Math and Logic Puzzles: Redux

[Who]

Why was I not told this existed?

Prove that the derivative of an odd function is an even function and that the derivative of an even function is an odd function.

Let f be a differentiable even function.
f'(x) = \lim_{h\to 0} [f(x+h)-f(x)]/h = \lim_{h\to 0} [f(-x-h) - f(-x)] / h = \lim_{h\to 0} [f(-x+-h) - f(- x)]/ - -h = \lim_{h\to 0} [f(-x+h) - f(- x)]/ -h = -f'(-x), thus f' is odd.
Let g be a differentiable odd function.
g'(x) = \lim_{h\to 0} [g(x+h)-g(x)]/h = \lim_{h\to 0} -[g(-x-h)-g(-x)]/h = \lim_{h\to 0} [g(-x-h)-g(-x)]/-h = \lim_{h\to 0} [g(-x+h)-g(-x)]/h = g'(-x), thus g' is even.

[Who]

Suppose you have 13 real numbers such that if you remove any one of them, you can divide the remaining 12 into two groups of 6 which each have equal sums. Prove that all 13 numbers are equal.

[Who]

Nice proofs, you got it.

[Who]

Spoiler: Above problem

Randomly generate 3 points, then randomly generate a 4th, then randomize which gets paired with which. If the 4th fits inside the triangle defined by the other three, no matter how you define the line segments they won't intersect. If it doesn't, then there is exactly one which it can be paired with to intersect the line defined by the other two. (Unless it lies on the line generated by two of them, which occurs with probability 0) The probability of it being paired with this one is 1/3. The probability that the fourth falls inside the area defined by the first three is the same as the expected value of the indicator variable, which is the same as the expected value of the area. Thus, the total probability is Expected area/3 = 11/(144*3).

[Who]

On Implosion's triangle problem:

Spoiler: Why I was wrong

Even if one isn't in the triangle formed by the others, that doesn't mean you can't make a triangle out of other points such that the 4th is in the triangle formed by the other 3. If you tried doing the thing I said then the lines would intersect past the edge of one of the segments

Spoiler: Fixed

If and only if you have four points such that no matter which point you choose, it is not in the triangle formed by the other 3, then it is possible to assign the line segments such that they will intersect. Proof of this: Make a triangle, extend the line segments into lines, note that the areas where the fourth point can't be are inside the triangle, and in the areas which meet the triangle at a vertex (As opposed to meeting the triangle at a side). If it's inside the triangle we're done, if it's in one of the areas which ends at a vertex, then that vertex is inside the triangle formed by the other 3.

If A is in the triangle BCD then B is almost surely not in ACD. (And neither is C nor D). In any set of four points, at most one of them can be in the triangle formed by the other 4. Proof of this: Either A or B must be further from line CD, the triangle defined by the closer one clearly cannot contain the further one, since all points in the triangle are closer to CD than the third vertex.

For i=1-4, let x_i be the event "the ith point is in the triangle formed by the other 3". Note that each x_i has probability 11/144. Also, all events are mutually exclusive. Thus, the probability that none of them happen is 1-4*11/144 = 100/144.

[Who]

In this specific case it doesn't have paradoxes because anything the super being could accomplish by predicting they could also accomplish by cheating (Have the box set up so that it automatically burns the money if you take both or some other similar thing). If we accept the super being's superness as fact, then the problem is the same as facing a cheating being, so obviously don't take the second box.

Other than the fact that it may be possible to use her to break quantum mechanics.

[Who]

(0!+0!+0!)! = 6
(1+1+1)! = 6
2+2+2 = 6
3!+ 3-3 = 6
(4-4/4)! = 6
5+5/5 = 6
6+6-6 = 6
7-7/7 = 6
(√(8 +8/8))! = 6
(√9)!+9-9 = 6

[Who]

I get the same result, also making some assumptions, but I feel reasonably confident in them.

How I get it:
Let A be the number of times a 1 is rolled minus the number of times a 4 is rolled, let B be the number of times a 3 is rolled minus the number of times a 6 is rolled, let C be the number of times a 2 is rolled minus the number of times a 5 is rolled. The event is now:
A+B>|C|
Also note that A, B, and C are identically distributed and approximately normal with mean 0 via CLT. Normalizing to make them variance 1 doesn't change the equation, so if we assume they are independent it's now
X>|Y| where X is normal with mean 0 variance 2, Y is normal with mean 0 variance 1. We then do basic integration (Set up the integral, note that it's the standard e^(-r^2/2) with r from 0 to infinity but over a sector with a weird angle) and end up with the answer scigatt gave. A,B, and C are obviously not independent, for example if A=N then that means that every roll was a 1 so B and C must both be 0. But they are [assumption] approximately independent because that is unlikely to happen.

[Who]

If the field has odd characteristic:
For any nonzero x, x and -x are distinct and square to the same value. Thus there are (|F|-1)/2 nonzero squares so there must be (|F|-1)/2 nonzero nonsquares, so let c be a no square then x^2-c is irreducible and thus extending by the root of that is a quadratic extension.

If the field is characteristic 2:
Apply the same argument but instead of x and -x take x and x+1, and instead of both have the same square we have that both have the same x(x+1). Thus there’s some c which is not of the form x(x+1), and we have that x(x+1)+c is irreducible.

[Who]

10^27

[Who]

1/2^27

6/35