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12:00 AM
I just ate some bread and 2 eggs.
-2
Q: SELF COMPLEMENTARY GRPAHS

Vamsi SaiA simple graph is called self-complementary if it is isomorphic to its own complement. Let G be a simple graph on n vertices. Prove that if G is self-complementary, then either n = 4t or n = 4t + 1 for some t belongs to N.

2 downvotes, with no comments to help a new user understand how to ask. This site is f**ked.
These people love clicking downvote because it makes them feel superior? I don't know.
 
The problem with this site is the massive number of questions like that, not the people downvoting them.
 
So what does the downvote serve? Nothing.
 
It removes the question from the public view, curating the site.
This is the entire point.
 
That is still a bad attitude to me.
This reminds me of the rap.
Black hat, black tie, black shoes with a black attitude
LOL
 
EricLawson gave this answer to his own question as an attempt to show by generalization the series he was trying to prove. (i did so in a different way in my answer).
1
A: Evaluating a series of hypergeometric functions

Eric LawsonSemiclassical has told me to post an answer to my own question using the technique i have explained to him, now for me to go any further i should state that the original function i am dealing with is: $ {}_2F_1(\frac{1}{2}, \frac{1-n}{2}; \frac{3}{2};\cos^2 (x)) $ therefore using the power ser...

does anyone recognize that series? it wasn't obvious to me how to proceed
 
12:12 AM
I really wished Lang wrote a book on algebraic topology. That's the only topic he did not write on for first year grad school.
I will try to bring him back from the grave.
 
@JasperLoy: i wonder what the proper invocation is for a dead mathematician
 
Yesterday, friend A treated me to dinner. Today friend B is going to treat me to lunch.
 
@RandomVariable is that explanation good enough?
 
the thing i was wondering about re: that integral
is it possible to subtract off that logarithmically diverging piece somehow? i.e. remove a piece from the integral which is explicitly log-divergent, and then have the rest be better behaved
 
12:24 AM
@Semiclassical did you see my added explanation?
 
@robjohn It might be a bit too good.
 
@RandomVariable too good? howso?
 
i did, but that proceeds a little differently than i had in mind. (a different regularization scheme, i guess)
i use the word 'regularization' because i'm reminded of some tricks seen in QFT integrals from time-to-time
(not that i'm actually very good at using them)
 
Blood smells like iron.
 
@JasperLoy It tastes like iron too.
 
12:26 AM
@PedroTamaroff does the sun burn your skin, too?
 
@PedroTamaroff Yes, chicken blood.
 
@robjohn I was just joking.
 
@JasperLoy okay... this is going nowhere good...
 
@robjohn You may delete it.
 
@robjohn Yes. I have to use a lot of sunscreen.
 
12:28 AM
@PedroTamaroff Enjoy your Bloody Mary, and Sue, and....
 
I like to eat KFC.
@robjohn This is going nowhere good...
 
@JasperLoy Blergh.
 
@JasperLoy Oh, it's a vampire joke...
 
I should have spent more time on competition math in school.
I never really did take part in many of those olympiads.
And by the time I went to uni, I had gone mad.
So in a way, I have never really seriously studied math.
But I will seriously study Calculus I on Jan 1, lol.
Mariah Carey is singing here this Fri.
@robjohn Do you think people good at competition math make better researchers?
 
@JasperLoy Nah.
 
12:36 AM
Seems that Tao and Perelman got golds at IMO.
@PedroTamaroff Then there is some hope for me, Professor Pedro.
 
@JasperLoy I don't know. There is a perseverance that is needed for research that is often missing from those who like competition math.
@JasperLoy they are both really bright and have that perseverance.
 
@robjohn I sure have perseverance. I have been trying to solve my mental problems for 15 years. I will put that in my CV.
 
Time for the park. BBL
 
@JasperLoy I'm not a professor.
 
@PedroTamaroff I'm guessing it is an alliterative joke
 
12:39 AM
I never heard of the Master Theorem until I came to this site.
It keeps popping up in questions.
 
@robjohn But why would you ask me if your explanation is good enough? Who am I to make such a judgement?
 
@RandomVariable You are indeed very good. However, be that as it may, explanations are not for the experts, necessarily.
 
"Many authors use the terms Fuchsian group and Fuchsian model interchangeably, letting the one stand for the other. The latter remark is true mostly of the creator of this page. Meanwhile, Matsuzaki reserves the term Fuchsian model for the Fuchsian group, never the surface itself." - creator of Wikipedia page on Fuchsian models
 
@anon What are you reading about, stranger?
 
hyperbolic geometry
 
12:42 AM
Curious. I never thought of you as a geometer.
 
according to a new post on arxiv, there is strong evidence the Euler product representation of zeta is conditionally convergent for Re(s)>1/2, and the truth of this would imply RH (although I don't see why they claim that implication is true).
 
@PedroTamaroff I have a ruler on my desk, so I am a geometer.
 
@anon =O
 
@anon ${\Huge\text{Geometry}}\leftarrow\text{hyperbole}$
 
@anon Link?
 
12:45 AM
I hope that RH can be solved by using calculus.
 
eww, no
 
Nobody thought that PC would be solved by PDE, lol. OK I am just crapping.
 
i'm disappointed that there's been no takers on my bounty for this:
18
Q: Using Fourier Series to compute sums

SuperAboundI have just started learning the basics of Fourier series and have some doubts about it. I am aware that Fourier series can be used to compute infinite sums. For example, $\zeta(2)$ and $\eta(2)$ can be evaluated by using the Fourier series expansion of $x^2$, where $x\in[-\pi, \pi]$. $$x^2=\frac...

 
@PedroTamaroff arxiv.org/pdf/1410.3520.pdf
 
@anon I remember your first name, lol. Tell me your last name some day.
 
12:47 AM
Pitt
 
OMG!
No way, I know it starts with *
1
Q: What type of series is this: k^(n) + k^(n-1) + k^(n-2) + k^(n-3) ...

MelodyI am wondering what type of series this this, where you have some constant (let's say 4) to the power of n, summed up where each new exponent keeps going n-1, n-2, n-3, n-4 ... and so on. So, 4^(n) + 4^(n-1) + 4^(n-2) + 4^(n-3) ... I thought it looked like a geometric series, but all I know abo...

So no upvote for my answer here? I am jealous, lol.
 
Hi @anon @Pedro et al
 
Hi @ted
 
@anon WHAT
@TedShifrin How's things?
 
@PedroTamaroff It is not true.
 
12:51 AM
@anon I'm liking the Pulp Fiction reference now.
 
Need to write my prob exam, @Pedro, after I grade my multi exam tomorrow
 
Apostol hardback is terribly expensive.
I just realised there are so many questions with bounty.
 
@TedShifrin Oh, I can I see your multi exam? I aced my last final.
 
@PedroTamaroff On what topic?
 
@JasperLoy Multivariable but on an easier level.
Plus some differential equations stuff.
 
1:05 AM
@PedroTamaroff The DE is only up to second order right?
 
@JasperLoy Hm, it is a nice introduction to many relevant topics.
For example, the Picard Littlewood theorem.
 
OK. I thought those would be in a separate ODE course.
 
1:22 AM
0
Q: Arithmetic Functions: Evaluate $ \sigma(210)$ and $d(63)$

usukidollEvaluate $ \sigma(210)$ and $d(63)$ I'm not sure if I got this correct, so here is my attempt. By Theorem 6.3, suppose we have $n=p_1^{\alpha 1}...p_s^{\alpha s}$, then $d(n) =(\alpha_1 +1)(\alpha_2 +1)...(\alpha_s +1)$ and $\sigma(n) =\frac{p_1^{\alpha_1+1}-1}{p_1-1}\frac{p_2^{\alpha_2+1}-1}...

 
 
1 hour later…
2:43 AM
How do linear fractional transformations (with real coefficients) act on RP^1?
 
2:58 AM
@Anon: It's the induced action of $GL(2)$.
 
3:09 AM
@TedShifrin !!!! hiiiiiiiii
 
3:36 AM
@TedShifrin oh
 
 
2 hours later…
6:01 AM
Someone invited me to the math room? How did that happen?
I thought invitation means user X and user Y only, hmm...
 
6:12 AM
Does amyone know group theory ?
I mean here in the chatroom
anyone*
 
Ask and never ask to ask. That is the rule of this room.
I have not slept for 22 hours. This is a record.
 
6:34 AM
@JasperLoy piker...
 
Perhaps this means a miracle is about to happen in my life.
Perhaps I will die in my sleep and be reborn in Germany instantly.
 
@JasperLoy I don't know if they'll let you keep your old avatar...
 
@robjohn Perhaps my new parents will call me Justin Bieber.
 
@JasperLoy Just don't leave your monkeys in strange airports.
 
Something weird is happening to my browser. Maybe I have been attacked.
I see weird patterns on top, as if I applied a theme.
 
6:59 AM
OMG, most of the German doctor's answers have 0 or less votes lol
 
still got a group theory question
 
Ask and never ask to ask.
It seems you do not understand this rule, lol.
@UserX Did you accidentally ignore me again? LOL
 
No. Can you help me?
 
Do you know the rule of this room? If you have a question, ask and don't ask to ask, lol. I am saying it for the third time.
Then anyone can answer you if you post the question in this chat.
 
Say we got some complex numbers that belong to one side of the comples plane defined by $y=\lambda x$ for some arbitrary $\lambda$
Then I want to prove that their sum can't equal $0$
I can prove it using some sums and complex numbers
But I wanna know if it is possible to prove it by showing that all their $\Re(z_k), \Im(z_k)$ doesn't form a group thus don't all have an addutive inverse
Is it possible?
don't*
 
7:09 AM
@UserX That doesn't make much sense... how do you intend to show it doesn't form a group without showing they don't have additive inverses in the first place?
 
Well, I want to show they don't have an additive inverse, group theory deals with those stuf
But I thought the problem could be collapses to show that these don't form a group somehow
collapsed*
anyway brb in 1 hour, got class
 
 
2 hours later…
9:08 AM
@anon You can realize $z \mapsto \frac{az+b}{cz+d}$ as translating $z$ by $\begin{pmatrix}a & b \\ c & d\end{pmatrix}$.
This gives you the action of $GL(2, \Bbb R)$ over $\Bbb C$.
Well the action over $\mathbf{H}$ would be more sensible.
@rehband Heya
@anon Probably you look at $1/\zeta(s)$, the convergence of the Euler product of which gives convergence of $\sum_{n = 1}^\infty \mu(n)/n^s$ and relate that somehow with $\mu(x) = O(x^{1/2+\epsilon})$, in turn equivalent to RH
 
Greetings
@rehband Prove it in one line (if you want to you can send emails to the guys from Harvard, MIT , Princeton for help and see if any is able to do that like that) $$\int_0^{\pi/2} \frac{x \cos^2(x) \cot(x)}{3+\cos(4x)} \ dx =\frac{\pi}{128} \log(9232+6528\sqrt{2})$$
 
@Chris'ssis Can you help me with something?
 
@BalarkaSen No (I don't want)
 
Why not :P It's an inequality. The zeta-type inequalities you used to work with.
I am having a hard time proving it.
 
@BalarkaSen Bad luck to you then. :-)
 
9:22 AM
Prove that for $r > 2$ this holds $$\frac{\zeta(r)}{\zeta(2r)}<\left(1+\frac{1}{2^{r}}\right)\frac{(1+3^r)^{2}}{1+‌​3^{2r}}$$
Weird.
I tried the Euler product but it didn't work.
@DanielFischer Morning.
 
Maybe some didn't get my point in one of the @Anastasiya-Romanova questions...
(let me find it)
12
Q: Prove $\int_{0}^{\pi/2} x\csc^2(x)\arctan \left(\alpha \tan x\right)\, dx = \frac{\pi}{2}\left[\ln\frac{(1+\alpha)^{1+\alpha}}{\alpha^\alpha}\right]$

Anastasiya-RomanovaWhen I showed to my brother how I proved \begin{equation} \int_{0}^{\!\Large \frac{\pi}{2}} \ln \left(x^{2} + \ln^2\cos x\right) \, \mathrm{d}x=\pi\ln\ln2 \end{equation} using the following theorem by Mr. Olivier Oloa \begin{equation}{\large\int_{0}^{\!\Large \frac{\pi}{2}}} \frac{\cos \left(\! s...

The last integral she reached can be done by using again the differentiation under the integral sign since you get a nice primitive for that. The rest is a piece of cake.
 
@BalarkaSen Afternonn. Trying to answer this question, or where did you come across that inequality?
 
Just saying ...
 
@DanielFischer Nope, I came across it on a different forum.
OK, if it's in MSE, then I give up.
Especially knowing that it was posted 21 days ago and yet is unanswered
 
9:39 AM
$$\LARGE \text{I HAVE SO MUCH FUN DOING MATH :-)}$$
 
Don't we all.
 
 
1 hour later…
11:04 AM
@Chris'ssis I like formula $(2)$ in my answer though... It unifies $\log(\sin(x))$ and $\log(\cos(x))$ and a continuum between.
 
11:26 AM
If $f$ is a convex function in $[0,2]$ then prove that $\int_{-2}^2 f(x) \mathrm{d}x \leq 2$
I have to prove this
I get that $f''(x) \geq 0 \forall x \in [0,2]$
 
11:41 AM
Why so quiet at this time of this day?
 
dunno. Trying to work this problem out..
 
The one above?
 
yea
 
What is a convex function?
 
I have developed suspicions that I'm trying to prove something incorrect
 
11:43 AM
Another way of saying parabola?
 
Have you heard the terms concave upwards or concave downwards?
No, it's not a synonym of parabola
 
Sorry, English isn't my first language
 
You check if the second derivative is always positive/always negative, or always geq 0/leq 0
 
Hessian test yes
 
Didn't know it was called that way
But yes
 
11:54 AM
Hessian is the Lagrangian capable second derivative test that can test non-degeneracy, but it is used mainly for checking calculus extrema in $\mathbb{R}^2$
I will answer your question in a few minutes, I just really need to eat something
 
12:10 PM
I'll be glad to see a solution.
 
Hi, I am back
Have you tried $f(x)=x^2$?
 
yea, I translated it wrong
It's concave downwards
not just convex
This clears the prob a little but now I'm stuck again
 
So like $f(x)=-x^2$?
 
yea but that's not a counterexample
 
Of course, that would be negative, I just meant the shape
 
12:15 PM
yes then
 
But $f(x)=-x^2 + 100$ is a counter example
 
lol yup
 
$f(x)= -x^2 + \frac{11}6 + k$ where $k$ is any positive value, unless I am still misunderstanding the question
 
@UserX There must be something missing in your premises. $f \equiv 1$ is convex, and its integral is $4$. You can have strictly convex functions arbitrarily close to that, so even strict convexity doesn't give you the conclusion.
 
@UserX Was the question still written down wrong?
 
12:21 PM
Don't think so
 
So all is good now :)
Hello @Studentmath
 
Heya @committing
 
@Studentmath Are you in Australia?
 
No, why?
 
No strong reason :). How are you?
@Iceboy This is your peak conversation time, why are you not talking tonight?
@IceBoy Also you have chatted at every time of every day, what is your sleeping routine like??
 
12:34 PM
@Chris'ssis Please post your answer to that OP. I wanna see it
@Chris'ssis Sorry I suddenly off from the chat yesterday. It was 2 a.m & my dad knocked my door room knowing I hadn't slept
 
Hi pal @Committingtoachallenge how are you? :-)
 
@Ice I am good friend, did you want to look at the drama here math.stackexchange.com/questions/982186/… ?
Just read the questions comments xD
 
@IceBoy I feel like a failure. What should I do?
 
@Committing fine, how about you?
 
@Ice User 'Tharindu' had his other accounts deleted from mass upvoting, he had three accounts all upvoting everything they each posted xD
@Studentmath I am fine thanks, just starting final study session(It is 10:44PM) so will got to sleep in a few hours
 
12:46 PM
@Committing about to sleep too here
 
@MatsGranvik just take a break and get away for awhile
 
Easier said than done.
 
I know, but it will work my friend
 
I feel that I am living in the wrong country.
But I can't tell you more because I am not allowed to.
 
yesterday, by Ice Boy
With the rise in popularity of the internet, math will soon have no national boundaries.
be patient it will eventually happen :-)
 
12:51 PM
@MatsGranvik Me too. There is nothing good about my country, except the food and the weather.
 
@Committingtoachallenge thanks for sharing that :D
 
@Ice :)
@Jasper Hello there!
 
@Committingtoachallenge Hi.
 
How are you today?
@Mats If I have worked out your country, is it ill advised to mention anything?
 
No that is ok. I live in Finland.
 
12:55 PM
I live in a country I rather not mention.
 
Oh it wasn't you, it was @Jasper that lives in S something?
 
@Committingtoachallenge Maybe.
 
How does one find the radius of convergence for the taylor series of $\frac{1}{1-x}$ at $x=a$?
It's easy at $x=0$
W|A comes up with$ |a-x|<|a-1|$
ohhhhh
nevermind
 
math.stackexchange.com/a/984192/168166 I just flagged another answer from our dear German doctor, lol.
 
@UserX The only singularity of the function is the pole at $1$, so the radius of convergence of the Taylor series is $\lvert a-1\rvert$.
 
1:01 PM
I do not want to offend him, but I really do not know whether he reads questions and emails properly or not.
Maybe I should start to stalk him and flag all his answers, lol.
@userx I hope I do not confuse you with my enemy I call X. Rest assured you are a friend and not an enemy, lol.
 
I thought everyone had limited flags
 
Yes, I did not say I will raise an infinite number of flags each day, lol.
 
the mods won't be delighted by your descision for sure. All my downvotes are on this same person though, so I approve.
 
@Anastasiya-Romanova I was referring to this one (that you may differentiate with respect to $\alpha$) $$I(\alpha)=\int_{0}^{\!\Large \frac{\pi}{2}} \frac{x\cot x}{\cos^2x+\alpha^2 \sin^2 x}\, \mathrm{d}x$$
 
Well, it is the mods' job to take care of flags. I will try to flag only things I think should be flagged, so the mods should not be unhappy.
Oh dear, the answer that does not answer the question now has an upvote. It makes me wonder if I read things wrongly myself, but I don't think so.
 
1:11 PM
Anyway, gtg, eat anf study
and*
 
@UserX Tip: Edit your chat message within a few minutes instead of post another line if you made a typo. Just click on the downarrow on the left to edit.
 
@Chris'ssis I made tons of stupid error today, so I think I should rest my brain. Maybe I'm just too tired after coming from school & course
 
@Anastasiya-Romanova For the first time in my life, I did not sleep for 23 hours.
 
You mean that you stayed awake for 23 hours?
Or you normally sleep 23 hours a night ;).
 
@Committingtoachallenge The first one.
 
Huy
1:23 PM
@JasperLoy: I did over 50 hours once. Then I fell asleep.
 
@Huy You are nuts.
 
Huy
@JasperLoy: I traveled from Vietnam back home to Switzerland with stopovers and I just couldn't sleep in a public space at the stopover.
 
@Huy I am glad you live in Switzerland. You are lucky.
 
Huy
I agree, I am lucky to live in Switzerland. Even more lucky to be born and raised here.
 
I think I am very unlucky to live in XXX. But still better than China or Saudi Arabia.
For the first time, my enemy X actually made a very good comment on the German doctor's post, lol.
@Huy Do you use American or British spelling?
 
1:32 PM
I don't wanna create a new post about the $-\frac{1}{12}$ thing so hopefully someone can answer me here. Why does the value when you evaluate the sum Ramanujan-style coincide with the analytical continuation of riemann zeta solution?
 
@Committingtoachallenge Shh, keep it a secret...
But @Committingtoachallenge do you agree with me that that post does not answer the question as stated?
 
@Anastasiya-Romanova Don't worry about that. Making mistakes is not the end of the world. I often do mistakes!
 
@JasperLoy Yep :)
 
@Committingtoachallenge And the German doctor responded with a comment showing that he still does not understand why his post does not answer the question.
 
@Anastasiya-Romanova OK, it was "to make mistakes" not "to do mistakes". You see, I made another mistake. ;)
 
1:40 PM
@Chris'ssis I was too busy to correct that, LOL.
 
@JasperLoy lol :-)))))
 
I must sleep early, girlfriend needs to get up early and I am too noisy, good night everyone(11:41PM here)
 
@Committingtoachallenge See you in your dreams.
 
@JasperLoy As always :)
 
@BalarkaSen Hi Balarka! Can you help me find a continuous surjective function $\mathbb{R} \to \mathbb{R}^2$? :P
 
1:44 PM
@Chris'ssis The grammar error part is so funny. Hahaha...
 
@Chris'ssis You've proved that in one line? :O
 
Huy
@JasperLoy: I prefer British spelling.
 
on average, i think american spelling has less letters
 
@rehband Yeah, sure.
@Anastasiya-Romanova Indeed. :D
 
@Chris'ssis Can I see it? :P
 
1:50 PM
@rehband No (however, I'm sure you have no doubt I'm right)
 
Huy
@IceBoy: Yes, but I think it looks weird.
 
@Chris'ssis OK. Of course I believe you
 
@Huy it just takes getting used to
 
Huy
@IceBoy: I don't get used to weird-looking stuff.
 
you don't have to "understand" it
:-)
 
1:53 PM
Anyway, which one is correct: $2\times3=2+2+2$ or $2\times3=3+3$?
 
both
 
both
 
ab=ba
 
@Huy Then you should have used travelled.
 
wait is this a trick question? Do we have to state decimal system or non-modular arithmetic or something?
 
1:56 PM
@Anastasiya-Romanova If anybody tries to tell you that one of the two were correct and the other not, they are wrong. Very very very wrong.
 
Huy
@JasperLoy: You are right. Thanks for pointing that out.
 
@DanielFischer So both are correct?
 
@Anastasiya-Romanova Aye.
 
@Anastasiya-Romanova Are you asking about definition?
 
@UserX No, it's just an elementary student question
 
1:59 PM
But, there are teachers in some parts of the world, who think differently, @Anastasiya-Romanova.
 
@JasperLoy What is the definition?
 
6 mins ago, by Ice Boy
ab=ba
 
@DanielFischer True! This question has been a national issue in my country recently
@IceBoy I'm not referring to commutative rules
@DanielFischer Could you enlighten me why both statements are correct?
 
@Anastasiya-Romanova Well, it depends on how you want to define it. I could define $2\times 3=3+3$ and $3\times 2=2+2+2$, and then show that because $3+3=2+2+2$, we have $2\times 3=3\times 2$ using geometrical arguments in elementary mathematics. Of course, this is not the axiomatic approach in higher mathematics.
 
Please...
 
2:03 PM
this is a property of equality 2+2+2 =3+3
 
My big bro tell to me that we can't see multiplication as a repeated addition
 
@Anastasiya-Romanova $$3+3 = 1\cdot 3 + 1\cdot 3 = (1+1)\cdot 3 = 2 \cdot 3 = 2\cdot (1+1+1) = 2\cdot 1 + 2\cdot 1 + 2\cdot 1 = 2+2+2$$
 
@Anastasiya-Romanova Well, you can, until you start to multiply by non-integers.
 
@Anastasiya-Romanova the substitution principle states an expression can be replaced by another expression that has the same value
6=6
 
@DanielFischer Wow! How come I didn't see it. it's very intuitive. A 2nd grader can easily understand with that explanation. Thanks...
 
2:07 PM
Wow, nobody thanked me for my brilliant exposition, lol.
 
@JasperLoy Wow! How come I didn't see it. it's very intuitive. A 2nd grader can easily understand with that explanation. Thanks...
...she is trolling us pal.
 
@JasperLoy He said "the repeated addition is the best way to introduce multiplication concept to kids"
 
@Anastasiya-Romanova I think so too.
 
@JasperLoy Your brilliant exposition looks like a professor explanation and an ordinary student is sometimes having difficulty to understand it
 
@Anastasiya-Romanova There is intuition in math, and there is rigour in math, and then we must reconcile the two. That is the highest level.
 
2:13 PM
@Anastasiya-Romanova How can the distributive property be more "intuitive" than substitution?
 
@JasperLoy He gave me counter examples like "how do you multiply 2 matrices or 2 vectors? Using repeated addition?" His argument seems good & I believe it (first time in my life)
@IceBoy I'm not trolling anyone. Sometimes I find myself couldn't answer easy questions
 
@Anastasiya-Romanova ok, ok i apologise :(
 
@IceBoy Well, maybe it's just my subjective judgement
 
no, you have a valid point
 
@IceBoy Could you explain again your substitution proof?
 
2:18 PM
6 = 1 + 1 + 1+ 1+ 1 + 1
group it any way you want
14 mins ago, by Ice Boy
6=6
 
@JasperLoy As you can see in my answers posted on MSE, almost all of my answers are very intuitive to me & I choose to avoid a rigorous part
 
an expression maybe replaced by another expression that has the same value
 
@IceBoy But your answer is explaining commutative rules in multiplication
 
we show that 4 + 2 and 6 have the same value by using the equals sign "="
 
math.stackexchange.com/a/984276/168166 Now we have a prof instead of just a German doctor, lol.
Chances are that his student posted his question on this site to ask for help.
 
2:26 PM
If this is true then hahahhaa
 
That is great!
 
If not then "type \text{meh}"
Dunno, my programing side took over for a second
 
Also @JasperLoy you know about Shifrin right?
 
@UserX Your programming side would have written if(this){...}.
 
@AlecTeal Yes, I know about Theodore Shifrin and also the composer Lalo Shiffrin, lol.
 
2:29 PM
^^
Exactly what language is type \text{meh} xD?
 
@UserX I think you misspelled programming.
 
Also Javascript doesn't count
 
@AlecTeal. Why?
 
@AlecTeal Hey, come on, implicit semicolon is tricky business.
 
Oh I see.
 
2:32 PM
Can't tell if sarcasm....
 
Unfortunately I have no degree. If I had, I would use username: Prof. Dr. Valentina Moy, lol
@UserX Where are you from?
 
@Anastasiya-Romanova I am in this world but I am not of this world.
I once got flagged in this room for saying the above, lol.
 
:-)
:D
 
@JasperLoy Why someone flagged a chat like that?
 
@Anastasiya-Romanova Someone who thinks everything he dislikes to hear should disappear.
@Anastasiya-Romanova I feel better these few days. I hope I can get well in a year or so.
 
2:39 PM
@JasperLoy Sometimes we easily lose our temper for simple reasons, I won't blame him for flagging your chat
@JasperLoy Yeayy! It sounds great!!
 
@Anastasiya-Romanova I hope the universities will still accept me despite my history.
 
@JasperLoy John Nash still got accepted in Princeton despite his schizophrenia.
 
Hi @Jasper @Anastasiya
 
Jeez!? What a complicated word (schizophrenia). Made some mistakes to type it until it's correct
@Sawarnik Hi too...
 
is the following identity familiar to anyone? i came across it while simplifying a result, and strongly suspect it's well-known
suppose i integrate a function F(x_1+x_2+...+x_n) over the positive region over R^n
up to a multiplicative factor, it seems this should be identical to s^n F(s) integrated over R_+
i almost feel like i've seen it in doing laplace transforms, but haven't found a solid ref yet
 
2:54 PM
@Semiclassical By "the positive region" you mean $(0,\infty)^n$? And it's a constant factor times $\int_0^\infty s^{n-1}F(s)\,ds$ then.
 
right, that's what i was after
i can see how it comes out from an appropriate change of variables
but i'd rather cite a known formula than reinvent the wheel
 
Right, and Fubini.
 
right
it seems irritatingly familiar
 
@Anastasiya-Romanova Many people confuse schizophrenia with split personality disorder.
 
@JasperLoy I can't distinguish it either
 

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