Posts for OmnipotentEntity

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Here's my full calculation: rendered version. EDIT: I figured out my mistake. I didn't substitute correctly and I dropped a critical factor of 1/2 in equation 14. now my result agrees with yours.
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That involves a hell of a lot of calculation, unless I am quite mistaken. You eventually get down to the following cubic: 6z^3 - 6z^2 - 3z + 1 = 0 With the other two variables being equal to: (-z + 1 +- sqrt(-3z^2 + 2z + 3))/2 And when you find the only real root of the cubic you have: z = 1/6 (2 + cbrt(44 + 6 sqrt(26)) + cbrt(44 - 6 sqrt(26))) And the exact solution for x^4 + y^4 + z^4 is extremely long, but it's approximately 3.793.
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Ferret Warlord wrote:
And here I was, after reading pwner's post, trying to use the tangent and secant functions for g and f and ended up with an even more convoluted integral. Thanks for the help, guys. Just FYI, The question that inspired this was, What two functions will trace an object moving along a hyperbolic path at a constant rate, much like how the trig functions trace a steady circle?
Oh neat! So it does! Did you need a method of calculating this using a computer? Or was this just idle curiosity?
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p4wn3r wrote:
sqrt(cosh 2u)du = dx So, if you could integrate sqrt(cosh 2u), you could invert this equation and write the answer. The problem is that the function is not elementary, so you'd need to use special functions.
This particular function actually has a pretty nice representation in special functions. integral(sqrt(cosh(2u) du) = integral(sqrt(1 - 2 sin2(iu)) du) = -i E(iu | 2) + C, where E is the Elliptic E function. So if we represent E(x | 2) as E2(x) then (where E2-1 is the inverse): u = -i E2-1(i(x + C)) So f(x) = cosh(-i E2-1(i(x + C))) and g(x) = sinh(-i E2-1(i(x + C))) If f(0) = 1 then (because arccosh(1) = 2k pi i where k element of Z) 2k pi i = -i E2-1(iC) -2k pi = E2-1(iC) C = -i E2(-2k pi) g(0) = sinh(-i E2-1(iC)) 0 = E2-1(-i E2(-2k pi)) E2(x) is non-zero for x != 0, so k = 0. So C = 0. So now we have an explicit, albeit difficult to use expression for f and g, namely: f(x) = cosh(-i E2-1(ix)) g(x) = sinh(-i E2-1(ix)) Plotted:
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The reason why the limit goes to sqrt(2)/2 is because the derivative equation requires that f'(x) = cos(u) and g'(x) = sin(u) for some u. This restriction also requires that f(x) = cosh(v) and g(x) = sinh(v) for some v. Because both derivatives cannot be zero, the functions must be either increasing or decreasing. Because of the first constraint, the functions need to approach each other asymptotically, so the derivatives have to also approach each other. The only point where sin(u) = cos(u) is at +- sqrt(2)/2. I don't believe that this differential equation has an elementary solution, but we can guess at what they might be related to. Because f(x) and g(x) are related to cosh(v) and sinh(v) and because they are asymptotically linear, this means that v must be a logarithmic-like function of x (in the limit, but not near 0). Because f'(x) and g'(x) are related to cos(u) and sin(u) and because they are asymptotically constant, this means that v must also be an asymptotically constant function of x that approaches pi/4.
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y = 1 is also a solution.
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From there it follows pretty easily that there exists real numbers that cannot be described in a finite amount of information. And that these numbers are the vast majority of all numbers.
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Warp wrote:
OmnipotentEntity wrote:
Were you looking for a uniform distribution on the number of face up cards? You didn't specify.
Well, I did talk about "an evenly-distributed random number between 0 and 52"
Ah, so you did. I apologize for missing that. Well, you can generate a random number between 1 and 64 using two distinguishable d8. Then if the number is between 1 and 52 flip that many cards over and shuffle normally. (If not regenerate the random number.) (EDIT: This seems to be equivalent to what Masterjun wrote.)
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Warp wrote:
OmnipotentEntity wrote:
Wouldn't just shuffling the standard way, but flip one half of the deck each time, work?
Then you would get half of the cards face up, not a random amount of them.
Mathematically, yes. However practically, there is some variation between the deck halves, and the deck halves may not actually be halves. So you'll get a normal distribution of face up and face down cards, where each card has a 50/50 chance of being face up (but just because each probability is 50/50 doesn't mean you'll get exactly 26 face up cards.) Were you looking for a uniform distribution on the number of face up cards? You didn't specify.
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Wouldn't just shuffling the standard way, but flip one half of the deck each time, work?
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FractalFusion wrote:
What is F/A analysis? Never heard of it before.
Took a class called "Engineering Economy." It had financial formulas. That one was called F/A, which as far as I've been able to gather stands for "Future Value given Annualized Payment."
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This is just a F/A analysis right? If then you just need to solve: 500 = ((1 + i)^N - 1)/i where i = 0.01, for N. Which gives 180.07 months, or 15 years.
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https://pastebin.com/ZYbscnCu Ripped from youtube video
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A simple way to prove that a triangle's internal angles are 180 degrees to simply use the fact that the external angles sum to 360 around any closed convex polyhedra, and the internal angle + the external angle is 180, and there's three such angles, 3*180 - 360 = 180. As far as I know, this doesn't use parallel lines at all. The sum of external angles can be proven by observing that in order to walk around a polyhedra you need to end up facing the same direction you began in. And the external angle is the change in heading at each corner. Whereas the fact that an internal angle + an external angle comes from the definition of an external angle directly. Actually, upon reflection it does, because it assumes that the heading direction does not change between beginning a side and ending it, which is rather antithetical to the notion of a straight line, but proving it requires constructing parallel lines. I'm leaving it just in case someone else had this idea, to help explain why it's wrong.
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Warp wrote:
Wikipedia defines Joule as "the energy dissipated as heat when an electric current of one ampere passes through a resistance of one ohm for one second." This seems to imply that a resistor of 1 ohm will always dissipate the exact same amount of heat regardless of what material it's made of. That feels surprising and counter-intuitive. Or, perhaps, it could be restated as: It seems to imply that there's a tight unique relationship between resistance and heat dissipation, regardless of the material.
That's not strictly true though. For instance, if you have an LED, the energy is dissipated as light. If you have a speaker, it's dissipated as sound. If you have a motor, the energy becomes mechanical work. And heat makes up only the rest of the energy. However, if we're talking about a passive resistive component that does no external work, then Wikipedia is correct. This comes simply from the conservation of energy. You can think of it in reverse. There isn't any reason for two materials to have the same heat dissipation when current is ran through it. Instead the amount material and its geometry determines how much energy is lost, and from this, the value of its resistance may be determined.
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Any n-bit number that ends with a 0 is even, whether it is positive or negative. If it ends with a 1 it is odd. Let us say for the sake of argument that we're dealing with 8 bit signed integers. Our two cases are: XXXXXXX0 * XXXXXXX0 and XXXXXXX1 * XXXXXXX1 Multiplying each of these through we get: XXXXXX00 for the even side. And the 0s tend to accumulate. For the odd side it's a bit more complicated. In the 1s digit we're left with a 1, obviously (odd * odd = odd); however, in the 2s digit we have the old 2s digit + old 2s digit. Because 0 + 0 = 0 and 1 + 1 = 0b10, this means that a 0 will live in the 2s place after multiplication. XXXXXX01 On the next iteration another zero appears in the new 4s place because it is equal to the 0b10*current 4s. The twos digit does not change because it is equal to 1 * 0 + 0 * 1. So we get an additional 0 XXXXX001 And so on. Eventually the zeros will pile up until it shifts them out of the precision of the integer regardless of how many bits it is.
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A minor remark. I just found out an updated version of this game was released in Jan 2017. Among other things, it increases Iji's walk speed. Perhaps now the previously actually impossible reallyjoel's dad difficulty is possible?
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If the numerical approximation is 1.1249976996... then the formula I have isn't too terribly bad after simplification, but you're correct that it's not invertible.
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Warp wrote:
(I understand that it's probably not possible to say "proof by contradiction is impossible for this number." I suppose what I'm asking is if there's some number that can be proven to not be rational, but for which there is no known proof by contradiction, or that proof is way too complicated to be feasible and there's a much easier way of proving it.)
Unfortunately, there is no real way to "access" irrationality directly. An irrational number is simply a number that is not rational, but a rational number is a number that can be expressed as p/q where p and q are integers. Because the definition of an irrational number is in terms of what it is not, then you have to prove that it's not a rational number, which means proof by contradiction. Other related properties of irrational numbers, such as a non-repeating decimal expansion, are consequences of this property and are themselves proven through contradiction (try it!). This means that the other standard argument to prove irrationality for numbers such as the Champernowne constant relies on proof by contradiction as well, albeit indirectly.
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I can diagonalize the matrix, but that requires fixing a value of h. I thought I could get a recurrence for h as well then somehow combine the two methods to get a single function in both f and h, but I couldn't figure it out. By continuing down this path I'm not suggesting that FractalFusion's answer is wrong or insufficient. (On the contrary, it's far and away more elegant than mine.) I was just trying to suss out why my answer wasn't working, and/or how to make it better.
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FractalFusion wrote:
Do you mean C(f+1,h) = 2 * C(f,h) + ~C(f-h, h)? This one instead would agree with your calculation C(6,2) = 2 * C(5,2) + ~C(3, 2) when f=5, h=2.
Yes, you are correct, I made a mistake copying from my notes.
FractalFusion wrote:
Some of my calculations don't seem to agree with this. For example: C(4,2)=8 (HHHH HHHT HHTH HHTT THHH THHT HTHH TTHH) C(4,1)=15 (everything but TTTT) ~C(3,1)=1 (TTT) 8≠(15-1)/2=7
That part may be entirely incorrect, as I didn't use or test out the h side in recursion, and I might have made a simple math error of some sort. Or it may simply be that the identity does not hold in certain cases. I never actually proved anything, I just saw a pattern and extrapolated. In this case: C(4, 1):
HXXX | True
THXX | True
?THX | True
??TH | True
Other | False
C(4, 2):
H(H)XX | True
TH(H)X | True
?TH(H) | True
(??TH) | False
Other | False
The assumption ? in the third lines take on different values between the lines. In C(4,1) the ? can only be T whereas in C(4,2) it can be T or H, which is where the missing combination came from. So it seems as if the relationship I gave earlier does not hold for any case where the patterns on the ? changes with the changing h, which is actually quite often. Hell, it does not even hold for the original example I gave (C(6,2) -> C(6,3) the ?s on the 4th line change from 3 to 4 possibilities.) I apologize for that and thanks for catching it.
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I approached the problem using a toy example. Let C(f, h) be the number of flip possibilities out of the 2^f that satisfy the condition that h heads appear in a row. Enumerating C(5,2) gives the following truth table (X is don't care, ? will be explained later).
HHXXX | True
THHXX | True
?THHX | True
??THH | True
Other | False
The ? marks are combinations that don't include the target sequence. So in the third line with one ? it can be either T or H. But the fourth line ?? can only be TT HT TH, not HH (otherwise it would be double counted in line 1.) Let ~C(f,h) = 2^f - C(f,h) for C(5,2) we have: 2^3 (first line) + 2^2 (second line) + 2 (third line ?) * 2 (third line X) + ~C(2,2) If we move from C(5,2) to C(6, 2) we get:
HHXXXX | True
THHXXX | True
?THHXX | True
??THHX | True
???THH | True
Other | False
Which can be examined line this:
HHXXX(X) | True
THHXX(X) | True
?THHX(X) | True
??THH(X) | True
(???THH) | True
Other | False
And so we have C(6,2) = 2 * C(5,2) (X on each line from 1-4) + ~C(3, 2) (last line) From here the recurrence is simply: C(f+1,h) = 2 * C(f,h) + ~C(f-(h+1), h) And then we can use the initial conditions: C(f=h, h) = 1 C(f<h, h) = 0 To find any given C(f, h) using a recurrence relationship. Now that we have a recurrence relationship and we can use eigendecomposition to find an explicit formula. First we have to get rid of the annoying constant: C(f+1, h) = 2 * C(f, h) - 2^(f-(h+1)) + C(f-(h+1), h) C(f, h) = 2 * C(f-1, h) - 2^(f-(h+2)) + C(f-(h+2), h) C(f+1, h) - 2 * C(f, h) = 2 * C(f, h) - 2^(f-(h+1)) + C(f-(h+1), h) - 2 * (2 * C(f-1, h) - 2^(f-(h+2)) + C(f-(h+2), h)) C(f+1, h) = 4 * C(f, h) - 4 * C(f-1, h) + C(f-(h+1), h) - 2 * C(f-(h+2), h) So for any given h we can construct a matrix A that when you multiply {C(f, h), C(f-1, h), ... C(f-(h+2), h)} against it you receive {C(f+1, h), C(f, h), ... C(f-(h+1), h)}, or the next iteration. Given that matrix, we can perform an eigendecomposition, let S be a matrix of the eigenvectors of A, and Λ be a diagonal matrix of the corresponding eigenvalues of A. Then SΛS^-1 = A (by some matrix math) Then because Λ is diagonal, it's very simple to take the nth power of and A^n = S Λ^n S^-1 (Because S and S^-1 cancel when multiplied.) I allowed Mathematica to chew on the condition where h=2, and it spit out: -(1/(4 Sqrt[11]))i (2 i Sqrt[11] (1+2^(3+f))+Root[-1-#1-#1^2+#1^3&,2]^(f-2) Root[704-10128 #1^2+42340 #1^4+#1^6&,2]+Root[-1-#1-#1^2+#1^3&,1]^(f-2) Root[704-10128 #1^2+42340 #1^4+#1^6&,3]+Root[-1-#1-#1^2+#1^3&,3]^(f-2) Root[704-10128 #1^2+42340 #1^4+#1^6&,6]) It's interesting that FractalFusion's answer does contain the x^3-x^2-x-1 pattern, but I'm curious where the strange x^6 formula comes from. You can construct a similar recurrence on h using the following: C(6,2)
HHXXXX | True
THHXXX | True
?THHXX | True
??THHX | True
???THH | True
Other | False
C(6,3)
HH(H)XXX | True
THH(H)XX | True
?THH(H)X | True
??THH(H) | True
(???THH) | False
Other | False
So we can say: C(f, h+1) = (C(f, h) - ~C(f-h, h))/2 With the initial conditions: C(f, h=0) = 2^f C(f, h=1) = 2^f - 1 So: 2 * C(f, h+1) = C(f, h) - 2^(f-h) + C(f-h, h) 2 * C(f, h) = C(f, h-1) - 2^(f-(h+1)) + C(f-(h+1), h) 2 * C(f, h+1) - 4 * C(f, h) = C(f, h) - 2^(f-h) + C(f-h, h) - 2 * (C(f, h-1) - 2^(f-(h+1)) + C(f-(h+1), h)) 2 * C(f, h+1) = 5 * C(f, h) - 2 * C(f, h-1) + C(f-h, h) - 2 * C(f-(h+1), h) C(f, h+1) = 5/2 * C(f, h) - C(f, h-1) + 1/2 * C(f-h, h) - C(f-(h+1), h) And the matrix can be set up in the same way, let's call this one B with eigenvector matrix T, and eigenvalue matrix M. Now I'm stuck. I have two matrices, and two degrees of freedom, but the matrix size of the f formula depends on the size of h, and vice versa. How do I put these together to find an explicit formula in two variables of C(f, h)?
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This is one I ran into on the Reddit a few days ago, and it's burrowed into my brain. The original question is: What is the probability that if I flip 15000 coins, at least 15 consecutive heads appears at least once. The answer is a very large irreducible fraction that's about 20.45%. I won't explain in case someone wants to attack this problem. My question is: If C(f, h) is the number of combinations for "f" flips and "h" consecutive heads. (So that P(f, h) = C(f, h) / 2^f, and in particular P(15000, 15) = 20.45%), is there an explicit formula (ie, not recursive) for C(f, h) in terms of f and h. I have a method of finding an explicit formula in f for a given h. But I don't know how to extend it to an arbitrary h.
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I apologize, unless it allows you to attack more often, it does the same damage as the mega buster. I thought I remembered there being a slight difference, but I was mistaken.
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Would switching to the Air Cutter speed up both Quickman and the Sniper Joe section?
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