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Messages  phenomist
1
« on: December 01, 2012, 12:23:53 am »
4 doors: Well, at least nobody's going to kill you for asking tricky questions. So let's ask them.
1) (Are you lying on this question) XOR (Is the safe door either door 1 or 2)? 2) (Are you lying on this question) XOR (Is the safe door either door 1 or 3)?
2
« on: December 01, 2012, 12:14:06 am »
3
« on: November 10, 2012, 05:51:03 pm »
5 mod 2 = 1 6 mod 2 = 0 7 mod 2 = 1 5 mod 2 = 1
4
« on: November 10, 2012, 03:46:49 pm »
I think you can just check the parity of the resultant product, so [[5][6][7][5]] = [[1][0][1][1]], as desired. This is because xor on bits is basically addition mod 2.
5
« on: February 27, 2012, 09:34:02 pm »
Was going to throw in L'Hopital's rule, but there's already an enlightening section on that on page 2.
It all depends on the context of the 0's, really, and simply putting "0" isn't giving the context. Hence its undefinability.
6
« on: February 02, 2012, 12:53:28 am »
Not sure if this works, but you could treat each exponent as a number and add/subtract.
7
« on: January 18, 2012, 02:31:34 am »
Also, ln(x)=ln(1/x)
If x is not between 1 and 1, take its reciprocal. (I'd also like to note that ln(negative number) is a bad idea.)
8
« on: December 01, 2011, 09:52:01 pm »
Well, the gamma function is a real extension of the factorial function used to define combinations, so I suppose it can be done. Only theoretical for now, though things like fractional derivatives are applicable in fields like fluid dynamics.
9
« on: December 01, 2011, 12:20:18 am »
More generally, the number of kdimensional subcomponents in a ndimensional hypercube is 2 ^{nk}* _{n}C _{k}. With the gamma function we can easily analyze noninteger dimension hypercubes
10
« on: December 01, 2011, 12:16:34 am »
If you need a closed form expression...
Note that {a_{i}}_{i=1} = {e_{i}}_{i=1} via symmetry, and likewise {b} = {d}. We can consider the finite state diagram a <> b <> c <> d <> e which is symmetric on c.
Now, we may reexpress each recursion: a_{n+1}=b_{n} b_{n+1}=a_{n}+c_{n} c_{n+1}=2b_{n}
Notably, this means we can express a and c directly in terms of sequence b. So it logically follows that we should try substituting those out in the recursion for b: b_{n+1}=3b_{n1}
The characteristic polynomial of this recursion is λ^{n+1}=3λ^{n1} => λ^{2}3=0.
In other words the roots are λ = sqrt(3) and sqrt(3). What does this mean?
Well, this means that we can express sequence b in terms of a linear combination of exponential functions, namely sqrt(3)^{n} and (sqrt(3))^{n}. In layman terms, b_{n}=A*sqrt(3)^{n}+B*(sqrt(3))^{n}.
It is not hard to calculate this; we already know that b_{1}=1 and (easily computed) b_{2}=2.
So we just solve a system of linear equations; 1=A*sqrt(3)B*sqrt(3) => sqrt(3)/3 = AB
2=3A+3B => 2/3 = A+B resulting in A=(2+sqrt(3))/6 and B=(2sqrt(3))/6.
Thus we have b_{n}=(2+sqrt(3))/6*sqrt(3)^{n}+(2sqrt(3))/6*(sqrt(3))^{n}.
But the million dollar question remains. We don't want the value of a_{n}, b_{n}, OR c_{n}; we want the sum!
Well, we add 2a_{n}+2b_{n}+c_{n}; the 2 is because we copied d and e over. We reexpress the a and c terms into b because we already have a form ready.
So we have Σ_{n}=2b_{n}+4b_{n1}=[(2+sqrt(3))/6*(2+4/sqrt(3))]*sqrt(3)^{n}+[(2sqrt(3))/6*(24/sqrt(3))]*(sqrt(3))^{n}
=[4/3+7sqrt(3)/9]*sqrt(3)^{n}+[4/37sqrt(3)/9]*(sqrt(3))^{n}
(Caveat; Σ_{1}=5, not 14/3; this occurs because the formulas for sequence b refers to a term 2 terms in advance, effectively drawing from the 1st term which is obviously invalid)
The sequence goes: 5,8,14,24,42,72,126,216,378,648,1134,...
Notice anything interesting? No?
I'll show it again. 5,8,14,24,42,72,126,216,378,648,1134,...
Ohey, every other term it triples! Partially due to how the structure of the recursion itself (with λ^{2}3=0 as a characteristic polynomial), but there in fact is a combinatorial interpretation! (Not all the algebraic sludgework QQ)
It all lies in PARITY. In fact, there are two mechanisms syncing together, digit parity and positional parity. While in the traditional sense, parity refers to evenness or oddness of numbers, since all digits are odd that's not going to be of much use. Instead we consider mod 4 parity, which helps separate the digits better (so 159 and 37). Positional parity refers to the evenness/oddness of the positional index of a particular string.
Well, as luck would have it, you're colorbound like a bishop on a chessboard! Congratulations. As a visual: ..13579 A■□■□■ B□■□■□ A = odd position B = even position
As it turns out, if you start at a black square, you're condemned to the color that you start on, as each consecutive digit means a position change, and since they differ by 2, it's also digit parity change! So the two colorchanging operations cancel out and you're back to square one.
This makes our task easier.
..13579 A■□■□■ B□■□■□ C■□■□■
Without loss of generality, assume that you start on a white square. As symmetry applies, both A3 and A7 have the same number of possibilities. Starting from A3 you can either go A3B3C3, A3B5C3, or A3B5C7. Starting from A7, you can get to C7 twice and C3 once. Adding them up, you round out to C3 receiving a path thrice, and C7 also receiving a path thrice. Hence the number triples every two turns.
The only thing is, that you can't really start on a black square and immediately expect a triple; the mechanism doesn't quite work like that. You have to wait until the second turn before you start tripling.
11
« on: November 30, 2011, 11:20:37 pm »
Are there plans to extend this to a "Ratings Per Post Scoreboard Automater"
12
« on: September 23, 2011, 07:32:14 pm »
Also note that f(1)=e
13
« on: September 23, 2011, 01:20:47 am »
Another method might be to note that both parabolas have the same curvature, only negated. Their vertex points are at (0,0) and (3,4). Hence, any line tangent to both parabolas must lie on the midpoint of their vertices (1.5,2), which can be easily shown by rotating the diagram.
Considering the tangent line at (a,a^2) has a slope of 2a, we can equate slopes: (a^22)/(a1.5)=2a => a^22 = 2a^23a => a^23a+2=0 => a=1 or 2.
These generate the desired lines, of slopes 2 and 4.
14
« on: September 21, 2011, 04:25:11 pm »
Whoa, conveyor belts!
15
« on: September 21, 2011, 01:06:53 am »
Wait, how would you quickly test primality for 120000digit numbers? If we could reliably do that, then let's go get that TiNspire key now.
