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

N3buchadnezzar

14:01

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

Studentmath

It's 4 years old

Integrator

Wow, @Anastasiya-Romanova !

Huy

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

N3buchadnezzar

@Huy It's mind googling.

Integrator

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!

N3buchadnezzar

14:05

@Integrator Is that bad? :p

Integrator

@N3buchadnezzar No Not at all but see this

N3buchadnezzar

@Integrator I am an Albatraoz

Integrator

googles Albatraoz

@N3buchadnezzar No idea what you meant!

N3buchadnezzar

@Integrator It's a song

Venus

@N3buchadnezzar Because of you

Integrator

14:12

@Venus Why don't you nominate yourself?

Venus

@Integrator Hey!? Is this for real??

N3buchadnezzar

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

Integrator

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

Venus

@Integrator I edit the title though :D

@Integrator Wait!? What is she thinking?

Integrator

@N3buchadnezzar What about @Anastasiya-Romanova?

N3buchadnezzar

14:14

@Integrator He is ok

Integrator

@N3buchadnezzar She :P

@Venus You edit the title????

N3buchadnezzar

@Integrator Iron man

Venus

@Integrator Yeah, I just did

Integrator

refreshing election page

@N3buchadnezzar Iron Man???

Venus

@Integrator What did you mean by refreshing?

Integrator

14:17

@Venus Oops.. Hitting refresh button!

Huy

Water is refreshing.

Integrator

@Venus Still 14 candidates! only!

Venus

@Integrator You should make it 15 :D

Integrator

@Venus I'm Not interested at all!

Venus

Me either

Integrator

14:22

@Venus That makes two of us! :P

Venus

Hi five again @Integrator :D

Integrator

@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 ...

Venus

@Integrator She answered that question too?

Haha

Let me see yaw

Integrator

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

Venus

Not it's -2, lol

She changes her ava anyway

Integrator

14:26

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

@Venus She removed pics from her profile :(

Venus

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

@Integrator You seems sad. Do you like her?

Integrator

@Venus Not really!

@Venus she's a kid!!!!

Venus

@Integrator yeah, she is a kid

Integrator

@Venus and I'm not :D

Venus

But your answer though.

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

@Integrator How old are you?

Integrator

14:31

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

evinda

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

Integrator

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

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

Venus

@Integrator Are you still a high school student?

Integrator

@Venus First year Engineering.

Venus

You're a freshman too

Integrator

14:33

@Venus That makes two of us!

Venus

Another hi five!

Integrator

@Venus what about You?

@Venus Hi five!

Venus

I'm a fresh woman :D

Integrator

@Venus Owww!

Hippalectryon

Hullo :D

Integrator

14:39

@Venus We've hit 15! Cheers!

The Artist

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

@Integrator Engineering? What type of engineering?

Integrator

@TheArtist Computer Science and Engineering!

The Artist

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

Integrator

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?

@TheArtist That's only one degree :P

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

The Artist

@Integrator Ohhh so one can't do this

@Integrator Ahhh I see

Integrator

14:44

@TheArtist Neither Mathematica!

The Artist

@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

Integrator

@TheArtist oh!

The Artist

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

Hippalectryon

Is there a generalized version of C-S ?

Cauchy-S

In arbitrary dimensions

And :O @DanielFischer is not coughing anymore

Mike Miller

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

Hippalectryon

14:49

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

The Artist

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

Mike Miller

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

Do you want more vectors involved?

Hippalectryon

@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

Mike Miller

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.

Hippalectryon

ok

Studentmath

14:57

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.

Mike Miller

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

Studentmath

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

Mike Miller

Oh, no, I misread.

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

Studentmath

Oh. That makes more sense..

sigh

I always feel stupid when I exercise on new subjects

Hippalectryon

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

Venus

15:06

@Integrator (+1) too for your answer!

Chris's sis

Greetings

Hippalectryon

@Chris'ssis :D

Studentmath

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

Mike Miller

@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

Balarka Sen

Short exact sequences.

Studentmath

15:22

@Mike will try the first generalisation, thanks!

Balarka Sen

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

Studentmath

Which is?

Balarka Sen

@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).

Studentmath

Woha, I will start with the first generalisation..

Balarka Sen

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.

Studentmath

15:39

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

Balarka Sen

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

Studentmath

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 $A$

Mike Miller

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

Studentmath

Who is Bill D?

Integrator

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

Balarka Sen

15:42

Bill Dubuque.

Hippalectryon

@Studentmath Bill Dates. He founded Dicrosoft

2

Huy

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

Mike Miller

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

Huy

@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.

Khallil

Did you manage to evaluate that integral, @Integrator?

Integrator

15:43

@Khallil Which integral?

Khallil

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

Balarka Sen

@Hippalectryon LOL

Studentmath

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

Mike Miller

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

Hippalectryon

@Studentmath :)

Mats Granvik

15:45

What does the InverseFunction command in Mathematica do?

Huy

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

Studentmath

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

Integrator

@Khallil No!

Khallil

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

Mike Miller

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

Balarka Sen

15:45

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

@Studentmath Why you may do what?

Studentmath

What I wrote above

Balarka Sen

I don't know what phi and psi are

Khallil

Don't you mean Bill Gates, @Balarka?

Studentmath

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

Khallil

:/

Huy

15:47

@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.

Hippalectryon

facebook.com/bill.mates

Balarka Sen

@Studentmath Why not work with the generalization?

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

Studentmath

I will, but want to get the basic first

Huy

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

Mike Miller

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

Balarka Sen

15:48

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

Studentmath

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

Balarka Sen

Bill Gates founded Gicrosoft then, I presume @Hippa

Hippalectryon

@BalarkaSen You didn't know that ?

Balarka Sen

Nope.

Studentmath

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

Balarka Sen

15:50

@Studentmath Heh? How is that an isomorphism?

How is that even a morphism?

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

Studentmath

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

Balarka Sen

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

Hippalectryon

@BalarkaSen @Twink wouldn't be happy

user134177

hi

user134177

i need help

Balarka Sen

15:56

everyone of us needs help

Huy

@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

Studentmath

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

Balarka Sen

Sure you can, @Studentmath

Studentmath

15:58

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

Ah, okay.

user134177

wants *

Balarka Sen

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

Gustavo Montano

16:11

Ello.

Nomination season is exciting!

Hippalectryon

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

Hat Man

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

Kelly Blunie

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

Balarka Sen

Yo @Gustavo

What's up?

Gustavo Montano

@BalarkaSen. How are you mate?

Not too much, you?

Balarka Sen

16:16

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

Gustavo Montano

Ahh yes. What to do when one is bored.

Hippalectryon

@KellyBlunie No need to do all of that

Gustavo Montano

Learn new things.

Balarka Sen

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

Hippalectryon

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

Gustavo Montano

16:17

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?

Balarka Sen

Not bad.

Gustavo Montano

Do you play online?

Balarka Sen

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

Kelly Blunie

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

Hippalectryon

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

Kelly Blunie

16:19

@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.

Balarka Sen

I mean computer.

Kelly Blunie

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

Balarka Sen

It's just the default Windows 7 chess.

Gustavo Montano

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

evinda

16:21

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

Balarka Sen

Heh.

Hippalectryon

@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 ?

Balarka Sen

Nope, @evinda.

Kelly Blunie

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

Balarka Sen

No idea about algebraic geometry.

Kelly Blunie

16:22

Yes it is suggesting that i should try that ,showing

Hippalectryon

@KellyBlunie Oh then it should be direct.

evinda

@BalarkaSen A ok... :)

Hippalectryon

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

Kelly Blunie

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

Because its define by an line equation

Hippalectryon

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

Daniel Fischer

16:26

@evinda The only flexes I know are these.

Alizter

eugh

I got 70% in a mechanics test today

I wrote no words

how silly of me

M.B.

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)

Khallil

Words are your friends, @Alizter!

^_^

Kelly Blunie

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

At same time

Hippalectryon

@KellyBlunie Oh I see

teadawg1337

16:32

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

Hippalectryon

@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

Hippalectryon

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

Which is another axiom

It's not really related

Kelly Blunie

What i have shown in 2?

So to prove that the subset is commutative i use 1

Chris's sis

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

evinda

16:35

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

Hippalectryon

@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 $(x_2+x_1,y_2+y_1)=a(x_2+x_1)+b(y_2+y_1)$ ? $ax+by=0$

Alizter

I vote pedro

Balarka Sen

Me too @Alizter

Alizter

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

teadawg1337

#Pedro4Prez

Alizter

16:39

@teadawg1337 its #pedro4mod

Balarka Sen

But none beats Dr. Sonnhard.

teadawg1337

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

Balarka Sen

And kick the integrals out, @teadawg1337

Kelly Blunie

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

Khallil

What's wrong with integrals, @Balarka?

Balarka Sen

16:41

We are anti-integral dudes.

Hippalectryon

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

Khallil

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

teadawg1337

@Khallil I believe he meant to tag @Kelly

Hippalectryon

@Khallil Auto tab

Khallil

Ah!

Thanks @teadawg1337. You da real MVP.

^_^

Hippalectryon

16:43

Maths Viper Prawn

Khallil

teadawg1337

@Khallil Most Vivacious Polynomial?

Integrator

I better leave!

Hippalectryon

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

Khallil

Have fun, @Integrator!

Almost, @teadawg1337. =P

teadawg1337

16:47

@Khallil Most Vicious Polyhedron?

Welcome back @Integrator

Khallil

MVP, @teadawg1337.

Integrator

@teadawg1337 Hi!

teadawg1337

brb, guys.

Hippalectryon

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

Kelly Blunie

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

Hippalectryon

16:51

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

Kelly Blunie

Can you explain what is wrong?

Hippalectryon

@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

Kelly Blunie

Yes that what though was true

Hippalectryon

Well then both are true

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

Silencer

Hi

Kelly Blunie

16:57

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.

Hippalectryon

@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 :

Transcript for

$Mathematics$ Mathematics