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Show that Aut(G) = 1 if and only if |G| = 1 or 2.

I have a proof (literally found it in my sleep), though it uses the axiom of choice. I'm curious about other approaches and if that can be avoided. Utterly unimportant.

Short version: I found an unexpected identity,

Floor(a/b)+Floor(2a/b)+...+Floor((b-1)a/b)

= Floor(b/a)+Floor(2b/a)+...+Floor((a-1)b/a)

where a, b>0 are integers. Is there a "nicer" way to write this sum? I'm particularly interested in the case when a and b are relatively prime. For example, restricting to such pairs and letting a=2^k, these sums are (a-1)(b-1)/2.

For context, with the Super Bowl today, I was wondering what scores were theoretically possible. You can score 2 or 3 points (among others), which is sufficient to score any non-negative integer except 1. If a, b>0 are relatively prime, the sums above represent the number of exceptions, i.e. the number of scores that are not theoretically possible in a game where you can score a or b points at a time. (If they are not relatively prime, a similar interpretation works, but it's complicated by the fact that infinitely many scores are unobtainable.)

This is a rather open-ended question. The number 1729 number seems to come up more than a four digit number should, so I thought it might be fun to collect a few interesting properties of it. The most famous is probably that 1729 is the smallest number that can be written as the sum of two positive cubes in two different ways. This question was inspired by a seminar talk where "1728" kept showing up (elliptic curves).

Show that

product from i=1 to infinity of

(1+x^(2i))(1+x^(3i)) (1-x^(4i-2))(1-x^(6i-3))

= 1.

Notes: This is a formal product; don't worry about convergence issues. I have a solution, but I feel I may have missed a simpler one. There is a combinatorial interpretation in the background I can give if anyone is interested.

Riddle 1:

A prison has 1000 inmates and a room with a light switch, initially off. Every day, a random prisoner is let into the room, and they may flip the switch if they wish. Once a prisoner believes every prisoner has been in the room, he may tell the warden. If he's right, all the inmates are let go, but if he's wrong, they're all killed. The prisoners are allowed to decide on a strategy before any of them are put into the room. How can they all survive?

(Assume the random prisoner choice never "gets stuck", that the prisoners live forever, that they all want out alive, etc.)

Riddle 2:

A prison has infinitely many inmates, one for each integer, and they're numbered. The warden is insane. One night, she decides the next day she'll line up all the inmates and randomly give them either a red or a blue hat. They won't be able to look at their hat or tell others about their hat, but each prisoner can instantly see every other prisoner's hat color. They will be taken in order into a secluded room, asked the color of their hat, and put back in line. The other prisoners are unable to hear their response. After every prisoner has had their turn, all those who answered wrong are killed, and the rest are set free.

The janitor overhears her plans and lets the prisoners know beforehand. They meet and decide on a strategy where no matter what the warden does, only finitely many of them will die. What is their strategy?

I have a solution to (1) and two solutions to (2), but more would be welcome, and at any rate I found these to be interesting so I thought I'd try sharing them.

...for t real. Feel free to do the case when t=1.

It's easy to do the integral from 1 to infinity with residue calculus, but I have no idea how to do it from 0 to infinity. Mathematica can only integrate it in terms of special functions (SinIntegral and CosIntegral). There's a chance the question just has a typo, but in case I'm missing something, I thought I'd ask.

Given m points on the surface of the unit hypersphere in Euclidean n-space, how can one determine if those points lie on a hemisphere (including the boundary, say)? It seems like this problem should have been studied already, but I'm unsure where to look.

Of course, it's helpful to think of the n=1, 2, and 3 cases. I actually have an algorithm in mind, though it's slow.

Darboux's theorem places heavy restrictions on the derivative of a function. In particular, it must satisfy the intermediate value theorem. A standard example of a differentiable but not continuously differentiable function is x^2 sin(1/x) (and 0 at x=0). Using this, a bump function, and a number of theorems concerning uniform convergence and differentiability, I can construct a function which is differentiable but discontinuous at points 1/n for integers n.

Can a derivative be discontinuous on a dense set? On a set of measure > 0? Are there any standard examples of particularly awful behavior in this vein?

http://en.wikipedia.org/wiki/Darboux%27s_theorem_%...

http://en.wikipedia.org/wiki/Smooth_function#Examp...

What's the most overcomplicated proof of a simple proposition you've seen or can think of?

An outline of the proof or a link to it is perfectly acceptable. The proposition should be simple (say early college level at most), but the proof can be as deep as you like.

Answering another question made me realize how poor my mental stock of completely overdone proofs is, hence this question.

Solving the equation

3^n - 2^m = +/- 1

for integers n and m came up as an aside (that I avoided through some tedious calculations) in this question: http://answers.yahoo.com/question/index?qid=201110...

We have the solutions

3^0 - 2^1 = -1

3^1 - 2^1 = 1

3^1 - 2^2 = -1

3^2 - 2^3 = 1

which are in fact the only solutions for 0 <= n, m < 1000. I suspect there are no other solutions, but I'm not sure how to prove such a thing, or look it up, though I feel certain it's been studied (and I even dimly remember reading about something similar).

More generally, of course, one can ask what the solutions of

p^n - q^m = +/- 1

are for p and q prime. However, the left side is even unless precisely one of p, q is 2. There's always a trivial solution p^0 - 2^1 = -1, so we may suppose q=2, n >= 1 and consider

p^n - 2^m = +/- 1

For n=1, the solutions correspond precisely to the Fermat and Mersenne primes, the infinitude of which is an open question in both cases. For n > 1, the only solution I've found is 3^2 - 2^3 = 1, using 2 < p < 1000 (prime), 1 < n < 100, 0 <= m < 1000. I suspect this is the only solution.

So, my question is: (1) solve 3^n - 2^m = +/- 1; (2) give a reference discussing the second form, p^n - 2^m = +/- 1 for n > 1 [I imagine a proof is too much to ask].

It is true that every Hermitian matrix* has an eigenvalue. In particular, every real symmetric matrix has an eigenvalue. This follows immediately from the fundamental theorem of algebra even in the real case: the characteristic polynomial has at least one (a priori complex) root; that eigenvalue is real since the matrix is Hermitian.

Question: Is there a proof that doesn't rely on the fundamental theorem of algebra (FTA)?

I've looked at 4 different proofs of this version of the spectral theorem, all of which use the above fact and the FTA: Wikipedia's; PlanetMath's (apparently; it actually just states the above result without reference, saying it's "well-known"); a linear algebra book of mine (which similarly just assumes the existence of such an eigenvalue); and a random blog's.

In a similar vein, can the above fact be used to deduce FTA? If so that would give a nice intuitive reason why the proof of the above seems like it has to run through FTA.

*That is, n by n matrix A over R or C such that A* = A where * denotes the (conjugate) transpose.

Take the real unit interval [0, 1]. For some fixed d > 0, you have an infinite sequence of intervals of the form [a, b] in [0, 1], where (b-a) >= d. The question is, does some (infinite) subsequence of these intervals have non-empty intersection?

I've shown that for each natural number n, the intersection of some n intervals is non-empty. More reasoning is needed to extend this arbitrarily high intersection to an infinite intersection, though, and I haven't been able to supply it. More than anything, I just need a fresh perspective.

Note: my actual situation is more complicated. The intervals [a, b] are actually (optionally closed, though I'd prefer if closedness didn't matter) Lebesgue-measurable sets with measure >= d. I imagine the situations are reasonably analogous, though, and using intervals cuts out the measure theory to make it much more accessible.

Given a function f: X -> Y and some subset A of X, the restriction f|_A : A -> Y is defined by f|_A (a) = f(a) for all a in A. This is a common bit of terminology. I was wondering if anyone knows of a vaguely common term for the opposite, in the following sense: given a function f : X -> Y and some set B containing Y, let f|^B : X -> B where f|^B (x) = f(x) for all x in X. Essentially, this is the composition of f with the natural inclusion map between Y and B.

I like the word "extension", but that's pretty commonly used. I like the notation f|^B since I don't think it's commonly used, and it parallels the usual restriction notation.

The power rule, d(x^a)/dx = ax^(a-1), can be proven very quickly using the properties of the exponential and natural logarithm functions. This is a beautiful but rather high level derivation. I was curious if anyone knows an "elementary" derivation.

Proving it in the case that a is a positive integer is an easy application of the binomial theorem and the definition of the derivative. It can be proven for a=-1 and the chain rule then gives negative exponents. For a=1/n, n a positive integer, it can be proven using the identity u^(n+1) - v^(n+1) = (u-v)(u^n + u^(n-1)v + ... + uv^(n-1) + v^n). From this and the previous cases with the chain rule, it's true for all rational exponents.

How do you extend this to real exponents? Is there some nice continuity argument, where (roughly) the formula holds for a dense subset and the derivative, as a function, is continuous, so the formula holds everywhere?

I looked in an analysis book for a proof but it just lists the exponential function one. I also briefly looked at Wikipedia, and it just stated the result without proof.

I want to solve the equation

3arcsin(5/(2R)) + arcsin(6/(2R)) + arcsin(8/(2R)) = pi

for R, where R is a positive real number. Of course R > 4 for the last term to be defined. I know that the solution is 5, and that this solution is unique [the derivative is negative, so the left side is monotonic]. However, I got this solution essentially from a numeric solver. Is there a nice way to find this solution?

I've thought of plugging both sides into the sine function, but then it turns into a long pseudo-polynomial problem, which doesn't seem helpful.

If it helps, this equation comes up in the following geometry problem. A pentagon is inscribed in a circle and has edge lengths 5, 5, 5, 6, and 8. What is the radius of the circle? An alternative computation of the radius is also acceptable.