i read in the course syllabus for ... advanced real analysis (a course that we have in university of Stockholm) that there are some applications on probability
Fourier once said. "Yesterday was my 21st birthday, at that age Newton and Pascal had already acquired many claims to immortality." It was 34 years later he'd be publishing his "Theorie Analytique de la Chaleur."
The dot product $A\cdot \vec{x}$ of a matrix $A$ with coefficients $a_1, a_2, \dots, a_n$ and vector $\vec{x}$ consisting of \textit{n}-tuple $(x_1, x_2, \dots, x_n)$ is denoted as $$ \sum_{i = 1}^m\sum_{j = 1}^n a_{ij}x_j $$.
In general, $A$ transforms the vector $\vec{x}$ into a new vector $B$ - the solution - via matrix multiplication, which is defined to be the dot product. In the case of two matrices $C$ with coefficients $c_1, c_2, \dots, c_s$ and $D$ with coefficients $d_1, d_2, \dots, d_s$, we write
your equation after "we write" doesn't use the $c$'s or $d$'s at all
and again surely $C$ and $D$ are arrays (matrices!) not lists
it also doesn't make sense to say "we write [expression]" without saying what we're writing - you have to put an equation down or say it's a definition of something
Hey guys, I found something cool. Mathjax also allows to write math in an other math lanugage, which is asciimath. This is not as complete as latex, but is easier and faster to write than latex. And maybe usefull for the chat here.
jsfiddle.net/4cqVD/1 Here is a chatjax script that allows both latex as asciimath. In this way you can write a matrix like this for example `[[1,2],[2,3]]`
This languages is not so complete as latex. But I can remember having chats here and wanting to type faster. This language is quite fast. \Bbb Z \times \Bbb Z becomes ZZ xx ZZ for example
So this script basicly allows both latex as this asciimath language. Which I thought is pretty cool :)
It's from Peter Jipsen who made this in 2005 I believe.
He was trying to make a language which is close as possible to the language mathematicians use when typing emails etc. and having no tools available to render it as math. After that he extended it with latex like notation, but that is typed faster.
But it started out with things like (1+3)/(2+3) that renders as \frac{1+3}{2+3}
But if I'm going to use this chatjax script, and nobody else, than this is kind of silly, as I'm the only wants that sees it. So @robjohn would you be willing to add this extended chatjax bookmarlet to your website ? You can find it here: jsfiddle.net/4cqVD/4 In short: It keeps working exactly the same, but it allows math to be typed also in an other lanuage (asciimath). See the jsfiddle for examples.
cocycle: in Weibel, intro to homological algebra, i am uncertain of the correct interpretation of the expression "Hom_R(Z_n, C)" for some chain of R modules C. i take this as the chain with degree k elements the R linear maps from the n cycles of C (i.e. ker d_n) to the degree k elements of C. is this correct, or should i be parsing Z_n as a chain itself?
This is exercise 1.1.4, page 3. The goal of the exercise is to prove that The nth homology of the cited expression vanishes => the nth homology of the chain C vanishes
I believe I have a proof (take the conanical Monic sending Z_n to C_n and use exactness of the given hom chain to show the existence of some map j: Z_n -> C_n+1 the image of C_n+1 under dj contains Z_n, hence H_n is zero), yet this assumes my interpretation is correct
Pedro, we can't simply do $(a,x)^{-1} = (a^{-1}, x^{-1})$ what would the inverse of $(1,0)$ be, that is the general argument so forget about that first form I posted, but look at:
in the definition of direct/inverse limits by universal property, where you've got A as the limit and B as some other group and they're both projecting onto members of the family of objects, is the induced morphism between A and B dependent on the index of the members, or does it have to be the same one no matter who you're projecting onto?
(I'd draw it but i'm useless at commutative diagrams in mathjax)
@AlexanderGruber consider $A_i:=C_2^i$ (direct product of $i$ copies of $C_2$ in AbGrp) with projections being the "take out all the outside coordinates" maps. if $G$ is some abgrp and you know $G\to C_2^i$ for all $i$, then you can construct $G\to\varprojlim C_2^i=\prod C_2$ by simply putting all of the coordinate functions together. this will be unique, and it depends on what the individual maps $G\to C_2^i$ are for $i=1,2,\cdots$
@robjohn I had quite a silly question. I have the inner product $$\int_{-1}^1 PQ$$ in $\Bbb R_2[X]$. I calculated $S^{\perp}=\langle 1+X\rangle^{\perp}=\langle 1-3X^2,X-X^2\rangle$ and I have to find the point there that minimizes $\int_{-1}^1(P-1-x^2)^2$ which is of course $\lVert P-Q\rVert^2$, $Q=1+X^2$. But this is minimized by $\lVert Q-\pi_{S^{\perp}}(Q)\rVert =\lVert \pi_{S}(Q)\rVert$, yes?
My textbook gives the comparison test for infinite series only for series with positive terms (if an and bn are sequences with an and bn \geq 0, then if an \leq bn, then such and such). Does it also hold for series with negative terms (if an and bn are sequences, then if an \leq bn, then such and such)? I don't see why not, but I'm just checking to be safe.
@anon yeah, whenever I take a concrete example it makes sense
but the definition was confusing me because it wasn't specified whether it was "for all $i,j$, a unique $\phi$ exists such that the diagram commutes" or "there exists a unique $\phi$ such that the diagram commutes for all $i,j$"
2nd one's a much stronger statement and is what i figured it was but i was nervous it could be the first one
Are these pairs of vectors orthonormal, only orthogonal, or only linearly independent? Change the second vector when necessary to produce orthonormal vectors.
$\begin{bmatrix}0.6\\0.8\end{bmatrix}$ and $\begin{bmatrix}0.4\\-0.3\end{bmatrix}$
They're orthogonal, since $(0.6)(0.4)+(0.8)(-0.3)=0.24-0.24=0$, but they're not orthonormal. I'm curious about if my math is right though to change them to orthonormal as per the instructions.
@AlexanderGruber in order to state the compabitility conditions you need to quantify over two variables, $i$ and $j$, taken from the index set. ("for all $i,j\in I$ with $i\le j$, $A\to A_i$ factors through $A\to A_j$ and $A_j\to A$). to state the existence and uniqueness of $B\to A$, you need to quantify over just one variable $i$ taken from the index set (for all objects $B$ and compatible maps $B\to A_i$ for all $i\in I$, there exists a unique $B\to A$ yadda)
in reality, the commutative diagram has all infinity of the arrows, not just two, but the diagram focuses on just two to get the point across (of compatibility)
in any case quantifying over $I$ once or twice doesn't really matter: in the end all you're saying is that the induced morphism depends on the total collection of morphisms
@AlexanderGruber then your diagram does not meet the condition set out by the "for all ..." inherent in the UP
although if you can define the morphisms on a cofinal subset of the index set there is a good sense in which that is "good enough" (since you can determine all of the other morphisms from those via the compatibility conditions)
@anon one more small thing you might be able to help me with: it's a terminology question. i'm learning about representable functors. a problem i'm given asks me to prove that in the category of torsion abelian groups, products are representable. do you know what that means? i'm not sure how a product can be representable, only definitions i've seen have made products objects (or triplets of objects w/morphisms)
I was thinking hom(A,B) was like $A^*\otimes B$, so the functor defined by "direct producting against $A$" would be naturally isomorphic to "apply $\hom(A^*,-)$
my understanding is that a direct product $\Pi$ of $X$ and $Y$ (equipped with projections $\Pi\to X,Y$) is such that any maps into $X,Y$ from a common source simultaneously factor through $\Pi$
@AlexanderGruber the number of objects we're quantifying over can be lessened, so the product need not be as "powerful." for example direct sum is coproduct in AbGrp but free product is the coproduct in Grp (the direct sum is not powerful enough for the UP here). again I am not sure about product, I only know this one example for coproducts.
it's probably trivial to those well-versed in UPs and such
@Pedro I understand the bigger picture, where we want to see if the points are within 2 epsilon of the limit. I am still having trouble finding the M given an epsilon.
If I want to find an orthonormal basis for $\mathbb{R}^{3}$, including the vector \begin{align*}\vec{q}_{1}=\frac{1}{\sqrt{3}}\begin{bmatrix}1\\1\\1\end{bmatrix}\end{align*}, do I have to use the Gram-Schmidt algorithm, or is there another way?
the only time a free product is abelian is when one of the factors is trivial and the other abelian
the coproduct (in Grp) of two nontrivial abelian groups will be nonabelian, whereas the coproduct (in AbGrp) of the same two groups will be their direct sum, which is abelian
Damn. Nothing like solving a tough problem to make yourself feel intelligent. Though I wonder if my solution was really that clever. I overlooked how simple it actually was at first. @PedroTamaroff @robjohn How would you guys proceed to solve this?
Find an orthonormal basis for $\mathbb{R}^{3}$, including the vector $\vec{q}_{1}=\frac{1}{\sqrt{3}}\begin{bmatrix}1\\1\\1\end{bmatrix}$?
Well, that's one.. I came up with $\left\{\frac{1}{\sqrt{3}}\begin{bmatrix}1\\1\\1\end{bmatrix},\ \frac{1}{42}\begin{bmatrix}1\\-5\\4\end{bmatrix},\ \frac{1}{14}\begin{bmatrix}-3\\1\\2\end{bmatrix}\right\}$
My solution isn't displaying on mathb.in so I'm not going to bother putting it on there... But lets just say I solved for two vectors $\begin{bmatrix}a\\b\\c\end{bmatrix}$ and $\begin{bmatrix}x\\y\\z\end{bmatrix}$ so that the whole set were pairwise orthogonal and then scaled down the two I found.
Where I really felt clever was where I came up with $\begin{bmatrix}-b-c\\b\\c\end{bmatrix}\cdot\begin{bmatrix}-y-z\\y\\z\end{bmatrix}=0\iff\begin{bmatrix}b\\c\end{bmatrix}\cdot\begin{bmatrix}2y+z\\y+2z\end{bmatrix}=0$. Then I just picked two nice $y$ and $z$ and solved for $b$ and $c$.
Though that may have been just blind luck as I picked $y=1$ and $z=2$ so that $2y+z=4$ and $y+2z=5$. $4b+5c=0$ was easy to solve then, but that wasn't by any means intentional.
This morning I created this one ...$$\lim_{n\to\infty}\left(\sqrt[\large2n+2]{\prod_{k=1}^{2n+2}\psi \left(\frac{1}{k} \right)}-\sqrt[\large2n]{\prod_{k=1}^{2n}\psi \left(\frac{1}{k} \right)}\right)$$
Let $a_1,a_2,\ldots,a_n\in \mathbb R$ and $f(x)=a_1\sin x+a_2\sin 2x+\ldots+a_n\sin nx$ such that $|f(x)|\leq|\sin x|$ for every $x\in \mathbb R$. Prove that $|a_1+2a_2+3a_3+\ldots+na_n|\leq1$.
Please prove without using of Derivation. Only by triangular equalities
