@user21820 So from what you guys discussed above, since "therefore" is given by $A\land (A\rightarrow B)$, it does look like we selected the row of the $A\to B$ truth table where "$A$ is true" ,and hence getting the same result as $A \wedge B$ for $A$ true
@Secret: Yes. It just means that the boolean algebra of classical logic does not completely capture English words like "therefore". Another common culprit is "even if".
I will not go even if you ask me a thousand times.
wait lets go back a bit. Is there a difference between selecting a row of $A\rightarrow B$ where A is true and simplifying an expression $A\land (A\rightarrow B)$ into $A\land B$?
Up until now I thought they were different. But in this case they give the same results. (the two methods that is)
@user400188 Oh I was more trying to make Secret's phrasing more accurate, since I said "Yes." without looking too carefully. As to your question here, yes because the semantics of "A and C" correspond to all worlds where both "A" and "C" are true, so you can first restrict to worlds where "A" is true and then see which of those also satisfy "C".
@LittleRookie "∈" is to make the statement precise. As x -> 0, we have x^2 ∈ O(x) but not x ∈ O(x^2).
Small question: Given a homogeneous linear recurrence relation with constant integer coefficients, is there an easy (i.e., arithmetical, implementing this in Mathematica here) method for determining its ordinary generating function?
Once you get used to it, it is extremely easy to use. For example you can easily substitute anything that goes to zero for x in the asymptotic expansion for ln(1+x), such as ln(1+x^2) ∈ x^2 + O(x^4) as x -> 0.
This is a "reverse" question of finding the asymptote of a function
Recently, I am interested in doing some sort of modelling which involve equations of the form
$$@(t)=1-f(t)$$
where $f(t)$ is required to satisfy the following properties
$$1^*:\left\{\begin{matrix}f(0)=0 \\ \lim_{t\rightarro...
Sometimes I wonder whether Big O notation can help simplify the expressions of limits, but I guess it only works for those that converge or asymptotes to something
tanh is a nice function to know, and it's called an activation function, because it is like a smoothed version of a step function and can be used to implement smoothed conditional expressions.
If I recall correctly they are the actual curves formed by a uniformly bendable rod when forced to obey point constraints (position and angle) and length constraints (arc length between points).
@LittleRookie Depends on what you mean by remainder. Certainly if you have only one term then it's the remainder all by itself. But if you're referring to say en.wikipedia.org/wiki/Taylor%27s_theorem, then the terminology is already determined.
Suppose f is the formal power series for sequence a[0..].. Namely, f(x) = a[0] x^0 + a[1] x^1 + ... So just plug that into the linear recurrence relation.
And cancel everything out except the first few terms of each row.
For example Fibonacci sequence is generated by a[n+2] - a[n+1] - a[n] = 0.
So compute formally f(x) - x f(x) - x^2 f(x), writing each term as a row and aligning the powers of x.
You should get everything cancelling except a triangular portion of monomials. Those would depend on your initial conditions, and that's the only place where the initial conditions come in.
@Danu Well there's always the horrible way of going about it, by writing out two generic 1-vectors w.r.t your basis and wedging to obtain a set of linear equations...
and then the point was that $K3\#\overline{\Bbb C P^2}$ is homeo to $3\Bbb CP^2\# 20\overline{\Bbb C P^2}$ by Freedman (using that the intersection form is the same) but not diffeo by SW invariants.
Since I think the former admits nonzero invariants while the latter doesn't because it has an embedded sphere which represents an infinite order homology class and has zero self-intersection.
Maybe there's a straightforward way to show that they're homeomorphic, by writing down a birational morphism between $K_3$ and something and using resolution of singularities...
Actually no, that wouldn't do. That wouldn't tell me if they're homeomorphic after blowing up once or twice or blowing up then blowing down or whatever
That's how they proved the h-cobordism theorem for dimension 4.
me neither
@Alessandro Another way to parse that is that there's an open subset of R^4 of the same topology as R^4 but not diffeomorphic to R^4. Crazy, eh?
What's nice is that there's some sort of an analogue happening for R^3. There's an open subset of R^3 of the same homotopy type as R^3, but not homeomorphic to R^3
@BalarkaSen I checked---the manifold I told you are correct, I think. You need this big theorem due to Taubes (on symplectic manifolds) to see that the former has nonzero SW invariants.
@DanielFischer Let $\mathscr{T}$ be the collection of subsets of $\Bbb R$ consisting of $\emptyset, \Bbb R$ and the rays of the form $(a, \infty)$, where $a \in \Bbb R$. I find that any bijective maps of $\Bbb{R}$ with the usual topology that are increasing are also homeomorphism for $\mathscr{T}.$ But $f(x)=-x$ is not a homeomorphism for $\mathscr{T}$ because $f((a,+\infty))=(-\infty,-a)$ which is not open.
I would like to prove that the set of homeomorphism of $T$ is of index $2$ in $\Bbb{R}$ with the usual topology. So I must write any homeomorphism for $\Bbb{R}_{\mbox{usual}}$ as two "class"
Meanwhile, I am currently procrastinating in Munkres, probably because I struggle between too lazy to do the exercise in Ch. 1 Sec 11 and I need to do those exercise to ensure I will not run into problems in well ordered sets in later chapters
It is used to show that the integral is in general a family of functions differ only by a constant. It's value is specified by the initial conditions and boundary conditions in differential equations
If you were still in engineering or science, you might came across them again as they are important in describing the dynamics of your construction and systems
I am aware of that. But I don't know how to set up a differential equation. I know what my integral is but I don't see how it relates to differential equations other than the constant of integration connection you now told me about.
@MatsGranvik Here's an example: Suppose you know that k(x) is your solution to e.g. the above differential equation I wrote. Then if you sub in y=k(x)+C where C is a constant, you will find it also satisfy the differential equation. However suppose now I give an initial condition that k(0)=r, where r is some given number, you will notice there's only one possible C that will work, thus you solution is k(x)+F where F is a number such that it satisfy k(0)=r also
So a differential equation is basically showing how y changes and y is your integral, which is all the changes in the system summed up and collected together
In fact, one can pick any known integral $\int f(x) dx = g(x)+C$, differentiate both sides by x, and it can be rearranged so that you have g and its derivatives forming an equation, that's is one of the example of a differential equation
the key is that derivative of constant evaluates to zero, thus there are a family of solutions for a given differential equation unless the initial condition is specified, which pick out one of the C
Need a quick sanity check: does something like $\frac{\mathbb{Z}}{\mathbb{Z}^4}$ make sense? Where $\mathbb{Z}^4$ is the free abelian group on 4 generators.
@JeSuis Not sure if I understand correctly what you want. The homeomorphisms with respect to $T$ are precisely the increasing homeomorphisms with respect to the standard topology. We get a map $\pi \colon \operatorname{homeo}(\operatorname{standard}) \to \operatorname{homeo}(T)$ by setting $\pi(f) = f$ if $f$ is increasing and $\pi(f) = -f$ if $f$ is decreasing.
@Secret I'm not sure if it's normal, but the context is that I'm trying to calculate homology groups. In the end, I ended up with a group generated by one generator, and another group generated by a countably infinite number of generators
Unfortunately, my group theory knowledge is still preliminary, I don't know of any theorems that can help me to check for normal subgroup of infinite groups quickly other than kinda brute force it.
I think in most situations(for example, in $S_n$ or $D_n$), proving by definition is too complicated because you have to calculate $gng^{-1}$ for every $n$ in $N$ and $g$ in $G$. To prove that all the left cosets are also right cosets is also too complicated because you have to find all those cos...
hmmm... in such context since $\mathbb{Z}^4$ is finitely generated, this makes computing $gxg^{-1}$ easier as there are really only 4 $x$ one need to worry about. However when there are countably infinite number of such $x$ then $\mathbb{Z}^{\aleph_0 ?}$ is not finitely generated, and that I have no idea
What I am suspecting is that $\mathbb{Z} \subset \mathbb{Z}^n$ for all finite $n$ though, thus free abelian groups might not be subgroups of $\mathbb{Z}$, let alone a normal subgroup
now that's a useful result since $\mathbb{Z}$ is also abelian. It then remains to check whether $\mathbb{Z} \supset \mathbb{Z}^n$ which I suspect to be no
@SimplyBeautifulArt It's easier to ask to compare the growth of functions. Asking which number is bigger is less meaningful and a bit harder. Though in the case of your contest, of course we're comparing numbers.
@LittleRookie Which step[s] do you want me to provide more justification?
> h^2/3 + O(h^4) ⊆ O(h^2) + O(h^2) = O(h^2) as h -> 0.
@PaulPlummer $\mathbb{Z}/n$ are the obvious ones, I don't recall the more wacky ones (if any). I only remember the theorem that allow the decomposition of a group into cyclic groups of prime powers applies to finite groups only
Given any variables x,y such that x -> 0: | If y in O(x^3): | | Let c > 0 such that [eventually] |y| <= c |x^3|. | | Note that [eventually] |x| <= 1. | | Thus [eventually] |y| <= c |x^2|*|x| <= c |x^2|. | | Thus y in O(x^2). | Therefore O(x^3) ⊆ O(x^2).
I am also suspecting $\mathbb{Z}$ is a subgroup of $\mathbb{Z}^n$ (defined as the free abelian group of n generators), not the other way around, since one can generate all the elements of $\mathbb{Z}$ by picking the words $ex \in \mathbb{Z}^n$ for all $x$ where $e$ is the identity
Need a quick sanity check: does something like $\frac{\mathbb{Z}}{\mathbb{Z}^4}$ make sense? Where $\mathbb{Z}^4$ is the free abelian group on 4 generators.
@Secret I'm not sure if it's normal, but the context is that I'm trying to calculate homology groups. In the end, I ended up with a group generated by one generator, and another group generated by a countably infinite number of generators
@LittleRookie: Note that I'm not using the very formal ε-δ definition of limits, because all you need is to be able to understand the reasoning I've presented, and it can be made rigorous by using sequences. Namely the above proof would expand to "Given any sequence x converging but avoiding 0, if y is a sequence in O(x^3) then ..."
If I take a set $A$ to be the complement of countable set the equality $A\cup U\cup V=\emptyset$ is impossible if $U$ and $V$ are non empty, so it means that one of them must be the empty set, so it's connected
And, assuming that $\{[a^n,b^m]\mid m,n\in\Bbb Z\setminus\{0\}\}$ has no nontrivial relations (which I'm sure it doesn't), that's the only way to write $ababa^{-2}b^{-2}$ in terms of those generators.
@BalarkaSen By the way, a while ago I mentioned something else in group theory, which I misstated. The real question is, if a group has $a^2=b^2$ for some elements $a$ and $b$, must it have an element of order $2$.
Can anyone please clarify something in this question? Given vector $A= (2,-1,-1)$ and vector $C=(0,1,1)$, $C$ rotates about $A$ with an angular velocity of 2 rad/s . Find the velocity of the head of $C$. As velocity is the cross product of angular velocity $w$ and $r$ and $w$ is in the direction of $A$, so the velocity should be a cross product of $A$ and $C$? If so, where does the angular velocity 2 rad/s fit in?
@AkivaWeinberger Is this question something you don't know? Is seems quite reasonable that $\langle a, b \mid a^2=b^2 \rangle$ would have finite quotient with the property you desire.
There is a nice fact about one relator groups that the only time you have torsion is when the relator is a proper power (after cyclic reductions and stuff like that)
Here is a "simple" proof utilizing GM-HM inequality.
Use GM-HM inequality on the set $\{\frac{a+1}{a},\frac{b+1}{b},\frac{c+1}{c}\} $ to get:
$$\left(\,\left(\frac{a+1}{a}\right)\left(\frac{b+1}{b}\right)\left(\frac{c+1}{c}\right)\,\right)^\frac{1}{3} \geq \frac {3} {\left(\frac{a}{a+1}\right)...
Well that took a while. Reading a paper, and they define subsurface projections, but the annulus case has a different definition than the other case, but they give no indication as to why, and I think I figured out why
Subsurface projections? If you remember what the arc and curve graph are, and have a subsurface you can "project" your curves in the curve graph of you original surface, to all the arcs and curves that are formed in the intersection of the subsurface with the curve
@BalarkaSen @MikeMiller @PaulPlummer Also, finite groups like that must have an element of order $2$, because suppose not. Then $a$ and $b$ must have odd order. Call the order of $a$, $n$. That means that $a=a^{n+1}=(a^2)^{(n+1)/2}=(b^2)^{(n+1)/2}=b^{n+1}$, which commutes with $b$. Thus, $a$ commutes with $b$, and $ab^{-1}$ is an element of order two.
@AkivaWeinberger I had essentially the same proof, although thought of it a bit differently: $\langle a \rangle \cap \langle b \rangle= \langle a^2 \rangle$, but if your group has no order two elements $\langle a^2 \rangle = \langle a \rangle =\langle b \rangle $ so $a$ and $b$ commute and we have a problem.
My original line of thinking was to define $c:=a^2=b^2$ and realize that it commutes with them both. And then a little bit of thinking of what happens with odd orders leads to that $a$ is a power of $c$, and then I think, "Oh, no!"
@BalarkaSen For the annulus case, you want to be able to keep track of Dehn twists, but if you are working on a surface, and you have a curve twisted up by a Dehn twist within an annulus, you can homotope that section and push the "Dehn twisted" part to some different annulus, losing all the information.
It's a grid; on even points, $a$ goes down and $b$ goes to the right, and on odd points, they're switched. (So, powers of $a$ lead you in a staircase-like path.)
@AkivaWeinberger Was that to me? I still don't really get how you start the first parts of the table. Like once I get the table started it makes more sense.
@Sie your initial columns for p and q are correct, just not the implication one. If you have F,F,T,T for p and T,T,F,F for q, then you'll be repeating two rows and your truth table will be incomplete
Lots of people explain how to set it up, in a sort of easy to remember way, without explaining why, but it doesn't really matter what order you do it in
Is there an easy way to prove that $$ \|f_r\|_{L^p} := \frac{1}{2\pi }\int_{-\pi}^\pi |f(r e^{i\theta})|^p \mathrm{d}\theta \ , \qquad (1 \leq p \leq \infty) $$
@BalarkaSen The fix is that you look at the cover corresponding to the cyclic subgroup from the annulus, which will be an open annulus, then you put a hyperbolic metric on it, so that you can define a boundary. When you lift a curve crossing through the downstairs annulus and you can still see the dehn twists upstairs even if you start to homotope downstairs
Basically in the annulus cover of your surface you can see dehn twists around the core of an annulus (the center circle) even if the dehn twist was homotoped on the surface, so it is no longer around that core, but a different core of a different annulus (but the cores are homologous)
@MikeMiller Suppose you embed a possibly knotted torus in S^3. To show that it bounds a handlebody on some side, can you do the folllowing?: This is a circle bundle over some fixed longitude in the torus. Each disk in the fiber bounds a disk by Schoenflies - fill it in. The result should be a fiber bundle; orientable disk bundles over the circle are trivial
actually maybe there is a thing to worry about. There's no canonical way to pick the disk the fibers bound, so it's not really well-defined. Can this be fixed?
I don't know if there is a convention, but in some cases you get that the image is open as a theorem (like some invariance of domain stuff), which may be the case for what you are trying to translate.
This is a "reverse" question of finding the asymptote of a function
Recently, I am interested in doing some sort of modelling which involve equations of the form
$$@(t)=1-f(t)$$
where $f(t)$ is required to satisfy the following properties
$$1^*:\left\{\begin{matrix}f(0)=0 \\ \lim_{t\rightarro...
Prove that every subgroup of an abelian group is a normal subgroup.
Here is a "simple" proof utilizing GM-HM inequality.
Use GM-HM inequality on the set $\{\frac{a+1}{a},\frac{b+1}{b},\frac{c+1}{c}\} $ to get:
$$\left(\,\left(\frac{a+1}{a}\right)\left(\frac{b+1}{b}\right)\left(\frac{c+1}{c}\right)\,\right)^\frac{1}{3} \geq \frac {3} {\left(\frac{a}{a+1}\right)...
