@ForeverMozart I know this, but overall, this is what I noticed over the time. Even if the tests I took maybe are not perfectly like the ones that are called to be real, the trend was always to get higher and significantly higher.
@I'manartist no ploylogs :( .. you've gotta be kiddin me .. well whatever can be managed by integration by parts of basic series transforms can be achieved by summation by parts too .. so I guess I shouldn't be surprised :) Seeing a first-hand calculation that way would be awesome though :D
A manufacturer has six distinct motors in stock, two of which came from a particular supplier. The motors must be divided among two production lines, with three motors going to each line. If the assignment of motors to lines is random, find the probability that both motors from the particular supplier are assigned to the first line.
The answer is $1/5$ - what I don't understand is where the $5$ comes from?
@Michael I've been told the probability is the number of outcomes in which the two from the particular supplier end up on the first line divided by the number of outcomes. However, I don't know how to calculate either of those things. The total number of outcomes would be $6$ in my mind, which is most likely wrong.
Yeah, this forum is amazing! I dislike asking questions when I've no reasonable attempt to show, though, which is unfortunately the case with this one.
According to Wikipedia the permutation matrix of $\pi \in S_n$ is given by $P_\pi$ with $(P_\pi)_{ij} = \delta_{\pi(i),j}$, which apparently results in $P_\pi P_\tau = P_{\pi \tau}$. But if I’m not mistaken we actually get $P_\pi e_i = e_{\pi^{-1}(i)}$, which results in $P_\pi P_\tau = P_{\tau \pi}$. Can anyone find whatever mistake I’m making?
@JulianRachman I have not read the citing paper thoroughly yet, though it seems to just be a "similar things have been done in..." citation, rather than an actual application of the cited paper.
(Not that this should be so surprising given the short time. It would have been really impressive if someone had found an application plus the time to write it up in such a short time)
even though the person citing us would probably have been able to, as he is a bit of legend by now
I am developing the quadratic sieve algorithm and I reached a new bottle neck: The matrix processing.
I been reading quit a lot about this topic and I found many solutions
Gaussian elimination: This perhaps the most common approach for this problem. It's running time is $O(N^3)$ above GF (2)
M...
@I'manartist Well in that case you have a short proof of $\sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{2^nn^2}$ too (the one I asked about on M.Se and you answered) :-)
@I'manartist Incredible!! :D I need that book now before my brain explodes in excitement :P .. will you keep a separate chapter on Euler-sums (in the book I mean)?
Consider the polynomial ring $R[x]$. $Z[x]_2$, an instance of $k Z[x]$, is a subring of $R[x]$, an ideal of $R[x]$. The task is to show that an ideal exist with some k>0 of the polynomial ring $R[x]$.
@r9m No, my book is structured in a different way. I don't have a chapter dedicated to the Euler-sum, I mean I don't have a chapter called like that, but I have all the crazy stuff from the Euler-sum area. :-)
@I'manartist so you are telling me that all the contents that you intended to be in the book have been added and what left to do are mere formalities of the publication-house? :)
@TobiasKildetoft For now I cannot think about any other approach to prove it in general. The variety V(x^k-x^{2k}), the ideal <x> with some k>0 -- perhaps, thinking.
I have NOT chatted much is what I meant to write. Sorry, but this is not a good time for me. I am getting ready for work and just got up. I may have to catch up with you all later. I was responding to r9m's 'hello'.
@MikeMiller: do you have any idea how it makes sense to talk about the translation length of a parabolic isometry? if we take the infimum of $d(x, f(x))$, this will always be zero for parabolic isometries (but not realized), so what could be meant by it? fix some point and then compare? does such a comparison even make sense?
or maybe just use the Euclidean metric and then it makes sense?
@TobiasKildetoft Consider $n,m\geq 1$ with $x^m=x^{m+n}$. With $k\geq m$, $x^{k+n}=x^k \forall k\geq m$. With $n\geq m$, $x^{2n}=x^{k+n}=x^k \forall k\geq m, n \geq m.$ With $k=n$,
I see that you started with rings, but is this not just slight variation of the proof that every finite semigroup has idempotent. math.stackexchange.com/questions/353028/…
Or maybe I should have linked to the proof at ProofWiki, which seems similar to what you were doing above: proofwiki.org/wiki/…
Consider $n,m\geq 1$ with $x^m=x^{m+n}$. With $k\geq m$, $\forall k\geq m \quad x^{k+n}=x^k$. With $n\geq m$, $\forall n \geq m \exists k\geq m \quad x^{2n}=x^{k+n}=x^k.$ With $k=n$, $\forall n \geq m \exists k\geq m \exists k=n \quad x^{2n}=x^n$ so
$$\forall n \geq m \exists k\geq m \exists k=n \quad x^{2n}=x^n$$
The proof which does not need any prerequisites is the same as hhh is trying to write down: Show that there is a cycle and play with it until you find an idempotent.
I recall that I found the way this proof is written down in the book by Hindman and Strauss quite nice. And also some answers in the post on main contain quite clever arguments.
@hhh There is no for all $n$ here. We have a fixed pair $(n,m)$ with a certain property (namely that $x^{m+n} = x^m$), and we need to show that we can also find such a pair with the added property that $n\geq m$.
$\forall n \geq m \exists k\geq m \exists k=n \quad x^{2n}=x^n$ is equivalent to $x^{2k}=x^k$ in some finite ring R $x\in R$ where the sufficiency has not been proved yet.
@hhh Well, if $R$ is a ring then $(R,\cdot)$ is a semigroup.
So if you have some result which is true for any semigroup/finite semigroup; then the same result holds for $(R,\cdot)$ if $R$ is any ring/finite ring.
It is a very fascinating question to me. Associate to each sequence $a=\{\alpha_n\}$ in which $\alpha_n$ is 0 or 2, the real number $$x(a)=\sum_{n=1}^{\infty}\frac{\alpha_n}{3^n}$$. Prove that the set of all $x(a)$ is precisely the cantor set@BalarkaSen
I can see that we have $3^n$'s in the denominator so intuitively this should be somehow related to the cantor set. But how should I prove it
So I have find that surjection and show that $x(a)$ is somehow related to the surjection(here I mean the surjection you talked about between the cantor set and [0,1]?@BalarkaSen
@Balarka A kind of observation: $$\sum_{n=1}^{\infty}\frac{\alpha_n}{3^n}=\frac{\alpha_1}{3}+\frac{\alpha_2}{3^2}+\frac{\alpha_3}{3^3}+...$$ . Now if I set $\alpha_2$=2 and the rest as $0$ I get $\frac{1}{3}$. Now similarly if I set $\alpha_1=2$ and the rest all $0$ I get $\frac{2}{3}$. This is just, when for the first time $(\frac{1}{3},\frac{2}{3})$ is removed from $[0,1]$. Similarly we can go for the other end points of the intervals.
@Balarka I am getting all the end points. Now I am thinking does the cantor set have points except the end points of the intervals? I would have to do something for them then
@DanielFischer hi, I have two linear maps $A,B$, if I define $B\circ A$ it's a bilinear form (we take three normeds space E,F,G). It's continuous and norm smaller than $1$. In my course they said that the norm is equal to $1$ but it's not trivial. Do you know how can I prove this ?
@BalarkaSen Well I have a feeling that all numbers in the cantor set have a relation with the number 2 but I have loads of physics left to do so I have to leave this for tomorrow
@Balarka there is this book recommendation that I want from you. What could be a good book for number theory that also deals a bit with transcendental number theory?(Ireland and Rosen(sure making some spelling mistake somewhere)?)
Ever heard of 'bordism invariance theorem' or something like that? It says closed $4n$-dimensional manifolds which occur as the boundary of an oriented closed $(4n+1)$-fold have zero signature.
(1.3, where 1 is best and 5 is fail, and 0.3 is the smallest increment possible)
I was most clueless on the following question: Let $X$ have finite fundamental group. Show that every $f:X\to S^1$ is nullhomotopic.
The induced map on $\pi_1$ is trivial and then I thought I could somehow just homotope the entire image using the homotopies on paths but I heard that one needed some little bit of covering space theory, and I have no idea how to do that.
I am developing the quadratic sieve algorithm and I reached a new bottle neck: The matrix processing.
I been reading quit a lot about this topic and I found many solutions
Gaussian elimination: This perhaps the most common approach for this problem. It's running time is $O(N^3)$ above GF (2)
M...
