Mathematics - 2014-07-18 (page 1 of 2)

bolbteppa

00:05

@pourjour For F^2 = U_1 +* W, if you want to get the point (x,y) let y + z = x and take (z,0)+*(y,y) = (z+y,y) = (x,y). You can do this for any point (x,y) in F^2 so hopefully you'll see it now.

2

Anybody know what a spinor is?
"Sure, it's just the square root of a geometry"

pourjour

00:56

@bolbteppa Hmm, I see

lol wut

02:01

tuw lol

Kaj Hansen

02:51

Any ideas as to why this might've gotten downvoted? It's entirely possible I made a mistake, but I'm not sure what... math.stackexchange.com/questions/870265/…

lol wut

03:03

possibly somebody didn't take a liking to the fact that you gave a good answer to a homework question

Kaj Hansen

@lolwut, I'm starting to think that might be it. I try not to give away everything, but sometimes that's difficult. I'm trying to think of how to cut it down.

Jasper Loy

04:07

@KajHansen You look like Matt Damon.

Why is there a game theory SE in creation?

lol wut

.
<--

Jasper Loy

What does that mean?

lol wut

Jasper Loy

Are you Bart?

lol wut

no, I am not that eyeglasses kid

Jasper Loy

04:12

Ah, he has vanished from SE.

lol wut

I will tell you this. My real name has appeared somewhere at some time yesterday.

Jasper Loy

Very mysterious.

Sush

Can someone please help me with Linear Algebra Done Wrong?

I can't understand why author has chosen "wrong" in its title.

r9m

@Sush There is a text called Linear algebra done right :-) ...

Sush

@r9m, there is another text called LADW See this

r9m

04:21

@Sush ya I know ... :P but I don't know why the author chose that title :P

Sush

We will have to use it as textbook in this semester. So I want to know what is 'wrong' there.@r9m

@r9m, ok!

I read its preface, but nothing is mentioned why is that wrong!@r9m

r9m

@Sush there is nothing wrong in that book as far as I can remember :P

Sush

@r9m, :D

So, why the title has "wrong"?

r9m

idk :| ..

Sush

;|

lol wut

04:27

linear algebra done like a boss by lol wut. look for it in bookstores near you

r9m

@lolwut XD ... you use the name lol wut even for your books ? :P

lol wut

yep!

significantly better sales than when I use my real name

r9m

XD !

lol wut

(which is so unfathomably gross that I shall not say it here)

r9m

fair enough :-)

Jasper Loy

04:46

What happened to Mr Eyeglasses?

Kaj Hansen

@JasperLoy, someone else told me that on here not too long ago, lol

Jasper Loy

@KajHansen Oh, might have been me as well. You are hot, lol.

Kaj Hansen

Why thank you :P

lol wut

Contemplates posting a picture of myself just to see who Jasper compares me with..

Jasper Loy

I shan't post my pic here, because I look fat =)

lol wut

04:54

That's an easy thing to fix.

Jasper Loy

I put on 10 kg the past year.

As a consolation I tell myself that is because of the meds I am taking.

lol wut

yeah, medications for psychological problems can be particularly problematic. how's your sleep?

Kaj Hansen

@lolwut, are you an older user under a new account? Same question @JasperLoy. Usually there aren't too many lower-rep users hanging out in chat.

Jasper Loy

Well, I sleep and eat irregularly, which is OK since I don't work now. I hope to get well enough soon, but that may never happen.

Kaj Hansen

(No offense of course)

Jasper Loy

04:58

@KajHansen I deleted all my old accounts. I delete accounts regularly, lol.

Kaj Hansen

Ah, that explains it.

Jasper Loy

Many people in this chat know me. You are considered a newbie, lol.

lol wut

@KajHansen i don't participate much on main

r9m

@JasperLoy I gained 8 kg in last 2 months .. I ain't joking :|

Kaj Hansen

Oh definitely. I just joined at the beginning of the summer thereabouts.

Something like that.

lol wut

05:00

prefers trolling a math chat room to answering/asking questions

2

Jasper Loy

I have deleted my account so many times and changed my username so many times that people no longer know who Jasper Loy is.

Kaj Hansen

Yeah, I can see how answering questions can get tiring.

Jasper Loy

Before I put on weight, I look like Justin Bieber, haha.

lol wut

If I regularly came up with very good questions, I would definitely be asking them here. Unfortunately, I only come up with questions that either (1) I can answer in a couple of days if not sooner or (2) are just plain too difficult leading me to lose interest quickly

Kaj Hansen

@lolwut, where are you with your education?

lol wut

05:07

the equivalent of a small child

Kaj Hansen

hahaha

Jasper Loy

I finished undergrad a decade ago. Hoping to go grad school when my mental illness is well enough...

Kaj Hansen

That's unfortunate @JasperLoy. I had really bad depression about 3 years ago. Flared up again a year and a half ago. It's finally gotten better recently.

Jasper Loy

@KajHansen I have OCD. Been suffering over a decade. Life has become meaningless. I just hope that my next life is better, since I believe in rebirth.

Kaj Hansen

Damn. I couldn't imagine what that's like...

skullpatrol

05:22

...unfortunately, I can do more than imagine.

But life goes on, as they say.

With or without meaning.

John Smith

05:53

In categories what does hom(A,-) denote? hom(A,B) is the set of all morphisms from object A to B. Is hom(A,-) the set of all morphisms from A (to any object, including itself)?

Anthony

Helllllllo?

Sawarnik

08:55

@Anthony Hello.

@KajHansen Oh. Why did you go into these depressions?

Anthony

@Sawarnik Hey.

Could you help me with my complex analysis homework?

Huy

$i^2 = -1$.

Anthony

:D

Huy

Does that help?

Anthony

No D:

Huy

09:02

What is it you're having trouble with then ?

Anthony

Showing the following

Huy

Can't you just use the theorem of residues and the standard estimate for line integrals?

Anthony

I haven't learned about residues yet :/

And I'm kinda hazy on this stuff, what standard estimate?

Huy

$$\left| \int_\gamma f \right| \leq L(\gamma) \cdot \max_{z \in \operatorname{im} \gamma} |f(z)|$$

I don't really remember to be honest. :(

Anthony

Whhhhhhat. What's L?

Huy

09:14

The length of the line.

But I think you need to work around the poles somehow first. That's why I was thinking of said theorem.

Anthony

There's just a pole at zero, right?

Huy

Yes.

I don't think that's a pole though, thinking about it.

Because of the logarithm.

Anthony

09:33

Oh.

Sawarnik

@Anthony I know nothing about it actually :D

Anthony

D:

@DanielFischer Might I bother you again?

Daniel Fischer

@Anthony I don't know. Might you?

Anthony

Please

PLEASE

0

Q: Bound in Complex Analysis

Can someone direct me towards the right way to approach this problem? Show $$\displaystyle \left|\int_{|z|=R} \frac{\log{z}}{z^2} dz\right| \leq 2\sqrt{2\pi}\frac{\log{R}}{R},\; \text{ for } R>e^{\pi}.$$ Edit: I'm considering the ML estimate, so L would be $2\pi R$ and I'm still debating M. The ...

This problem. @DanielFischer

Daniel Fischer

09:49

Hm, @Anthony, the ML estimate doesn't give you the desired factor of $2\sqrt{2\pi}$, it only gives you $2\pi$. So we need to look at the integral in more detail. Nothing obvious seen so far, however.

Anthony

Ugh.

Oh shoot.

I have a small typo.

Huy

@DanielFischer: I don't quite remember how to solve these kind of problems. How would I compute the residue of $\log z / z^2$? It's an essential singularity at $0$, right? But the logarithm doesn't have a Laurent series, no?

Anthony

@DanielFischer I ammended it, it was the Principle Log.

On the left.

Daniel Fischer

@Huy It's a branch-point, there is no residue. Only isolated singularities have residues.

Anthony

I guess that doesn't really matter.

Huy

09:51

Oh right.

Daniel Fischer

@Anthony The choice of the branch-cut matters, the chosen branch for that cut doesn't, since $$\int_{\lvert z\rvert = R} \frac{dz}{z^2} = 0.$$

Anthony

Oh no, I was saying it didn't matter that I forgot to write Log, as opposed to log.

Ugh I have no idea how to do this problem...

Anthony

10:09

Well damn, I got really close.

Daniel Fischer

@Anthony You can explicitly evaluate the integral. That's easy in this case. You get a strictly better result, though, and I can't see a natural way to get the given bound.

@Anthony I looked at your .png. You misread the bound. The factor is $2\sqrt{2}\,\pi$. Then the ML estimate plus Julian's answer gives you the result.

Anthony

Oh.

Crap.

I'm sorry.

That makes so much more sense.

@DanielFischer Thanks for your help.

Karl Kronenfeld

10:33

I can't help but laugh. It appears that someone is downvoting all of my answers in which I am a smartass (e.g. this)

Alizter

@KarlKronenfeld maybe he thought of the same snappy answer and he is jealous.

Karl Kronenfeld

@Alizter I figure someone believes that my answers are bringing down the value of the site (not that those particular questions are any good) and wants to express it through downvotes.

skullpatrol

the system will catch'em

Karl Kronenfeld

I don't care if he/she is caught. I just found it funny which answers were targeted.

Daniel Fischer

@skullpatrol At the rate of one downvote per four days?

skullpatrol

10:45

@DanielFischer Pardon me, I didn't know the rate.

@KarlKronenfeld my answer was down voted on that answer too >8(

skullpatrol

12:21

@lolwut here :-)

Mats Granvik

12:45

Anyone good with Mathematica and q-factorials? I would like to know how to program this sequence with the description given. https://oeis.org/A127926

The Pari program is a bit terse to understand:

(PARI) {a(n)=if(n==0, 1, polcoeff(1-q- sum(k=0, n-1, a(k)*q^k*prod(j=1, k+1, (1-q^j)/(1-q+q*O(q^(n-k))))), n, q))}

Huy

13:42

What is the convention of the notion $C(I)$ for some closed interval $I$? Is it $C^0(I)$, i.e. the space of continuous functions on $I$?

Jasper Loy

I got 18 votes for a one line answer.

N3buchadnezzar

@JasperLoy Was it "LOL" ?

And hey, wht did you do with all your points and answers?

Jasper Loy

@N3buchadnezzar Oh I deleted my old accounts. This account is two days old.

N3buchadnezzar

meh, go retire

Hippalectryon

@JasperLoy why do you keep deleting them ?

Jasper Loy

13:51

@Hippalectryon I just feel like doing it.

N3buchadnezzar

Does your old answers dissapear? Some might still find them usefull

Jasper Loy

@N3buchadnezzar They remain, so there is no real loss.

N3buchadnezzar

@JasperLoy floss I think you wanted to say floss

Hippalectryon

14:40

asdasdasd00

Alizter

phew its 33 degrees here

but no sea so it sucks

r9m

@Alizter celcius ?

Hippalectryon

woot forget my above message it's an error i didnt even see i posted it

Alizter

14:56

@r9m yes

Too humid

I can survive ~45 in Turkey but its dry.

Vrouvrou

15:13

Hello, please the definiton of $\lim_{u\rightarrow+\infty}G(t,u)=-\infty$

is $\forall M>0 ,\exists R>0 $ such that $|u|\geq R \Rightarrow G(t,u)\leq M$

or $G(t,u)\leq -M$ ?

please

Mats Granvik

The number of divisors of $n$ are to $n$, as the sum of the divisors $n$ are to the odd numbers.

Chris's sis

Greetings

Vrouvrou

can someone answer me ?

@Chris'ssis please can you help me

Balarka Sen

@MatsGranvik what do you mean by "are to"?

Mats Granvik

grammar mistake, I left out the word "of".

The number of divisors of n are to n , as the sum of the divisors of n are to the odd numbers.

Balarka Sen

15:21

i still don't get what you're trying to say.

Mats Granvik

I can speak to you in Mathematica.

Balarka Sen

oh noes. can't you speak in mathematics instead of mathematica =)? i don't get what you mean by "are to".

r9m

@Alizter its 30 - 35 °C here :| with humidity that I can't describe :P

Balarka Sen

@r9m heh heh

Mats Granvik

@BalarkaSen It is an analogy.

Alizter

15:24

@r9m Where are you?

Balarka Sen

@MatsGranvik like?

r9m

@Chris'ssis Greetings great SIS :D

Balarka Sen

@Alizter Top-secret.

Chris's sis

@r9m Hi :-)

Balarka Sen

Only the secret agents of area 51 knows about where he is, @Alizter.

r9m

15:24

:D

Mats Granvik

Natural numbers in a really complicated way:

Clear[t, n, k, a, b, x, y];
nn = 16;
t[n_, 1] = 1;
t[n_, k_] :=
t[n, k] =
If[n >= k,
Sum[t[n - i, k - 1], {i, 1, n - 1}] -
Sum[t[n - i, k], {i, 1, n - 1}], 0];
a = Table[Sum[t[n, k], {k, 1, nn}], {n, 1, nn}]

Balarka Sen

@MatsGranvik I can't understand a word of it.

r9m

@Alizter 22.5667° N, 88.3667° E :P

N3buchadnezzar

Mmmmm

Balarka Sen

@r9m star. best location ever.

r9m

15:26

@BalarkaSen Indeed :D

N3buchadnezzar

$\lfloor \sqrt{n} + \sqrt{n+1} \rfloor = \lfloor \sqrt{4n+2} \rfloor \ , \quad n \in \mathbb{N}$

Daniel Fischer

@Vrouvrou Since you consider $u\to +\infty$, it must be $u > R$, not $\lvert u\rvert > R$. (The latter would be for $\lim_{\lvert u\rvert\to\infty}$.) And since you want the limit to be $-\infty$, and you imposed $M > 0$, it must be $\leqslant -M$. So the definition is $$\bigl(\forall M > 0\bigr)\bigl(\exists R > 0\bigr)\bigl(\forall u > R\bigr)\bigl(G(t,u) \leqslant -M\bigr).$$ That is if you consider one fixed $t$. Otherwise, you must throw a $\forall t$ in.

Mats Granvik

It is a recurrence. It says that if you sum n+1 ones and subtract them with n ones, you 1 one. Then sum the ones together and you get n.

The analogy is that if you change the summation index in the same procedure you get the number of divisors of n.

Balarka Sen

@N3buchadnezzar Meh.

@MatsGranvik n+1 "ones"? what ones?

N3buchadnezzar

@BalarkaSen Meh

Balarka Sen

15:29

also, what do you mean by "1 one"?

Mats Granvik

1+1 - 1 =1
1+

posted to early

r9m

@Chris'ssis hi :D .. you said you were cooking sth for me yesterday :-) ..

Balarka Sen

@Mats erm. 1 + 1 - 1 = 1. Yes. So?

Mats Granvik

Yes that is the invention. @BalarkaSen

Balarka Sen

1 + 1 = 2 and 2 - 1 = 1. So that's a fact and a pretty trivial one. Am I missing something?

Mats Granvik

15:31

Same program letting k be the summation index instead of n

Clear[t, n, k, a, b, x, y];
nn = 16;
t[n_, 1] = 1;
t[n_, k_] :=
t[n, k] =
If[n >= k,
Sum[t[n - i, k - 1], {i, 1, k - 1}] -
Sum[t[n - i, k], {i, 1, k - 1}], 0];
a = Table[Sum[t[n, k], {k, 1, nn}], {n, 1, nn}]

gives number of divisors of n.

Balarka Sen

Sorry, I don't understand mathematica.

Mats Granvik

I will take it from start.

Chris's sis

:16657863Yeah ... :-) (I've been busy with some stuff)

Mats Granvik

(1)-(0)=1
(1+1)-(1)=1
(1+1+1)-(1+1)=1
(1+1+1+1)-(1+1+1)=1
(1+1+1+1+1)-(1+1+1+1)=1
(1+1+1+1+1+1+)-(1+1+1+1+1)=1
...
forever

Then

1=1
1+1=2
1+1+1=3
1+1+1+1=4
1+1+1+1+1=5
1+1+1+1+1+1=6
1+1+1+1+1+1+1=7

...

the natural numbers.

Then let k be summation index instead of n, and you get the divisors of n.

r9m

@Chris'ssis 'kay .. no rush ! :-)

Balarka Sen

15:36

@MatsGranvik what is k? what is n?

@N3buchadnezzar I have a problem for you.

Additive number theory.

Chris's sis

@r9m I think the question you asked yesterday might be tackled by a clever application of the trapezoidal rule.

Mats Granvik

@BalarkaSen Think of the Pascal triangle, the way the next column is equal to the partial sum of the n previous row terms in the previous column.

r9m

@Chris'ssis Nice .. okay! :D ... (I have an nice solution too :D .. ) :Note: problem created by me ! :-)

Chris's sis

@r9m Can you share it? It's even great if it's created by you. :-)

Mats Granvik

Then subtract with the n previous row terms in the same column. The result is lower triangular matrix with all ones, and row sums the natural numbers.

Then instead of taking n previous row terms, take k previous terms. The result is the matrix equal to 1 if k divides n and 0 other wise.

Balarka Sen

15:41

@MatsGranvik I can't visualize what you're saying

It's too vaguely defined.

Mats Granvik

This is what you get if you add instead of subtract: https://oeis.org/A175105
Modulo 2 it is equal to 1 if k divides n, 0 other wise.

Balarka Sen

so what's the recurrence if you subtract or whatever you want to do? can you write that up in latex?

i think i will get it better that way.

r9m

@Chris'ssis sure .. but my solution is a bit indirect :) .. I related it to an integral (that I took from one of China math's questions while I was trying to answer it) after which it becomes easier .. the rest is original :P .. I intend to ask it on the main :D .. to see new approaches :-)

Mats Granvik

http://math.stackexchange.com/questions/268159/m%c3%b6bius-function-from-random-number-sequence

$$\begin{align}
T(n,1) &=1, \\
\text{for } k>1, T(n,k) &= \sum\limits_{i=1}^{k-1} T(n-i,k-1) -\sum\limits_{i=1}^{k-1} T(n-i,k)
\end{align}$$

@BalarkaSen

Chris's sis

@r9m Well, this is a kind of research, and it's a very good thing. While studying things, one may discover really great things. Who would have ever imagined that Au-Yeung series may compute in one line? I did that due to personal research.:-)

Mats Granvik

15:46

number of divisors of n

add a "1" in the recurrence and you get sum of divisors of n.

r9m

@Chris'ssis OH !! :D .. Au Yeung series .. what is that ? :-) (google search shows it's the name of a Hong Kong film actor :P)

Chris's sis

9

A: A closed form for the sum $\sum_{n=1}^{\infty}\left(\frac{H_n}{n}\right)^2$

I believe the answer you're looking for is in this Wikipedia article : The following identity was first conjectured by Enrico Au-Yeung , a student of Jonathan Borwein, using computer search and the PSLQ algorithm, in 1993 : $$\sum_{k=1}^\infty \frac{1}{k^2}\left(1+\frac{1}{2}+\frac{1}{3}+\c...

r9m

@Chris'ssis oh !! sorry :( How could I forget about that ?! (slaps @self) :|

Chris's sis

:D

Balarka Sen

@Chris'ssis You are claiming that you can compute it in one-line?

Chris's sis

15:50

@BalarkaSen Sure (For doing that, I use a result that may be proved elementarily).

Balarka Sen

Do you mind sharing? I am interested.

Chris's sis

@BalarkaSen I'll share it, but not now. I wanna put it in my book.

r9m

Nope ! .. publish it first .. :D

Balarka Sen

Yeah. Just publish it.

N3buchadnezzar

who cares teddy bears?

:p

Balarka Sen

15:51

@Chris'ssis Darn the book. Publish a paper.

@N3buchadnezzar I have a problem for you.

r9m

@Chris'ssis the answer sos440 provided there is one of the reasons why I started active participation in this site ... instead of just lousily copying my assignments :P

N3buchadnezzar

@r9m :P

Balarka Sen

@r9m I like Cody's method better.

Cody (a.k.a. Galactus) has loads in his bags when it comes to Euler sums.

r9m

all the answers in the link are fantastic and highly illuminating :D

Balarka Sen

@N3buchadnezzar Ahem. Are you ignoring me?

Chris's sis

15:54

@BalarkaSen I like much the idea of having my own book that is going to be a collection of marvellous questions with marvellous solutions.

Balarka Sen

@Chris'ssis You can still have a book after publishing some results to your paper, you know.

You can have a result both in a book and in some paper.

Chris's sis

@BalarkaSen Some papers are about to be published, but I didn't do that alone (some special series are treated there).

r9m

300 - This is Integralia .. :P

Chris's sis

@r9m :D

N3buchadnezzar

@BalarkaSen Maybe, baby

Alizter

16:06

@r9m Of course. I visit that place regularly.

Mats Granvik

Does this sequence have a closed form? http://oeis.org/A177510
starting: 1, 1, 2, 3, 5, 8, 14, 25, 46, 87, 167, 324, 634, 1248,...

skullpatrol

@N3buchadnezzar

r9m

@Alizter :O !!

Alizter

@r9m Look outside.

I am waving

Can you see me?

Balarka Sen

@N3buchadnezzar $T$ be the set $\{\lfloor n\log(n) \rfloor : n \in \Bbb Z\}$. $T + T$ be the collection of elts obtained by adding two (not necessarily distinct) elts of $T$. Is every integer greater than $25$ is in $T + T$

r9m

16:08

@Alizter :P lol .. the place where I am ,.. hehe .. you can't possibly wave from outside and expect me to see you :P

Alizter

@r9m Oh. I think the Google Earth starship is invisible to mortals.

@r9m Your neighbour hood is very flat.

r9m

@Alizter ya .. why ? :P

Balarka Sen

@Alizter locally flat domains?

Alizter

@r9m See!

r9m

@Alizter What the hell is that ? :P

Alizter

16:11

@r9m Your coordinates alpha leader.

Balarka Sen

haha

r9m

@Alizter mother of god :P LOL .. you mean my nbd is a god forsaken barren land ? :P

Balarka Sen

missiles ahead! dive dive dive!

Alizter

@r9m Why do you have a massive pin on your house?

r9m

@Alizter so that I can poke you when you come nearby :P

Alizter

16:15

r9m

@Chris'ssis Version 2: Evaluating : $$\lim\limits_{n \to \infty} \left(\left(\sum\limits_{k=1}^{n}\dfrac{1}{k}\tan\dfrac{k\pi}{2n+1}\right) - \dfrac{2\log n}{\pi}\right)$$ :-)

Chris's sis

@r9m These ones look nice. :-)

r9m

16:31

@Chris'ssis :D This is much healthier :P

Jasper Loy

17:30

This chat is dead.

Argentina might have won Germany if @PedroTamaroff played.

Mats Granvik

@JasperLoy What is the q analog of the matrix inverse of the Pascal triangle?

Jasper Loy

@MatsGranvik I do not know. I am only a banana.

Mats Granvik

I don't know either, but this should be a good candidate:
$$\begin{array}{lllllllllll}
1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\
-1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\
0 & -1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\
0 & 1 & -1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\
0 & -2 & 2 & -1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\

row sums are the Möbius function and the second column is found in the oeis by Paul Hanna.

Jasper Loy

@MatsGranvik So what is your favourite area of math?

Mats Granvik

@JasperLoy It is pretty narrow.

@JasperLoy When you expand a Taylor series does it always result in a formula for the coefficients in the series?

Shisui

18:44

I've got a quick question.

Polynomial functions belong to the larger class of functions called rational functions.

What class of functions do the rational functions belong to?

(It was mentioned in Spivak that rational functions belong to a larger class of functions that are thoroughly studied in calculus.)

blue

well, many

"those with Laurent series expansions"

or much bigger still, "meromorphic functions"

Shisui

Any bigger than that?

blue

bigger than meromorphic? that's pretty big. you go any bigger and you fail to be differentiable

Shisui

Why is that?

blue

do you know what meromorphic means?

Shisui

18:49

Nope.

blue

then you should start there...

it means "everywhere complex differentiable, except for countably many poles with no accumulation points." basically, the subject of all of complex analysis.

Jasper Loy

anon knows everything

Shisui

That sounds very interesting.

blue

I suppose you could go to a larger class of real differentiable, but then you don't have all the derivatives you want

Jasper Loy

@blue Are you feeling blue?

blue

18:55

are you?

Jasper Loy

I always am, until the day of complete liberation.

I am still trying to look for the perfect calculus book. Rudin's PMA does not have the techniques and applications of calculus...

@blue Is there any calculus book you recommend?

blue

for what porpoises?

collection?

Jasper Loy

For aceing the GRE.

Eg, Rudin does not even talk about Lagrange multipliers.

He also does not talk about maxima and minima of functions of several variables.

PMA is very, very incomplete...

Shisui

What's the GRE?

blue

well, the vector calc book I did was by susan colley, and I liked that. I believe it had lagrange multipliers.

Jasper Loy

19:01

@Shisui It's a test you take to apply to grad school in US

Shisui

Cool!

Jasper Loy

@blue Currently my best choice is Apostol's Calculus 1 and 2.

Shisui

If s(x) = 0 when x is irrational and 1 when x is rational, what is s(x+y) when y is rational?

Jasper Loy

@blue Can you access ffilms.org?

@Shisui Depends on what x is.

blue

the website loads for me, if that's what you mean

Shisui

19:08

So if x is rational, then x+y is rational so s(x+y) = 1.

Jasper Loy

@blue It doesn't load for me. I should try using an American VPN...

blue

@JasperLoy s(x+y)=s(x) when y's rational

Shisui

If x is irrational, is it necessarily true that x+y is irrational (for rational y)?

blue

@Shisui if y and x+y are rational, then (x+y)-y is rational

hence y rational and x irrational implies x+y is irrational

Jasper Loy

@Shisui Can you access ffilms.org?

Shisui

19:09

Oh, ok. You've worked backwards.

That means that s(x+y) = s(x) for rational y.

Hold on.

Oh, yea I understand it.

I could've done it by contradiction too right?

Suppose x is irrational and y is rational and their sum c is rational.

y = f/g and c=h/i

x + f/g = h/i which means that x = h/i - f/g = (hg-fi)/gi which is rational, but we've said that x is irrational so that means that c is irrational, right?

@blue

blue

there is really no point in writing down explicit fractions

Shisui

Oh yea, x = rational - rational = rational but we've said that x is irrational so the contradiction leads us to x+y being irrational.

Jasper Loy

I have an enemy on this site. I upvote all the answers to a question except the enemy's.

Chris's sis

@robjohn How are you doing? You're silent these days ... :-)

Jasper Loy

19:24

@Chris'ssis Let us know when your book is published!

Chris's sis

@JasperLoy OK ;)

robjohn

@Chris'ssis Busy dealing with flags and questions. I was at UCLA on Wednesday. How are you?

Jasper Loy

@robjohn I was there too in my dreams, lol.

Chris's sis

@robjohn Pretty creative these days. Now I'm playing with some series and writing a couple of proofs.

robjohn

@JasperLoy At UCLA or answering questions?

Jasper Loy

19:30

@robjohn At UCLA, lol. Have you spoke to Ethan recently? We shared some secrets.

robjohn

@Chris'ssis I haven't seen your recent series.

@JasperLoy No, how is he?

Jasper Loy

@robjohn Not good, but I won't share the details here.

robjohn

@JasperLoy Is he still applying to colleges, or did he get accepted?

Jasper Loy

@robjohn Well, he is re-applying next year.

robjohn

@JasperLoy Okay. I haven't spoken with him recently.

Jasper Loy

19:34

@robjohn I was thinking of applying to all the 10 UC's and nothing else.

robjohn

@JasperLoy with common applications, that might not be too hard.

Jasper Loy

@robjohn I have not looked at all the UC's, but so far I like UCLA, UC Berkeley and UCSC. Also, somehow I think that I will like California.

Chris's sis

@robjohn This one is not a series created by me, but it's a nice series I try to compute right now. Prove that
$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2} \left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots +\frac{1}{n+n}\right)=\pi C-\frac{33}{16}\zeta(3)$$

Pedro Tamaroff

19:58

@blue

blue

yes

Transcript for

$Mathematics$ Mathematics