« first day (1542 days earlier)      last day (1917 days later) » 

6:00 PM
How much is cos 150? -root3/2?
@UserX
 
@Huy Ahah, so you can show that a more general identity is true for the difference quotient and then take the limit. nicely done! i wonder if there's any overlap between that identity and the 'identity of Feynman' mentioned in the Wilcox(?) paper
 
@UserX choose a textbook with one of the words: intuitive, elementary, introductory in the TITLE
 
@Sawarnik I dunno, am I wolframalpha or your hand to draw a unit circle?
I never learnt the angles, I always unit circle them
 
Yeah, I just did that.
 
if i were forced to do things by hand alone, i'd remember that 150 deg. and 30 deg. are supplementary angles
 
6:03 PM
@robjohn did you see this one? $$\lim_{n\to\infty}\sqrt{n}\int_0^{\pi/2}\sin^n(x) \underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt{2+\cos(x)}}}}}_{n - \text{radicals}} \ dx$$
 
and that cosines of supp. angles only differ by a minus sign
that just leaves cos(30 deg) which is fine
 
@Chris'ssis does the nested radical contribute anything more than a constant to the integral?
 
@robjohn lol, well said. :-)
 
@robjohn what happened to the map of the world to pin our locations?
 
@IceBoy Yeah that was nice :O
I would like it pinned again.
 
6:06 PM
@IceBoy that nested radical behaves like $2$.
 
@robjohn thanks
 
@robjohn who did this?
 
is it no longer possible to move the pins?
 
@UserX I don't remember who did the map.
@MikeMiller I believe it is, but only move your own.
 
6:08 PM
He had waaay too much free time
 
@robjohn that was all I intended to do. I can't seem to do it, though.
 
@UserX Alex Gruber
 
@IceBoy Yeah.
 
ah - there's an edit button you have to click
 
@UserX Why? :O
@robjohn Can we pin it again? It used to be fun.
 
6:10 PM
@MikeMiller good, you found it. I was just about to mention that.
 
What's the most connected field to set theory?
 
@UserX the map was "set up" by Alexander Gruber, but people just pinned their own locations to the map.
 
@UserX Like you can pin yourself too :)
 
I just did
So yea
Guys clear me something out; are the netherlands holland?
 
yes
 
6:17 PM
Hi @Semiclassical
I meant if that condition is true, then its got to be equilateral. @semi
 
@Chris'ssis $\sqrt{2\pi}$ (Used Beta function and 2 for the nested radical)
 
@UserX You live in Kallithea?
 
@robjohn Yeah, indeed. :-)
@robjohn Yeap, this is exactly my way too.
 
Elassona
Damn I just downloaded a 80 mb book and it was 38 pages. What the fuck
 
@UserX That happens with picture books :-)
 
6:21 PM
Haha. Where's the 80 MB then?
@robjohn Oh, I see :D
I definitely lack the skills of @Jasper [sincerely] :(
 
@Sawarnik which skill is currently noticeably lacking?
 
@robjohn Well this, math.stackexchange.com/questions/860989/… I could have easily got 100 points out of it.
 
0
A: Is this a topology?

UserXWell, this is a definition but giving out the answer is never good so this is the hint; Let $X$ be a non-empty set. A set $\tau$ of subsets of $X$ is said to be a topology on $X$ if; 1)$X$ and the empty set belong to $\tau$ 2)The union of any(finite or infinite) numbers of sets in $\ta...

I'm kinda unsure if this is an answer
Should I remove it?
 
@UserX Tiny town this :D
 
@Sawarnik you knew the answer... what skill is lacking there?
 
6:29 PM
@UserX That's just a definition.
 
@BalarkaSen read the question
 
@robjohn Of answering low fruits and getting the 100 points ... Jasper would have posted this as answer but I did it as a comment with 10 votes instead of 100 rep.
 
Anything more than the definition is giving out the answer
 
Your answer is essentially MPW's answer.
 
It requires some thinking on the questioners part though
 
6:32 PM
@robjohn Besides, is this fine, or I should trim down the answer? math.stackexchange.com/questions/989297/…
 
May be. I dunno. Picking out the best answer from the collection of answers of a trivial question is rather a hard decidability problem, @UserX
 
I wonder why people are so amazed about this answer (that I like it much , but I don't see where is that so crazy difficulty ...)
For instance, let me show you something
 
@Chris'ssis Not everyone are as brilliant as you.
 
The core of $I_3$, that is $$\int_{0}^{1} \frac{ \{ \mathrm{Li}_{3}(x) + \mathrm{Li}_{3}(-x) \} \log(1 - x^{2})}{x} \, dx$$
 
I am working on group : $G=(Z/7Z)^*$, if I look at $<2mod7>$ I have 2 and ? I
 
6:36 PM
can be done in one line, real methods only
 
It's all the power of 2mod7 right in $G=(Z/7Z)^*$?
 
@MarcGato $U(7) \cong \Bbb Z/6\Bbb Z$
 
@Chris'ssis can you post it on that mathsharing website?
 
@BalarkaSen $U(7)$ ??
 
$(\Bbb Z/7 \Bbb Z)^\times$, in other words.
@MarcGato What do you mean?
I don't even understand your ill-formatted vaguely-stated question.
 
6:41 PM
okay, I did not know this notation : I don't understand example 2.2 here math.uconn.edu/~kconrad/blurbs/grouptheory/cyclicgp.pdf
$<a mod 7>$ is the set of all power of $a mod 7$ right?
 
@BalarkaSen He's asking if 2 generates the group.
 
Oh. That's false, @Marc
2 only generates {1, 2, 4}. For examples, U(7) contains 3 which is not in <2>
@MarcGato Yes, <a> denotes the cyclic group generated by a.
In particular it means a group generated by a single element 'a', but if it is finite, it is cyclic (prove it). It's an obvious fact, but it's better to get used to it.
 
Holy shit since I started learning set theory I started understanding most questions on the newest ones instead of 1 out of 4
 
And hello @MikeMiller
 
@BalarkaSen ok thanks, yes I know that: I juste have some difficulties to understand :
2 only generates {1, 2, 4}
 
6:48 PM
what is the difficulty you are having?
2^n modulo 7 is either 1, 2 or 4. You can prove that if you're not convinced.
 
@BalarkaSen I have a proposal, it's not a joke: let's close the eyes and compute $$\int_0^1 \frac{\operatorname{Li_3}(x) \log(x+1)}{x} \ dx$$ or $$\int_0^1 \frac{\operatorname{Li_3}(x) \log(1-x)}{x} \ dx$$
both can be done by easy real methods.
 
Ok you answered my question, I have to look at 2^n modulo 7 : I was not able to write correctly $<a mod 7>$ : so is the set of power of a modulo 7 ?
 
I can't do that, I am not that skilled, @Chris'ssis
@MarcGato Yes.
 
@BalarkaSen cool
 
@MarcGato Isn't that notation already explained in the very beginning of the book?
"In a group $G$, we denote the (cyclic) group of powers of some $g \in G$ by $\langle g \rangle = \{g^k : k \in \mathbf{Z}\}$"
 
6:53 PM
Words.
 
@Sawarnik Free groups?
 
sure is g^k, so i was thinking (ok is silly but with a false example i was lost..) like $(2mod7)^k$
 
@MarcGato Well 2^ks are all 2^k modulo 7 in U(7)
 
@BalarkaSen Ok, when we write 2 modulo 7 what is this?
 
what is what?
 
6:58 PM
$2^n modulo 7$
 
@MarcGato You mean you want to know what "2^n mod 7" means?
 
@BalarkaSen Yes, is the rest of the division algorithm?
 
It's the residue left after dividing 2^n by 7.
I'd really suggest a fundamental book on number theory before diving on to abstract algebra, @Marc
 
@BalarkaSen Yes it would be great, but I have some course on abstract algebra without studies number theory... :(
Do you have a good reference?
 
The chapter on divisibility by Montgomery-Zuckermann (and further if you want to, it'd help in algebra, actually).
 
7:03 PM
thank you
 
Blah.
 
I just found the same integral here
21
A: A Challenging Logarithmic Integral $\int_0^1 \frac{\log(x)\log(1-x)\log^2(1+x)}{x}dx$

Omran KoubaLet the considered integral be denoted by $I$. Our starting point is to reduce the number of logarithms of different arguments in the integrand. Thus, using the fact that $6ab^2=(a+b)^3-2a^3+(a-b)^3$ we obtain \begin{align*} 6I&=\underbrace{\int_0^1\frac{\log x}{x}\log^3(1-x^2)dx}_{x\leftarrow \s...

Kouba is very good, at least this proof is magnificent. (it's hard to beat)
Again, this is what I consider to be work to admire, good to learn, to keep in mind. All is easily done, even a kid can understand the proof.
 
@BalarkaSen How can I find the maximum order of an element in the symmetric group $S_5$? I know the size of $S_5$ is $120$.
 
@MarcGato Note that order of product of two cycles is lcm of the order of the two individually. Use this.
 
@BalarkaSen Ok thanks
 
7:17 PM
My guess is that even $$\int_0^1 \frac{\log(x)\log(1-x)\log^4(1+x)}{x}dx$$ has a nice closed form (but not sure about the integral with power of $3$).
 
@Chris'ssis where do you find these integrals? Risch's algorithm states that even the antiderivative of this has a closed form I think
 
@Chris'ssis At least in terms of Euler Sums...
 
@UserX It simply came to mind now after seeing the previous integral. It's not hard to guess that.
@robjohn Yeah.
 
@BalarkaSen I found 6, now I am reading this property : for each divisor of the size of a (finite) cyclic group, there is exactly one subgroup of that size. Does the subgroup necessary exist ?
 
@robjohn I was also looking at that $$\int_{0}^{\frac{\pi}{2}} \theta \log^{3} (\tan\theta) \, d\theta $$
from the last post downward here integralsandseries.prophpbb.com/topic225.html
This reminds me of something ... @Anastasiya-Romanova used in a post
Oh, not Anastasiya-Romanova but Tunk-Fey
6
A: Integral $\int_0^{\pi/2} \theta^2 \log ^4(2\cos \theta) d\theta =\frac{33\pi^7}{4480}+\frac{3\pi}{2}\zeta^2(3)$

Tunk-FeyFrom Table of Integrals, Series, and Products Seventh Edition by I.S. Gradshteyn and I.M. Ryzhik equation $3.631\ (9)$ we have $$ \int_0^{\Large\frac\pi2}\cos^{n-1}x\cos ax\ dx=\frac{\pi}{2^n n\ \operatorname{B}\left(\frac{n+a+1}{2},\frac{n-a+1}{2}\right)} $$ Proof Integrating $(1+z)^p z^q...

 
7:32 PM
@Chris'ssis how could you make such a mistake?
 
and perhaps I might also use some of the SuperAbound's technique there.
@robjohn Maybe it's not a big mistake since they are from the same family (and possibly the same person?) ... :-))))
 
@Chris'ssis I don't know...
 
@robjohn I wonder why Tunk-Fey is not active anymore ... (I hope it has nothing to do with my inspection of his work) :-)))
This one math.stackexchange.com/questions/981650/… is also in Ovidiu's book if I'm not wrong, maybe not as a problem but as a result he used it.
 
@MarcGato 6 is correct, yes. Lagrange maybe?
I'm a bit busy to look at that, sorry.
 
This one si also a funny question math.stackexchange.com/questions/983044/…
 
7:38 PM
@BalarkaSen No problem, you have already help me a lot today ;)
 
hehe, I did these integrals some time ago ...
7
A: Inverse Trigonometric Integrals

M.N.C.E.For the first one, \begin{align} \int^1_0\frac{\arctan^2{x}}{x^2}{\rm d}x =&\color{#BF00FF}{\int^\frac{\pi}{4}_0x^2\csc^2{x}\ {\rm d}x}\\ =&-x^2\cot{x}\Bigg{|}^\frac{\pi}{4}_0+2\int^\frac{\pi}{4}_0x\cot{x}\ {\rm d}x\\ =&-\frac{\pi^2}{16}+4\sum^\infty_{n=1}\int^\frac{\pi}{4}_0x\sin(2nx)\ {\rm d}x\\ =

M.N.C.E.=Tunk-Fey? :-) Who knows ...
 
@Chris'ssis I got some interesting integrals you might like
Somewhere on my "holy shit I'll never manage to solve these" file
 
@UserX OK
I need to check the last proof there ... (but not now)
Oh. let me check the generating function at least ...
 
@Chris'ssis $$\int_{\Bbb {R}_{\geq 0}} \frac{\cos x^2-\cos x}{x} \mathrm{d}x$$
 
@UserX This is one from the "holy shit I'll never manage to solve these" file?
 
7:45 PM
Yea. Is it trivial?
 
@UserX Sure, it's trivial.
 
Here's another one; $$\sum_{\Bbb {N}_{\geq 1}} \frac{1}{n} \int_{2n\pi}^{\infty} \frac{\sin z}{z} \mathrm{d}z$$
Well they may be trivial for you, they're not easy for me at the least
 
lol
 
@UserX you might like to know I posted this one some time ago right here - I don't remember if someone provided with a solution.
 
I got some sums too, along with limits
Interested?
 
7:50 PM
Sure.
 
@Mick Hey, after a long time..
 
$$\displaystyle{\sum_{n=1}^{+\infty}\frac{\left(\frac{3-\sqrt{5}}{2}\right)^n}{n‌​^3}}$$
@Chris'ssis $$\displaystyle{\int_{0}^{+\infty}e^{-x}\ln^2x\,dx}$$
Is the last one here somewhere?
It seems that there would be many ways but all have problems
 
@UserX the use of the trilogarithm identitites combined with the numerator where you have a solution to $x^2-3 x+1$.
As in the case of golden number, you need to cleverly make use of $x^2-3x+1$.
@UserX what is my last creation in the spirit of this style? Wait a few seconds ...
(I did such a thing like 2,3 weeks ago?)
 
Got a question on a change of variables someone did, can you explain me how you would think that?
 
@UserX see above (it's done in the spirit of your penultimate question)
 
8:02 PM
@Chris'ssis for $$\displaystyle{\int_{0}^{2}\frac{\ln (x+1)}{x^2-x+1}\,dx}$$ he applies the following change of variable ;$\displaystyle{x=-1+\frac{3}{u+1}}$ and of cource as usual with these weird change of variables it makes the integral trivial
How could anyone see that change would be so magic beforehand?
 
@UserX Well, there are some generalizations where you can use such tricks to nicely finish these integrals. I created such ones some time ago. I can try to find one.
 
@Chris'ssis Do you have any books(yours or someone else's) that introduces someone to tricks used on these integrals?
 
@UserX I can give you such a paper.
 
@Chris'ssis looking forward to it
 
(it's in my native language, there is no English version - but this shouldn't be a problem I think)
 
8:10 PM
 
@Hippalectryon Do you agree with the fact I stated above about the integral @UserX asked me? $$\int_{\Bbb {R}_{\geq 0}} \frac{\cos x^2-\cos x}{x} \mathrm{d}x$$ I said it's trivial.
 
@Chris'ssis Why are you asking me ? I'm not a pro :/
 
@Hippalectryon $$\LARGE{\text{YOU ARE A VERY GOOD PROFESSIONAL!!!}}$$
:-)
 
No I'm not :c How can I be a pro at 16 xD
 
lol ;)
 
8:20 PM
Why don't you believe me :c
 
Shame on me, I'm the only one with no math background :'(
 
What do you even mean by 'no math background' ?
You've been doing maths for years
 
@Hippalectryon For years? No ... Just look at the questions I was asking a bit ago ...
 
For more than one year
 
@Hippalectryon OK THEN :-)
 
8:25 PM
If you were asking questions, then you were doing maths :)
 
That's why is trivial ...
$$\int_{\Bbb {R}_{\geq 0}} \frac{(\cos x^2-e^{-x^2})-(\cos x-e^{-x})+e^{-x^2}-e^{-x}}{x} \mathrm{d}x=$$
$$\underbrace{\int_{\Bbb {R}_{\geq 0}} \frac{\cos x^2-e^{-x^2}}{x} \mathrm{d}x}_{\displaystyle 0}-\underbrace{\int_{\Bbb {R}_{\geq 0}} \frac{\cos x-e^{-x}}{x} \mathrm{d}x}_{\displaystyle 0}+\underbrace{\int_{\Bbb {R}_{\geq 0}} \frac{e^{-x^2}-e^{-x}}{x} \mathrm{d}x}_{\displaystyle \frac{\gamma}{2}}=\frac{\gamma}{2}$$
Q.E.D. (D.U.I.S. for the first 2 integrals)
@UserX ^^
OK, let me put it in my notebook, I wanna add it to my book.
 
@Chris'ssis lol
 
hi @Hippalectryon
 
@Alizter @Chris'ssis still won't believe I'm 16 :/
 
8:42 PM
@Hippalectryon I think @Alizter is shocked, he can't even speak after reading you :-)))
 
Lol :)))))))))))))
 
Hi @DanielF ... Had fun today ... Proved weak law of large numbers and derived variance of binomial r.v. painlessly :)
 
Hi @Ted. Good. I hope some of your students also had fun with it. And learned it ;)
But remind me, @Ted, what was the weak law of large numbers?
 
The top half seem to be getting it ... Not the bottom. Test 2 next week ...
 
8:50 PM
Half ain't bad.
Although, that depends on how badly the bottom half doesn't get it.
 
The mean of $n$ identically dist ind r v cinverges to the mean in measure ...
 
Falls out of Chebyshev immediately, once one knows variance of the sum, etc.
 
Hey Ted
 
Never having thought about or learned this stuff, I now wish I'd taken it in college.
Hi @UserX
 
8:53 PM
Why are there so many (either obvious or hidden and revealed with the right substitution) depressed cubics questions lately on MSE?
 
They need prozac? :)
 
WLLN is really helpful for getting intuition about probabilistic things in the real world, I think...
 
I don't have time to look at most ... Only to observe all my downvotes :)
 
I prefer SLSN, @Mike.
 
@TedShifrin There's no reason to believe Prozac will help. Every cubic reacts differently to the various medications.
@DanielF It's certainly extremely important to know that law
 
8:56 PM
Yeah, @Mike, what is super-cool is that the standard deviation of the average of $n$ identical is the standard deviation over $\sqrt n$. I tried to motivate that, but it's sort of magic :)
I'll do SL and Central Limit at the end. Have to do joint distributions for a few weeks first.
 
@TedShifrin why do people downvote you and treat you as an enemy? Past failed students or what?
 
I don't think students ... I think people.who have a grudge ...here.
All my students who I know are on here are good friends :)
 
Has anyone seen a summation with a term of the form i(i+1)...(i+k) before?
 
$k$ fixed? @Ty
 
k could be any positive integer
 
8:59 PM
Sum over $i$ with $k$ fixed?
 
yes
just wondering if there's a name for it
 
Do I count as a student?
 
So it comes from differentiating $x^{i+k}$ a bunch of times ...
 
Central limit theorem is also good
 
Only if you hate me, @Mike. :) I don't think you and @Pedro count.
 
9:08 PM
@TedShifrin Check invite?
 
@Chris'ssis I was eating. Forgot to do that today.
 
@Alizter In general do you prefer healthy food?
 
@Chris'ssis Yes, usually. I usually eat just healthy food once or twice a day.
 
@Alizter and practice some sport every day?
 
I eat Iceberg lettuce every day
@Chris'ssis I try to do sport at least 3 times a week. I walk every day if that counts
I am coding today for the first time in a while.
 
9:21 PM
@Alizter Language?
 
@Sawarnik C#
 
Ok.
 
Finally got my PC!
time to try out Mathematica 9.0
Dsolve[9 y + 7 y'[x] + y'''[x] == 0, y[x], x] why doesn't this work?
lol nvm
 
9:37 PM
You need to give it initial conditions, for starters, @UserX, plus DSolve
DESolve?
 
Dsolve is the command
 
maybe you don't need initial conditions ... I think it puts in constants ...
 
and I want the general solutions
 
Well, you definitely need to capitalize the Solve
use the generous help menu
 
Ohhh that solved the problem
 
9:41 PM
You're welcome.
 
@TedShifrin!
 
10:06 PM
This is too silent,
 
Indeed.
 
Hey @TedShifrin, what's an introductory book on set theory, emphasising on the understanding of the theorems (as much as it can) that goes up to ordered sets, ordinal numbers etc?
Or anyone here in general
 
@UserX Hammack is good.
And available online, as you can see.
For further set theory (axiomatic stuff) you'd probably like Handbook of Mathematical Logic.
 
@Balarka It's going into too much detail and it's too huge for me at the time. I can't spend time to read a 300 page set theory book
 
@UserX Just do the exercises.
 
10:12 PM
@BalarkaSen without reading the theory? o.O
 
Skim through them. I believe much better mathematics is done by just doing exercises.
 
@BalarkaSen you definitely misunderstand me. I don't know Set Theory at all.
 
Oh. Then I don't believe you're going to find anything < 300 pages, sorry.
 
10:30 PM
@Pedro!
 
@BalarkaSen Heerroror.
 
@PedroTamaroff You got cold?
 
@BalarkaSen No it is ridiculously hot nowadays.
 
@PedroTamaroff I am trying to generalize the fact that any two closed set in a metric space can be separated by open sets by replacing sets by spheres. Noted that this is possible for convex sets in R^n and proved it. So I am now trying to prove this for normed vector spaces. I told prof whether a more general extension was possible, he referred to geodesic metric spaces.
So having lots of fun.
@PedroTamaroff We'd soon be freezed to death in this part of the world.
 
@BalarkaSen But then it is not always possible.
 
10:34 PM
@PedroTamaroff Right, it is not.
It is, in metric spaces where the notion of convexity is defined.
Prof told me he would tell about geodesic metric spaces next when we meet. So interested.
 
@BalarkaSen It is not always possible to separate arbitrary disjoint closed set by balls.
Balls are bounded.
Take an unbounded set.
 
Oh right.
Compact.
Otherwise two parallel copies of R sitting in R^2 would be counterexamples.
 
Hi, I want to prove that a group of order 15 is cyclic (without Sylow's and Cauchy's theorem). Arguing by contradiction I can write $G$ as a union of disjoint set $A,B$ and $\{1\}$ with A is the set of elements of order 5 and B is the set of elements of order 3. I just proved that two subgroup of prime order, the intersection is trivial. How can I continue?
 
@Pedro Balls needn't be bounded.
 
@MikeMiller What's your definition of bounded?
Mine is "it lies inside a ball."
 
10:40 PM
@Pedro Bro.
The ball of infinite radius.
:D
 
@MikeMiller There's no such thing.
Balls have finite radius.
>=)
 
@BalarkaSen You won't be able to prove it for normed vector spaces. You need nice norms for that.
 
Maybe your balls do.
 
double entendre, @Mike?
 
@DanielFischer Hello Daniel-senpai.
 
10:42 PM
Hola @Pedro.
 
@DanielFischer Do you know how to show that $$\int_x^{\infty}e^{-t^2/2}dt\geqslant e^{-x^2/2}(x+x^{-1})^{-1}?$$
I reduced it to showing that $$\int_0^{\infty}e^{-\alpha t^2/2-t}dt\geqslant (1+\alpha)^{-1}$$
Here $\alpha,x>0$.
 
Off-hand, I don't.
 
@robjohn said he knows, but he's away.
@DanielFischer How are things for you?
 
So-so. The weather is getting rotten.
 
@DanielFischer OK. Normed vector spaces over R.
 
10:47 PM
@DanielFischer Oh, how so?
 
@UserX in the same style as I proceeded above, I got the generalization $$\int_{\Bbb {R}_{\geq 0}} \frac{\cos(x^n)-\cos (x)}{x} \mathrm{d}x=\frac{n-1}{n}\gamma, \space n\ge 2$$
 
Got to sleep. Bu-byes.
 
@BalarkaSen Doesn't suffice. Consider two translates of the unit ball with respect to the $\lVert\cdot\rVert_1$-norm in a $\lVert\cdot\rVert_\infty$-normed $\mathbb{R}^n$ for $n > 1$.
@PedroTamaroff Well, it's autumn.
 
@DanielFischer OK, then so classifying the norm also adds to the list of problems.
 
If the norm is induced by an inner product, it works. Probably a uniformly convex norm is enough, but I'm not sure.
Well, I'm going to bed. Night.
 
10:56 PM
@robjohn I think I'll also include in my book the generalization above since all is done elementarily, easily. The readers will love that.
I used copy/paste for $\int_{\Bbb {R}_{\geq 0}}$, I wouldn't write things like that.
 
@PedroTamaroff No, balls CAN have infinite radius.
 
@BalarkaSen Well that's plain ludicrous.
 
@Pedro I mean, it just means we allow one more ball than you do.
Or rather we have more balls than you
 
Leave me outta this ...
 
@BalarkaSen Defend your beliefs!
 
11:07 PM
@PedroTamaroff That was a wrong link. Let me find out the right one/
 
@BalarkaSen My balls have finite radius.
Speaking of balls, where is Sarah?
 
Oh noes @PedroTamaroff. Simmons actually assumes balls to be of finite radius.
Page 64, ex 4, here
I was wrong.
@Mike That's precisely why I don't prefer to call them balls.
Spheres. Open spheres.
 
That's not preference... that disagrees with modern notation.
 
@MikeM Whatever. I don't want my terms to have double meaning. Especially with something like balls.
Even disks are good.
 
Good :)
 
11:19 PM
Well, above I can also use some $s>1, s \in \mathbb{R}$ ...
 
Balls can have finite radius and yet feel infinite ...
@Balarka: Spheres are always the boundaries of balls.
 
Simmons use different terminologies
 
Well, that's bad, but his book is old.
 
OK, disks then.
 
What is his definition of sphere?
 
11:23 PM
But NEVER balls.
@TedShifrin Open sphere. $\{y \in X : d(x, y) < r\}$
 
Disks tend to be closed balls now ... Unless you say open ...
Ugh, he's using sphere for ball. We don't do that in the modern world.
 
I don't give a damn to the terminologies. I am very sorry to say this.
ducks to an incoming smack
 
Well, one has to communicate with people, so it matters.
 
bah.
 
Otherwise you have to define terms in every conversation.
You're overruled.
 
11:25 PM
mathematics don't need communication. it can be done in one's head.
 
Fine, just don't talk to me or @Mike.
 
If you are going to dogpile me for terminologies, OK.
oops. I need to run. have to sleep.
 
Thus, I can also easily compute $$\int_0^{\infty} \frac{\cos(x^{\alpha})-\cos(x^{\beta})}{x} \ dx$$ hmmm, this I also need to add to my book ...
 
Night.
 
In all my work I only use real methods, no complex analysis method involved.
The same here
No touch on complex analysis methods. This is going to be added to my book.
 
r9m
11:35 PM
@PedroTamaroff using CS/Jensen we have $\displaystyle \left(\int_{x}^{\infty} te^{-t^2/2}\,dt\right)^2 \le \left(\int_{x}^{\infty} t^2e^{-t^2/2}\,dt\right)\left(\int_{x}^{\infty} e^{-t^2/2}\,dt\right)$ and since $\displaystyle \int_{x}^{\infty} t^2e^{-t^2/2}\,dt = xe^{-x^2/2}+\int_{x}^{\infty} e^{-t^2/2}\,dt$, we have $\displaystyle \frac{2e^{-x^2/2}}{x+\sqrt{x^2+4}} \le \int_{x}^{\infty} e^{-t^2/2}\,dt$, the RHS is greater than the one required to prove :)
 
Cool beans.
 
11:51 PM
@Pedro @Ted \o
 
r9m
@PedroTamaroff here's an alternative argument .. brought to you by Lord Unreasonable @Chris'ssis
 
@PedroTamaroff Yes?
 
@r9m lol, Lord Unreasonable? :-)
 
@r9m That is a bit tighter on the upper end and looser on the lower end than what I got yesterday
 
r9m
@robjohn yesterday ?! oops I was absent sir ! link plz ! :-)
 
11:58 PM
18 hours ago, by robjohn
@PedroTamaroff I can show that $$ x\int_x^\infty e^{-t^2/2}\,\mathrm{d}t \le e^{-x^2/2} \le\left(x+\frac1x\right)\int_x^\infty e^{-t^2/2}\,\mathrm{d}t $$
 

« first day (1542 days earlier)      last day (1917 days later) »