Am I right in saying: (Complex Analysis) If our f is holomorphic in some region D, but we have some closed contour C that goes around some hole somewhere inside (but not included in) D, then we cannot use Cauchy's result to conclude the integral is 0, since f is hence not holomorphic on and inside C. (Since we're only given that f is holomorphic in D and not the 'hole')
1) Let $f$ have an antiderivative $F$ in a region $D$ containing a $\underline{closed \,\,\,contour}$ $C$. Then
\begin{equation}
\int_Cf(z)dz = 0
\end{equation}
2) Also, Cauchy's theorem says: If f is $\underline{holomorphic}$ inside and on a $\underline{closed \,\,\,contour}$ $C$ then
\begin{eq...
I actually hail from the physics side of things, but its MP so that could be a compromise
My object of interest is the case where I have a product of many many rational expressions in some variable
So I already covered the case where I have more zeroes in the denominator, I found something like this:
Anyway, I'd also like to find something like this variant of the Heaviside rule but for polynomial long division
Finding reading material on this question seems to be a rather difficult task, I was hoping maybe someone here knew how to divide a general polynomial by another general polynomial, ofc assuming these polynomials don't have the same coefficients
And assuming that the polynomial in the denom P_D has a smaller degree than the one in the one in the numerator
@robjohn Hello sir! Can you please give me a proof that $$f(x)= \begin{cases} 0 & if~x~is~irrational \\ 1/q & if ~x~is~of~the ~form~p/q~(it ~in ~simplest~form~and~p,q \in \mathbb Z , q\neq 0 \end{cases}$$ ($0\lt x\lt 1$) have a limit zero at any $0\lt a \lt 1$.
Here is the proof given by Spivak
But I understand your proofs quite easily, so if you have time can you please write up an elegant proof (like you do). It’s a request not a demand.
@Knight The idea is simple: If you fix the denominator, there a discrete amount of rationals so the smallest distance from $a$ to the nearest such one can't be too small
@Knight For any $q$, there are fewer than $q^2+q$ rational numbers with denominators less than or equal to $q$ within $1$ of $a$. Find the closest of these $\le q^2+q$, say at a distance $\delta$ from $a$, of course, not counting $a$ if $a\in\mathbb{Q}$. For any $|x-a|\lt\delta$, you have $f(x)\lt\frac1q$.
This is a $\delta$-$\frac1q$ proof, instead of $\delta$-$\epsilon$
I gave it some thought and used this argument: First I show that |z-a| = |conj z - conj a| Then I replace |z-a| in the eps-del definition, and there I have it...?
Did you see the new ratio? It gives zeta zeros. The limit as such is a bit slow to compute because it requires so much symbol manipulation by the computer. But never the less I think there is actually hope that this could lead somewhere, since by a "global" derivative formula for the Riemann zeta function
(that I told Leaky Nun about earlier in chat), one can cross out all derivatives by nesting recursively the limit for the derivatives to any degree derivative.
By applying the `FullSimplify` command in Mathematica to right hand side we get: $$\zeta (c)^3 \left(\frac{1}{\zeta (s)}-\frac{3}{\zeta (c+s-1)}+\frac{3}{\zeta (2 c+s-2)}-\frac{1}{\zeta (3 c+s-3)}\right)$$
Which appear to have alternating binomial coefficients in the numerator.
?
??
1=1 1=1+1-1 1=1+1+1-1-1 1=1+1+1+1-1-1-1
...
"Can you do addition?" the White Queen asked. "What's one and one and one and one and one and one and one and one and one and one?" "I don't know," said Alice. "I lost count." Through the Looking Glass.
@Thorgott then i have the eps-delta definition of the limit of some function g(z) as z approaches some a, written in terms of the conjugates of both z and a instead. So... I suppose that translates to the shorthand syntax of [lim z-> a]{something} being the same as [lim conj z -> conj a]{same thing}
Here you go. Let $e_1,\dots,e_n$ be an orthonormal frame (working on an open set, of course), and let $e_1$ be the chosen element of the unit tangent bundle. Then we have the dual coframe $\omega_1,\dots,\omega_n$ and we have $de_1 = \sum\limits_{j=2}^n \omega_{1j}e_j$, so the volume form on the fiber is $\omega_{12}\wedge\dots\wedge\omega_{1n}$.
Thus, the volume element of the whole bundle is $\omega_1\wedge\dots\wedge\omega_n\wedge\omega_{12}\wedge\dots\wedge\omega_{1n}$.
You might complain that $de_1$ doesn't make sense. But this is how you should understand this intuitively. I can put the canonical connection on the tangent bundle and do $\nabla e_1$, or I can do it all in terms of differentiating the $\omega_i$.
I've never done this before, so its all new to me :'). Mathematica is unbelievably useful for engineering. It's really great for long derivations where you want to know at least what ur shooting for
Oh, Mathematica is highly useful for all of us. Since I was retired and couldn't legally get departmental copies anymore, I had to give in and buy it for myself.
Computations (often for answering MSE questions when I don't want to do the whole thing out by hand), but I've always used it for graphics (especially for differential geometry). Most of the pictures in my geometry notes are from Mathematica, and I have all sorts of animations written and programs to solve DE that can't be solved other than numerically and then plot them (e.g., geodesics on surfaces).
I have a Mathematica primer I wrote for my diff geo students. I can send it to you if you're interested.
@TedShifrin thanks! It's interesting because, with the problems you've assigned me in the past, I don't think any of them involved mathematica. What role do you think programming plays in math education?
@Threnody yeah, because a sequence $(z_n)_n$ converges to $a$ iff $(\overline{z_n})_n$ converges to $\overline{a}$
user434058
5:49 PM
Let $\mathbf r$ be a function of $q_1,q_2,\dots,q_n,t$, where $t$ is time. I will be referring to total time derivatives by a dot over the symbol. Now, how do I prove that$$\frac{\partial \dot{\mathbf r}}{\partial \dot{q}_j}=\frac{\partial \mathbf r}{\partial q_j}$$ where $j\in\{1,2,\dots, n\}$?
user434058
@TedShifrin Hi, how do I prove the above identity? :-)
@Ted Hmm, I'm not seeing why that is the volume form on the fiber. The volume form can be expressed in the frame as $\sum_{i=1}^n(-1)^ie_i\omega_1\wedge...\wedge\widehat{\omega_i}\wedge...\wedge\omega_n$, but I don't see why that is $\omega_{12}\wedge...\wedge\omega_{1n}$.
@Stan: You weren't doing things that lend themselves to graphics or complex computations.
@Thorgott: What you wrote doesn't actually make sense in general; it's a vector-valued form, not a form. Think about the unit sphere. $e_1$ is your position vector, and $e_2,\dots,e_n$ give an orthonormal basis for the tangent space at $e_1$.
@FakeMod: If you're not doing Hamilton's equations, this looks very suspicious.
user434058
@TedShifrin I encountered this while learning Lagrangian mechanics.
You have $f(q,t)$, where $q$ is a (vector) function of $t$. So when you write $\dot f$, this is $\sum \partial f/\partial q_j\, \dot q_j + \partial f/\partial t$.
@TedShifrin I suppose the most i've had to use wolfram products is double checking my math to make sure i don't have a sloppy algebra error or something when doing derivations
@Knight Think of the rationals with denominator $q$ in an interval of length $1$ (mod $1$): $\frac0q,\frac1q,\frac2q,\ldots,\frac{q-1}q$. There are $q$ of them. Summing for all denominators up to $q$ gives $\frac{q^2+q}2$. Adding those on the other side, gives $q^2+q$. Of course, there will be overlaps where the numerators and denominators have common factors, but that only reduces the number.
Mathematica can be used pretty much in the same way Excel spreadsheet formulas work. The hard part is to learn how to use speed ups with operators such as the hashtag operator #. Michael De Vlieger in the OEIS is skilled at using those Mathematica operators.
I would say that the intersection of tasks that a spreadsheet is suited for and those Mathematica (or other similar tools) are suited for is pretty close to just the basic calculator
Well, you can certainly make Table work like you were using Excel. But often the things I put in the Table are far more intricate than Excel could handle.
And for quite a lot of stuff I used version 3 which was supplanted in 1997 (because the Chevie package for dealing with Hecke algebras had not been ported to version 4)
Only after doing a ton of coding and computations did I learn about the PyCox package for SAGE which would have been much more suitable since it was actually kept up to date
It should be $\sum_{i=1}^n(-1)^ide_i\vert_p\omega_1\vert_p\wedge...\wedge\widehat{\omega_i\vert_p}\wedge...\wedge\omega_n$ (which makes sense after identifying the tangent spaces as subspaces of $T_pM$) as volume form on the unit sphere in $T_pM$.
@Thor: You keep going back to the formula for the volume form for $S^{n-1}\subset\Bbb R^n$ in terms of coordinates. But when you're working with a manifold, you have no such coordinates. That's why I'm doing everything in terms of frames for the tangent space.
The frame does give us coordinates. $de_1\vert_p,...,de_n\vert_p$ is a global chart for $T_pM$ when we view it as a manifold in its own right and then contraction with a unit normal field can be made to work. I am trying to understand how you arrive at your volume form, but I neither get that on its own nor do I get why it's the same as my volume form.
No, what you're writing down doesn't make sense on an abstract manifold. I don't even understand how you're getting a chart for $T_pM$. It's the $e_i$, not their derivatives, that give a basis for the tangent space.
To differentiate, you need a connection.
Do you agree that if we take $S^{n-1}\subset\Bbb R^n$ and choose an orthonormal frame for the tangent space locally, namely $e_2,\dots,e_n$, then wedge product of the dual coframe elements gives the volume form?
If we write $x$ for the position vector on the sphere, then we're writing $dx = \sum\limits_{j=2}^n \eta_j\otimes e_j$ if $\eta_j$ is the dual coframe. I'm using $e_1=x$ in the setting we were doing, and then writing $\eta_j = \omega_{1j}$, because when you're used to the moving frame set up, this is telling you how $e_1$ twists toward $e_j$.
If we dualize to the unit cotangent bundle instead, we can cook up a $1$-form which eats a tangent vector at a covector and spits out the value of the covector at the underlying tangent vector (after projecting to base). This is the tautological $1$-form $\alpha$, and my volume form would be $\alpha \wedge (d\alpha)^{n-1}$, the one coming from the contact structure on $T^*_1 M$.
@TedShifrin I might be able to get academic versions, but I think that the Personal version is about as expensive and so I haven't bothered getting the academic versions.
Mine's 12.0. I did upgrade and when I did, I lost all Microsoft and Adobe products I had purchased ten years ago. So now if I have to update the pictures in any of my books (for which I used Illustrator) I'm up the creek.
@Thor: The point is that to get a coframe on the tangent bundle (rather than on the base manifold) you need to know how the frame is twisting as you move around.
Especially since I may never need to use it again. I did find Inkscape, which is free. It seems to have most of the tools that Illustrator did, if only I can figure out how to use them.
I was already angry at Adobe for telling Apple that it had to stop working on QuickDraw GX or Adobe would stop producing Mac applications. That cost me my job at Apple.
QuickDraw GX made it extremely easy for third party developers to mimic things that only PS or Illustrator could do. So rather than trying to do something great using QDGX, they said "stop it!"
I am not sure if there has been healthy competition on anything involving any of the major software companies for quite a while (maybe apart from offering cloud solutions)
OK, @Thor. Officially, I should have used $\pi^*\omega_j$ to put the base coframe up on the unit tangent bundle, but I didn't want to complicate things.
@Thor: What I'm talking about also works naturally on Grassmannians/Stiefel manifolds. And, with universality, you can do CGB just by reducing to the tautological bundle on the Grassmannian.
@Tobias @robjohn: How would I make my own cloud? I guess I have storage space somewhere due to my internet subscription. But I certainly don't have any 'server' other than my own desktop.
Mainly because the alternative is most likely that it gets hosted on a couple of dedicated servers with not much scaling potential and which will be horribly tricky to get access to for stuff like CI/CD
@TedShifrin I am not deep enough into the details that I would know how to get started
@TedShifrin I have a server that I am going to use for that purpose. A private cloud for saving things. One thing I hate about clouds is that "synching" is mostly a black box and you don't always have control over what gets deleted when you simply want to remove it from a particular device, or having it come back when you've deleted it.
I will simply have a repository of things I want on the server. That way I can free up space on my phone, etc, and still have access if I should want it later.
BTW, @robjohn, when you were talking about your friend and his adviser at Princeton, I had originally assumed it was math. But when I read your link, I discovered it obviously was not.
Ok so suppose $A\subseteq\Bbb R$ is uncountable, let $A'$ denote the set of limit points of $A$, we want to show that $A\setminus A'$ is countable, right?
No, $A\setminus A'$ is the set of isolated points of $A$, I want to show that if $A$ is uncountable, then uncountably many of its points are not isolated, but for that I need $A\setminus A'$ to be small (a priori we might have $A'=\varnothing$ otherwise)
\neq was fine, we are trying to show that some point of $A$ is a limit point of $A$
Lucas wants to prove that if a subspace of the reals is made of isolated points then it is at most countable. I want to prove that if a subspace of the reals is uncountable then many of its points are limit points, {1/n} is not problematic
ok, this may not be what Alessandro was going for, but you can argue as follows by contradiction: a) if there were an uncountable set consisting of isolated points, you would be able to find uncountably many disjoint open intervals in $\mathbb{R}$ (why?) b) this is impossible (why?)
Re: unit tangent bundle. I don't even see why this viewpoint is consistent with the more abstract one. Let $SM$ be the unit tangent bundle and $SO(TM)$ be the oriented orthonormal frame bundle. We have projections $\pi\colon SO(TM)\rightarrow SM$ and $p\colon SM\rightarrow M$ and our frame gives a section $s\colon U\rightarrow SO(TM)$. There are forms $\tilde{\omega}^i_j$ on $SO(TM)$ such that $s^{\ast}(\tilde{\omega}^i_j)=\omega^i_j$ (not only for this, but for any frame). It can be checked that $\tilde{\omega}_1^2\wedge...\wedge\tilde{\omega}_1^n$ is vertical and $SO(n-1)$-invariant, so p…
My $\omega_1^j$ are the connection forms relative to the moving frame. Surely those are forms on $M$? $e_1,...,e_n$ is a frame on $U\subseteq M$, so the coframe $\omega_1,...,\omega_n$ are $1$-forms on $U$ and then $\omega_1^j$ are defined by $d\omega_1=\sum_{j=2}^n\omega_1^j\wedge\omega_j$.
So it's best to think of everything up on the complete frame bundle. I'm choosing a local section of the frame bundle of $M$ as a bundle over $SM$. The $(n-1)$-form I wrote down depends on $e_1$ but is independent of the choice of $e_2,\dots,e_n$.
Isn't that similar to what I described above? You have connection forms on $SO(TM)$ and then $\omega_1^2\wedge...\wedge\omega_1^n$ is $SO(n-1)$-invariant and pushes forward to a form on $SM$
yeah, what I was wondering was if whether pushing forward the connection forms one has on the frame bundle to $SM$ gives the same forms as when pulling back the connection forms one constructs on an open set via moving frames to $SM$
but I now realize that obviously won't work, cause the former are intrinsic and the latter depend on a choice of moving frame
You can only do the latter if you choose (locally) a fixed moving frame. We can't do that in this situation. You want the Haar measure of the gauge group in there :P (to mix metaphors).
You mean the non-base part? I explained that three times already. This is how you intrinsically construct the volume form on the unit sphere. Or indeed on any Riemannian manifold. You take the wedge of the orthonormal basis covectors.
