@Studentmath you could try to enumerate the family of graphs in the 2nd half of this paper. it's pretty interesting, though I hear the author is a d-bag who advertises it shamelessly.
I need help to understand the probability derviation of a paper. Please help me. For the following, please only treat $|h_{R,B}|^2$ and $|h_{A,R}|^2$ as random variables (other parameters can be treated as constants). Those two random varibles are i.i.d. such that they are exponential distributed...
suppose we have $nk$ variables in $n$ linear equations and $n^2$ homogenous quadratic equations over an algebraically closed field, and $1\le k\le n$. is there a simple argument that there is at most one solution?
(I am assuming that there is at most one solution)
more intro books to abstract algebra would restrict themselves to groups, rings and fields, with very little order theory
you would need a text on universal algebra specifically, but that would still probably be too broad. best to just pick up a book that's specifically about order theory or lattices.
I have proven $\Bbb C[G]\cong\bigoplus_{V\in{\rm Irr}(G)}V\otimes_{\Bbb C} V^*$ (both as $\Bbb C$-algebras and left/right $\Bbb C[G]$-modules) and $\sum_{g\in G}\chi_V(g)\chi_W(g^{-1})=|G|\delta_{VW}$ with basically no character theory at all. I am wondering how to leverage these two facts to prove the multiplicative identity of the $V$-isotypical summand of $\Bbb C[G]$ is given by $\frac{\dim V}{|G|}\sum_{g\in G}\chi_V(g^{-1})g$.
@Bob I have forgotten which video it is. But roughly the derivation I remember is. He takes the derivative of the Geometric series somehow repeatedly and arrives at the exponential function.
Kind of like this in Mathematica: Clear[x] Series[1/(1 - x), {x, 0, 12}] D[%, x] D[%, x] D[%, x] D[%, x] D[%, x] D[%, x]
% means the previous lines output
Given that you know that this remains unchanged: D[E^x, x] D[%, x] D[%, x] D[%, x] D[%, x] D[%, x] D[%, x]
D[%,x] is the derivative of the previous output
The point I guess is that you need not take the geometric series, you can take any taylor power series and arrive at the exponential function through repeated derivatives, given that E^x is a function such that its derivative is E^x unchanged.
no, wait, now I said too much. Probably not any power series. The geometric series it should be.
If $$A= A_1 \oplus A_2 \oplus \cdots \oplus A_n$$, how can one prove that $B \subseteq A $ iff $$B=B_1 \oplus B_2 \oplus \cdots \oplus B_n, B_i \subseteq A_i?$$
I'm not sure if this is actually true. Obviously the $\cdots \Rightarrow B \subseteq A$ is easy.
@Sush I had never heard of "order theory" before. I just looked it up on Wikipedia and it just seems to be the study of order relations in many branches. I don't think it belongs to any branch; it is sort of branch-orthogonal.
I'm going through some old functional analysis lecture notes and I don't remember what the following notion is supposed to mean: $$H_0^1(\Omega) := \| \cdot \|_{H_0^1} - \operatorname{clos}(C_0^{\infty})$$ I know it is supposed to be a Hilbert space but I have no idea what the right side of the equation tells me. I know how the norm on the space is defined: $$\| u \|_{H_0^1}^2 = \int_\Omega | \nabla u | ^2 \, \mathrm dx$$ This Hilbert space apparently is helpful for solving the Dirichlet problem using Riesz' theorem.
@DanielFischer let $a, b$ be rationals. When covering $[a, b] $ with $a+k(b-a)/n$ for all $n$ and relevant integer $k$, I'm pretty sure that some rationals in $[a, b] $ are left out, but I can't prove it. The rationals are countable, so I can't get any argument based on that
Ok, so $z = x + iy$ and $x = \frac 1 2 ( z + \overline z)$. So from the first equation $\frac {\partial z} {\partial x} = 1$ hence $\frac {\partial x} {\partial z} = 1/1 = 1$, but from the second equation $\frac {\partial x} {\partial z} = \frac 1 2$, treating $z$ and $\overline z$ as independent variables.
@r9m Well, those solutions should motivate us, not stop us doing things. I mean, you need to think that you'll be able to do such things and even greater ones during the time. :-)
@r9m in general, things will flow naturally, you only need to work, learn the stuff you like. Just think about it: what should I say about myself that I have no background in mathematics? I don't discourage myself because of that. Moreover, I wanna become like Ramanujan and even more ... (in terms of integrals, series and limits)
@r9m Never give up your dreams! Fight the impossible and put it down every single day! :-)
@LTS For functions of several variables, you can't just take the reciprocal of $\frac{\partial x_i}{\partial y_j}$ to obtain $\frac{\partial y_j}{\partial x_i}$. You need to invert the Jacobi matrix of the coordinate change, $$\begin{pmatrix} \frac{\partial x_i}{\partial y_j}\end{pmatrix}^{-1}_{1\leqslant i,j \leqslant n} = \begin{pmatrix} \frac{\partial y_j}{\partial x_i}\end{pmatrix}_{1\leqslant j,i\leqslant n}.$$
@BalarkaSen I knew @Chris'ssis is yet to write a book .. so I was asking if he/she had it on a paper .. paper and books entirely different idea .. they are not comparable :|
@r9m Have you ever seen this one? $$\lim_{ n\to \infty} \sqrt{n}\left(l-\sqrt{ 1+ \sqrt{2 + \cdots +\sqrt{n}}}\right)^{1/n}$$ where $l$ is the limit of the nested radical.
Hi, are there any mods around with superpowers? I'm trying to read the (in my browser corrupted) third comment under my answer here. Perhaps if one inserts some extra spaces (or perhaps introduce more $$'s), the comment (or at least parts of it) becomes readable.
@robjohn If an author says to assume that $f(a+z)$ can be expanded in the form $f(a+z) = c_{0}+c_{1}e^{-z} + c_{2}e^{-2z} + \ldots$, what sort of functions is he referring to?
@r9m I just had my official oral exam. At the end, the guy asked "what are, in your opinion, the most important theorems in analysis and linear algebra?" :/
@robjohn The author describes a method to evaluate a certain class of definite integrals. But you need to know the function you're dealing with could in theory be expanded in such a form. It's unclear if $(z+a)^{-n}$ could be expanded in such a form?
@robjohn gp-pari has a sumalt command, that rather quickly sums alternating series, or even things that are almost alternating, up to a given precision (took a minute for it to sum a 'rather complicated' series up to 100-digit precision). does mathematica have such a command?
Let's be frank here : Integrals and the connections between them is usually a deep subject to study and is related to transcendental number theory. I have just learned about some of them and I was simply awestruck.
For example, google [Kontsevich + Zaigier + Ring of Periods]
iirc, it's the number of functions from the empty group to the empty group, and thus there is only a single such function - the empty function, thus, set theory wise, $0^0=1$
I wonder if I hope for sirens in my test tomorrow or not. If it goes well I guess not, but otherwise it could be helpful..
