lets say you are tracking a point that meanders through a mesh-like region, in discrete jumps. however you keep refining the mesh such that the point looks as though it is moving continuously through the space
+1 for a very good addition to my and others' list-the Shifrin notes are a gem online and I wasn't aware of the Kleiman notes! — Mathemagician1234Mar 3 '13 at 19:58
From the elementary inequality $$\frac{\sin x}x\le\frac{2+\cos x}3,$$ we get with $x=\pi/6$ easily $\pi\ge\frac{18}{4+\sqrt{3}}=3.1402\ldots$
Proof of the inequality (elementary, though not obvious): let $$f(x)=\frac{\sin x}{x(2+\cos x)}.$$ In order to prove $f(x)\le\lim_{x\to+0}f(x)$, we prove $...
So I'm still thinking about fibers of morphisms. This time of the morphism of schemes induced by $k[x]\hookrightarrow k[x,y]/(xy)$. By looking at the picture the fibers over $(x-a)$, $a\neq 0$ and the fiber over $(x)$ should be different, indeed I got $\operatorname{Spec} k$ for the former and $\operatorname{Spec} k[x]$ for the latter. I don't know how to interpret those results geometrically though
$f$ is singular on the whole circle $C = \{y^2 + z^2 = 1\}$ in the $xy$-plane, and $f|_C$ itself is singular at the pair of antipodal points $\{(1, 0), (-1, 0)\} \subset C$.
Away from those two points $f$ is a fold singularity near $C$, and those two points are cusps.
If $X$ is a prevariety there is a standard way to associate a scheme $t(X)$ to it. Is it true that if $X$ and $Y$ are prevarieties over $k$ then $t(X)\times_k t(Y)=t(X\times Y)$? It looks like it's true to me
A topological space $X$ with a sheaf $\mathcal O_X$ such that there is an open cover $U_i$ of $X$ such that $(U_i,\mathcal O_{|_{U_i}})$ is isomorphic to an affine variety (an irreducible algebraic set in $k^n$ with its structure sheaf)
Basically something that looks like a variety locally
The usual example of a prevariety which is not a variety is the line with two origins
I'm interested in the above statements for varieties as well
Because to get $\Bbb P^1$ you glue together $\Bbb A^1$ and $\Bbb A^1$ along the map $\Bbb A^1\setminus\{0\}\to\Bbb A^1$ given by $t\mapsto 1/t$, to get the line with two origins you use the map $t\mapsto t$
prevarieties are really just a slight generalization of varieties as algebraic subsets of $k^n$ with a structure sheaf. But you can associate a scheme to every prevariety
If someone brought up the words 'finite group representation' why would your mind immediately go to the concept of averaging a numerical function over the finite group?
The existence of a positive definite Hermitian form is very useful to decompose your representation once you have produced a proper sub-representation; I don't believe it can be used effectively for producing sub-representations. However, in practice, the averaging idea used to produce the scalar...
is there a proof that for algebraically closed field $k$, $ax^2+bx+c$ has double roots iff $b^2-4ac = 0$ that does not do casing on $\operatorname{char}(k)=2$?
Besides, there's a host of theorems that are true for characteristic not 2 but are false for characteristic 2. So any such technique would only work on a some special class of statements
Because to get $\Bbb P^1$ you glue together $\Bbb A^1$ and $\Bbb A^1$ along the map $\Bbb A^1\setminus\{0\}\to\Bbb A^1$ given by $t\mapsto 1/t$, to get the line with two origins you use the map $t\mapsto t$
The crucial difference between the two schemes is that $\Bbb P^1$ has no nontrivial maps to $\Bbb A^1$, but the line-with-two-origins obviously has one.
@AkivaWeinberger luv it but (lol) somehow he didnt use word fractal a single time in article. o_O ... found another 2009 paper on Riemann + Collatz fractals lately by King luv it you might find something interesting (33 refs to "fractal") empslocal.ex.ac.uk/people/staff/mrwatkin/zeta/king2010.pdf
@LeakyNun In general, a polynomial has a repeated root iff $\prod_{i\ne j}(r_i-r_j)=0$, and (at least in the monomial case) that gives you the discriminant. (It's symmetric in the roots so it becomes a polynomial in the coefficients)
The nonmonomial case would be the homogenous version (homogenification? I dunno) of the monomial case, probably
i have this one Michael Mitzenmacher, Eli Upfal-Probability and Computing_ Randomized Algorithms and Probabilistic Analysis-Cambridge University Press (2005) @vzn
@Theantomc that sounds pretty advanced. youre undergrad right? what year? did you figure out the R^2 E(x) problem yet? suggest intro level stats for that type of thing
yes i m degree in CS. We do few point in probability, just about some initial stuff. We study just basic probability and when i change course i find Markov and other inequality that left me stupid @vzn
in a book i find this E(yi - zi)^2 = E(yi^2 ) + E(zi^2 ) + 2E(yi zi) = Var(yi) + Var(zi) + 2E(yi)E(zi)
@Theantomc yes those are basic properties of E(x) as outlined on the wikipedia page try to understand/ prove the basic ones & then youre in good/ better shape for understanding YFs proof
in that calculation 2E[(x−1/2)2]+E[x−1/2]E[y−1/2]=2E[(x−1/2)2] in don't understand why E[x−1/2]E[y−1/2] will be deleted, seem that the result is 0, but i dont understand why, for me the 2 expectation are the same value
Let $q$ be a power of a prime and $n\in \mathbb{N}$. I want to calculate the dimension of the image of the linear map $\theta : \mathbb{F}_{q^n} \rightarrow \mathbb{F}_{q^n}$, $\theta (\beta)=\beta^q-\beta$.
For that we have to find a basis of the image, right? But how can we determine a basis? Could you give me a hint?
@vzn maybe i caught the point. Because the mean is area of the rectangular , in my case the rectangular are based in 0 and from -1 to 1 this area is 0. I see that can calculate this expression , in 0 to n is (n+1)/2 so i hope that i m right
@LeakyNun We have $$\ker (\theta)=\{\beta \in \mathbb{F}_{q^n}: \theta (\beta )=0\}=\{\beta \in \mathbb{F}_{q^n}: \beta^q-\beta=0\}$$ Are there $q^n$ elements in the set?
> Behind it all is surely an idea so simple, so beautiful, that when we grasp it - in a decade, a century, or a millennium - we will all say to each other, how could it have been otherwise? How could we have been so stupid for so long? - John Archibald Wheeler
@Theantomc its basically "expected value of a balanced distribution". it works for gaussian or uniform (your case) or ... try to look it up/ prove it using a property out of a intro stats book or page. it doesnt seem to be on the wikipedia pg, seems an oversight en.wikipedia.org/wiki/Expected_value
aha found it for you For a random variable following this distribution, the expected value is then m1 = (a + b)/2 and the variance is m2 − m12 = (b − a)2/12.en.wikipedia.org/wiki/Uniform_distribution_(continuous) ... this is a general property of "balanced" or "(left/ right) symmetric" distributions. try proving it using calculus... it presumably has some name somewhere...
i have to prove that for a sequence $(u_n)$ in $(\mathbb{R},|.|)$ if $\lim_{n\to\infty} u_{2n}=l_1$ and $\lim_{n\to+\infty}u_{2n+1}=l_2$ then $l_1$ and $l_2$ are the only adherent value for $(u_n)$ . I SUPPOSE THAT THERE IS AN OTHER ADHERENT VALUE $l$it means that there exists a subsequence $(u_{\varphi(n)})$ such that $lim_{n\to+\infty} u_{\varphi(n)}=l$ how to continue please ???
So if you have a sequence that converges to something other than $l_1$ and $l_2$, then it can't have infinitely many even values and it can't have infinitely many odd values
