Why do we care about outcomes when talking about random variables but we care about events when talking about probabilities? A similar question was already asked on the site, but I am not satisfied with the answers.
The point is precisely that for line bundles $E\otimes E^*$ is always trivial. I wanted to give an OP an example of a nontrivial example for rank $\ge 2$ (in the real category).
No, I was remembering the usual thing for defining Pontryagin classes by complexifying the bundle and playing that game.
The argument was: for S^2, rank >=3 bundles are determined by $w_2$, so that if we show $w_2 = 0$, we see that the bundle is trivial. I was using this to argue that E o E is trivial, for E = TS^2.
Now I am trying to construct a bundle E over a space M so that E o E is non-trivial. It suffices to construct one with a non-trivial characteristic class.
Write L for the bundle det(TM). (L + R) o (L + R) = L o L + 2(L o R) + R o R = 2(L o R) + 2R is what I should have written, as the tensor square of a line is trivial
It should be direct that $TS^2 \otimes_{\Bbb R} TS^2$ is trivial because the clutching function is $S^1 \to \text{GL}_4(\Bbb R)$, $f(\theta) = R_{2\theta} \otimes R_{2\theta}$. There should be some explicit nullhomotopy.
So recall that two PL knots are equivalently if they can be connected by a finite sequence of "triangle moves" --- the operation that takes a linear triangle in 3-space with one edge along the knot, and replaces that edge with the other two, or the inverse of this operation
It then suffices to show that one triangle move can be done, at the level of diagrams, via the Reidemeister moves
And that sounds like the kind of thing amenable to a fiddly proof
the Reidemeister moves are actually all immersion moves except moves of type I, right? That's the "pinching a self-loop infinitely tight until it vanishes" thing
Hm. If $f : M \to N$ is an immersion with $\dim M = m$, $\dim N = n$, then I can arrange so that $M^k \to N^k$ ($X^k = X \times \cdots \times X$ k-fold product) is transverse to the diagonal $\Delta_N \subset N^k$ (modify so that $f^k \pitchfork \Delta_N$ and then replace $f$ by $f|_{\Delta_M}$, and argue by stability that our new $f^k$ is also transverse to diagonal).
The diagonal has codimension $nk - n$, so the set of $k$-ple points of $f$ manifests as a submanifold of $M^k$ of dimension $mk - (nk - n)$
Except this allows multiplicity on the coordinates, which I don't want. So to find the true set of $k$-ple points, I have to throw out the set of $(k-1)$-ple points.
@MikeMiller I agree with you @nbro Mathematical definition is different from what we use in AI, because the AI doesn't needs to follow that crap approach, So, I'd go in favour of what Tensorflow guide has written.
@AbhasKumarSinha It is not true that math isn't required for artificial intelligence. You can apply AI methods without knowing the details. However, you cannot do research without really knowing math. In my case, I have an intuition behind the mathematical concepts that I need, but I wanted to know more about measure theory to really understand probability theory.
@AbhasKumarSinha I think they are being used inconsistently, in the context of AI, because many people involved in AI do not have a sufficiently good background in math. Often people talk about distributions and they refer to pmf or densities, or maybe they talk about random variables without really knowing it's a measurable function.
@nbro They have But, they often change the definitions and notations to make things simple for AI, because AI and maths both have different approaches.
@zacts ML is subset of AI
@zacts Machine Learning has more mathematics
@nbro I have not seen any kind of such people.
@nbro If you see good journals, they often have more mathematical skills than mathematicians itself.
@AbhasKumarSinha This is completely not true. The usual proofs in ML are relatively easy (and involve only basic calculus and linear algebra) and someone, like me, who's got a background in CS could have potentially written them.
@AbhasKumarSinha Show me a proof, in the context of machine learning, that is really hard to understand. Please, do not even cite AIXI. I am familiar with it. Probably and AFAIK, it's the machine learning (more specifically, reinforcement learning) concept that involves move mathematical concepts than any other ML concept.
Assume we have a large number of proofs in first order predicate calculus. Assume we also have the axioms, corollaries, and theorems in that area of mathematics in that form too.
Consider the each proposition that was proved and the body of existing theory surrounding that specific proposition ...
Lol, just kidding , please don't take it seriously, we actually need more and more mathematicians in AI, to make better systems to work to develop even better versions of AI
> In geometry, Boy's surface is an immersion of the real projective plane in 3-dimensional space found by Werner Boy in 1901. He discovered it on assignment from David Hilbert to prove that the projective plane could not be immersed in 3-space.
3 occurs in the usual definition of $RP^2$ as the set of lines in $R^3$. That is, the quotient space of $R^3-0$ that identifies $x\sim cx$ for all nonzero $x\in R^3$ and nonzero real $c$. The homeomorphism $(x_1,x_2,x_3)\to(x_2,x_3,x_1)$
for example induces a threefold symmetry of $RP^2$.
oh what
it just comes from translating the coordinates?? lol
Take an immersed surface $M$ in $\Bbb R^3$ and make it generic so that you only have transverse double and triple points, locally modelled on two planes in $\Bbb R^3$ intersecting at a line and three planes intersecting at a point. The curves of double points are immersed curves in $M$ which intersect in triplets at the triple points
If there are two connected components of double point curves in $M$, one can tube them up by the following procedure: Consider the configuration $\{z = 0\}, \{y = 1\}, \{y = -1\}$ in $\Bbb R^3$. Simply tube the planes $\{y = 1\}$ and $\{y = -1\}$ togather. The two disjoint lines of double points are appropriately tube thereof
If there are two triple points in $M$, one can run them over similarly; consider $\{z = 0\}, \{y = 0\}, \{x = 1\}, \{x = -1\}$. Tube the planes $\{x = 1\}$ and $\{x = -1\}$ appropriately so that the triple points at $(1, 0, 0)$ and $(-1, 0, 0)$ are removed.
The self intersection locus changes like so
These two procedure comes at the cost of changing the topology of $M$ by attaching a handle.
This does not change the Euler characteristic mod $2$. So operate until you have an immersion of $M$ in $\Bbb R^3$ with a self intersection locus consisting of a single connected immersed curve with either no triple point (in which case my curve's embedded) or a single triple point.
If $M$ is orientable break the immersed curve into embedded loops, the local structure of the immersion along which is $S^1 \times \mathsf{X}$ where $\mathsf{X}$ is literally the letter X, and tube it up to look like $S^1 \times \mathsf{)\,(}$
Eh, scratch the last sentence. I don't actually think that makes sense.
Should go the other way around. If there are no triple points, then I have a big embedded curve on $M$ along which the immersion has double points. If it's orientable, the immersion looks like $S^1 \times \mathsf{X}$ there, in which case tube it up to look like $S^1 \times \mathsf{)\,(}$. If it's not orientable it might look like two Mobius bands intersecting at the center circle but that's also ok because then it's like the mapping torus of $+ \to +$ given by rotating by 180
There is an obvious cylinder which bounds its boundary
And these tubing procedure also leaves Euler characteristic mod 2 invariant, so we have obtained an embedding of $M$ in $\Bbb R^3$, forcing $M$ to be orientable
So if $M$ admits an immersion in $\Bbb R^3$ with no triple points it has to have $\chi \equiv 0 \pmod 2$. $\Bbb{RP}^2$ never admits one since $\chi(\Bbb{RP}^2) = 1$.
$\Bbb{Z}[C_2^2]$ gets its $C_2$-module structure via $\sigma(\sigma_0,\sigma_1)=(\sigma\sigma_0,\sigma\sigma_1)$
Oh I was still typing up the setup
So after the identification, writing $C_2=\langle \sigma\rangle/\langle \sigma^2\rangle$, we see that $\sigma$ acts by $xy$ on $\Bbb{Z}[x,y]/(x^2-1,y^2-1)$.
Now, let $\Bbb C^*$ be a $C_2$-module, by complex conjugation
I want to find all maps $\text{hom}_{C_2}(\Bbb{Z}[C_2^2],\Bbb C^*)$
These seem to be determined by where I send $x$ and $xy$, which seems odd
But it may just be notational confusion
Since $$\sigma\cdot 1 = xy,\quad \sigma\cdot y = x,\quad \sigma\cdot x = y,\quad \sigma\cdot xy = 1$$
Where I had previously identified $$(\sigma^0,\sigma^0)=1,\quad (\sigma^1,\sigma^0)=x,\quad (\sigma^0,\sigma^1)=y,\quad (\sigma^1,\sigma^1)=xy,$$ when moving from $\Bbb{Z}[C_2^2]$ to $\Bbb{Z}[x,y]/(x^2-1,y^2-1)$
If you were doing $\Bbb{Z}[x]/(x^2-1)\to\Bbb C^\times$, it is determined by where you send $1$, and you have $\psi(1)=c\in\Bbb C^\times$, and $\psi(a+bx)=ab|c|^2$
Sometimes I take solace in:
"Young man, in mathematics you don't understand things. You just get used to them."
- John von Neumann
It seems to me that some of the art is "if-this-then-that" kind of stuff, but there's a whole bunch more that basically comes from the intuition you get from basic...
Every explanation about scaling a 2D vector, or equivalently having a line segment PQ on cartesian plane and then find a point R on the line PQ satisfying PR/PQ = r (fixed given r) starts with that one specific case in the picture. A formula for the coordinates of R is then given for that case.
However, that is also the only case that is covered. The cases whereby the slope is non-positive and the line segment PQ is vertical are not shown to share the same formula. Also, the case of point Q being the head of the vector is not proved.
> Another way to phrase identity is it is a strictly singular thing and the act of picking out precisely said thing among other things. Then, sameness is when said should be strictly singular thing turns out to have nonzero multiplicity.
For $k \geq 0$, let $X_k = $ the set of all integers $x\in \Bbb{Z}$ such that:
$$
x^{2^k} - 1 = q_1 \cdots q_r,
$$
for $r \gt k + 1$ and $q_i$ are each an odd prime.
Then $X_k \subset X_{k+1}$ by multiplying by $x^{2^k} + 1$ which is always at least one prime in factor.
$$
x \in X_k, y \in X_k ...
Is this a correct rephrasing? If $x^{2^k}-1$ is factored into a product of $r\geq k+1$ odd primes, then it is impossible that $x^{2^k}+1$ is less than any one of the primes that appear in this factorisation
@ShineOnYouCrazyDiamond I don't know, if I knew, I could parse it obviously lol (and the problem wouldn't exist)
It sort of will work because each time $xy$ is not of that form, then $xy \pm 1$ is another twin prime pair
Suppose the twin prime conjecture is false. That is there are only finitely many primes $p, q$ such that $x^2 - 1 = pq$ for some $x \in \Bbb{N}$.
This means that there exists $N \in \Bbb{N}$ such that for all $x \gt N$ we have that $x^2 - 1 = (x - 1)(x+1)$ is necessarily the product of at least $3$ primes, not necc. distinct, primes.
Since $x, y \gt N$ we have $xy \gt N$. Thus the set of all non-solutions to the twin prime problem that are greater than $N$ forms a semigroup. In fact, this semigroup must indeed be $(N, \infty)$ itself by construction, since if $x \in (N, \infty)$ is a s…
@yh05: You can draw a picture (in this case with positive slope), but talking about $\overrightarrow{PR} = c\overrightarrow{PQ}$ with $c>0$, $c<0$, etc., is totally general.
Can anyone help me out with the binomial expansion of \frac{2}{7\left(2x+1\right)}+\frac{6}{7\left(x-3\right)}+\frac{3}{\left(x-3\right)^2}, up to x^{2}?
My answer - \frac{1}{3} - \frac{84}{189}x + \frac{693}{567}x^{2}
If you put your math expressions inside dollar signs, it'll typeset for us in MathJax.
The numbers involved look absolutely horrendous. I don't know why you're calling this a binomial expansion. So you want the second-order Taylor polynomial of that rational function at $0$.
$\frac{2}{7\left(2x+1\right)}+\frac{6}{7\left(x-3\right)}+\frac{3}{\left(x-3\right)^2}$, up to $x^{2}$? My answer - $\frac{1}{3} - \frac{84}{189}x + \frac{693}{567}x^{2}$
3 occurs in the usual definition of $RP^2$ as the set of lines in $R^3$. That is, the quotient space of $R^3-0$ that identifies $x\sim cx$ for all nonzero $x\in R^3$ and nonzero real $c$. The homeomorphism $(x_1,x_2,x_3)\to(x_2,x_3,x_1)$
for example induces a threefold symmetry of $RP^2$.
Sometimes I take solace in:
"Young man, in mathematics you don't understand things. You just get used to them."
- John von Neumann
It seems to me that some of the art is "if-this-then-that" kind of stuff, but there's a whole bunch more that basically comes from the intuition you get from basic...
For $k \geq 0$, let $X_k = $ the set of all integers $x\in \Bbb{Z}$ such that:
$$
x^{2^k} - 1 = q_1 \cdots q_r,
$$
for $r \gt k + 1$ and $q_i$ are each an odd prime.
Then $X_k \subset X_{k+1}$ by multiplying by $x^{2^k} + 1$ which is always at least one prime in factor.
$$
x \in X_k, y \in X_k ...
