Mathematics - 2014-12-07 (page 3 of 6)

beginner

10:02

where do series come up in application?

is there application for knowing that $\sum \limits_{n=1}^\infty \cfrac1n$ is harmonic and diverges?

Balarka Sen

@beginner yes. that primes are infinite can be concluded from the divergence of harmonic series.

Committing to a challenge

because the spacing can become further and further apart but still not stop?

or an actual big proof

Balarka Sen

actual proof, not big though

Committing to a challenge

do we only care about primes for cryptography and personal interest?

Balarka Sen

not really.

beginner

10:05

where is the proof

Balarka Sen

@beginner Consider the function $\zeta(s) = 1 + \frac1{2^s} + \frac1{3^s} + \frac1{4^s} + \cdots$

beginner

so at $s=1$ we have harmonic series

Balarka Sen

Now note that $\frac{1}{2^s} \zeta(s) = \frac1{2^s} + \frac1{4^s} + \frac1{6^s} + \frac1{8^s} + \cdots$

3

Committing to a challenge

Riemann zeta function. Getting powerful here xD

Balarka Sen

Thus, $(1 - 1/2^s)\zeta(s) = \zeta(s) - \frac1{2^s} \zeta(s) = 1 + \frac1{3^s} + \frac1{5^s} + \cdots$ (the even terms are cancelled out)

@Committingtoachallenge yes so you better pay attention at the proof

Committing to a challenge

10:09

Finally get to learn it, next, the polylog :).

Balarka Sen

now, $1/3^s(1-1/2^s)\zeta(s) = \frac1{3^s} + \frac1{9^s} + \frac1{15^s} + \cdots$

beginner

did you accidently miss or on purpose $\frac{1}{4^s}$

Balarka Sen

Thus, $(1-1/3^s)(1-1/2^s)\zeta(s) = (1-1/2^s)\zeta(s) - 1/3^s(1-1/2^s)\zeta(s) = 1 + 1/5^s + \cdots$ (the 3-multiple terms are cancelled out)

@beginner typo

beginner

ok

Balarka Sen

So, if you continue doing like this, you'll get that $\prod_p (1-1/p^s) \cdot \zeta(s) = 1$, where the product runs through primes

Committing to a challenge

10:11

So you are removing all the non-primes like a sieve

Balarka Sen

yes.

thus, $\zeta(s) = \prod_p (1-1/p^s)^{-1}$

beginner

oh wow cool

Balarka Sen

now, if we naively sub in $s = 1$, we get $1 + 1/2 + 1/3 + \cdots = \prod_p (1-1/p)^{-1}$

if there are finitely many primes, the product converges, but the sum diverges, contradiction

2

so there must be infinitely many primes

Committing to a challenge

I see, thank you, now the polylog :)

Balarka Sen

there you go. the fundamental connection of riemann zeta with primes, which actually leads to the motivation for riemann hypothesis (roughly speaking)

beginner

10:14

i dont get why subbing in $s=1$ means anything when we removed it

Balarka Sen

@Committingtoachallenge polylogs are boring stuff. try studying more about riemann zeta

@beginner what do you mean?

Committing to a challenge

Haha, alright.

beginner

oh nevermind i see

thank you teacher

Balarka Sen

i'm not a teacher

beginner

you are my teacher

Balarka Sen

10:15

i don't want to

beginner

you are my surveyor

Balarka Sen

i'd rather not

beginner

part time

unpaid

Balarka Sen

gives up arguing

beginner

other than big appreciations

Balarka Sen

10:17

@beginner rather than talking nonsense why don't you study more about riemann zeta function

beginner

isnt that jumping?

Balarka Sen

and i'd appreciate if whoever starred the message to unstar it

beginner

wasnt me i promise and to prove i wil star it

Balarka Sen

@beginner no. studying something from scartch isn't jumping.

beginner

but i havent done series really before so it is jumping

Balarka Sen

10:18

what the. serial starring is annoying, whoever is starring

beginner

is it bad cause of the typo?

Committing to a challenge

It is Integrator, he liked my swearing that got me banned.

Balarka Sen

@beginner well series isn't really much hard stuff

beginner

yeah it looks fun

i think kaj gave me the harmonic as a practice problem

Balarka Sen

try proving that the harmonic series diverges, yes

it's important

beginner

10:20

and chris's gave me $\sum \limits_{n=1}^\infty \frac{1}{n(n+1)}$

Integrator

@beginner I guess, it telescopes!

beginner

i saw a proof of the harmonic which spoiled it

Balarka Sen

@beginner why don't you replace $1$ by $n+1 - n$?

beginner

for chris's, and the $1$ on the bottom?

Balarka Sen

1 on the numerator

@Committingtoachallenge are you familiar with complex analysis?

beginner

10:25

it made a $\pm \sum \limits_{n=1}^\infty \frac{1}{n+1} + \sum \limits_{n=1}^\infty \frac{1}{n(n+1)}$

Committing to a challenge

I haven't really done any complex analysis, next semester I will be doing Functional&Complex analysis(+3 other courses)

beginner

dont tell me ill get it

Balarka Sen

@Committingtoachallenge OK. If you knew a bit complex analysis, I had something which might have given you some motivation for studying the zeta function.

Swapnil Tripathi

Hello everyone! Does anyone know about when the edited questions are bumped?

Committing to a challenge

@BalarkaSen Thanks Balarka, interesting stuff, I'll check it out.

Balarka Sen

10:28

Have fun.

Committing to a challenge

@INtegrator stop please

Enough starring @Integrator

Balarka Sen

let's just ping a mod

Integrator

@SwapnilTripathi What happened?

Balarka Sen

@robjohn

Committing to a challenge

@Integrator Unstar them please

Swapnil Tripathi

10:30

hey @Integrator. I was wondering when do the questions bump to the active ones! I was reading a post on meta yesterday. Does editing the answers bump the question too?

Integrator

@SwapnilTripathi As soon as you edit them, I guess

Swapnil Tripathi

Does editing the answers bump the question too? @Integrator

Integrator

@SwapnilTripathi Yes

Swapnil Tripathi

Ok. So I'd refrain from editing old questions (and answers)!

Integrator

@Committingtoachallenge @BalarkaSen I swear it's not me!

Committing to a challenge

10:33

@Integrator You deny it now, really?

Integrator

@Committingtoachallenge wait!

Committing to a challenge

@Integrator You let yourself be accused while chatting to someone else and not respond to me.

Integrator

@Committingtoachallenge See that!

Committing to a challenge

Interesting computer clock mr editor.

Four minutes off time :\

Integrator

@Committingtoachallenge I live in india!

Committing to a challenge

10:35

It should have said 4:04, but it said 4:00

Swapnil Tripathi

Me too @Integrator. :P

Integrator

@Committingtoachallenge Wait again!

Committing to a challenge

Oh you are overlaying the current clock?

Swapnil Tripathi

@Integrator which OS is it?

Integrator

Committing to a challenge

10:36

Yet the starring stopped, which the troll doing it would definitely revel in continuing now.

Integrator

@SwapnilTripathi Windows 8.1 !, Chrome in Windows 8 mode!

Balarka Sen

let's not feed the troll

Integrator

@Committingtoachallenge should I give you my account credentials?

Swapnil Tripathi

@Integrator (O.o) ...ok!!

Integrator

@Committingtoachallenge I'm pinging @robjohn

Swapnil Tripathi

10:39

I'm being targeted now!

Integrator

@SwapnilTripathi It's not me! Really!

Swapnil Tripathi

Haha! Unstarred.. :D

Committing to a challenge

I have to go

Swapnil Tripathi

lol. Smart @Committingtoachallenge (removed)

Integrator

@Committingtoachallenge Mods can see deleted posts and deleted comments! I think a room owner or mod can see deleted messages also!

@SwapnilTripathi You can edit old posts if required when Mse is not much busy!

GBeau

10:42

@KajHansen I think B2 was pretty obviously ln(4/3), but I'm not certain my justification was adequate

Swapnil Tripathi

Hey. How was Putnam :P @GBeau

GBeau

It went well, actually

I think I definitely surpassed my goal

I think it helped that I didn't have a lot of pressure on myself

There was at least one question that I'm fairly certain I got full or near-full points on

Swapnil Tripathi

@Integrator Yes. It is quiet in SE nowadays! I flagged a post an hour ago and it is still not down!

GBeau

(that is, in addition to the correct answer, I believe I gave a very rigorous justification)

Swapnil Tripathi

Good to hear. :) @GBeau

I studied Abstract Algebra from Gallian and it had a special exercise which had question from various competitions (Putnam being one of them). Saying that the questions were tough would be an understatement!

GBeau

10:47

I was surprised, actually

I was able to solve several more, but I can't say my justification was for sure adquate

B2 (what I mentioned above), was asking for the max value of $\int_1^3 \frac{f(x)}{x}$ if $\int_1^3 f(x)=0$, and $f(x)$ $-1\leq f(x)\leq 1$ between 1 and 3

as I mentioned, I think it's pretty easy to reason intuitively that the max must be ln(4/3)

Peter Sheldrick

DO you guys get stuff like this? math.stackexchange.com/questions/1055378/…

GBeau

it's just saying f(x) is between -1 and 1, not f(x)-1

Peter Sheldrick

i really try to help within my means but the question just goes dead

Swapnil Tripathi

@GBeau wow! I'm bad at it, anyways.

GBeau

I reasoned at the time that the max would be given by the piecewise where f(x)=1 from 1 to 2 and f(x) = -1 from 2 to 3

and then you can just do the integral to reach ln(4/3)

but you also need sufficient justification that this is larger than all other possible f(x)

Integrator

10:59

@SwapnilTripathi I guess it was an answer! Because you cannot flag a question now! You can only close them!

@SwapnilTripathi And it takes about 5 reviews to remove an answer, Not sure though

robjohn

11:18

@BalarkaSen Did you wish to say something?

evinda

Hi :) Could someone explain me why it stands that $\{ n \} \in n \cup \{ n \}$, knowing that $x \in A \cup B \rightarrow x \in A \lor x \in B$ ?
$$n \text{ is a natural number}$$

robjohn

@evinda In general $\{n\}\not\in n\cup\{n\}$ unless there are some other conditions we are not seeing.

Balarka Sen

@robjohn I was actually trying to ping your attention towards the serial starring that just made the starpanel explode.

Committing to a challenge

E.g. Integrator's stars

robjohn

@Committingtoachallenge How do you know whose stars they were?

Committing to a challenge

11:34

@robjohn They started as soon as he arrived, and he has starred me a few times before for fun

robjohn

@BalarkaSen I assume the stars were removed

Anonymous

0_0

Committing to a challenge

4 are still there, unless I need to refresh

No, 4 are there lol

You can unstar things?

Or you just deleted the posts themselves

robjohn

The ones remaining look as if they might be starred justifiably

Committing to a challenge

Balarka's $\zeta$ was starred because of the typo

Anonymous

11:38

@Committingtoachallenge How's studies going?

Committing to a challenge

@Ashwin Alright I suppose, although today was only 2.5hrs of the main texts + 40 min of another unrelated text

So not great, but I did finally workout again properly

And my sleep patterns are acceptable again

Anonymous

@Committingtoachallenge I will be posting today's work on my blog

Committing to a challenge

Looking forward to it :)

Anonymous

@Committingtoachallenge :D I got to go now.Bye!

robjohn

@Committingtoachallenge no more typo... :-)

Committing to a challenge

11:39

@Ashwin Cya later

@robjohn Oh very nice, thanks

I didn't know you could do that

evinda

@robjohn I am looking at the proof of the sentence: For any natural numbers $n,m$ it holds that $n'=m' \rightarrow n=m$.

Proof:

Let $n'=m'$ for some natural numbers $n,m$.

Then $n \cup \{ n \}=m \cup \{ m \}$.

Therefore, $\{ n \} \subset m \cup \{ m \} \rightarrow n \in m \cup \{ m \} \rightarrow n \in m \lor n=m \rightarrow n \subset m$.

In the same way we get that $m \subset n$.

So, $m=n$.

How did we get that $\{ n \} subset m \cup \{ m \}$ ?

Committing to a challenge

so $n=n'$?

Or that is prime as in the derivative

evinda

@Committingtoachallenge $n'=n \cup \{ n \}$

Committing to a challenge

Might wanna edit that last line with the backslash before subset

robjohn

@evinda Ah, so the unseen conditions are that you are constructing integers, so these sets are not general sets.

evinda

11:47

@robjohn Is it maybe like that?

$$n \cup \{ n \} \subset n \cup \{ n \} \rightarrow n \cup \{ n \} \subset m \cup \{ m \} \rightarrow \{ n \} subset m \cup \{ m \} \rightarrow n \in m \cup \{ m \}$$

Hi @anon!!! Do you have maybe an idea about the following exercise? chat.stackexchange.com/rooms/19204/p-adic-fields

Chris's sis

Greetings

robjohn

@Chris'ssis Good morning (4AM here)

Hakim

Hello @Chris'ssis

Chris's sis

@Hakim Not that bad, thanks. You?

@Hakim Hi

Hakim

@Chris'ssis I'm fine, thanks. :-)

robjohn

11:57

@Chris'ssis or early, depending on your point of view

Chris's sis

@robjohn Indeed. How many hours do you usually sleep per night?:-)

@Hakim Great! :-)

robjohn

@Chris'ssis sleep... is that a requirement?

Chris's sis

@robjohn I think so, for a good health I think it's recommended to sleep 6-8 hour per night.

@robjohn There was a period, like 3 months when I couldn't sleep for more than 3-4 hours. That was a bad period, definitely. Happily now all it's fine.

Pedro Tamaroff

@KarlKronenfeld You'd have to assume the Hilbert ring is a domain.

robjohn

@Chris'ssis I don't think I usually sleep more that 4 hours a night. Sometimes I get 6

Chris's sis

12:05

@robjohn I might guess you consume tons of coffee. :-)

robjohn

@Chris'ssis I don't drink coffee... I don't like it. I occasionally have some tea, but only 2 or 3 cups on a cold day.

Chris's sis

@robjohn Well, on longer periods of time, 4 hours only is not very good I think. I suppose you're used to that.

@robjohn I don't like it either (although sometimes I use some, but not for staying awake).

Integrator

Seems like nobody is interested in voting on @robjohn answer!

Chris's sis

@Integrator Where?

Integrator

Poor Mod !!! +1

@Chris'ssis $$\int_0^1\arctan(\sin x)\,\mathrm dx$$

N3buchadnezzar

12:09

Anyone have mathematica here? Looking for a quick way to factor this $-5x^4+19x^3+77x^2-120x+18$ into two quadratic polynomials. Any ideas?

Integrator

@N3buchadnezzar I do!!

@N3buchadnezzar I don't know syntax!! :(

N3buchadnezzar

=(

Integrator

But it yields true for IrreduciblePolynomialQ[18-120 x+77 x^2+19 x^3-5 x^4]

Chris's sis

I'm thinking to buy this one (at my next salary)

Intel Core i7-5960X @ 3.00GHz

cpubenchmark.net/…

I'm sick & tired of the CPU I presently have. Actually, I'll buy a new computer, and probably I'll give this one to my cousines, it's good for games.

N3buchadnezzar

@Integrator I thought all quartic polynomials could be written as a product of two quadratic polynomials

Integrator

12:19

@Chris'ssis It's a beast

Chris's sis

@Integrator Yeap, I know. :-)

Venus

Any idea to approach this integral
$$\int_0^\pi\frac{\sin^2x}{1+(1-a\sin^2x)^2}dx$$

Chris's sis

@Venus What is the range of the value of $a$?

Venus

$a\ge0$

Chris's sis

@Venus Just a bit, I'm almost done.

Venus

12:26

@Chris'ssis I am waiting patiently

Chris's sis

Done

Integrator

@robjohn That wasn't me who starred all those posts! Today I starred only one post! And that is... oops... deleted!

@robjohn today according to my timezone UTC+5:30

robjohn

@Integrator I removed some stars, but the comments are still there.

Chris's sis

@Venus There seems to be a problem though ...

Integrator

@robjohn But the point is I wasn't the one who was serially starring everything!

Wow @robjohn haven't down-voted yet!

Venus

12:34

@Chris'ssis What problem?

robjohn

@Integrator I wasn't assuming anything about the starring fiend...

Integrator

@robjohn Thank god!

robjohn

@Integrator I'd rather comment and attempt to have people correct what is wrong, than to downvote.

3

Integrator

@robjohn That's how a mod ought to be!

Chris's sis

@Venus Well, let me fix a possible typo, My work is done.

Daniel Fischer

12:36

Some probably well-meaning but misguided individual seems to just have serial-upvoted me :(

Integrator

@DanielFischer How many time?

Daniel Fischer

Five.

Just wanted to push me over the next thousand, I think.

Integrator

@DanielFischer You shouldn't worry about that!

@DanielFischer unless it's not something like this!!!

Daniel Fischer

@Integrator I don't worry about that, I just prefer it when people upvote only the answers they just happen to come across (and find good) rather than going to the profile to see which answers they might upvote.

Integrator

@DanielFischer Just for record, I've starred your message!

Daniel Fischer

12:44

@Integrator As long as you don't start serial-starring my messages ...

Integrator

@DanielFischer I think i should star that one too!

Daniel Fischer

Two is okay, at three it becomes critical.

Integrator

@DanielFischer I cannot resist, I better leave!

Daniel Fischer

I can resist everything - except temptation.

Chris's sis

@Venus $$2 \left(\sqrt{\frac{1}{\sqrt{2}}-\frac{1}{2}} \pi +\frac{\left(\frac{1}{8}+\frac{i}{8}\right) \pi \left(\frac{2 i \sqrt{2-(1-i) a}}{a+(-1-i)}-\frac{2 \sqrt{2-(1+i) a}}{a+(-1+i)}\right)}{\sqrt{2} a}\right)$$ that can be further simplified.

Venus

12:47

@DanielFischer Are you busy? You ignore some of chats to you Y_Y

beginner

@chris'ssis can i have a new simple series please

Daniel Fischer

@Venus I miss some pings sometimes without intentionally ignoring them. Did I miss one of yours?

Pedro Tamaroff

Friggin rain

Chris's sis

@Venus $$\frac{\pi \left(\frac{i}{\sqrt{2-(1+i) a} a}-\frac{i}{\sqrt{2-(1-i) a} a}+2 \sqrt{\sqrt{2}-1}\right)}{\sqrt{2}}$$

beginner

i love rain @pedro

i miss it

Venus

12:48

@DanielFischer You missed lots of my chats

Daniel Fischer

@Venus Oh, when?

Pedro Tamaroff

Are you a farmer?

Chris's sis

@beginner OK

beginner

@PedroTamaroff no

Daniel Fischer

@PedroTamaroff Perhaps just moved to Atacama. I'd miss the rain too if I lived there.

Venus

12:50

@Chris'ssis Wait...

Chris's sis

@Venus I checked that and it works numerically.

beginner

it must rain a heap where you guys are

i get rain a few times a year and it is so nice on a tin roof

Pedro Tamaroff

Daniel I like your new picture.

Venus

@DanielFischer chat.stackexchange.com/transcript/message/18942595#18942595

237

A: Is the following matrix invertible?

Find the determinant. To make calculations easier, work modulo $2$! The diagonal is $1$'s, the rest are $0$'s. The determinant is odd, and therefore non-zero.

How to find determinant matrix using modulo 2? Could you enlighten me @DanielFischer? Thanks.

Chris's sis

@beginner This one is very nice $$\sum_{n=1}^{\infty} \frac{1}{(n+1) n!}$$

beginner

12:52

@Chris'ssis yay thank you

Chris's sis

@beginner Welcome :-)

Daniel Fischer

@Venus Ah, okay. If you ping me while I'm having breakfast, don't expect me to respond ;)

Venus

@Chris'ssis How did you get this?

@DanielFischer OK fine

Daniel Fischer

@Venus You don't really want/need to compute the determinant exactly, you just want to see whether it is $0$ or not.

And in this case, we can easily see that the determinant is nonzero because it is an odd integer.

beginner

@Chris'ssis ohh i think i see, the $n!$ comes in from the previous multiplication, i think i will get it soon

Venus

12:56

@DanielFischer So Andre's answer only proves that the matrix has a non-zero determinant & therefore it's invertible?

beginner

ohhhh woops it is $\frac{1}{(n+1)!}$ i never seen that before

Daniel Fischer

@Venus Yes, exactly.

Chris's sis

@beginner Try to be creative. One of the things you need to have in mind when playing with series are the magic words "telescoping sum". :-)

Committing to a challenge

@Chris'ssis ohhh the last one was one of those, cool thank you :)

i enjoy these aswell

Daniel Fischer

And reducing the matrix entries modulo $2$ just makes it easier to see that the determinant is odd, hence nonzero, @Venus.

Chris's sis

12:57

Factorial

In mathematics, the factorial of a non-negative integer n, denoted by n!, is the product of all positive integers less than or equal to n. For example, The value of 0! is 1, according to the convention for an empty product. The factorial operation is encountered in many areas of mathematics, notably in combinatorics, algebra, and mathematical analysis. Its most basic occurrence is the fact that there are n! ways to arrange n distinct objects into a sequence (i.e., permutations of the set of objects). This fact was known at least as early as the 12th century, to Indian scholars. Fabian Stedman...

@Committingtoachallenge Welcome ;)

Committing to a challenge

@Chris'ssis Probably one of my worst areas in Math xD

Chris's sis

$$\sum_{n=1}^{\infty} \frac{1}{(n+1) \cdot n!}$$

@beginner ^^^

Venus

@DanielFischer Suppose that the entries matrix are random numbers, so we can find whether its determinant zero or not by using a certain modulo?

beginner

oh i meant i hadnt noticed $(n+1)n!=(n+1)!$ hehe

Chris's sis

@beginner :D

$$\sum_{n=1}^{\infty} \frac{1}{n \cdot (n+1) \cdot (n+1)!}$$

@beginner ^^^ This is a bit more advanced, but still easy, you can do it. OK

beginner

13:01

i havent finished the last one yet hehe, thank you for more

Daniel Fischer

@Venus If the determinant is nonzero (and the entries are integers), then you can detect that the determinant is nonzero by looking at it modulo some $p$. It's not always evident which $p$ works well, but it's always easy to check modulo $2$ and fairly easy to check modulo other small moduli ($3, 4,5,\dotsc$). So unless one sees something specific, one checks the small moduli. At some point it becomes less work to compute the exact determinant without taking a modulus, so one needs to spend

a little thought on how far it is worth to pursue the small moduli.

beginner

@Chris'ssis ill get them both within an hour hopefully :) then i have to leave

Chris's sis

@beginner It's good to know that $$\lim_{n\to\infty} \left(1+\frac{1}{1!}+\frac{1}{2!}+\cdots +\frac{1}{n!}\right)=e$$

@beginner try to make use of that

beginner

@Chris'ssis $\sum \limits_{n=0}^\infty \left(\frac{1}{n!}\right)=e$

Chris's sis

@beginner yeap

beginner

13:05

?

ok thank you

so i have $\sum \limits_{n=1}^\infty \frac{e}{n+1}$

which is harmonic and diverges?

wait ill get it

Chris's sis

@beginner You have $\displaystyle \sum_{n=1}^{\infty} \frac{1}{(n+1) \cdot n!}$ where how can you rewrite $(n+1) \cdot n!$? You said above.

The Artist

@Chris'ssis do you have an integral exercise for me? :) Something that's hard, but using high school finishers knowledge.

beginner

what title can i give you chris'ssis? can i call you series trainer :-)?

or series lecturer

Chris's sis

@TheArtist $$\sum_{n=0}^{\infty} \frac{1}{C_n}$$ where $C_n$ is Catalan's number

@beginner No need for a name.

The Artist

@beginner Ramanujan ;)

Integrator

13:10

@TheArtist $$\int_2^3\frac{\mathrm dx}{x\ln (x+5)}$$

The Artist

@Chris'ssis Okay xD

beginner

@Chris'ssis ramanujan the series lecturer

i have to give you a name cause Ted is professor, Balarka is unpaid teacher

The Artist

@Integrator there is a problem with the latex ?

beginner

kaj can be big bro

Chris's sis

@beginner hehe, I need no name.

Venus

13:11

@DanielFischer In the given problem, using mod 2, we get 4x4 'identity' matrix, right? But how can Andre know that the determinant is odd? I know that determinant of 'identity' matrix is always 0.

Integrator

@TheArtist Not now!

beginner

@Chris'ssis big sis then

Daniel Fischer

@Venus No, the determinant of the identity matrix is $1$.

The Artist

@Integrator Okay :) yes

Venus

@DanielFischer I meant 1, sorry. Typo

Chris's sis

13:13

@beginner are you done with the series? :-)

Integrator

@TheArtist Let's see if I can replace Mathematica with you!

beginner

@Chris'ssis not yet it is hard hehe

Integrator

Lol, I'm capped for today!

Daniel Fischer

@Venus Okay. So we know the determinant of (the matrix modulo $2$) is $1$. But the determinant of (the matrix modulo $m$) is (the determinant of the matrix) modulo $m$.

beginner

big sis was it $e-2$?

Venus

13:19

@DanielFischer So we can conclude that its determinant is odd because the determinant of (the matrix modulo 2) is 1, not 2?

Integrator

@Venus See if you can solve this

beginner

@Chris'ssis yay it is $e-2$

Daniel Fischer

@Venus Right. And if the determinant of the reduced matrix had been $2$ (or, $0$, modulo $2$), then we wouldn't know whether the original matrix had nonzero determinant or not.

Then we could reduce modulo $3$ to see what that gives.

Chris's sis

@beginner Yeah.

beginner

now to do $\sum_{n=1}^{\infty} \frac{1}{n \cdot (n+1) \cdot (n+1)!}$

N3buchadnezzar

13:22

@Integrator Simple way to show that $$ \int_a^b = - \int_b^a $$ ?

beginner

@N3buchadnezzar because it inverses both minus signs

Integrator

$$\int_2^3\frac{\mathrm dx}{x\ln (x+5)}=-\int_3^2\frac{\mathrm dx}{x\ln (x+5)}$$ then?

Venus

@DanielFischer Are you saying that if matrix mod $n$ is not equal to $n$, then we can conclude that the matrix has a determinant (an odd one)?

@Integrator No. I can't.

beginner

@N3buchadnezzar because -[a-b]=[b-a]

N3buchadnezzar

@beginner duuuh. Thanks

beginner

13:25

@N3buchadnezzar or i might be wrong i dont know

@N3buchadnezzar it might be wrong cause absolute values and stuff like that

Daniel Fischer

@Venus Not necessarily odd. But if the determinant of the reduced (modulo $n$) matrix is not $\equiv 0 \pmod{n}$, then the determinant of the original matrix is not divisible by $n$, and hence nonzero (since $0$ is divisible by everything).

Integrator

@beginner $$F(x)=\int f(x) dx \implies F(a)-F(b)=\int_a^b f(x) dx$$

Daniel Fischer

@N3buchadnezzar It's the definition of $\int_b^a$ for $a < b$, isn't it?

beginner

what is a negative factorial like $(-1)!$?

N3buchadnezzar

@DanielFischer Deppends onhow you define it does it not. Eg Rieman or Lebesgue etc.

Venus

13:27

@DanielFischer Oh, I get it now. Basically, this method is only to know whether the matrix has a determinant or not, right? OK, thank you so much sensei... ^^

Integrator

@Venus That was fun!

beginner

@Chris'ssis are negative factorials $1$? like $(-1)!=(-2)!=1$?

Daniel Fischer

@N3buchadnezzar For Lebesgue, we only have $\int_{[a,b]}$, not $\int_a^b$. Although one uses the latter notation a lot.

N3buchadnezzar

$$ \int_a^b f(x) \,\mathrm{d}x = F(b) - F(a) = - \Bigl[ F(a) - F(b) \Bigr] =- \int_b^a f(x) \,\mathrm{d}x$$

Chris's sis

@beginner read here mathworld.wolfram.com/GammaFunction.html

@beginner Did you get such a value in your calculations to my series? If so, something is wrong there.

beginner

13:34

@Chris'ssis i did: $\frac{1}{n(n+1)(n+1)!}=\frac{1}{\frac{((n+1)!)^2}{(n-1)!}}=\frac{(n-1)!}{((n+1)‌!)^2}$ but i did a typo before so i think this is going in the right direction so far

$$\sum \limits_{n=1}^\infty \frac{1}{n(n+1)(n+1)!}=\sum \limits_{n=1}^\infty \frac{(n-1)!}{(n+1)!^2}$$

@chris'ssis i have to go now :(. i will finish your series before we meet again :)

Chris's sis

@beginner There is a nice way to finish it.

beginner

is it using the gamma function?

Chris's sis

@beginner No.

beginner

can i get rid of the squared somehow?

bye @Chris'ssis @BalarkaSen

Balarka Sen

i'm still here

Chris's sis

13:41

@beginner maybe it helps to try simple tricks like writing $1=n+1-n$

Balarka Sen

@PedroTamaroff Do you have some good group theory problem?

Huy

@Venus: I'm sorry, I just returned from lunch with my parents. I see you solved your problem though. :)

Chris's sis

@beginner bye

Balarka Sen

I'm bored off from algebraic topology :P

GBeau

so many n factorials on A1 on the putnam yesterday :P

it asked to prove that the nonzero reduced coefficient numerators were always 1 or prime in the taylor series about x=0 of (x^2-x+1)e^x

Venus

13:44

@Huy How dare you leave a girl in trouble? -_-"

Huy

-_-''

Pedro Tamaroff

@BalarkaSen Maybe.

Balarka Sen

Fire away.

Pedro Tamaroff

PEW PEW PEW.

Now you're dead.

Balarka Sen

Don't take everything I say so literally, @Pedro

Huy

13:46

@Venus: What subjects are you most interested in so far, in mathematics?

Pedro Tamaroff

@BalarkaSen OK, I'll start with an easy one.

Suppose $G$ is a nonabelian group of order $p^3$. Then $G/Z(G) = [G,G] = C_p^2$.

Balarka Sen

You gotta be kidding me. That's basic stuff.

$Z(G)$ is either of order $1$ or $p$ or $p^2$. It can't be $1$, by just enumeration using the Cayley formula.

Venus

@Huy Almost everything that I can easily understand. If that's too complicated, I'd rather to not care a damn

Balarka Sen

If $Z(G)$ has order $p^2$, then $G/Z(G)$ is cyclic of order $p$.

But in that case, $G$ is abelian.

So $Z(G)$ must be of order $p$, hence $G/Z(G)$ must be of order $p^2$.

It's not cyclic, so it can only be $\Bbb Z/p \times \Bbb Z/p$ by classification of groups of order $p^2$.

@Pedro ^

Chris's sis

$$\int_0^1 \arctan(x) \log^3(x) \ dx=\frac{21}{64}\zeta(4)+\frac{9}{16}\zeta(3)+\frac{3}{4}\zeta(2)-\frac{3}{2}\pi‌+3\log(2)$$

@Venus ^^^ (it just came from my research)

Can we find an easy way to evaluate it? This time "easy" really means easy.

Venus

13:58

@Chris'ssis I'll try later. Now, I have to study for exam

That's not easy btw

Chris's sis

@Venus I give you my word I can finish it in one line. (using a result of mine)

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$Mathematics$ Mathematics