Am I being an idiot: one defines $K$-rational points in projective space as $\Bbb P^n(K) := \lbrace [x_0, \dots, x_n] \in \Bbb P^n : \text{all}\ x_i \in K\rbrace$. The author then says "Note that if $P = [x_0, \dots, x_n] \in \Bbb P^n(K)$, it does not follow that each $x_i \in K$."
perhaps he means to say that $x_i$ is not well-defined as it depends on a choice of representative, whereas $x_j/x_i$ is independent of the choice of representative
(the condition $x_i\neq0$ is of course independent of choice of representative too)
Is there actually a method for converting from a recursive relation to a function or non-recursive relation?
Thing is, I can keep creating error functions that will supposedly continue to increase in accuracy because I have a gudermannian approximation. I just store the difference of a true hyperbolic function and a circular function approximation over and over infinitely many times and you get an infinite precision transcendental function through infinite recursion, or at least in theory :)
Imagine using the same e^x to approximate the function you're trying to approximate by approximating e^x using an approximation that uses e^x...
The recursive relation computes a set of infinitely many error functions that approximate a circular function, say cos(x), that then approximates the Gudernmannian function and goes back to the original hyperbolic functions, cos(gd x) of course being sech(x). so you generate the approximated hyperbolic, subtract it from the exact sech(x), and then repeat the cycle infinitely many times.
I need to read the proof that the definition of profinite group is equivalent to a Hausdorff compact t.d. group but I remember it being long and boring but I feel like it's something I should know
So I guess in a way, I did find an infinite precision closed form of the circular functions that converges more quickly than other infinite series as a recursive function...
Hey, is there like some gigantic list of trig composition identities somewhere that I can use? The one on wikipedia is rather sparse.
I need something that makes a known closed form function from a circular function with the constraint that the argument to the circular function must not be another circular function (or its inverse).
But you can see my idea here, right? These are "one level of nesting"... if I knew the error function for any single one of these, I could make the error zero: desmos.com/calculator/w4wecmcpws
So I mean... if one of you knows how (without parametric equations) to make a circle using only one level of nesting rather than something like cos(arcsin(x)), I'm all ears.
<in old man voice>: Y'know, I've been in this industry for 51 years out of my whole life since I was 1 year old... I've ne'er seen in all those 51 years <some significant object of interest> in all my life!
I thought you were going to go for something like that
1. Just reverse engineer a closed form from the infinite series. 2. Profit.
If I can convert anything to an infinite series, why is there not an operation to convert back from an infinite series?
(To a closed form on the reals)
Also would also be nice if I could just, y'know... do something like xsinh(1)/(1+x^2) and get a circular function, but the infinite series doesn't yield to that nor does it plot a circular function.
(2k+1)! varies, so I guess it is a matter of finding an infinite series where the denominator/coefficient is constant which is also a valid definition of the circular functions. Same of course obviously goes for (-1)^k.
Now if I could find a suitable function that maps linear and continuous rotation to the perimeter of a square, that would be great. I already have sine and cosine of a square.
Question: can you make the Archimedean spiral using rectangles or square, or no?
Can't I get "hyperbolic square functions", map from hyperbolic functions to the hyperbolic square functions, then map from hyperbolic square functions to square functions and then to the circumference of a circle to get the circular functions?
Let $\mathcal{C}$ be a category and $I\in\mathcal{C}$. What do the monomorphisms in $\mathcal{C}/I$ look like? Obviously, if a morphism in $\mathcal{C}/I$ is a monomorphism in $\mathcal{C}$, it is a monomorphism in $\mathcal{C}/I$, but I don't see whether the reverse implication holds.
I don't really see a way of tautologically conjuring up the necessary commutative triangles
Right, well I'll be heading off to bed now. I bet you there's some way to map things to the Euclidean plane using the square's sine and cosine from other geometries... even from complex, if only in part, by defining a square in that plane, doing whatever mappings you want, and then putting it on a square in the Euclidean plane.
> “There was a seminar for advanced students in Zürich that I was teaching and von Neumann was in the class. I came to a certain theorem, and I said it is not proved and it may be difficult. Von Neumann didn’t say anything but after five minutes he raised his hand. When I called on him he went to the blackboard and proceeded to write down the proof. After that I was afraid of von Neumann.” ― George Pólya
Locally profinite means every neighbourhood of the identity contains a compact open subgroup; closed subgroups of locally profinite groups are locally profinite and open subgroups are clopen, and locally profinite groups that are compact are just profinite groups, so by this definition, every open neighbourhood of the identity contains a profinite group
As a "stand alone" statement, "Das brauch ich" und "Ich brauch das" are basically the same, but I think "das brauch ich" is more common and sounds more natural
On the other hand, after something like "ja" you have to say das brauch ich
But it doesn't mention this isomorphism is renaming so I assume for arbitrary
@Astyx you can mention this but I think let's not care about minor error
I don't have a lot of time so I will write my precompiled question from my brain
"Isomorphic groupoids are essentially same" I think if a group $\bar{G}$ constructed by renaming elements then they are Isomorphic and are essentially same. But above statement means arbitrary Isomorphic groupoid are essentially same. Or may be I am wrong since I couldn't see two arbitrary groupoid are same. The property of Isomorphic groupoid IMO says it isn't necessary for Isomorphic groupoid to have same elements and binary operation thus renaming is the option remaining.
"Groupoid isomorphic with with a groupoid F is indistinguishable from a suitable renaming of $F$ as far as the elementsand the way they multiply are concerned" I think the Isomorphic used in the groupoid Isomorphic with a groupoid $F$ and it's renaming should be same.
And that isomorphism is renaming is renaming one. Or else I don't understand how can arbitrary isomorphism make the above statement true.
Your sentences are unreadable. An isomorphism is renaming elements in a way that the structure is preserved, like Astyx is said; so you can identify isomorphic mathematical objects. Be it groupoids, or whatever.
You cannot prove that an isomorphism is a structure preserving bijective because that's what it is by definition. You could find specific examples of isomorphisms and prove that they satisfy those properties
It doesn't. You can relabel the elements of $G$ by those of $\bar{F}$ in a specific manner, compatible with the binary operator (so that (label of a*b) = (label of a)*(label of b))
You could for instance change the symbol representing the integers 0-9 to random scribbles, but as long as they still have the same relationships re addition and multiplication you haven't created a new object, you've just relabelled the points
@Stupidquestioninc Look at simple cases of groups like S_3, D_3, GL_2(F_2) and others. You are studying a structure and you are saying two structures are the same if they behave the same way. It doesn't matter if I call one structure hell and the other heaven
All these years later, and SE still hasn't implemented a "side chat" feature for questions.
There's a reason why moderators still end up deleting the comment section and moving everything to a private chat room leaving behind only a reminder: "The comment section is not a chatroom..." etc.
$\overline{F}$ is group constructed by relabelling elements of $F$ and by this we mean $f\theta=\overline{f}$ where f is element of F and f bar is element of F bar . Where theta is mapping which maps F to F bar obeying bijection and homomorphism.
@Stupidquestioninc What you said cannot be correct, because according to your first comment $\theta$ is a map $\theta :F \to \overline{F}$ whereas according to your second comment $\theta$ is a map $\theta : F \to G$.
@Stupidquestioninc Right, so what you said makes sense only if $\overline{F} = G$, equal as sets. Elements of $\overline{F}$ are the same as the elements of $G$, on the nose. No relabelling required.
You're correct.
But that doesn't give any insight into what's happening, we just walked in a circle.
The book is trying to explain how relabelling is the same thing as an isomorphism
Specifically, if you have an isomorphism $\theta : F\to G$, then you can relabel every element of $F$ by an element of $G$ by taking the image through $\theta$
This preserves group structure and shows than, up to a relabelling of the elements, $F$ is the same as $G$
I will define the following groupoid. $V_4 := \{1, a, b, c\}$, multiplication defined by $1 * a = a * 1 = a, 1 * b = b * 1 = b , 1 * c = c * 1 = c$, $a * a = b * b = c * c = 1$, $a * b = b * a = c$.
(1) Verify this defines a groupoid. In particular, check for three things: $V_4$ has an identity, an inverse and a partially defined multiplication. Draw the multiplication table for $V_4$. Bonus: In fact, demonstrate $V_4$ is a group
I will also define the following groupoid. $\Bbb Z_2 \times \Bbb Z_2 := \{(0, 0), (1, 0), (0, 1), (1, 1)\}$, where multiplication is defined by adding componentwise, but you always set $2 = 0$. So $0 + 0 = 0, 0 + 1 = 1 + 0 = 1, 1 + 1 = 0$. This is called addition modulo 2. Computer scientists do this all the time.
(2) Ditto for $\Bbb Z_2 \times \Bbb Z_2$. Verify everything about $V_4$ in exercise (1) for this guy.
@BalarkaSen hehe no identity no inverse no is groupoid multiplication table i will show you wait for a mom v_4 is not group by along the fact no identity and inverse
question: assume that $u: \mathbb R^2 \to \mathbb R$ is differentiable at $(x_0, y_0)$, with continuous partial derivatives. how do I prove that there's a linear order approximation of $u$ in terms of $u_x$ and $u_y$? i.e., how do I prove that there are positive reals $\Delta x$, $\Delta y$, $\epsilon_1$, $\epsilon_2$ s.t. $u(x_0+\Delta x, y_0+\Delta y) = \Delta x \, u_x(x_0, y_0) + \Delta y \, u_y(x_0, y_0) + \epsilon_1 \Delta x + \epsilon_2 \Delta_y$?
should be an easy analysis exercise but it's just a step in a "Cauchy-Riemann + $C^1 \implies$ differentiable" proof, so I don't think I should be wasting a lot of time on that
@Secret I remember being in high school having my own ideas about division by zero. My idea was rather simple, although I haven't actually thought about it much since then: let x/0 be a number in the set of "mirrored" reals such that (x/0)/0 = x, and 0/0 = 0.
Could we consider the statement that the Earth is Cube is the negation of the Earth is round?
