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12:18 AM
^^ allow the commutative property and you have hug
 
12:55 AM
Good night :)
 
1:42 AM
Evening! How can we simplify (x,yz)/(x,y)(x,z)?
 
 
2 hours later…
3:41 AM
@DonLarynx oh crap. I'll beat you to 100
 
4:37 AM
I'm thinking about bounty-ing math.stackexchange.com/questions/1102688/…
cos I would really like to see a well worked answer to it.
Is it worth bountying more than 50 to encourage a better answer?
 
 
5 hours later…
9:09 AM
12
A: Equation of a rectangle

J. M.Based on Raskolnikov's answer here, one can build an implicit Cartesian equation for a $2p \times 2q$ rectangle: $$\left(\frac{x}{p}\right)^2+\left(\frac{y}{q}\right)^2=\sec\left(\arctan\left(\frac{x}{p},\frac{y}{q}\right)-\frac{\pi}{2}\left\lfloor\frac2{\pi}\arctan\left(\frac{x}{p},\frac{y}{q}\...

what is a 2p x 2q rectangle? ^^
Another generation question related to this chat room...anything special we've to do to render mathjax stuff here... like for e.g. $2p \times 2q$
^ *general ... typo my bad...
 
9:41 AM
@deostroll A rectangle with sides of length 2p and 2q, I guess.
 
 
1 hour later…
11:02 AM
How are you @JasperLoy?
 
@Committingtoachallenge Not good. My condition got a bit worse since Dec, and it is still bad.
 
@JasperLoy That isn't good. What condition specifically?
 
@Committingtoachallenge Well, OCD mainly, but also some depression and trauma.
 
@JasperLoy Was it worsened by not starting your studies, or the studies weren't started because it worsened?
 
Have you thought about going back to the doctor pal @JasperLoy?
 
11:05 AM
Or neither
 
@Committingtoachallenge The latter, of course.
@skullpatrol Maybe, we'll see.
 
Hey @Skull You don't seem to be as active here since you stopped being Ice boy
Or am I confusing you
 
You are correct. @Committingtoachallenge
 
@skullpatrol I miss the old display picture xD
 
11:07 AM
@JasperLoy I am on phone internet(limited) so I can't click that, what is it
 
@Committingtoachallenge Just a documentary on John Nash.
 
@JasperLoy You may have just caused immense procrastination with that
 
@Committingtoachallenge With what?
 
I have a beautiful mind on my computer and I have been wanting to watch a movie for weeks, but without internet I couldn't.
I had totally forgotten I had it on here
 
@Committingtoachallenge It's a docu, not the movie.
 
11:11 AM
@JasperLoy But I was reading the Wiki page and it mentioned it haha.
 
You shouldn't be watching things about a decent into madness @JasperLoy you want the exact opposite, don't you my friend?
 
@skullpatrol No, it helps me. It does.
 
@Jasper $\text{Jasper adds link} \implies \text{AC reads wiki page} \implies \text{AC watches Movie referenced}$
 
@Committingtoachallenge OK. I read the book, watched the Hollywood movie, and also watched the docu now.
 
How? @JasperLoy does it help you.
 
11:14 AM
And my coffee is cold and old :\...(I'll still drink it)
 
@Committingtoachallenge $\left(\left|\frac{|x|}p+\frac{|y|}q\right| +\left|\frac{|x|}p-\frac{|y|}q\right|-2\right)^2 +\left(\frac{|x|}p-1\right)^2\left(\frac{|y|}q-1\right)^2=0$
 
@skullpatrol Well, that I am not alone in my struggles with mental illness, and that miracles are possible.
 
@robjohn My coffee?
 
@Committingtoachallenge The movie is not exactly what happened in real life. It is quite different.
 
They put him in an institution @JasperLoy
 
11:16 AM
@skullpatrol Yes, I know.
 
@Committingtoachallenge nah... I was trying to give an alternate equation for a rectangle.
 
@JasperLoy and drugged him up
 
@robjohn You didn't tag anyone so I had no idea aha
 
@robjohn Did you succeed?
 
I wish I had a window to do Math on
 
11:21 AM
@Oxinabox My personal opinion is that it is best to start with lowest possible bounty. Since if there is now answer and you want to put a bounty on the same question again, you have to increase the amount of the points. (You can find pointers to some relevant info on bounties here and some of the questions tagged might be of interest for you.)
 
"John, you watched a mugging, that's weird..." - I didn't remember that line ahahaha
 
11:37 AM
@JasperLoy I have now...
 
@robjohn Is it a publishable result? =)
 
@JasperLoy oh, yeah... behold, a rectangle!
 
12:01 PM
Talk to you @JasperLoy later my friend. Think about going back to the doctor, ok?
 
12:47 PM
@ModdedBear No thanks.
 
1:02 PM
If one uses the "answer your own question" feature, then the question does not appear before one has also supplied the answer, right?
(I want to point out the possibility to someone, but I wanted to make sure it worked as I expect)
 
1:13 PM
Hi @DanielFischer!!! Could we prove by induction the correctness of Quicksort, that is the following?

Quicksort(A,p,r)
  if (p>=r) return;
  q=Partition(A,p,r);
  Quicksort(A,p,q-1);
  Quicksort(A,q+1,r);
 
1:44 PM
Last night I came up with a really good combinatorial proof that ${n+a-1 \choose a-1} = \sum _{k=0}^{\left\lfloor n/2 \right\rfloor} {a \choose n-2k}{k+a-1 \choose a-1}$. It's one of those ones that just really makers you feel smart when you get it.
Makes*
Also, here's hoping that rendered properly since I typed it on my phone
 
How would I prove that $x^8-x^7+x^2-x+15$ has no roots $\in \mathbb{R}$?
 
2:02 PM
Find 8 complex ones
I.e., factor it into linear terms in C and look at what the roots are. If you're allowed to use technology anyway.
If not you'd be in for a hell of a lot of synthetic division
Although there are probably some tricks I don't know so there's probably a better way.
 
@SamuelYusim Nope, cant do that
there's probably a trick to this
 
If not then it's sure an interesting problem
Okay so one thing worth noticing is if it has no real roots it must be a product of 4 quadratics with coefficients in R
But that in itself is horrifying
 
:D
@SamuelYusim and the quadratics dont have roots in $\mathbb{R}$. right
 
Of course. If a poly has no roots in R neither will any of its factors
Because then the original actually would, a contradiction
So if I were you I'd start long dividing that polynomial by quadratics with friendly coefficients until I came up with a better idea
 
that probably isnt the way Im supposed to do it, but Ill try it anyway for funsies
 
2:23 PM
Wow I had forgotten how sad 'A beautiful mind' is, thanks @Jasper lol
 
2:34 PM
I am sorry for whatever I said to offend you @Venus, I actually can't recall what it was, but you were on ignore, so I imagine I said something rude
 
@Huy Just stopping by to say I love this version ;)
@Committingtoachallenge No worries, I've forgotten it long time ago :-)
 
@Venus Haha that's good :)
 
@Committingtoachallenge No one's at mistake anyway. Nice to make friends here :-)
 
@Venus Yep it sure is, except it is a little distracting at times :P
 
@Committingtoachallenge Hey, I gotta go. See ya...
 
2:38 PM
@Venus Okay talk later :)
 
@Committingtoachallenge Sure... :-)
 
3:00 PM
today was a good day
I solved a problem in my sleep
and it turned out to be right
normally when I think that happens it turns out to be wrong.
 
hello
i am new here. can i ask questions here?
 
I don't know you, but I would assume you can. I recommend you try at least.
 
okay
Let f be irreducible in F_p[X] for a prime p. How to show that f divides X^(p^n)-X if and only if deg(f) divides n?
 
@Committingtoachallenge Well, I think in some ways my life is sadder than his.
 
@Epsilondelta can't help you :(
 
3:06 PM
:(
 
3:17 PM
@DanielFischer @robjohn When we have the recurrence relation: $T(n) \leq T \left( \frac{9n}{10}\right)+T\left( \frac{n}{10}\right)+cn$, does the recursion ends when $\frac{9n}{10^i}=1$ or when $\frac{n}{10^i}=1$ ?
That's the recursion tree:
 
@Epsilondelta Ask on the main site, QED.
 
3:36 PM
@evinda I think you meant "... does the recursion ends when $\frac{9^{\color{red}{i}n}{10^i} =1$ or $\frac{n}{10^i} = 1$."
 
@Oracle Yes, that's what I have meant. Do you have an idea?
 
I think the second one converges faster.
 
Do we take the greater value of i or the smaller one? @Oracle
 
I don't understand what you are asking.
 
@Oracle Do we solve for i to check from which of these relations we find when the recursion ends?
 
3:41 PM
$\frac{n}{10^i}<\frac{9^in}{10^i}$. And the conclusion follows.
@evinda Aren't you learning Introduction to algorithms, Cormen...
 
@Oracle Yes. So de take the greater of the numbers and solve for i?
 
@evinda, yes to what?
 
@Oracle to find when the recursion ends?
 
2 mins ago, by Oracle
@evinda Aren't you learning Introduction to algorithms, Cormen...
@evinda what is your answer?
 
@Oracle Yes
 
3:47 PM
@evinda I knew it. I can read minds.
 
@Oracle :p
@Oracle So, do we look for the greatest number and having found it we solve for i?
 
@evinda You study computer science, don't you?
 
No @Oracle
What's with you? @Oracle
 
@evinda What else do you learn?
 
@Oracle Set theory, data structures, algebra........
 
3:48 PM
@evinda What degree and department are you in?
 
@Oracle Applied mathematics. you?
 
@Oracle I thought you were a mindreader! all these questions
@evinda guten Tag :D
 
Guten Tag @Alessandro :)
 
@Alessandro I ask questions just for fun. I already know the answers. For example, I knew the answer to evinda's question.
 
Then tell us also something about you @Oracle
 
3:53 PM
@evinda I'm THE oracle.
 
Ok.... :D
 
I don't study pure or applied mathematics.
 
But...? @Oracle
 
@evinda I do not study computer science either. Also I haven't read introduction to algorithms by cormen.
 
@Oracle Are you a student?
 
3:57 PM
Yes
 
@Alessandro Hattest su heute Deutschunterricht?
@Oracle physics?
 
No
 
@Oracle Chemistry?
 
@evinda Ja, ich habe jeden Tag (von Mo bis Fr) Unterricht
 
@Alessandro Achso
 
3:58 PM
@evinda Doesn't make sense. I don't study chemistry.
 
@Oracle I wanted to right du
@Oracle Hint?
 
@Oracle evinda misspelled it, google gets it right when "su" is repleaced with "du"
@Oracle economics?
 
@Alessandro That makes sense. I don't major in economics!
 
@Oracle I'm running out of guesses, computational biology?
 
@Alessandro Are you from Germany? @evinda Are you from Poland?
 
4:05 PM
@Oracle I'm Italian, but I live in Germany at the moment
 
4:20 PM
Does it always hold that $\log_a n>\log_b n$, knowing that $a<b$ ?
 
4:30 PM
@evinda $\log_{1/2}(4)=-2$ and $\log_{2}(4)=2$. I think it holds if $1<a<b$
 
Can anyone tell me how good is this book? flipkart.com/elementary-number-theory-7th-english/p/…
 
hello,I have post a interesting analysis problem,can see math.stackexchange.com/questions/1106888/…
I hope someone can disscuss
 
@Sawarnik I haven't read this book, but it is mentioned in this question, togheter with a lot of others number theory book that you may find interesting math.stackexchange.com/questions/329/…
 
Ah.
 
@robjohn,and @DanielFischer
 
4:42 PM
@Alessandro Thanks, I should order it then. Only affordable book that suits my need .. and the link shows its a good book :)
 
@evinda Also if $a<b<1$
 
@Alessandro So does it only hold for these two cases?
 
@Sawarnik There are other books a lot more upvoted, but it does seem to be good. I'd wait for the opinion of someone who has actually read it though
@evinda $\log_{1/2}(1/2)=1$ and $\log_2(1/2)=-1$ so it holds also for $a<1<b$, depending on $n$. Actually I'm starting to doubt of what i wrote before!
 
5:00 PM
@math110 yes?
 
can you see this analysis problem?
4
Q: Prove the $f(x)$ is continuos in open interval?

math110Today,I found a interesting problem: if $$F(x,y,z)=\begin{vmatrix} \cos{x}&\sin{x}&f(x)\\ \cos{y}&\sin{y}&f(y)\\ \cos{z}&\sin{z}&f(z) \end{vmatrix}\ge 0$$ for all $x,y,z$ of an open interval $I$ for which $x<y<z<x+\pi$. show that: $f(x)$ is continuous in $I$ and has finite left...

this problem is from china analysis book,The authors only said it is creat by Hardy,
 
5:44 PM
Why is this proposition $$x \leq y \Rightarrow x=y \lor x<y$$ is not valid in classical logic ?
 
6:32 PM
@Choups314 "classical" logic ?
 
6:46 PM
Sorry I meant in Intuitionistic logic
 
@Choups314 I have never done that kind so i'm not sure, you might want to ask on main.
 
on main ?
 
the main site
 
Ho ok ;)
 
Hello,@robjohn
6
Q: Prove the $f(x)$ is continuos in open interval?

math110Today,I found a interesting problem: if $$\begin{vmatrix} \cos{x}&\sin{x}&f(x)\\ \cos{y}&\sin{y}&f(y)\\ \cos{z}&\sin{z}&f(z) \end{vmatrix}\ge 0$$ for all $x,y,z$ of an open interval $I$ for which $x<y<z<x+\pi$. show that: $f(x)$ is continuous in $I$ and has finite left-handed a...

 
6:59 PM
@math110 continuous
 
we can prove $f$ is convex?
if we can prove $f$ is convex,I can solve it
if we $f(x)(z-y)+f(y)(x-z)+f(z)(y-x)>0$ then jack explain is convex
But this form,I can't prove it
 
@Choups314 Where in France do you study ?
 
Hippalectryon: In Rennes, Chateaubriand high school
 
Oh ok
 
You are from france ?
 
7:06 PM
Indeed
 
@Hippalectryon didn't know you could do this
 
@MikeMiller Me neither
 
Thank you,@robjohn,your answer is very nice!
 
hell yeah, bikes!
 
r9m
@AlexanderGruber san ??why :'( ..
 
7:08 PM
 
r9m
@math110 nice question !
 
@r9m Why what ? It's obviously missing personal work !
 
@math110 I had written the answer a while ago, but since so many others had gotten close to the answer in comments, I waited to see if any of them would answer. Since none did, I posted mine.
 
Yeah,Thank you
 
r9m
hmm .. I came late at the party :| whatever ..
 
7:11 PM
if $f(x)\cos{(z-y)}+f(y)\cos{(x-z)}+f(z)\cos{(y-x)}>0$ we also $f$ is convex?
 
Greetings
 
@Chris'ssis Hello
 
@Hippalectryon How is it going?
 
Fine, what about you ?
 
@Hippalectryon Preparing other problems for my book. I think I'll have around 300 problems (limits, series and integrals).
 
7:25 PM
:D
 
@Hippalectryon Does it sound good? I'd put more but it's about many pages and I might not be able to publish my book otherwise ...
 
@KajHansen When does it hold that $\log_a n>\log_b n$, knowing that $a<b$ ?
 
Yep !
 
@Hippalectryon OK :-)
The greatest challenge to me now is to publish my book ... :-)
 
@Chris'ssis Don't forget to tell me when it's available !!
 
7:30 PM
@Hippalectryon Do you realize how may professors will initially say it's not possible that someone without any math background publish such a book? Some will say it's something wrong there and they will try to explain that. (this message to remain for the future)
:-)
 
I don't see why it would be "wrong"
 
@Hippalectryon Maybe it's not the proper word in English for what I meant.
Something is rotten in the state of Denmark! I mean for many this will not make any sense. How on earth someone without any math background publishes such a book? lol :-))))) (it will be fun!)
:-)
 
hai guise
 
@iwriteonbananas Hai
 
friday friday gotta get down on friday
@Chris'ssis you will create a global media outrage!
2
 
7:34 PM
@iwriteonbananas :-)))
 
it'll be the new 9/11
 
@iwriteonbananas lolllll :-)
 
GUY WITH NO MATH BACKGROUND PUBLISHES BOOK ON LIMITS SERIES AND INTEGRALS
5
 
@iwriteonbananas Wait, no one said anything about a "GUY". I'm just a sis. :-)
 
dayumm gurrl, are u single tho?
 
ewww reddit
that's for lazy people
 
7:55 PM
@DanielFischer Could I ask you something about Countingsort?
When we have for example the array arr[] = { 10, 6, 8, 2, 3 } we get an array count={1,1,2,3,3,3,4,4,5,5}. What do the numbers of the array count represent and how can we use them?
Hey @Huy
Could I ask you something? When does it hold that $\log_a n>\log_b n$, knowing that $a<b$ ?
 
@r9m for math110's question?
 
8:49 PM
@evinda count[d] tells you how many numbers with last digit $\leqslant d$ are in the original array. By that, you know where to place every number in the sorted (according to last digit, ignoring anything before that) array.
 
@Arkamis q_q indeed
 
user136984
I love Toruses!
 
@Toroidal :D
 
9:29 PM
It brings me joy to pass so many review tests.
 
user136984
Does anybody know where I might happen to find the 200-page paper by Andrew Wiles where he proves the semistable case of the Taniyama conjecture?
 
user136984
It might stop me from having strange dreams about elliptic curves which put me in a strange mood for the rest of the day...
 
I guess it's friday. That badly formatted MATLAB question remains unclosed.
On a normal day, even a perfectly valid MATLAB question is closed within minutes.
 
9:51 PM
@DanielFischer Suppose that $A$ is normal operator. Can you tell me why $R(|A|) \perp N(A)$, where $R$ is image, $N$ kernel and $|A|=\sqrt{A^{*}A}$
 
@Cortizol $\lvert A\rvert$ is self-adjoint. What follows for the relation between $R(\lvert A\rvert)$ and $N(\lvert A\rvert)$?
 
@DanielFischer umm, $N(|A|)=R(|A|)^{\perp}$
 
@Cortizol And how are $N(\lvert A\rvert)$ and $N(A)$ related?
 
@DanielFischer Are they equal?
 
@Cortizol What do you think?
 
9:58 PM
@DanielFischer I think so. Maybe we can use polar decomposition $A=V|A|$ where $V$ is partial isometry?
 
@Cortizol That, or $\lVert Ax\rVert^2 = \langle Ax, Ax\rangle = \dotsc$
 
@DanielFischer $\|Ax\|^2=\| |A| x\|^2$, i am going to kill myself...
 
@Cortizol Okay, if you must, but wait a couple of decades.
 
@DanielFischer Okay, so, for $R(|A|) \perp N(A)$ we don't need actually that $A$ is normal. But is it $R(A) \perp N(A)$ for normal $A$?
 
Morning.
 
10:05 PM
Morning @Mike.
 
@DanielFischer Maybe I know, I think $N(A)=N(AA^*)$ and $\overline{R(A^* A)}=\overline{R(A^*)}$, something like that, and than to use $AA^*=A^*A$ because $A$ is normal.
@DanielFischer I was right (almost...): math.stackexchange.com/questions/235092/…. Thank you for helping me
 
10:46 PM
@r9m I think there's pretty obviously missing context there. "How to prove (equality)" is not a good question. (We don't know what techniques he is familiar with, whether he wants to use imaginary numbers or not, etc. There are many details he could improve the question with.)
 
Does anyone here familiar with Diophantine equation?
 
11:44 PM
That's a fairly broad topic, @Ilya_Gazman Basically, every mathematical problem in the world can be encoded into a Diophantine equation :)
 

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