Hi all! can someone points me where i can find a reference for the fact that the kernel of a nowhere zero differential 1 form $\alpha$ on a manifold $M$ is a subbundle of $TM$?
@LeakyNun Theorem: You can't control what people believe. Proof: Belief + Proof = Truth. But, Proof - Control = Faith. Therefore, Truth + Belief = Rightness, so Belief = Nothing in absence of Truth, and Control + Belief = Dogma. Q.E.D.
The key to this is that the ACTUAL sum of all the hats (mod 100) must be one of the numbers from 0 to 99.
Whichever number it is, that player will have guessed his/her hat color correctly.
A corollary: no one else will have guessed correctly. Only the person whose player number matches the act...
I'm sure of my math, but trying to write a symbolic proof in MathJax is distracting (as I'm not 100% familiar with it yet) and so I'm wondering if I left any steps out due to constantly fiddling with formatting....
I have a question: Is it important that a random variable has values?! for example if we have coin, then we have Heads or Tails. So, how do you work in this case!
and this -discrete- random variable is completely defined by the pairs $(100, \frac{1}{2}), (3,\frac{1}{2})$, where the first term is the possible outcome, and the second term is the probability
I know that it hasn't been shown that $\pi$ and $e$ are algebraically independent over $\mathbb{Q}$. However it must be easier to show that the fields $\mathbb{Q}(\pi) $and $\mathbb{Q}(e)$ are not the same. Yet, I have no ideas on how to approach this problem. Does anyone can show a proof or give...
$\pi$ and $e$ both have a binary expansion that is non repeating
What is the minimum criteria for a sum or product of numbers to fail to produce a recurring decimal?
Suppose I have irrationals $x=\sum_{k=1}^{\infty}\frac{a_k}{2^k}$ and $y=\sum_{k=1}^{\infty}\frac{b_k}{2^k}$ where $a_k,b_k \in \{0,1\}$. Then the sum is given by $x+y=\sum_{k=1}^{\infty}\frac{a_k+b_k}{2^k}$ and their product is given by $xy=\sum_{k,l=1}^{\infty}\frac{a_kb_l}{2^{k+l}}$
Since $x,y$ is irrational, then the sequences $a_k,b_k$ does not contain periodic components
If $x$ and $y$ can be decomposed into $w+z$ where one of these is rational (hence has a recurring binary expansion), then $x$ or $y$ must contain some subsequence that is recurring.
The first thing to make sense of, is for the two possible case of summation (with or without carryovers) what kind of map does that corresponds:
Take the number $0.001_2$ as example. This can be written as the expansion $\sum_{k=1}^{\infty}\frac{c_k}{2^k}$ where $c_k =(0,0,0,1,0...)$
Now, if I pick some sequence $d_k$ where $k\neq 4$, then the sum $s_k = c_k+d_k$ will have a 1 at the position $k \neq 4$
This means, suppose a binary sequence $a_k$ has the $j$th component = 1, and we are interested in computing the sum $a_k+b_k$. If $b_j=0$, then $s_j=1$
However, the more interesting (and in some sense harder to understand) behaviour are carryovers:
Suppose we have binary sequences $a_k, b_k$ and that the ith position is =1. Then $s_i =0, but s_{i-1} = 1$ if $s_{i-1} = 0$
The trick is then how to formulate these two phenomenon in terms of some kind of shift mapping, so that the structure of carryovers in binary expansions can be better tracked
We currently knew the following: A binary sequence < 1 will have $a_0=0$ and $a_{n>0} \in \{0,1\}$
In particular, $a_j \in \{0,1\}$. Suppose $a_j = 0$. Then 3 things can happen:
$\forall j \in \Bbb{N},(a_{j-1},a_j)+(b_{j-1},b_j)$ is given by the recursion relation \begin{align} (0,0)+(0,x) & =(0,x),x\in \{0,1\}\\ (0,1)+(0,1) & =(1,0)\\ \end{align}
Now for all rationals $x \in \Bbb{Q}$ the following will hold as the binary expansion is recurring:
Hi! I'm working on the bootstrap percolation model for a school project, and I wonder if you had any suggestion about software to do this: -I want to make a simulation of the model (like this: http://mathworld.wolfram.com/images/gifs/bootperc.gif). -I want to make some illustration (such as https://upload.wikimedia.org/wikipedia/commons/7/7b/Bond_percolation_p_51.png).
About birthday problem. If we want the probability that 4 people share the same birthday, then we should do the following: Pr[ 4 people share the same birthday] = 1- {event where everyone has different birthday + event where all pairs have different birthday + event where all 3 people have different birthday} Is that right!
Given two binary strings $a_k$ and $b_k$. It's sum $s_k=a_k+b_k$ is given by the following algorithm:
1. Construct the sequence $s_k(a_k,b_k)$ which notate the common position of ones in $a_k$ and $b_k$
2. Subtract each element of $s_k$ by one entrywise to get the sequence of positions of carryovers $c_k$
3. Construct the sequence $u_k$ by taking the union of the positions of ones that are unique to $a_k$ or $b_k$
4. If $u_k$ and $c_k$ have no common entries, then we are done. Else subtract those common entries by 1 to produce another carryover position sequence $c_k^{(2)}$
5. Repeat step 4 until termination due to $c_k^{(n)}$ and $u_k$ have no common entries
i don't understand this concept. for instace: a={1,2,3,4,5,6}, the number of functions from a to a that are copying each number of a to one of it' diviors is $2^3*3^2$?
it seems to be true, since only 4 and 6 can be divided, and 1,2,3 are the only natural results of the division, so in a={1,2,3,4,5,6} the number of function from a->a that are copying each number of a is $3^2*2^3$
(cont.) Need to do some chemistry, thus will do some maths later. But a preliminary conclusion of the above investigation of $\pi$ and $e$ is that, due to the behaviour of carryovers in binary being basically a shift mapping of a sequence of common ones, it is possible the solution to the (ir)rationality of $\pi + e$ lies on the probability of finding some binary subsequence in both $e$ and $\pi$ and whether their intersection (common ones) is mostly contributed by $\pi$ or $e$
Hello!! At the following picture we have that $e^{j2\pi ft }=\cos (2\pi ft ) +j\sin (2\pi ft )$. Is the sinus equal to 0, since $2\pi ft$ is an integer multipli of $\pi$ : $e^{j2\pi ft }=\cos (2\pi ft ) +j\cdot 0=\cos (2\pi ft ) $
@robjohn In one of your recent answers, you asserted that $$\int_{-\infty}^{+\infty}f(g(x))\,\mathrm{d}x=\int_{-\infty}^{+\infty}\sum_{g(x)=\alpha}\frac1{\left|g'(x)\right|}\,f(\alpha)\,\mathrm{d}\alpha$$ without explicitly placing any conditions on $f(x)$ or $g(x)$. But aren't some conditions needed?
what are some of the practical applications of Stirling Number of the second kind? I somehow remember somewhere it was in counting functions, but dont remember clearly
Let $Res(f,g)$ be a resultant of two polynomials $f(x)$ and $g(x)$.
Is it true that resultant does not change under a linear change of coordinates
$x\mapsto x+\alpha$?
Thanks a lot!
I'd show that if $\alpha$ is algebraic over $F$ then $F(a)$ is an extensionnof finite degree. Then you show that finite degree extension are algebraic, so $\alpha^2$ is also algebraic. You can do the same with $F(\alpha,\beta)$ to get sums and products
Instead of using the whole algebraic closure which is a bit of a cannon to shoot a fly I'd consider the smallest of its subfields containing the algebraic numbers we're interested in
@RandomVariable This is a change of variables combined with the inverse function theorem. As long as things are integrable, it should be fine. You can require that $f\in L^1$ and that $|g'(x)|$ does not vanish to make sure that everything is integrable.
31.3 Theorem: A finite extension field $E$ of a field $F$ is an algebraic extension of $F$.
Proof: We must show that for $\alpha \in E$, $\alpha$ is algebraic over $F$. By Theorem 30.19 if $[E:F] = n$, then $$1, \alpha, \cdots, \alpha^n$$ cannot be linearly independent elements, so there exist $a_i \in F$ such that $$a_n \alpha^n + \cdots + a_1 \alpha + a_0 = 0,$$ and not all $a_i = 0$. Then $f(x) = a_n x^n + \cdots + a_1 x + a_0$ is a nonzero polynomial in $F[x]$, and $f(\alpha)=0$. Therefore, $\alpha$ is algebraic over $F$.
@robjohn I'm probably looking at this all wrong, but for $g(x) = x + \tan x$, I get $$\int_{-\infty}^{\infty} f(g(x)) \, dx = \sum_{n=-\infty}^{\infty} \int_{(n-1/2)\pi}^{(n+1/2) \pi} f(g(x)) \, dx = \sum_{n=-\infty}^{\infty}\int_{-\infty}^{\infty} f(\alpha) \left(g_{n}^{-1}(\alpha) \right)' \, d \alpha$$ where $g^{-1}_{n}(\alpha)$ is the inverse of $\alpha = g(x)$ on $\left((n-1/2)\pi, (n+1/2)\pi)\right).$
I don't understand how applying the IFT could possibly lead to the result if we don't first interchange the order of integration and summation, which is why I asked you about conditions.
The key to this is that the ACTUAL sum of all the hats (mod 100) must be one of the numbers from 0 to 99.
Whichever number it is, that player will have guessed his/her hat color correctly.
A corollary: no one else will have guessed correctly. Only the person whose player number matches the act...
Let $Res(f,g)$ be a resultant of two polynomials $f(x)$ and $g(x)$.
Is it true that resultant does not change under a linear change of coordinates
$x\mapsto x+\alpha$?
Thanks a lot!
