Mathematics - 2012-12-01 (page 2 of 2)

@KaliMa the curve $ab=X$ approaches the horizontal and vertical axes as asymptotes - the further along you go, the closer the curve becomes to the axis, but never actually touches it

user40730

@AlexeiAverchenko well maybe this makes it more clear: I have a convergent sequence in $\mathbb R$. so this sequence is bounded. Now I wanna consider its subsequences. are they bounded?

KaliMa

but i mean at the X threshold, does the whole thing intersect at X to a point, or stop off early/truncate and become a hard line like in the graph

Alexei Averchenko

4:55 PM

yeah

because they are convergent

KaliMa

i assume it gets cut off hard

Old John

@KaliMa sorry - not sure what you mean by "at the X threshold"

user40730

@AlexeiAverchenko but without having that its subsequences are convergent you can't say it?

Alexei Averchenko

sure i can

KaliMa

when a or b = X

Alexei Averchenko

4:57 PM

any subset of a bounded set is bounded

KaliMa

the cutoff point that should have nothing past it

user40730

yeah and the same for the sequences...

Old John

@KaliMa sorry - that is not making any sense to me

user40730

okay

user40730

thx

Alexei Averchenko

4:58 PM

np

Old John

back later folks

MSEoris

5:47 PM

hey

sonicboom

6:28 PM

If you have $x \in A$. And then it turns out A is the empty set, can you say this is a contradiction as this is an impossibility?

a Dangerous Idea

6:59 PM

@sonicboom It just means x can not have any values because there are no members in the empty set, right?

sonicboom

Well the question seems to me to be 'Can we have a point x in the empty set'. Which would be a contradiction as it's wouldn't be empty then!

a Dangerous Idea

But x is a symbol used to represent members of the set.

sonicboom

I should have said 'Can we have an element x in the empty set'.

I think I should've mentioned that $x \in X$ and also $x \in A$. So when A turns out to be the empty set it is a contradiction as we can't have x in X and x in the empty set.

a Dangerous Idea

7:14 PM

Yes, 'Can we have an element in the empty set'. Is a contradiction.

But x is a symbol used to represent members in the set.

It is the members that are in the set, not x.

sonicboom

Yes an x is equal to a member of the set X. The set X being the entire space. So we can't have a member of the set X in the empty set.

anon

putnam = halfway done

second half in T minus 35 min

Old John

@anon how was first half?

anon

There was an easy one I got right. Another one I BS'd on. Other four I got nada.

ZERO

Guys, I have a simple question. Say I have an equation like a^x = a^3. I know it's common sense that allows me to say that x = 3, but how would you prove that?

Jayesh Badwaik

7:26 PM

what is nada?

Old John

@ZERO lake logs of both sides

anon

in the positive reals? logarithm with base a. or, show x->a^x is an increasing function and hence injective.

nada = nil = zilch = $\varnothing$ = 0

Jayesh Badwaik

ahh

in India, nada means the thread which is strung through the pyajamas to tighten them. :P

anon

There was a hard one that basically says the affine group of $\Bbb F_p^n$ has an element for which the cyclic group generated by it acts transitively.

Old John

@JayeshBadwaik I was going to ask what it meant in Indian :)

Jayesh Badwaik

7:28 PM

@OldJohn Indian!! Actually, what I said is in Hindi.

Old John

(I use "Indian" loosely !! _ know there are loads of languages)

Jayesh Badwaik

:-) I know. Actually, it reminded me of a joke by Russell Peters with the same idea.

Old John

@JayeshBadwaik I have a book somewhere entitled "Sanskrit in 30 Days" - I bought it just for the insane title :)

a Dangerous Idea

nada is spanish

ZERO

@OldJohn, could you show me?

Old John

7:30 PM

@ZERO show what?

the book?

ZERO

I would imagine that it would look the same even if you took the logs of both sides

Ilya

@OldJohn Taking into account 21.12.2012, it would be hard to learn Sanskrit even using that book

Jayesh Badwaik

@OldJohn Hahaha. Sanskrit in thirty days! I don't know Sanksrit even an iota BTW. My brother took it for six years though.

@Ilya No... Not again!! This is driving me crazy!! (Everyone is crazy about it)

Old John

ah - if you take logs, you get $x\log a = 3\log a$ (assuming something about $a$) then divide by $\log a$

You need $\log a \ne 0$ - but you can probably get that or look at it as a special case

a Dangerous Idea

@OldJohn Is this the book?

ZERO

7:35 PM

Ack, I feel like a total noob. What are the '$' for and '\'?

@aDangerousIdea Yes!

@ZERO LaTeX markup

afk - must eat

later

afk - exam

7:36 PM

good luck

Ilya

and off they went

a Dangerous Idea

7:54 PM

to see the wizard of Oz

Old John

8:43 PM

sheesh - quiet here tonight

Nimza

9:01 PM

@Ilya ho-ho! What a person! :)

user19161

@OldJohn Nice new picture. Very sexy.

Bean

Does anybody know about Kolmogorov Complexity?

Old John

@WillHunting thought it about time I showed my face

user19161

@OldJohn Yes, considering you are almost at 6k!

Old John

@WillHunting creeping closer :)

and creeping up on the generalist badge, I think

user19161

9:12 PM

Wow @amwhy you are getting closer to 10k! =)

user19161

@anon Good luck! I am with you in spirit.

Nimza

Hm, how to show that $\int\limits_{0}^{\infty} f(x) dg(x) = \langle f(x), g'(x) \rangle$ for Lebesgue-Stiltjes integral?

Peter Tamaroff

10:17 PM

@Nimza What are the hypothesis on $f$ and $g$?

Nimza

@PeterTamaroff $f \in \mathcal{D}(0,\infty)$, $g$ is arbitrary nondecreasing

Peter Tamaroff

If $g$ is differentiable, then $$\int_0^\infty f dg =\int_0^\infty f g' dx=\langle f,g'\rangle$$

@Nimza What is $\mathcal D(A)$?

Nimza

@PeterTamaroff test functions, $C^{\infty}(0,\infty)$ with compact support

Peter Tamaroff

@Nimza What is a test function?

Nimza

@PeterTamaroff base functions on which distributions are defined

Peter Tamaroff

10:20 PM

@Nimza Oh, OK. Distributions are "über function" right?

Nimza

@PeterTamaroff generalized functions, functionals - do you mean that by über function?

Old John

really neat answer

Peter Tamaroff

I just bought "Integral, measure, derivative" and "Elementary complex and real analysis" by Shilov.

Nimza

@PeterTamaroff I didn't read that

Peter Tamaroff

@Nimza über means "over"

Nimza

10:23 PM

@PeterTamaroff aha, but what does mean "over-function"?

Peter Tamaroff

@Nimza I usually use it to mean "super" or "supra".... like übermensch

Nimza

@PeterTamaroff :)

Peter Tamaroff

@OldJohn that looks likea mug shot =P

"I just bought "Integral, measure, derivative" and "Elementary complex and real analysis" by Shilov." @OldJohn Do you know those?

(I also bought Arkham City and a XBOX360 controller for windows, but that's classified procrastrination)

Old John

@PeterTamaroff It is ... a shot of a mug, yes :)

Not sure I know Shilov's book - but I have a few books with titles very much like that :)

Peter Tamaroff

@OldJohn I like unification!

Old John

10:29 PM

Has anyone seen the comment following my comment here?

@PeterTamaroff me too

Peter Tamaroff

@OldJohn HJAAHAHAH yes, I upvoted it

Barcelona won 5 to 1

And Real Madrid is playing right now

Old John

@PeterTamaroff It is actually a continuation of an email conversation will WJ - I imagined that the rest of MSE would be baffled by it :)

Peter Tamaroff

@OldJohn Hahhaah =P

Gotta go.

Bye byes,.

user19161

@PeterTamaroff You might like his Elementary Functional Analysis too.

user19161

@old You choosing to use the side view makes it look very artistic. I wonder if you are looking into the past or the future, or both.

user19161

10:37 PM

@gus Your school is starting soon, yay!

Old John

@WillHunting At the time I was actually looking at a camel I was about to ride ... nothing so symbolic as past and future :)))

user19161

@OldJohn Perhaps I should take one facing the left, then we can face opposite directions!

Old John

@WillHunting facing apart or facing together? :)

user19161

@OldJohn That's exactly what I was thinking. Great minds think alike!

Old John

@WillHunting :)

Argon

10:45 PM

@OldJohn Is that you in your avatar?

user19161

@Argon Hello Aaron!

Argon

@WillHunting Hi Jasper!

How are you?

Old John

@Argon yep

Argon

:) You aren't that old!

Old John

60 is old enough :)

Argon

10:50 PM

Until 120, as they say!

Old John

that photo was 2 weeks ago on holiday in the middle east - in case anyone was thinking it was taken 20 years ago :)

speedy eating

Argon

HAHAHAHA!

user19161

@Argon Bad, as usual.

Argon

@WillHunting "Bad, as usual"? That's not good!

user19161

@Argon Haha, don't you know some of my secrets already?

Argon

10:58 PM

@WillHunting Yes, but still!

user19161

@Argon I hope some kind of miracle will happen soon.

Argon

There can be miracles when you believe.

-Mariah Carey

user19161

Haha, the voters clearly did not understand my answer. math.stackexchange.com/questions/248817/…

user19161

Of course I omitted some details, but I think they are not aware that that is a correct answer.

Old John

@WillHunting Yep - +1

@WillHunting am I right here?

Limitless

11:12 PM

@WillHunting, I am wanting to give a Part 3 to the Hint 2 found here, but I can't figure out how to integrate the integrals with the quadratics in the denominator.

@WillHunting, any tips?

user19161

@OldJohn Yes. Of course, it need not be injective if you allow a point to be represented twice!

Old John

@WillHunting Yeah - I am not totally sure of the exact conditions required - but I thought I needed to correct the impression given in the comments :)

user19161

@Limitless Hmm I can't remember my antiderivatives, but maybe the integrals involving D and E can't be computed in that manner of splitting. One may need to split it up otherwise. Would be good if you try to work it out yourself too.

Limitless

@WillHunting, I have tried everything from Weierstrass substitution to finding if it can match the form $\frac{1}{a^2+u^2}$. It just won't play nicely.

user19161

@Limitless You mean the integral involving D and E?

Limitless

11:22 PM

@WillHunting, indeed.

Argon

Partial fraction decomposition, then integrate each fraction.

Limitless

Nope

user19161

@Limitless Ah I used to do plenety of these integrals in high school. Let me find something to help you.

Limitless

The roots of the denominator are in $\mathbb{C}$.

I don't know complex analysis, @Argon.

Argon

Do you know partial fraction decomposition?

@Limitless It needs no CA

Limitless

11:23 PM

@Argon, CA?

Argon

@Limitless Complex analysis

Limitless

@Argon, how to do you use partial fraction decomposition when the denominator does not factorize over $\mathbb{R}$???

Argon

Let's see

Limitless

Broaden my horizons, @WillHunting! Broaden them, please. :P

Argon

Multiply out denominators to get

user19161

11:25 PM

@limitless What is D and E specifically?

Limitless

@WillHunting, I don't know. I was trying to write this without giving that so that the asker could fill in the details. I'm not trying to post a full solution.

Argon

$$6x^5+x^2+x+2 = (x^2 + 2x + 1)(2x^2 - x + 4)(x+1)\left(\frac{A}{(x+1)^3}+\frac{B}{(x+1)^2}+\frac{C}{x+1}+\frac{Dx+E}{2x^2-x+4}+F\right)$$

user19161

One can use the trick that the antiderivative of f'(x)/f(x) is ln |f(x)|.

Limitless

@Argon, $2x^2-x+4$ has roots $x_{+,-}=\frac{1}{4}\pm \frac{i\sqrt{31}}{4}$. I don't think partial fraction decomposition works here.

@WillHunting, this applies to the first integral, yes?

user19161

One can also try to complete the square in the denominator of the fraction.

Argon

11:28 PM

@Limitless Expand, then then group the powers

It should still work

Old John

does it work if you just multiply out all the brackets and equate coefficients?

Limitless

Hmm, @WillHunting, it would appear that doesn't work on the first integral involving $D$: If you let $f(x)=2x^2-x+4$, then $f'(x)=4x-1\ne x$.

Argon

@OldJohn Yep, this is what I mean

It should work

Limitless

@Argon, I'm not trying to post a full solution. I don't understand what you're saying given that fact.

Old John

I agree - if you have the right expressions, then it has to work :)

user19161

11:30 PM

So for example, if I have $\frac{2x+3}{x^2+x+1}$, I can split it up into $\frac{2x+1}{x^2+x+1}$ and $\frac{2}{x^2+x+1}$.

Argon

@Limitless You simply clear denominators and then group together all the terms of equal powers!

Limitless

@Argon, I'm trying to tell you: I'm not trying to solve for the variables $A,B,C,D,E,$ and $F$.

I'm trying to give a hint as to how to evaluate the following integrals: $\int \frac{1}{2x^2-x+4}dx$ and $\int \frac{x}{2x^2-x+4}dx$. I am sorry this was not clear.

user19161

After that in the second fraction above I can complete the square as $(x+\frac{1}{2})^2+\frac{3}{4}$.

Old John

@Limitless for those integrals, the standard method is to complete the square in the denominator

and then use a trig substitution

Charlie

hmm

Limitless

11:34 PM

@OldJohn, how do I do that? I'm sorry that I don't know how.

I feel like I should know this, but it's not like completing the square in Algebra II.

Old John

$\int \frac{1}{2x^2-x+4}dx = \int\frac{1/2}{x^2-1/2x+2}dx = $

$\int\frac{1/2}{(x-1/4)^2+15/16}dx$ then put ...

user19161

@limitless See my steps above and this article. en.wikipedia.org/wiki/List_of_integrals_of_rational_functions

Limitless

@OldJohn, okay. I see what you're doing there.

Argon

@Charlie hmmm.........................................

Old John

$x- 1/4 = \sqrt{15}/16\tan\theta$

Charlie

11:37 PM

@Argon hmmmm...........................................

Argon

@Limitless I see. For #2, logs will work, no?

@Charlie hhhhhhhhhhhhhmmmmmmmmmmmmmmmmmmmmm.....................

user19161

Also, don't attempt the integral questions until you are really sure you can compute them!

Limitless

@WillHunting, why? It's always a great learning process.

Charlie

@Argon hahahahahahhah

user19161

@Limitless Hmm OK. I mean you should post only when you are sure you can do it yourself, but it's alright.

Limitless

11:39 PM

@WillHunting, Calculus is one of those subjects where I'm never entirely confident in what I'm doing since I just started learning it.

Things are harder than they seem.

user19161

@Limitless Also, I got your email.

Old John

@Limitless I have been doing calculus for 42 years and still feel like that sometimes

Limitless

@WillHunting, wonderful.

Thanks for the help, guys. I'm really trying to get something productive done this weekend. I can't get anything done on the weekdays.

Charlie

@Limitless can anybody?

:P

Limitless

@Charlie, true story. The school I attend is worthless sometimes.

Old John

11:45 PM

every day seems like a weekend when you have retired :)

6

@Limitless correction to my earlier substitution: should probably be $x- 1/4 = \sqrt{15}/4\tan\theta$

Limitless

@OldJohn, I think you messed up here: $\int \frac{\frac{1}{2}}{x^2-1/2x+2}dx=\int\frac{1/2}{(x-1/4)^2+15/16}dx$. I am getting
$$\begin{align}
\int \frac{1}{2x^2-x+4}dx&=\int \frac{\frac{1}{2}}{x^2-\frac{1}{2}x+2}dx\\
&=\frac{1}{2} \int \frac{1}{\left(x-\frac{1}{4}\right)^2+\frac{31}{16}}dx.
\end{align}$$

Old John

@Limitless yeah - $31/16$ - sorry

it is late here, and I am tired :)

so the substitution will be $x- 1/4 = \sqrt{31}/4\tan\theta$

you should end up with $\sec^2\theta d\theta$ cancelling out $\tan^2\theta+1$ on the bottom of the fraction

Limitless

@OldJohn Interesting. We can sub that in and simply change $dx$ to $d\theta$?

And yeah, I noticed that. :)

Old John

no - $dx$ becomes $\sec^2\theta d\theta$ within a rational multiple

to be precise: $\sqrt{31}/4 \sec^2\theta d\theta$

Limitless