Mathematics - 2014-03-25 (page 1 of 2)

Pedro Tamaroff

01:05

@seaturtles

Jesus Salas

01:27

Hi, sbdy know how to convert a Graph isomorphism problem to a Subset Sum problem?

Surya

Hi @Pe

Hi @PedroTamaroff would you mind helping me with this math.stackexchange.com/questions/725491/…

I am spent like 2 days on it :\

Pedro Tamaroff

01:46

@Surya Let me try.

@Surya You already have a decent answer, the formatting is bad, that's all.

Surya

02:09

@PedroTamaroff thank you so much

Pedro Tamaroff

Well I didn't do anything at all!

Surya

:)

Pedro Tamaroff

@KarlKronenfeld Hai. I have developed a masochist behaviour which consists of looking at results involving matrix algebra.

Ted Shifrin

LOL@Pedro

Karl Kronenfeld

ew

Stop forgetting my accounts...

02:13

Can anybody tell me how to solve a right triangle given a point and degrees?

Karl Kronenfeld

hi @TedShifrin

Ted Shifrin

hi @Karl

Pedro Tamaroff

@Stopforgettingmyaccounts... SOHCAHTOA. One of those works.

@TedShifrin @KarlKronenfeld Now I am looking at the special case of Bruhat's Decomposition Theorem for ${\rm GL}(n,k)$.

Stop forgetting my accounts...

@PedroTamaroff Thanks for trying, but I know about it. I know everybody thinks it's the golden goose, but there is something else I need to do first.

Pedro Tamaroff

Apparently it is a big-ass theorem.

Ted Shifrin

02:14

GL?

Pedro Tamaroff

Yas.

Mike

Hi @TedShifrin

Ted Shifrin

heya @Mike

Mike

I was gonna answer that Leray cover question earlier but you beat me to it.

Surya

@PedroTamaroff my masochism involves doing problems from spivaks calculus book

Stop forgetting my accounts...

02:15

Can anybody help me here? (easy points)
http://math.stackexchange.com/questions/725585/solve-triangle-given-point-and-angle

Ted Shifrin

I've hardly answered anything in months, so now I answered a few

Pedro Tamaroff

@Surya Spivak is good!

Mike

I got the max rep for a day today from answering a couple LHF.

Ted Shifrin

I even answered a Spivak question the other night (it was actually one of the ones I wrote :P)

Pedro Tamaroff

@TedShifrin That's cheating! Hehehehehe.

Ted Shifrin

02:16

I still had to stop and think about it from the beginning, @Pedro.

Surya

@TedShifrin I remember reading your name in the preface :)

Ted Shifrin

Yup, @Surya ... Fraid so.

Karl Kronenfeld

@Stopforgettingmyaccounts... Are you sure that's all of the information you were provided?

Pedro Tamaroff

@TedShifrin I just realized I never read the prologue. =P

Ted Shifrin

@Mike, I also answered a diff geo question about non-Riemannian connections that was causing consternation

Pedro Tamaroff

02:19

My Spanish version says, roughly: Most new problems are due to Ted Shifrin.

Stop forgetting my accounts...

@KarlKronenfeld the coordinate specified is in reference to the origin of a 2D grid, and yes, that's it.

Mike

@TedShifrin You're answering better questions than I am.

Karl Kronenfeld

@Stopforgettingmyaccounts... Well, where's the origin in your figure?!

I can assume the origin exists, I cannot locate it.

Ted Shifrin

well, @Mike, it's all a competition, you know.

Stop forgetting my accounts...

@KarlKronenfeld somewhere, it really doesn't matter where you draw it. The origin is at (0,0) as it always is. The point (-1, 2) is relative to it. That's all (really).

Karl Kronenfeld

02:20

omg

@stop the problem is that I could draw the triangle really really tiny in my representation of the plane and get the exact same figure as if I drew an enormous replica.

All I need to do is put that one vertex at (-1,2)

Stop forgetting my accounts...

@KarlKronenfeld the bottom of the triangle is on the x-axis. Yeah... I should've probably mentioned that.

Karl Kronenfeld

..

Pedro Tamaroff

@KarlKronenfeld I'm reading dis.

Mike

@TedShifrin I like Lee's book... it's an easy read, and I can easily supplement it with Warner.

Pedro Tamaroff

It includes a proof that ${\rm PSL}(2,F)$ is always simple, for $|F|>3$.

=O

Ted Shifrin

02:23

yeah, Lee's style is quite pleasant.

Mike

In retrospect, Warner is fine. I dunno what problem I had with it before.

Stop forgetting my accounts...

Alright, everybody. I figured it out. I also learned that I should definitely not pursue a career in mathematics, lol.

Ted Shifrin

Neither should I, @Stop

@Mike: I hope you work on some of my more interesting problems I sent you :D

Karl Kronenfeld

I'm starting to read Hartshorne pretty seriously, and I've already realized that I don't have a strong enough background in commutative algebra.

Ted Shifrin

yeah, you need to review Atiyah-Macdonald seriously or Eisenbud-Harris

Karl Kronenfeld

02:41

@TedShifrin I'll probably take a look at Eisenbud-Harris, thanks

Mike

@TedShifrin I haven't worked on any yet. Once I finish the equivalent of ch1 in Warner, I'm going to work through the equivalent of problems elsewhere.

@KarlKronenfeld Did you do the problems in AM?

Karl Kronenfeld

@Mike no, that's something on my todo list though

Mike

that was my problem when I went through it, I think

I just read the chapters and there's so much more content in the exercises

Ted Shifrin

one only learns math by doing lots of exercises ... the more interesting the better

Karl Kronenfeld

I haven't really read the chapters either lol

Mike

02:43

that would do it to you yes

Karl Kronenfeld

I've mainly worked from Matsumura's text

Ted Shifrin

I think the other two are better, @Karl

Mike

don't know it

Ted Shifrin

I looked at Matsumura when I was in grad school, but I never bought it

Mike

"Bought it" as in $ or as in "believed in/agreed with"

Karl Kronenfeld

02:44

I don't think his explanations are all that great.

Anyway, I have to take off. Thanks.

Sush

Suppose $x$ and $y$ are positive integers, $x>y$, and $3x+2y$ and $2x+3y$ when divided by 5 leaves remainder 2 and 3 respectively. it follows that when $x-y$ is divided by 5 the remainder necessarily is 4.

How?

I can see that remainder should be -1.

Pedro Tamaroff

@Sush -1+5=4.

TA-DA!

Sush

@PedroTamaroff, why did you use that fact?

Pedro Tamaroff

Think about it.

Sush

Ok, i got it, sorry.

When we divide, we try to keep the number as big as the dividend or less than it, so remainder can't be nagative, right?

Pedro Tamaroff

03:02

YAS.

Sush

If $a,b,c$ and $d$ satisfy the equations:$$a+7b+3c+5d=0$$$$8a+4b+6c+2d=-16$$$$2a+6b+4c+8d=16$$ $$5a+3b+7c+d=-16$$ then $(a+d)(b+c)$ equals $-16$.

How?

eXtremiity

03:22

Interesting enough, you have 4 equations. Can one hypothesis that if I add/subtract the equations SUCH THAT the RHS is 16, then the resulting LHS should yield (a+d)(b+c)?

Clearly if you add all 4 equations together, you will get this result.

I think I am wrong though. I can't seem to show it.

Sush

@eXtremiity, Thank you!

eXtremiity

Did you get it out?

I don't think I am right.

Sush

@eXtremiity Summing the second and third equations we get a+b+c+d=0. Summing the first and fourth, 6(a+d)+10(b+c)=−16. This is a system in two variables .

He taught me!

eXtremiity

Hmmmm.

Sush

@eXtremiity, the number of different solutions (x,y,z) of the equation x+y+z=10 where each of x, y and z is a positive integer is 36.

How?

eXtremiity

03:37

Btw, the first question you stated. Extremely interesting.

You can treat (a+d) = x and (b+c) = y , and indeed you can solve for them individually.

Sush

@eXtremiity, i think we can take 1 to 10 for x and accordingly y and z. but, how that accordingly takes us to 36?

eXtremiity

I think this can be done using the Axiom of Choice.

Indeed, since, $x,y,z$ are all positive integers, then each $x,y,z \in (0,10]$

Sush

Oh soory, I think we can't take 10 or even 9 as we have to take only positive inegers.

@eXtremiity

eXtremiity

Ahh yes, you are right.

Or 8 ?

If one number is 8, this implies that one other number has to be 0.

Oh no, there are only 3 integers we're looking fro

So yeh, $x,y,z \in [1,7] \in \mathbb{Z}$

I guess you could count your way to your answer.

r9m

03:57

@eXtremiity hello :)
do you know any sites (other than torrents) to download Naruto (and Shippuden) in good quality ?

eXtremiity

Hey r9m. I do not download my episodes. I simply stream them. To answer your question though, Pirate Bay delivers them in 1080p.

r9m

@eXtremiity can't use piratebay ... our insti banned torrent downloads .. =( .. but where do you stream them ?

eXtremiity

Huh? What do you mean?

r9m

@eXtremiity where do you watch it ?

eXtremiity

I use, cc-anime.com

Islands

04:01

Can someone help me show that if I is a radical ideal, then I:J is radical. ( : being quotient operation for ideals)

r9m

@eXtremiity do you know about narutobase.net/Naruto-Anime-Downloads.html ... its available in 720p here ..

didn't know about it till yesterday .. :)

eXtremiity

Oh that is awesome. I still stream though. I don't usually keep these files on my computer.

r9m

@eXtremiity you have good internet connection .. but our wifi is very slow ,., usually can't stream them online ..

eXtremiity

I see.

r9m

and on top of that torrents is banned .. disaster !!

eXtremiity

04:08

Where are you from?

r9m

India ..

eXtremiity

So what do you usually do to stream these videos?

r9m

I used ZBigZ like sites to download them .. then download them (at super slow speed :( )

eXtremiity

Hmmm, perhaps download them while you're sleeping.

r9m

ya

eXtremiity

04:12

Or studying.

r9m

same :)

eXtremiity

:]

r9m

you watch other animes ?

eXtremiity

No, but I am going to start getting into some.

Samurai Shamploo is one that I am keen to watch.

r9m

okay .. I watch a boxing anime .. Hajime No ippo .. thats not bad at all :D

eXtremiity

04:14

Cool :D !

r9m

@eXtremiity bye .. gtg :D

eXtremiity

Cya @r9m !!!

Islands

Can someone help me show that if I is a radical ideal, then I:J is radical. ( : being quotient operation for ideals)

eXtremiity

@Sush.

Interesting enough, the answer is $8+7+6+5+4+3+2+1$.

Jessy Cat

05:24

How come Swarnik never comes in here anymore? He's always only off in his little private room :(

Good night, everyone!

Sush

06:11

@eXtremiity, please explain it!

why 8+7+6+5+4+3+2+1?

Sush

06:22

Can someone please answer this: "A,B and C are three commodities. A packet contains 5 pieces of A, 3 of B and 7 of C costs $24.50 . A packet containing 2,1 and 3 of A, B and C respectively costs $17.00 The cost of a packet containing 16, 9 and 23 items of A,B and C respectively is......" and answer is $100

But i can't understand how the answer is derived. it is 3 variables in 2 equations! so isn't the info lacking?

Sush

07:03

@skullpatrol, HI!

skullpatrol

Hi :-)

Vrouvrou

08:14

hello

please what is the difference between a Fredholm operator and a Fredholm map ?

eXtremiity

08:33

@Sush. Well, it follows from the multiplication principle. Some like to call it the axiom of choice. The multiplication principle states that if you can put $a$ number of things in one set, $b$ in a different set and $c$ in a set different from the other 2, then the number of combinations you can make is $ a \times b \times c$.

So I keep $x$ fixed. I.e, let $x = 1$. Next choose $y$. You can choose $8$ things for $y$. This results in $z$ also being fixed. Thus the number of combinations you can make is 8 IF $x$ is fixed as 1.
Now, say $x$ is fixed as 2. Well then, $y$ can only be 1,2,3,4,5,6,7. I.e. there are only 7 things for $y$. $z$ stays fixed as follows. You continue this pattern and you have
8+7+6+5+4+3+2+1. Furthermore one may think about the possibility of rearranging the order. But the way I have described the combination is one that accounts for all combinations of all orders. Therefore one should be conv…

Sush

@eXtremiity, thank you, so much, sir!

eXtremiity

You are welcome.

Sush

I would like to give some of my reputation to you.

eXtremiity

Haha, is that possible?

Sush

I don't know !

By the way, why is $z$ fixed?

@eXtremiity

eXtremiity

08:43

Ok, lets do a few examples

( * ) , ( * ), ( * ). <- These are the sets.

Sush

yes.

eXtremiity

(1), (1)...therefore z has to be ?
(2), (1)...therefore z has to be ?
(3), (1)...therefore z has to be ?
(4), (1)...therefore z has to be ?

As you can see, $z$ has to be fixed in a way to ensure that the sum is equal to 10.

I guess one can say that $z$ can only be one number as a consequence of $x$ and $y$.

Sush

(1), (1)...therefore z has to be 8, am i right?

eXtremiity

That is correct.

But $y$ can be [1,8]

So under the multiplication principle think of it like this:
(Keep $x$ fixed. Thus, I can only put 1 number here) , ( I can put any numbers from [1,8]. Thus I can put 8 numbers here), (As a consequence of the previous 2, I can only put the REMAINING number to satisfy the equation).

Hence $ 1 \times 8 \times 1$

Sush

Let me try! we are assuming x=1 and so 1+y+z=10 so z=9-y and as z can take values 1 to 8, so y can be 1 to 8.

@eXtremiity

Right?

eXtremiity

08:49

Well, given that equation, yes.

But the multiplication principle is where you have to be heading towards in order to generate the number of combinations of $x,y,z$.

skullpatrol

09:22

Hi @Sush :D

Sush

09:33

@skullpatrol, Hi!

@skullpatrol, are you a student? Will you please help me with US academic system?

I have heard that courses there are not that tough, but they let them grow in practical life, right?

Vrouvrou

09:51

please what is the difference between a Fredholm operator and a Fredholm map ?

Sush

@Vrouvrou, ask your question to Robjohn. He may help you. add @ and then robjohn to ask him.

Vrouvrou

@Sush thank you

@robjohn please what is the difference between a Fredholm operator and a Fredholm map ?

Sush

@Vrouvrou, you are welcome. He has always been the helping hand when i lose my hope.

robjohn

@Vrouvrou why do you think there is a difference? I think that they can refer to the same things.

Vrouvrou

ok thank you so there is no diffrence

Sush

10:11

@robjohn, for any two sets $S$ and $T$, let $S\delta T=(S\cup T)-(S\cap T)$.Let A,B,C be sets such that $A\cap B\cap C=\phi$ and numbers of elements in each of $A\delta B$,$C\delta B$ and $C\delta A$ equals Then the numbers of elements in $A\cup B\cup C$ equals

But I think that should be the least number as there may be some elements in some intersection, e.g. $A\cap B$, right?

robjohn

@Sush I think some of your message is missing. However, $|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|-|B\cap C|-|C\cap A|+|A\cap B\cap C|$

@Sush This follows from the Inclusion-Exclusion Principle

In fact, the formula above is on that page

Sush

for any two sets $S$ and $T$, let $S\delta T=(S\cup T)-(S\cap T)$.Let A,B,C be sets such that $A\cap B\cap C=\phi$ and numbers of elements in each of $A\delta B$,$C\delta B$ and $C\delta A$ equals 100.Then the numbers of elements in $A\cup B\cup C$ equals 150 as my book says.

Ilya_Gazman

Hi guys. I think the question I just asked is not clear enough. Can you help me please?

0

Q: Calculate average formula of random numbers

I have array of $N$ random numbers between $\min$ to $\max$. I am building a $2$ dimension graph where $Y$ dimension are the array numbers, and $X$ dimension are neutral numbers from $0$ to $N$. How can I find the formula of straight line that will be at minimum distance from all the points on t...

linear-algebra

Sush

@robjohn, i corrected my question. please let me know why is that just 150 and not bigger?

for any two sets $S$ and $T$, let $S\delta T=(S\cup T)-(S\cap T)$.Let A,B,C be sets such that $A\cap B\cap C=\phi$ and numbers of elements in each of $A\delta B$,$C\delta B$ and $C\delta A$ equals 100.Then the numbers of elements in $A\cup B\cup C$ equals 150 as my book says.

Let here red, blue and yellow parts have 50 elements each and green also has 25 elements. So, isn't $A\cup B\cup C$ has 150+25 elements?

@robjohn, hope i am clear now.

@JasperLoy, hello!

Jasper Loy

10:42

@Sush Hi, your circles are very colourful, lol.

Sush

Ya! will you please answer me? i can't understand why 150 is the answer. I think 150 is just the minimum digit.

@JasperLoy

robjohn

11:05

@Sush $|A\Delta C|=125$ and $|B\Delta C|=125$

Sush

11:34

@robjohn, thank you so much:)

Ilya_Gazman

12:03

Hello

Sush

Hello!

Ilya_Gazman

Hi, do you know Linear regressions?

Sush

Suppose that A,B and C are sets satisfying $(A-B)\Delta(B-C)=A\Delta B$ then $A\cap B=B\cap C$ .

Ilya_Gazman

Cool

Sush

But i get A=C. how am i wrong?

Ilya_Gazman

12:06

Sorry I don't kno

Sush

i learnt LR but am not good at

Ilya_Gazman

know*

Sush

@Ilya_Gazman, its ok!

Ilya_Gazman

;)

Sush

do you kno or know* that you can edit your posts? (lol:))

Ilya_Gazman

12:07

Hi @Vrouvrou! Do you know Linear regressions?

Yeah forgot about that...

Sush

do you have some question about Linear regressions?

@Ilya_Gazman

Ilya_Gazman

yeah

0

Q: Calculate Linear regression segment

I have array of random numbers. How can I calculate linear regression segment? I am interested in finding the exact formula so I be able to use it in my work, please help me finding this formula with the next declarations: $N$ - the number of random numbers in the array $S$ - the sum of the num...

linear-algebra statistics

I just want a simple formula

Sush

OK. if you can use [enter description](link) then that will not look cumbersome

@Ilya_Gazman

Martin Sleziak

Is there any logical reason why there were several changes of the tags on this question by the same user? In fact all suggested edits by that user so far have been edits which left question with only one tag.

Sush

Please answer this

@MartinSleziak

@MatsGranvik please help!

Does someone know set theory?

Martin Sleziak

12:25

Cannot you somehow use $A-B=A\triangle (A\cap B)$ and the associativity of $A\triangle$?

It seems that your assumption is equivalent to $A\triangle B\triangle (A\cap B)\triangle (B\cap C)=A\triangle B$.

Not sure whether it least somewhere, just the first idea that came to mind, @Sush

I think $X\triangle Y=X$ implies $Y=\emptyset$.

Using this and the above equality you get $(A\cap B)\triangle (B\cap C)=\emptyset$ and you are done.

Stefanos

@Sush @sush what is \Delta?

Martin Sleziak

I suppose $\Delta$ is notation for symmetric difference.

BTW I think that would have been an acceptable question for main. (Maybe you would get other solutions.)

Leaving for lunch, see you later!

Sush

12:47

@MartinSleziak, i think you mean $A-B=A- (A\cap B)$

Sorry, have your lunch:)

Martin Sleziak

@Sush No, I meant $A-B=A\triangle (A\cap B)$.

This is true, too. And it seems to be suitable for the problem you stated.

Since thenyou can use other properties of $\triangle$.

By definition $A\triangle (A\cap B)=[A-(A\cap B)]\cup[(A\cap B)-A]$. Since the latter set is empty, you get $A\triangle (A\cap B)=A-(A\cap B)=A-B$.

Sush

@MartinSleziak, Thank you!

Jasper Loy

I got another downvote. Someone here must be angry with me for something, lol.

user134853

good after noon everybody :)

Sush

@MartinSleziak, how does $P\triangle Q=\emptyset$ imply $P=Q$?

Jasper Loy

12:59

@robjohn I got a downvote yesterday and another downvote today on old answers, lol. I think I have made an enemy.

Martin Sleziak

Suppose $P\ne Q$. This means either that there is some $x\in P$ such that $x\notin Q$, or the other way round. (There exists an $x$ such that $x\in Q$ and $x\notin P$.)

What can you say about $x$ and $P\triangle Q$ then?

***********

Or you could use associativity: If $P\triangle Q=\emptyset$, what can you say about $(P\triangle Q)\triangle Q$?

user134853

@JasperLoy i up vote you ;)

Martin Sleziak

(The first approach seems to be more elementary, but the second one seems more elegant to me.)

Sush

@MartinSleziak, trying!

Jasper Loy

@user134853 I see you are a new member.

user134853

13:02

yea

Jasper Loy

This downvoter must be crazy to cast exactly one downvote on me each day.

Unfortunately, I have no suspects.

user134853

i understand it, i got downvoted once only because i told someone that he did not understood my question.

Sush

I can say that $x\in (P\cup Q)\cap(Q'\cup P')$

@MartinSleziak

am i right?

Martin Sleziak

No, if $x\in P$ and $x\notin Q$, then you get $x\in P\setminus Q$ and, consequently, $x\in P\triangle Q$. Which means $P\triangle Q\ne\emptyset$ and you get a contradiction.

The same works in the case $x\in Q$ and $x\notin P$. (They are two different cases.)

Sush

Can I solve this problem using venn diagram? As I can't use such high-level reasoning in the exam!

@MartinSleziak

Martin Sleziak

13:08

There was no high level reasoning used at any step. But let's try a different approach.

You know that $P\triangle Q=\emptyset$?

user134853

Ok now for some math,
http://math.stackexchange.com/questions/725679/claims-on-the-basis-of-lim-limits-x-to-inftyfgx-l/725767?noredirect=1#725767
it is not mine but i have been trying to solve it by myself and i am stuck a bit,
i really wanna know the answer because it is exactly the last subject i was studying

Martin Sleziak

What can you say about $(P\triangle Q)\triangle Q$.

user134853

it is Calculus limit

Martin Sleziak

@Sush Why don't we continue in this room so that this is not interrupted by other conversations?

Sush

I can say that it is Q

user134853

13:12

@JasperLoy how are you in calculus?

Jasper Loy

If I get another downvote tomorrow, I will get a mod to investigate it, be warned!

3

user134853

there is no reason not to

skullpatrol

@JasperLoy investigate what?

Jasper Loy

@skullpatrol My downvotes, lol. See my rep history.

skullpatrol

@JasperLoy people are allowed to have opinions, no?

user134853

13:16

he got a devote war on him

Jasper Loy

@skullpatrol Yes, but one yesterday on an old answer, and today again, so it is suspicious.

@user134853 Moreover, I have not cast any downvotes myself.

skullpatrol

@JasperLoy If it is a serial down voter the system will catch them pal

Jasper Loy

@skullpatrol There are ways to do it without the system catching, lol.

skullpatrol

@JasperLoy there are ways to not let it bother you also.

Jasper Loy

It is possible that the downvoter think I should not answer closed questions.

However, when I answered it the question was not closed yet, lol.

And there is no way to answer if it has been closed.

skullpatrol

13:21

@JasperLoy All I'm saying is don't take this voting business so seriously.

Jasper Loy

@skullpatrol Yeah I am not, don't worry, lol.

skullpatrol

@JasperLoy It is all electronic 1's and 0's anyway :-)

Jasper Loy

@skullpatrol Yes, and 100k users are not necessarily good in math, lol.

skullpatrol

Yep.

Martin Sleziak

@MartinSleziak The same user is discuss in SO chat.

Ben Porat

14:00

Can someone help with a calculus limit problem?

Sawarnik

14:44

@BenPorat Like? It means that I would like to see the problem but there is zero chance I might help you :v

eXtremiity

14:57

@BenPorat. I am on the same lines.

If it clicks, it clicks :p .

Ben Porat

15:15

@Sawarnik @eXtremiity math.stackexchange.com/questions/725679/…

that is the problem not even mine but i have to know how to do it

because that is the lest subject i was studying and my situation is not good if i don't know how to answer it

eXtremiity

∨ Means or, correct?

Ben Porat

yea

that one you need to disproof with 2 Dirichlet functions

eXtremiity

He claims it is correct.

Anyway, I do not know how to do it

Ben Porat

Yea i saw, but it is not, i am stuck on 3 and 4
thanks for your time @eXtremiity :)

eXtremiity

Apologies for not providing more help.

Ben Porat

15:24

No need to apologie, atleast you tried

Ramana Venkata

15:39

What is the equivalent statement for $\lim_{x \rightarrow \infty} f(x) = \infty$

in terms of the $\epsilon-\delta$ definition i.e.

robjohn

@JasperLoy I got several last week. Perhaps they choose someone new each week :-)

Sawarnik

@robjohn Too sad the mods cant do anything :p But anyways, why doesn't that come to stop? Dont they realize that their rep is dwindling?

robjohn

@Sawarnik It only costs a point here and there. Not too hard a habit to support.

Sawarnik

@robjohn Hmm, that means established users are doing this?

robjohn

@Sawarnik could be. Remember it takes 125 rep to downvote.

@RamanaVenkata for any $N\gt0$, there is an $x_N$ so that for all $x\ge x_N$, $f(x)\ge N$.

Hawk

15:53

@robjohn do you feel there should be a system to see who has downvoted so that they can be contacted when it seems to be irrational?

Ramana Venkata

@robjohn Thanks

robjohn

@Hawk It kind of ruins the anonymous voting idea.

@Hawk but it would really be nice if people did comment when they downvoted. They don't need to indicate that they did indeed downvote, but at least some reason why the downvote was cast would be nice.

Sawarnik

@RamanaVenkata Reading you blog, mathstack.wordpress.com/2011/07/22/… . I agree with most parts.

Hawk

@robjohn yes, that is true...but not many people follow that...do they?

Sawarnik

@robjohn Right, whenever people click the downvote button, they should be made to write a small note describing why did so, and some specific people could access that.

robjohn

15:57

@Hawk it depends. If the downvote was for no good reason, they usually don't comment.

Hawk

@robjohn yes, but aren't there any way to counter this? should there be review for downvotes too?

@Sawarnik no, if that is the case, then anynymous voting is hampered as robjohn said.

Sawarnik

Why does downvoting need to be so anonymous? I don't understand that.

Hawk

@Sawarnik There are many busy and highly skilled people on math.se who cannot spend so much time to comment along with downvoting because there are many cases when downvoting is obvious due to the low-standard.

Sawarnik

@Hawk Atleast for some newer users? But yeah, its practically not good to install a review or ... But atleast I think just mods should be able to see who downvotes? Why does downvoting need to be so anonymous, again?

Hawk

@Sawarnik yes, mods ability to see who downvoted can be thought upon, and also there can be a system where users with reputation below a specific mark have to explain the downvote

But, definitely, when new users explain their downvoting, there might be a chance of heated discussion upon how justified the downvoting was.

Sawarnik

16:09

@Hawk Ok, me has a q for you. Show that there is no root of $x^{35}+\dfrac{20205}{2+x^{17}+\cos^2x}=100$.

Hawk

@Sawarnik looks good, thanks :D

@robjohn would you please check this problem, and please see if you can spare some time to provide an elegant and elementary solution to this problem. Also, offer your comments, if the question should be closed or not, as I have already received 3 close votes on the question.

@robjohn math.stackexchange.com/questions/726046/…

Sawarnik

@Hawk Will Viete s formulas help? Just guessing so dont take it seriously.

Hawk

@Sawarnik for which question, mine or yours?

Sawarnik

@Hawk For yours of course. And shouldnt the q be, If λ is any real root of the polynomial...?

Hawk

@Sawarnik if that had been the case, then it would be a very trivial problem. See my attempt, you will understand

Sawarnik

16:18

@Hawk Oh, but then what does $|c|<1$ actually mean for some complex c?

Hawk

@Sawarnik if $c=a+ib$ then $|c|=\sqrt{a^2+b^2}$

@Sawarnik So, it means $\sqrt{a^2+b^2}<1$

Sawarnik

Oh, ok.

@Hawk Are you trying my q?

Hawk

@Sawarnik I have got the basic idea, and now searching for some pen and paper to execute the idea.

Hawk

16:44

@Sawarnik I think it is done.

Sawarnik

@Hawk How?

Hawk

@Sawarnik Derivative, double derivative to show the minimum value never reaches -20005

@Sawarnik Though I accept my solution is neither comprehensive nor elegant

Sawarnik

@Hawk Well here is the actual q, math.stackexchange.com/questions/70838/…

Hawk

@Sawarnik admittedly a far better solution than mine and much more elegant. thanks

Sawarnik

@Hawk Ok, you want some geometry?

Hawk

16:54

@Sawarnik how do you find these questions? they are quite old yet good...

@Sawarnik no geometry, please...they are nightmares and i would not be needing them for my entrance.

Sawarnik

@Hawk By clicking some upvoted links on the related questions column of good questions.

Oh. Which enterance? JEE?

Hawk

@Sawarnik no, i dropped out of engineering for math...don't you know? I have said that before.

Sawarnik

@Hawk Some trig proofs? [Not like the usual ones.]

Hawk

I didn't see the (removed) msg...what did you post there?

@Sawarnik yes I don't have any problem with trig

Sawarnik

Asking you which univ you are aiming at?

Some real $x,y,z$ satisfy:

$$\frac{\cos x + \cos y + \cos z}{\cos(x + y + z)}
=
\frac{\sin x + \sin y + \sin z}{\sin (x + y + z )}
= p$$

The prove that: $\cos (x + y) + \cos (y + z ) + \cos (x + z) = p$

Transcript for

$Mathematics$ Mathematics