letterX wrote:Stirling's approximation: know it!
Yesssssss. Perfect.
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Macbi wrote:How many essentially different game theoretic scenarios are there with 2 players, each choosing between 2 options?
Yakk wrote:hey look, the algorithm is a FSM. Thus, by his noodly appendage, QED
Yakk wrote:hey look, the algorithm is a FSM. Thus, by his noodly appendage, QED
letterX wrote:... you... wouldn't call that useful? It's pretty much the reason we have logarithms.
Quaternia wrote:Consider the following two person logic game.
[imath]\forall x \exists y P(x,y)[/imath]
Where [imath]x \in[/imath] {a,b} and [imath]y \in[/imath] {c,d}.
Player 1 picks the x, Player 2 picks the y. Player 1 wins if the Decidable Property P is false, Player 2 wins if the property P is true.
This is a two-player, two-option game, as Player 1 can only choose either a or b, and Player 2 can only choose either c or d. (Xor, the or is exclusive)
There are as many games of this type as there are decidable properties that involve some sort of relation between two objects in the logic used! (in a not-too-weird logic that has quantifiers as well as some other concepts that I used implicitely)
Eebster the Great wrote:I think pretty much the reason we have logarithms is
1. inverse exponential
2. integral of 1/x
3. multiply-add
The third is becoming much less significant as time goes on.
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ameretrifle wrote:Magic space feudalism is therefore a viable idea.
jestingrabbit wrote:The word "proof" is a noun that denotes a series of propositions that establish some truth. The word "prove" is verb that denotes the act of assembling a proof.
That is all.
rat4000 wrote:Am I right?
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void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}
doesn't P(winning on a roll) ≠ P(two sixes)?rat4000 wrote:I win if at least two of the dice are sixes
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void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
jestingrabbit wrote:Ask yourself what is a simple set of properties that uniquely describes the Fibonacci numbers. Does this new sequence share these properties?
I expect you'll want to use the fact that [imath]{n-1 \choose r-1} + {n-1 \choose r} = {n \choose r}[/imath] somewhere.
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enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};
void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
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l0 = [a, b, c, d, e, f, g, h, i, j]
l1 = f(l0) = [c, e, h, f, b, g, d, i, j, a]
l2 = f(l1) = [h, b, i, g, e, d, f, j, a, c]
...
l30 = l0 = [a, b, c, d, e, f, g, h, i, j]
The 62-foot tall statue of Jesus constructed out of styrofoam, wood and fiberglass resin caught on fire after the right hand of the statue was struck by lightning.
meatyochre wrote:And yea, verily the forums crowd spake: "Teehee!"
diabolo wrote:I believe the number of iterations needed is equal to the least common multiple of the lengths of the orbits, in this case lcm(5,2,3) = 30.
This works on all the random examples I tested but I'd like to know if there is a theorem or something (other than my own "yeah it feels right and it works so far") confirming this?
rat4000 wrote:Well, here I am again with yet another question not worthy of its own thread. I'm trying to work my theoretical knowledge of integrals into a practical problem.
Say I have a rocket in a vacuum without gravity. The rocket will accelerate at 2m/s^{2} for 5 seconds, then its fuel will run out. How far will the rocket have gone from its starting place after ten seconds?
Well, the velocity is the antiderivative of acceleration, so if the acceleration is a constant a(t) = 2m/s^{2}, the velocity will be v(t) = ∫a(t)dt = 2t m/s (+c, which we set to zero). The position is the antiderivative of velocity, so s(t) = ∫v(t)dt = t^{2}m (+c, which we set to zero). Then the rocket will have gone 5^{2} = 25m after 5 seconds and will be moving with a speed of 2*5=10m/s, so after another 5 seconds it will have gone another 50m, or 75m in all.
Is there anything wrong in what I wrote?
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