Mathematics - 2014-08-21 (page 1 of 4)

Balarka Sen

12:00 AM

Try standard texts, like Hardy-Wright.

MrWho

@BalarkaSen Yeah, but at first, she was a Number theorist.

I don't know, what leaded her to Geo

Balarka Sen

books by people who become geometer after studying number theory are most dangerous for health.

Khallil

$$ \int_{0}^{\frac{\pi}{2}} \dfrac{\log (\sec (x))}{\tan (x)} \text{ d}x = -\int_{0}^{\frac{\pi}{2}} \dfrac{\log(\cos(x)) \cdot \cos (x)}{\sin(x)} \text{ d}x$$

Balarka Sen

she is a hard-core hyperbolic geometer. sheeesh.

MrWho

@robjohn When do you get Field Medal?

Balarka Sen

12:03 AM

i gotta run. need some serious sleep.

MrWho

@BalarkaSen Bye, good night

Khallil

@BalarkaSen Peace out, man!

Jasper Loy

12:27 AM

@MrWho He is too old to get it.

MrWho

@JasperLoy Really?

@JasperLoy He has the potential, though.

Jasper Loy

@MrWho Just draw a circle with centre at the point the angle is formed. Then consider the ratio of the arc subtended by the angle and the circumference of the circle. Multiply this by 360 degrees and you get your angle. This is independent of the radius of the circle.

MrWho

@JasperLoy How did they understand that the degree of the circle is $360$, why not $400$?

skullpatrol

12:49 AM

There is no way to really "understand" a choice made by convention, you just get used to it :-)

MrWho

1:09 AM

@skullpatrol I knew it is a convention, wanted to be sure.

Alex J Best

1:41 AM

@MrWho there is a system called gradians where the circle is divided into 400ths, but most people use the standard 360 degrees.

Gradian

The gradian is a unit of plane angle, equivalent to 1⁄400 of a turn. It is also known as gon, grad, or grade. One grad equals 9⁄10 of a degree or π⁄200 of a radian. In continental Europe, the French term centigrade was in use for one hundredth of a grad, and the term myriograde was in use for one ten-thousandth of a grad. This was one reason for the adoption of the term Celsius to replace centigrade as the name of the temperature scale. == History == The unit originated in France as the grade, along with the metric system. Due to confusion with existing grad(e) units of northern Europe, the name...

MrWho

2:53 AM

Where is the best place on SE to start solving simple and decent analytic geometry + integrals?

is there any category for the integral and question levels?

@robjohn

@TedShifrin Hi Ted

@skullpatrol

Jasper Loy

4:02 AM

@MrWho I thought you have enough of these in your calculus books already.

skullpatrol

5:00 AM

Did you see the you tube clip I posted about math research @robjohn?

robjohn

5:40 AM

@skullpatrol No... where was that? I can look in the history...

@skullpatrol Is it one of these?

@skullpatrol The most recent is very true.

skullpatrol

@robjohn yes, the most recent one :-)

he's from Princeton also

:-OMG! four pages of videos

rehband

6:02 AM

Hey guys

skullpatrol

hi

rehband

I'm confused about when a matrix is positive/negative definite

A symmetric matrix is positive definite $\iff$ all leading principle minors are positive, right?

Do we have a necessary and sufficient condition for a symmetric matrix to be negative definite?

For example is the following matrix negative definite? $$\left( \begin{matrix}-6&2\\2&-2\\ \end{matrix} \right)$$

skullpatrol

Have you read this?

rehband

Ok, so the odd principal minors need to be negative and the even ones positive for a matrix to be negative definite

Alec Teal

6:19 AM

Hey just quickly, can someone confirm this comment: math.stackexchange.com/questions/903855/…

Mike Miller

6:39 AM

@AlecTeal What are you referring to

Alec Teal

The $f^{-1}$ intersection thing Daniel asserts

Mike Miller

yikes

looks gross

Alec Teal

You see what I say in the next comment

I have proved that if you have $f^{-1}$ instead of $f$ it's not a subset any more, but an equality there.

I want that confirmed.

Mike Miller

I am afraid I am too much of a coward to do the deed, and those symbols scare me

Balarka Sen

Hello, @Mike.

Mike Miller

6:45 AM

mystical greetings

skullpatrol

symbols scare you? :)

hi btw

Alec Teal

Hi

Mike Miller

symbols are terrifying

Balarka Sen

symbols are terrifying if they can't be visualized.

@AlecTeal the "sheet" word quickly draws attention to anyone familiar with a little algebraic topology. =P

Alec Teal

Haha

What I don't get @MikeMiller and @BalarkaSen and @skullpatrol is why some real analysis homework will get 70 views, 5 answers and 10 upvotes, but that, a genuine not homework question gets little attention

Balarka Sen

7:02 AM

which one are you referring to?

Alec Teal

The question I linked the comment to

Balarka Sen

I see 23 views, 0 answers and 1 upvote.

skullpatrol

well, homework questions are on tests

Jasper Loy

7:31 AM

@AlecTeal There is nothing to get. It is all random.

Jasper Loy

7:43 AM

@MikeMiller Do they usually teach topological vector spaces in the first year grad analysis sequence, or just Banach and Hilbert ones?

Mike Miller

Usually Banach is focused on

skullpatrol

@MikeMiller did you see the math research clip i posted?

Mike Miller

Nope

Balarka Sen

I think I have an allergy for the math in wikipedia. They disfigure some facts in a sophisticated manner in the process of briefing the theories; usually hard to catch but doesn't fool the eyes of the one who studied the topic rigorously.

Mike Miller

@skullpatrol I was kind of hoping it would be a variety of people answering the question

Balarka Sen

7:57 AM

Even though not more popular than wiki, I much like mathworld in this business. They swiftly maintain the "whereof one cannot speak, thereon one must be silent" policy.

Mike Miller

Wittgenstein fan, @Balarka?

Balarka Sen

Sure am

Mike Miller

You're too young to be reading stuff like that. It'll mess you up, make you a logician. :)

Balarka Sen

=P

I have self-discipline about everything except MSE chat, don't worry.

robjohn

Jeez... I was chastised for answering a question in a similar (but I think better) way as someone else. I had started before they posted, and didn't see their post until I posted. Excuuuse me!

I deleted soon after I posted, when I read the other answer.

Mike Miller

8:01 AM

Ooh, I hit 5k

skullpatrol

congrats

Balarka Sen

@Mike virtual beers

robjohn

@MikeMiller Congratulations!

Mike Miller

Thanks, @robjohn

@BalarkaSen I'll virtually practice moderation

robjohn

@MikeMiller drink virtually responsibly.

2

Mike Miller

8:03 AM

:)

robjohn

friends don't let friends chat drunk...

skullpatrol

drunks chat the most

robjohn

when people are stupid drunk, they get designated drivers. I think when people are stupid in comments, they should get designated writers.

skullpatrol

what would be the breathalyzer test for a stupid comment?

robjohn

@skullpatrol can they resist replying to sensible retorts?

skullpatrol

8:10 AM

@robjohn I agree.

(sorry, I couldn't resist :)

Alec Teal

stackoverflow.com/questions/25420638/… no reward for efforts!

Mike Miller

@robjohn why was this bumped? not a complaint (though I don't like trivial edits to bump questions), but I can't tell why it's up again now... nothing seems to have been edited

robjohn

@MikeMiller the ~~spilling~~ spelling was fixed (accept -> except)

Mike Miller

ah, the edit hadn't appeared for me until I refreshed. thanks.

Balarka Sen

@robjohn how did you do that strikethrough thingy?

skullpatrol

8:17 AM

--- strike ---

no spaces^

Balarka Sen

~~strike~~

ah, cool

i plan to use, misuse and abuse this.

skullpatrol

it is a ~~cool~~ fun feature :)

Martin Sleziak

8:43 AM

It is mentioned in the faq, but I did not know about possibility of using ~~strikethrough~~ in chat. I suppose that <s>strike</s> <s>strike</s> does not work in chat, although it does work on the main.

Jasper Loy

9:47 AM

Where is Ted? I would like to ask him about DG.

N3buchadnezzar

Heya

Jasper Loy

@N3buchadnezzar No meat, no pudding.

N3buchadnezzar

Let a function from the set $ \{ f \in C^1 (\mathbb{R},\mathbb{R}) \, ; \ f(0)=0 \}$ to the set $C(\mathbb{R},\mathbb{R})$ be defined by $f \mapsto f'$. Show that this function is a bijection

Any hints? I know I have to prove that the mapping is surjective and injective, but not exactly sure how to start >^.^>

Karl Kronenfeld

10:19 AM

@N3buchadnezzar You could also give an explicit inverse. :)

Balarka Sen

@Karl!

Karl Kronenfeld

Balarka?

Balarka Sen

It's been some time since you visited here.

Karl Kronenfeld

~14 days

Balarka Sen

Right.

@Karl Did you see my new formulation of galois theory in the context of groups?

Karl Kronenfeld

10:32 AM

@BalarkaSen Hm

no

Balarka Sen

here goes : define Aut_H(G) to the auts of G which fixes H pointwise (G is a group and H <= G). the main theorem is that if N <= H <= G and N is a char subgroup of H which is in turn a char subgroup of G then there is a left-exact sequence 1 --> Aut_H(G) --> Aut_N(G) --> Aut_N(H)

the auts stuffs are a mimicry of galois groups and the char subgroups are a mimicry of galois extensions.

@Karl i won't bother giving a proof if you believe what i claimed

Karl Kronenfeld

What's the last homomorphism?

Balarka Sen

you mean the third

Karl Kronenfeld

I only enumerate nontrivial ones, confusingly.

Balarka Sen

it's the restriction map Aut(G) --> Aut(H) itself restricted to Aut_N(G)

ps : well defined, as H is char to G

Karl Kronenfeld

10:44 AM

Oh, I missed that somehow

Then, I don't need to see your proof.

Balarka Sen

that's all the proof is about, man

Karl Kronenfeld

About reading your hypotheses carefully?

Balarka Sen

which hypothesis?

Karl Kronenfeld

H char G is the one I missed initially..

Balarka Sen

no, no, i mean there is nothing more to the proof than to produce the last morphism.

@KarlKronenfeld yeah, essentially so

what d'you think of this? proof's pretty elementary, but i am curious that something like galois theory which is defined over fields (an abstract object with terribly many structure) can be also defined over groups -- an object with far fewer structures.

essentially the action of Aut stuffs over the groups are precisely some kind of galois action.

Chris's sis

11:15 AM

Greetings

skullpatrol

Greetings my friend

Chris's sis

@skullpatrol Hey, how are you doing? :-)

skullpatrol

@Chris'ssis Fine thanks, how are you?

Chris's sis

@skullpatrol I was trying to arrange a bit my collection of problems (some thousands). This is going to be a hard task.

skullpatrol

@Chris'ssis it sounds very time consuming...

Chris's sis

11:22 AM

I also work on some new problems & do some research.

@skullpatrol I wish to publish a fantastic problems book.

skullpatrol

nice

Chris's sis

@skullpatrol I mean to be more than an usual book, to inspire you and make you fall in love with the limits, series and integrals.

skullpatrol

@Chris'ssis I found out yesterday on youtube there is a now a "Calculus for dummies" published with 1001 worked out examples.

Chris's sis

The reader to say at almost every page something like: "wow, that's amazing".

@skullpatrol Where? Do you have the link?

skullpatrol

►

Chris's sis

11:31 AM

@skullpatrol I see. This is the kind of problems I wanna add to my books (an example).

@skullpatrol The proof is elementary, but tools used are incredibly powerful.

skullpatrol

interesting

nullgeppetto

Hi all! Question: Could we define a space $\mathbb{R}^{n} \times \mathbb{R}^{n\times n}$ for which we may define some algebraic operations (is this term correct?), such as summation, multiplication, etc? I am thinking something like the set of complex numbers, $\mathbb{C}$... Please help!

Balarka Sen

11:52 AM

Thought this would be fun to read

wait a sec isn't A_n a bit too obvious? Q(s_1, s_2, \cdots, s_n)/Q(\sqrt{\delta})?

oops i mean Q(s_1, s_2, \cdots, s_n)/Q(\sqrt{\delta}, a_1, a_2, \cdots, a_n)

Khallil

12:08 PM

Now I know how to strike through text and all that jazz.

Thanks, @MartinSleziak!

Math is ~~hard~~ worth the struggle.

Testing out the link.

Balarka Sen

@Khallil $|\{X \subseteq B : |X| \leq 1\}| = |B| + 1$

Khallil

Hmm, I don't get why that's true.

Balarka Sen

Try with an example. $\{1, 2, 3\}$. Subsets are $\varnothing, \{1\}, \{2\}, \{3\}$

Khallil

Oh, I forgot about the empty set.

Balarka Sen

Duh

I have been telling you that for years

Khallil

12:17 PM

Haha! It's one of those things that I'll eventually shake off.

Balarka Sen

@Khallil Given a set $|A| = n$, what is the cardinality of $A + A = \{x + y : x \in A, y \in A\}$?

$A \subset \Bbb N$ and is finite, btw

otherwise it doesn't make sense =P

@Khallil What makes you think so much?

Khallil

Sorry! I was just making my breakfast.

The wonders of Weetabix.

wat

Click the link!

no wish to do that

12:25 PM

It's a type of cereal.

You don't have to if you don't want to.

But do it!

Urm, your question.

Balarka Sen

click

ugh

Khallil

What's wrong with Weetabix?

Balarka Sen

erm i don't get quite well with cereal breakfasts but nevermind, just look at the question.

Khallil

Looking!

Well, if I had the set $A=\{1,2,3\}$.

Balarka Sen

no, no, no.

Khallil

12:28 PM

I'm just taking an example!

Balarka Sen

i wouldn't try looking at examples.

it's no use.

you have to believe that it's of no use. don't start with examples.

Khallil

Fine. Well, for each $x \in A$, there'd be $n$ other elements to choose for each $x+y \in A+A$.

There are $n$ $x$ to choose from, so ...

Balarka Sen

yes

Khallil

You'd have $\underbrace{n \times n \times \dots \times n}_{n \text{ times}}$ elements.

Which means that $|A+A| = n^n$?

Balarka Sen

not quite.

how are you multiplying all that stuff?

Khallil

12:31 PM

With multiplication.

Balarka Sen

no, why are are you doing it?

Khallil

Erm, well ...

Balarka Sen

look, how many elts of A do you have?

Khallil

$n$

Balarka Sen

so how many ways can you pick $x$ from them?

Khallil

12:34 PM

$n$ ways

Balarka Sen

any how many ways you can pick $y$ from them?

Khallil

$n$ ways

Balarka Sen

so how many ways can a set of $n$ people and another set of $n$ people shake hands bijectively?

Khallil

Oh!

Hold on, I was going to say $n^2$ but what does bijectively mean?

Is it that one person can only shake hands with the same person once and that they can't shake hands with their own group?

Balarka Sen

yes

Khallil

12:37 PM

I made a little edit. They can't shake hands with their own group, right?

yes

Cool, so it's $n^2$.

no it's not.

Awww, wat?

each person from set1 shakes hands with only one person with set2.

that's what bijective meant, no?

Khallil

12:38 PM

Oh, right.

Balarka Sen

wait no.

i think you were right.

Khallil

I was?

When?

At the very beginning?

Balarka Sen

now i am confused gah. let me unpin some tabs from the top bar

they are distracting me.

ok, so you want $\{x + y : x, y \in A\}$

$x$ and $y$ are not necessarily distinct.

Khallil

That's what I thought.

The 'not necessarily distinct' part.

Balarka Sen

ah

so the answer is?

Khallil

12:43 PM

That means that for each $x$, there are $n$ other elements, $y$, to pair it up with.

We have $n$ of these such $x$.

So it's $n^2$?

Balarka Sen

chuckles that was fun

you are essentially right, but not quite. i was deceiving you.

Khallil

Oh come on. Your deceptive tactics are turning my brain into mush.

Balarka Sen

what you wrote is the cardinality of $A \times A = \{(x, y) : x, y \in A\}$

try an example.

Khallil

Which is what I wanted to do from the start.

Balarka Sen

$\{1, 2\} + \{1, 2\} = \{1+1, 1+2, 2+1, 2+2\} = \{2, 3, 4\}$

@Khallil As I said, I was deceiving you.

Khallil

12:46 PM

$A=\{ 1,2,3 \}$ means that $A+A=\{ x+y : x, y \in A \}$.

Balarka Sen

the whole problem with your abstract thinking is that addition is commutative =P

N3buchadnezzar

Hmmm

Khallil

$\begin{aligned} \{1,2,3\} + \{1,2,3\} & = \{1+1, 1+2, 1+3, 2+1, 2+2, 2+3, 3+1, 3+2, 3+3\} \\ & = \{ 2, 3, 4, 3, 4, 5, 4, 5, 6\} \\ & = \{2, 3, 4, 5, 6\} \end{aligned}$.

$\therefore |\{1,2,3\} + \{1,2,3\}| = 5$

Balarka Sen

heh

Khallil

Oh, I see.

Commutative being?

Balarka Sen

12:49 PM

a + b = b + a

N3buchadnezzar

$ \displaystyle I_n = \int_{[0,1]^n} \frac{\log(1+x_1 \cdots x_n)}{1-x_1\cdots x_n} \,\mathrm{d}x_1 \cdots \mathrm{d}x_n = ?$

Khallil

Oh, I see.

But addition is commutative.

Balarka Sen

that's what i said

Khallil

Oh, I reread it. I added a few words to what you said by accident!

Balarka Sen

you can't treat $A + A$ like $A \times A$

FYI, @Khallil, it's a problem in additive combinatorics to count the number of elts in $A + A$. You can give it a go if you want, it's not that hard, but careful with your flow of thoughts.

Khallil

12:51 PM

Because ordered pairs are ordered whereas addition doesn't need to be ordered, it's commutative.

Balarka Sen

yup

Khallil

That's a neat question.

Stuff like that always trips me up.

I'd say my weakest point is combinatorics.

N3buchadnezzar

@Khallil Not girls?

Khallil

I've not read up on, nor have I practiced combinatorics at all.

Balarka Sen

@Khallil Don't ignore counting. It's everywhere needed in mathematics.

Khallil

12:52 PM

I won't! I have up until now, but I won't ignore it any longer!

N3buchadnezzar

dodges into a pack of girls

Jasper Loy

I don't know why Rudin's PMA is so popular, there are several problems with it and it is so expensive.

N3buchadnezzar

Probably shoulda washed mine, smells like R. Kelly's sheets
But shit, it was nine.nine dollars! (Bag it)

Jasper Loy

@N3buchadnezzar I am talking about the hardback of course. But really there are problems with that book. For example, it defines the multiple integral in terms of the iterated integral instead of proving it from the usual definitions.

N3buchadnezzar

Indeed. And the book always uses the shortest proofs possible. Although it is a great book to use as a refference when writing papers and alike ^^

Jasper Loy

1:06 PM

@N3buchadnezzar I am slowly beginning to prefer Lang's books on all topics.

In fact, I am currently thinking of getting 10 of Lang's books...

N3buchadnezzar

:p

Jasper Loy

math.stackexchange.com/questions/905066/… talks as if we know what books she is referring to, lol.

Anyway I have guessed and answered.

N3buchadnezzar

she read your grandpas books
she looks ridiculous

Khallil

N3buchadnezzar

uuuugh

Khallil

1:18 PM

Hahahahaha! I love animated gifs like that one.

Mr. Alien

ohh but he looks more normal though.. I look like a chimp ;)

Khallil

Sodium, I think you look pretty similar. You don't look like a chimp, trust me!

Mr. Alien

:p

Huy

1:31 PM

Do the brackets $[\cdot]$ usually denote the floor or the ceiling function?

Khallil

I thought that $\lfloor ... \rfloor$ denote the floor function and $\lceil ... \rceil$ denoted the ceiling function. Anyway, I've seen $[ ... ]$ denote the floor function more often than not.

user161954

Hm, if I have an equation system like: (x+y)^2 = something (x-y)^2= something else, is it OK to take square roots of both equations and solve the corresponding equation system, or should one be more careful?

Khallil

Be more careful.

$(x+y)^2 = a \implies x^2 + 2xy + y^2 = a$
$(x-y)^2 = b \implies x^2 - 2xy + y^2 = b$

user161954

Isn't it an equivalence?

Khallil

Then I'd try to isolate $xy$ on it's own and rearrange for $x$ or $y$ to sub back in.

What do you mean by 'it' and 'an equivalence'?

user161954

1:36 PM

But taking square roots only gives implication

Khallil

Oh, sorry. I'm really lazy with notation.

user161954

Where you write \implies.

Ah, but x or y could be zero,

Khallil

Very true.

Do you have a specific example in mind?

Or is this just a general query?

nullgeppetto

Where should I look for spaces(?) of the form $\mathbb{R}^{n} \times\mathbb{R}^{n\times n}$? I am not sure where these things belong to... Could you give a hand? thanks!

user161954

I mean, surely (x+y)^2=a implies (x+y)= \pm sqrt(a), shouldn't one be able to apply this to the system of equations and afterwards check the solutipns?

No, x and y are arbitrary.

Khallil

1:41 PM

Yea, I'm sure you could but it'd be length as you'd need to consider each of the cases $\pm a$ and $\pm b$.

user161954

So four cases, really?

Khallil

Yea, but as I said. I'm lazy.

^_^

Huy

(x+y)^2 = 1
(x-y)^2 = 1
x+y = 1
x-y = 1
y = 0
x = 1
or vice versa
but you lost the -1, 0 solutions

Khallil

Did you see my reply to your $[...]$ question, @Huy?

Huy

@Khallil Yes it makes sense I think

in the context

Khallil

1:49 PM

$(1) \qquad (x+y)^2 = 1 \implies x^2 + 2xy + y^2 = 1$
$(2) \qquad (x-y)^2 = 1 \implies x^2 - 2xy + y^2 = 1$
Subtracting $(2)$ from $(1)$ gives $4xy = 0$ which means that either $x$ or $y$ are equal to $0$.
$x=0 \implies y^2 = 1 \implies y = \pm 1$
$y=0 \implies x^2 = 1 \implies x = \pm 1$.

My solutions are $(0,1)$, $(0,-1)$, $(1,0)$ and $(-1,0)$. Is that right, @user161954?

Chris's sis

@robjohn I'm celebrating an infinte product by creating another infinite product. I should do this more often.

Khallil

Continue until infinity. ^_^

Infinite products and sums are such a neat idea, especially when they converge. It's mind-blowing!

Transcript for

$Mathematics$ Mathematics