Mathematics - 2014-07-27 (page 1 of 3)

Khallil

02:03

Thanks for the positive rep, @Hakim.
http://math.stackexchange.com/questions/530779/integral-of-int-0-pi-2-frac-sqrt-sin-x-sqrt-sin-x-sqrt-cos-x/799678#799678

^_^

Kaj Hansen

03:02

Hello everyone.

skullpatrol

Hello.

Kaj Hansen

"3 hours later". lol, been quiet tonight it seems.

skullpatrol

yep

over 9000 hours later ...

:D

Kaj Hansen

Whenever I see that in the chat, I'm reminded of the "[amount of time] later..." things in Spongebob episodes, hahaha

2

Vibhav Pant

05:21

Hello

skullpatrol

06:04

Hello

r9m

06:23

konnichiwa :)

china math

hello,@r9m

r9m

@chinamath hi

china math

en I have inequality problem ,have you interesting it?

r9m

sure :) .. I can try

china math

0

Q: How prove this $x_{1}+x_{2}+\cdots+x_{n}<\frac{5}{3}$

Let $x_{1},x_{2},\cdots,x_{n}\ge 0$ with $$x_{i}x_{j}\le 4^{-|i-j|}$$ for $i, j = 1, \dots, n.$ Show that $$x_{1}+x_{2}+\cdots+x_{n}<\dfrac{5}{3}.$$ My idea: since $$(x_{1}+x_{2}+\cdots+x_{n})^2=\sum_{i=1}^{n}x^2_{i}+2\sum_{i<j}x_{i}x_{j}=2\sum_{i\le j}x_{i}x_{j}-\sum_{i=1}^{n}x^2_{i}...

inequality

It is said this problem a few people solve it

It's South East Mathematical Olympiad

r9m

06:38

@chinamath $(\sum\limits_{i=1}^n x_i)^2 = \sum\limits_{i=1}^n \left( x_i (\sum\limits_{j=1}^n x_j)\right) \le \sum\limits_{i=1}^n \sum\limits_{j=1}^n \dfrac{1}{4^{|i-j|}}$

china math

@r9m,and then? Thank you

Vibhav Pant

Is the equation $\sin x + \cos x = 1$ equivalent to the differential equation $y+y' = 1$? (Here $y=\sin x$)

china math

then How find this sum $\sum\limits_{i=1}^n \sum\limits_{j=1}^n \dfrac{1}{4^{|i-j|}}$?

@VibhavPant,yes

r9m

@chinamath that is not too difficult I guess ^_^

china math

without loss of we can Assmue that $i>j$?

but this sum $>\dfrac{5}{3}$

r9m

06:48

oh ! i see ..

Vibhav Pant

Well, then I tried solving this to get $x = -\ln |1-y| \implies y=1-e^{-x}+C$

And since $\sin\pi/2 + \cos\pi/2 = 1$, I get $\pi/2 = -\ln|1-1|$

Which seems wrong to me

r9m

@VibhavPant ? what do you mean ? $\sin x + \cos x = 1$ has countable number of solutions .. its equivalent to solving $\sin \left(x + \frac{\pi}{4}\right) = 1/\sqrt{2} = \sin \frac{\pi}{4}$ .. but solving the diffn eqn $y'+y = 1$ is an entirely different thing :|

Vibhav Pant

@r9m I tried reducing it to a differential equation

since $\mathrm{D}\sin x = \cos x$

r9m

@VibhavPant what is the difference between solving a equation and solving a differential equation ?

Vibhav Pant

I guess differential equations have infinite solutions if the initial value is not specified

(Im still new to the topic, I started it yesterday)

Chris's sis

07:00

Greetings

china math

@r9m,Jack give this answer

1

Q: How prove this $x_{1}+x_{2}+\cdots+x_{n}<\frac{5}{3}$

Let $x_{1},x_{2},\cdots,x_{n}\ge 0$ with $$x_{i}x_{j}\le 4^{-|i-j|}$$ for $i, j = 1, \dots, n.$ Show that $$x_{1}+x_{2}+\cdots+x_{n}<\dfrac{5}{3}.$$ My idea: since $$(x_{1}+x_{2}+\cdots+x_{n})^2=\sum_{i=1}^{n}x^2_{i}+2\sum_{i<j}x_{i}x_{j}=2\sum_{i\le j}x_{i}x_{j}-\sum_{i=1}^{n}x^2_{i}...

inequality

r9m

@chinamath okay :)

china math

are you understand this answer?

I don't why $x_{1}=1/4$

r9m

I'll look at it later :)

Chris's sis

07:19

@r9m I have a new question proposal

$$\int_0^{\infty} \left\{\dfrac1{x+1}\right\} \left\{\dfrac1{x}\right\}\left\{\dfrac1{x-1}\right\}\mathrm dx$$

@chinamath look that this one

$$\int_0^{\infty} \left\{\dfrac1{x+k}\right\}\cdots \left\{\dfrac1{x+1}\right\} \left\{\dfrac1{x}\right\}\left\{\dfrac1{x-1}\right\}\cdots \left\{\dfrac1{x-k}\right\}\mathrm dx$$

Balarka Sen

07:40

@VibhavPant Maybe maybe not

Depends on what the equation is.

@KajHansen heh

china math

@r9m,can you see it?

I fell this answer have somewrong,

Balarka Sen

@r9m Hardly any difference if you are solving an algebraic equation.

For any polynomial in $\Bbb Z[x]$ there is a corresponding differential equation of order the same as the degree of the polynomial over the differential extension $\Bbb C(x)$

china math

@r9m,we let $x_{1},x_{2},\cdots,x_{n}$ is $\dfrac{1}{4^k},\dfrac{1}{4^{k-1}},\cdots,\dfrac{1}{4},1,\dfrac{1}{4},\cdots,\df‌rac{1}{4^k}$

it is clear $\lim_{k\to\infty}(x_{1}+x_{2}+\cdots+x_{n}}=\dfrac{5}{3}

and this example such this condition

Balarka Sen

08:20

Anybody here knows how to read/convert a dvi file?

Vibhav Pant

There are tools for dvi to pdf IIRC

Balarka Sen

OK, nevermind, I just found a pdf version of the document.

Daniel Fischer

@BalarkaSen Any good viewer handles dvi as well as pdf and ps files. Install a good OS and use e.g. Okular.

Balarka Sen

So I'd have to install a new OS and download a good viewer to see a dvi file? I'd rather pass...

WAT

writelatex.com/1262185tdqxpr#/3065150

I thought I was saving my notes in there ^

Who the hell replaced the whole comm. alg. content by rapid French and analysis all over?

2

r9m

08:37

@chinamath you are right ! :) nice !!

china math

@r9m,so I want see this problem solve it

r9m

@BalarkaSen C(-_-)D (((.. (>| I hear ya !

@chinamath sqing opened this thread !

china math

I think maybe this inequality is hard,becasue In china IMO team problem have prove x(1)+x2+……+x(n)<2

and I can prove it,but for 5/3,it's hard I think

Balarka Sen

snif my comutative algebra notes. I was saving them for weeks.

china math

09:13

@r9m，this inequality have one solve it,can you understand? [math.stackexchange.com/questions/879305/…

I think his hint is also can't solve it

r9m

@chinamath It is difficult to read from the image .. sorry

china math

oh,Thank you all the same

G.T.R

10:12

@BalarkaSen I added the whole tex source

N3buchadnezzar

10:47

Alizter

11:18

@N3buchadnezzar That's dark.

Alizter

11:31

@Chris'ssis $$\int\limits^\infty_0 \int\limits^\infty_0 \left\{\frac1{xy-1}\right\}\left\{\frac1{xy+1}\right\}\mathrm dx\ \mathrm dy$$

N3buchadnezzar

@Alizter <3

Chris's sis

@Alizter you might begin by letting $xy=t$

@Alizter This integral diverges though.

N3buchadnezzar

@Chris'ssis What if the lower limits where $1$?

Alizter

@Chris'ssis Really?

Chris's sis

@Alizter Yeah.

Alizter

11:41

Wierd mathematica found a result

oh wait no

Chris's sis

@Alizter Which is that one?

Alizter

when i increased precision

636 -> 1.35x10^17

Yeah that thing diverges

Chris's sis

@Alizter lol :-)

@Alizter Indeed.

Alizter

Maybe a plot will help me find something interesting

N3buchadnezzar

@Alizter Try it with 1 instead of zero as the lower limits?

Alizter

11:45

throws up

Khallil

I've got a quick question, guys. I need to inductively prove that $T_n < n$ for $n \geq 1$, $n \in \mathbb{N}$.
$$ T_n = \dfrac{1}{2} + \dfrac{2}{3} + \dots + \dfrac{n}{n+1} $$

Alizter

That thing is howible

N3buchadnezzar

Howible!

Khallil

So far, I've found that $T_{k+1} < k+1 \implies \ T_k - k < \frac{1}{k+2}$.

Since I've assumed that $T_k < k$, $T_k - k < 0$.

How would I go about combining those two to finish the proof?

@Alizter That looks cool!

Alizter

Well assume $T_n < n$ we have $T_{n+1}=T_n+\frac{n+1}{n+2}<n+\frac{n+1}{n+2}$

Now show that $n+\frac{n+1}{n+2}<n+1$ and you are done

Khallil

11:50

How'd you get that?

(I'm referring to the $n + \frac{n+1}{n+2}$ on the RHS of your first inequality.)

Alizter

Well $T_n<n$

That is what we assumed

Khallil

Yep.

Alizter

and if we show that this implies $T_{n+1}<n+1$

then we can use induction

Khallil

I'm not following.

Alizter

Rearrange the inequality $\frac{n+1}{n+2}<n+1\implies\frac{n+1}{n+2}<1$ which is true.

Then show a base case and it is proven

Hakim

11:53

@Khallil If we can show that $T_n\lt n\implies T_{n+1}\lt n+1$ then we're done.

Khallil

Oh, I get it.

I ended up with $n+2>0$ which is always true for the original restriction of $n \geq 1$.

^_^

Alizter

There you go :)

Khallil

Thanks @Hakim and @Alizter!

Hakim

You're welcome

N3buchadnezzar

So we have

$$T_n = \frac{1}{2} + \left( 1 - \frac{1}{3} \right) + \cdots + \left( 1 - \frac{1}{n+1} \right) = \frac{1}{2} + (n - 1) - \sum_{k=3}^{n+1} \frac{1}{k} $$

Proving $T_n < n$ is the same as proving $\sum_{k=3}^{n+1} \frac{1}{k} < \frac{1}{2}$

Alizter

12:13

@N3buchadnezzar overkill?

N3buchadnezzar

@Alizter Of course :p

Khallil

12:35

@N3buchadnezzar I think I had a similar way of doing it.

$$ T_{k+1} = T_{k} + \dfrac{k+1}{k+2} $$ $$ \begin{aligned} T_{k+1} < k + 1 & \implies \displaystyle \sum_{r=1}^{k} \left( 1 - \dfrac{1}{k+1} \right) + \dfrac{k+1}{k+2} < k + 1 \\ & \implies k - \sum_{r=1}^{k} \dfrac{1}{k+1} + \dfrac{k+1}{k+2} < k + 1 \\ & \implies \sum_{r=1}^{k} \dfrac{1}{k+1} > \dfrac{-1}{k+2} \end{aligned} $$

Since $k \geq 1$, it's "pretty obvious" that this is true.

@N3buchadnezzar

It might actually require another induction to be shown true.

@Alizter what do you think?

N3buchadnezzar

@Khallil Yeah

We are brushing up a local "pub" which is driven by students for students. Run on charity. Things are improving, we made a decent surplus last year to use on renevations =)

imgur.com/NUhNFvl,qgCcaZ2,C5h2rdo,tdNlAbl ^^

kwak

13:10

@N3buchadnezzar Doesn't it smell smoke? :<

N3buchadnezzar

@kwak smoke?

kwak

@N3buchadnezzar the thing many people unfortunately do

a pub often allows smoking

oh a students pub, sorry I didn't read fully

N3buchadnezzar

Komrades do not smoke

kwak

yep

Khallil

Don't drink and derive guys.

That's how the proof of 1 being equal to 2 came along.

N3buchadnezzar

13:13

Just drink cheap vodka and watch Twitch plays pokemon

Khallil

YES!

kwak

smoke causes mutations in your brain :s, not very accurate, but that's how I picture the danger of it

pollution in general does, I guess

N3buchadnezzar

Last year we consumed about 40 beer crates a week in the first two starting weeks of uni.

kwak

you'd need a tank

N3buchadnezzar

Yeah :p

I just found these i.imgur.com/iOWBcMI.jpg and i.imgur.com/NS0NK0D.jpg. Heh

skullpatrol

13:29

@kwak I smoked for many years as a teenager and it did not mutate my brain.

N3buchadnezzar

@skullpatrol Yes. Yes it did.

skullpatrol

@N3buchadnezzar No. No it didn't.

@kwak marijuana and harder drugs mutate the brain

kwak

@skullpatrol I'd trust an MRI or something tangible

skullpatrol

@kwak there is not enough known about something as complex as the human brain to trust anything pal :-)

N3buchadnezzar

@skullpatrol I am joking :p

skullpatrol

13:37

@N3buchadnezzar I know, I know...

:D

N3buchadnezzar

Everyting I say Is a joke on, life universe and the darkness

kwak

@skullpatrol some years after quitting, did you feel "smarter" or it was the same?

skullpatrol

the same

kwak

how long ago did you quit?

skullpatrol

>10 years

but I went back to it 5 years ago for a year then quite again

it is highly addictive

nicotine has the same chemical structure as heroine

kwak

13:46

hmm I'm convinced you can have better reaction times, be "witter" without smoking, all my friends who were really good in Math didn't smoke, but maybe I'm wrong

nicotine is the only good thing in it, the rest of compounds are carcinogen, but fortunately after quitting things can repair I guess

skullpatrol

remember that is FIRST hand smoke, drawing it deep into my lungs

kwak

@skullpatrol just like secondhand smoke, with heating additionally

and a high concentration

skullpatrol

there is a HUGE difference with second hand smoke

kwak

Hmm, I think many people consider secondhand smoking like an aerosol spray, it "vanishes" magically after a few centimeters. I believe there are quite many carcinogen particles flying around (but you know my problem about it :) )

skullpatrol

nah, that's different

if you smoke a cigarette you are sucking the smoke deep down into your lungs

the chemicals get directly into your blood

and you feel the buzz

kwak

13:53

@skullpatrol as much as if you were another person in the same room, hermetically closed

less the heat

Alexander Gruber

smoking builds lung character.

skullpatrol

@kwak the actual smoker is willfully sucking up the smoke

Alexander Gruber

highly recommended

kwak

@AlexanderGruber you're one :/ ?

Alexander Gruber

@kwak Sure am

kwak

13:55

@skullpatrol I was just talking about the consequences, in both cases it will finish in the lungs, hot smoke or cold smoke

Alizter

@Chris'ssis How are you?

Alexander Gruber

@kwak when you smoke a cigarette, you draw the smoke through a filter

that is what the yellow/brown part is of a cigarette, near the smoker's mouth

Chris's sis

@Alizter thanks, I'm working. And you? :-)

Alexander Gruber

second hand smoke is not drawn through a filter so it contains more bad stuff

skullpatrol

@kwak The smoker WANTS the smoke in the lungs

kwak

13:57

@AlexanderGruber So this is a bit less bad than the sidestream smoke, that's what you want to tell me?

oh ok, that's why I don't appreciate at all being near smokers, like near a polluting plant

Alexander Gruber

@kwak right. i choke when i get a whiff of my own second hand smoke.

Alizter

@Chris'ssis I am investigating a new integral

Chris's sis

@Alizter Great! :-)

kwak

@AlexanderGruber So the secondhand smoke is for persons downwind smokers :(

Alizter

@Chris'ssis When you solved my integral for 0 to infty did you handle 2 to infty and 0 to 2 or both at the same time?

Chris's sis

14:00

@Alizter I treated these cases separately. The case when $x$ from $2$ to $\infty$ is just a piece of cake.

skullpatrol

@kwak just think about the difference in INTENTION of the smoker and the nonsmoker.

who WANTS the smoke in their lungs and who doesn't?

N3buchadnezzar

@skullpatrol nobody? It casuses long cancer, amongst a myriad of other bad things.

skullpatrol

@kwak if I take shallow breaths, then the second hand smoke will not get into my system

kwak

@skullpatrol but when you get past a smoker, you take a part of it right? non voluntarily, but you still take a bit

Chris's sis

@Alizter You ought to come up with a proof there since it's your integral. :D

skullpatrol

14:07

@kwak yes, a small fraction of what the actual smoker is trying to take deep into his lungs

kwak

@skullpatrol shallow breathes, well, it exposes less, but still a bit again, it's continuous

skullpatrol

@kwak do you drink alcohol?

kwak

@skullpatrol I could (more than smoking) but really not often

but digestive process is different than respiratory process

skullpatrol

@kwak How about exercise?

kwak

@skullpatrol I do some frequently, I'm rather too skinny than obese

skullpatrol

14:16

@kwak do you wanna start another room?

maybe some other time, I gotta go

sorry

kwak

@skullpatrol ok later if ypu want

Balarka Sen

14:30

@G.T.R tex source?

Balarka Sen

14:44

2

@N3buchadnezzar

N3buchadnezzar

@BalarkaSen Yo

Balarka Sen

Hullo

Khallil

15:06

Hullo smurt @BalarkaSen

Balarka Sen

Herro, @Khallil

Khallil

^_^

Alizter

15:38

@Chris'ssis I am working on it

Chris's sis

@Alizter That's good. :-)

Alizter

@Chris'ssis I am trying to deal with some sums on the 1-2 case

Chris's sis

OK

Alizter

@Chris'ssis I am getting stuck however

@Chris'ssis I cannot get any further from $\displaystyle \sum_{n=1}^\infty\int^{n+1}_n\frac{t-n}{t(2t+1)}\mathrm dt$

Chris's sis

@Alizter $ \ dx$? Be creative ... invent, create the way ... :-)

Alizter

15:45

I tried partial fractions but they usually mess up convergence. :(

Chris's sis

@Alizter Well, that integral is easy. The part (just) a bit more difficult comes after that.

Sara Sharp

Solve the following THREE variable linear program using Excel. MAXIMIZE X1 + 2X2 + 3X3; subject to the constraints: X1 + X2 + X3 <= 9; -1X1 + 2X2 + 5X3 <= 15; X1 >= 0; X2 >= 0. The optimal solution value for this linear program is:

a. The optimal solution value is 17.

b. The optimal solution value is 15.

c.The optimal solution value is 20

d.The optimal solution value is 19

N3buchadnezzar

@skullpatrol The only time the word incorrectly isn't spelled incorrectly is when it is spelled incorrectly

5

Balarka Sen

Is this answer really appropriate on that question? Should I remove it from there and post it in here?

I was misguided by the title. Thought it was another one of those big-list questions.

Alizter

16:01

@Chris'ssis I made a sub so that I could move the integral away and I get this $\displaystyle \int_0^1\mathrm dw\,w\left(\frac1{w+1}+\sum^\infty_{n=2}\frac{(-1)^n}{w+n}\right)$

Balarka Sen

Anyway, I'll leave it there for now.

Gotta goes.

Alizter

That series is weird.

Bye @BalarkaSen

Balarka Sen

Or perhaps I'll delete it.

Chris's sis

@Alizter I wouldn't say that.

$$\sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n+x}=\log(2)-\psi(x)+\psi(x/2)+\frac{1}{x}$$

Alizter

@Chris'ssis Where does this come from?

Chris's sis

16:07

@Alizter It may be derived in more ways. One way is to make use of the Weierstrass representation of the gamma function.

skullpatrol

@N3buchadnezzar When you say "spelled incorrectly" are you referring to the meaning of the word or the spelling of the word?

N3buchadnezzar

@skullpatrol both

skullpatrol

@N3buchadnezzar I normally spell it it

N3buchadnezzar

it

skullpatrol

yes

Chris's sis

16:20

$$\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \frac{(-1)^{i+1}}{i+2^j}=1-\gamma$$

@Alizter it's newly created.

Alizter

@Chris'ssis That is some nice symmetry

skullpatrol

16:36

Use–mention distinction

The use–mention distinction is a foundational concept of analytic philosophy, according to which it is necessary to make a distinction between using a word (or phrase) and mentioning it, and many philosophical works have been "vitiated by a failure to distinguish use and mention". The distinction is disputed by non-analytic philosophers. The distinction between use and mention can be illustrated for the word cheese: Use Cheese is derived from milk. Mention "Cheese" is derived from the Old English word "cyse". The first sentence is a statement about the substance called cheese; it uses the word...

Alizter

@Chris'ssis What about $$\sum^\infty_{i=1}\sum^\infty_{j=1}\frac{(-1)^{i+j}}{2^i+2^j}$$

Chris's sis

@Alizter For my series I used the identity previously posted.

Alizter

@Chris'ssis naughty. ;)

Chris's sis

:D

Alizter

oooh that telescopes nicellyy..

Chris's sis

16:45

@Alizter Yeah ... :-)

Alizter

@Chris'ssis I managed to squeeze out $1-\gamma$ so ill assume the rest goes away ;)

Chris's sis

@Alizter Yeap.

Alizter

@Chris'ssis Do you have any ideas for my one?

sarah

@Alizter Hmm that one looks very interesting

Hakim

17:02

@Chris'ssis I've calculated numerically the integral $\left\{\tfrac1{x+1}\right\}\left\{\tfrac1{x-1}\right\}$ from 0 to infinity and compared it to your value $\ln\pi$ and it was off by 0.694486

Alizter

@Hakim But the osscilations may be messing it up?

Chris's sis

@Hakim I see. Well, I think you were wrong somewhere. You need to carefully treat the case when $x$ from $0$ to $1$.

Hakim

@Alizter That's possible, through I increased the precision

Alizter

@Hakim mathematica?

Hakim

@Alizter ye

Alizter

17:04

The osscillations will make it too inacurate.

Symbolic is the way to go.

And because Mathematica cannot we need more people to confirm log pi

Hakim

@Alizter Okay let me repeat the calculations with increased precision

@Chris'ssis right

Alizter

@Hakim Increase the working precisions aswell

Chris's sis

@Hakim there you have fractional part of a negative number. Now, there are 2 definitions of the fractional part of a negative number.

@Hakim mathworld.wolfram.com/FractionalPart.html

I guess your different result comes from this point.

sarah

That integral is really weird around 0

Chris's sis

@Alizter For doing this, I might think of turning that double sum into a single sum by cleverly using some knowledge of number theory.

Alizter

17:27

@Chris'ssis How?

Chris's sis

17:41

@Alizter before treating the alternating case, did you think of $$\sum^\infty_{i=1}\sum^\infty_{j=1}\frac{1}{2^i+2^j}$$?

Vibhav Pant

Im trying to solve the differential equation $y'-3y=e^{2x}$, where y=0,x=0. If we let $f$ be a solution, my book says that $f(x)=e^{-A(x)}(f(0) + \int^x_0e^{A(t)}e^{2t}dt)$, where A(x)=$\int^x_0 (-3) dt=-3x$

sarah

@Chris'ssis I didn't really look into it because Mathematica said ComplexInfinity.

However I can see that it is either converging or very very very slowly diverging

i bet the former

Vibhav Pant

....and I get the correct answer

Alizter

@sarah @Chris'ssis Yeah mathematica gave me the same answer

Vibhav Pant

Guess I was wrong somewhere before

sarah

17:48

I will have a look

Alizter

@Chris'ssis What kind of trick are we talking about?

sarah

i gotta go bye

skullpatrol

later

Alizter

@sarah bye

Chris's sis

@sarah @Alizter it converges.

Alizter

17:58

@Chris'ssis Yes I just tested it.

@Chris'ssis Mathematica doesn't play nice with this one

Also do you reckon it is worth getting Mathematica 10? @Chris'ssis

Chris's sis

@Alizter I have Mathematica 8.0, but I don't know how things work in Mathematica 10.

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$Mathematics$ Mathematics