Assume you contain no constants, then you also can't contain any polynomials whose coefficient of $t^n$ is non-zero, because otherwise taking that derivative barely enough times would kill it, then keep on going, and you see that this invariant subspace is just $0$
[Random] Inspired from looking at the way the calculation jobs are arranged: Sequence of indefinite length $\{a_0,a_1,\cdots a_n\}$, where $n \in \Bbb{N}$ and is unknown
@Daminark Note that if you contain a linear polynomial $at + b$ with $a \neq 0$, you have to contain all the constants (because $a$) and also $t$ (subtract an appropriate multiple of $a$ so that you get containment of $at$)
I think the technique will be something like that. Say you have a polynomial $f(x)$ of degree $n$. Write $f(x)$ as $a_n x^n$ plus a linear combination of it's derivatives.
Inductively prove you are containing $x^k$ for all $k \leq n$
So now, take a compact metric space and a group $G$ of homeomorphisms. Show that there's some non-empty, closed, $G$-invariant subset $X'$ on which $G$ acts minimally
About the only thing I can add is that, by the magical power of Google, I came across the following sentence in a preprint: "Choose, by a Zorns lemma argument, a closed G-invariant subset Z of X such that the restriction of the action on Z is minimal."
I'm annoyed at myself for not knowing how to answer it. Just because I know what Hamiltonian mechanics is doesn't mean I know how to answer questions about sympletic manifolds :/
I'm tempted to table this problem until office hours
And work on the other stuff
I'm nearly at the stage where I can prove that if $\{S_k\}_{k=1}^n$ is a partition of $\mathbb{Z}$, then some $S_k$ contains arbitrarily long finite arithmetic progressions
@Daminark is closure of a $G$-invariant set also $G$-invariant? say $A$ is the set and $z$ a limit point so that there is a sequence $z_k \to z$ with $z_k \in A$. Then for any $g$, $gz_k \to gz$, right? so $gz$ is also a limit point of $A$, isn't it?
I'm working on the lecture, there's a reason I'm spotty on this chat
Just that it's taking me a long time for me to figure out some stuff that ended up being easy so I'm kinda breaking off from the work for occasionally in hopes that I'll figure it out as I get back to it
@Daminark Do you know an example of a minimal set S of a G-action on X such that orbits of points in X are neither (1) all of S nor (2) dense in X? (maybe forget about it if it's not relevant to your lecture)
@Daminark I have to check if I am parsing it correctly (I'm having trouble moving from foliations lingo to this), but here's what I'm trying to say. Consider Z/n acting on S^1 generated by rotating 2pi/n; then S = nth roots of unity is a minimal set, and closure of every point is all of S. Consider Z acting on S^1 generated by rotating 2pi*sqrt(2) angle; then S = S^1 is the only minimal set.
an S which is neither of those (closure of none of the points is all of S, S $\neq$ X) is called an "exceptional minimal set"
these are the most interesting examples of minimal sets imo. if you want i can refer you to read a few pages of Candel-Conlon (the foliations textbook i have been reading) to check out an example. although it's probably not too relevant to whatever you are studying in dynamics
Consider a max heap tree with $100$ elements and a node from the same level is chosen randomly among all valid balanced heap trees with $n$ nodes. What is the probability that it is the $36^\text{th}$ smallest element______ .
My attempt:
(Please, find definition of Binary Heap on Wiki: https...
If $f$ is a continuously differentiable function and $f \rightarrow a$ , for some $a \in \mathbb{R}$ as $x \rightarrow \infty$,then $f^{'}(x) \rightarrow 0$ as $x \rightarrow \infty$ ? is this true or $f^{'}(x) \rightarrow b$ for some $b \in \mathbb{R}$ ? I thought that as $f(x)$ tends to $a$ so it is constant htere after and hence the derivative would be zero as $x \rightarrow \infty$ , is this reasoning correct?
Funny enough, for that counterexample, even though the limsup and liminf are + and - infinity respectively, you kind of can say that $f'(\infty) = 0$ :P
@LeakyNun Either extend the domain to $S^1 \cong \mathbb{R} \cup \infty$, or (which is probably more common) to the Riemann Sphere $\mathbb{C}_\infty$.
What does it mean by $f_i$'s a are fractions over an algebraic number field $K$?
That $f_i$'s are elements of the type $\frac{a(x)}{b(x)}$ where $a(x)$, $b(x)$ are polynomials in variable $x$ where coefficients of $x$ are elements of $K$?
[Chemistry] The program still refused to work, have been going back and forth with the technicians many times already, fixing as many errors we can find
If after this batch and it still not working, it is possible this error trouble shooting will become so frustrating that even literature review will be comparably more relaxing
There is good news through, one calcualation have worked and it predicts computationally that it got a complex with a rare one carbon bonding mode for that given type of ligand
Any idea why this answer is accumulating downvotes? https://math.stackexchange.com/questions/2353725/is-it-possible-that-37-12k-for-some-k/2353733#2353733
The following question is not the question I am interested in. It is the question that is inspired from this question that I am interested in:
The question is the following:
Consider $a_i,m,l \in \Bbb{N}, m,l$ mixed, and $i \in \Bbb{N}$. Find the conditions given $m$ such that there is a unique sequence $(a_i)$ that satisfy the following:
$$\sum_{i=1}^l a_i =m$$
(We can then easily see that the question that inspired this more general number theory question is the special case where $m=30,l=3$ and $a_i \in S \subset \Bbb{N}$)
Well, I am curious on how the possible choice of integers will be restricted and in what way as $a_i$ get specified one by one. Rough observations will suggest if the number is big enough ,then you have only few possible choices for the remaining $l-1$ entries.
As for the orignal question that I took from somewhere, pretty much what you guys said as otherwise 3 odds always give an odd number no matter the combination
NB I suspect the general question may have something to do with this:
In number theory, the specialty additive number theory studies subsets of integers and their behavior under addition. More abstractly, the field of "additive number theory" includes the study of abelian groups and commutative semigroups with an operation of addition. Additive number theory has close ties to combinatorial number theory and the geometry of numbers. Two principal objects of study are the sumset of two subsets A and B of elements from an abelian group G,
A
+
B
=
{
a
+
b
:
a
∈
...
though I need to check whether sum of squares fall in this domain
what I am suspecting about the general question is that if $m$ is some special numbers such as primes, perfect squares etc., then if $a_1$ is quite large, then there are far fewer possible choices for the remaining $l-1$ entries compared to when $m$ is not special, but that's just a guess, based on how in many domains of maths, symmetries tend to reduce the number of "allowed configurations" or allowed solutions
The general question that is inspired from this is probably not so trivial though. I think we will need to split $m$ in cases where e.g. it is prime, it is square, it is cube etc.
In mathematics, arithmetic combinatorics is a field in the intersection of number theory, combinatorics, ergodic theory and harmonic analysis.
== Scope ==
Arithmetic combinatorics is about combinatorial estimates associated with arithmetic operations (addition, subtraction, multiplication, and division). Additive combinatorics is the special case when only the operations of addition and subtraction are involved.
Arithmetic combinatorics is explained in Green's review of "Additive Combinatorics" by Tao and Vu.
== Important results ==
=== Szemerédi's theorem ===
Szemerédi's theorem is ...
$\widehat{qwertyuioplkjhjgfdsazxcvb}$ does not work. The realy wide hats cannot be done via mathjax https://tex.stackexchange.com/questions/100574/really-wide-hat-symbol
set union is like including all of the elements of set A and set B like (omg sorry if my latex is rusty) $$ A \cup B$$ let A = [1,2,3] and B=[4,5,6] then the set union would be [1,2,3,4,5,6]
@Mann My last example is somewhat cheating, but you can still look at say $A = \{(x, y) \in \Bbb R^2 : x^2 + y^2 \leq 2\}$ and $B = \{(x, y) \in \Bbb R^2 : x^2 + y^2 \geq 1\}$. That's a 2 dimensional generalization.
@Mann Sorry. I misspoke. $A$ and $B$ are both connected in that example. What I wanted to say is $A = \{(x, y) : 1 \leq x^2 + y^2 < 2\}$ and $B = \{(x, y) : x^2 + y^2 \leq 1\} \cup \{(x, y) : x^2 + y^2 \geq 2\}$
Here is my attempt to prove the Well-ordering principle, i.e. Any non-empty subset of $\Bbb N$, the set of natural numbers has a minimum element.
Proof: Suppose there exists a non-empty subset $S$ of $\Bbb N$ such that $S$ has NO minimum element.
Define $A = \left\{n\in \Bbb N : (\forall s\in S...
Oh, hello, everyone. I wanted to ask a logic question that wasn't answered in math stack exchange. It is a universal algebra question. What is an explicit identity or set of identities to generate the universally valid equations of the structure $(R,+,*,r)$ where $R$ is the set of real numbers and $r$ is an algebraic constant?
Ring axioms seemed to be a good candidate for the structure $(\Bbb{R},+,*,r)$ but without any more details on how $r$ behaves, it is unsure if there are other possibilities
I am afraid this does not quite answer my question. I need an identity explicity within the signature $(+,*,r)$ For example, if $r$ was 1, we would need at least the identity $(x * 1) = x$. A natural follow-up question would be the same thing with regard to the signature $(+,-,*,r)$. — user107952Jul 4 at 17:38
what? you only said $r$ is algebraic, if you are seeking for identities that are specific for each $r$ then you essentially need to run through each cases e.g. if $r$ is 2 and the structure has no 1 nor 0, you would get an extra identity for instance $r$ will be the generator of an ideal
Anyway, I think I have some mistake on my thinking, guess other abstract algebra people will better answer this question
Side note: Is it possible to have a ring like structure involving the reals but with torsion elements?
Is it true that a subgroup $G$ of $GL(n,\mathbb{C})$ is a parabolic subgroup (i.e. contains a Borel subgroup that is, a maximal connected solvable subgroup) if and only if it is simultaneously block-triagonalizable, i.e. there exists a matrix $T$ with $TGT^{-1}$ block-triangular?
Yes, I am considering only a single specific algebraic $r$. Every algebraic number is the zero of a polynomial with integer coefficients. In terms of such a polynomial, I need a specific identity. No subtraction allowed, only addition and multiplication.
No, there are negative elements. It is just that subtraction is not in the signature of the algebra.
My conjecture is that every algebraic number except 0 needs a single identity besides commutativity, associativity, and distributivity of addition and multiplication, and 0 itself needs 2 identities, namely $(x*0)=0$ and $(x+0)=x$
I mean of course there are other identities, but I believe that is a minimal generating set.
If the underlying set are the complex numbers, then any polynomial can be factorised into linear factors by the fundamental theorem of algebra, which means if any polynomial contains $r$ as a root, then it must have the factor $(x-r)$.
however, since you are working in the reals, for any algebraic number $r$, there may exists more than one irreducible polynomial in the reals that has $r$ as root. For example, $x^2-4=0$ and $x^2-x+2=0$ will both have a real root $r=2$. That would give you at least two identities to specify $r$ for example
Every algebraic number is uniquely determined by a monic polynomial of deg $m$, its minimal polynomial
If $r=\sqrt{2}+\sqrt{3}$ then your conjecture will ran into trouble since its minimal polynomial is $x^4-10x^2+1$ which has constants $-10$ and $1$, but you want a constant free identity
So, are those conjectures about 0 and 2 and other algebraic numbers correct? It is a bit tricky because I am not allowing subtraction in the equations, at least not in this particular question.
Oh, I see.
But there could be a way to re-express such a polynomial so that it is constant free except for r.
$2$ works because the ideal of even integers are generated by $2$, thus you have this identity $x*2=x+x$ unrelated to its algebraicness
but if you want an identity that is constant free and can uniquely specify algebraic numbers, then you need to find the subset where the minimal polynomials have only powers of x and powers of r in it
That means, you need the set of minimal polynomials for each $r$ where the $x^0$ term can be decomposed into sums or difference of powers of $r$ and powers of x.
and for the first point (existence of a number as sums and difference of powers of r), it is already in number theory territory thus I have no idea
What I am not sure however is whether all algebraic numbers that have minimal polynomials that are not constant (in the context of universal algebra) free can always be expressed as sums, powers and nth roots of algebraic numbers which does
but even then, that's still not what your conjecture is looking for (because you only allow one $r$). So my preliminary conclusion is, there's only a subset of algebraic numbers that will satisfy your conjecture, and its identity is given by its minimal polynomial with terms rearranged such that all - becomes + at both sides
And one of the most common weird idea (which underlies nearly all quantum mysticism) is the false identification of entanglement with nonlocal signalling