00:00 - 16:0016:00 - 21:00

4:00 PM
Hmm $$\frac{\log 0.5}{\log 10} = - \frac{1}{1 + (\log 5) / (\log 2)} \approx \frac{1}{1 + (\log 4) / ( \log 2 )} = \frac{1}{3}$$

30-digits.
You get only 1, lol

@N3buchadnezzar: You come in third! Congratulations!

Don't count me.
I haven't even done the calculations.
Nebu is second.

@BalarkaSen: He gave me a third, so I give him a third. Doesn't that sound fair?

4:03 PM

Daniel used memorized table values, Balarka gave a numerical method requiring computers.

My methods don't need calculators.
I can get them easily by hand.

No no no, not allowed

Take, for example, Euler's method.

grabs pencil, and makes toothpaste out of it

4:04 PM

@N3buchadnezzar: mmh, and you used your brain. Okay, you receive the antecedent of the ratio you gave me.

Nebu deserves 1st prize.
I did no calculations and Daniel used sheer memory.
Wait, why not use Newton-Raphson out of $10^x = 2$?
Yeah, that'd be a lot faster

@BalarkaSen Indeed
Today I tried to implement a root finding algorithm for polynomials, harder than expected.

@BalarkaSen: That's what I said. I gave him 1st place. (The antecedent of the ratio he gave me)

4:07 PM
@N3buchadnezzar $$\sum_{n=1}^\infty \frac{1}{n^{s}} <> 1 + \sum_{a=2}^\infty \frac{1}{a^{s}} + \sum_{a=2}^\infty \sum_{b=2}^\infty \frac{1}{(a \cdot b)^{s}} + \sum_{a=2}^\infty \sum_{b=2}^\infty \sum_{c=2}^\infty \frac{1}{(a \cdot b \cdot c)^{s}} + \sum_{a=2}^\infty \sum_{b=2}^\infty \sum_{c=2}^\infty \sum_{d=2}^\infty \frac{1}{(a \cdot b \cdot c \cdot d)^{s}} + \cdots$$

I have memorized a few values of logarithms, and roots. But mostly I try to simply memorize the first taylor expansions. Good enough for everyday occurances, like getting mugged

I can do taylor in my head.
It's nothing hard.

Give me the first 3 digits of $\log 3$, or I will take your money and kick this puppy
@BalarkaSen Cumbersome, not hard.

@MatsGranvik Wouldn't that be "="?

@BalarkaSen
:14185814
@BalarkaSen I mean I have been thinking if this could be used to evaluate the zeta function.

4:09 PM
@MatsGranvik It cannot be.
That mu-formula is slowly converging.
And mu is completely randomistic.
Fast accelerations exist for zeta (and computing nudge-nudge-wink-wink zeta zeros)
@N3buchadnezzar Prove that $$\sum_{k=1}^n \frac{(-1)^{k+1}}{k} \binom{n}{k} = \sum_{k=1}^n \frac1k$$

@BalarkaSen: Than can you atleast show me the Newton-Raphson way of doing it?

@Nick Sure.

@BalarkaSen Doesn't that work without the sums?

But I need to fetch pen and paper.

never mind.

4:14 PM
@MatsGranvik Does it?

@BalarkaSen Why would you need pen and paper?

I was wrong.

Sure, I have time (Although my net connect might end any moment, it's very unreliable) but I will see it sometime and be amazed by it nonetheless

@N3buchadnezzar Well, I need to approximate the $10^k$ for $k \in \Bbb R$ list.
And by hand! (EEK!)

@BalarkaSen "and computing nudge-nudge-wink-wink zeta zeros" But no formulas for zeta zeros.

4:16 PM
There are approximations.

10^x = -1/2 so
$g(z) = x - \frac{10^x-1/2}{10^x \cdot \log 10}$

Why not simply $10^x = 2$
I leave the work to Nebu.

The negative sign :p

@BalarkaSen Is there a better approximation than LeClaire's
?

@N3buchadnezzar Ah. But you can manage that. $10^{-x} = 2$
@MatsGranvik I am not sure.

4:21 PM
@skullpatrol Are you here?

Andre LeClaire's zeta zero approximation with a European Excel spread sheet formula:
Your eyes will probably hurt now.

@BalarkaSen Which is the same as $10^x = 1/2$ as i did, geesh :p

Too much approximation questions here.

times 2*PI()*EXP(1) of the reciprocal

@N3buchadnezzar It is $2\pi$. The residue at the upper singularity is $\frac1{2i}$ and the residue at the lower singularity is $-\frac1{2i}$. Since the contour circles the upper singularity counterclockwise and the lower singularity clockwise, each contributes $\pi$.

4:26 PM
@robjohn
13 mins ago, by Balarka Sen
@N3buchadnezzar Prove that $$\sum_{k=1}^n \frac{(-1)^{k+1}}{k} \binom{n}{k} = \sum_{k=1}^n \frac1k$$

@robjohn thanks for looking at my question :-)

@robjohn Thanks

A star and an unstar
:14186025 You are wrong. "There is no black hole" - Stephen Hawkings.

@skullpatrol The opposite of star is rats, lol.

4:35 PM
Hullo, @Sawarnik

Hello
Do we know wheter $sin(e)$ transcendental or irrational?

@Sawarnik Not yet proved to be transcendental, AFAIK

I have decided that Balarka is a genius.

@JasperLoy Is it a thing to be decided?

@Sawarnik You know not what I mean.

4:36 PM
@BalarkaSen But it is irrational?

@Sawarnik Of course.

@Sawarnik Yes.

@PedroTamaroff you missed @PedroForquesato :D

@skullpatrol Who's that?

@PedroTamaroff dunno, he said nothing

4:39 PM
Of course it is, why I behave like a idiot sometimes?

$a_0=0$ and
\begin{align} a_n &=\sum_{k=1}^n\frac{(-1)^{k-1}}{k}\binom{n}{k}\\ &=\sum_{k=1}^n\frac{(-1)^{k-1}}{k}\left[\binom{n-1}{k}+\binom{n-1}{k-1}\right]\\ &=a_{n-1}+\sum_{k=1}^n\frac{(-1)^{k-1}}{k}\binom{n-1}{k-1}\\ &=a_{n-1}+\frac1n\sum_{k=1}^n(-1)^{k-1}\binom{n}{k}\\ &=a_{n-1}+\frac1n \end{align}

@Sawarnik Maybe you are one, lol.
2

Exactly

@JasperLoy Yes, I know. It deserves a star.

Everything comes from $F(n) = F(0) + \sum_{k=0}^{n} \Delta F(k)$
Lemme see $\sin(e) + i\cos(e) = e^{ie} = \left ( e^e \right)^i$ So if $e^e$ is algebraic, then it's transcendental by Lindeman-Weierstrass, but then $\sin(e)$ and $\cos(e)$ are algebraically dependent so $\sin(e)$ have to be transcendental. If $e^e$ is transcendental, then the LHS can be either irrational or transcendental. If transcendental, done. If irrational then clearly $\sin(e)$ is irrational.

4:41 PM
6 mins ago, by Jasper Loy
I have decided that Balarka is a genius.
Me too.

So in either case $\sin(e)$ is irrational.
But that can be done without Lindeman-Weierstrass, I presume.
Can it be, @Pedro?
Perhaps you though of something like that, huh?
Diophantine approximation would play probably.

@BalarkaSen $\sin(e)+i\cos(e)=ie^{-ie}$

I am getting increasingly disgusted by religions X, Y and Z.

Oh, that should be $\cos(e) + i\sin(e)$
But that does not change the result in any case.
=)

no. but it's good not to leave these things to be picked at later

4:44 PM
Yes, of course.
Typos are the real enemies of a mathematician

followed closely by Physicists :-)

(removed)

@skullpatrol Is it to be decided?

I think I should ignore Sawarnik.

(r(e(m(o)v)e)d)

4:49 PM
@JasperLoy Why?

(removed)

23 secs ago, by skullpatrol
(r(e(m(o)v)e)d)

ignores Sawarnik

@JasperLoy Ok, then. Congratulations.

I'm going to ignore myself.

4:52 PM
(removed)

Attached
Yup, nets back. Whoopie. I'm pretty sure it's gonna die again.

@nick I am getting a copy of all the suttas.

@JasperLoy what's a sutta?

@Jasper: Great but don't turn out killing white snakes!
If you know what I mean

@robjohn Otherwise known as sutra. A discourse of the Buddha.
@Nick I am thinking of becoming a vegetarian soon.

4:55 PM
@robjohn He s a Buddhist.

Hi @MattN.

@JasperLoy is sutta an alternate spelling or a typo?

@nick Let me show you the series I am getting, hold on.
@robjohn Sutta is in Pali while sutra is in Sanskrit.

@JasperLoy ah, I see. Thanks.

Come on Japser, waiting for the series. TV not power, I assume.

4:57 PM
@JasperLoy You know sanskrit?

@nick Here, these 6 books. wisdompubs.org/collections/…

Duh. I thought a TV series.

@BalarkaSen No, I don't.
@MattN. I like The Wonder Years.

@JasperLoy Utsabe baesane chaiba, dhurbhikshe rashtrabiplabe

@JasperLoy Looks good. Never heard of it before though.

4:59 PM
@BalarkaSen What does that mean?

@BalarkaSen I only know English and a bit of Chinese.

We recently ran out of Person of Interest and then sort of by chance found Castle.

Rajwadware swasane cha, ja thishtathe swa bandhaba

@JasperLoy: Yay! That sounds awesome. I haven't read anything remotely buddhist in a while. (other than math, ofcourse)

@BalarkaSen Look likes you have learnt a shloka.

5:00 PM
Also, if you didn't get the white snake reference

@Sawarnik Yes.

@Nick I watched some white snake drama series before.

@BalarkaSen Not expected of you. But seriously, how could you understand that language!

I normally get bored half way through season 2. Like e.g. Breaking Bad got me so bored that I stopped watching and same with Lord of the throne craft..err I mean Game of thrones.

@Nick if you consider $5:4$ even odds, yes.

5:02 PM
Person of Interest is unusually entertaining even though the acting is not always great.

@JasperLoy: Yes but you probably haven't seen Jet Li as a jerky monk.

@MattN. Hello there.

@PedroTamaroff Hi!
This isn't even funny. What is this?

@MattN. I have no idea.

@Tamaroff: What now? What now is we cut open your belly and release the only uncorrupted human being in existence.

5:05 PM
why
whyy

I no longer follow this conversation.
Maybe I never did...

@MattN. Maybe no one ever did

Maybe.

Maybe it's maybeline

Fetus in fetu

5:08 PM
Doctor fetus

Someone get a coffin for this conversation.

Has any of you here watched Summer Storm? Nice German movie about gays.

Nope. Never even heard of it.

@MattN. I saw Jack Johnson live yesterday night.
Be be cool.

@PedroTamaroff Oh kool. How do you know I know him?
: D
But I do.

5:09 PM
@MattN. You're a young female.

Not personally.
@PedroTamaroff So are you.

@MattN. Yes.
Mike is a cat.
@JasperLoy is a banana.

So if you are on the same bus maybe we could... ?

@MattN. ...?

ooops
Have to go watch the next episode of castle. Girlfriend has come back from snooze. bbl

5:14 PM
@MattN. #envy

@PedroTamaroff I have no gf too.

@JasperLoy Can you reverse your decision?

1

How can I prove that $40/29<\sqrt{2}<42/29$ given $20<29/\sqrt{2}<21$? I did lot of approaches I obtained in one that: $$\dfrac{29}{42}<\sqrt{2}<\dfrac{29}{10}$$ but couldn't do something other Thank you.

Wow, nobody wants to upvote me there, SAD PANDA.

@JasperLoy No bragging for votes here, mean panda.

@pedro What do you think of my answer above?
I think it's time for me to delete my account, lol.

5:29 PM
@JasperLoy Not good infront of other answers.

Well, since nobody appreciates it, I shall delete it...

@JasperLoy It is not the case that nobody appreciates, its the case that I dont appreciate it because you unfriended me.

if $\Sigma=UDU^T$ is the population covariance matrix, how can we generate normal samples from this covariance structure? Let $z_{i,j}$ be in $N(0,1)$.

6:13 PM
The Fourier-Dirichlet series at the critical line of this matrix is asymmetric:
http://oeis.org/A191898
But the matrix is symmetric. I want it to stay symmetric even if I calculate the Zeta zero spectrum.

6:37 PM
its on cotour integration
contour*
no1?

7:24 PM
@PedroTamaroff Don't envy, we've just watched the last episode.
Of season 1.
: )
But for today no more castle. : (
If it was up to me I'd watch the 6 seasons all in one go and then go into a sleep-deprivation coma.
@DanielFischer Are you around?

@MattN. Sort of.

@DanielFischer Why only sort of?
Watching a film while answering questions? : )

Nah, digesting. Had a good dinner.

Me too. Lasagne. What did you have?

Hmm, whaddayacallit? Bohneneintopf, if you can translate that. Doesn't sound like much, but tastes really good.

7:34 PM
It sounds tasty.
"stew" maybe?

green bean stew?

@Mike!

I thought of writing that, @Matt, but I don't really know what a stew is in Britain/Ireland, so I was cautious, maybe it's not like that.

(I don't want to say 'bean stew' since that usually means white beans)
Hi @TedShifrin

@Ted!!!

7:36 PM
Hi @Daniel!

@DanielFischer You are right of course, it's not the same.
Wow. Like a scene from a film where Ted and Daniel are running towards each other in slow motion.
Ted!!! ... Daniel!!!

that's a bit much

@MattN. That's not slow motion, we are slow.

Never mind me. I am reading too much into it.
@DanielFischer : D

Not quite Almodovar here!

7:38 PM
@Mike I agree. Too much drink : D

Where's my drink?

@TedShifrin No, more like "Almodo-bar" : )

Too much or too little.

@TedShifrin I don't know. You tell me.

Grumph.
Good question for you guys: If $f(x,y)=f(y,x)$ is $C^1$, what's a sufficient condition for the critical points to be on the diagonal?

7:42 PM

I knew that psychically.

I feel like I'd need pen and paper for that one (Jacobian stuff?)

Chain rule tells you the critical point set is invariant under the involution.
But most "real world" problems give rise to symmetric solutions.

The question being "chain rule with what", the map $(x,y) \mapsto (y,x)$?

Yes.

7:45 PM
Which is why I was thinking Jacobians... change of variables. But in retrospect that's nonsense,

Nonsense?

The only places I know for using the Jacobian (determinant) are change-of-variables for integrals and stuff about smooth inverse (inverse function theorem, Sard's theorem, &c)

Oh, determinant is nonsense here, yes.

I'm always glad to know I'm talking nonsense.

Derivative, not gradient. And be careful where you evaluate and in what order you compose.

7:52 PM
Yes, those reasons are why those messages are deleted.

Does it hold that if $f(x,y) = f(y,x)$ then $d/dx f(x,y) = d/dy f(x,y)$? I think yes.

Not sure what you mean by derivative in this context.

Now do I need to delete in kind?

These are functions $\Bbb R^2 \rightarrow \Bbb R$, so the natural notion of a derivative is the gradient, no?

No @Matt

7:53 PM
@Mike Critical points is where the partial derivative vanish.

@TedShifrin Too bad. Can you give a counterexample?

Greetings

the involution maps $\Bbb R^2\to\Bbb R^2$!

What are "the involution maps"?

7:53 PM
@TedShifrin I said 'derivative of the RHS', meaning gradient of the RHS, meaning...
@MattN. $(x-y)^2$

No, "maps" was a verb there

I don't know what an involution is.
@Mike Thank you! That's good.

A map that equals its inverse.

Ah.

Here, the linear map switching $x$ and $y$.

7:56 PM
I sort of thought of $f(x,y)$ as a map $\mathbb R^2 \to \mathbb R$.

@TedShifrin So we should have $\nabla f(x,y) = J_I\nabla f(y,x)$.

Yes, that is.
Yes @Mike.

Then $f(x,y) = f(y,x)$ makes it mirrored at the perpednicular plane through $x=y$

@TedShifrin From here I need to use paper, so I'm out to study for my art history final.

See ya @Mike.

7:59 PM
@TedShifrin I always feel sad when I see an old MO answer of Thurston's/

Good luck with the art history.

Yeah, but the greats die, too ...

He was relatively young.
Not young, but relatively so

@robjohn I just attended a question where I used the fact that $$\lim_{N\to\infty} (2N+1)\prod_{n=1}^{N} \left(\frac{(2n-1)}{2n}\right)^2=\frac{2}{\pi}$$

Older than me ...

8:00 PM
So you just need to make sure you don't die by 65 :)

@robjohn this can be used to that previous question I showed you last days. So, I have 3 solutions there so far. :-)

80 is a good number.
You made it past 27, so you might as well go all the way.

@Chris'ssis that the sum diverges?

After cancer and heart disease, I dunno about 80.

@robjohn sure. I showed you my proof, didn't I?

8:01 PM
@Mike Thurston posted on MO? I didn't know that. He was at Princeton when I was a grad student.

I have a lecture to prepare.

@Chris'ssis yes

@TedShifrin What subject.
I am going to propose to my girlfriend.

@robjohn Yes, fairly regularly.

To watch another episode of Castle : D

8:02 PM
MAA lecture for undergrads ... Something on optimization problems and geometry.

@Mike I think that Igor Rivin was a Thurston student.

Ew. Sounds applied.

@MattN. I've vowed to watch the entirety of LOST by 2015.

Haha @Matt

LOL@MattN.
You don't know Ted well enough :)

8:03 PM
I don't know him at all : )

@robjohn that limit is in fact that Wallis product (just written in another form).

@Mike I don't know it but I've heard of it.
Let's have a look on imdb...

It's on Netflix.

Does anyone know Ted well enough? @Mike

@TedShifrin Ted does.

8:04 PM
Oh, sounds ok. But only 6 seasons! Why do you need so much time? : D

"The second proof of Koebe uses holomorphic functions, i.e. the Cauchy-Riemann equations, and some topology."
I might want to look at that.
@MattN. I'd rather not bingewatch it... I have better things to do, like waste my time in here.

I was just going to say that. : P

And qualifiers in grad school :P

Life is one big waste of time.

@Chris'ssis I think that is how I showed that the sum diverged. I forget what I used when you asked.

8:06 PM
So is optimisation stuff not applied maths?

LOL

Not I guess.

@TedShifrin this second theorem must have been the one my professor was talking about to me... with "dipole green's functions"

Not necessarily. I'm thinking of neat max/min problems with geometric solutions, including symmetry principles (hence my earlier question).
It's been centuries since I've thought about uniformization proofs, @Mike.

@TedShifrin So... that's you?

8:10 PM
Um, no, @Matt.

How do I prove that the limit as x tends to c of (x^c-c^x)/(x^x-c^c) exists?

What have you tried? @user112495

Got to go afk. Later!

Bye @Matt

@Ted Shifrin I've written down the epsilon delta definiton of a limit, but from there I can't make any progress.
@TedShifrin I've written down the epsilon delta definiton of a limit, but from there I can't make any progress.

8:16 PM
Forget $\delta$-$\epsilon$. Find the limit!

@TedShifrin I can evaluate the limit itself, I just can't prove it exists.

How did you evaluate?

@TedShifrin I used l'Hopital rule

@TedShifrin and I got (1-log(c))/(1+log(c))

8:19 PM
I've never seen this problem before. Cool. What makes you want a formal proof?

@TedShifrin I have been given a problem by my tutor that says to show the limit exists and to find its value.
@TedShifrin It doesn't say to use epsilon delta, i'm just not sure how else you can prove limits exist.

By applying L'Hôpital correctly (verifying hypotheses carefully)!

@TedShifrin So I need to show the limit of f'(x)/g'(X) exists (along with the other conditions)?

Yes, which you've done.

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