Mathematics - 2016-09-05

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I do math art

00:03

@r9m It's late here and in 4 hours I have to get up.

I'm out to take some sleep.

r9m

@Idomathart good night! :-)

I do math art

@r9m I cannot answer you the same! Good luck with your exam! ;)

r9m

@Idomathart :D thanks!

I do math art

@r9m btw, I was asking myself now if any in this country work at this hour on crazy integrals and series. :-)

r9m

@Idomathart well if there are chances are they wouldn't meet you is low :P .. they are working so late .. naturally they'd be exhausted next morning .. and hence chances of a decent conversation that might eventually lead to math tends to 0 :P

I do math art

00:11

@r9m :-))) true

r9m

well I'm only joking :P .. I suppose if one of 'em breaks the cycle of hard work maybe they can spot another 'en road :)

I do math art

@r9m :D btw, do you have in mind a math area you didn't study intensively and that you'd like to study it like that in the future? Something attractive. Maybe you give me ideas.

r9m

@Idomathart well I haven't learnt so many things ,, measure theory, Functional analysis, differential geometry (among things I look forward to learning) ..

I do math art

@r9m OK

@r9m Hope to seriously learn complex analysis as a first objective after I finish some stuff here. I have more books I have to study in detail.

Jonathan Richard Lombardy

Hello everyone, I am new here. I am starting to learn real analysis and algebra, and was hoping to get some book recommendations for self study. I am sorry if I violated any rules.

r9m

00:20

@Idomathart ya .. and that complex analysis .. :) I like it ..

@JonathanRichardLombardy well if you are doing a first read .. I'd suggest Bartle -> Apostol (Mathematical analysis) + Rudin (principles of mathematical analysis) in that order for analysis :) ..

Algebra .. maybe Herstein along with Dummit and Foote for group thory .. and don't ask me about linear algebra .. Hoffman Kunze maybe?

Pavel

00:45

Hello, I was wondering, who proved the fact that for every positive integer n there is a gap of n consecutive non-primes? Does this theorem have a name?

1 hour later…

Jonathan Richard Lombardy

02:03

@r9m @r9m thanks, will try them for analysis, I am reading Algebra by M. Artin any opinions?

2 hours later…

r9m

04:10

@JonathanRichardLombardy Artin's algebra is nice too ..

Guacho Perez

04:23

Hello

I have a question regarding multiple integrals and their bounds after a change of variables that no one has answered yet and is causing me trouble

Guacho Perez

05:08

How can I determine the bounds on multiple integrals after a change of variables or if I wanted to change the order of integration without diagrams?

Most often I see professors drawing a picture to find the bounds, but there will be times when pictures are not availabe (4 D space), what then? Is there a systematic method for determining the bounds without recurring to diagrams or spatial intuition?

1 hour later…

Martin Sleziak

06:28

Since I would like to get the chat room for bounties started, it seems reasonable to advertise it here. Here is link to the chat room and here is the relevant post on meta.

morbidCode

Hi, can I ask for help here? I don't think it would be a valid question in the chat.stackexchange site

Tobias Kildetoft

@morbidCode yes, this is often a good place to ask for help, though obviously whether people can help you depends on whether the right people are present.

morbidCode

06:44

@TobiasKildetoft thanks. I am currently learning graphs and slopes and I am wondering if it can be explained in a way that you don't need to look at the graph and the coordinates. My problem is I am blind so I really want some resources that can help me learn these things on my own. Is that possible?

Tobias Kildetoft

@morbidCode That is a very good question that I unfortunately have no idea about.

morbidCode

That is fine. I do have a more important question though. Here in
https://mitpress.mit.edu/sicp/chapter1/node22.html
and footnote 61
https://mitpress.mit.edu/sicp/chapter1/footnode.html#2458
there is a section regarding the newtons method for finding square roots. The problem is I absolutely don't understand how it works. I have a basic understanding of derivatives from asking a question in the site, but I absolutely can't apply my understanding to this algorithm. Do you mind taking a look? How can I go about learning these things?

Tobias Kildetoft

@morbidCode That site seems to be much more focused on how to do these things algorithmically, rather than how they actually work

arctic tern

when you say how, do you mean why?

Tobias Kildetoft

@arctictern same thing in this context (well, essentially)

morbidCode

07:02

@TobiasKildetoft sorry I'm unclear. I think I don't need to know how it works, I think knowing "how to do these things algorithmically" as you said is enough. Can you explain to me what the algorithm is doing to get the square root of a number?

Tobias Kildetoft

@morbidCode Unfortunately not. The code is just too different from the various languages I know (or at least the ones I remember)

Balarka Sen

Hi @Tobias, @arctic.

Tobias Kildetoft

@BalarkaSen Hi

morbidCode

@TobiasKildetoft I think there is a formula here in footnote 61 mitpress.mit.edu/sicp/chapter1/footnode.html#2458 can you explain this formula?

Tobias Kildetoft

07:18

@morbidCode Not without more context than provided there, no. Possibly the wikipedia page will be more help

morbidCode

@TobiasKildetoft I see. But do you know what field of math is required to learn these things?

Tobias Kildetoft

@morbidCode As I recall, Newton's method for square roots uses some fairly simple calculus together with a good idea.

morbidCode

@TobiasKildetoft I will try to learn calculus as soon as I can. Thanks!

Martin Sleziak

@morbidCode Maybe you can find some advice at matheducators.SE. I have seen that they had some posts about teaching blind or visually impaired students.

morbidCode

@MartinSleziak thanks. I'll look into that

2 hours later…

Tobias Kildetoft

09:12

Just received an email inviting me to send my papers to Journal of Mathematics Research. Claims to be index by MathSciNet, but was in fact only index for about a year, then stopped being indexed like 3 years ago.

SteamyRoot

It's not on Beall's list, unless you're talking about the International Journal of Mathematical Research & Science

Tobias Kildetoft

@SteamyRoot The publisher is on the list (he only lists individual journals when the publisher is not on the list)

SteamyRoot

Ah...

Well, at least their website seems somewhat credible for a publisher

The one I mentioned has a hilarious website: ijmrs.org

Tobias Kildetoft

Sure, they also did manage to get indexed for about a year. No idea how terrible their papers are

jeez, the scrolling news thing is just awful

Agawa001

09:56

@r9m you bugger i thought i m crosseyed when i saw the double occurence of ur name

mmm or maybe a bug

Noah P

10:15

How come the prime numbers don't become more infrequency at higher ranges?

r9m

@Agawa001 where? I didn't understand what you mean by double occurence of my name?

SteamyRoot

11:09

@NoahP they do? The number of primes $\leq n$ is approximately $n / \ln (n)$.

1 hour later…

Agawa001

12:32

@r9m just a bugging bug

any discoveries yet ?

1 hour later…

Guacho Perez

13:58

How can I determine the bounds on multiple integrals after a change of variables or if I wanted to change the order of integration without diagrams?
Most often I see professors drawing a picture to find the bounds, but there will be times when pictures are not availabe (4 D space), what then? Is there a systematic method for determining the bounds without recurring to diagrams or spatial intuition?

1 hour later…

arctic tern

15:05

@GuachoPerez you're basically asking for a systematic way of figuring out the image or preimage of a region under a differentiable map

I guess one thing to say is to think about limiting cases (boundaries), but I don't know what there is to say in complete generality

Balarka Sen

16:01

Hi @Krijn

Krijn

Hey @Balarka!

Balarka Sen

What's up?

Krijn

My class isn't cancelled because enough people showed up :D!

Balarka Sen

Cool

Krijn

What are you doing today?

Balarka Sen

16:03

A bit of topology, a bit of complex analysis.

Krijn

You've been doing that for some months now, right?

Balarka Sen

I guess.

arctic tern

entire hours throughout the month

Balarka Sen

i am a slow learner

Krijn

Today in class we discussed the Artin Kern (Artin kernel in english, probably) and it reminded me of some user in this chat

Balarka Sen

16:06

hehe nice pun

arctic tern

takes too long to figure out pun

Balarka Sen

"artin kern", "arctic tern"

arctic tern

I know, just saying it took me more time than I expected.

Balarka Sen

ah

arctic tern

was wondering who in chat was like Artin

Basem Fouda

16:31

hi

anything interesting here?

user227867

16:46

@JonathanRichardLombardy Hello Johnathan. I am very glad you asked about books. I recommend Protter and Morrey's A First Course In Real Analysis and Jacobson's Basic Algebra 1 for you. Good luck.

amWhy

17:02

@JonathanRichardLombardy Hi, Jonathon! What did you use (text) for calculus? If you feel strong in your understanding of Calculus, Apostle might be a good next step. For a beginner in algebra, Fraleigh's test is a great start, from which you'll have a great basis to jump to Dummit and Foote.

@JonathanRichardLombardy There are a lot of suggestions for self study on the main site, too. Just search each topic, and you're sure to come across many helpful resources and perspectives.

DemCodeLines

17:21

20 + 8 mod 6...In this expression, what's the correct order of operations?

Do you add first and the find the remainder of 28 and 6? Or do you find the remainder of 8 mod 6 first, then add it to 20?

arctic tern

17:53

@DemCodeLines if a mathematician is using mod as a binary operation (rather than the usual congruence relation), then I'd bet they'd intend it as (20+8) mod 6.

we like to do operations with residue classes, and that's what that would be representing

A---B

18:23

How can i approximate $\sum^{10,000}_{k=2} {1\over\sqrt{k}}$ without integration ?

arctic tern

any particular reason you want to approximate a (well, essentially a) Riemann sum without being aware that's what it is?

A---B

What is Riemann sum ?

arctic tern

(a) something you can google (b) something you should learn about when you learn integration (c) something you would need to know in order to realize at all that the sum can be approximated with integration

which leads me to the question, how did you know that sum can be approximated with integration without knowing what a riemann sum is?

or, is your unfamiliarity with integrals the reason you're asking for a way to do it without integrals?

A---B

the or is the reason

arctic tern

I see

it might be interesting if there were a integral-free way, but you seem to know about limits and analysis to some extent so you might find learning some calculus to be worthwhile in and of itself

A---B

18:34

It was oly question which i can't do. when i asked my teacher she told me that we need to use integration but also told not to delve too much in it, or i will confuse things.

i know integration but not riemann sum.

never heard of it :(

I do math art

That's a pretty easy thing to approximate ...

(just back)

Just think of how you can write, say,

$$\sqrt{k+1}-\sqrt{k}$$

arctic tern

that's a good way

Agawa001

@Idomathart heya wats new ?

I do math art

and note that the form above taken as a summand leads immediately to a telescoping sums which we love to work with.

A---B

partial fractions

?

arctic tern

18:38

@Idomathart pedagogically, I'd use "simple" rather than "easy," because "easy" could be interpreted as "easy to find" instead of "easy to understand," and the former would be condescending or downputting to a lot of students \teachermode

depends on audience I guess

I do math art

@arctictern Well, in English they don't make much difference to me, but I got your point. :-)

arctic tern

@A---B artist-person is hinting at rationalizing the numerator of $(\sqrt{k+1}-\sqrt{k})/1$, then using $\sqrt{k}\approx\sqrt{k+1}$

since $$\sqrt{k+1}-\sqrt{k}=\frac{1}{\sqrt{k+1}+\sqrt{k}}\approx \frac{1}{2\sqrt{k}} $$ we can rewrite $\sum 1/\sqrt{k}$ as $\approx 2\sum(\sqrt{k+1}-\sqrt{k})$ which is telescoping

I do math art

Yeap

A---B

wooo thank you very much, @Idomathart and @arctictern

I do math art

user image

@Agawa001 hey! Not the best day (or one of them) of my life here. How about you? :-)

Agawa001

18:45

@Idomathart neither me, i had more promising days

I do math art

@Agawa001 hehe, I do too!

Agawa001

maybe i m getting old

I do math art

@Agawa001 better think positively :-)

Agawa001

perhaps, lack of social motivation too

mm i need to bring latex back

and always refering to robjohn official latex sponsor in chat

I do math art

@Agawa001 May bad days are in general related to something in mathematics that is not as I would like, or anything related to it.

Green tea and mountain honey to raise up my spirit! ;)

BBL

Agawa001

18:54

lol, i know the effect of coffee rather better for my case

I do math art

@robjohn hey! You missed a lot. Today I was preparing to ask you something, then I went to go jogging, and now I see you around (after many days I think).

robjohn

@Idomathart How are things?

I do math art

@robjohn My research is still as I wish it to be, but I think I'm pretty slowed down with the publishing part ... (too much waste of time for publishing a book)

Annoying.

@robjohn Why did you miss so much? I was thinking you might be in some vacation, but I remember that you entered here also in those periods, every year ...

MATH ASKER

quick question

How many integers between 1000 and 9999, inclusive, contain no fives?

I don't think I understood the question? So no fives like in the end?

How do i solve this problem?

I do math art

Let me write an infinite series

MATH ASKER

19:07

? I don't getit

Jake1234

[adjoint functor]https://en.wikipedia.org/wiki/Adjoint_functors

hey guys, would someone please help me understand the universal morphism definition? If I take the "left adjoint functor" , and I do the construction, obtaining some functor $G$, does that $G$ satisfy the definition of a "right adjoint functor", I would have thought so, but I don't see how I would show that.

Tobias Kildetoft

@Jake1234 Yes, if $G$ is right adjoint to $F$ then $F$ is left adjoint to $G$ (and vice versa)

Jake1234

yes I realize that, but I don't see how it follows from the definition.

Tobias Kildetoft

@Jake1234 depends on which of the equivalent definitions you pick

Jake1234

The universal morphisms one

Tobias Kildetoft

19:10

I don't recall how one does it. You should be able to find it in most standard references, such as Mac Lane

It may be easier to first show equivalence to one of the more symmetric definitions

I do math art

@MATHASKER This you can do alone too. Initially I was thinking of a nicer question and a bit harder.

MATH ASKER

@Idomathart I don't know how to do it though?

Jake1234

@MATHASKER I understand it as "a number between 1000 and 9999, such that non of the digits are 5" - the possibilities are - 1st digit 8 (9 possible numbers on the first digit, you can't choose 0), then 9 on 2nd-4th digits. So you have 8*9^3 ?

@TobiasKildetoft Hah that's good advice, I don't understand why I didn't look in some book first...

I do math art

@MATHASKER Initially I was referring to the meaning of

$$\biggr \lfloor \frac{n}{5} \biggr \rfloor+\biggr \lfloor \frac{n}{5^2} \biggr \rfloor+\biggr \lfloor \frac{n}{5^3} \biggr \rfloor+\cdots$$

MATH ASKER

@Jake1234 I don't understand 9999-1000=8999 so out of that they can't be fives right?

Jake1234

19:16

I don't know what you're saying. You have all those numbers, {1000, .... , 8999} and you're asking how many there are of those, that have no 5 in any of their digits.

MATH ASKER

Oh so thats what the question was asking, so the number can't be like 4052?

because it has a 5 right>

Jake1234

If we asked the same question for numbers from 10 to 19, including 19, we would have a set {10, ... , 19} (10 elements), and 1 of them would contain a 5 - 15. So there's 10 numbers, 9 don't contain 5, 1 does.

Yeah that's how I understand it.

MATH ASKER

OH

Oh so for this one because its (1000...8999) and because in each 1000-11000 it contains a digit that has one 5 out of the 10 right

Up until this point I get it but why would I mulptiply 8*9^3 @Jake1234

I do math art

@robjohn I also had to reinstall all my stuff on one of the computers- blue screen - and nothing worked anymore. Pretty ruined day.

Jake1234

Try to look at all the possible numbers between 1000 and 8999, that don't contain a 5.

First digit can be either 1,2,3,4,6,7,8,9 - that's 8 numbers. The second digit can 0,1,2,3,4,6,7,8,9 - 9 numbers, same for 3rd and 4th digit. So all the possibilities are 8*9*9*9

robjohn

19:27

@Idomathart That is bad. any idea what happened?

MATH ASKER

Ohh okay

I do math art

@robjohn That's wierd, really, it's not even connected to internet, and on it I keep all my mathematical stuff. I use it only for this. I don't see why I got blue screen out of a sudden. After reinstalling OS, things seems to be just fine. All data is fine and saved since I have more back-ups. On its HDD all data remained intact.

MATH ASKER

Now i get it

thanks

Evinda

Hey!!! Does anyone of you speak turkish?

I do math art

@robjohn I meant all important data which is mathematical stuff.

@robjohn I should have seen me as I reacted when I saw the blue screen in front of my eyes :-)))

Eventually all is fine and back to normal.

@robjohn What was also interesting is that Win XP CD didn't have active the repair option anymore, so I couldn't do much. Then I tried some stuff with the repair console, but it didn't work.

user227867

19:39

@Idomathart You misspelled 'weird', LOL.

I do math art

@robjohn an expert on some site said that in these conditions there isn't much you can do about it.

user227867

@Idomathart Time for you to upgrade to Windows 10, LOL.

user227867

@robjohn I hope everything is fine after the fire, finally.

I do math art

@JasperLoy JASPER IS BACK!!!

@JasperLoy How is it going lately?

user227867

@Idomathart I did not make any videos lately. Tomorrow I will go and meet a friend. I am OK, not too good, not too bad.

I do math art

19:42

@JasperLoy Give me some more songs, new ones. :-)

user227867

@Idomathart Soon. Good luck in publishing the book!

I do math art

@JasperLoy Thanks!

user227867

@Idomathart I am going to bed. Good night!

I do math art

I'm taking some tea now and barely could write. GN

MATH ASKER

In a group of 20 automobiles, 7 fail safety inspection. How many ways can 5 cars be selected from the 20 so that all fail the safety inspection?

I'm really confused about this question

I tried doing 7p5 first

and subtracted that from 20P5

I don't know if that how i was supposed to do it

or can i do it like this 13p15

and subtract the answer from that with 7p5

Oh wait I can't do 13P15

I do math art

19:49

@MATHASKER This is also a very simple question. Just do some efforts, even if you fail, get the logical way on your own.

MATH ASKER

@Idomathart is it? I'm really confused about this question

robjohn

@MATHASKER is that $\frac{15!}{2!}$?

MATH ASKER

15!/2! @robjohn?

robjohn

@MATHASKER is that the permutation of 15 things taken 13 at a time?

MATH ASKER

@Idomathart how do i do this problem? can you explain it to me?

@robjohn I don't know

I do math art

19:52

@MATHASKER As the previous problem this is also the same as difficulty. Just think of it and get the solution on your own.

robjohn

@MATHASKER what does 13P15 mean?

MATH ASKER

@robjohn that was just a thought i got, I couldn't do that because how would i choose 15 out of 13

I've tried it but I can't, even a little bit of hint would help @Idomathart

Ok so I tried to do this

20 C 7 to find the possibility of failed inspection cars

than I did this 20 C 5 to find the possibility of 5 cars chosen out of the 20

than i subtracted them

and got 62,016

Did i do it correctly @robjohn @Idomathart ??

I got it its 21

I do math art

20:14

@MATHASKER If the question is well understood, the only way you can form groups of 5 such cars is by considering the cars with that fail.

You cannot consider 4 cars ok and say add one that doesn't fail. No matter that you have 20 or more cars.

MATH ASKER

I just did 7 C 5 to get 21

Oh

I do math art

@robjohn Maybe you considered the question differently?

@MATHASKER How about selecting 10 cars, 5 that are good and 5 that fail?

MATH ASKER

I don't know, I'm doing my hw so thats where these questions are coming from

I do math art

20:33

27027

MATH ASKER

whats that?

robjohn

@Idomathart I don't know what the question was. I only saw 13P15, and assumed it meant $\frac{15!}{2!}$, not $\frac{13!}{(-2)!}=0$.

I do math art

@robjohn Oh, OK.

@robjohn I also wondered if the word select could be interpreted differently in English in this context, also considering a certain numbering of cars.

MATH ASKER

@robjohn The question was In a group of 20 automobiles, 7 fail safety inspection. How many ways can 5 cars be selected from the 20 so that all fail the safety inspection?

what was 27027 though? @Idomathart

I do math art

@MATHASKER The answer to my problem above.

MATH ASKER

20:46

Oh the 10 cars one?

Balarka Sen

I think it's 7C5: the number 20 irrelevant here. But combinatorial word problems are always confusing - what if they didn't specify the 7 cars among the 20, in which case the answer would be 20C7 * 7C5?

Hi @TedShifrin.

Ted Shifrin

Hi @Balarka ... I see we're trying to count

Balarka Sen

Yeah, just complaining that I never learnt to properly read a combinatorics problem.

Ted Shifrin

Well, you're always good at complaining.

Balarka Sen

Hopefully I'll fix that this year: got this stuff in high school.

Ted Shifrin

20:49

And are we selecting a subset or an ordered set of 5. I would interpret it as combinations, not permutations.

I think the answer is $\binom 75$. The set of 7 that fail inspection is presumed fixed.

MATH ASKER

Yes I also thought it was combination because it doesn't look like its a dependent proble

Balarka Sen

Right, didn't interpret it as permutations.

MATH ASKER

Among 14 radios, 5 have broken dials. If 4 radios are selected, how many such selections are possible in which 2 radios have broken dials?

For this I tried to do the following

Balarka Sen

I was just wondering what if someone means "there are 7 cars which fail safety inspection out of 20, and 5 of them are chosen so that all of them fail safety inspection. How many possibly configurations can there be?"

In which case it's be 20C7 * 7C5, yup?

Ted Shifrin

No, @Balarka. I don't read the problem as being that we can randomly selected any 7 to fail inspection.

MATH ASKER

20:52

7 C 2 = 21 because that the possibility of it not being broken

Ted Shifrin

Huh? @MathAsker

MATH ASKER

and than i did 5 C 2 to find the possibility of it being broken

and I multiplied those to get 210, was i correct?

@te

@TedShifrin ? 7 c 2 because there are 7 radios that are not broken

Ted Shifrin

You're not doing possibilities/probabilities here. But I assume they mean precisely 2 have broken dials, not at least 2. So you have to pick 2 from 5, yes, but then you have to pick the remaining 2 from how many?

14-5 = 7 ?

MATH ASKER

right?

ya so i did 7 choose 2

Ted Shifrin

Whoa ... 7 + 5 = 14 ??

You have more serious counting problems, my friend.

I do math art

20:55

@BalarkaSen I have no idea how that question might ever make sense. What you mean by configuration there?

user189740

Hey,

I have a problem in a stats course I was given. I was given the probability density function define as

$f(x) = \frac{c}{x}$ where $0 \leq x \leq 3$ and $0$ otherwise.

First we are asked to find $c$. Normally I would assume the function equals 1 when integrated from 0 to 3 and use that to solve for $c$, but in this case when I integrate I get infinite.

Any ideas on how I can tackle this?

MATH ASKER

oh wait its 9 C 2

I'm sorry I was in a hurry @TedShifrin

Ted Shifrin

@byteofthat: I bet there's a typo and they mean $1\le x\le 3$, or something like that. You're right, of course.

There you go, @mathasker :P

Balarka Sen

@TedShifrin Both interpretations come to mind when I read the problem :S

Ted Shifrin

You have to pay attention to details like that, though, when you're doing counting problems.

MATH ASKER

20:57

so the answer would have been 360 ways

user189740

@TedShifrin it would not be the first time my stats prof made a mistake in our handouts. I'll check with him tomorrow.

user189740

Glad to hear I wasn't missing something obvious.

Ted Shifrin

Nope, you're totally correct, @byte.

MATH ASKER

ya true, I was just wanted to finish my homework fast

Balarka Sen

@Idomathart That is, you aren't given the 7 cars which fail the safety inspection. It can be any 7 of the 20.

This isn't implicity said in the problem, I agree. Hence why I said I need to learn to read these problems properly.

Ted Shifrin

20:58

I had a student who took 6 college courses from me, @mathasker, and he ended up with C's when he should have had B's or A's ... because he didn't spend enough time on homework and he didn't read test problems carefully, even though I begged him to read them 5 times and check his work 5 times before turning in tests.

I do math art

@BalarkaSen Then you have probabilities in the game. I was talking about exact calculations knowing the exact good and bad parts.

Ted Shifrin

@Balarka: I read it with real-world logic. The 7 cars that fail are a particular 7, not a random 7.

Balarka Sen

@Idomathart Probabilities? No, I mean, you have a factor of 20C7 in that case.

MATH ASKER

@TedShifrin Oh ok will do now

Balarka Sen

@TedShifrin Right.

Ted Shifrin

21:00

Anyhow, I didn't check what 36*21 is, but I'm pretty sure it's not 360.

Let's see: 36*20 = 720 and 720+36 = ...

MATH ASKER

Why would it be 21? wouldn't I do 5 C 2

Ted Shifrin

@Mathasker: You need to pay attention to detail.

MATH ASKER

to find the possibility of 2 radio dials that are broken

I do math art

Now why would I ever want to solve such problems again here when that little text of the problem is interpreteted in many ways like a religious text?

Ted Shifrin

Oh, you're right, you got me all confused with your arithmetic error. It is 5C2. So 360 is right :P

On that note, I'm leaving before I get more confuzled. Have fun counting!

MATH ASKER

21:01

Oh okay I still have one more qustion

though

Ted Shifrin

Hrumph. Yes?

MATH ASKER

If replacement is allowed, how many 3-digit numbers can be formed using the digits 6,7 and 8?

Balarka Sen

@Idomathart Too used to interpreting stuff in many ways because of ambiguous word problems from high school. Don't mind me.

MATH ASKER

so for this I just tried doing 3! but looks like I don't understand the question

Ted Shifrin

OK. What's your answer, @Mathasker?

MATH ASKER

21:02

If replacement is allowed? what does that mean, maybe I could solve it if i knew that

Ted Shifrin

No, no. For example, you have 678, 666, 786, 877, and so on ... How many total 3-digit numbers?

I do math art

@BalarkaSen OK. It's fine.

Ted Shifrin

It means you can reuse the digits.

Balarka Sen

@MATHASKER It means you can use the digits again and again.

Ted Shifrin

Without reusing the digits, your 3! is correct.

MATH ASKER

21:04

Oh I knew the problem meant I could use the number again and again

but i was confused i guess

Ted Shifrin

So how many choices for the first digit?

MATH ASKER

for the first digit its 3 choices

Ted Shifrin

How many choices for the second digit?

MATH ASKER

so would it be 3*3*3

same

Ted Shifrin

Bingo.

MATH ASKER

21:04

But when would i stop

multiplying>

Ted Shifrin

You just did. How many digits are there?

MATH ASKER

3*3*3 * 3*3*3 * 3*3*3 ?

Ted Shifrin

AGHHHHHH.

MATH ASKER

three digits

Balarka Sen

lol

MATH ASKER

21:05

ohhh I thought it was infinite

Ted Shifrin

But you didn't think that!

MATH ASKER

ohh ok thanks a lot

Ted Shifrin

Sure thing. Just stay calm and think your way through the problem logically. You'll do great.

MATH ASKER

Ok, will try

Ted Shifrin

That's all one can ask. :)

I do math art

21:10

@MATHASKER By simple logic you're done with any such question, after you have carefully read the question, of course.

Balarka Sen

Reading the problem correctly is an important part of combinatorics, alongside logical thinking.

Alyosha

If $(W,\omega)$ is a symplectic manifold, is there an obvious way of extending $\omega$ to the symmetric product $\text{Sym}^n(W)$ of $W$?

robjohn

@MATHASKER Your choice will not change the 7 that failed the inspection, so the question is asking how many ways can you choose 5 cars from the 7 that failed, that is $7C5$. Order doesn't seem to matter in this question.

Alyosha

If $W=S^2$ then $\text{Sym}^nW=\mathbb{CP}^n$ and the Fubini Study form is a (I think it might be the most natural) symplectic form on $\mathbb{CP}^n$, but it's unclear whether that arises in a natural way from the form on $W$.

I do math art

@robjohn The word select made me think if the order should be considered or not.

Balarka Sen

21:13

@Alyosha Is symmetric product always a manifold, even?

I doubt.

Think about $\Bbb R^2$ modulo $\Bbb Z/2$ switching the two factors.

Alyosha

Is the dimension not usually $n \cdot \dim W$?

Or $n-1$ perhaps.

Balarka Sen

Dimension in what sense? (I gave an example in which case it's not a manifold, namely, second symmetric product of $\Bbb R$)

Well, a manifold with boundary.

Alyosha

I don't know when it's a manifold, my knowledge is fragmentary.

I'm looking it up.

If $W$ is a Riemann surface then it's a manifold.

Balarka Sen

Why?

Alyosha

Unclear- it seems hard to find references.

Apparently it also has a complex and symplectic structure.

I do math art

21:38

@robjohn what do you think about my new username?

I change it once in a while :D

Actually I thought it would be much better to use something that to represent my style of doing mathematics.

@BalarkaSen don't you feel the need to change your username once in a while? Unless you maybe use your real name.

Anyway.

Have to prepare some stuff here.

BBL

Balarka Sen

@Idomathart It's my real name.

I do math art

@BalarkaSen OK

The Pointer

22:05

Hi everyone, can someone please look at the simple question in this thread regarding multivariable calculus and help me out: math.stackexchange.com/questions/1915858/…

I have a horrible understanding of trigonometry and I'm struggling to grasp a very simple concept

I do math art

@robjohn have you tried that elementarily? I think you did it or at least it seems like that to me for some reason. $$\sum_{k=1}^{n} (-1)^k \left(\frac{1}{k+1}+\frac{1}{k+2}+\cdots+\frac{1}{k+n}\right)\binom{n}{k}$$

rorty

any advice for not forgetting old mathematics? does anyone use spaced repetition programs like anki?

amWhy

@rorty Teach it! (tutor...etc). Not familiar with anki. Tell me more?

I do math art

@robjohn actually let it starts from $k=0$ $$\sum_{k=0}^{n} (-1)^k \left(\frac{1}{k+1}+\frac{1}{k+2}+\cdots+\frac{1}{k+n}\right)\binom{n}{k}$$

@robjohn with integrals it is pretty straightforward.

robjohn

@Idomathart I was just about to ask that

@Idomathart no, about starting at $k=0$

I do math art

22:20

@robjohn Sorry, I misunderstood you. Yeap, it starts from $k=0$.

amWhy

@rorty Seriously, tutoring or teaching, or helping out in chat here can really help solidify what you know and don't want to get rusty, and/or refresh what you've studied previously.

@robjohn (quick distraction) Hi meanie! You should "carve" yourself into a mean pumpkin next month!

Okay...I'm gone....

robjohn

@Idomathart It seems to be $\mathrm{B}(n,1)-\mathrm{B}(n,n+1)$, if I haven't made a mistake.

I do math art

@robjohn Yeap, kind of!

@robjohn Did you do it elementarily? (maybe if avoiding beta function and rearranging the stuff a tiny bit)

robjohn

@Idomathart From the form I gave, I used the Beta function

But it was only for integer arguments, so it could be done with integration by parts

$$\frac1n\left(1-\frac1{\binom{2n}{n}}\right)$$

I do math art

Yeap, that looks fine.

robjohn

23:27

@amWhy Perhaps...

amWhy

@robjohn Love it!!

I do math art

@robjohn I'm going to generate that using one of my master theorems. Trying that later.

Agawa001

23:51

@Idomathart dontu go dodo ?

I do math art

@Agawa001 lol. I'm preparing to some sleep, it's pretty late, and there is less time to sleep, again, like 4 hours.

Agawa001

@Idomathart it is so calm in night, so better for me and whoever wants a quiet environment to work in to go nocturnal

going owls

like dracula in transilvania

I do math art

@Agawa001 Yeah, but losing many nights can ruin your health on long term.

@robjohn lol, cool gravatar!

Agawa001

lol

the mean peach

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