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Sep
9
comment Is there a closed-form equation for $n!$? If not, why not?
@John Gietzen: Nice quote. But, as all computer scientists know, "any combination of sums, products, powers, exponential functions, or logarithms with a fixed number of terms will not suffice to express $n$," either! (In our usual number systems, the number of terms needed to express a positive integer $n$ is asymptotically $\log(n)$.)
Sep
9
answered Is there a closed-form equation for $n!$? If not, why not?
Sep
9
comment Area of a recursive function
@IVlad: In general there's no reason to restrict to [-1,1]. But the problem statement contains many indications that this is the intention, including (a) the implicit assumption that f_n actually has an area, (b) your friend's assertion that the (limiting) area is 1, (c) the loose statement of the problem. All these lead one to consider the conditions under which the problem might make some sense and have a nontrivial solution. In this case, there is a clear difference in behavior of the f_n on [-1,1] compared to the rest of the reals. Finally, yes, it's correct to say the area is -infinity.
Sep
8
comment Terminology for handling probabilities with partial knowledge
They are called "personal probabilities" in the literature on Bayesian statistics, which is the field where the possibility of different probability assessments among different observers is considered.
Sep
8
comment Area of a recursive function
[-1,1] is the only interval where it's interesting to integrate. Beyond that interval, you can see that (at any fixed point x) the values of f(x) eventually get larger and larger in size, almost doubling each time, and are very negative. Note, too, that the integral of even f_0 itself diverges. Thus your series isn't even defined unless you limit the domain of the f_n. Whoever told you the area was 1 must have been assuming the domain of f is restricted to [-1,1].
Sep
8
comment Is this a known algebraic identity?
This is a brilliant idea. @Moron: Both sides give the probability of discovering segments 1, 2, ..., n in order. The left side enumerates all the possible ways in which an n+1st segment could be accounted for. The expressions must be equal by the law of total probability. If you were to add any finite number of new segments, you would have to sum over all possible ways in which they could be inserted in the order. This would be algebraically the same as repeatedly applying the identity for inserting one new segment.
Sep
8
answered Area of a recursive function
Sep
7
comment Can we slice an object into two pieces similar to the original?
@muad: Boxes aren't strictly convex.
Sep
6
comment Mental card game
@Dan: I simply did not interpret the problem the way you apparently intended, due to an ambiguity in English. Where you wrote "without any person learning additional information beyond that which can necessarily be inferred from n, k, the identity of the winner, and the value of that person's own card," I understood "that person" to refer to the immediately preceding person, namely the "winner." I hope you weren't the one who downvoted my response, which was offered in a constructive spirit and--whether wrong or right--suggests alternative approaches to analyzing the problem.
Sep
4
comment How to solve DE that relate values of derivatives at different points?
Differential-difference equation. (Google it.)
Sep
4
comment Is there a (deep) relationship between these various applications of the exponential function?
(4) and (5) are related to combinatorial limits and the Central Limit Theorem, whereas (1) - (3) are mutually connected in obvious ways. So the challenge comes down to finding a deep link between any of {1,2,3} and {4,5}. One possible avenue of investigation is via the "characteristic function" (Fourier transform) of the Gaussian, which is a key tool in most proofs of the CLT. Another promising route is via the Heat equation, diffusions, and random walks.
Sep
4
comment Evaluating a convergent improper triple integral over the unit sphere
Additional response to Américo's comment: Integrals like these arise in the solution of the Schrodinger equation for central forces (e.g., one-electron atoms). This is a quantum mechanical system with a large group of symmetries. (The full group is SO(4), but that's a subtle result; in any event the group SO(3) acts in the obvious way on the coordinates.) A deep way to solve and explore such highly-symmetric systems is to study the representations of the symmetry group. That provides an algebraic--and extremely practical--perspective on the solutions of PDEs and on computing related integrals.
Sep
4
comment Evaluating a convergent improper triple integral over the unit sphere
Interesting question. Most of Calculus amounts to algebraic manipulation of expressions according to rules--sum rule, product rule, chain rule for derivatives; integration by parts, substitution, etc., etc. The rules themselves of course obtain their meaning from analytical considerations, traditionally encapsulated in a few basic theorems about derivatives of polynomials and elementary functions, the Fundamental Theorem of Calculus, and some limit theorems. Thus, what is new to you here most likely is the notation, which highlights the algebraic patterns in this problem.
Sep
4
answered If $(a^{n}+n ) \mid (b^{n}+n)$ for all $n$, then $ a=b$
Sep
3
comment Evaluating a convergent improper triple integral over the unit sphere
Integration by parts can reduce [n,0,0;a] ultimately to [0,0,0;-1], which gives the surface area $4 \pi$. That makes the solution entirely algebraic, showing directly why we should expect integrals of the form [i,j,k;a] to be rational multiples of $4 \pi$ whenever $a$ is rational. This approach generalizes to other dimensions, too.
Sep
3
comment Evaluating a convergent improper triple integral over the unit sphere
@Américo, @Moron: you guys are kind. I'm sorry to make you wait; I thought it would be fun to post a solution of a different nature.
Sep
3
answered Evaluating a convergent improper triple integral over the unit sphere
Sep
3
comment Proving that this sum $\sum\limits_{0 < k < \frac{2p}{3}} { p \choose k}$ is divisible by $p^{2}$
This is identical to a problem that appeared on the Putnam exam a couple decades ago.
Sep
2
comment What is your favorite proof that $e^{ix}$ has a period of $2\pi$?
Well, yes, but then it's a tautology. Now you have to prove that the lcm of the periods of sine and cosine equals $2 \pi$!
Sep
2
comment Evaluating a convergent improper triple integral over the unit sphere
The spherical coordinate conversion for z is not correct.