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

user40730

4:09 PM

short question: a convergent sequence is bounded. so does every bounded sequence have a bounded subsequence? does it apply for all subsequences?

Old John

4:45 PM

@KaliMa Not sure why yellow is the area you want - why not the red part as well?

KaliMa

a<=b ?

Old John

OK - I had forgotten that was one of your inequalities

KaliMa

i dont know if my bounds are right because when i try in wolfram it stops well below X

if you go into wolfram and type "plot -100<=-ab<=100 and a<=b" it seems to graph and stop before 100 on the axes

why is this?

is it supposed to go on forever and taper off at 100?

or is the graph just ending early

tinyurl.com/bq22j4t

Old John

The graphs for $ab=X$ etc should go on - not sure why W|A has stopped them there

Alexei Averchenko

@p05 you mean convergent?

KaliMa

4:51 PM

do they taper off to points?

user40730

@AlexeiAverchenko you mean: every convergent sequence has bounded subsequences?

Alexei Averchenko

no...

Old John

@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 '\'?

Old John

@aDangerousIdea Yes!

anon

@ZERO LaTeX markup

Old John

afk - must eat

a Dangerous Idea

later

anon

afk - exam

a Dangerous Idea

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

@OldJohn Ah. So you take $x=\sqrt{31}/4 \tan \theta+1/4$, and $dx=(\sqrt{31}/4 \tan \theta+1/4)'d\theta$ as per usual. Makes much more sense.

Old John

11:58 PM

I prefer to work out $dx/d\theta$ from the substitution $x- 1/4 = \sqrt{31}/4\tan\theta$

you get $dx/d\theta = \sqrt{31}/4\sec^2\theta$

Limitless

@OldJohn Typo

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$Mathematics$ Mathematics