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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?
 
4:45 PM
@KaliMa Not sure why yellow is the area you want - why not the red part as well?
 
a<=b ?
 
OK - I had forgotten that was one of your inequalities
 
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
 
The graphs for $ab=X$ etc should go on - not sure why W|A has stopped them there
 
@p05 you mean convergent?
 
4:51 PM
do they taper off to points?
 
user40730
@AlexeiAverchenko you mean: every convergent sequence has bounded subsequences?
 
@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?
 
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
 
4:55 PM
yeah
because they are convergent
 
i assume it gets cut off hard
 
@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?
 
sure i can
 
when a or b = X
 
4:57 PM
any subset of a bounded set is bounded
 
the cutoff point that should have nothing past it
 
user40730
yeah and the same for the sequences...
 
@KaliMa sorry - that is not making any sense to me
 
user40730
okay
 
user40730
thx
 
4:58 PM
np
 
back later folks
 
5:47 PM
hey
 
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?
 
6:59 PM
@sonicboom It just means x can not have any values because there are no members in the empty set, right?
 
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!
 
But x is a symbol used to represent members of the set.
 
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.
 
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.
 
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.
 
putnam = halfway done
second half in T minus 35 min
 
@anon how was first half?
 
There was an easy one I got right. Another one I BS'd on. Other four I got nada.
 
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?
 
7:26 PM
what is nada?
 
@ZERO lake logs of both sides
 
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
 
ahh
in India, nada means the thread which is strung through the pyajamas to tighten them. :P
 
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.
 
@JayeshBadwaik I was going to ask what it meant in Indian :)
 
7:28 PM
@OldJohn Indian!! Actually, what I said is in Hindi.
 
(I use "Indian" loosely !! _ know there are loads of languages)
 
:-) I know. Actually, it reminded me of a joke by Russell Peters with the same idea.
 
@JayeshBadwaik I have a book somewhere entitled "Sanskrit in 30 Days" - I bought it just for the insane title :)
 
nada is spanish
 
@OldJohn, could you show me?
 
7:30 PM
@ZERO show what?
the book?
 
I would imagine that it would look the same even if you took the logs of both sides
 
@OldJohn Taking into account 21.12.2012, it would be hard to learn Sanskrit even using that book
 
@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)
 
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
 
@OldJohn Is this the book?
 
7:35 PM
Ack, I feel like a total noob. What are the '$' for and '\'?
 
@aDangerousIdea Yes!
 
@ZERO LaTeX markup
 
afk - must eat
 
afk - exam
 
7:36 PM
good luck
 
and off they went
 
7:54 PM
to see the wizard of Oz
 
8:43 PM
sheesh - quiet here tonight
 
9:01 PM
@Ilya ho-ho! What a person! :)
 
user19161
@OldJohn Nice new picture. Very sexy.
 
Does anybody know about Kolmogorov Complexity?
 
@WillHunting thought it about time I showed my face
 
user19161
@OldJohn Yes, considering you are almost at 6k!
 
@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.
 
Hm, how to show that $\int\limits_{0}^{\infty} f(x) dg(x) = \langle f(x), g'(x) \rangle$ for Lebesgue-Stiltjes integral?
 
10:17 PM
@Nimza What are the hypothesis on $f$ and $g$?
 
@PeterTamaroff $f \in \mathcal{D}(0,\infty)$, $g$ is arbitrary nondecreasing
 
If $g$ is differentiable, then $$\int_0^\infty f dg =\int_0^\infty f g' dx=\langle f,g'\rangle$$
@Nimza What is $\mathcal D(A)$?
 
@PeterTamaroff test functions, $C^{\infty}(0,\infty)$ with compact support
 
@Nimza What is a test function?
 
@PeterTamaroff base functions on which distributions are defined
 
10:20 PM
@Nimza Oh, OK. Distributions are "über function" right?
 
@PeterTamaroff generalized functions, functionals - do you mean that by über function?
 
I just bought "Integral, measure, derivative" and "Elementary complex and real analysis" by Shilov.
 
@PeterTamaroff I didn't read that
 
@Nimza über means "over"
 
10:23 PM
@PeterTamaroff aha, but what does mean "over-function"?
 
@Nimza I usually use it to mean "super" or "supra".... like übermensch
 
@PeterTamaroff :)
 
@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)
 
@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 :)
 
@OldJohn I like unification!
 
10:29 PM
Has anyone seen the comment following my comment here?
@PeterTamaroff me too
 
@OldJohn HJAAHAHAH yes, I upvoted it
Barcelona won 5 to 1
And Real Madrid is playing right now
 
@PeterTamaroff It is actually a continuation of an email conversation will WJ - I imagined that the rest of MSE would be baffled by it :)
 
@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!
 
@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!
 
@WillHunting facing apart or facing together? :)
 
user19161
@OldJohn That's exactly what I was thinking. Great minds think alike!
 
@WillHunting :)
 
10:45 PM
@OldJohn Is that you in your avatar?
 
user19161
@Argon Hello Aaron!
 
@WillHunting Hi Jasper!
How are you?
 
@Argon yep
 
:) You aren't that old!
 
60 is old enough :)
 
10:50 PM
Until 120, as they say!
 
that photo was 2 weeks ago on holiday in the middle east - in case anyone was thinking it was taken 20 years ago :)
 
HAHAHAHA!
 
user19161
@Argon Bad, as usual.
 
@WillHunting "Bad, as usual"? That's not good!
 
user19161
@Argon Haha, don't you know some of my secrets already?
 
10:58 PM
@WillHunting Yes, but still!
 
user19161
@Argon I hope some kind of miracle will happen soon.
 
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.
 
@WillHunting Yep - +1
@WillHunting am I right here?
 
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!
 
@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.
 
@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?
 
11:22 PM
@WillHunting, indeed.
 
Partial fraction decomposition, then integrate each fraction.
 
Nope
 
user19161
@Limitless Ah I used to do plenety of these integrals in high school. Let me find something to help you.
 
The roots of the denominator are in $\mathbb{C}$.
I don't know complex analysis, @Argon.
 
Do you know partial fraction decomposition?
@Limitless It needs no CA
 
11:23 PM
@Argon, CA?
 
@Limitless Complex analysis
 
@Argon, how to do you use partial fraction decomposition when the denominator does not factorize over $\mathbb{R}$???
 
Let's see
 
Broaden my horizons, @WillHunting! Broaden them, please. :P
 
Multiply out denominators to get
 
user19161
11:25 PM
@limitless What is D and E specifically?
 
@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.
 
$$6x^5+x^2+x+2 = (x^2 + 2x + 1)(2x^2 - x + 4)(x+1)\left(\frac{A}{(x+1)^3}+\frac{B}{(x+1)^2}+\frac{C}{x+1}+\frac{Dx+E}{2x^2-x+4}+F\right)$$
 
user19161
One can use the trick that the antiderivative of f'(x)/f(x) is ln |f(x)|.
 
@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.
 
11:28 PM
@Limitless Expand, then then group the powers
It should still work
 
does it work if you just multiply out all the brackets and equate coefficients?
 
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 $f'(x)=4x-1\ne x$.
 
@OldJohn Yep, this is what I mean
It should work
 
@Argon, I'm not trying to post a full solution. I don't understand what you're saying given that fact.
 
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 $\frac{2}{x^2+x+1}$.
 
@Limitless You simply clear denominators and then group together all the terms of equal powers!
 
@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 $(x+\frac{1}{2})^2+\frac{3}{4}$.
 
@Limitless for those integrals, the standard method is to complete the square in the denominator
and then use a trig substitution
 
hmm
 
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.
 
$\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
 
@OldJohn, okay. I see what you're doing there.
 
@Charlie hmmm.........................................
 
$x- 1/4 = \sqrt{15}/16\tan\theta$
 
11:37 PM
@Argon hmmmm...........................................
 
@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!
 
@WillHunting, why? It's always a great learning process.
 
@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.
 
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.
 
@Limitless I have been doing calculus for 42 years and still feel like that sometimes
 
@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.
 
@Limitless can anybody?
:P
 
@Charlie, true story. The school I attend is worthless sometimes.
 
11:45 PM
every day seems like a weekend when you have retired :)
6
@Limitless correction to my earlier substitution: should probably be $x- 1/4 = \sqrt{15}/4\tan\theta$
 
@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
$$\begin{align}
\int \frac{1}{2x^2-x+4}dx&=\int \frac{\frac{1}{2}}{x^2-\frac{1}{2}x+2}dx\\
&=\frac{1}{2} \int \frac{1}{\left(x-\frac{1}{4}\right)^2+\frac{31}{16}}dx.
\end{align}$$
 
@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
 
@OldJohn Interesting. We can sub that in and simply change $dx$ to $d\theta$?
And yeah, I noticed that. :)
 
no - $dx$ becomes $\sec^2\theta d\theta$ within a rational multiple
to be precise: $\sqrt{31}/4 \sec^2\theta d\theta$
 
@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.
 
11:58 PM
I prefer to work out $dx/d\theta$ from the substitution $x- 1/4 = \sqrt{31}/4\tan\theta$
you get $dx/d\theta = \sqrt{31}/4\sec^2\theta$
 
@OldJohn Typo
 
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