$a_n=\big(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+\dots + n}\big)$ How can i find $\sum_{n=1}^{15} (a_n)$?

$x$ can't be zero (because, when $x=0$, the left-hand side of that equation doesn't exist). So we can divide both sides by $x^2$.

You can also see it's necesssity if you consider that $1/z$ should have equal-but-opposite residues at 0 and infinity

That gives us $f(1/x)=0$.

@Fawad How?

Nvm

In any case, we have $f(1/x)=0$, which means either $1/x=p$ or $1/x=q$

That means $x=1/p$ or $x=1/q$.

We can plug those into $x^2f(1/x)$ again to double-check that it does, in fact, equal zero when $x$ equals $1/p$ or $1/q$.

Yes.

Thus, $1/p$ and $1/q$ are the roots of $x^2f(1/x)$, so they're the roots of $cx^2+bx+a$.

Thanks. Bye

@BalarkaSen $\frac 1{2\pi i}\int_\gamma f(w) \text{d}w$

@Alessandro Where $\gamma$ is a contour around $0$ (where you assume your pole is). Now sub in $w = 1/z$ so that $\gamma$ becomes a contour around infinity.

What do you get?

I mean, why do we take the residue in $0$ of $z^{-2}f(\frac 1z)$ to get the residue at infinity of $f$ rather than the residue in $0$ of $f(\frac 1z)$

oh, ok, I see

derp

:)

hi chat

hi

anything fun today?

hey

It's sort of funny because $f$ need not have a pole at infinity, yet can have residue at infinity.

pole?

You can always define it as the sum of the other residues

residues?

cOmPlEx AnAlYsIs

lol

oh

reading about it just now

@Semiclassical can i get your opinion about the question i wrote above

@EricSilva are you trying to spongebob meme?

It's like real analysis, except functions are actually nice rather than trying to backstab you with counterexamples to every plausible sounding assertion

@arctictern spongebob meme?

Doctor , it hurts between my legs.

Well your husband pole is leaving too much residue

It's like real analysis but actually algebra

@EricSilva yeah, the mocking spongebob meme

oh

spongebob?

oh wow I've never seen this before in my life

it's sweeping the meme world at the moment

I'm always late on new meme trends

@Waiting yea, I haven't been very motivated these days, that's why I'm not around a lot!

Where $k(x,y)$ is $1-x, y\le x$, and $1-y, y \ge x$

Is there a neat way to show it is diagonalizable? Or even better, to find the orthonormal basis?

Or representation, more so..

oh, it should be $\prod_{n=1}^{15} (a_n)$

@ShaVuklia no pain no gain!

No money no funny!

anyone wanna look at the problem: $a_n=\big(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+\dots + n}\big)$ Find $\prod_{n=1}^{15} (a_n)$?

@AbdullahUYU first, note that the terms being summed in $a_n$ are the reciprocals of 1,3,6,10,15,etc

Do you recognize that sequence?

Reciprocating AGPs

yes, i have already realized that

Okay. Do you know the formula for that sequence?

i dont know the one has reciprocals

just find the sum of AGP

I guess first

I'm not asking for that yet

oh

What formula generates the sequence 1,3,6,10,...

i know sum of sum of first $n$ positive integers can be observerd on pascal's triangle

$n(n+1)/2$

isn it? @Semiclassical

Right. So $a_n =1+1/3+\cdots + 2/(n(n+1))$

yes

Now, this actually has an easy sum. To see that, try splitting $2/(n(n+1))$ into two fractions

hmm wait

oh

oh

spongebob meme is not good unfortunately

can u help me?

Who lives in a nateapple under the see!!!

$2/n-2/(n+1)$

oh, something cancels out

Right. If we apply that to each term in $a_n$, we get $$a_n=(2/1-2/2)+(2/2-2/3)+(2/3-2/4)+\cdots + (2/n-2/(n+1))$$

What happens?

@MikeMiller Hi Mike. Is there a ""link"" between a symmetric compact convex set with non-empty interior and a lattice $\mathcal{L}$ of $\Bbb{R}^n$?

oh

i wrote what happens :)

Sounds like a cool question but it's out of my pay grade.

@MikeMiller That meme does suck, you're right.

Be more specific

I don't understand the meme.

pay grade ?!

I guess it's like when someone mimics you

when you say something to them.

I guess somebody connected that face with that moment

Do you know what I mean balarka?

Yeah but I still don't get it. ¯\ _(ツ)_/¯

question becomes $\sum_{n=1}^{15} (2+\frac{2}{n+1})$

Well it's pretty stupid :)

What's a meme?

2 minus, but otherwise yes

memeyouyou

yes, i write it wrong

Mmkay. Now, the first term is easy to sum

I was reading the problem: Show that there doesn't exist a space $X$ such that $X\times X$ is homeomorphic to $S^{2}$, the 2-dimensional sphere.

oh, it should be $\prod_{n=1}^{15} (2-\frac{2}{n+1})$ @Semiclassical

pardon me

With algrebraic topology is it possible to use "basic" Homology ? Mayer-Vietoris? Excision theorem ?

@Studentmath in general integral operators are compact and self adjoint if they satisfy some symmetry condition, then you can apply the spectral theorem, if you're actually asking for a specific orthonormal eigendecomposition im not gonna work it out

Do you mean, they wanted you to do $a_1a_2\cdots a_{15}$?

yes

@JeSuis Is that X x X the cross product of the space with itself?

i misspelled it

So apparently things can have non-integer dimension...

fractals man

can't live with em

Mmkay. In that case, it helps to write your result for $a_n$ as one fraction

@Daminark Hausdorff dimension?

Yeah

We haven't started them yet, Marianna just gave a quick "This is what we'll do next class and why it's insane"

Stuff's weird.

Especially when the dimension is integral, for example the Sierpinski tetrahedron has dimension $2$.

@BalarkaSen I came up with this a couple hours ago!

yay

Also it's weird that this is the only quarter of the 3 in which we actually finished everything we intended to, last quarter we lost some calendar days and in first quarter we were trying to cover so much

Like the final class is optional

it seems not terms cancels easily, any thoughts?

remember back when I used to think all every ses splits? :)

@Steamy O lawd

that's bc you take the class and then start reading books and realize she hasnt actually given any rigorous proofs @Daminark

even weirder is that you can have curves in $\Bbb R^2$ with positive $2$ dimensional Lebesgue measure @Dami

As I said, take your result for $a_n$ and write it as one fraction

I mean so, her geometric measure theory stuff was definitely modulo a lot of technical details

*all

@Daminark you ever tried Ricci blow?

@SohamChowdhury Cool.

Not all, there were proofs we had which were mostly complete

oh, got it

Even then I don't /really/ mind that to be honest

"mostly"

@AlessandroCodenotti Jordan curves, yup.

I'm skeptical

It was much more fun than, for example, if we had actually shown that Lebesgue measure satisfied the regularity properties for Caratheodory extension

Details are important

it turns out that answer is $2^{11}$

I mean proofs are overrated imo, but still Marianna makes you feel like she proved things and then doesn't prove things.

Like I'm willing to take certain things on faith that things could be made rigorous so that we could focus more on the ideas. Part of it is that I completely glaze over at epsilon detail in analysis

That sounds like religion

You can even have compact subsets of $\Bbb R^2$ whose border has positive $2$-dimensional Lebesgue measure

imo that's actually less weird than the dimension thing

@Daminark Also, broccoli and cauliflower aren't completely 3-dimensional

@Daminark sure, that's fine, but at some point if stuff doesn't get very technical you can't solve any problems

@JeSuis Are you allowed to use homotopy groups?

@JeSuis When is X homeomorphic to $S^2$ ?

Yeah. And you see why? @AbdullahUYU

$\pi_2(X \times X) = \pi_2(X) \times \pi_2(X) = \Bbb Z$. But $\Bbb Z$ is not square of any group (why?)

yeah, i calculated it myself

if you don't mind can we continue to solve problems? @Semiclassical

@Zee Religion is when you have faith based on unverifiable personal/spiritual experience, this faith of mine is, I know this has been done rigorously and I'd rather get to cool things than spend a week staring at Greek symbols

Sure, though I don't know how fast I can answer

@BalarkaSen Is this because it is a two-dimensional manifold?

@BalarkaSen yes

@Dodsy $S^2$ is a 2-dimensional manifold, yes.

arf you mean $\pi_n$ ?

And @EricSilva I guess, I dunno, like in Schlag's quarter we were being more careful about things and I was just not really engaged at that point

No I mean $\pi_2$. $\pi_2(S^2) \cong \Bbb Z$.

Higher homotopy groups of $S^2$ are rather complicated.

time is no proplem and i have to say that you are good at finding the right way to solve problems

@BalarkaSen No i mean, I thought for X to be homeomorphic to a circle it must be a compact connected one dimensional manifold.