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1:17 AM
> Theorem 1 (Zarach). It is consistent with ZFC- that ω1 exists but is singular and hence that a countable union of countable sets can be uncountable.
Proof. Consider the model W as constructed in the L ́evy collapse of אω above. Note that every cardinal אVn is collapsed in V [Gn ] and hence in W , but אVω remains a cardinal in every V [Gn ] and hence also in W . Thus, ω1W = אVω , which has cofinality ωinV andhenceinW,aswitnessedbythesequence אVn |n<ω ∈V ⊆W. So W satisfies that ω1 is singular. In particular, W satisfies that ω1 is a countable union of countable sets, as ω1W = {אVn | n ∈ ω}.
Pretty much saying that $\omega_1$ under the model W is the same as $\aleph_{\omega}$ under the model $V$
NB ZFC^{-} is ZFC-P in the above arxiv
 
1:29 AM
Some axiomatic systems based on ZF to be tested:
System 1: Extensionality, Regularity, Specification, Pairing, Union, Replacement, Infinity, Existence of an inaccessible cardinal, dependent choice (optional)
System 2: Extensionality, Regularity, Specification, Pairing, Union, Replacement, Infinity, (weaker and/or more generalised version of powers axiom to be figured out), dependent choice (optional)
System 3: Extensionality, Regularity, Specification, Pairing, Union, Replacement, Infinity, (weaker and/or more generalised version of powers axiom to be figured out), Existence of an inaccessible cardinal, dependent choice (optional)
Goal: To preserve as many notions of functional analysis and real analysis, quasiinfinite sets and allowing a richer construction of uncountable sets
Actually wait a minute, what does an inaccessible cardinal even mean without powerset...
 
 
1 hour later…
2:53 AM
morning
 
3:50 AM
Afternoon nap dream: Reading a wikibook with a dark grey background. One of the pages talks about generalising to infinity and there's a paragraph saying something along the lines of there is no way to get more elements unless (forgot), followed by a set builder notation
$$\{f |f(k) \mapsto f(forgot)|(forgot) \mapsto (forgot) | (forgot)\}$$

The article then ends with "But what if there exists something that can compute every computing matter, that consciousness is (forgot) real"

I then click the link to the next page, which talks about something titled "Conscious Fir" and the 1st paragrap
Of all my dreams, these afternoon dreams are the ones that are most potentially informative on how to solve problems. However they are so hard to recall thus most of the time, I only remember less than 1% of it
 
anyone know basic vector math / physics?
 
Depends on what type of vector/phyiscs
 
velocity, movement, 2d plane, applying force, changing directions
 
What is the concept in your problem you have trouble with. Remember SE prefers conceptual, not "how to do this problem" questions
 
Conceptually, if I know the current $x,y$ (position) of a moving object, as well as its velocity $V_x, V_y$ -- but I want the object to ultimately arrive at some point $x_t, y_t$
to get it to arrive at this point I am having trouble understanding how to change its velocities based on how hard i can strike the object
 
3:59 AM
Since you mentioned it is a vector problem, the first step is to resolve the motion into x and y components. So you want to be able to given your initial position and velocity, compute the acceleration required to allow you to move some displacement to your final position.
 
the difficulty is the initial vx vy and the fact that I do not have unlimited force
I can't just immediately strike the object with the right force and angle to immediately change its vx vy to make it go towards the target point
sometimes i can if vx, vy is slow enough
but sometimes best i can do is try to get it as close as possible to the target but I don't know how to do this
 
Is your given velocity a function of time? If it is, you can simply integrate that expression for the velocity, plug in the initial velocity to obtain the arbitrary constant and then solve the final position
since velocity is the derivative of displacement and displacement is final position-initial position
 
no, i'm asking a different question than that
 
Does your object has a mass. If yes you might be able to use conservation of momentum to solve it.
...Actually I think I might need some context without the full problem. What kind of motion will that object be undergoing, is it standard scenarios like projectile motion, simple harmonic motion or something else?
 
mass is 1 for simplicity
imagine a billiard ball moving across a huge table
and at time $t=0$ you know the position and velocity of the ball
you want to send it towards the hole
you can strike it once with some maximum force F from any angle
score is determined by distance to hole at time $t=1$
maximum score is obviously hitting the ball directly into the hole but sometimes not possible
if the ball is already going very fast
and your max force is not enough to completely change the ball's course
but i am not sure how to solve for the cases where you can't 100% change the velocity so it's moving along the vector pointing towards the hole
 
4:18 AM
Ok, so that means, choose a coordinate frame, you have some mass with initial position and velocities $(\mathbf{x}_i,\mathbf{v})$ and your hole has some position vector $\mathbf{x}_f$. There is a maximum to the force applied which means the magnitude of the acceleration vector has a maximum, and the angle of hitting determines the relative proportions of y and x components of the acceleration. So in other words, you want to solve for $a$ the following vector equation:

$$\mathbf{x}_f - \mathbf{x}_i=\int_0^1 \mathbf{v} + \mathbf{a}t dt$$
Actually, to do this more generally:
$$\text{Minimise} \left( \mathbf{x}_f + \mathbf{x}_i+\int_0^1 \mathbf{v} + \mathbf{a}t dt\right)$$
So if there is an exact solution, the final distance (given by the magnitude of displacement since all the motions are linear) of the ball from the hole is exactly zero, otherwise, it will be some finite value
 
4:55 AM
@ArukaJ What matters is the impulse you can impart ot the object, not the maximum force. Given any finite force, if you can apply it for long enough you can make the velocity what ever you want.
So the question is, under what conditions can you not change the velocity enough so that the ball arrives at the desired point?
If you can arbitrarily change the direction of the force, then it doesn't matter that you can only apply a force of a fixed magnitude. Eventually you can always get it to the point you want.
 
I think the problem might be an undergraduate problem, thus the impulse is expected to be short for an elastic (billard ball) interaction
(and he does mentioned the object is a billard ball)
 
You still need some kind of impulse-scale though. Basically you're saying its acting like a delta-function force in Newton's 2nd law. A sudden 'kick'. But you need some dimension-ful coefficient to sit in front of that delta function, and that coefficient will be an impulse-scale.
Giving a maximum force doesn't help. Because if you apply any finite force over a shorter and shorter time scale, the impulse imparted goes to 0 as the timestep approaches 0
 
I don't know the right word for it, I have the ability to impart a new velocity to the object
 
But again, acceleration by itself doesn't help
 
thorugh vf = vi + a
 
5:01 AM
that equation isn't dimensionally correct
v has units of m/s, a has units of m/s^2
 
it's for a game so the physics aren't going to be real-world here
i don't know the 100% correct terminology, i just know that i can pick a direction and force and i can change the velocity
for example if it's on the origin, not moving
 
Ok so the word youre looking for isnt force or acceleration, its impulse
 
and i strike it from the left with force 10
 
Impulse is a change in momentum
 
then i get vf = 0 + 10*1 = 10 velocity
towards the right
(mass is 1)
and in this case the position also updates to 10 as well (as opposed to what we'd expect with 1/2 at^2)
so a bit "simpler" in this regard
iirc impulse is F*t
 
5:03 AM
Yea what you actually mean isnt that you strike with a force of 10N. You mean that you strike with a force, such that the impulse is 10 m/s
 
and since i'm interested in t=1 timescales it's equal to force
and mass is 1 so acceleration = f as well
 
You;re making lots of approximations
and not syaing that you're making them
and tis leading to your language being imprecise
 
i mentioned up front that mass is 1, and that i was interested in t=0 to t=1
only major simplification is how position is updated between the two times, and it's minor (doesn't really change the nature of the question)
 
Thats true, but if you actually apply a force of magnitude F over t=1timeunit, the position is changing over the whole time
 
@KevinDriscoll Yeah, that's how undergraduate billard ball type problems behave. I am not sure if these problems actually have an impulse scale in mind though, but they do basically act like delta distributions. I guess the only thing we can say is the impulse is some small but finite value short enough to minimic a kick
 
5:06 AM
i understand that in practice the position is changing the whole time
 
Youre also making the approximation that 1 tiemunits is 'short enough' that you don't have to update the position during the application of the force
 
and that's normally where the vf = vi * t + 1/2 at^2 stuff comes in
 
and thats pretty important because it dignificantly changes the nature of the problem
 
but here i'm doing away with that aspect for simplcity's sake
 
So anyway, the point being that you have some maximum impulse you can apply to the object. And so in that case if you don't ahve enough impulse to actually make the object get to the required point, then you have to make the new velocity as close as possible to the velocity that would get it to the specified point
So if you draw the initial velocity and the velocity hat would make it pass through the specified point as vectors
 
5:10 AM
yes
 
Then the 'best' you can do with just 1 'kick' will be to apply the maximum impulse if the direction of (velocity that would get to the specified point) - (initial velocity)
That gets you a final velocity vector that is as close as posible to the desired velocity
and since after this kick its just constant velocity motion, getting the velocity as clsoe as possible will get the position as clsoe as possible as well
 
hmm... I was not aware that one can do away with positions and simply work at the level of velocity vectors using impulses
 
its because you just have constant velocity motion after the 'kick'. Everything just goes in a straight line.
 
So as long as you get it moving along the line closest to the one you want, itll end up at the position as close as possible
 
5:15 AM
I tossed this together to try to illustrate
ball's at origin, red vectors represent current motion (+ the horizontal and vertical decomposition)
black hole is the ideal destination point, blue vector is pointing from current position to hole
if i had unlimited impulse i could use paralellogram method to just figure out where to hit the ball and how hard and it'd go straight into the hole next time unit
 
Doesnt even have to be the next time unit
As long as you get the new velocity ot be along the blue vector, itll go into the hole eventually
because constant-velocity motion is just motion along a straight line
 
yes (my goal is to get it to that hole as fast as possible)
next time unit technically i can apply another hit, but this leads to this current subproblem of getting it to the hole each hit as close as possible
as this translates overall to getting it into the hole as fast as possible
 
Sure, you just want to get the new velocity to be along the blue vector, with as large a magnitude as possible
 
right
that is where i am struggling
since it also depends on magnitude of red relative to my max impulse ability
 
So then just take a vector that represents the maximum kick (AKA maximum impulse) you can apply, put its tail at the tip of the current velocity vector
and then move its tip until it touches the blue line
because $v_f = v_i + v_{kick}$
 
5:21 AM
i tried that, e.g. i'll make a new picture here
 
The new velocity vector will point along the blue line
 
i think my problem was also the angling
 
Once yuve got the picture draw, you can use basic trig to say what the angle is
probably using some law of cosines or something
Ok, so the case where you do have enough impulse to get along the blue line is solved. You just carry out that procedure i mentioned, so let me think again about the case where you don'tahve enough
 
imgur is being difficult
so let's say that green vector is my max impulse
 
Oooooh, wait. The fact that you can do subsequent hits and want to find the mimumum total time actually might make this more complicated. I have to think more carefully
 
5:24 AM
this is what i would do if I could, for sure, make the ball move along the blue line
(this is just the paralellogram method), my problem is what i do if i can't "reach the blue vector" with my green vector
 
Yea, no this is right for when you ahve enough impulse to move the velocity. because at each time step, you're maximizing the distance the ball travels toward the hole
So you're minimizing the total time, so thats fine
 
i am assuming that by constantly re-applying this same method each hit, it minimizes overall time
since it's minimizing distance to hole each time
i don't know how to prove that but it seems right to me(?)
 
It is right
 
but if the green vector can't reach the blue one, i'm at a loss
 
So to do it when you don't ahve enough, you know that eventually you want the ball velocity to be along the line connecting the current position and the hole.
 
5:27 AM
i guess that's the best i can do
just hitting it back along the red
basically trying to slow it down along its current path repeatedly until i can make the green connect?
or something else?
 
If you want to do it in one kick but the green vector can never align with the blue one, then the next best option is to minimise their difference
 
Ya I agree with Secret, this was my original answer
Any velocity other then the one that minimizes the distance will lead to the ball being farther away fromt eh hole that the closest velocity in the next tiem step
and that extra distance menas itll take longer to get it there eventually
The problem is there is no fixed velocity you're minimziing with respect to
because the velocity can be ANY velocity along the blue line
AH! But any velocity that gets the ball to the hole in 1 time step is equivalent
So you can take the 'desired' velocity to be the minimum velocity that gets the ball to the hole in 1 time step
and then you want to new velocity to be as close to that velocity as possible
 
i may not always be able to do it in one kick, i can always do it in finite kicks but I'm generalizing to a single kick because whatever optimizes one kick will optimize the whole thing
by just doing it repeatedly
 
Yea we understand
 
Yeah, any sequence of velocities that can approach or get into the hole will be equivalent to one single velocity vector that is the optimal one because of how vector addition works
(only the initial and final value matters)
 
5:34 AM
right
if i had unlimited impulse ability i can just one-shot the ball into the hole
 
Im not sure you udnerstand what Secret is saying
 
by just drawing a vector from the red vector tip to the hole and just applying that to the ball
?
 
Hes saying it doesnt matter what vector along the line connecting the ball and the hole you choose, as long as you apply the maximum impulse to the ball and get the new velocity to be as close as possible to ANY velocity along the blue line
Then youre doing the optimal thing
 
not sure i am following the logic there
it seems possible to just max-impulse it to any point along the vector infinitely but never hitting the hole
 
Draw a picture. Imagine the ball is going some direction and it will take at least 3 kicks to get it into the hole
And do the 3 kicks 2 different ways
 
5:38 AM
he said any sequence of velocity sums that get to the hole is the same as one single vector due to how summation works, and that's fine / i agree with that
but we can do 5+5 = 10 but we can also do 1+1+1+1+1+1+1+1+1+1=10
former is more "optimal" than the latter
(to use an example)
 
1 second
 
yes we can "reorder" the kicks
vector sums are commutative
n! ways to arrange n kicks
 
I actually took Secret to be pointing out a stronger conclusion, but Im checking it now
Its kind of funny because the way you've setup your 'physics' if the ball starts at the origin, then the position and velocity vectors are always parallel
 
because you just move to the end of the velocity vector at each time step
 
5:46 AM
I think I need to reread the chat transcript...
 
@Secret If you're saying the two sets of green arrows sum to the same thing, i agree/understand, but one of them is using more green arrows than the other, is what i mean
in my definition "optimal" means as few green arrows as possible
or are you saying something more than that?
 
Actually, I think I start to recall how billard ball games work. You have a billard ball that travel at some initial velocity in a straight line, you then hit it at some angle, the ball's final velocity after the kick is the sum of its initial velocity plus the velocity of the kick (picture updating..,)
 
yes, the velocities update in sum
 
So yes you can end up with a scenario where one kick is not sufficient to brought it to the hole if the ball travels too fast
 
yes
 
5:52 AM
Let me update the picture first
 
in the event that green impulse is too small i almost wonder if this is the optimal decision:
 
It isn't but im struggling to articulate to you why
 
was the earlier approach wrong too?
 
Nope thats totally right
 
making the green vector connect to blue from tip of red
 
5:56 AM
Notive youre doing something very different here
 
so as we make green smaller and smaller it gets closer and closer to the origin
as we drag it up the blue line
 
It works, but its not optimale
 
so what if green arrow = red?
in magnitude
 
If green = red then your proposal would have the ball stop after 1 time step, so you wouldnt get any clsoe ror farther away from teh hole
 
(Complicated diagram incoming!)
 
5:58 AM
But I claim that if instead you aling the green arrow so that it points from the end of the red to the hole, then you'll actually get closer @ArukaJ
 
For the one kick case (I will be explaining it shortly)
 
so always aiming the green vector towards the intended destination no matter what?
with max impulse?
 
Depends onw hat you mean by aim
You dont aim the green arrow toward the hole from where the ball is
you aim the green vector toward the hole from where the end of the current velocity is
 
Basically, the blue vector is the displacement of the ball's current position to the hole's centre, the red vector is the initial velocity, the green vectors are the possible directions of impulses, and the dark red vectors are the final velocities after the kick. The green vectors are bounded by a dotted circle which means all the possible directions with maximum impulse is given by that circle. This particular scenario is produced so that the ball will never get into the hole in a single kick
 
6:00 AM
i mean like:
 
Hold on I clearly need ot think more carefully about this
 
For all possible velocities after the kick, the trajectory of the ball is bounded by the cone shown. The dotted circle centred on the ball (the tail of the blue arrow) marks the t=1 point where the ball must stop. Therefore in order to get to the hole as close as possible, you need to compute:
 
are there just three green vectors for demonstration sake
(we could have just drawn one?)
trying to understand the diagram
 
yes. In reality there are uncountably many
 
ok
 
6:03 AM
Yea here's the thing. After each time step, the ball always ends up at the point given by vi + vkick. So during each time tep you want to maximize the distance you travel toward the hole. So that means always pointing the green arrow toward the hole.
So yes, our original solution fo the case where you can reach the blue line in 1 time step actually is wrong.
 
To get to the hole as close as possible, compute Intersection p (t=1 circle, cone (which is a function of the circle of maximum impluse), circle centered at hole)
 
You always want to point the green vector from the end of the initial velocity toward the hole
 
Since the t=1 circle is fixed as a function of the final velocity, the intersection is unique for each direction. In particular, if the radius of the circle centered on the hole goes to zero, it corresponds to the case where you can one shot into the hole
 
I'm with you on everything in that diagram so far up until the t=1 part
let me re-read
 
(clearing clutter in diagram)
 
6:07 AM
Notice in Secret;s diagram that for the circle bounding all of the maximum-impulse arrows there is 1 point thats closest to the hole
 
i don't undertstand the t=1 circle thing
you could have a green arrow pointing straight back at the red
 
That is the point where the circle intersects the line connecting the hole and the initial velocity vector
 
and the ball would be inside the dotted circle at t=1 would it not?
 
It would end up on the circle
Becuause the ball always travels to the end of the final velocity vector after each time tep
 
6:08 AM
and in fact one of his green arrow / final velocity arrows are inside the circle
 
given how you've set up your physics
 
t=1 marks where the ball will end up after the kick 1 second later
 
unless he means the smaller dotted circle
as the t=1 circle
ok that must be it
 
The smaller circle shows you all possible maximum impulses
 
i'm still not following this big dotted circle
if the impulse is that small how are we saying the ball can be at any point around that big circle
 
6:10 AM
Ya I dont see that either
In fact, the ball will end up at some point on the smaller circle
 
that's what i mean
wouldnt the small dotted circle represent all possible t=1 positions of the ball?
 
Ok let me try again:
 
Indeed it does
 
if we do nothing, the ball ends up at the tip of the red vector
 
and so you ant to pick the point on that smaller circle thats closest to the hole
 
6:11 AM
if we apply a max kick in some direction, it goes to the small circle on the border
 
And that point is precisely the point where the circle intersects the line connecting the hole and the tip of the initial velocity vector
 
ahhhh
that's really interesting
 
which implied you want to make the green arrow point from the tip of the red arrow toward the hole
 
this whole time i was assuming we had to compensate for current velocity more complicatedly
 
Yes your understanding is right Kevin. Since all the arrows I have drawn are already velocities, everything happens between t=0,t=1 so the t=1 circle is redundant
 
6:15 AM
@Secret wait are you saying in your diagram the green arrow is optimal?
i'm still not understanding those other circles
 
Now it's optimal
That is the direction you need to kick to get the ball closest to the hole
 
ok that makes more sense to me
 
As kevin said, the small circle marks all the positions at t=1
 
Im shocked I didnt see that that was the right answer earlier. It took sitting down with a pencil and paper and actually drawing it out
I did a whole one on my pad here tha took like 7 time steps ot get to the hole
 
i'm trying to expand his picture
and i'm ending up going in a spiral
is this right?
 
6:26 AM
Does your table have friction: That is, does your ball come to a stop at t=1?
 
No its frictionless
 
ok
 
i can use friction though,, it would just decrease the size of the next velocity vector by some scalar amount
 
@ArukaJ I am not 100% sure, but I think this is right. I think the way youve drawn these particular velocities you will always overshoot the hole
 
so instead of mirroring my vector it would be slightly shorter
 
6:28 AM
For example try to draw the same picutre, but instead of doing the thing we said was optimal
 
the picture above is using the "optimal" approach of always hitting towards the hole
 
Try and rotate the initial velocity so that it falls along the blue line. I think youll see you pass the hole before that can happen
 
i need to break out the paper and pencil
hard to use graphics
 
What I mean is, try drawing the green arrows so that they point toward the closest point on the blue line from where teh initial velocity is
and I thin youll STILL overshoot the hole
 
i think it also depends on how hard it's being hit
like we can always go sub-max, too
it's just we can't go more than max
 
6:30 AM
Ya goign submax here though will definitely lead to shooting past the hole though
 
although submax would, yeah
same problem
 
I think, but am not 100% sure that actually you can ALWAYS hit it at a maximum, as long as the ball always goes in teh hole when it moves over it
 
If the ball were to skip over the hole if its goign too fast itd be a different story
 
This is some really shitty convergence speed...
(Note I have not take account riccochet of walls, so...)
 
6:32 AM
Ya im sure theres some way of calculating how the number of steps scales with the velocity in multiples of the max or something
The worst-case scenario is you're headed directly away from the hole I imagine
 
I think intuitively, the first thing I will do if the ball travels too fast (> max impulse) is to hit it in the opposite direction so it slows down, then the problem becomes one where there is a one shot solution to the hole
 
That was Aruka's instinct too
and as a human thats probably how Id do it too
You could try that method with the drawing you already did. And hopefully if were right it takes more than 8 steps
 
oh secret already made something like this and more accurately LOL
 
(Faster strategy coming shortly)
 
but if i am always hitting towards the hole then I am spiraling into it with many moves
ill try our earlier strategy
 
6:40 AM
Ya Im open to the idea that if youre going to overshoot this isnt the fastest
Its not 100% clear in my mind that you cant sacrifice some distance on one step to get more on the next
 
This is a two stage strategy:
Stage 1 (Initial velocity > Maximum impulse and there exists no one shot solution): Hit the ball so that it maximise the decrease in its velocity and also bring it closest to the hole's direction)

Stage 2: (One shot solution to the hole is found) Pick impulse that get to the hole
 
damn it you beat me to it again (I was trying to drop the impulse at 90 degree triangle to blue vector)
 
Ok ya so there is a thing about if youre going to overshoot, you want to go as close to the blue line as possible
There will still be velocities large enough though that even if you try this, youll stil overshoot
Ya doing it perpendicular to the blue vector isnt always going to be optimal. Sometimes to get it to not overshoot youll only have to go a little ways further toward the blue line from the locally optimal solution from earlier
 
how are you drawing these so quickly btw
what are you using?
 
power point
 
6:47 AM
and the nice clean arrows and everything
 
I let my earlier thoughts about the single-step problem cloud my judgment
The 1-step answer is easy about whats optimal
The multi-step problem is much harder
I'd bet like $100 though that someone worked all this out, and probably 200+ years aho
 
The "hit it until it is slow enough, then one shot hole" strategy
only works without friction
 
in practice i do have some friction but it's applied at the end of the turn
 
6:50 AM
Not sure which one is more optimal though
 
now i'm totally lost though as to what heuristic to use
 
There are many solution strategies, but it is the game mechanics that determines which strategy to use
 
If you can get the ball directly in the hole without overshooting by always applying to force 'toward the hole' then that will be faster than anything else
 
it's easy enough to get it to the hole if my impulse can one-shot it
 
If there's riccochet, the problem can become chaotic...
 
6:52 AM
Depends on the shape of the boundary
 
Typical billard ball tables are rectangles
 
But the idea of chaotic doesnt really make sense when you can apply an impulse at each time step
So for a rectangle, the dynamics are integrable
 
I see
 
Any regular polyhedron is integrable.
The circle and the ellipse are also integrable I think
But if you cut the circle into 2 equal arclength halves, then glue the halves together using line segments
That system is chaotic
 
do you think making 90 degree triangles with the original blue line makes it optimal?
 
6:54 AM
@ArukaJ Not always
For example, if the ball is only going to sliiiiiightly miss the hole using the 'force towar dhte hole' algorithm
then you only have to point the force sliiiiiightly away from the hole to get the optimal path
going 90 degrees to the blue line would be too much
 
ahh
 
So this idea occurs to me and Im not sure if its right
 
maybe this is a brachistochrone curve? googling around
 
ya I thought about it, it probably is
it cant be exactly that though
because this isnt a constant force
 
a choppy approximation thereof
 
6:57 AM
But related tot he idea that occured to me, it WOULD be a brachistochrone if you used a constant force
so what you have is some piecewise-linear approximation to that
ya exactly
its a bit more complicated problem though because yo can change the direction of the force
so you could find the solution for a constant magnitude force where you can change the direction
it wont be exactly a brachistochrone
and then your solution here is a piecewise-linear approximation to that optimal result
 
Hi, I want to know how to calculate the margin for some ratio. Suppose that we have have achieved a ratio of 2.5 and the best is 2. So we have 20% margin. I want to know how to calculate this margin percent? Is it as following: 2.5-2/(2)*100 = 25%. It is not exactly 20% so this is why i want to check here
 
The constant force, changing direction problem I dont think would be toooooo bad using calculus-of-variations/least-action. But its not trivial. Again, I bet someones worked it out
@YOUSEFY Thats kind of ambiguous. It depends on waht you mean by margin
For example, if 2 represents your costs, and 2.5 the revenue, then you'd say you have a profit margin of (2.5-2)/2.5 = 20%
 
@KevinDriscoll Agree! this is why I don't understand how he calculate it. I read this paper and says "and the best known integrality gap of the LP is 2 [8], leaving a 20% margin for improvement." arxiv.org/abs/1412.0681
 
But if instead 2 represents last quarters revenue, and 2.5 this quarters revenue, then to find the 'percent increase' youd want (2.5-2)/2 = 25%
 
(i gtg but thanks for the input @KevinDriscoll and @Secret )
 
7:04 AM
@ArukaJ Good luck. Sorry it turned out to be more complicated than I realized at first
 
@KevinDriscoll "2.5" represents my best achievement while 2 represents that I can reduce 2.5 to 2 by some "unknown technique".
 
@YOUSEFY I mean hes saying the lower bound is 2 and the current best known is 2.5. So you can decrease teh current best known factor by 20%. Or one could say, the current best known factor is 25% larger than the optimal one
That is 20% is the maximum possible percent decrease from the current best
So by using the term 'margin' its clear enough that theyre referring to the margin with reference ot the current best, rather than with respect to the optimal
 
@KevinDriscoll you are exactly right. 2 is a lower bound and 2.5 is so far an upper bound. Yea, you are right about that margin is for current best, not to the optimal.
 
Honestly, you could say it either way
And which one you choose is just a conventional choice
As long as its clear in context what you mean
 
@KevinDriscoll I understand that to calculate the maximum possible percent decrease from current best is (2.5-2)/2.5=20%. which is (gap between lower bound and upper bound)/best upper bound = margin for improving!
Really thank you @KevinDriscoll for answering this question. Best!
 
7:23 AM
No problem, good luck
Time for me to go to sleep
 
Have a good sleep! here in Russia we are in afternoon! we are already in good time to study!
 
7:40 AM
Hello, Does anyone here know why in the end of the following answer, the uniqueness of limits implies that $\{f(x_j)\}\to f(x)$ math.stackexchange.com/questions/2535356/…
 
$\mathsf{CRing} \mathcal{CRing} \mathscr{CRing} \mathbf{CRing}$
 
Hey, I could really use some help understanding this post: math.stackexchange.com/questions/417365/…
In particular, could someone help me understand lemma 2?
(symplectic stuff)
 
 
2 hours later…
9:36 AM
[Random]
$S=\omega$
$S \subsetneq f(S)$
$T=f(S)-S$
$g : f(S) \mapsto f(S)$
$g(x) = \begin{cases}x,x\in S \\ ?,x \in f(S)-S \end{cases}$
 

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