Mathematics - 2014-12-09 (page 3 of 7)

« first day (1588 days earlier) ← previous day next day → last day (3436 days later) »

2:01 PM

Any reason this has recieved so many views? math.stackexchange.com/questions/9075/integration-of-1-sinxcosx

It's 4 years old

Wow, @Anastasiya-Romanova !

@N3buchadnezzar: Because engineers google a lot before their exams.

@Huy It's mind googling.

Look at the title, No $\LaTeX$

@N3buchadnezzar Well, I can say for sure that it was featured in hot network posts, No $\LaTeX$ in title!

2:05 PM

@Integrator Is that bad? :p

@N3buchadnezzar No Not at all but see this

@Integrator I am an Albatraoz

googles Albatraoz

@N3buchadnezzar No idea what you meant!

@Integrator It's a song

@N3buchadnezzar Because of you

2:12 PM

@Venus Why don't you nominate yourself?

@Integrator Hey!? Is this for real??

@Integrator Because a moderator must be good at integration ^^

@Venus Yup! Don't tell anybody, I'll vote for you!

@Integrator I edit the title though :D

@Integrator Wait!? What is she thinking?

@N3buchadnezzar What about @Anastasiya-Romanova?

2:14 PM

@Integrator He is ok

@N3buchadnezzar She :P

@Venus You edit the title????

@Integrator Iron man

@Integrator Yeah, I just did

refreshing election page

@N3buchadnezzar Iron Man???

@Integrator What did you mean by refreshing?

2:17 PM

@Venus Oops.. Hitting refresh button!

Water is refreshing.

@Venus Still 14 candidates! only!

@Integrator You should make it 15 :D

@Venus I'm Not interested at all!

Me either

2:22 PM

@Venus That makes two of us! :P

Hi five again @Integrator :D

@Venus Hi five! :D

-1

A: 2014 Nominations for moderator on Math.SE

So far the nomination page reveals that female users are less likely to nominate theirselves or to be nominated by the other users, so I decide to take this opportunity to nominate myself as a moderator. Please, do not see this as a gender issue. I am not a feminist and I consider being a female ...

@Integrator She answered that question too?

Haha

Let me see yaw

@Venus Shhhh... We shouldn't say much!

Not it's -2, lol

She changes her ava anyway

2:26 PM

@Venus It was $0$ when I up-voted it!

@Venus She removed pics from her profile :(

Look what have you just done? You make her votes minus. :D

@Integrator You seems sad. Do you like her?

@Venus Not really!

@Venus she's a kid!!!!

@Integrator yeah, she is a kid

@Venus and I'm not :D

But your answer though.

What did you mean by "not really!"? ^^

@Integrator How old are you?

2:31 PM

@Venus I was 14 few days back, now 18!!!

Hi @robjohn Now, I found that the intersection multiplicity is $8$.. Could you take a look at my post? :) math.stackexchange.com/questions/1052007/…

@Venus I shouldn't say much!! :P

@evinda robjohn does that? let me check his tags!

@Integrator Are you still a high school student?

@Venus First year Engineering.

You're a freshman too

2:33 PM

@Venus That makes two of us!

Another hi five!

@Venus what about You?

@Venus Hi five!

I'm a fresh woman :D

@Venus Owww!

Hullo :D

2:39 PM

@Venus We've hit 15! Cheers!

@Integrator I thought I got it, but I didnt. Carelessness :/ how do u do it ?

@Integrator Engineering? What type of engineering?

@TheArtist Computer Science and Engineering!

@Integrator That's great, a double degree....may I ask, which do you find more interesting ?

5

Q: Evaluate $\displaystyle\int_2^3 {\text{d}x\over x \log(x + 5)}$

I have this integral to solve. I have tried both integration by substitution and integration by parts but couldn't solve it. $$\int_2^3 {\text{d}x\over x \log(x + 5)}$$ How do i solve this?

integration definite-integrals

@TheArtist That's only one degree :P

@TheArtist It's B.E (Comp Sci and Engg). B for Bachelor :(

@Integrator Ohhh so one can't do this

@Integrator Ahhh I see

2:44 PM

@TheArtist Neither Mathematica!

@Integrator over here computer science comes as BSc, so they never put it under BE...which means sadly if one wants to do both, he has to do a double degree which takes 5-5.5 years

@TheArtist oh!

@Integrator that's y I assumed that your doing a double degree

Is there a generalized version of C-S ?

Cauchy-S

In arbitrary dimensions

And :O @DanielFischer is not coughing anymore

@Hippalectryon Can you state the version of CS you want to generalize?

2:49 PM

@MikeMiller The "basic" one. $\left<x\mid y\right>\le ||x||\cdot ||y||$

@Integrator jus curious, how is that question off topic? (U flagged it :p )

@Hippalectryon Right, I'm not quite sure what you mean by arbitrary dimensions, then.

Do you want more vectors involved?

@MikeMiller Sorry that was unclear. I meant

yeah

more vectors involved

@MikeMiller I also flag as "off topic" just for the "This question is missing context or other details" reason

Then I dunno. Of course we don't have an 'inner product of three vectors', but maybe we can get some inequality where the LHS is <x,y> + <x,z> + <y,z> or something.

ok

2:57 PM

Quick question

The homomorphism between $Hom(Z\times Z, A)$ and $A\times A$ which is defined as $\eta (\psi)= \psi (1,1)$ is isomorphism, right?

Nah, nope.

Can't be, silly question.

@Studentmath You've got them in the wrong spots.

It should be $\psi(1,1) \times \psi (1,1)$, no?

Oh, no, I misread.

@Studentmath \psi(1,0) \times \psi(0,1)

Oh. That makes more sense..

sigh

I always feel stupid when I exercise on new subjects

@Studentmath Don't worry, you'll feel strong when you've mastered it :D

3:06 PM

@Integrator (+1) too for your answer!

Greetings

@Chris'ssis :D

Think I will try to prove it is isomorphism, sounds like a nice practice

@Studentmath More generally, Hom(A \times B,C) = Hom(A,C) \times Hom(B,C)

You can generalize this to infinite stuff, too; infinite direct sums (in the first variable!) commute with Hom

Short exact sequences.

3:22 PM

@Mike will try the first generalisation, thanks!

I meant there is also a generalization of Hom(A x B, C) \cong Hom(A, C) x Hom(B, C)

Which is?

@Studentmath Something like if A, B (which contains a copy of A), C \cong A/B and D are groups, then Hom(D, A) is inside Hom(D, B) and there is a surjective morphism from Hom(D, C) to Hom(D, B)/Hom(D, A).

Woha, I will start with the first generalisation..

Sheesh. It's hard to intepret short exact sequences in pure english...

@Studentmath There is also a third generalization.

Which consists of a lot of modules stuff.

Whoops I meant abelian groups above.

3:39 PM

I don't get just one thing - why is $[\phi(1,0)\psi(1,0)]\times[\phi(0,1)\psi(0,1)]=[\phi(1,0)\times \phi(0,1)][\psi(1,0)\times \psi(0,1)]$

@Studentmath What is \phi? What is \psi?

I mean, I get it intutively, but not why can I do that

Well, starting from the most basic case - say homomorphisms from $Z\times Z$ to $A$

Wow, I had no idea Bill D was once a moderator...

Who is Bill D?

@TheArtist I didn't flagged it, I voted to close it as off-topic! And so did other 5 people.

3:42 PM

Bill Dubuque.

@Studentmath Bill Dates. He founded Dicrosoft

2

@MikeMiller: What? I thought you've been around longer than I have.

@Huy I've been around just over a year.

@MikeMiller: I know. I checked your profile after you said you had no idea Bill D was once a moderator and realised you've only been around for such a little time.

Did you manage to evaluate that integral, @Integrator?

3:43 PM

@Khallil Which integral?

$$\displaystyle \int_{2}^{3} \dfrac{1}{x \log(x+5)} \text{ d}x$$

@Hippalectryon LOL

Hah, I am soon here for a year. @Hippa :D

@Huy Do you know what happened between him being a moderator and him being suspended? Because that's a big change of status.

@Studentmath :)

3:45 PM

What does the InverseFunction command in Mathematica do?

@MikeMiller: Isn't it possible to read about it anymore?

@Balarka anyhow, any simple reason why I may do that? (since that's what tells me it is homomorphism at all)

@Khallil No!

I don't think it has an elementary antiderivative, @Integrator. The best I could do was find the Maclaurin expansion and evaluate that!

I went through old meta posts once, but they were (probably intentionally) light on details.

3:45 PM

I didn't know Bill Mates was the one who founded Microsoft though.

@Studentmath Why you may do what?

What I wrote above

I don't know what phi and psi are

Don't you mean Bill Gates, @Balarka?

Homomorphisms from $Z\times Z$ to $A$, for example

:/

3:47 PM

@MikeMiller: I don't recall exactly and don't want to spread false information. I remember that he was involved in a public dispute in comments where he wasn't exactly decent.

facebook.com/bill.mates

@Studentmath Why not work with the generalization?

You are looking at Hom(A \times B, C)

I will, but want to get the basic first

@MikeMiller: There was most likely more going on, before and maybe also after, but I don't know the details either.

That's sufficient for me, and noble of you to be wary about what you say, @Huy. Thanks.

3:48 PM

It's equally hard to work with Hom(Z \times Z, C) and Hom(A \times B, C) @Studentmath

Well okay, let me see if I can get that better

Bill Gates founded Gicrosoft then, I presume @Hippa

@BalarkaSen You didn't know that ?

Nope.

In that case I assume I will define the isomorphism as $\psi(a,b) \to \psi(a) \times \psi(b)$

3:50 PM

@Studentmath Heh? How is that an isomorphism?

How is that even a morphism?

$\phi$ is an element of $Hom(A, C)$, not $\phi(a)$

Actually, I will shut up and try to work it out before I speak :P

Think straight about the elements of Hom groups. They are morphisms, not the groups elements.

@BalarkaSen @Twink wouldn't be happy

user134177

hi

user134177

i need help

3:56 PM

everyone of us needs help

@BalarkaSen: No, just the two of you.

user134177

1

Q: Let ($\mathbb{R}^n$,*) be a division algebra over $\mathbb{R}$, $n>1$, $\Rightarrow n$ is even

I want to prove: Let ($\mathbb{R}^n$,*) be a division algebra over $\mathbb{R}$, $n>1$, $\Rightarrow n$ is even. I'm stuck. My thoughts are: I want to use the hairy ball theorem and I want to construct a vector field such that the vector field vanishes nowhere: $v:S^{n-1}\to \mathbb{R}^n$, so $...

algebraic-topology differential-topology

user134177

i'm not sure what simon want to hear

If $\phi$ is homomorphism from $A$ to $C$ and $\psi$ from $B$ to $C$ why won't I just define $\eta(\phi,\psi)=\phi \times \psi$?

As the isomorphism

Sure you can, @Studentmath

3:58 PM

From $Hom(A,C)\times Hom(B,C)$

Ah, okay.

user134177

wants *

See, that wasn't so hard after all @Studentmath.

Gustavo Montano

4:11 PM

Ello.

Nomination season is exciting!

I wonder if [old name]CareBear will go for it

@Hippalectryon see chat.stackexchange.com/transcript/message/18968623#18968623

Hi there
Can anyone help me in this question https://math.stackexchange.com/questions/1059071/proving-a-set-v-is-a-vector-space-in-one-of-the-axiomsis about vector spaces.i already have an answer but is missing one detail for me.
The detail is the question i ask in the post,that is what answer is correct for setting the axiom true.
If anyone could help i would apreciate

Yo @Gustavo

What's up?

Gustavo Montano

@BalarkaSen. How are you mate?

Not too much, you?

4:16 PM

Not bad. Nothing to do, bored off, so playing chess.

Gustavo Montano

Ahh yes. What to do when one is bored.

@KellyBlunie No need to do all of that

Gustavo Montano

Learn new things.

It's been some time. I'm rusty at the useful openings.

@KellyBlunie $V$ is a subspace of $R^2$ right ?

Gustavo Montano

4:17 PM

I have taken the time to learn HTML/CSS/JS. Interesting to say the least. I never learn't how to play chess. How is it?

Not bad.

Gustavo Montano

Do you play online?

Not really. I'm just playing at my machine.

Yes but the text ask me to verify the remaining axioms, and i not sure what is the rigth answer i think is n.2

@KellyBlunie Oh wait I didn't read your request here right. What exactly is your question ? I only see one answer to your question ...

4:19 PM

@Hippalectryon yes

Gustavo Montano

Your machine.

Interesting, I recently made an AI for a board game.

I wonder what the code is for your machine.

I mean computer.

I am trying to prove, show that all axioms are true

It's just the default Windows 7 chess.

Gustavo Montano

if ("Barlarka Sen" wins) blowUp();

4:21 PM

Hey @DanielFischer @anon @BalarkaSen
Do you maybe have an idea how we could find the flexes?

http://math.stackexchange.com/questions/1059418/flexes-of-cubic-curve

Heh.

@KellyBlunie But why do you need that ? You just need to show that $V$ is a sub-vector space of $R^2$ ! Or, is your book explicitly asking for all the axioms ?

Nope, @evinda.

In particular this axiom u+v=v+u where they are vectors

No idea about algebraic geometry.

4:22 PM

Yes it is suggesting that i should try that ,showing

@KellyBlunie Oh then it should be direct.

@BalarkaSen A ok... :)

@KellyBlunie Addition is commutative on $R^2$ right ?

Yes but i am not sure what holds true in my question 1 or 2 (answer)

Because its define by an line equation

@KellyBlunie Well you're just taking a subspace of $R^2$, so the commutativity on $V$ is inherited from the commutativity on $R^2$

4:26 PM

@evinda The only flexes I know are these.

eugh

I got 70% in a mechanics test today

I wrote no words

how silly of me

Quick Q: is the subset sum problem still NP-complete if we assume that all elements of the set are distinct?

It seems like it surely has to be

But I know that particular problem can be solved in certain special cases

(by solved I mean computed in polynomial time)

Words are your friends, @Alizter!

^_^

@Hippalectryon but how do i show that .i have showned 2 ways and both can not be rigth

At same time

@KellyBlunie Oh I see

4:32 PM

@Kelly What's the problem you're having trouble with?

@KellyBlunie The first way is good

user134177

1

Q: Let ($\mathbb{R}^n$,*) be a division algebra over $\mathbb{R}$, $n>1$, $\Rightarrow n$ is even

I want to prove: Let ($\mathbb{R}^n$,*) be a division algebra over $\mathbb{R}$, $n>1$, $\Rightarrow n$ is even. I'm stuck. My thoughts are: I want to use the hairy ball theorem and I want to construct a vector field such that the vector field vanishes nowhere: $v:S^{n-1}\to \mathbb{R}^n$, so $...

algebraic-topology differential-topology

@KellyBlunie In the second way, you assume that $A+B$ is in $V$

Which is another axiom

It's not really related

What i have shown in 2?

So to prove that the subset is commutative i use 1

@robjohn I don't see you on this list math.stackexchange.com/election

4:35 PM

@DanielFischer A ok... :D inflections is an other word that is used... :p

@KellyBlunie Wait it's even weirder

@KellyBlunie I think I got your error

@KellyBlunie You wrote $B+A=(x_2+x_1,y_2+y_1)=a(x_2+x_1)+b(y_2+y_1)$

@KellyBlunie How do you get $(x_2+x_1,y_2+y_1)=a(x_2+x_1)+b(y_2+y_1)$ ? $ax+by=0$

I vote pedro

Me too @Alizter

He is the most honest and most active (review wise)

#Pedro4Prez

4:39 PM

@teadawg1337 its #pedro4mod

But none beats Dr. Sonnhard.

@Alizter In due time, he shall take over the world :P

And kick the integrals out, @teadawg1337

@Hippalectryon That equal its not supose to be there should be a ,

What's wrong with integrals, @Balarka?

4:41 PM

We are anti-integral dudes.

@KellyBlunie Then what exactly are doing in your way (2) ?

I'm very confused. What do you mean, @Hippa?

@Khallil I believe he meant to tag @Kelly

@Khallil Auto tab

Ah!

Thanks @teadawg1337. You da real MVP.

^_^

4:43 PM

Maths Viper Prawn

user image

@Khallil Most Vivacious Polynomial?

I better leave!

@KellyBlunie You're saying $f(a)=f(b)$ hence $a=b$

Have fun, @Integrator!

Almost, @teadawg1337. =P

4:47 PM

@Khallil Most Vicious Polyhedron?

Welcome back @Integrator

MVP, @teadawg1337.

@teadawg1337 Hi!

brb, guys.

@KellyBlunie Basically, what you wrote is true but your reasoning is wrong

@Hippalectryon sorry for the delay, i have corrected the post can you go see what i meant in my 2 answer?

4:51 PM

@KellyBlunie I have seen your edit, hence my last post

Can you explain what is wrong?

@KellyBlunie Let $f$ be our linear function ($ax+by=0$), you're saying $f(A)=f(B)$ so $A=B$

That is only true because a linear function is injective

Yes that what though was true

Well then both are true

They're just two ways that lead to the same result

Hi

4:57 PM

Are both way correct solutions to prove u+v=v+u?
Now thinking back isnt the axiom just refering only to the vectors and not the subspace itself.
And i would only yave to use an equation if only if there was an equatition definition addition in this subpace,i seen acouple of this maybe because this i am confused.

@KellyBlunie For a+b=b+a it's indeed on the vectors

@KellyBlunie I don't really see why you're confused though

@KellyBlunie It's like saying the following :

« first day (1588 days earlier) ← previous day next day → last day (3436 days later) »

Transcript for

Dec '149

$Mathematics$ Mathematics

Associated with Math.SE; for both general discussion & math qu...

join this room
about this room

00:00

06:00

12:00

18:00

all times are UTC

$Mathematics$

site design / logo © 2024 Stack Exchange Inc; legal

mobile