Mathematics - 2013-08-29

Alizter

12:00 AM

@PeterTamaroff Dihd Ih Ihntehruhpt youh?

skullpatrol

Try a separate room :)

Ted Shifrin

Oh, so not at the level of my book, then. ... Remember that you're doing something like $k$-dim volume (or flux) of a $k$-dim manifold in $\BbbR^n$.

Alizter

I don't need it solved. I have solved it. I just like to see how other people do it

Peter Tamaroff

@TedShifrin Oh?

Ted Shifrin

If you read my stuff, you'll see that $dx\wedge dy$ in, e.g. $\mathbb R^3$, computes the signed area of the projection on the $xy$-plane when you plug in two vectors.

Peter Tamaroff

12:03 AM

@TedShifrin Right, yes, I have read that.

@TedShifrin How can we write $$\int_a^b f\circ\gamma\;\;\lVert \gamma'\lVert$$ as the integral of a form over a chain?

Ted Shifrin

So $f$ and pullback computes it for parametrized $k$-dim thingies.

That actually is NOT a pullback of a $1$-form. $ds$ is orientation-independent, so is not a form. It's a measure or a density. In complex geometry, such things (area, etc.) are given by pulling back glibal forms and integrating!

Global ... I hate typing on my iPad.

Peter Tamaroff

@TedShifrin Yes, I had the hunch.

So what is the integral of a form is $$\int_a^b f\circ \gamma\;\cdot\;\gamma'$$

Ted Shifrin

You're doing the integral of $f$ wrt arclength on the curve param by $\gamma$.

Peter Tamaroff

@TedShifrin That is the first, where "$ds=\lVert \gamma'(t)\lVert dt$".

I mean the other one where we take the dot product.

Ted Shifrin

For oriented hypersurfaces, you can get the "area" as a form on the ambient $\BbbR^n$. See my discussion of area form of a surface and line integrals (work — drat physics) in my book.

I'll check back later ...

Peter Tamaroff

12:14 AM

@TedShifrin Section?

skullpatrol

...thanks for dropping by :)

Ted Shifrin

12:41 AM

@Peter: 8.3 and 8.4 ...
Thanks, @skull.

Peter Tamaroff

@TedShifrin OK.

@TedShifrin Now I have to sing. Heh.

Ted Shifrin

What are you singing?

Peter Tamaroff

@TedShifrin I'm practicing. I started taking singing lessons a few weeks ago.

skullpatrol

Opera?

Peter Tamaroff

Nope, just scales.

Ted Shifrin

12:44 AM

Never a dull boy :) classical or popular for your goal?

Peter Tamaroff

@TedShifrin I like Andrea Bocelli, which is kinda Pop Opera =P

Ted Shifrin

Shades of The Who :)

Peter Tamaroff

@TedShifrin Ah?

Ted Shifrin

Nah, he's just popularizing opera, right? The Who were famous for their 60s rock opera Tommy.

Peter Tamaroff

@TedShifrin Oh! I like the Who.

Ted Shifrin

12:47 AM

My youth :) Generally, I'm a classical music fanatic.

Peter Tamaroff

@TedShifrin It must have been awesome to grow up with those bands.

Now we have crap.

Ted Shifrin

LOL ... I was a folk music fan then ... Judy Collins, Joan Baez, Tom Rush, etc.

Kevin Driscoll

Not everything we have now is crap!

Ted Shifrin

Joni Mitchell ...

This argument won't be for me :)

Kevin Driscoll

I'll admit it: I think Kanye West's last 2 albums showcase serious artistry

and thats from a genre that most consider just derivative

Ted Shifrin

12:51 AM

No pun intended in here ...

I know ... It's an integral part of the genre :)

skullpatrol

musical calculus

Ted Shifrin

There's actually lots of math in music ... Even a bit of group theory.

Kevin Driscoll

@TedShifrin I just got the joke

skullpatrol

the music of the spheres

Peter Tamaroff

@TedShifrin I gave a book about music and math to my brother.

It is quite a nice book.

skullpatrol

1:06 AM

:-)

Peter Tamaroff

@skullpatrol Yeah, I was too lazy to correct it.

skullpatrol

@PeterTamaroff Never be lazy on mount Olympus :)

Peter Tamaroff

1:25 AM

@skullpatrol Heh.

Peter Tamaroff

1:39 AM

Does anyone know italian here?

Alizter

1:58 AM

so tempted to write babadehboobedi

Peter Tamaroff

@KarlKronenfeld Yao, yao.

Kevin Driscoll

@PeterTamaroff You speak Spanish so you basically understand Italian. A friend of mine got a C in his college Italian class speaking almost exclusively Spanish

Peter Tamaroff

@KevinDriscoll Yeah, but I want to know what the thing says precisely.

Kevin Driscoll

@PeterTamaroff Do Italians ever do things precisely?? :-P

Peter Tamaroff

@KevinDriscoll Dunno.

The phrase is:

"Questa notte ho fatto un sogno, so sogna de la mia bella, che la dormiva sola."

So something like "This night I had a dream, a dream of my beauty, that slept alone"?

Kevin Driscoll

2:08 AM

Seems about right, but I don't really knwo

Questa -> esta notte-> noche sogno -> sueno

Its funny how similar it is

Peter Tamaroff

@KevinDriscoll Questa is "Esta", "This".

"notte" is "noche".

Kevin Driscoll

isn't that what I said? its what I meant

Peter Tamaroff

"sogno" is sueño, yes. In fact the ñ (I think) was originally written as "gn"

Kevin Driscoll

I see, I put in some spaces that didn't actually come through in the message

Peter Tamaroff

@KevinDriscoll OH, derp.

I read it differently.

LOL.

I read your arrows like "esta notte-> noche sogno"

"noche sogno -> sueno"

and so on...

Ted Shifrin

2:11 AM

Rehi :)

Kevin Driscoll

Ya, silly chat client deleting my extraneous spaces

@TedShifrin 好久不見

Ted Shifrin

Well, I know some of 5 languages, but that's not one of 'em

Kevin Driscoll

Oh haha, sorry my keyboard was set to Mandarin

I meant to say "hello"\

Ted Shifrin

Uh huh :)

Peter Tamaroff

@MarianoSuárez-Alvarez You around?

Kevin Driscoll

4:39 AM

What was that site where you enter a number and it tells you what exact combination of objects it is close to?

skullpatrol

6:24 AM

Talk about asking a question in a vacuum^ :)

hi @anon how are you?

anon

meh

skullpatrol

:-|

anon

interesting, the skull is talking to me

skullpatrol

we have talked before pal

anon

indeed

Chris's sis

7:38 AM

Greetings,

Here is an old unanswered question. I find it totally impressive! math.stackexchange.com/questions/366266/finding-the-limits

(a huge amount of beauty)

Chris's sis

9:36 AM

The beauty of some things is hard to measure.

@robjohn maybe it's hard to imagine, but this morning I computed that limit mentally (another solution).

robjohn

9:59 AM

@Chris'ssis the log sum?

Chris's sis

@robjohn yes.

@robjohn did you manage to take a look at my (first) proof?

robjohn

@Chris'ssis I have gotten part way through

Chris's sis

@robjohn ok

skullpatrol

I like this approximation for e: $\frac {685}{252}$

eXtremiity

Hi can, anyone give me a hand with complex analysis

?

Karl Kronenfeld

10:18 AM

@eXtremiity Probably not, but if you asked the question, I would know for certain.

eXtremiity

I just have trouble seeing the whole mapping part of the topic.

For example I have |z-1|=1and i want to see its image under the mapping w=(1+i)z-2

It's asking me to, what, transform the points of z defined by the first equation according to the new equation w.

So the first equation tells me that z are all the points lying on a circle with centre (1,0) and radius 1.

I broke down w=(1+i)z-2 into two equations:
a=(1+i)z
b=a-2.
and ive decided to find 'a' first followed by 'b'.

Karl Kronenfeld

ok

eXtremiity

I can't see the transformation of 'a'.
I see that every complex number produced by 'a' will be all points z defined on the circle i described above rotated anti-clockwise pi/4. but how on earth do I seee this

"how do i draw it"

Karl Kronenfeld

It also scales

eXtremiity

What do you mean?

Karl Kronenfeld

10:24 AM

Write $1+i=\sqrt 2e^{i\pi/4}$

eXtremiity

Oh yes, the mod of (1+i) is sqrt 2.

Ok yes, scaled.

Karl Kronenfeld

You may as well assume $a=\sqrt 2z$ when given a circle.

This is a variation of what you already said; I am just confirming it.

eXtremiity

Ummm..

Well given that each z is defined from |z-1|=1, I want to see the transformation a=(1+i)z.

That's what I was tryingt o say.

Karl Kronenfeld

You can first multiply by $e^{i\pi/4}$ and then multiply by $\sqrt 2$. So first you rotate the circle, and second you expand the circle by $\sqrt 2$.

eXtremiity

Yes. I understand. So, (hope I'm not rushing you or going ahead of what you're getting to) how does $e^{i\pi/4}*z$ look like?

what'sup

10:29 AM

hi smart guys

Karl Kronenfeld

It moves z around the circle by the angle $\pi/4$.

eXtremiity

Oh god. So you're telling me the circle itself doesn't move ?

Karl Kronenfeld

That first thing, $e^{i\pi/4}$, does not move the circle anywhere. ;)

eXtremiity

And that makes sense, far out.

Why am I so stupid. sigh.

In addition, of course, the circle has radius \sqrt(2) now?

Karl Kronenfeld

After applying the transformation $a$, yes.

eXtremiity

10:32 AM

Excellent, so next is b.

and that's just a vertical shifting of two units, correct?

Karl Kronenfeld

horizontal shifting of two units toward $-\infty$

eXtremiity

****. yes, REAL part.

Ok, the answers have

Karl Kronenfeld

I'd say you just need more practice with the geometry of the complex numbers. (Visualizing multiplication by various constants)

eXtremiity

hold on, ill thank you soon, just trying to confirm the answers.

so, the final answer would me

|z+1|=sqrt(2)

Karl Kronenfeld

Yes.

eXtremiity

10:39 AM

The answers say

|w+1-i|=sqrt 2, where w=(1+i)z-2

sigh. I have no clue what these answers are saying.

However, I must leave now and I want to say thank you for all your help ^_^.

Karl Kronenfeld

They're still using the original $z$, since they want to follow the transformation more "locally".

eXtremiity

Very much appreciated @KarlKronenfeld. Have a good night/day.

Karl Kronenfeld

I believe if you plot the solution you will get the same picture.

@eXtremiity You're welcome, bye.

eXtremiity

Mmmmm, unsure of what that means,but I understood the transformation, and I'm sure |z+1|=sqrt(2) would still be an accepted answer in an exam, i..e (hopefully).

Thanks again ! cya

user87637

11:34 AM

-1

Q: What is Problem of Representation Theory?

Classify all representations of a given group G, up to isomorphism. I think for arbitrary G, this is very hard! I shall concentrate on finite groups, where a very good general theory exists. I think there must be another problems!!!

user87637

WTF is he posting???

user87637

Oh man he just edited it to bump it again...

skullpatrol

Ask him to clarify.

user87637

Someone did already.

skullpatrol

Be more specific.

user87637

11:42 AM

He even posted an answer to his own question, not sure what is going on.

user87637

Did you see his answer?

skullpatrol

Yes.

Looks like a trap...

user87637

Feels a little trollish to me...

skullpatrol

That's what I meant "a troll" trap :-)

A trap set by a troll to trap other trolls.

If that makes sense.

Dominic Michaelis

12:04 PM

hi

skullpatrol

Hi! how are you?

Dominic Michaelis

I am fine

Chris's sis

12:59 PM

@robjohn I'll show you immediately my $2$nd solution that is somewhat related to it.

(I write in this moment the last details)

robjohn

I'm not sure that helps at all, now that I look at it.

Dominic Michaelis

I think this is a nice calculus exam question

skullpatrol

@Jasper Be careful pal...

Chris's sis

@robjohn are you there?

@robjohn you may find the proof in the same place as the previous one. I hope you really like this one! :-)

robjohn

@Chris'ssis yes

@Chris'ssis I will look now

Chris's sis

1:14 PM

@robjohn OK

robjohn

@Chris'ssis got it.

Chris's sis

@robjohn perfect :-)

what'sup

hi guys

@robjohn @Chris'ssis what is the question you were talking about ???!!!

Chris's sis

@robjohn at any rate, I need to improve my wording in English, but the essence of the proof is right there.

Chris's sis

1:29 PM

@what'sup have you seen this amazing question? math.stackexchange.com/questions/366266/finding-the-limits

@what'sup it's one of the most beautiful things one may see in calculus. (imho)

skullpatrol

@Chris'ssis Imho $e$ is the most beautiful number in calculus.

Chris's sis

@skullpatrol sure, it's definitely a beautiful one.

skullpatrol

@Chris'ssis Did you like my rational approximation to it?

Chris's sis

@skullpatrol definitely (+1) :-)

skullpatrol

:D

Chris's sis

1:38 PM

:D

skullpatrol

@Chris'ssis Have you seen it before?

Chris's sis

@skullpatrol to be honest, I don't remember ... (I learned so many things in the last period of time ... and forgot many others)

skullpatrol

I wonder why nobody forgets $\frac{22}{7}$ as an approximation for $\pi$?

Martin Sleziak

Did you know that SE sites have Greatest hits page?

I have only learned about that page now.

skullpatrol

@MartinSleziak nice, thanks for sharing :-)

Dominic Michaelis

1:44 PM

@Chris'ssis you surely like this

Chris's sis

@DominicMichaelis nice +1. One question: is it possible to look at it, do nothing and say precisely what is its limit?

Dominic Michaelis

@Chris'ssis without doing anything I don't know. I looked at it and know what to do, with daniels comment you can do it even much faster

Chris's sis

1, 1, 2, 5, 15, 52, 203, 877, 4140, 21147, 115975 (Bell numbers)

relate all to Dobinski's formula :D

Dominic Michaelis

@Chris'ssis right (time e everytime)

Chris's sis

@DominicMichaelis :D

what'sup

1:50 PM

@Chris'ssis it's a good question thank you

Dominic Michaelis

@Chris'ssis but well for that i need glasses

Chris's sis

@DominicMichaelis do you like this question here? math.stackexchange.com/questions/366266/…

@what'sup welcome anytime!

what'sup

:-)

skullpatrol

@MartinSleziak Did you see the

"View greatest hits you
participated in."

option on the right hand side?

Dominic Michaelis

@Chris'ssis looks like integrals

but right now i don't see of which function.

tomorrow is my last exam

Chris's sis

2:34 PM

@DominicMichaelis wish you have a good luck with your last exam!

skullpatrol

3:16 PM

Hi @MarianoSuárez-Alvarez how are you?

Sujaan Kunalan

Does anybody have any idea why I can see the LaTex code but it doesn't render?

(on the main website)

skullpatrol

@SujaanKunalan Have installed this?

Sujaan Kunalan

I have not, but it used to render. Has there been some sort of update for MathJax?

skullpatrol

not that I know of

Chris's sis

4:19 PM

@robjohn I'll try to come up with a 3rd solution based upon your hint (Riemann sums). The only problem is that I'm a bit sleepy now ... (at any rate, I'll try to finish all by going the way you suggested)

robjohn

@Chris'ssis I think something like that should work. I should flesh it out myself.

Chris's sis

@robjohn the thing that annoyed me when you gave me the hint was that I didn't think at this possibility at all. I still wonder why.

Riemann sums usually have a strong word to say when dealing with such limits.

IvanMatala

5:17 PM

hey guys

how did '1' become '2' ??

Peter Tamaroff

@IvanMatala Aliens.

Or maybe the CIA covering some Alien situation.

@IvanMatala Do you know what the $\delta$ does?

Usually $\int_{\Bbb R}f(t)\delta(t-t_0)dt=f(t_0)$ IIRC:

Jayesh Badwaik

@PeterTamaroff Wassup?

I see you are chewing up Spivak Manifold. Nice.

I am starting on it now.

Peter Tamaroff

@JayeshBadwaik Ah, cool.

@JayeshBadwaik What have you read alrady?

what'sup

hi

skullpatrol

hi

Alizter

5:28 PM

hi

Jayesh Badwaik

@PeterTamaroff Not much since I left the chat. I have done some measures now. Reading the Rudin's Book on Real and Complex Analysis.

IvanMatala

yeah the delta is like obeying sifting property

it has only 1 at t = 0

so when you multiplied it with e^(bla bla).. then you will only have non zero value at t = 0

Jayesh Badwaik

@PeterTamaroff You?

what'sup

how are you doing @Alizter @skullpatrol

skullpatrol

@what'sup Fine thanks, how are you?

what'sup

5:31 PM

fine thanks @skullpatrol

i'm now watching the greatest hits

skullpatrol

@what'sup of?

what'sup

math.stackexchange.com/questions/greatest-hits

IvanMatala

@PeterTamaroff thanks

Peter Tamaroff

@JayeshBadwaik Mostly Apostol and Rudin. I should re-read some bits of Rudin, since they're quite awesome

skullpatrol

4 hours ago, by skullpatrol

@MartinSleziak Did you see the

"View greatest hits you
participated in."

option on the right hand side?

what'sup

5:34 PM

yes

Jayesh Badwaik

@PeterTamaroff Yes. Baby Rudin? Apostol is quite good too. I was reading the book "Cauchy Schwarz Master Class" till now.

Peter Tamaroff

@JayeshBadwaik Aha.

Jayesh Badwaik

Nice book on inequalities.

what'sup

@PeterTamaroff @JayeshBadwaik give me some names of good books

Peter Tamaroff

@what'sup NOT ENOUGH INFORMATION FOR A SIGNIFICANT ANSWER.

what'sup

5:38 PM

@PeterTamaroff i know just a moment .

@PeterTamaroff title : calculus By : Tom apostol

for ex

Peter Tamaroff

@what'sup That is a good book.

what'sup

@PeterTamaroff sure

@PeterTamaroff but i couldn't download it on the (computer)

Peter Tamaroff

@what'sup Try harder.

what'sup

ok do you have any links @PeterTamaroff to download the book

Peter Tamaroff

@what'sup Don't really know now. Make a search.

what'sup

5:42 PM

ok it needs windjvu right ??

Peter Tamaroff

@what'sup Yep.

what'sup

@PeterTamaroff thnx

ok (good topology books) @PeterTamaroff

????

Peter Tamaroff

@what'sup Munkres, Mendelson, Kelley.

what'sup

ok i'll make a search @PeterTamaroff

sorry (abstract algebra ) ???

user87637

@what'sup Topology is huge. What sort?

user87637

5:47 PM

@PeterTamaroff There is general, algebraic and differential topology for starters.

Peter Tamaroff

@Jasper Do you know any good book that can complement's Spivak's succinctness?

what'sup

thanks for the help good bye now :-)

$$ \infty - \mathrm{uncountable } $$ thanks .

user87637

@PeterTamaroff There is Analysis on Manifolds by Munkres.

Jayesh Badwaik

@PeterTamaroff A good book is "Multidimensional Real Analysis" by Duistermaat and Kolk (2 parts)

Peter Tamaroff

@Jasper Ah. I trust Munkres. Seems to be a solid guy.

@JayeshBadwaik OK.

user87637

5:52 PM

@PeterTamaroff I like none of his books.

Peter Tamaroff

@Jasper ORLY?

user87637

@PeterTamaroff Yes, really.

Peter Tamaroff

@Jasper Why not?

user87637

@PeterTamaroff His topology is too long-winded. His algebraic topology does not treat homotopy theory. His differential topology is too thin. His manifolds book is too thick and does not treat abstract manifolds. QED.

Peter Tamaroff

user87637

5:55 PM

All Munkres books are terrible. I like Bredon's Topology and Geometry which covers general, algebraic and differential topology all in one.

user87637

For general topology, I like Willard and Kelley. I dislike Hatcher's algebraic topology as it is too handwaving.

user87637

I like Lee's Topological manifolds as it has the only good treatment of both curve classification and surface classification in the entire textbook literature.

skullpatrol

@Jasper Could you post your top book choices on your profile again please?

user87637

@skullpatrol You can see my blog now, dude. Check profile for link.

user87637

I only list the holy ones. There are too many good ones, and too many bad ones. You should really choose for yourself.

skullpatrol

6:00 PM

Oh? I didn't know you were a blogger now :-)

Btw, I like this shade of blue.

user87637

@skullpatrol Dodgerblue.

Peter Tamaroff

@Jasper I prefer France blue.

user87637

7:24 PM

@PeterTamaroff I prefer Pedro.

Peter Tamaroff

@Jasper So, what is it with you? Are you studying?

user87637

@PeterTamaroff No, I will be studying earliest next year.

Peter Tamaroff

@Jasper But aren't you reading on your own or something?

user87637

@PeterTamaroff No, I will do that next year. I am too unwell to study anything now. And I might never ever study math again...

Peter Tamaroff

@Jasper Unwell? Oh. What is it?

user87637

7:30 PM

@PeterTamaroff What I told you, those are serious things you know...

user87637

I am already taking meds but they don't seem to be of much use...

Peter Tamaroff

@Jasper Right. But didn't you, how do I say it, got off it?

user87637

Life is full of tragedies, some people suffer all life and then die.

user87637

@PeterTamaroff I am still trying to become functional again. Right now, I am a useless thing.

user87637

The only math I do now is 1+1=2 on MSE, LOL.

skullpatrol

7:32 PM

I apologize for being "insensitive" towards your illness before :(

user87637

I saw you visited my blog @skull

user87637

Hmm, what should my next post be?

skullpatrol

@Jasper Yes, dodger blue just like you.

user87637

I have many things on my mind, I think I will write them on my blog in a restricted way.

user87637

By not naming countries, organisations or people nobody can sue me for defamation.

user87637

7:40 PM

@skullpatrol Oh come on, why do you reference that? LOL

user87637

He is probably busy. I remember he was organising a conference.

user87637

@skull Are you majoring in math or sth else?

skullpatrol

I am only a banana.

user87637

@skullpatrol Haha, don't steal my lines!

skullpatrol

I learn from the best :D

user87637

7:43 PM

Charlie has not been coming to this chat eh?

skullpatrol

Look! a flying tiger :O

user87637

And a cat, LOL.

user87637

That makes it two cats, since a tiger is a cat.

skullpatrol

Hi @David

Alizter

8:54 PM

What is this question saying?

skullpatrol

@Alizter Can it be proven that the 3 lines intersect at one point?

Alizter

@skullpatrol Thank you but it looks like somebodies already had a go at it

Alizter

10:03 PM

Can anybody rearrange this trig expression to make $\phi_2$ and $\lambda_2$ subjects? $$\sin^2\left(\frac d{2r}\right)=\sin^2\left(\frac{\phi_2-\phi_1}2\right)+\cos(\phi_1)\cos(\phi_2)\sin^2\left(\frac{\lambda_2-\lambda_1}2\right)$$

robjohn

10:22 PM

Here is how I finished off my ideas
**Lemma:**
$$
\lim_{n\to\infty}\frac{\log(n)}{n}\sum_{k=2}^n\frac1{\log(k)}=1\tag{1}
$$
**Proof:** Note that
$$
\int_2^x\left(\frac1{\log(t)}-\frac1{\log(t)^2}\right)\,\mathrm{d}t=\frac{x}{\log(x)}-\frac2{\log(2)}\tag{2}
$$
By L'Hospital,
$$
\lim_{x\to\infty}\dfrac{\int_2^x\frac1{\log(t)^2}\,\mathrm{d}t}{\int_2^x\frac1{\log(t)}\,\mathrm{d}t}
=\lim_{x\to\infty}\frac1{\log(x)}=0\tag{3}
$$
By $(2)$ and $(3)$, we get that
$$
\lim_{x\to\infty}\frac{\log(x)}{x}\int_2^x\frac1{\log(t)}\,\mathrm{d}t=1\tag{4}

Peter Tamaroff

@robjohn Death by $\log,\log$. I can accept that.

Alizter

@robjohn clap. clap. clap. Well played Sir.

$$r=\theta^{\sin \theta}$$ speech bubble :)

Karl Kronenfeld

I must apologize. :( I spent half of the time wrongly thinking the given circle was centered at $0$. Of course, multiplication by $e^{i\pi/4}$ does not hold the circle in place if the center is nonzero. In this case the transformation $a$ dilates everything by $\sqrt 2$ and rotates everything by $\pi/4$.
The dilation transforms the circle in the following way: it makes the circle have radius $\sqrt[4] 2$ and it moves the center from $1$ to $\sqrt 2$.
The rotation keeps the radius the same but moves the center from $\sqrt 2$ to $\sqrt 2e^{i\pi/4}=1+i$.

Oops, I meant radius of $\sqrt 2$, when I wrote $\sqrt[4] 2$.

Alizter

11:00 PM

What is $$\max_{x<0}\Re\left(x^{1/x}\right)=^?$$

and $$\lim_{x\to0^-}\Re\left(x^{1/x}\right)=^?$$

Alizter

11:40 PM

meh im done

robjohn

@Alizter This depends on which branch of log you use.

If $log(x)=\log(|x|)+i\pi$, then $\mathrm{Re}\left(x^{1/x}\right)=\cos(\pi/|x|)e^{-\log(|x|)/|x|}$

and in which case it oscillates wildly between ever increasing bounds as it approaches $0$ from below.

So the $\sup\limits_{x<0}$ would be $+\infty$ and the limit does not exist.

Transcript for

$Mathematics$ Mathematics