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 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
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?
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.
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.
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.
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
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…
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...
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?
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.
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.
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.
@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
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.
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 $F$?
$\mathbb Z \times \{0\}$ 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.
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
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 $y_2(x)$...
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
@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.
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.
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 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.
@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 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
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).
@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.
@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 ?
@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 ?
@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 $y=C_1y_1(x)+C_2y_2(x)$...
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
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
@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
@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...
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.
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?
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.
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.
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.
