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12:00 AM
mm, presumably you mean a level set of this function. expressing it locally as a graph, in informal terms, means writing one 'variable' (subject to the constraint F = 0, or whatever condition you are using to define the cylinder) as a function of other 'variables.'
at least that's how i think of it.
 
yes I do mean level set
 
in particular examples you can do this with algebra, the theorems show conditions where you can do it even if nice algebra is not available.
i don't think anyone would need general facts of functions defining manifolds to analyze a cylinder, although i understand you took that as an example.
 
@copper.hat. By applied I mean something that is not too theoritical :/
 
Also the idea of being expressed as a "function" versus not being function. For this I refer to the expression $x^2 + y ^2 = 1$. So around the points $\pm 1$ we can't write this expression as a "function" of the other variable unless we restrict the domain. and in this example taking it to higher dimensions the same comes about to an extent. So since this isn't a "function" on the whole domain what do we call it? Aren't all hypersurfaces a function of their variables?
 
Is there anything wrong with answer here :/
-1
A: Can someone help me check my understanding of subspaces?

AvraThanks for your question. A subspace is made-up with a set of vectors (called bases) that have to be independent from each other, otherwise, they will be multiple from each other. Once you have a set of vectors making up a subspace, they should be independent from each other, that is if you take ...

That is what I remember from my linear algebra class :/
 
12:06 AM
@Avra you don't need bases to talk about subspaces and you also don't need an inner product. (Also orthogonal is not the same as linearly independent)
 
it introduces some stuff that is superfluous. that the things defining a subspace be linearly independent (not required, although the vectors in this problem are), and the dot product (always present in R^n but not always required for analysis).
 
Yes that is what I do understand about the manifold definition @leslietownes . So if we are not able to express this level set as a graph, how do we "express" the level set?
 
the function you get from the theorem is generally only locally defined and may depend on choices.
 
I understand what is needed to check to determine if it is or isn't a manifold. I'm just trying to wrap my head around the concept in relation to the hypersurface, whatever it may be. Because the hypersurface is described in terms of a function
 
@LukasHeger. Sure! I just got things mixed up then :/
I deleted my answer {:-|
 
12:09 AM
what do you want to do? is the ultimate task to verify whether a level set of a function is or is not a manifold? that is a local question.
 
what do you mean by "want to do"?
 
thanks lesile. Got it %100
 
Let me see if describing the picture in my head/ looking at the illustration from Ted's book helps: I got the hypersurface, then some tangent plane to that surface at the point $a$ where we are examining things.....In this neighbourhood I can express this "special set of points (i.e manifold)" as a $C^{1}$ function.

What's going on outside of this neighbourhood? Say I can't express any set of points outside of the neighbourhood as $C^{1}$ functions. How would I be able to describe these parts of my hypersurface?
I think I've articulated it the way I'm seeing the gaps in my understanding...
I'm gonna read the wiki article and see what happens
 
unit-length closed interval is an interval of length 1?
determines the smallest set of unit-length closed intervals that contains
all of the given points!
By sorting points in nonincreasing order, we can take smallest point and build unit interval [xi, xi+1] (unit interval). Following this, we can cover all points!
I don't know what does the previous even is trying to say
The solution as you see is straightforward, so what is the point behind it please?
How this is the smallest set that covers all points if we are basically taking UNIT intervals around each smallest point we pick from the set points? If nothing makes sense to you, I will exit the room :/
 
1:04 AM
dc3 i think it's fairly common for the functions you get from the implicit function theorem not to be globally defined and not to exhaust the points on a manifold. there are probably topological characteristics that distinguish graphs of functions of n variables from general n-manifolds
with most of the exciting stuff probably happening in how various patches of a thing fit together
 
1:21 AM
i'm about to mark my 10th year on this hellsite
 
 
1 hour later…
2:33 AM
i'm 6mo behind you.
yesterday my son returned :-(, my wife got in an accident on the way back (she's ok, car not so) :-( & my friend's son returned from is 1.7k mile tout of ca.
 
2:46 AM
ooof. i hope the car thing isn't too expensive.
 
2:59 AM
cars can be replaced :-)
 
when i first moved back here, my wife bought a car, and someone ran a red light and totaled it before i could even see the thing.
daughter came home and said "i told miss fuentes [a teacher] all about the dead heron." this dead bird she saw at the duck pond. she's been obsessed with it. "miss fuentes said it was sick." great. "am i sick?" not so great.
 
:-).
kids are better at dealing with reality than adults.
 
3:19 AM
my daughter is something of a celebrity at day care. kids mob her when we drop her off, and when we pick her up. i don't know why.
the teachers use her as an example, i think because they don't want anyone else breaking a limb there. "let's all walk slowly so we don't break our legs."
 
teaching fear seems to the thing nowadays.
 
my mother in law said she didn't like how my daughter was jumping on her bed in a recent instagram post. feel free to unfollow, mother in law. :)
she has to regain that muscle function somehow.
 
the problem with social media is that it is effectively permanent. in laws should know better...
 
3:35 AM
i just don't see how i can collect on any personal injury suit if the injury isn't on camera.
2
 
:-).
 
 
3 hours later…
6:16 AM
@Koro that's exactly the same proof pattern as rudin's proof of the single variable chain rule btw
 
which proof @shin?
the question you posted today?
 
yeah
 
no, i think.
i'll check again.
 
with regards to the question you asked about $\epsilon(i) \to 0$ as $i \to \infty$
you use the fact that if a fraction has a limit at infinity, it implies that given a side of the fraction there's an error bound on the other side at all $n$
 
yeah right! The proof is along the same lines :) @shin: Try to prove limit rules using these ideas. :)
 
6:20 AM
yeah i'm trying haha, it looks nicer than epsilon-delta proofs
 
I mean if $\lim_{x\to a}f(x)=L$ then we may note that $f(x)= L+o(x)$, where $o(x)\to 0$ as $x\to a$. etc.
:)
 
how long does drafts stay saved on the ask a question thing
also aren't both $|A^{-1}| = |A|^{-1}$ and $(A + B)^{-1} = A^{-1} + B^{-1}$ incorrect formulas
 
@Koro yeah exactly, but for fractions it actually gives you an error bound for both sides of the fraction, not just for the entire fraction
 
6:45 AM
@shin: to make your proof work, did you try Cauchy criteria?
 
huh no, i'll have to try that
i've been stuck on how i could find a bound for $\sum_{i = 1}^n b_i(L_1 + \epsilon(i)$)
 
That is, For any $m<n$, $a_m+a_{m+1}+...+a_n= (b_m+...+b_n)L_1+\sum_{i=m}^n \epsilon(i)$
 
ohhhhh very nice
 
By choosing large $m$, I think the desired result may be achieved.
 
niiiice
yeah i think that works
 
6:48 AM
But you'll have to be careful with that $\epsilon(i)$ term
But that's not a problem because by choosing $m$ large enough it can be made arbitrarily small.
 
hm. we'll need to prove that $\sum_{i=m}^n \epsilon(i)$ converges
 
hmm, I think it may not converge. Consider $\epsilon(i)=\frac 1i$
But I believe that there should be some way to fix it.
 
@Koro $\lim \limits_{n \to \infty} \sum_{i=1}^n b_i + \lim \limits_{n \to \infty} \sum_{i=1}^n \epsilon(i) = \frac{L_a}{L_1}$. suppose either limits diverge. then we get a contradiction
 
you should start indexing $i$ from $i=m$
 
the equality holds for all $i$!
right?
$\lim \limits_{n \to \infty} \sum_{i=1}^n b_i(L_1 + \epsilon(i)) = \lim \limits_{n \to \infty} L_1\sum_{i=1}^n b_i + \epsilon(i) = L_1 \left(\lim \limits_{n \to \infty} \sum_{i=1}^n b_i + \sum_{i=1}^n \epsilon(i) \right) \iff \lim \limits_{n \to \infty} \sum_{i=1}^n b_i + \lim \limits_{n \to \infty} \sum_{i=1}^n \epsilon(i) = \frac{L_a}{L_1}$
 
7:06 AM
the first equality seems wrong: $b_i$ should multiply $\epsilon (i)$ also?
 
oh, you are very right
 
so this equality is also wrong.
19 mins ago, by Koro
That is, For any $m<n$, $a_m+a_{m+1}+...+a_n= (b_m+...+b_n)L_1+\sum_{i=m}^n \epsilon(i)$
 
it's completely mistaken hehe my bad
 
So we have $\sum_{i=m}^n b_i \epsilon(i)$ in RHS instead of $\sum_{i=m}^n \epsilon (i)$
 
we can do instead:
$\lim \limits_{n \to \infty} \sum_{i=1}^n b_i(L_1 + \epsilon(i)) = \lim \limits_{n \to \infty} \sum_{i=1}^n b_iL_1 + b_i\epsilon(i) = \left(L_1\lim \limits_{n \to \infty} \sum_{i=1}^n b_i + \lim \limits_{n \to \infty}\sum_{i=1}^n b_i\epsilon(i) \right) \iff \lim \limits_{n \to \infty} \sum_{i=1}^n b_i + \frac{1}{L_1}\lim \limits_{n \to \infty}\sum_{i=1}^n b_i\epsilon(i) = \frac{L_a}{L_1}$
now suppose one of the two limits diverges, we get a contradiction
 
7:13 AM
You can't take the limit inside unless you know the limits in parentheses exist.
 
it has to exist because it is always equal to $L_a$, right? i may be wrong
 
no, I mean $\lim (x_n+y_n)=\lim x_n+\lim y_n$ if both limits on RHS exist. Right?
in this case, you don't know if $\lim_{n\to \infty} \sum_{i=1}^n b_i$ exists.
 
hm you're right
we still prove they exist the same way: suppose they don't. then it equals $\frac{L_a}{L_1}$ or $L_a$, depending on at what point you begin contradiction. But this is a contradiction: neither can diverge if they equal a constant
right?
 
@shin: $n+(1-n)=1$ and you may note that both sequences $x_n=n$ and $y_n=1-n$ diverge.
 
hm very true good example
 
7:20 AM
But I believe your proof can be fixed.
 
7:34 AM
@Koro
ok, we use: $\lim \limits_{n \to \infty} \sum_{i=1}^n b_i(L_1+\epsilon(i)) = L_a \iff \sum_{i=1}^n b_i(L_1 + \epsilon(i)) = L_a + \mu(n)$ with $\mu(n) \to 0$ as $n \to \infty$. We get: $L_1\sum_{i=1}^nb_i + \sum_{i=1}^n b_i \sum_{i=1}^n \epsilon(i)= L_a + \mu(n)$ so $\left( \sum_{i=1}^n b_i\right)(L_1+\sum_{i=1}^n \epsilon(i))= L_a$ and $\sum_{i=1}^n b_i = \frac{L_a + \mu(n)}{L_1+ \sum_{i=1}^n\epsilon(i)}$. The last form guarantees the limit exists whether or not $\sum \epsilon(i)$ converges.
 
meanwhile, I also got an idea to prove this.
 
show
 
We have $\frac {a_i}{\epsilon(i)+L_1}=b_i$ and since $\epsilon (n)\to 0$, there exists an M such that for all $i\ge M$ we have $|\epsilon (i)|\lt \frac {L_1}2$
Using this we have $\sum_{i=M}^n \frac {a_i}{\epsilon (i)+L_1}\ge \frac 2{3L_1} \sum_{i=M}^n b_i$
So by limit comparison, the result follows. But I think you wanted to avoid limit comparison.
@shintuku no, that's not correct. $\sum b_i \epsilon (i)\ne (\sum b_i)(\sum \epsilon (i)$
 
it's distributivity of summation, no?
 
It's not true @shin.
$(1.2+2.3+3.4)\ne (1+2+3)(2+3+4)$
 
7:46 AM
hehe
you're right
oh, right it's only true for $\sum \sum b_ie(i)$
thanks hehe
 
8:02 AM
@Koro i think i'll use your proof, it is way clearer than mine and actually works hehe
 
8:43 AM
But then you really don't have to introduce $\epsilon(n)$. You can conclude by using $\epsilon,\delta$ definition and limit comparison. :)
as both the methods (the one I stated above using $\epsilon (n)$) and limit comparison are more or less the same.
 
 
2 hours later…
10:35 AM
Is it possible to square the circle if we consider other geometries? Does this problem even make sense?
 
Does the fundamental class not have to be a generator?
 
I found an article which says it's possible in hyperbollic and elliptic geometry for certain radiuses and angles of square, and the construction of square and circle don't depend on each other
 
In general setting (compact connected orientable manifold)
 
11:24 AM
@robjohn I have been looking at the Dominated Convergence Theorem (DCT), as given here for example. If it is applicable, then for sufficiently small $h(n)$ (i.e. large $n$), the remainder, viewed as a sequence of functions $g_{h(n)}(u)$ could be replaced by the converging function. Since it is $o(h^m)$ (presumably uniformly in $u$), the converging function seems to be $g(u)=0$.
Although this is a constant, for the purpose of moving the remainder outside the integral it would maybe be more sensible to end up with a function only depending on $h$, if that would make any difference. Maybe you were alluding to the DCT.
 
12:09 PM
Hey guys! I just wanted to ask a quick question as a sanity check.

I have been asked to expand $e^{-x^2}$ in the neighborhood of x = 0 to three terms plus remainder (writing the remainder in Lagrange's form). A simple enough question, as you just have to calculate a bunch of derivatives.

From calculating the derivatives, it's clear the odd powers of x will vanish in the Taylor expansion. Thus, in order to get a Taylor polynomial with three terms we need to calculate up till the 4th derivative. However, since the 4th degree and 5 degree Taylor polynomial are equal, it made sense to me to c
However, the textbook claims that the error should be:
$$\frac{x^6}{3!}e^{-\theta^2 x^2}$$
I am 98% sure the textbook is wrong on this one, but just in case I'm being a numpty I wanted to get an outsider's opinion
 
12:40 PM
Some more food for thought, @robjohn.
If $\frac{g_{h(n)}(u)}{h^m}$ converges uniformly to $0$ (by the assumption that the Taylor expansion occurs uniformly in $u$), then so does the remainder $g_{h(n)}(u)$. Since by the definition of uniform convergence, given an $\epsilon$, there is an $N(\epsilon)$ that does not depend on $u$; that is, there is a function $y(h)$ such that $$\left |\frac{g_{h(n)}(u)}{h^m}\right|<y(h)<\epsilon.$$ $y(h)$ exists in order to solve $y(h)<\epsilon$ for $n$ in $h(n)$ to obtain $N(\epsilon)$. Multiplying the inequalities above by $h^m$ and attaching $<\epsilon$ on
Thus $g_{h(n)}(u)$ is bounded above by some function only depending on $h$. Since $g_{h(n)}(u)$ is unknown, I'm unsure though if it is bounded above by some function only depending on $u$ (denoted $g(x)$ in the link above). This is a requirement for the application of the DCT.
It is maybe clearer to avoid it, but I have been somewhat inconsistent with writing out that $h$ is a function of $n$, i.e. $h=h(n)$.
 
 
1 hour later…
2:12 PM
0
Q: Topology and Geometry, 1 simplex is 2 boundary

monoidaltransformconsider page 172-173 in Topology and Geometry by Glen Bredon. In particular, Lemma 3.1. The result is: If $f,g$ are paths in $X$ such that $f(1)=g(0)$ then the $1-chain$ $f\cdot g - f - g$ is a boundary. Here, $f\cdot g$ is the concatentation of the paths $f$ and $g$. He proves the result as fol...

 
 
1 hour later…
3:34 PM
 
3:52 PM
@DanielAdams $\land$ is the logical symbol for 'and'
$\lor$ is the logical symbol for 'or'
often in programming languages, & and && are used for bitwise and logical 'and', while | and || are used for bitwise and logical 'or'.
 
yet another gratuitous use of symbols from logic
i think i used wedge and vee the few times i had to use those symbols, as a form of protest against using them. i'm not sure if it's the same symbol.
 
11 Days till Halloween. I'll be waiting in the pumpkin patch like Linus, with my blanky, waiting for the Great (albeit Mean) Pumpkin! But I'm okay with a Mean Scare Crow, or a Mean Gourd. Hint... HINT... @robjohn ??
 
scary strawberries are the #1 most requested costume this year. bigger than 'squid game.'
 
4:09 PM
@leslietownes Wow. I don't know if @robjohn will want to appear here as a mean strawberry. As you can imagine, the Mean pumpkin/Gourd is a natural fit with his orange attire. I'll search for it. I'm quite sure I saved the image file of his Mean Pumkin costume.
 
why strawberry costume, why not dragon fruit costume? @amWhy
 
my daughter came into my bedroom this morning and i said hi, can i have a hug? she said "no." then she said "i like your dresser" and left.
koro this is an interesting idea. my daughter is obsessed with dragons.
 
dragon fruit
 
what if we just put a dragon mask on a not-that-great homemade strawberry costume
my daughter asks for 'dragon stories' every night. they're made up stories about a dragon who does whatever she did that day.
 
Aren't dragons kind of like unicorns, i.e., imaginary critters? I wonder if reincarnation of dinosaurs over the history led the human imagination to recall/imagine their existence, in the form of dragons and unicorns??
 
4:17 PM
or a pokemon costume? :)
 
i'm not sure she could wear a pokemon costume to her day care. they have a ban on commercial characters. it's part of some kind of anti-capitalist indoctrination.
 
Whatever happened to witches and pirates, and ... Great (Mean) Pumpkins? ;-)
 
hmm, I see so commercial characters are not allowed.
 
witches and pirates promote harmful gendered stereotypes.
 
@leslietownes If scary strawberry is so popular, you can bet it's likely a commercial character on some streaming network.
 
4:20 PM
she could wear a big orange box with a crow hat, and be a mean squarecrow.
 
then some fruit costume will be nice to promote healthy diet or something?
 
@leslietownes Fine, but not Great (and mean) Pumpkins!
 
despite the ban on commercial characters my daughter has learned most of the characters and plot of 'frozen' from an older girl at school.
 
5:08 PM
@LukasHeger The former one "should" be $\pi^{\mathrm e}$ and the later one "should" be $(\varpi^e) = (p)$.
 
i didn't get that at all. i must have a huge gap in my education.
 
@DavidChoi Use the T.P. and remainder for degree 2 for $e^x$ and plug in $-x^2$ for $x$.
 
 
1 hour later…
6:31 PM
@leslietownes There is a school in Seattle that cancelled its Pumpkin Parade and the costume day because they said it was rascist. It was claimed that "As a school with foundational beliefs around equity for our students and families, we are moving away from our traditional 'Pumpkin Parade' event and requesting that students do not come to school in costumes."
 
if a pat h $f$ is path homotopic to a boundary, then must it also be a boundary?
 
halloween does seem to give some people very bad ideas, but at least we can rest easy that no gourds were unduly caricatured.
 
They didn't even ask parents, or students, how they felt. They just decided that this would be "woke".
 
just another point in the column of social media making everyone dumber vs. rationality.
 
Around here, I don't think that anyone thinks Halloween is anything but fun. There are so many houses decorated.
 
6:42 PM
we have some good spots in our neighborhood.
 
7:17 PM
'Afternoon, guys. Can I get some feedback on how to improve this and also checks for accuracy? incongruous-yew-7c4.notion.site/…
 
@leslietownes Hah! Even Mean Gourds are "woke"!!
@robjohn I do not mean to tell you what to do... I just have enjoyed very much, as have a lot of others, your periodic "holiday spirit." Perhaps I should leave you to your own creative whims. Maybe it stops being fun fun for you, when requests are made. In any case, on this site, it is woke to have fun.
@leslietownes It's sad that the our local Human Society needs to send out warnings about keeping one's black cat safe and indoors, during this time of year.
 
nobody's getting livvy.
 
@leslietownes Nor Shai !!
 
7:39 PM
Our cats never go outside. Too many coyotes here.
 
@robjohn Actually, the average life span of more or less strictly indoor cats is almost 8 times the average lifespan of indoor cats that can come and go. Smart move, really, for kitties sakes'
 
i've seen some coyotes around here, but mostly closer to the airport across several freeways. never in our neighborhood.
livvy is never going to know what the outdoors are like. she's too busy sleeping on one of her multiple cat beds.
 
@leslietownes and some beds are heated, to boot! :-)
 
she loves her heated bed.
we have a heated bed, a bed shaped like a black cat head (you crawl into the mouth to sleep in it), a bed shaped like a cupcake, various items of upholstered furniture, and our daughter's bed. some days she sleeps in all of them, one after the other,
it's good to be a domesticated cat
 
8:04 PM
@leslietownes At first, given your use of "we", I was having a hard time imagining that your and your wife have so many beds! ;D
@leslietownes Indeed!
 
hah. i did mean cat beds. if i had a heated bed, i'd never get up.
 
@leslietownes Tell me about it. I spend so much time at my desk, that Shai owns most of the upholstered furniture. One he reserves for our interactive play. He catches toys, and bats them, very well. The furniture, when he's being "spunky" is his playground. Else his beds, except for his beds by the heat registers.
But anyone in CA doesn't need a heated anything. ;P Perhaps in Northern CA. But then again, look at Texas, last winter.
 
a lot of freezing goes on in our house, mostly via stares.
 
8:33 PM
Heh here are soooo quiet
 
That tends to be a good thing
I think I've removed most of the mistakes and inaccuracies from what I've written down so far, but it isn't quite "there" yet from what I'd like.
Still need to figure out how to get rid of $2^{\log_2(|m|)}$ in umod2pN.2.false.b.
 
9:18 PM
@monoidaltransform what does it mean for a path to be a boundary
 
probably that the singular chain asscociated to the path is $0$ in $H_1$
if that's the intended interpretation, then the answer is yes by Hurewicz
 
ahh a boundary in that sense, I was thinking topologically
yes then that's the answer
 
simpler than that, it's just that homotopic => homologous
 
You need path-homotopic to know it’s a cycle.
 
my homotopies between paths are always path-homotopies, but yes, sloppy language on my part
 
9:33 PM
Not criticizing, just commenting for monoidal.
 
9:57 PM
@LukasHeger was machst du jetzt?
 
writing up how to fix an error in SGA3
 
@LukasHeger have you heard of "(er) kömmt"? it's apparently dialectal / archaic
 
10:12 PM
yeah I heard it
iirc Marx used it
"Die Philosophen haben die Welt nur verschieden interpretiert; es kömmt drauf an, sie zu verändern."
 
10:32 PM
\sum_{\ell=1}^{u-1} b_{u+1} how do I convert this into dot notation?
 
10:54 PM
@PM2Ring got a SageMathCell question for you: do you have any idea how to make it do insert instead of overtype when you click into the script? i keep getting little things wrong and have to delete the entire line
(actually, maybe a more pertinent question: where tf is Insert on my laptop keyboard)
oh, there it is
figured it out, disregard
 
vello what is the type of b?
 
@leslietownes b doesn't have a type
I just need to write it in dot notation
 
i mean
 
but what is b, as a data structure? does it have entries and if so how many? do you want \ell to feature in the sum somehow? at the moment it's just (u-1) b_{u+1} because all of the terms being added are the same
 
as written, all you've got is a sum
and yeah, just adding the same element b_{u+1} over and over
 
11:10 PM
sum ell=1..u-1 b_{ell+1} would be a sum of the u-1 terms b_2 + b_3 + ... + b_u; if you wanna write this as a dot product it may help to know what b is, so we know what we can or cannot dot it with
 
@leslietownes there is no data type for b \ell is just a letter "l" in latex
 
this feels like a "what does the color red taste like" question to me
inasmuch as it doesn't seem well-formed
 
if we don't know where b lives it's hard to be sure what we can dot it with. maybe you're asking for a dot product of whatever b represents with a sequence of u-1 copies of 1 in it and zeros elsewhere?
i can think of reasons why doing this might be desirable in programming, if you have primitives for dot product type operations and you don't want to reinvent the wheel, but if ell and u are changing it might not be faster than just computing the sum some other way
 
it has nothing to do with programming, I have to convert the summation into dot product. I don't understand how to do it since the index of the summation mentions upper bound instead of the lower bound
 
do you have the original problem?
 
11:18 PM
the thing we're wrestling with is that the question seems to presuppose a thing called b that has entries, but we don't know how many entries, so we don't know where it lives
in a world where (1,0) dot (1,0) and (1,0,0) dot (1,0,0) are different dot products (even if they result in the same thing), this is a problem
 
right now, the problem would literally just be finding another way to represent "b_{u+1} + b_{u+1} + b_{u+1} + ... + b_{u+1}"
which seems dubious
 
i'd look at the definition of b and pin down exactly what about it we want to sum
 
@Semiclassical yes exactly I need to represent it in another way
 
that's a very strange problem to be given
for one, the index seems to be doing exactly nothing
 
I know but I need it
 
11:22 PM
the simplest way i can think of (aside from something correct but silly) is to define two vectors. first one is B=(b,b,b,...,b), second is U=(1,1,1,...,1)
and then B.U is just the sum of the elements in B
it's a bit silly here, though, b/c you can just write B=bU
 
b_{u+1} + b_{u+1} + b_{u+1} + ... + b_{u+1} is it the other way of writing it? why did the index stay the same?
 
this is where i was wondering if maybe ell+1 was intended in the index
 
because your index of summation is ell, not u
 
looks a little more 'dot-producty' that way
 
ell is the lower bound
 
11:26 PM
that's not the point
 
index is u+1
but index doesn't contain the lower bound
 
i can't make sense of what you're saying. \sum_{k=1}^n b_k = b_1+b_2+b_3+...+b_n
 
that is why, I am confused
 
but \sum_{k=1}^n b_n = b_n + b_n+b_n+...+b_n
the index on b is irrelevant if it's not the index being summed
 
it is going to be summed until infinity?
 
11:28 PM
no
the upper limit is n
summation of k from k=1 to k=n
 
the number of times of the difference between upper and lower then?
I understand how summation works but I don't understand the specific case which I mentioned above
 
if you don't understand that "\sum_{\ell=1}^{u+1} b_{u+1} = b_{u+1} + ...+ b_{u+1}" then i'm a bit dubious about said understanding
if the summand was b_\ell, then you'd get "\sum_{\ell=1}^{u+1} b_{\ell}=b_1+b_2+\cdots+b_{u+1}"
but you insist that it's not that, so all you have is b_{u+1} taken u+1 times. nothing more or less
 
@Semiclassical I see what you mean
@Semiclassical okay got it
in my case it is b_u-1 so is it going to be = b_{u-1} + ...+ b_{u-1}
 
is it b_{u-1} or (b_u)-1
 
actually it is \sum_{l=1}^{u-1} b_{u+1}
b_{u+1}
 
11:38 PM
okay
so you'll have u-1 copies of b_{u+1}
 
but the upper bound is {u-1}
 
why not u-2 ?
 
take u=5. then you've u-1 = 4 and thus ell=1,2,3,4
so that's u-1 = 4 terms
 
got it
Okay, I think I got how to solve it. Thanks a lot!
 

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