∑_{k=0}^{m} \binom{n − 1 + k}{k} = \binom{m + n}{n}

Count the number of (m + n)-paths from (0, 0) to (0, 0) using steps (1, 0) and (0, 1) on ℤ_{m} × ℤ_{n}. Let k be the row of the first return to column 0.STOP

\usepackage{youngtab}
\young(1,5,3)

STOP X_{1}, …, X_{n}

E(max_{i} X_{i}) =

∫_{0}^{∞} E(max_{i} X_{i}| ∑_{i} X_{i} = t)\frac{d}{dt}P (∑_{i} X_{i} ≤ t)dt =

∫_{0}^{∞} tE(max_{i} X_{i}| ∑_{i} X_{i} = 1)\frac{d}{dt}P (∑_{i} X_{i} ≤ t)dt =

E(max_{i} X_{i}| ∑_{i} X_{i} = 1)E(∑_{i} X_{i})

If P (X_{i} ≤ t) = 1 − e^{−t} for all i, then E(max_{i} X_{i}) = ∑_{k=1}^{n} \binom{n}{k} \frac{(−1)^{k+1}}{k} and E(∑_{i} X_{i}) = n

Furthermore the lengths X_{i} are the lengths of a partition of [0, ∑_{i} X_{i}] into n parts chosen uniformly at random.

Thus E(max_{i} X_{i}) = \frac{1}{n} ∑_{k=1}^{n} \binom{n}{k} \frac{(−1)^{k+1}}{k} for the lengths X_{i} of a uniformly random chosen partition of [0, 1] into n parts.

Todo: simplify ∑_{k=1}^{n} \binom{n}{k} \frac{(−1)^{k+1}}{k} using EKHAD

(Nothing above is properly justified.)STOP

Let P (f) be the Newton polytope (or -gon) of the polynomial f.

P (fg) = P (f) + P (g)

(Minkowski sum). For the non-obvious inclusion, argue that the lex-largest monomial is the product of the two lex-largest monomials of the factors (and thus a single product and not a sum of products) and that lex-order can be defined in any direction, not just in an axis-parallell manner. STOP

When asking questions with yes/no answers, one should ask questions whose answers have probability as close as possible to 0.5 of being yes (or no) when varying over all possible situations.

When there are several answers, the largest probability of any particular answer should be as small as possible.

(Probabilities conditioned on the answers received so far.)

Question: how bad can the greedy solution be? (taking the best question at each point, without considering how constrained questions one will be limited to in the future because of one’s choice.)STOP

The graph of the result of identifying x and −x in the dodecahedron is the Petersen graph. Of course.