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5:00 AM
This can be rearranged slightly to get $$\dfrac{1-e^{-n}}{n(1-e^{-1/n})}$$
So the question is then to take the limit of this as $n\to\infty$.
 
Zee
What a sexy limit
 
How did you rearrange it
I dont quite see the rearrangement
 
Well, $\dfrac{1-e^n}{e^n} = e^{-n}-1$
 
Ok got it
 
And $\dfrac{e^{1/n}}{1-e^{1/n}}=\dfrac{1}{\frac{1}{e^{1/n}}-1}=\dfrac{1}{e^{-1/n}-1}‌​$
Which, upon flipping the signs on top and bottom, combines to give the result I said.
Ok?
 
5:04 AM
yes
 
Ok.
Now, the behavior in the numerator is easy; as $n\to\infty$, it just goes to $1-0=1$.
The bottom is the harder nut to crack: $\lim_{n\to\infty}n (1-e^{-1/n})$.
Easiest way to do that, though, is to rewrite the limit as $\lim_{t\to 0}\frac{1}{t}(1-e^{-t})$ where $t=1/n$
 
Which is just a difference quotient, so the limit equals $\dfrac{d}{dt}e^t\Big|_{t=0}=1.$
 
whats a difference quotient
I forgot
 
The thing you use in calculus to define a derivative.
 
5:08 AM
Where's the difference quotient
I can't really see it
 
In this case, I'm doing $f'(x)=\lim_{t\to0}\dfrac{f(x)-f(x-t)}{t}$
(the more typical version is with f(x+t)-f(x), but this version is equivalent)
I'm taking $f(x)=e^x$ to make this work.
So I should probably also have written it as $\dfrac{d}{dx} e^x|_{x=0}=1$.
Oh well.
 
Ok I get it
 
Right.
 
Thats a nice result
 
Yeah.
 
5:11 AM
But would take too long in a competition
 
So this approach does work, and (to my mind) is more straightforward.
But yeah.
 
Did you figure out the riemann sum to integral or no
 
It's straightforward but tedious, as one of my textbooks would say.
Nope. :/
and I doubt I'll be able to tonight.
Might be better as a question for the main site, i.e. "In what sense is this sum a Riemann sum for the stated integral?"
 
Right
Thank you!
 
Glad to be of help :)
 
 
2 hours later…
7:30 AM
A book I'm reading says that g(x)=4x^3 +3x^2 −6x−1=0 can have 3 distinct roots, or 1, but not 2. Why not 2? Why isn't (x+a)^2(x+b) also a possibility?
 
I have a question that's asking me to take a 3 x 4 matrix and use the index $i, j$ of each entry to calculate the entries position in a 1 x 12 matrix.

I found the following formula which I think is correct:
$n = i\times Ncol + j$ where i is the row number, j is the column an entry belongs to and Ncol is the total number of columns (also starting index from 0).

So that's an equation to find n in terms of i and j. But the next part of the question asks me to find equations for i and j in terms of n such that I can get the original position of the 2d matrix back and this is confusing me.
 
Hey everyone!
Yo @Astyx finally beat paper-io
 
Hi, congrats
You have fulfilled your destiny
What will you do with your life now ?
 
Not really too sure tbh
 
Who is ?
existential crisis incoming
 
7:33 AM
But yeah pulling off the loop around the board but once it worked, it worked, unlike the strategy of just filling up the board
I never beat 75% with that
existential crisis already here but I'm still ignoring it
 
Neither did I, it takes too much time and concentration
I guess that's the way to go :p
Nothing better to wake you up in the morning than the sound of a construction machine destroying the street in front of your house
 
Arthur Dent seems almost to prefer it.
 
Press F to pay respects to the street
 
Usually it's birds that don't know how to sing
Actually they have just started since the machine stopped
 
"Birds can you please get your lives together kthx"
 
7:39 AM
Yeah kinda
Magpie make really horrible noise
And so fing loud
 
Ohi
What's up ?
 
Hm.... my roof ? :D
 
leaves
Actually this might be a better reaction
Jokes aside, how're you doing ?
 
Great, but I have my exams next week
 
7:45 AM
So do I, heh
Same place
 
These exams are just there to make sure you didn't fully drop off during first year right ?
 
Yep, but they also count to rank people for the choice of stages abroad in third year
 
Exams, at which level? Undergrad, grad, etc?
 
@Daminark I guess undergrad (I'm never too sure how to convert the French system to the English one)
 
7:49 AM
I see, I didn't know they had special exams of that sort then
I was thinking of it as quals almost
Usually my mapping is Lycee is high school, the next thing is undergrad, after that is grad
 
@Daminark Oh, it's an exam specific to my school. Not nationwide
 
I see
 
After Bacalauréat, you can do two year of Classes Préparatoire to then join an École d'ingénieur (for 3 or 4 years)
Some schools make this "classes préparatoires" thing something intern to their schools while others rely on the state to form their students who then take competitive exams to apply to the schools
@Semi @Steamy Turns out 10W isn't that much, and that the problem in lifting a ball with light is that, because of the intensity spectrum of the laser, the ball is not in a stable position and thus moves to the left/right
Searching in the transcript of this chat made me find this message
 
8:27 AM
Hi anyone here?
If I have 360/x = 200. How do I remove 360 the numerator?
So I have to divide 360/x with /360 and 200 / 360 in order to get rid of 360?
 
@sockevalley yes
 
I'm suppose to give an example of an function with the period 200 degrees and amplitude of 2,5. I figured it to to be 2.5sin1.8x. I don't however understand why it is sin and not cos. @hi
@Hippalectryon
 
Well, do you know that sin(x+pi/2)=cos(x) ?
Basically, sin and cos are the same function, they're just shifted by pi/2. So both answers are correct @sockevalley
 
Never seen sin(x+pi/2) = cos(x). Probably above my course since they don't seem to mention it in my textbook. OK, in that case that the answer should be both cos and sin but they choose sin in my textbook answer.
 
9:07 AM
@Astyx It's indeed not much, but I'm pretty sure you can make things of similar weight levitate using electromagnets with less power
Of course, that does require them to be made of the right material and such, but still
 
That's true. It feels counterintuitive we are able to lift objects with light though, which is what bugs me
But heh, science I guess
 
On the one hand, it feels weird because you're shooting massless things at it
But, massless stuff still has momentum :P
 
Yeah, it boils down to $E = h\nu$
Or rather $p = {h\over \lambda}$ actually
There are videos on youtube of this guy who built 100W laser guns just because he could
 
 
1 hour later…
10:38 AM
0
Q: How to decide $36^\text{th}$ smallest element in max-heap tree of $100$ elements?

Mithlesh UpadhyayConsider a max heap tree with $100$ elements and a node from the same level is chosen randomly. What is the probability that it is the $36^\text{th}$ smallest element______ . My attempt: Somewhere, it explained as: $P = (1/7 \cdot 0)+(1/7 \cdot 1/2)+ (1/7 \cdot 1/4) + (1/7 \cdot 1/8) + (1/7...

Please, take a look.
 
11:01 AM
hi, i there is a problem with this question :
i need to prove that for each $x \notin \Bbb Z$ the sum $\sum_{n=2}^{\infty} \dfrac{cos(2\pi n x)}{log(n)}$ converges. i think this is not true. am i right?
 
cfp
11:54 AM
I'm having a lot of difficulty in attracting any interest to my questions. See e.g. math.stackexchange.com/questions/2335798/… and math.stackexchange.com/questions/2338180/… . Do you have any suggestions? Would questions like these be better on Math Overflow?
 
@Liad Dirichlet's test + cos is real part of compex exponential
 
@SteamyRoot so you saying the claim is true ?
 
I'd think so, yes
 
@Astyx well,that's basically the idea behind a solar sail
 
@SteamyRoot what i was thinking is that $cos(2pi n x) = cos(x) $ so the sum is $cos(x) \sum 1 /log(n)$ and this sum diverges . where am i wrong?
 
12:04 PM
You're wrong in saying that $\cos(2\pi n x) = \cos(x)$
I mean, the former depends on $n$ which is going to have a huge influence.
 
huh... what a stupid mistake
 
with the big difference being that solar radiation is presumably not anywhere near as coherent as a laser source
 
The factor $2\pi$ doesn't matter much - but it's there to make the series converge when $x \notin \mathbb{Z}$
If you just had $\cos(nx)$, the condition would be $x$ not a multiple of $2\pi$
 
yea , now that i found my mistake i can work on proving that it converges
 
12:24 PM
Any idea about diophantine equation $xy = z (x^2-xy-y^2)$
x & y are coprime
 
@Semi And also solar sail work because there is no opposing force
What I meant is that it seemed strange than light is powerful enough to actually compensate another force such as gravity
 
It's often said that gravity is the weakest force, though :P
 
Eh, gravity is pretty weak all things considered. It'll depend on the mass of the object, though
 
In a certain way, it'd be surprising if you couldn't levitate something using light.
But it would be very surprising if you could easily launch something into orbit using light
 
12:30 PM
My point
 
i need to find $\sum 1 / n \ ^ 2$ using $f(x) = x \ ^ 2$ (Fourier series) , in the interval $[-0.5,0.5]$ but im getting that the coefficients is zero because $\int_{-0.5}^{0.5} x ^ 2 e \ ^{-2\pi in x }dx = 0$ where is my mistake ?
@Astyx where :P
 
@liad that integral doesn't equal zero
 
^
 
^
 
12:32 PM
<
 
<
 
v
(What am I doing with my life ?)
 
By the power of guessing based on what the solution should be, I claim these integrals will be $\frac{1}{2n^2 \pi^2}$
 
^^vv<><>ab
 
it's ba
 
12:34 PM
^
 
vv^vv^^vv
Good morning
 
Ohi Akiva
 
Still disappointed that it didn't unlock the final chapter in MGSV
 
Afternoon
 
12:35 PM
m @BalarkaSen
 
h @Astyx
 
Given two Zariski-closed subgroups $G$ and $H$ of $GL(n,\mathbb{C})$ as embedded into $\mathbb{C}^{n^2+1}$ via $A \mapsto (A,\det(A)^{-1})$. If $G$ and $H$ are isomorphic as abstract groups, are they isomorphic as algebraic groups, i.e. is there an isomorphism $f \colon G \to H$ such that $f$ and $f^{-1}$ are defined by polynomial functions? If not, what is a counterexample?
 
$\int_{-0.5}^{0.5} x \ ^ 2 e \ ^ {-2 \pi i nx} = \dfrac{e \ ^ {-2pi i nx } }{-2 \pi n } x \ ^ 2 + \dfrac{1}{\pi i n} \int_{-0.5}^{0.5} x e \ ^ {-2 \pi i n x} dx $
first expression is between -0,5 and 0.5 and is zero , and the second one evaluates to zero too
 
h r u @BalarkaSen
 
Again, no
On both counts
 
12:38 PM
where is the mistake ?
 
@Astyx g
 
$e \ ^ {- \pi nx } = e \ ^ {\pi nx } $
 
Actually, I'm being too quick. First one may be fine
 
@Liad Wut
Absolutely not.
 
Second one is definitely wrong, though
 
12:39 PM
why ? cos (-x) = cos(x)
 
@Liad it's a real number
$e^{-1} \ne e^1$
 
cosine is the real part of the complex exponential
the imaginary part is the sine, for which that doesn't hold
 
yea and the imaginary is zero
huh...
again stupid mistake.
 
Real exponential is not the real part of the complex exponential :P
2
 
yesterday, by Leaky Nun
Should we put "ΑΓΕΩΜΕΤΡΗΤΟΣ ΜΗΔΕΙΣ ΕΙΣΙΤΩ" in the chatroom description? @Daminark @TedShifrin @Fargle
this got 5 stars now
 
12:41 PM
@SteamyRoot You wish ...
 
although @AkivaWeinberger has dissented
@Astyx que penses-tu?
 
sorry we just learned about the $e \ ^ i$ and im doing mistakes.
 
And I'm abstaining from it
 
I just don't want to be kicked out from the chat for not knowing geometry :(
 
cfp
Hi All. Now there's a bit more activity, I hope you forgive me reasking this: I'm having a lot of difficulty in attracting any interest to my questions. See e.g. math.stackexchange.com/… and math.stackexchange.com/… . Do you have any suggestions? Would questions like these be better on Math Overflow?
 
12:42 PM
I don't get it @LeakyNun
 
I'm against in on the account that this chat is, well, already geometry-focused enough without it :P
 
@Astyx what do you think about my suggestion (on the starboard and quoted above)
 
It's "Let no one ignorant of geometry enter here"
 
What do these greek words mean ?
 
12:43 PM
Also, AFAIK, there aren't really any contemporary sources that this was on Plato's academy
 
I'm a bit surprised at the lack of interest in those questions as well @cfp
 
it is alleged to have been on the entrance of Plato's academy @Astyx
 
What does it mean ?
 
1 min ago, by Akiva Weinberger
It's "Let no one ignorant of geometry enter here"
 
cfp
@Semiclassical Good to hear! So Math Stackexchange is where they belong, rather than Math Overflow? I never know quite where the line is.
 
12:44 PM
@Akiva It doesn't say anything about letting people ignorant of geometry stay
 
My guess, though, is that it's because random matrix theory is definitely a specialized subject
 
Now it depends what you mean by geometry
And by ignorant too
 
wait , why dont we have $e \ ^ {- \pi n } = e \ ^ { \pi n ]$ ? both equals $cos (\pi n)$
 
So while it's entirely appropriate for MSE it may not be reaching the right audience
 
you're getting too technical @Astyx it's just a quote of historical and mathematical interest
 
12:46 PM
They don't equal $\cos(\pi n)$ at all
 
Again, noo
 
I know, I'm doing it on purpose
 
@Liad $e^{i\pi n} = \cos(\pi n) + i \sin(\pi n)$, nothing more, nothing less.
 
Ah... why?
 
$\operatorname{Re} e^{i x} = \operatorname{Re} (\cos(x) + i \sin (x)) = \cos(x)$
 
12:47 PM
ok but $sin(\pi n) = 0$
 
@Liad only when $n$ is an integer
 
cfp
Ahh OK. I guess I just have to wait and hope that someone appropriate stumbles across them. Thanks for your comments.
 
There's not a single real exponential in there
 
$n$ is a n integer
 
well, $\cos(x) = \frac{e^{ix}+e^{-ix}}{2}$ also
 
12:47 PM
@LeakyNun $= (-1)^n$
 
@Astyx well...
what is $n$?
 
That's what I was getting at re: the first term you wrote out from integration by parts being zero
I think you were actually right on that.
 
The only thing you do have in that trend, is like, $\left| e^{a+bi}\right| = e^a$
 
Integer, unless you're being very deceitful
 
But the remaining integral from integration by parts can't be discarded in this manner
 
12:48 PM
so why you saying that $e \ ^ {-\pi n i} \ne e \ ^ {\pi n i}$ . sorry for bugging you but i want to understand
 
Stop dropping the i
 
The exponential is a monotonic function
 
@Liad because $1 \ne e^{2\pi i n}$
 
$-\pi n < 0$ hence $e^{- \pi n} < 1$
 
You need that in your exponentials if you want to get any connection to sine or cosine
 
12:50 PM
and $\pi n > 0$ hence $e^{\pi n} > 1$
All for $n > 0$, of course
 
what is $n$? a real number? complex?
 
Integer
 
oh, then $1 = e^{2\pi i n}$
 
$e \ ^ { - \pi n i } - e \ ^ {\pi n i} = 0$
 
and then $e \ ^ {-\pi n i} = e \ ^ {\pi n i}$
 
12:51 PM
Second, the statement is that $e^{i n \pi}=e^{-i n \pi}=(-1)^n$ for integral n
 
right
 
this is what i got in the left exp. of what i wrote
 
then where is the error?
 
That's true but only because the argument of the exponentials is i * pi
 
huh... @SteamyRoot did you talk about $e$
 
12:52 PM
If you have an $x$ in there too, that turns stuff non-integral.
 
as the exponent function?
 
It will not be true if you replace n with something non-integer
 
Well, e to the power something is the exponential function, yes.
 
$e \ ^ {\pi i n } = cos(\pi n ) + sin (\pi n) i$
 
Yes...
 
12:54 PM
do you agree that $e \ ^ {- \pi i n } - e \ ^ {\pi n i } =0 $ ?
 
Yes!
 
So on what are we arguing about?
 
You claimed that $e^{\pi n} = e^{- \pi n}$
 
the first term of what i wrote earlier does evaluates to zero
@SteamyRoot sorry i forgot the $i$
 
Right
Either way, if you do your partial integration, the non-integral term will indeed vanish
But what you're left with inside your integral is still $e^{2 \pi i n x}$
And this $x$ makes it so, in general, $e^{2 \pi i n x} \neq e^{-2 \pi i n x}$
 
12:57 PM
ok so im left with $\dfrac{1}{\pi i n } \int_{-0.5}^{0.5} x e \ ^{-2 \pi in x}dx$
 
Missing an $i$ in the exponent
 
For that, do integration by parts once more.
 
again . i keep forgetting it after the $i$ of \pi :P
yea im about to do that
we have another term that becomes 0
 
People sometimes put the i in front of the 2 to make that more obvious
That may help you avoid forgetting it
 
we have $\dfrac{1}{\pi i n} \dfrac{1}{2 \pi i n} \int_{-0.5} ^{0.5}e \ ^{- 2\pi i nx}dx$
 
1:00 PM
Why does the first term become $0$ ?
 
same reason as before
 
Last time you had an $x^2$.
Now you have an $x$.
 
we have $x$ multiplying it
 
$x^2$ gets rid of the signs, $x$ doesn't.
 
Huh...
i was about to say that what im left with is zero
 
1:01 PM
The integral you're left with, is going to be some multiple of $\sin( \pi n)$, which will vanish.
 
@SteamyRoot With one very important exception
 
Well, you should always treat the $n = 0$ case separately in Fourier stuff :P
 
Lolyes
Also, nice Freudian slip there (threat instead of treat)
 
There is no evidence!
 
ok im getting $-\dfrac{cos (\pi n )}{\pi \ ^ 2 n \ ^ 2}$
 
1:04 PM
Pfffft
 
(there totally is, because there's a history button, but shhh)
 
the result is correcT?
 
Please write out the entirety of what you get from the second integration by parts. Right now it's not clear which part you're referring to
 
I'm pretty sure the entire integral should be that thing, multiplied by $-1/2$
 
That moment when your superiority complex is so advanced you start to threat numbers
 
1:06 PM
I tried scaring $\pi$ into being algebraic but it didn't work :(
 
Foolish fool! $n=0$ is always a threat
 
That's an irrational threat
 
ok i have $\dfrac{1}{\pi i n} (\dfrac{2 cos(\pi n)}{- 2 \pi i n})$
which equals $\dfrac{cos(\pi n)}{\pi \ ^ 2 n \ ^ 2}$
 
No, n=0 is entirely rational :P @Astyx
 
no, it's an empty threat
 
1:08 PM
threat by vacuity
 
@SteamyRoot you got this result too?
 
I swear that if you don't exist, I will hurt you very bad
 
Tell that to sin(n*pi)/n ! @Justwinbaby
 
"So if I do exist you won't hurt me, and if I don't exist you can't hurt me. Seems pretty win-win."
 
1:11 PM
@Liad I still think you're missing a factor $2$ in the denominator.
 
right!
 
I might be wrong, though.
 
you are not
i forgot about the $x = 1/2$
 
user84215
Is there any proof, without using variational techniques, of the fact that a complete surface with positive Gaussian curvature is compact?
 
ok, so for $x = 0$ we have $0 = \dfrac{1}{12} - \dfrac{1}{4 \pi \ ^ 2} \sum \dfrac{1}{n \ ^ 2} + \dfrac{1}{\pi \ ^ 2} \sum \dfrac{1}{(2n-1) \ ^ 2}$. how can i make the last term nicer?
 
1:25 PM
Adding and removing the even terms that are missing
Which you can then easily simplify (even numbers are nicer than odd ones because they factor easily)
 
@Astyx exactly what i just did
:P
 
In your last message ?
I meant $\sum {1\over (2n-1)^2} = \sum {1\over n^2} - \sum {1\over 4n^2}$
 
yea i turned it into $\sum 1/n \ ^ 2 - \sum 1 / (2n) \ ^ 2$
 
I don't think I follow
But nvm
 
i wrote exactly what you did above it
 
user84215
1:28 PM
Is there any proof, without using variational techniques, of the fact that a complete surface with positive Gaussian curvature is compact?
 
But not in this chat ?
@aminliverpool If we didn't answer at first it means no one in the room knows the answer, there is no need to repost your question
We don't answer when we don't know otherwise this room would be chaos
 
Hey guys! how many rotations of order 4 does the octahedron have? I can only see 3
 
I think there are only 3
 
apparently there's supposed to be 6
 
If there are 3 rotations of order 4, there are 6 symmetries of order 4
If I'm not mistaken
 
1:36 PM
270 is also 90
 
The only order 4 rotations are the ones around the axes joining opposite pairs of vertices, not?
So 3
 
what do you mean if there are 3 rotations then there are 6 symmetries
 
A rotation preserves orientation, a symmetry does not
If you take the reflexion of you octahedron through a mirror, you get a new octahedron that isn't the rotation of the first one
(non formally)
 
so what's an example of those 3 symmetries that I didn't count?
ohhh you mean any rotation of order 4 composed by a reflection that cuts the octahedron into pyramids?
 
1:42 PM
@BalarkaSen yes that's what I thought. I'm given a table that says type of axis, order, number. For the type "passing through edges" and order 4, the number is 6. I thought it would be 3
 
Actually that's just order 2
 
Hi
 
But if you rotate that by $\pi\over 4$ around the $AF$ axis, you get something of order 4
Hi @Dattier
 
Comment se sont passer tes concours ?
 
Pas encore finis
 
1:43 PM
Yes, it should be an axis going through opposite vertices, not opposite edges.
 
@Astyx : Tu as été admissible pour l'ENS ?
 
J'ai passé les oraux des mines hier et avant-hier, j'ai l'X la semaine pro, l'ENS la semaine encore après et centrale en dernier
Oui
 
@BalarkaSen yes sorry meant to say vertices
 
Bon courage
 
Merci !
 
1:44 PM
Tu as des nouvelles de @GranDom
 
@Astyx what you have drawn is a reflection right?
 
@GRANDODOM
 
Oui je l'ai vu sur Paris il y a quelques temps
Yes @John11
LeGrandDODOM
 
Lui aussi est admissible aux concours qu'il a voulut
 
Lui a déjà passé les concours l'année dernière je crois
 
1:46 PM
Ah, c'est lui qui à l'X ? dans la gendarmerie ?
pour son service
 
Non ça c'est Hippalectryon
 
Ah, d'accord
 
D'ailleurs en fin de compte je pense aller à l'X plutôt qu'aux ENS
 
Je pense que c'est mieux
 
Pourquoi ?
 
1:47 PM
surtout si tu veux avoir une carrière hors ensaignement (lol faute involontaire mais qui dit bien ce que je pense de la situation)
 
Mm c'est surtout que je ne suis plus sûr de vouloir absolument faire des maths ma profession
 
ENS c'est avant tout l'enseignement même dans les Universités
 
Dans le sens que je n'ai aucune idée de ce que la recherche en maths est, vraiment
 
Comme je te l'ai dit, je fais de la recherche en raisonnement humain, et soit je suis complétement singlé (ce qui n'est pas à exclure) soit en faisant cela, je contribue de manière significative à facilter la vie de mes compratiotes
 
Enfin bref, je suis un peu trop fatigué pour parler de ça maintenant
 
1:53 PM
Ok
Is it true that $$g\in C^2([0,1],\mathbb R), 8(\max(g)-\min(g))\geq \min(g'') $$ ?
 
More geometry puzzles: $ABC$ be a triangle, and $E, F, D$ points on $AB, BC, AC$ such that if $P$ is the point of intersection of $AF, EC, BD$ then $\angle EPB = \angle BPF = \angle FPC = \angle CPD$ and $EP = FP = DP$. Calculate $\angle BAC$.
 

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