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1:19 AM
@AMDG $$\sum_{n=1}^k\frac12k(k-1)=\frac12k^2(k-1)$$ and $$\sum_{n=1}^k\frac12k\left(k^2+1\right)=\frac12k^2\left(k^2+1\right)$$
@robjohn Thank you, but I meant the finite sums whose closed forms are these expression.
Good evening
I am wondering why LU decomposition is useful.
@AMDG Just take the difference.
@robjohn How do you mean?
@Bob For (computer) computational reasons.
1:22 AM
$$\frac12k(k-1)-\frac12(k-1)(k-2)=k-1$$ so the sum of $k-1$ is $\frac12k(k-1)$.
is LU decomposition used when you have multiply systems of linear equations and the only thing that is different is the constants?
@robjohn As in $\sum_{n=1}^k (n - 1)$ or just $k - 1$ itself?
Multiple … yes, that certainly can be useful.
@TedShifrin That makes sense to me. Thanks. I hope you have been well.
Because by sum I mean solving $\sum_{n=1}^k f(n) = \frac 1 2 k (k - 1)$
1:27 AM
The point is that back-substitution and forward-substitution needed to work with systems are very computationally efficient.
@AMDG my examples are for the first one. The second one is just division by $k$
You solve $Ly=b$ and then $Ux=y$ very efficiently.
@AMDG So my computation shows that $f(n)=n-1$ works
@robjohn Ah, nice. Also, looking at the equation I put forward, what you said makes sense now.
@TedShifrin That makes sense
1:29 AM
Sometimes things I say may accidentally make sense :)
@TedShifrin I find the things you say normally make sense
I believe you were an outstanding math professor
Aw, thanks. ;)
@robjohn Alright but where did the factor of $(k - 2)$ come from?
good night
1:36 AM
$\Gamma$ is the coolest greek letter
Actually now I'm just wondering where $\frac 1 2 (k - 1) (k - 2)$ comes from at all
i'm here doing proofs, being like, let $\Gamma$, i.e. Death's soul-reaping scythe, be a set of propositions
And the Russians get to use it all the time
sans serif even
ah yes
sans serif it is a brutalist appartment block in some exsoviet city center
This is probably readable to most people in here, at worst by easily guessing: Гамма Функция
1:39 AM
brutalist appartment block-amma functsya
Ты не смешно smh)
lol anyways it's literally just Gamma Function in Russian.
To top it off, hand-written Le looks like lambda and typed Le looks too much like $\pi$.
Hello math chat. Does anyone know, offhand, if it's possible to encode the statement "these two circles don't intersect" in the roots of a polynomial? I've been breaking my mind on this idea and I'm pretty sure the answer is no, but who knows (I was thinking of trying to make it so that the distance between the circles is greater than the sum of their individual radii, but that's an inequality so doesn't have roots)
@Rithaniel Why not? Look at $f^2+g^2$ ?
@Rithaniel What, you mean like defining the alphabet as $[1, 26]$ and then just $\prod_{n=1}^{m} (x - f(n, m))$ and $m =$ non-whitespace character count?
@TedShifrin Yeah, I was looking at something like $(x-u)^2+(y-v)^2-r_1^2-r_2^2-n^2$ with a second polynomial $nm-1$ to prevent $n$ from being zero, but this encounters issues because it has complex solutions. What I really need is some way of separating the real components of the circles, because complex circles always intersect
1:53 AM
No, take one polynomial $f$ which equals $0$ on the first circle and the other polynomial $g$ and do what I said.
Ah, okay, so something like, for a circle with radius $r$ centered at $(u,v)$, you get $f(x,y)=(x-u)^2+(y-v)^2-r^2$, and then $g(x,y)=(x-t)^2+(y-s)^2-q^2$?
Then, if $f^2+g^2$ has a root at $(x,y)$, then $(x,y)$ is on both circles. But how do I write a polynomial such that it is only zero if the circles $(u,v,r)$ and $(s,t,q)$ have no real overlap?
That’s impossible. Open/closed nonsense.
Alright, then yeah, that's what I was afraid of
Oh well, I just can't do the trick I wanted to do, then
2:12 AM
@AMDG $\color{#C00}{f(k)}-\color{#090}{f(k-1)}=\color{#C00}{\frac12k(k-1)}-\color{#090}{\frac12(k-1)(k-2)}$
Alright, but why?
Can you add that up?
Look up telescoping series for some inspiration
I'm starting to think I should have taken a break many hours ago.
Then again, I never know when to stop
Thanks, but I think my brain is thoroughly fried for now. I'll have to look at that later.
Something tells me the closed form I'm after for this simple factoring of numbers is going to involve a transcendental function, but then again...
If you take the logarithm modulo 1 over the rationals, we get the "fraction" part of the logarithm. All integers on the line $y = \ln x - \lfloor\ln x\rfloor$ share the same factors, yet differ by factors of $e$ linearly, and by reason of its repeating, I would expect a solution of some kind through the complexes, but I'm not sure how best to model that through the complexes, especially since I've not really worked much with them.
2:34 AM
why not take a break and cure your platonism with a good SEP article plato.stanford.edu/entries/wittgenstein-mathematics
If I try getting the same sort of thing using the complex exponential itself, the results are questionable to me because whether I use cosine or sine to line up the peaks with powers of two, I'm not sure how the other part, imaginary or real respectively, should look.
> that the only genuine propositions that we can use to make assertions about reality are contingent (‘empirical’) propositions, which are true if they agree with reality and false otherwise
> truth is conformity of subject to object
There is nothing new under the sun
Modern philosophy is literally reinventing the wheel, except it isn't a wheel, it's something else.
truth != true propositions
Eh, I'm not getting into another one of these discussions with you quite frankly. Sorry.
That being said... what would be a canonical representation of any arbitrary periodic waveform that is discontinuous?
only a true stallion has the stamina for philosophy
Like I suppose I could just do something like solve $e^{i\pi f(t)} = x + i\ln x - i\lfloor\ln x\rfloor$ but like... would that be a sort of canonical or natural representation of it from the complexes?
Some set of spirals perhaps that uniquely passes through all the points on the line $y = \ln x - \lfloor\ln x\rfloor$ as a surface existing in Euclidean space $\Bbb R_x \cdot \Bbb R_y \cdot \Bbb C\setminus \Bbb R$ and whose intersection with the real x-y plane is the solution set.
If I just go to the linearly interpolated version, it simplifies to a set of cones.
Well I mean more that in such case, the slopes reduce to constants.
3:33 AM
I'm a bit confused why yesterday it was said that thinking about monotonic subsequences of rationals is the wrong intuition. It is totally an easy way to see the convergence visually and there seems to be nothing wrong with solving the problem that way :P
You are misstating yesterday’s discussion.
1 hour later…
4:54 AM
do i agree with ted? i think i do. (more ducks at the pool today. i even got to swim with a duck.)
@leslietownes Wow!
1 hour later…
6:23 AM
leslie’s duck dynasty
6:42 AM
In Hatcher's theorem 1.8, why is $[f_r] =0$ for all $r\ge 0$?
because the family {f_r} is a homotopy of paths? look at proposition 1.2
7:25 AM
Yes, thank you so much. :-)
7:55 AM
I'm trying to show that the integral from a to b of a function f is the same as the integral from a+c to b+c of the function f(x-c). In summary, I've tried to argue in the following way:

Say the integral is A. Let \epsilon>0 be given. Since f is integrable over [a,b], there is an arbitrary partition P_1 with mesh size less than \delta>0 such that the approximating Riemann sum S_1(P_1, f(x-c)) is within \epsilon of A. Then I defined a partition P_2 by adding c to each of the elements of P_1 and showed that the approximating sum S_2(P_2, f(x-c)) would be equal to S_1(P_1, f) and hence it wou
8:06 AM
P_1 isn't arbitrary; it depends on epsilon, and comes either from your definition of riemann integrability or a theorem about it. once epsilon is fixed, you're definitely not reasoning over the set of all partitions of [a,b] here (although you could certainly establish a one-to-one correspondence between the set of all partitions of [a,b] and [a+c,b+c] and maybe even approach the problem that way).
if your'e trying to say, well, there's some mesh size such that all partitions of [a+c,b+c] with some mesh condition do blah blah, yes, you're leaving something out, or not phrasing it in a way that makes it clear that's what you're doing.
but the idea is the right one.
books vary in precisely how they define riemann integrability and the value of the riemann integral, but you ought to be able to get it out of what you're noticing. given any partition P of [a,b], you can write down a corresponding partition P' of [a+c,b+c] with the exact same mesh size, and the riemann sum of f with respect to P is the same as the riemann sum of the translated f with respect to P'. and, maybe crucially and maybe not depending on your definitions, vice versa.
so, exactly the same sets of real numbers that end up being riemann sums, and exactly the same behavior of elements of these sets under refinement of partitions.
@leslietownes So would saying that "given any partition P of [a,b], we can write down a corresponding partition P' of [a+c,b+c] with the exact same mesh size" be enough before I show that the sums are the same or should I state this another way?
Can anyone please validate the above solution? I think the line marked with a green ink in the 2nd picture, is an incorrect simplification from the above expression...
2 hours later…
10:32 AM
Found this on one of my favorite blog
The author solved the problem using graph-theoretical methods
It's shocking that there $2n+1$ many different colors for any given $n$. I thought
there're only $7$ :)
11:32 AM
Weierstrass gap theorem: Let $M$ be a compact Riemann surface of positive genus $g$, and let $P\in M$ be arbitrary. Then there are precisely $g$ integers $1=n_1<n_2<\cdots<n_g<2g$ such that there does not exist a meromorphic function $f$ on $M$ which is holomorphic on $M\setminus\{P\}$ with a pole of order $n_j$ at $P$. This shows Mittag-Leffler theorem cannot be generalized to compact Riemann surfaces?
11:59 AM
I don’t understand
the proof that $\pi_1 (S^1)= \mathbb Z$
I've never heard of this gap theorem, but the right generalization of Mittag-Leffler to compact Riemann surfaces is the Riemann-Roch theorem
Get used to it first, understanding comes later.
@Thorgott Maybe? But this gap theorem shows it cannot be generalized fully.
depends on what "generalization" means to you
@user2236 Those are the synonyms
12:06 PM
In particular, the following part: Let $p: R\to S^1$ be the covering map $s\mapsto (\cos 2\pi s, \sin 2\pi s)$. Consider the map $f: \mathbb Z\to \pi_1(S^1, (1,0))$ defined as $f(n)= [w_n], w_n:I\to S^1: s\to (\cos 2\pi ns, \sin 2\pi ns)$. I want to show that f is a homomorphism.
the conclusion is simply not true in general, but the Riemann-Roch theorem precisely measures the failure for the conclusion to hold
I mean the literal generalization. existence of meromorphic function with given poles and orders. That gap theorem shows the generalization is infinitely false.
This is done by the following step which I don’t understand: $f(n+m)=[w_{n+m}]=[p\circ (w_m’t_m w_n’)]= [w_m . w_n]$, where $w_m’(s)=ms$ is a path in R, $t_m(s)=s+m$
I don’t understand the last equality.
in fact, I don’t even understand how $w_m.w_n$ is even defined. It seems to me that it makes no sense (it’s not concatenation nor composition).
Then, we somehow have $[w_m.w_n]=[w_m][w_n]$.
Oh it’s concatenation and it makes sense because w_n’s are loops based at (1,0)
Still, I don’t understand the third equality above. What happens to the covering map?
It seems that Hatcher forgot to put this proof in the book.
Because Hatcher proves only surjectivity and injectivity of f.
Proof for lifting of paths is so complex.
12:35 PM
@Thorgott But I don't think it measures precisely. It's inequality, not equality.
I mean not the theorem itself but what it measures
12:47 PM
@onepotatotwopotato true that
12:57 PM
The Riemann-Roch theorem is an equality. I think Riemann first proved the weaker inequality and then Roch provided the correction term, but theres a 50/50 chance I've mixed up their roles in the history.
You are correct, since Roch was Riemann's student.
@Koro He remarks on it on p.29. There's no proof as there isn't really a need for one. It's a consequence of the discussion of reparametrization/associativity that precedes that section.
I'm more confused by what $w_m^{\prime}t_mw_n^{\prime}$ is supposed to be
@anak yeay
inequality I mean the inequality appears in the vector space $L(\mathcal{U}) = \{f\in \mathcal{K}(M):(f)\geq \mathcal{U}\}$ for divisor $\mathcal{U}$. The dimension Riemann-Roch measure is $L(\mathcal{U})$.
@Thorgott ok.I don’t understand how :(. $w_n’$ is a lifting of path $w_n$ and the product is composition of maps hence giving a path in R from 0 to m+n.
@leslietownes thinking more about it, f_r are loops based at 1 but how is it known that f_r is path homotopic to f_0?
If this is known, then the question is answered in toto.
He studied under Riemann for a while, but switched universities before his dissertation, apparently.
I also just realized both Riemann and Roch died of tuberculosis.
At least according to Wiki.
@Koro so $t_m$ is a path from $m$ to $m+1$? surely not what you want.
1:12 PM
Can TB be cured now?
@Thorgott $t_m$ translates inputs by m.
$t_m: R\to R: s\to s+m$
I know, but you're not writing what you mean to write
@Koro Usually, yes. With a very long course of antibiotics---I think you have to take the drugs for 6 months?
1:28 PM
Time, patience, and effort
Given an arbitrary function $f: \mathbb R \to \mathbb R$ with graph $F$, can we always find a set of complex numbers $C$ and a function $g: C \to \mathbb C$ such that the image of $g$ equals the graph $F$?
Should be retitled "my review of this algebra book".
yes shin
cool, thanks Thorgott. i guess also it would be more accurate to say graph of the image of $g$, right?
No, that would be wrong
1:36 PM
hm ok thanks. i'm just thinking, the image of $g$ is a set of complex numbers, while the graph $F$ is a set of ordered pairs of real numbers
The "equals" in your original statement should be understood as "is the same under the bijection $x + iy \leftrightarrow (x,y)$"
ah, there you go. thanks anak
From that, it should be straightforward to come up with a proof of your statement.
1 hour later…
2:46 PM
I wonder why $S^n-\{p\}$ is contractible, p is a point on the sphere.
Can someone help me understand this notation? $\mathbb Z \times \{0\}$
Is it just $\Bbb Z$ embedded into the real line?
3:03 PM
@Koro the space is homeomorphic to a certain other space that you're well aware is contractible
try n=1 and n=2
$\mathbb Z \times \{0\}$ is the cartesian product of $\Bbb Z$ with $\{0\}$ therefore it's expressing just a bunch of points on a line. You get points like $(-2,0) , (-1,0), (0,0),(1,0), (2,0)$
So therefore, $S=\Bbb R^2 -(\mathbb Z \times \{0\})$ is just taking $\Bbb R^2$ and deleting these points. The result is a plane with countable punctures.
3:50 PM
i strongly urge people not to wrap their heads around things, the rigid structure of the skull would make this an excruciatingly painful experience
4:16 PM
There is the corpus collosum where one could insert something with a minimum of pain
is there an analytic definition of a pivot column or row?
something like $\vec x$ is a pivot column/row if and only if $\vec x = ?$
A good [question] (https://math.stackexchange.com/q/4669953/977780)
from topology but going to closed soon, I guess.
Looks like someone’s homework with zero effort.
@shintuku No. it’s all a matter of context in the matrix.
dang, alright thanks
4:35 PM
Can anyone say, what they mean by the term linearly independent in the phrase " the two solutions of this differential equation is linearly independent" ?
I found it's usage often, but never got to know the real meaning
It’s a linear algebra term. Neither is a constant multiple of the other.
I think it means two solutions of a differential equation, say $y_1(x)$ and $y_2(x)$ are linearly independent if $y_1(x)$ do not depend upon $y_2(x)$...
@TedShifrin So what's the meaning ?
It means they’re really different solutions giving you different information.
it just means $c_1y_1(x) + c_2y_2(x) = 0$ iff $c_i = 0$, so there is no way to construct $y_1$ and $y_2$ from one another through addition and multiplication
@shintuku Ohh...ok, with this meaning in my mind, let's see if the thing that I stumble upon makes sense...
@shintuku You mean $c_1$ and $c_2$ are arbitary constants, right ?
4:50 PM
let $c_1$ and $c_2$ be arbitrary given constants. then the above follows
@Franklin Just think of it as I told you. Neither is a constant multiple of the other.
The functions $\sin 2x$ and $\sin x\cos x$ are NOT linearly independent. The functions $xe^x$ and $x^2$ are linearly independent.
$\text{neither}=k\text{(the other)}$
@robjohn I see you've preemptively adopted an Easter icon. And amWhy isn't even here haranguing you!
@TedShifrin I changed from the St Patrick's Day clover to the Easter Egg right after St Paddy's Day. I added the avatar to the animation of one of my answers.
5:07 PM
@Thorgott For n=1, we have R. Thanks :)
@robjohn ROFL. applause
For n=2, we should have plane (R^2) if there is any justice left in this world.
I wrote a custom ColorFunction that renders the image.
What's more, in any dimension, you can write down the "obvious" homeomorphism.
@robjohn You clearly have too much time on your hands!
@TedShifrin It seemed as if it would be a useful thing to have.
5:13 PM
@TedShifrin I think you mean stereographic projections.
That makes sense :).
So there were 3 questions in a quiz I took recently. The time duration was 30 min. 1st question was to calculate homology groups of $\Delta^n$ with all faces of same dimension identified.
The chain complex will be $…Z\to Z\to…$. And ith homology group will depend upon i even or odd.
@Koro Stereographic projections preserve angles (at least on $S^2$).
I have not investigated how that generalizes to higher dimensions
Since you measure angles in (tangent) planes, it generalizes just fine.
I forgot to write that $H_i (\Delta^n)=0$ for all i >n. This is true because there are no i cells (i>n) in the simplex.
I lost some marks because of that.
5:19 PM
That's an interesting question, @Koro. Torsion depends on dimension.
Offhand it's not obvious to me that the result is independent of how one identifies the faces (i.e., orientations).
The second question was to find relative homology groups $H_n(X,A)$ where X=$S^1\times S^1$ and A is a finite subset of X. I solved this one but forgot to consider the case when A is empty set. So I lost some marks here also.
@TedShifrin it doesn’t matter I think. I thought of delta 2 (triangle) for example.
$C_n=$ free Abelian group generated by n dim faces. Since all faces of same dim are identified, this is just Z.
I was thinking about that. It seems to me that $\partial_2$ can be either $\times 3$ or $\times 1$.
The third question was to prove that $R^n - $ finite points n>2 is simply connected.
I actually don’t know Van Kampen theorem so I skipped this one so got 0 for this.
Of course you were supposed to know Van Kampen.
I don't think the simplex problem is well-defined. If I take the disk with two boundary semicircles, the way I identify the two semicircles changes the surface.
I am thinking about your $\partial_2$ comment.
5:29 PM
We've actually talked about this before. You get either the projective plane or the sphere.
Paging @Thor for an official ruling.
So I think that image of partial 2 is homomorphic to Z.
It’s difficult to type in an iPad.
It's either $\Bbb Z$ (in which case $H_1=0$) or $3\Bbb Z$ (in which case $H_1 = \Bbb Z/3$). So it matters.
It's quite possible I'm missing something here, but I don't see it.
Because $\partial_2: C_2\to C_1$. $\partial_2 ($face)= boundary of the face= one edge.
No, you have to look at the sum of the three oriented $1$-simplices of the $2$-simplex.
Yes, they are identified (either forwards or backwards).
yeah, they are identified so we don’t have to worry about that I think.
5:36 PM
I disagree. Identification requires saying how you identify.
Reread my comments about sphere/projective plane and think.
Oh I see.
Apparently, some order is tacitly assumed when identifications in a simplex are spoken of.
Although, I’m not aware of such conventions.
Tacitly? The "obvious" tacit assumption would say that I get $3$ for the boundary map.
I'll wait for @Thor to comment, but I think I'm right about the ambiguities here.
Does this appear as an exercise in Hatcher? If so, no one's ever asked me about it before.
Let me check.
Re our discussion, see exercise 2 on p. 131.
Yes, it is there.
Exercise 2.1.9
5:42 PM
@TedShifrin Ok, that's a real simpler explanation. Actually, let me elaborate the issue as it might make it more understandable : I was studying about the "Method of Frobenius" and there while using that method, we essentially, obtain an indicial equation and get two roots corresponding to which, we get two solutions. The point is, that the book says, the power series solutions so obtained might not be linearly independent. But then I encountered a lemma :
@TedShifrin yes, I understood your point.
Yeah, I found it too, Koro. I stand by my confusion.
Now, I was working with this problem:
The solution given was:
This is toooooo much. What's the point?
I already hate this stuff.
@TedShifrin I am sorry! Sorry! I didn't see this
5:46 PM
We don't want to read pages of this crap.
And yes. This ends this
@TedShifrin 😂😂 I wish, I could say this as well...
I taught this stuff back in 1974, and I have never thought about it since.
I don't see what the issue with linear independence is.
@TedShifrin To be really short and frank with these crap is : I didnot understand why they did not write the general solution $y=C_1y_1(x)+C_2y_2(x)$ and wrote instead $y= C_1y_{11}(x) + C_2y_{21}(x)$. Is this because, they essentially wanted the two solutions to be linearly independent ? Is $y_1(x)$ and $y_2(x)$ not linearly independent? If so, then why ?
That was the whole point of posting these long pages of the solution
the rest of the indexes might be functions that are linearly dependent?
@TedShifrin Wow!!!
5:49 PM
They're talking about arriving at the solutions in two different ways, I think, so they're referring to different possible pairs of functions.
Yes, they're talking about $y_1$ and $y_2$ as two different general solutions that you get by using different roots. Each is a combination of two functions. This is not consistent with your use of notation (or theirs in general, I suspect).
I would say that their notation is not good.
@shintuku So they want to express the solution in this form : $y=C_1y_1_r(x)+C_2y_2_r(x)$ where $y_1_r,y_2_r$ are linearly independent?
@TedShifrin Cant agree more, really!!!
@TedShifrin But how do they know, that $y_1(x)$ and $y_2(x)$ are not not linearly independent?
They are both turning out to be the general solution of the same problem. This is NOT the usual use of the notation. I just explained that.
You have to throw out your usual use of that notation.
I'm not going to keep saying it.
@TedShifrin Which notation?
I think I have studied from this book.
In 2015 or 16
@Koro Clearly I get different results if I wrap the boundary of the $2$-cell once around the $1$-cell and if I wrap it three times. In the first case, it's just a usual disk. In the second case, it's a non-manifold.
5:55 PM
Time flies.
@Koro it's from SL Ross
You could try E Coddington’s books also.
@TedShifrin This is true. Yes, I now get, what you meant by the notation. But these are not my concern. @shintuku My main concern is: I think the solution they denoted by $y_1(x)$ and $y_2(x)$ , well, they didn't write the general solution of the given differential equation as
$y=C_1y_1(x)+C_2y_2(x)$ because my mind says, that $y_1(x)$ and $y_2(x)$ are not linearly dependent and they want to express the general solution in terms of linearly independent solutions. But my question is, how did they know, that they are not linearly independent ?
Van Kampen is from Holland.
@Franklin they probably checked/verified this fact without informing you.
Holland = Netherlands
@Koro That's probably, the only reason, I can think of and that might be true as well :)
6:04 PM
@Franklin what are $y_1$ and $y_2$? Maybe I missed something.
@robjohn They are the solutions of the given differential equation. It's denoted like that in the long pictures I posted above.
So, they are given as independent. Is your question how they know that there are two linearly independent functions that satisfy the DE?
I believe I explained it above, but Franklin is not taking the time to understand what I wrote.
Van Kampen died at the age of only 33. :(
@Koro But then that they finally express $y= C_1y_{11}(x) + C_2y_{21}(x)$. And that as I take it, $y_{11},y_{21}$ are linearly independent: This fact, might also have been used and verified previously without informing me ?
@robjohn No, that's the point. They are not given as independent...
@robjohm they are writing $y= C_1y_{11}(x) + C_2y_{21}(x)$, and not $y=C_1y_1(x)+C_2y_2(x)$ because they want the symbolic coefficient of $C_1,C_2$ to be linearly independent ? I dont get why $y_1(x)$ and $y_2(x)$ are not linearly independent and $y_{11}(x)$ and $y_{21}(x)$ linearly independent ?
6:12 PM
@Franklin Look at the expansions of $y_1$ and $y_2$. They are not linearly dependent.
@robjohn You truly mean linearly dependent in your comment, right ? If this is so, then I completely agree with you. My point is : $y_1(x)$ and $y_2(x)$ are linearly independent as well . And we can write $y=C_1y_1(x)+C_2y_2(x)$...
"not linearly dependent" and "linearly independent" are pretty close
I think they are just using different pairs of functions to write the general solutions. Nevertheless $y_1(x)$ and $y_2(x)$ and ($y_{11}(x)$ and $y_{21}(x)$) these all pairs are linearly independent among them , right ? @robjohn
I feel the mathmeatical definition is lacking
of arc length
It appears so
6:17 PM
because we only approximate arc length from below not above unlike the riemann integral
so the defintion is lacking
would anyone agree with me or disagree?
@robjohn ?
arc length is the sup of the line segment approximations. Since the line segments are the shortest path between the points they connect, this is fine.
That is, the line segment approximation of any partition of the curve will be less than the arc length.
@Shinrin-Yoku Haven't we already discussed this at length a few weeks ago?
To be explicit, area under a curve is defined to be the Riemann integral. Arclength of a curve is not defined by an integral.
That is fine, but when you take the sup of the line segments you can only get a lower approximation, by the same logic we should only take the lower riemann integral and not the upper one @robjohn
@robjohn was this intended to me ?:)
6:32 PM
@Franklin Yes.
@TedShifrin ^, also we did discuss it, and the conclusion was that we can come up with a better defintion for convex functions, but i wonder why no books ever mention it? they just guve the seemingly lacking def
I just realised that that is problematic
Why make a special definition when it only applies to a small class of curves?
Remember that these notions are important outside the setting of a freshman calculus class.
why have a lacking def for a larger class? we can also have have only the lower riemaan integral and we would get a larger class @TedShifrin
yet we approxiamte from both ends in the riemann case
@TedShifrin ok, now I get the thing you were trying to make me understand, that I wrote, in my last comment to @robjohn ? I think you hinted upon what I wrote in my previous comment, right? Correct me if I am again mistaken...
@Shinrin-Yoku I do not follow. Length of a curve makes sense much more generally than in the context of smooth curves and integrals in calculus.
6:37 PM
But the definition of length is unsatisfactory...
@Franklin I've stopped paying attention to this. I was just answering your specific question since robjohn is otherwise occupied.
@Shinrin-Yoku Who are you to say that?
By that logic we could forget about upper riemann sums and only care about lower ones....
No. You have no logic.
The line segment approximation only gives a lower approximation. I want an upper one.
You are ignoring exactly what I said earlier. Arclength is NOT defined by an integral. Area IS defined by a (Riemann) integral.
Why do you want one?
6:39 PM
@TedShifrin ok ok ...Thanks to you and robjohn....yes it must be boring. Hehe!
I know it is defined by sup of line segment approximations. I am not talking of integrals @TedShifrin
I already explained to you that there is no way to give arbitrarily good upper approximations for most curves.
Ok, so there is no good way, so why is the definition useful>
What are you talking about?
Why do you need an upper approximation for it to be a "useful" notion?
You are obsessed with this for some reason I cannot understand. Why?
Imagine I wanted to measure the length of a curve in real life, the lower approimation would not suffice I would need an upper one tooo @TedShifrin
6:42 PM
for any upper approximation there is a lower approximation that is as good as it
not at all @shintuku
Why do you need that? What you do need to know is that the curve actually is rectifiable. So I agree that to know that you need some upper bound on the polygonal approximations. But it does not need to be a good one.
Van Kampen is such a cool name.
So what is the easiest example of a non-rectifiable curve (a curve without finite length) that you know, @Shinrin-Yoku?
an upper bound on the polygonal approximation is not enough, since it need not be an upper bound on the length of the curve
6:47 PM
@shintuku what’s the lengthiest name that you’ve heard of ?
No. I said approximations.
You need theorems that tell you conditions for curves to be rectifiable.
@TedShifrin What approximation?
For me, it is Avogadro’s full name.
@Koro What do you call a bowl full of Avogadro's number of avacados?
6:49 PM
Sorry I dont understand @TedShifrin
Lorenzo Romano Amedeo Carlo Avogadro, conte di Quaregna e Cerreto
is the full of Avogadro
are you saying because all physical curves are locally convex/concave it does not matter?
To know a curve is rectifiable, you need to prove that there is a single number that is greater than all (lower) approximations. Your approach would require you to find arbitrarily "good" upper bounds. That's infinitely harder. I'm asking you if you've thought about curves that are NOT rectifiable. You might want to do that.
What are "all physical curves"? Ones that are $C^2$?
Even then they don't have to be locally convex/concave, do they?
6:53 PM
where should I start with a proof that the nonpivot columns of a matrix are linear combinations of the pivot columns?
i would say a "good" definition of arclength allows us to compute the length of a physical curve without any worry... in this definition we have the problem that we ony have lower bounds on upper bounds @TedShifrin
*not upper bounds
Of course there may be a proof that for locally convex curves the def works and since all physical curves are such we are good
@Shinrin-Yoku We know that any $C^1$ curve is rectifiable, and we know more general things than that. I do not believe you will come up with a superior definition.
@shintuku is it known that col rank = row rank?
I do not agree that "all physical curves" are locally convex. And when you switch from convex to concave, all things can go wrong, so your definition will be in trouble.
Ok. Thanks for your help @TedShifrin
6:56 PM
no dimension proof @Koro
@shintuku This proof is in any good book (or in my lectures on line).
You need to use vectors in the nullspace to give a proof.
why are not physical curves locally convex/concave @TedShifrin?
and why does the shift from convex to concave cause problems in finding an upper bound?
Because you can have very rapidly oscillating "physical curves," whatever those are.
Because you cannot give a reasonable definition that will work when you have both. You have to chop the curve into convex pieces and concave pieces and then add the lengths together. If you have to chop into more than a finite number of pieces, you're in trouble.
hm, seems like I found a proof that uses the nullspace. what would lead you to think of using the vectors in the nullspace?
for a given interval we only need to chop into finitely many pieces @TedShifrin
since it is locally convex/concave
6:59 PM
Because vectors in the nullspace give you linear combinations of the columns that result in $0$.
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