Mathematics - 2013-11-10 (page 2 of 2)

Alizter

13:05

@Chris'ssis Here is one: Prove that $\zeta(3/2)$ is irrational

robjohn

@MarkS. did you want that answer marked not CW?

Mark S.

Sorry, I meant that I answered a question which is not CW, but accidentally clicked to make my answer CW. Is that worthy of flagging (I don't want to abuse the flag system)?

robjohn

@MarkS. The normal route would be to flag the post. However, I have changed it since you asked.

Mark S.

Thanks!

Chris's sis

@Alizter This reminds me of my last question of this kind: Prove that $$\sum_{n=1}^{\infty} \frac{1}{2^{n^2}}$$ is irrational

@Alizter I did it by contradiction argument.

Pedro Tamaroff

13:12

@Chris'ssis That should be easy.

Much like proving $e$ is irrational.

Chris's sis

@PedroTamaroff Indeed

Alizter

13:28

@PedroTamaroff But there is no contradiction with that argument! Don't you know! There has always been an integer between 0 and 1!

Pedro Tamaroff

@Chris'ssis A nice proof would be to interpret it in base two. No period whatsoever.

Thus, irrational.

Chris's sis

@PedroTamaroff yeah. Good point.

user96977

13:46

a homomorphism preserves multiplication $f(a)f(b) = f(ab)$, but how does it follow that $f(a^{-1}) = f(a)^{-1}$ ?

Alizter

@TruthSerum what about if $a=b$ then $f(a)^2=f(a^2)$

then you could stretch that over $a^n$

user96977

thanks, good example. but does it generalize? for negative exponents especially @Alizter

Alizter

If $f(a)f(b)=f(ab)$ then $f(a)=f(ab)f(b)^{-1}$

mikhailcazi

Whoa, did not know SE had chat rooms. Hey everyone!

Jack M

hi

user96977

13:57

i'm not sure. i think maybe $f(ab)f(b^{-1})$...

Alizter

$\implies f(a)^{-1}=f(ab)^{-1}f(b)$

Jack M

@TruthSerum $f(a^{-1})f(a)=\ ?$

Alizter

we hope its 1

Jack M

it is 1

user96977

that must be one. but why does a homomorphism preserve inverses?

Alizter

13:59

thinking

now if we multiple by $f(b)^{-1}$

we get $$f(a)^{-1}f(b)^{-1}=f(ab)^{-1}$$

Jack M

are you guys just trying to prove that $f(a^{-1})=f(a)^{-1}$, or something else?

Alizter

yes

user96977

@JackM yes

Jack M

well then there's no need to involve any kind of $b$

inverses are something that involve a single element

Alizter

I think i have it

$f(a^{-1})\ne f(a)^{-1}$ let $b = a^{-1}$ the $\implies f(b) \ne f(a)^{-1}$ then mult by $f(a)$ we get $f(a)f(b)\ne f(a)^{-1}f(a)$

$\implies f(ab) \ne 1$

Jack M

14:07

it's a bit confusing to "let $b=a^{-1}$"

just write $a^{-1}$, no need to give it a new name

no?

Alizter

$\implies f(a/a)\ne 1$

user96977

Thanks for your effort @Alizter :)

Alizter

Then $f(1)\ne1$ which would contradict your homomorphism

Jack M

as long as you've proven that $f(1)=1$, yes

user96977

is $f(1) = 1$ not by definition of a homomorphism?

Jack M

14:09

I suppose you could

Alizter

$f(1)f(1)=f(1)$ divide by $f(1)$ to get $f(1)=1$ our homomorphism being $f(a)f(b)=f(ab)$ therefore this is true

user96977

in general there could be more than one $a$ s.t $f(a)=1$

user96977

namely the kernel of the group

Alizter

@TruthSerum Rigorous enough?

Jack M

no, I believe you can just use $f(ab)=f(a)f(b)$ as the definition

the fact that $f(1)=1$ follows from that

Alizter

14:11

which is what i said

Jack M

yes indeed

I missed that

user96977

looks ok @Alizter

Alizter

@JackM Congratulations on your 5 votes. Indeed $\pi^2$ is the area of a unit circle ;)

user96977

$f(a)f(a^{-1}) = f(a a{^-1}) = 1$ which suggests that $f(a^{-1})$ is the inverse of $f(a)$

Alizter

@TruthSerum Yep

now try and prove for $a^{-n}$

user96977

14:23

not today :)

user96977

someone says that i need to prove the cancellation law too

Jack M

the cancellation law is probably the very first thing you should prove about groups

user96977

yes, it comes first in my book

user96977

it seems almost redundant in it's simplicity

user96977

If $a,b,c \in G$ and $ab = ac$, then $b = c$

Jack M

14:26

if you think it's redundant, try proving it if $G$ is a monoid instead of a group

user96977

i only said it seems that way. i've never studied monoids

Jack M

a monoid is basically a group that might not have inverses

the point is that it can't be proven for monoids, because you need inverses to make it true

user96977

sounds terrible

Jack M

monoids aren't very interesting

user96977

ah i see. perhaps that will help me remember the proof :)

Jack M

14:28

an example of a monoid is the set of strings, with the operation being concatenation

user96977

i see. it requires he existence of inverses

Jack M

$(abc) + (defgc) = (abcdefgc)$

user96977

ah i see

Jack M

hmm, that's interesting. You actually CAN cancel in that particular monoid

so, bad example

well, you can't cancel if you allow infinite strings, for instance

$(a)+(aaaa...)=(aaa)+(aaaa...)$ but $(a)\ne(aaa)$

scratch that, I'm not even sure it's a monoid if you allow infinite strings

Alizter

15:34

@robjohn I updated my question do you see any immediate patterns in the series'?

Nick

15:56

@Charlie: Thank you for digging through all that undelightfulness to get me that. You're right it is somewhat boring to do but it's good exercise for basic algebra. Tell me something interesting to turn my attention to.

Nick

16:12

Guys, I need help understanding some basic maths.

Can you help?

Pedro Tamaroff

@Nick Maybe.

Nick

@Pedro: math.stackexchange.com/questions/561236/…

Pedro Tamaroff

@Nick Ah, sorry. Dunno about statistics!

Enjoys Math

Either one, @Nick, it's statistics!

Nick

XD I have no clue what you guys have against statistics. I only put that tag in there so moe people would see it.

It has something to do with Arithmetic Means and Harmonic Means.

But it's fundamentally physics.

@Pedro: Well, can you tell me something interesting then? I'm on a quest to expand my knowledge on things. Anything, Anything at all.

Chris's sis

16:20

I just received by e-mail this question: Find the smallest $a\in \mathbb{R}_{+}^{*}$ such that
$$ \{a\}+\left\{ \frac{1}{a} \right\}=1$$
$\{x\}$ - fractional part of $x$

Nick

@Chris'ssis: I have two questions. (1) What's the star supposed to mean and (2) Do the curly braces have any special meaning?

Chris's sis

@Nick "*" means "without $0$"

Nick

@Chris'ssis: aww, that's neat, Ok, but I still don't get what the curlies mean.

Charlie

@Nick what kind of thing?

Chris's sis

@Nick the fractional part of the number. en.wikipedia.org/wiki/Fractional_part

@Charlie Hello! Have you brought some good news? :-)

Charlie

16:28

@Chris'ssis hi Chris!

@Nick I read your question

Nick

@Charlie: Anything Mathematical! :D and yeah, I'm that much of an idiot!

Charlie

@Chris'ssis I don't know, it's a beautiful day today

@Nick it's a good question

Chris's sis

@Charlie glad to hear that

Nick

@Charlie: um, which one are you talking about?

Vrouvrou

hello

i need help for this please

0

Q: Question on relative homology

i have that $H_p(X,Y)$ is isomorphic to $Z_p(X,Y)/(B_p(X)+C_p(Y))$, where $Z_p(X,Y)=\lbrace \sigma\in C_p(X), \partial\sigma\in C_{p-1}(Y)\rbrace$ and i want to deduce that $H_0(X,Y)$ is the free module generated by the path connected components of $X$ that do not contain points of $Y$ but i d...

Charlie

16:31

@Nick speed average

@Chris'ssis and you?

Nick

@Charlie: Ah, well, do you have an answer?

Chris's sis

@Charlie A bit depressive today. I think that's because it's a rainy day.

robjohn

@Alizter I see nothing.

Chris's sis

(I'm not inspired at all, this is not my day - less productive)

robjohn

@Chris'ssis I love rainy days

Charlie

16:35

@Chris'ssis :(

Chris's sis

@robjohn sometimes I love them too.

Nick

@Chris'ssis: Why?

Charlie

@Nick I have an explanation, wanna hear?

Nick

@Charlie: Yes!

Charlie

@Nick here it goes: it's a concern in statistics to find a gdoo estimator to a certain parameter

@Nick your parameter is the average speed

Nick

16:39

and my estimator is either distance or time.

So, this is a problem in statistics ... not Physics? (...wierd)

Cortizol

What substitution to use for this integral
$$\int_1^{\infty} \frac{\arctan x}{x^2 \sqrt{x^2-1}}$$

Alizter

@Cortizol First parts

Charlie

@Nick yes, cool huh? You need an estimator that is consistent and not biased

L

Nick

@Charlie: Um, how are my guys biased?

Hypem

Hi everyone ! Need your help on this one :
http://math.stackexchange.com/questions/560564/direct-sum-and-tensor-product-of-two-representations-of-a-group
Thanks !

(Direct sum and tensor product of two representations of a group)

Alizter

16:43

@Cortizol Then make the sub $x=\sec \theta$

Charlie

@Nick it must follow certain properties

@Nick we have variance and expected value

Nick

@Charlie: Maybe it could be a fault with my logic at where the motion is halved. Algebraically both my equations are right, maybe it depends on the circumstances.

Charlie

@Nick just to ilustrate

Nick

@Charlie: Off-topic question.: What model is your phone?

Charlie

@Nick I am not sure , but usually things are calculated in function of time, not space

@Nick samsung galaxy y

Nick

16:51

@Charlie: Bravo, you seem to be able to do a lot with it!

Charlie

@Nick ;)

Sohaib I

Hello

I got a question

How do we calculate the covariance matrix?

stattrek.com/matrix-algebra/covariance-matrix.aspx

This above given link is incorrect right?

Charlie

@Nick but the mse chat is quite disturbing on mobile

Nick

@Charlie: I can understand what that must be like.

Charlie

@Nick :)

@Nick what is bothering you, nick?

Nick

17:03

@Charlie: A crumby processor that can't handle chrome.

and the ding sound mse makes. So many other sounds in the world. Why did they have to choose the one that sounds like a repressed fart.

Charlie

@Nick HAHAHA

Chris's sis

@Cortizol I just finished your question. Maybe you wanna try the differentiation under the integral sign.

Nick

@Charlie: Or maybe it just sounds weirder on my laptop's speaker.

sonicboom

Just a quick check of something if someone wouldn't mind. On the interval (0, 1). I believe the sequence of functions f_n(x) = x/(nx + 1) converges to the zero function uniformly as x/(nx + 1) < 1/n and 1\n goes to 0 as n goes to infinity. Does this look correct?

leo

@sonicboom indeed

Chris's sis

17:08

Try this $$I(a)=\int_1^{\infty} \frac{\arctan (a x)}{x^2 \sqrt{x^2-1}} \ dx$$ and then differentiate with respect to $a$. The rest is piece of cake.

leo

I've never seen such amount of correctness

Charlie

@Nick turn it off :)

leo

@Nick

Nick

@leo

leo

@Nick can't ping you more than twice

Sohaib I

17:11

@robjohn Help me man :P

Nick

@leo: I can't ping you more than once

robjohn

@SohaibI what's up?

Charlie

@Nick did you turn the ping off?

Nick

Oh, ping. So, that's what it really sounds like.

Sohaib I

@robjohn Read my question. Please :P

@robjohn This one: chat.stackexchange.com/transcript/message/12106615#12106615

Charlie

17:13

@Nick it sounds like an arrow

Alizter

@Chris'ssis I get $$\arctan x \ln\left(\frac{x+1}{\sqrt{x^2-1}}\right)-\int\frac{\ln\left(\frac{x+1}{\sqrt{x^2‌-1}}\right)}{x^2+1}dx$$

Nick

Wait a second guys, the area of a unit circle is pi. Not 7 pi^2 . What is Jack M. talking about?

Alizter

@Chris'ssis I either messed up or this is an evil question

@Nick He made an error. So we are rubbing it in. Go like it.

Charlie

@Nick he got confus

robjohn

@SohaibI I have never worked with that matrix. I don't know if that definition is standard or not.

Nick

17:17

@Alizter: The more a person likes it, the wronger it gets. XD

Charlie

@Nick :) is there any thingy would you like to learn ?

Nick

@Charlie: Nothing specific, I'm in a mood for anything right now!

Sohaib I

@robjohn Oh alright. Thanks a lot man!

Charlie

@Nick }:)

Alizter

@Chris'ssis I just discovered a major flaw in the question you asked. The one with the answer $\zeta(3/2)$

Chris's sis

17:22

@Alizter what is that one?

Charlie

What did you say @nick ?

Alizter

If n is a perfect square it doesnt evaluate as nicely as the product of the factors are not $n^{\tau(n)/2}$ as $\tau(n)$ will be odd for a square

hmmmmmmmmmmmmmmmmmmm

Nick

@Charlie: Sorry, that must be very annoying.

@Charlie: Do you know something about Fibonacci numbers?

Chris's sis

@Alizter you mean that the product of the divisors isn't always $n^{\tau(n)/2}$?

Alizter

Yes

Charlie

17:30

@Nick I think everyone knows a bit about it

Alizter

@Chris'ssis Because of perfect squares

Nick

@Charlie: Ok, Why is the sum of 10 terms of it equal to 11 times the 7th term.

Charlie

@Nick 11 , 7 ...prime numbers, curious . I don't know why. Do you know if it repeats with another terms?

Nick

You would not believe the number of people that I fooled into thinking I'm an arithmetic genius with that.

It works with any 10 numbers generated by the fibonacci rule.

The rule being, $ a_n = a_{n-1} + a_{n-2}$

Chris's sis

@Alizter did you check that, for instance when $n=4$?

Charlie

17:34

@Nick that's cool, I didn't recall that one, but I don't why, but I honestly don't know why

@Nick last semester I gave a small lecture on the RH

Alizter

@Chris'ssis Wait don't worry I just realised the tau root eleminiates it

Nick

@Charlie: RH ..? Right-Hand? Red-Head? ...

Chris's sis

@Alizter you said there is a major flaw ...

Alizter

in my argumet

;)

Chris's sis

@Alizter ok :-)

@Alizter you was about to ruin my day :-)

Charlie

17:37

@Nick riemann hypothesis

Alizter

@Chris'ssis hhehe

Nick

@Charlie: I've heard of the Riemann-zeta function but like the Taylor series, I'm going to postpone learning about it till ...well, until I have to. Is it cool, the hypothesis I mean?

Charlie

@Nick the ideas that surrounds it, imho, are nicer than the statement of the hypothesis

Nick

@Charlie: "The Riemann hypothesis implies results about the distribution of prime numbers" - That is so cool

Charlie

@Nick indeed

What???

Nick

17:54

@Charlie: What is $\coprod$ thing supposed to do?

Disregard any post I remove. I remove it because when I post it, I realize that I posted sheer idiocy. (Yeah, I have that rare condition that I say things before I think about them)

Charlie

@Nick coprod ?:/

Nick

@Charlie: If you like primes, here's something I made. khanacademy.org/cs/prime-number-visualization/1843966980

Also, how do you link like that?

Charlie

@Nick your not the only one ;)

@Nick [write blablabla](linkyoururl)

@Nick hehehehe

Nick

18:14

@Charlie: A glitch using Prime numbers

Cool, the link thing works.

Enjoys Math

@Nick: isPrime(n-2)||isPrime(n+2)

you don't need the second check as you're in an if that already checked that

I code a little too

What does it do? print the twin primes or something

Charlie

I code a bit too, but don't ask me:D

Nick

@Enjoys: It checks if the value is prime and that part was just overkill because I have this inane tendency to think that if it isn't fried, it ain't dead.

Charlie

XD

Nick

The life of 2*pi*r

Enjoys Math

18:21

@Nick what does the picture pixel color and location represent?

Charlie

@nick

Don Larynx

Show that $x^2 + x$ can be factored in two ways in $\Bbb{Z}_6[x]$ as the product of non-constant polynomials that are not units.

Answer: Zero is not a unit. One way is $x(x+1)$, because $2(2+1) = 0$. Another is $(x^2 + x)(1) because $2^2 + 2 = 0$. Correct?

PaulRS

18:53

@DonLarynx $1$ is a unit polynomial.

even more, it is a constant!

Expand $(x-2)(x-3)$.

sonicboom

19:14

The sequence of functions f_n(x) converges pointwise on the interval [0, 1]. But if the domain is say [0, 2] does it still converge pointwise to some function?

Don Larynx

That gives me $x^2 - 5x + 6 = x^2 + x$. But how am I supposed to find that other than using intuition?

Chris's sis

hehe, that's lovely (related to my previous question): wolframalpha.com/input/?i=frac%28x%29%2Bfrac%281%2Fx%29%3D1+

Enjoys Math

That's totally random

lol

What's $frac(x)$?

Alizter

@Chris'ssis Find the number of smallest $|z|$ that satisfies $\left\{\frac{1}{z}\right\}=\frac{1}{\{z\}}$

Chris's sis

19:30

It's clearly the question needs some restrictions there, it's just the way I received it.

Alizter

@Chris'ssis Let me rephrase. Let $z$ be a solution to the equation $$\left\{\frac{1}{z}\right\}=\frac{1}{\{z\}}$$ find the smallest value for $|z|$.

Chris's sis

$$a+\frac{1}{a}=\lfloor a\rfloor+\lfloor\frac{1}{a}\rfloor+\{a\}+\{\frac{1}{a}\}=\underbrace{\lfloor a\rfloor+\lfloor\frac{1}{a}\rfloor+1}_{\in \mathbb{Z}}$$

Then $a^2-ka+1=0, k \in \mathbb{Z}$. Hence it's easy to find $a$ that is $a=\frac{k\pm\sqrt{k^2-4}}{2}$.

Vrouvrou

hello , can someone help me

Alizter

@Chris'ssis Do you reckon this converges? If so what? $$\sum_{n\in\Bbb N} \frac{\sin n}{\Gamma(\sin n)}$$

Vrouvrou

1

Q: Question on relative homology

i have that $H_p(X,Y)$ is isomorphic to $Z_p(X,Y)/(B_p(X)+C_p(Y))$, where $Z_p(X,Y)=\lbrace \sigma\in C_p(X), \partial\sigma\in C_{p-1}(Y)\rbrace$ and i want to deduce that $H_0(X,Y)$ is the free module generated by the path connected components of $X$ that do not contain points of $Y$ but i d...

algebraic-topology homological-algebra homology

Alizter

19:43

@Chris'ssis What about this? $$\sum_{n=0}^\infty \frac{\tau(n)!}{\sigma(n)!}\longrightarrow^{\quad}\sqrt{2}$$

Chris's sis

@Alizter this is nice

Alizter

@Chris'ssis This is hypothetical though. It looks like it I am not sure how to prove it though

@Chris'ssis At n=100 things don't look too good

Chris's sis

@Alizter Your series starts from $n=0$?

Alizter

noooooo 1 sorrryyy

Chris's sis

@Alizter ok

Alizter

19:49

@Chris'ssis Interesting note $\sum^\infty_{n=1}\frac1{\tau(n)}\longrightarrow\infty$

Chris's sis

@Alizter not that interesting. It comes as a consequence of a previous series I posted here.

Alizter

@Chris'ssis This one looks like a Fourier series gone wrong :P

wolframalpha.com/input/…

Chris's sis

@Alizter you may only consider the series in primes, but since there are infinitely many primes, the series clearly tends to $\infty$ (at denominator you'll have 2).

@Alizter I was also wondering last days about a very fast way to prove that $$\sum_{n\ge 1} \left(\frac{\tau(n)}{\sigma(n)}\right)^2$$ converges.

Alizter

use that formula with the square and the series and the cube thing

Chris's sis

@Alizter that one I posted here :-)?

Alizter

20:01

@Chris'ssis I think so

PaulRS

Hi

Alizter

$$\int_0^\infty e^{-\tau(\lfloor x\rfloor)}dx$$

@Chris'ssis

PaulRS

@Alizter Very divergent...

Alizter

@PaulRS The correct term is evil

Chris's sis

@Alizter did you consider what happens in $[0,1)$?

PaulRS

20:12

(should actually start from $x=1$ too, what is $\tau(0)$)

Chris's sis

@PaulRS that was my point too above.

PaulRS

@Chris'ssis yes

@Alizter The asymptotic estimate is evil?

Alizter

@PaulRS This world is

PaulRS

xD

Alizter

@Chris'ssis What about $$\int_0^1 \frac{\Gamma(x+1)}{\lfloor x\rfloor!}$$

Chris's sis

20:16

@Alizter it doesn't make too much sense to me. Is there a typo?

Alizter

@Chris'ssis No thats it

Chris's sis

@Alizter Ah, you changed it.

Alizter

@Chris'ssis Yeah the other one was a divergent typo

Chris's sis

@Alizter you might want to know that $\int_0^1 \Gamma(x+1) \ dx$ is a question I posted here a long time ago. Let me check that.

(btw, it works nice by reflection formula or digammma function - integration by parts)

PaulRS

I suppose you swap the integration order or something of the sort, right?

Alizter

20:20

@Chris'ssis Just realised no matter what $x$ the denominator is going to be 1 in this bound

so there fore it just evalutes to your integral

PaulRS

@Chris'ssis Are you sure you are not confusing it with $\log \Gamma(x+1)$ ?

Chris's sis

@PaulRS Ah, yeah. My bad. It's with $\log$

PaulRS

@Chris'ssis An idea would be to write $\Gamma(x+1)$ as an integral, and swap the order of integration... might work

Chris's sis

@PaulRS right. hehe, I had in my mind the idea there is a log in front.

Anyway, I wanna find that post.

@PaulRS btw, did you meet before the series I posted above?

PaulRS

@Chris'ssis $\sum_{n\ge 1} \left(\frac{\tau(n)}{\sigma(n)}\right)^2$ <- this one?

Chris's sis

20:28

@PaulRS yes

PaulRS

@Chris'ssis No, hadn't seen it before.

Your problem is finding a nice proof of its convergence

?

Chris's sis

@PaulRS ok. Only interested in proving its convergence at the moment.

@PaulRS yes

PaulRS

@Chris'ssis The idea would be that $\tau(n) / n^\delta \to 0$ for each fixed $\delta > 0$... (in particular pick $\delta=1/4$).

Chris's sis

@PaulRS I see.

PaulRS

@Chris'ssis Can you prove that statement?

Chris's sis

20:41

@PaulRS I have it in a textbook. So, $\tau(n)=\mathcal{O}(n^{\delta})$, and $\tau(n)<M\sqrt[4]{n}$. Since $\sigma(n)\ge n$, then $\left(\frac{\tau(n)}{\sigma(n)}\right)^2 <\frac{M^2}{n\sqrt{n}}$

PaulRS

@Chris'ssis Ah, good, that is the exact same idea. I assume then that you are looking for something else

Chris's sis

Jes, I'm horrible at typing fast.

@PaulRS yeah, any alternative way is welcome :-)

Alizter

$$\int^\beta_{-\beta}\frac{dx}{\sin^2\beta+\cos^2(x)}$$

@Chris'ssis

Nick

@EnjoysMath: Whoops, fell asleep. The prime pixels are highlighted. The pixels on screen have been numbered from 0ish to 500ish . The colors repressent the type of primes.

Chris's sis

@Alizter if I'm not wrong, the answer to the previous integral is $\displaystyle \frac{\log(2\pi)}{2}$

Alizter

20:56

@Chris'ssis Nice how did you get that?

Nick

@Charlie: XD, I fell asleep and I had this amazing dream. I dreamt that I proved Legendre's Conjecture using mere induction.

Chris's sis

@Alizter I started out by integration by parts ...

Alizter

@Chris'ssis I see what do you think of my new integral
?

Chris's sis

@Alizter apparently it looks nice, but there might be some unpleasant surprises. Is this the correct form?

Alizter

good I need to ermm great the typo is fixed things around here ;)

Vincent Tjeng

21:07

hi guys, does anyone know of methods to calculate the sum of a column in an inverse matrix without computing the matrix itself?

basically, I want to solve the matrix equation A.x=b analytically, where A is a sparse matrix and b is the column vector {1, 0, 0, ..., 0}

Alizter

@Chris'ssis Its a useless integral :(

Chris's sis

@Alizter by the way, can you finish the evaluation of $\int_0^1 \Gamma(x+1) \ dx$ by using that little hint?

Alizter

im tired night all

Chris's sis

@Alizter I hope you can.

@Alizter really? :-( I wanted more questions.

@Alizter you might love this one $$\int_1^2 \frac{e^{x}-e^{2/x}}{\sqrt{x^3+2x}} \ dx$$

Charlie

21:23

@Nick :D

N3buchadnezzar

@Chris'ssis What is the trick ? :p

Chris's sis

@N3buchadnezzar I let Alizter guess it. I'm sure he/she has already fallen in love with it. :-)

N3buchadnezzar

I hate parts :p

@Chris'ssis Ah!

I am not a clever man

You pulled that integral on me before!! Fool me once shame on you, fool me twice shame on me.

Chris's sis

@N3buchadnezzar :-)

N3buchadnezzar

21:41

@Chris'ssis Divide and conquer + $u=2/x$

Chris's sis

@N3buchadnezzar btw, do you have this one in your e-book? $$ \int_0^1 \Gamma(x+1) \ dx$$

N3buchadnezzar

DUUH

@Chris'ssis $$\iint_S B(x+1,y+1) \mathrm{d}A$$

Where $S = [0,1]\times[0,1]$ ;-)

$$\int_0^1 \Gamma(x+t)\mathrm{d}x = \frac{1}{2}\log (2\pi) + t \log t - t$$

That is about all the Gamma integrals I know by heart.. =)

I meant $\iint_S \log \Beta(x,y) \mathrm{d}A$ sorry..

Chris's sis

@N3buchadnezzar Ah, is there a $\log$ in front?

N3buchadnezzar

Yeah

Chris's sis

@N3buchadnezzar Better later than never. (referring to the update) :-)

N3buchadnezzar

21:50

Should be a log in front of the previous one aswell

The log gamma integral is a nice one. Also I have never seen the beta version anywhere before.

Here math.stackexchange.com/questions/433868/…

Chris's sis

@N3buchadnezzar these integrals flow naturally. (+1 for your question)

N3buchadnezzar

$\arctan x + \arctan y = \arctan\left( \frac{x + y}{1 + xy}\right)$ Is this correcT?

I can not for the ife of me remember the signs in the last expression

Sanchez

22:06

1 - xy

Chris's sis

I DID A HUGE MISTAKE! (1 week I won't touch any mathematical book - far too overloaded)

Alexander Gruber

$\hat{\mathbb{Z}}\cong \prod_p\mathbb{Z}_p$.

never has the chinese remainder theorem been so complicated

Chris's sis

@Alizter my answer above wasn't correct. It's definitely time for a looonnng break. I do elementary mistakes.

N3buchadnezzar

@Chris'ssis I can not think straight either. I keep making sign errors.

anon

@AlexanderGruber I think adelic strong approximation is an even more complicated version!

N3buchadnezzar

22:21

@Chris'ssis I know I have asked this before, but could you quickly check whether the integrals stated in my post here are correct? math.stackexchange.com/questions/430276/…

I have done the calcutions for the first one a few times, and keep getting $$ \frac{\log^2 2a^2}{a^2}\bigl[ \arctan n - \arctan m \bigr] $$

Faraad

anyone want to talk real analysis

Adobe

I need your help guys on those two questions.

2

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