

7:16 AM

No?

(Or the teddy knows him well...)

Matt N.

I don't know and I don't want to know but all he seems to be doing on this site is high school maths. Unless he's writing a comment of the form "what have you tried" / "why don't you read wikipedia and get some intuition" / etc

:4314631 Exactly.

@BenjaminLim Now you're my friend : D

Kannappan Sampath

@MattN I see.:)

Matt N.

Yes.

I think there should be a mechanism on this site that only allows people to comment on topics that they actually understand.

Kannappan Sampath

Then, I cannot post Arturo's message to the newcomers here, because I don't understand most of things too.

Matt N.

Of course I won't. : )

Ok, bbl! : )

Mike

7:25 AM

I think this book I recently checked out assumes some knowledge I don't have. Any idea what a "free product" is?

Kannappan Sampath

@Mike of what?

free product of what?

(of groups is quite common.)

Mike

The book mentions that some group can be constructed as the free product of 2 generators.

Foool

Hey guys!

Kannappan Sampath

Hi @Foool

Foool

Hey Kannappan!

I am have riddle that's seems interesting

1 + 1 = 3
2 + 2 = 3
3 + 3 = 3
4 + 4 = 2
5 + 5 = ?

My friend gave me some more terms:
5 + 5 = 3
6 + 6 = 2
7 + 7 = 2
8 + 8 = 2
9 + 9 = 4

Kannappan Sampath

7:33 AM

So, now what should we evaluate?

anon

5+5, apparently

oh, ...

Foool

The solution.

anon

you mean the intended pattern?

Foool

Yep, that's right.

Kannappan Sampath

BTW, the new gravatar...@anon How does this look?

anon

7:36 AM

For now, it makes me confuse you with Benjamin :-)

Oh no, it's spoonwood bait. Shudder.

Kannappan Sampath

!!!!! So uniqueness is lost!!!!

I seriously wonder why people don't understand they can be pleasant on the internet.

Read his comments, they would just turn me the other way around---Now, I don't want to help him realise anything.

Anyway, there are other users who would take care of this question. It's time for lunch here.

Bye people.

anon

@Kanna If you're talking about Spoonwood, lots of users haven't been pleasant to him. His being obtuse is pretty grating though.

Kannappan Sampath

@anon who is this spoonwood.

I was talking about Ranadheer

anon

here's a representative sample :P

user19161

So much drama again.

user19161

7:47 AM

@Foool 5+5=10. QED.

Kannappan Sampath

@JasperLoy where..

BBL.

Matt N.

@Mike Here.

user19161

@KannappanSampath I mean what you and Anon are discussing now.

Mike

@Matt Thanks. I'll look at that some more once I'm awake. I hope this book is going to prove useful.

I swear, if I have any brain cells left that haven't exploded by the time I figure out this class of problems, it will be a miracle...

Kannappan Sampath

8:24 AM

How do I enter matrices in W| A?

Mike

What's "W| A"?

Kannappan Sampath

Wolfram Alpha

Mike

Well, I never really used the site, but I found this: wolframalpha.com/examples/Matrices.html

Kannappan Sampath

@Mike Thank you. : )

Benjamin Lim

@KannappanSampath Please change your gravatar, it is very similar to mine!

Mike

8:36 AM

What in the world is a gravatar?

Kannappan Sampath

@BenjaminLim Will do later OK?

Mike

Ah well, looks like I'm not accomplishing much staying awake. Good night everyone.

Rajesh D

11:39 AM

Why do i see less people in here now a days?

where have the "usual suspects" gone?

2

user19161

12:00 PM

@RajeshD Who are the usual suspects? There are usually fewer people during the weekends.

user19161

@RajeshD "Nowadays" is one word, not three.

Rajesh D

Hi @Jasper

i was just word playing

N3buchadnezzar

12:29 PM

Heya, I know you guys hate those modular questions, but yeah. I will try to just ask this one thing. I am trying to help my friend with a few problems

The problem is to show that $50^{8}+1$ divides $13$. My first hunch was to use euler eg $50^{12} \equiv 1 \pmod{13}$, any hints on how to proceed from here?

Now I feel silly 50^8 + 1 does not divide 13, how silly of me.

Kannappan Sampath

1:10 PM

Hah! Capped!

user19161

@KannappanSampath Congrats!

Kannappan Sampath

Thank you.

user19161

@N3buchadnezzar I think you mean to say that 13 divides that. When we say a divides b we mean b is divisible by a.

N3buchadnezzar

1:26 PM

@JasperLoy Gah, English! bah!

user19161

@N3buchadnezzar Yeah you will get used to it when you read more math books in English.

N3buchadnezzar

I actually know these things, it s just the internet melt down my brain.

As of now all my mathbooks are in English, although my philosophy and physic book are still in Norwegian. At least for this term.

user19161

@N3buchadnezzar physics

N3buchadnezzar

Singular?

skullpatrol

yup with an "s".

user19161

1:34 PM

@N3buchadnezzar The subject is called physics, not physic.

user19161

Similarly we have mathematics, not mathematic.

N3buchadnezzar

Perhaps I should take a course in English as well. :$

skullpatrol

1:46 PM

@N3buchadnezzar Latin physica, plural, natural science, from Greek physika, from neuter plural of physikos of nature, from physis growth, nature, from phyein to bring forth

user19161

2:04 PM

@N3buchadnezzar :$ is new to me.

N3buchadnezzar

@JasperLoy See here. 12th from the top.

skullpatrol

embarrassed?

N3buchadnezzar

Indeed, does none of you still use MSN? I feel so old now!

user19161

@N3buchadnezzar I never used it!

N3buchadnezzar

!

Kannappan Sampath

3:11 PM

@JasperLoy \'

Zhen Lin

it seems like all the people who used to use MSN are now on Facebook chat or whatever...

N3buchadnezzar

Never!

leo

does $$\frac{1}{x+\frac{1}{n}}\to \frac{1}{x}$$ uniformly?

Matt N.

3:29 PM

@leo On what? Have you looked at the graph(s)?

leo

@MattN in $[0,1]$

Henning Makholm

@leo If 0 is in the domain, then the function you suggest as a limit does not exist at all!

Matt N.

@leo One would have to work out the details to verify this but looking at the graphs makes it seem that the answer is no because 1/x has a singularity at zero and the f_n don't so 1/x is always very far away from f_n at zero.

leo

@HenningMakholm yep, you right

@MattN I see

thanks

Matt N.

Did you get the domain wrong?

Henning Makholm

3:33 PM

Even if we correct the domain to $(0,1]$, @Matt's argument still applies, suitably rephrased.

Matt N.

Yes but I was thinking of $[1,2]$ or $[1, \infty)$ or something like that.

: )

Matt N.

3:58 PM

Call for votes. (Look at the "answer".)

N3buchadnezzar

Is the trapezoid rule given as $$ T(n) = \frac{b-a}{2n}\sum_{k=1}^{n} f\left( \frac{b-a}{2}k + \frac{1}{2}\right) $$ ?

Where $ x \in[a,b] $ and we use $n$ sub-intervals.

Henning Makholm

@MattN The linked "answer" is identical to a deleted MSE answer on that very question.

Matt N.

@HenningMakholm I see.

Henning Makholm

Which strongly suggests an inference that "Belsa Zarkin" is the same person as "Wolfgang Mueckenheim".

Matt N.

@HenningMakholm Yes, Asaf and I have been discussing this just now.

But we don't want to jump to conclusions. : )

Henning Makholm

4:06 PM

It's more dignified to "suggest inferences" than to "jump to conclusions".

Matt N.

Yes. : )

leo

Do you know some interesting application of group theory to biology, microbiology or medicine?

Hi all!

@robjohn SHoo!

hey

4:17 PM

@robjohn Hello there : )

@N3buchadnezzar eh?

Hi.

bbl

@robjohn "Fee-fi-fo-fum, I smell the blood of a squared Chat-owner"

robjohn

@N3buchadnezzar That does not look like the trapezoidal rule since usually the trapezoidal rule weights the endpoints half as much as the interior points.

It doesn't use the middle point of the interval

afair

@N3buchadnezzar See here

N3buchadnezzar

4:23 PM

@robjohn I figured it out, I left my books at school and trying to figure things out from my notes.

robjohn

It averages the values at the endpoints rather than use the midpoint.

N3buchadnezzar

It seems I quoted the midrule instead, silly me!

robjohn

@N3buchadnezzar okay, sorry for the spam :-)

N3buchadnezzar

@robjohn I do not think you are the one who has to apologize for that :$

robjohn

@Henning: have you capped yet? ;-p

I see you're 30 points away

Henning Makholm

4:27 PM

@robjohn It's slower than yesterday, but decent for me these days, I think.

robjohn

I have a busy day today, I don't think I will be able to work enough here to get many points. Life has been busy recently.I am 40 points away from 23456, however :-)

Henning Makholm

185 it will be. I have to go. Seeya.

robjohn

@HenningMakholm that's what you think :-)

Oh, look... @Henning has capped.

I have to go and deal with things. bbl

leo

4:48 PM

someone

?

Transcript for

$Mathematics$ Mathematics