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What's the optimal way to encode the following constraint in conjunctive normal form?
Exactly 3 out of n propositional variables is allowed to be true

Is it of this kind?
At least 3 out of n propositional variables is allowed to be trueAt most 3 out of n propositional variables is allowed to be true
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Bullet51

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Are the functions 0 and e^x the only functions that are its own derivative?
And are there any functions that are its own derivative's derivative?
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gameoflifemaniac

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Any scalar multiple of e^x (like 2*e^x) is also its own derivative. Furthermore, these are all also their derivatives' derivatives. Examples of functions that are not their own derivative but that are their derivatives' derivatives would be the hyperbolic cosine and sine.

77topaz

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More generally, functions that equal their n-th derivative will be composed of terms of the form Ce^kx, where C is a constant and k is an n-th root of unity.
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BlinkerSpawn wrote:More generally, functions that equal their n-th derivative will be composed of terms of the form Ce^kx, where C is a constant and k is an n-th root of unity.

So, that n-th root can be complex?
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gameoflifemaniac

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gameoflifemaniac wrote:
BlinkerSpawn wrote:More generally, functions that equal their n-th derivative will be composed of terms of the form Ce^kx, where C is a constant and k is an n-th root of unity.

So, that n-th root can be complex?

Yes — z^n = 1 has exactly n solutions which lie evenly spaced around the unit circle. For example, the fourth roots of unity are 1, i, -1, and -i, which form the vertices of a square centered around 0; the third roots of unity are 1, -(1/2)+(√(3)/2)i, and -(1/2)-(√(3)/2)i, which form an equilateral triangle centered around 0.
x₁=ηx
V ⃰_η=c²√(Λη)
K=(Λu²)/2
Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt)

$$x_1=\eta x$$
$$V^*_\eta=c^2\sqrt{\Lambda\eta}$$
$$K=\frac{\Lambda u^2}2$$
$$P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$

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Aidan F. Pierce

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gameoflifemaniac wrote:
BlinkerSpawn wrote:More generally, functions that equal their n-th derivative will be composed of terms of the form Ce^kx, where C is a constant and k is an n-th root of unity.

So, that n-th root can be complex?

Yes, in particular cos(x) = (1/2)e^(ix)+(1/2)e^(-ix) and sin(x) = (-i/2)e^(ix)+(i/2)e^(-ix).

Macbi

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Macbi wrote:Yes, in particular cos(x) = (1/2)e^(ix)+(1/2)e^(-ix) and sin(x) = (-i/2)e^(ix)+(i/2)e^(-ix).

Yes, and those are examples of functions that equal their fourth derivative (but not their first, second, or third).

77topaz

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What would the tilings {1,∞} and {∞,1} look like?
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muzik

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muzik wrote:What would the tilings {1,∞} and {∞,1} look like?

{∞,1} would be a half-plane (and not exactly a tiling, considering the {,1} means the tiling consists of a single shape)
The structure of {1,∞} is a more difficult question, because it can't be a dual tiling to {∞,1}. If you were to take, say, the edge graph of a {n,∞} tiling you would then have a pattern with an infinite number of segments (one-sided polygons) meeting at each vertex in the hyperbolic plane, but these don't constitute a tiling in and of themselves because the constituent (one-sided) polygons share no edges.
Under the strictest definition of a tiling, a {1,∞} tiling would be an infinite number of edges between two vertices (or the dual tiling to {∞,2} embedded in the Euclidean plane)
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A silly question from a silly pony---

I was looking at the slides for Harrison's recent "[i]Let's make set theory great again!/i]" talk (more out of curiosity than anything else); on page 31 (that is 10.1), he talks about, among other things,

[o]ther convenient ‘magic’ like using symmetries, transferring results via isomorphisms, homotopy equivalence or elementary equivalence (Urban’s Ultraviolence Axiom) is done by theorem proving, not the foundations.

...Urban's Ultraviolence Axiom? (Ultraviolence? Ultraviolence? Or just autocorrect gone wild?)

I imagine that "Urban" here is Christian Urban, but Google knows nothing of said axiom. Neither does Arxiv, not that I checked very thoroughly. Any pointers, tips, links...?
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Apple Bottom

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Apple Bottom wrote:(Ultraviolence? Ultraviolence? Or just autocorrect gone wild?)

It may be "univalence".
Well, autocorrect has gone wild.
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Bullet51

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Bullet51 wrote:
Apple Bottom wrote:(Ultraviolence? Ultraviolence? Or just autocorrect gone wild?)

It may be "univalence".
Well, autocorrect has gone wild.

That makes sense. Thanks!
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Apple Bottom

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What curve has the property that an object rolling on it has constant speed?
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gameoflifemaniac

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gameoflifemaniac wrote:What curve has the property that an object rolling on it has constant speed?

Assuming there's no friction, simply a horizontal surface.

77topaz

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77topaz wrote:
gameoflifemaniac wrote:What curve has the property that an object rolling on it has constant speed?

Assuming there's no friction, simply a horizontal surface.

That is true as long as the object is a perfect sphere of uniform density. Other cases would require a different solution. For a given object to roll on a frictionless surface at a constant speed, you need a surface that keeps the center of mass a constant distance above the ground.
Gamedziner

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77topaz wrote:
gameoflifemaniac wrote:What curve has the property that an object rolling on it has constant speed?

Assuming there's no friction, simply a horizontal surface.

But a sphere rolling on a horizontal surface with some initial velocity will slow down. I figured out the curve will have decreasing curvature. The curve would look roughly like this:
What the curve would roughly look like
img.png (2.77 KiB) Viewed 3819 times

But what curve is it exactly?
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gameoflifemaniac

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gameoflifemaniac wrote:What curve has the property that an object rolling on it has constant speed?

gameoflifemaniac wrote:But a sphere rolling on a horizontal surface with some initial velocity will slow down. I figured out the curve will have decreasing curvature. The curve would look roughly like this:
img
But what curve is it exactly?

Just a wild, wild guess but...
Brachistochrone?
sorry, I don't really know
this is just a wild guess
If you're the person that uploaded to Sakagolue illegally, please PM me.
x = 17, y = 10, rule = B3/S23b2ob2obo5b2o$11b4obo$2bob3o2bo2b3o$bo3b2o4b2o$o2bo2bob2o3b4o$bob2obo5bo2b2o$2b2o4bobo2b3o$bo3b5ob2obobo$2bo5bob2o$4bob2o2bobobo! (Check gen 2) Saka Posts: 2732 Joined: June 19th, 2015, 8:50 pm Location: In the kingdom of Sultan Hamengkubuwono X ### Re: Thread for Non-CA Academic Questions If we ignore air-resistance, friction generally doesn't depend on velocity. So you just need a constant downward slope so that the constant friction cancels out the constant force of gravity. EDIT: I'm being stupid. It doesn't even matter if friction does depend on speed, because the thing is going at a constant speed anyway. So even with air-resistance the answer is going to be a constant slope. Last edited by Macbi on August 18th, 2018, 12:57 pm, edited 1 time in total. Macbi Posts: 581 Joined: March 29th, 2009, 4:58 am ### Re: Thread for Non-CA Academic Questions gameoflifemaniac wrote:What curve has the property that an object rolling on it has constant speed? gameoflifemaniac wrote:But a sphere rolling on a horizontal surface with some initial velocity will slow down. I figured out the curve will have decreasing curvature. But what curve is it exactly? The only reasons spheres slow down while rolling on horizontal surfaces are friction and air resistance/drag (mostly friction). You can only maintain a constant speed by rolling the sphere down a line/curve with a slope that causes the speed gained by gravity to be equal to the speed lost by friction and drag, even if that loss is zero. If the forces of gravity, friction, and drag all stay constant, maintaining a given speed requires a perfectly straight line, as changing the slope at any point changes how much energy, and therefore force, is transferred away from the sphere to the surface. A curve has changes in slope, meaning you would need changes in gravity, friction, and/or drag to keep a constant speed using one. Gamedziner Posts: 606 Joined: May 30th, 2016, 8:47 pm Location: Milky Way Galaxy: Planet Earth ### Re: Thread for Non-CA Academic Questions I've probably asked this before, but is there a name for the following type of geometry, and has it been studied in any detail?: Take two points anywhere on a plane, labelled A and B. If the straightest possible path from A to B must intersect itself at least once, the plane can be said to be "XXXX". Bored of using the Moore neighbourhood for everything? Introducing the Range-2 von Neumann isotropic non-totalistic rulespace! muzik Posts: 3217 Joined: January 28th, 2016, 2:47 pm Location: Scotland ### Re: Thread for Non-CA Academic Questions muzik wrote:I've probably asked this before, but is there a name for the following type of geometry, and has it been studied in any detail?: Take two points anywhere on a plane, labelled A and B. If the straightest possible path from A to B must intersect itself at least once, the plane can be said to be "XXXX". You'll have to give your definition more precisely. I can't think of anything that could have a property like this. If a path from A to B intersects itself at C, then there is a shorter path that goes from A to C and then directly to B. So the shortest path from A to B can't intersect itself. In most kinds of geometry the shortest path is considered straight. Macbi Posts: 581 Joined: March 29th, 2009, 4:58 am ### Re: Thread for Non-CA Academic Questions muzik wrote:I've probably asked this before, but is there a name for the following type of geometry, and has it been studied in any detail?: Take two points anywhere on a plane, labelled A and B. If the straightest possible path from A to B must intersect itself at least once, the plane can be said to be "XXXX". The geometry would have to be limited in directional movement, like lines tracing the paths of photons reflecting off mirrors. Gamedziner Posts: 606 Joined: May 30th, 2016, 8:47 pm Location: Milky Way Galaxy: Planet Earth ### Re: Thread for Non-CA Academic Questions Gamedziner wrote: muzik wrote:Take two points anywhere on a plane, labelled A and B. If the straightest possible path from A to B must intersect itself at least once, the plane can be said to be "XXXX". The geometry would have to be limited in directional movement, like lines tracing the paths of photons reflecting off mirrors. Or we could have negative length lines WADUFI Sarp Posts: 128 Joined: March 1st, 2015, 1:28 pm ### Re: Thread for Non-CA Academic Questions Is this identical to a sphere? x = 120, y = 114, rule = B3/S2351bo$52b2o$54b2o$56bo$57bo$48bo8bo$49b2o7bo$51b2o6bo$10b97o$10bo42bo7bo44bo$10bo42bo6b2o44bo$10bo41b2o5bo46bo$10bo40bo5b2o47bo$10bo39bo5bo49bo$10bo38bo5bo50bo$10bo37bo5bo51bo$10bo42b2o51bo$10bo95bo$10bo48b2o45bo$10bo48bobo44bo$10bo48b2o45bo$10bo48bobo44bo$10bo48b2o45bo$10bo95bo$10bo95bo$10bo95bo$10bo95bo$10bo95bo$10bo95bo$10bo95bo$10bo95bo$10bo95bo$10bo95bo$10bo95bo$10bo95bo$10bo95bo$10bo95bo$10bo95bo$10bo95bo$10bo95bo$o9bo95bo$bo8bo10b2o83bo$2bo7bo8b2o85bo$3bo6bo8bo86bo$4bo5bo8bo86bo$4bo5bo7bo87bo$5b2o3bo6bo9b3o76bo$7b2obo5bo10bobo76bo$9b2o4bo11b3o76b4o$10bo4bo11bobo76bo3bo$10b2o3bo11bobo60b2o13b2o4bo$10bobobo75bobo11bobo5b2o$10bo2b2o75b2o11bo2bo7bo$10bo79bobo8b2o3bo8bo$10bo79b2o9bo4bo8bo$10bo89bo5bo9b2o$10bo89bo5b3o8bo$10bo88bo5b2o2bo7bo$10bo87bo5bobo2bo8bo$10bo87bo4bo2bo2bo9bo$10bo86bo5bo2bo3bo$10bo85bo5bo3bo4bo$10bo95bo4bo$10bo95bo$10bo95bo$10bo95bo$10bo95bo$10bo95bo$10bo95bo$10bo95bo$10bo95bo$10bo95bo$10bo95bo$10bo95bo$10bo95bo$10bo95bo$10bo95bo$10bo95bo$10bo95bo$10bo95bo$10bo95bo$10bo95bo$10bo95bo$10bo95bo$10bo95bo$10bo95bo$10bo95bo$10bo48b3o44bo$10bo48bobo44bo$10bo48b3o44bo$10bo48bobo44bo$10bo48bobo44bo$10bo95bo$10bo95bo$10bo95bo$10bo53bo41bo$10bo52bo42bo$10bo51bo43bo$10bo51bo43bo$10bo50bo44bo$10bo49bo45bo$10bo48bo46bo$10bo48bo46bo$10bo47bo47bo$10b97o$58bo$58bo$57bo$57b2o$59bo$60bo$61b3o$64b4o$67b2o!
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