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20:04
@TobiasKildetoft Care to explain? The way I see it now is that $S_3\cong H\subset S(S_3)\cong S_6$. It makes sense to me that if we just fix 4,5,6, we get $S_3$. So I don't see why we don't have $S_3\cong\{\sigma\in S_6\mid\sigma(4)=4,\sigma(5)=5,\sigma(6)=6\}$.
I do not understand. (Does that suggest a proof?)
I mean, there are of course more than one domain whose boundary is the same as a given disk in the plane. Namely, the disk itself and the exterior complement.
@ShaVuklia I did not say that subgroup was not isomorphic to $S_3$. I said that was not the one you get from Cayley's theorem, which gives you a very specific subgroup.
@ShaVuklia refer to either Steamy's demonstration or my demonstration of how we find the subgroup given by Cayley.
@TobiasKildetoft It's the subgroup of the permutations $S_6$?
20:07
@ShaVuklia What do you mean by "the"?
@ShaVuklia $S_6$ has a million subgroups
There are multiple subgroups of S_6 which are isomorphic to S_3, embedded differently.
[Random]
Prerequisites:
The Hardest Logic Puzzle Ever is a logic puzzle so called by American philosopher and logician George Boolos and published in The Harvard Review of Philosophy in 1996. Boolos' article includes multiple ways of solving the problem. A translation in Italian was published earlier in the newspaper La Repubblica, under the title L'indovinello più difficile del mondo. It is stated as follows: Three gods A, B, and C are called, in no particular order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter...
@Tobias but cayley doesn't give a subgroup?
it just says that there is a subgroup
@ShaVuklia it does.
Refer to either Steamy's demonstration or my demonstration of how we find the subgroup given by Cayley.
20:09
@ShaVuklia It gives an injective homomorphism which is basically the same
Fatalism is a philosophical doctrine that stresses the subjugation of all events or actions to fate. Fatalism generally refers to any of the following ideas: The view that we are powerless to do anything other than what we actually do. Included in this is that humans have no power to influence the future, or indeed, their own actions. This belief is very similar to predeterminism. An attitude of resignation in the face of some future event or events which are thought to be inevitable. Friedrich Nietzsche named this idea "Turkish fatalism" in his book The Wanderer and His Shadow. That acceptance...
It doesn't just say there is a subgroup; it explicitly identifies the subgroup.
At least the proof of Cayley's theorem does.
so you want me to specifically list all the elements of this subgroup?
Warning: The following content based on the hardest logic puzzle ever contains rant
@LeakyNun Originally I was gonna be like "Come on there can't be a million subgroups", but then I remembered that 720 elements is a lot
20:10
@Daminark I made that figure completely up.
@ShaVuklia yes
Suppose you are a patient affected by coughing with 4 possible scenarios:
I've literally already done that
$S_6$ has 1455 subgroups.
$S_3\cong\{\sigma\in S_6\mid\sigma(4)=4,\sigma(5)=5,\sigma(6)=6\}$
Lolol
20:11
@ShaVuklia I said, the subgroup you get by fixing $4,5,6$ is not the subgroup given by Cayley
1. Chronic: For every second, you cough
I + many other people said
yea well I give up
actually I never said because other people already said.
I don't understand what you're asking me here
20:11
@BalarkaSen I want to prove that there is a single domain whose boundary is the union of two given circles.
2. Dormant: For every second, you will not cough
@ShaVuklia Do you remember how Cayley's theorem is proved?
59 mins ago, by SteamyRoot
If $1 = (0,0), 2 = (1,0), 3 = (0,1), 4 = (1,1)$, then you map, say, $2$ to $(1 2)(3 4)$ because $2 \cdot 1 = (0,0) + (1,0) = (2,0) = 2$, $2 \cdot 2 = (1,0) + (1,0) = (0,0) = 1$, $2 \cdot 3 = (1,0) + (0,1) = (1,1) = 4$, $2 \cdot 4 = (1,0) + (1,1) = (0,1) = 3$
@ShaVuklia Apply the construction of Cayley's theorem
To show that $S_3$ is a subgroup of $S_6$
3. Random: For every second, there's a chance that you will or will not cough
20:12
That's what he's asking
4. Wicked: For every second, the occurence of coughs occured in a way to maximally make you suffer
@LeakyNun But it inly has 56 subgroups up to conjugation.
@Emolga By circle you mean a Jordan curve on the plane? And you mean two disjoint circles?
@TobiasKildetoft I see
The logic question, is to devise a suitable mindset, medication, environment etc. ,such that no matter what the nature of the cough is, you minimise the suffering
20:13
I also remember that somehow $S_6$ is one of the only two $S_n$ groups with non-trivial outer automorphisms
@BalarkaSen I mean just a plain old circle, i.e. I ask about the boundary of an annulus
Ok, so we have $f\colon S_3\to S(S_3)\colon\sigma\mapsto\epsilon_\sigma$.
What we know: We know from the solutions to the hardest logic puzzle ever and its extension, the solution to 1-3 exists
@ShaVuklia right, and what do the elements map to?
However: Is there exists a solution to a wicked cough?
20:14
I haven't given any element yet?
so e.g.:
More philosophically, is metafatalism possible?
$\epsilon_{(12)}(13)=(12)(13)=(132)$
Why is the above contain a rant: A wicked cough prevent me to get sleep last night and I was forced to wake up at 6:00
they map to just some other permutation
@ShaVuklia yes, so what is $\epsilon_{(12)}$ as an element of $S_6$?
20:16
I donno, (12)?
@ShaVuklia hardly
Use my method to find the elements
1 hour ago, by Leaky Nun
$$\begin{array}{c|c}
\cdot&e&a&b&c\\\hline
e&e&a&b&c\\\hline
a&a&e&c&b\\\hline
b&b&c&e&a\\\hline
c&c&b&a&e
\end{array}$$
1 hour ago, by Leaky Nun
So $a$ is the permutation that sends $(e,a,b,c)$ to $(a,e,c,b)$
1 hour ago, by Leaky Nun
i.e. $(1,2)(3,4)$
oh yea, I can see it's not (12)
but I donno, I don't really feel sharp anymore.
I'll try tomorrow
‎To see that my question is not trivial, see that this is false for general domains: en.wikipedia.org/wiki/Lakes_of_Wada
@ShaVuklia alright
@Emolga True, but for just union of two concentric round circles it's very easy.
That's why I was asking whether you had Jordan curves in mind.
The point is the annulus that bounds the two concentric round circles has disconnected complement and each of the connected components contain exactly one circle.
20:23
Are $\Bbb F(a,b,c)$ and $\Bbb F(a,b)$ isomorphic?
For a domain to have boundary exactly the two circles, it would necessarily have to be included inside the annulus.
nvm, any two bases must have the same cardinality
What's a domain again?
20:40
@EricSilva What are you doing today
well im getting paid to study up a bit on geometric flows
so far it's pretty interesting stuff, reading up on maximum principles for heat-type equations
how bout you
ah, cool
i was hoping someone would give me a push so i can finally read up on Jacobi fields
oh dude jacobi fields are very useful
they're a nice way to express a lot of variational things
My ears perk up a bit whenever I hear the word 'variational'
(b/c Lagrangian mechanics = principle of least action etc etc.)
right lots of the geometers here use those things as examples but idk anything abt mechanics at all so sucks to be me
but i spent basically the whole summer studying variational problems in geometry so i think that stuff is p great
20:45
@Eric would you want to tell me a story
@BalarkaSen did you have any chance to think about my question? :P
@Balarka maybe, what story
oh I mean the Jacobi fields story. I don't really know what they are or what I can do with them
@Danu Ah now that you say it I remember you pinged me something. I didn't think it was something I could be able to answer though
Sorry for that
I'm back and asking stupid questions again.
0
Q: Show that $\frac{1}{1-x}$ is equal to its Taylor Series on $(-1,1)$

ALannisterI need to show that the function $f(x) = \frac{1}{1-x}$ is equal to its Taylor Series (about $a=0$) on $(-1,1)$. Thus far, I have found the Taylor Series: $\frac{1}{1-x} \approx \sum_{k=0}^{\infty}x^{k}$, and since $f^{(n+1)}(x) = \frac{(n+1)!}{(1-x)^{n+2}}$, the Lagrange Error Formula gives us ...

Let $R$ be a commutative ring. Is it true that $R/\langle ab \rangle \cong R/\langle a\rangle \times R/\langle b \rangle$? @TedShifrin
20:47
I have loads of holes in my knowledge about Lagrange error and yet I'm expected to teach it. Not good.
oh well basically they encode the relationship between geodesics and curvature. If you try to study the exponential map on a Riemannian manifold and understand, for example, its singularities, Jacobi fields just pop up @Balarka
of course this is all in do Carmo :P
@ALannister: Teaching is truly the way to fill holes and acquire deeper understanding. But ask a colleague or us if you need to.
Hi Eric, Semiclassic, Balarka, Danu, world.
Hi @Ted
Just did @TedShifrin. Posted a link to my question up there.
20:49
Hi @Ted
@Leaky: Not in general, but if there's a hypothesis, yes.
I got a copy of Bishop & Crittendon lately (huzzah for office moves and the resulting slew of free books)
What's the rundown on it?
@TedShifrin polynomial rings over fields?
Then Captain Obvious posted "it's a geometric series". Well, duuurrrrr! But I can't just cop out and use that! The students don't know what a geometric series is yet! We have to prove this the long way.
@EricSilva I see. I know it's in doCarmo, I just get stuck in a rather unproductive loop sometimes where I can't decide if I should start reading or not.
20:50
my knowledge of it is limited to "oh hey it's called geometry of manifolds"
the chapter is p easy iirc
No, no, you need a requirement on $a$ and $b$. Play around with $R=\Bbb Z$ for a while.
@ALannister: You're on the right track. You want the largest possible denominator ... or at least a sup of such values for $z\in (-1,1)$. So what is that?
@ALannister student is learning Lagrange whatever and don't know what a geometric series is.
knowing what a geometric series is != being able to prove it via Lagrange error
There is a more explicit way to do this problem, by the way, since you're playing with geometric series. You have an exact formula for the error for a geometric series.
20:52
@LeakyNun we do it backwareds at my school.
@ALannister interesting
And I'm getting a whole lot of answers that shsow how incredibly smart the people posting are, but aren't helping me at all.
(the error is itself a geometric series, isn't it?)
@Leaky: Is $\Bbb Z/\langle 4\rangle \cong \Bbb Z/\langle 2\rangle \times \Bbb Z/\langle 2\rangle$?
@TedShifrin obviously not :P
20:53
What is the most abstract/minimalistic/whatever of the existence of the closed form of the geometric series?
@Secret prove the formula for each partial sum
@Semiclassic: $\dfrac 1{1-x} = 1+x+x^2+\dots+x^n + \dfrac{x^{n+1}}{1-x}$.
@Secret and then just limit (sum to infinity is defined as the limit of the partial sums)
So when does it work for integers, @Leaky?
20:54
$S=1+xS$ is how I always view it
@TedShifrin the largest possible denominator is 2^{n+1}
@TedShifrin when $a$ and $b$ are coprime
Right. The sup is that, and that's all you need to get an upper bound on the error. :)
That sounds like [REDACTED]
@LeakyNun Right, I see
20:54
Right, @Leaky. Do you know an ideal-theoretic way of stating coprime?
@Semiclassical SShhhh. That's precisely what this is.
@TedShifrin trivial intersection
There is no such thing, @Leaky.
oh, right
LOL @redacted .. or is the recatted?
I know that $\lim_{n \to \infty}\frac{|x|^n}{n!} = 0$, is $\lim_{n \to \infty} \frac{|x|^{n+1}}{(n+1)} = 0$ as well?
20:56
@Balarka I remember Neves gave me a lot of applications of Jacobi fields in comparison geometry if you think that stuff is interesting it might motivate you. I quite like it, personally and it's coming up a lot in the flow stuff that im being paid for right now.
Without hte factorial
I should've done "That sounds [an old kind of television]" :P
@ALannister just use sandwich
though the fact that CRT counts as old breaks my brain a little, since I grew up with them
What do you know about $x$ here, @ALannister?
20:56
It's between -1 and 1
@EricSilva I can try some of that
So, |x|<1
Note that $2^n/n$ does most likely not go to $0$.
Exponentials beat out polynomials, right?
20:57
What do you mean $2^{n}/n$? Where does that come from?
I'm answering your question for general $x$.
But x isn't general. x has absolute value smaller than 1
You need to sit down and develop intuition for these basic precalculus-y things.
@Balarka Cheeger and Ebin is a nice book with some cool stuff in it
So when we raise it to an integer power, it gets smaller.
20:57
I understand that. You asked a question up there ^^^.
Oh, I see. Well I meant in the context of this thing here.
@ALannister This was a general question to which you should know the answer.
I'm going to tell you to sit and think a bit more before you ask.
Hi @TedShifrin
But when x is between -1 and 1, it goes to the denominator!
In terms of teaching Calc II (and even precalculus), you need to get your students to have a feel for exponentials >>> polynomials >>> logs ...
20:59
@TedShifrin intuitively, $a$ and $b$ are coprime if $a=hm \land b=hn \implies h=1$, i.e. the only ideal that contains $a$ and $b$ is the ring itself.
I don't like what you just said.
me!
See!
21:00
Oo.
Ted hates me confirmed
@ALannister Please don't say stuff like this when you're teaching.
I said hi to you ages ago, @Danu.
I know
I won't. I'm just sasying it on here because I'm typing and so I want to say it in the quickest way possible
I had a lame question that I posed to Balarka earlier---mind if I borrow some of your attention too?
21:00
Go ahead, @Danu.
@EricSilva I'll check out that book, thanks
But, for $-1<x<1$, $|x|^{n+1} < 1$, right?
Sure, @ALannister, of course. Indeed, $|x|^n\to 0$, if we need it.
2 days ago, by Danu
@BalarkaSen Here's a simple question: I have seen several people writing the following: $SO(6)/U(3)=SU(4)/S(U(3)\times U(1))=\Bbb C\mathrm P^3$. The first equality sign is what I don't understand: The denominator is unchanged because $U(3)\cong S(U(3)\times U(1))$, with the isomorphism of Lie groups given by $A\mapsto (A,\det A^{-1})$, but the top is weird. $SU(4)$ is the double (universal) covering of $SO(6)$, so I don't understand how this can work. Do you understand what might be going on?
@BalarkaSen I agree that the domain $D$ is contained in the annulus $A$. To show that it is equal to it, I think that we can consider $D$ in the subspace topology of $A$ and since it is not equal to the whole space it must has a boundary, which is also a boundary considering $D$ in the plane, contradiction.
21:03
I did have a question earlier which (and you might know best, @Ted)
I happened to get a copy of Bishop and Crittendon for free a few days ago.
[Random] Computing some random sum:
Any opinion on it? I haven't opened it up yet.
$$\sum_{n=1}^m\frac{n}{e^n}$$
Summation index: $i$ or $n$? fixed
@Secret it's an arithmetic-geometric sum
21:05
@Danu: The person with the quickest answer for such things will be @anon. I don't have any idea why $SU(4)$ should be the double cover of $SO(6)$. I didn't think that $\text{Spin}(6) = SU(4)$.
That does happen to be true
it's one of those accidental iso's
I only deal with 3 dimensional real numbers.
who's anon, by the way?
@Danu @anon
My brain is saying "look at the Dynkin diagrams"
but uh
21:06
ArcticTern among others
not sure that's the best way
@Leaky: Did you figure out the answer to my ideal question?
oh, okay
\begin{align}
\frac{1}{e} & \\
\frac{1}{e}+\frac{2}{e^2} & = \frac{e+2}{e^2}\\
\frac{1}{e}+\frac{2}{e^2}+\frac{3}{e^3} & = \frac{e^2+2e+3}{e^3}\\
...
\end{align}
He's had about 10 names, @ALannister.
21:07
8 mins ago, by Leaky Nun
@TedShifrin intuitively, $a$ and $b$ are coprime if $a=hm \land b=hn \implies h=1$, i.e. the only ideal that contains $a$ and $b$ is the ring itself.
But never the Mother of Dragons.
but, the dynkin diagram corresponding to $\mathfrak{su}(4)$ is $A_3$ i.e. 3 dots in a row
I have another question @Ted, more linear algebra flavored
@Leaky, that's wrong. $\langle 2\rangle \cap \langle 3\rangle = \langle 6\rangle$.
What does the operator $\ominus$ mean, if $\ell$ is a line contained in a two-plane $D$, what does $D\ominus \ell$ mean?
21:08
@TedShifrin the ideal that contains $a$ and $b$, i.e. join, not intersection
how do I canonically find some complementary thing
(it's supposed to mean something like "the other line")
oohhh, sorry ... $\langle a \rangle + \langle b \rangle = \langle 1 \rangle$. Yes.
$\sum_{n=1}^m \frac{n}{e^n} = \frac{1}{e^m} \sum_{n=1}^m ne^{m-n}$
and the dynkin diagram corresponding to $\mathfrak{so}(6)$ is $D_3$
but the diagram for D3 is the same as for A3 (need at least n=4)
I've never in my life seen that symbol, @Danu.
21:10
:(
so $\mathfrak{su}(4)$ and $\mathfrak{so}(6)$ are isomorphic Lie algebras.
This paper I'm reading uses it without comment
Seriously, I have never seen it before.
...I'm not sure where I was hoping to end up, but uh
oh well
why not orthocomplement
21:11
@MikeMiller I'm confused cause I'm not sure I should be using the metric
That's because of some context which I will type out in a bit
(I'm getting a phone call)
I probably don't want to read that context
ok
I support using metrics against their will
lel
$(p\langle a\rangle,q\langle b\rangle) \mapsto (pb+qa)\langle ab\rangle$ right @TedShifrin
21:13
hmm, how Bezoutian
@Leaky: Usually you use the elements you use to get $1$ to give the isomorphism.
Maybe Ted wants to know, so here I go
This occurs in the description of some flag manifolds
@TedShifrin that makes sense
@TedShifrin Which dovetails nicely with Bezout's theorem, come to think of it
and one of them, $G_2/T^2$, is described as follows. Take the imaginary octonions, complexify and bilinearly extend the inner product (inherited from $\Bbb R^7$) and the cross product.
21:15
Ayup.
@Semiclassical even in other rings? @TedShifrin
Now consider an isotropic complex line $\ell$.
@Leaky: Google Bezout domains.
\begin{align}
1\\
e+2 & = (e+1)+1\\
e^2+2e+3 = (e^2+e+1)+(e+2) & = (e^2+e+1)+(e+1)+1\\
...\\
e^{m-1}+2e^{m-2}+\cdots + (m-1)e+m & = \sum_{n=1}^{m}\sum_{k=1}^ne^{n-k}
\end{align}
thanks @TedShifrin
@Secret you're overcomplicating everything. It's an arithmetic-geometric series.
21:16
Lemma: The subspace $\ell^a=\{x\in \operatorname{Im}\Bbb O\otimes\Bbb C\mid x\times \ell=0\}$ is three-dimensional and also isotropic
Then $G_2/T^2$ is the space of pairs $(\ell,D)$, where $\ell\subset D\subset \ell^a$
(I don't care really about proving this or anything)
I'm not following all this, @Danu, of course.
:'( OK
I should have learned way more about octonions than I ever did.
If one really wants to do $\sum_{n=1}^m n e^{-n}$ from scratch, consider what happens when you multiply it by $e-1$.
oh, octonions
21:17
Now the point that bothers me is the following
@Secret what Semi said.
"Take the imaginary octonions..." Waiter, I'll have that check please...
They say that $D\ominus \ell$ defines another (isotropic) line and that the map $(\ell,D)\mapsto D\ominus \ell$ defines a fibration with fiber $\Bbb CP^1$
So, I somehow need to obtain a line in a canonical way such that one such line is associated to a $\Bbb CP^1$'s worth of pairs
So $\ell$ is isotropic, which means it's orthogonal to itself with respect to the complexified inner product.
Yes
Exactly like the oriented 2-planes you taught me about
(the map $(\ell,D)\mapsto D\ominus \ell$ is a fibration over the quadric)
21:20
Yup.
What is special about $D$ here?
The cross product with $\ell$ vanishes, and $D$ is isotropic
(cross product also bilinearly extended, of course)
Let me ponder a bit.
So maybe something like $\ell=\langle e_1+ie_2\rangle$, $D=\ell\oplus\langle e_2+ie_1\rangle$ would be an example...
No.
I don't even see how to work with this :P
$(e-1)\sum_{n=1}^m n e^{-n} = \sum_{n=1}^m (e-1)n e^{-n} = \sum_{n=1}^m ne^{1-n}-ne^{-n}$ ....uh
how does that help?
Define $e_1=I,e_2=J,e_3=K,e_4=L,e_5=IL,e_6=JL,e_7=KL$
21:25
@Secret split into two sums
tweak with the indices
@TedShifrin so $\Bbb Z[x]$ isn't a PID, so although intuitively $2$ and $x$ are coprime, their join isn't $\langle 1\rangle$.
Then $\ell=\langle e_1+ie_2\rangle$, $D=\ell\oplus \langle e_4+ie_7\rangle$ works.
Correct, @Leaky.
Or write it out term by term prior to multiplying by $e-1$
@Danu: Is there an $\mathscr m$ so that $\ell$ and $\ell\times\mathscr m$ span $D$?
Because all basis vectors are isotropic, and $e_1\times e_4\times e_5=e_2\times e_7$
@TedShifrin In my example, doesn't seem so...
Since the only candidate would be $e_4$ for the first term but $ie_2e_4\neq ie_7$
21:28
They've got to define this somewhere. This is nuts.
it is fucking nuts
I hate this paper
Honestly, the whole paper is 0 explanation
Look in some references in the paper?
Most of the lemmas are proven in the following fashion: "To prove this, it suffices to check this special case (see [13]); that this suffices follows from [21]), so we obtain our claim"
Hence my suggestion to look at references :P
Regrettably none of the other papers really go in depth. They give the basic definitions needed for some of the early lemmas but nobody really cares about these flag manifolds or $G_2/T^2$ specifically so there is not octonionic description of it in the papers.
21:30
I wonder if Bryant uses that notation.
So the background is there but this $\ominus$ only appears for the one bit specifically about $G_2/T^2$ in the paper so this is not in the refs
it's a paper by Wood (and Svensson)
I've also found some slides of Svensson online, where once again he uses $\ominus$ without definition
Seriously?
Wood also has some things online but his stuff seems focused completely on some other parts so there's barely any mention of this part
The only thing I can think of having seen is symmetric difference of sets.
Yeah, that's all I could find too
21:32
I dunno. I can't help.
But I don't see how to apply it in this setting
Perhaps the metric after all, as Mike suggested
But I don't see how to see that ther's a CP^1's worth of pairs that define the same $D\ominus \ell$... Doesn't seem too obvious
But when you have isotropic things, you don't get the right orthogonal complement.
Oh damnit yes
Makes no sense
WTF
what else can one use?!
But you can still choose a complement. This is like normal forms for indefinite quadratic forms.
Of course you can choose a complement, but how to do it consistently for all pairs?
21:34
I have a bit of a silly question. If x is in a module M, then it's clear that xM is a subset of M, right?
Locally, there should be a consistent way to get the normal form for the quadratic form. Not globally.
I don't even know what it means, @User203940.
Modules don't have a multiplication.
Sorry for my ranting by the way
I've been stuck trying to read this stuff for too long :P
You didn't behave objectionably. I too would be angry.
On to bigger and better things... :P
Let me know if/when you find out.
21:39
Where can I learn about the cohomology groups of $BSO$? It seems to me that it will give me a clue about the claim that the Euler class of an oriented rank 3 bundle is the Bockstein of $w_2$.
The reason I believe it is this comment on MO.
I don't know this stuff. It might be in Mosher/Tangora, which I recommended to you. Or probably in Milnor's writings somewhere.
Also, ARE YOU SEEING THIS, Del Potro vs. Thiem?!
26 unforced errors, 0 winners
Hey there everyone!
Del Potro has gotta be injured, or deeply ashamed :P
I want to go watch. I've had guests for 4 days and have missed most of the tennis. I'm leaving chat to watch.
Hi/bye Demonark.
21:41
It's the worst match ever
Just wait for the next one :P
I'm sorry Shalapolov lost.
Yeah
He's the new hope!
Bye for now!
Plays like Federer used to
though I don't like his face, he looks a bit arrogant to me :P OK cya!
$\sum_{n=1}^m (ne^{1-n}-ne^{-n}) = \sum_{n=1}^m ne^{1-n} - \sum_{n=1}^m ne^{-n} = \sum_{k=0}^{m-1} (k+1)e^{-k} - \sum_{n=1}^m ne^{-n} = 1+ (m-2)\sum_{k=1}^{m-2}e^{-k}+me^{-m} = 1+ (m-2) \frac{(1-e^{-(m-3)})}{e(1-e^{-1})}+me^{-m}$
Some symmetry exists: All terms except the first and last term get subtracted and become a geometric series
21:45
right
So: $$\sum_{n=1}^m ne^{-n} = 1+\frac{1-e^{3-m}}{(e-1)^2}+me^{-m}$$
$$\sum_{n=1}^{\infty} ne^{-n} = \lim_{m\to \infty}1+\frac{1-e^{3-m}}{(e-1)^2}+me^{-m} = 1 + \frac{1}{(e-1)^2}\lim_{m\to \infty} (1-e^{3-m}) + \lim_{m\to \infty} me^{-m} = 1+\frac{1}{(e-1)^2}$$
check: If $m=1$, that becomes $$1+\frac{1-e^2}{(e-1)}^2+e^{-1}=1+\frac{1+e}{1-e}+\frac{1}{e}=\frac{2}{1-e} + \frac{1}{e} \approx -0.796$$
but it should just be $\sum_{n=1}^1 n e^{-n}= 1/e\approx 0.368$
So something's not quite right
Hmm.. let me check again the number of $\frac{1}{e^n}$ terms in the middle after the subtraction...
The way I'd look at it:
\begin{align}
(e-1)(e^{-1}+2e^{-2}+3e^{-3}+\cdots+me^{-m})\\
=1+(2-1)e^{-1}+&(3-2)e^{-2}+\\ &\cdots +(m-(m-1))e^{-m+1}+(-m)e^{-m}\\
=1+e^{-1}+e^{-2}+\cdots +e^{-m+1}-m e^{-m}
\end{align}
blah, need to do align
\begin{align}
& 1 + 2e^{-1} + 3e^{-2} + \cdots + me^{-(m-1)} + 0\\
-) & 0 + e^{-1} + 2e^{-2} + \cdots + (m-1)e^{-(m-1)} + me^{-m}\\
= & 1 +e^{-1} + e^{-2} + \cdots + e^{-(m-1)} - me^{-m}
\end{align}
21:55
which correctly returns $1$ when $m=1$, thank goodness
we still disagree about a minus sign at the end.
kk
so that should end up as $\frac{1}{1-1/e}-me^{-m}$
ah, got wrong on the minus sign in the past workings
it's pretty tedious.
nice thing about that $me^{-m}$ is that it obviously goes to 0 as $m\to \infty$, leaving only the first term
Hmm, one thing makes me curious is the shifting that is going on in the summand $ne^{-n}$, I wonder if there is some umbrella term that describe all series that has such symmetry...
Here's a more generic statement.
Suppose $A(x)=\sum_{n=0}^\infty a_n x^n$.
(If you only want finitely many of them to be nonzero, just take $a_n=0$ for sufficiently large $n$)

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