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$$
\sum_{k=0} ^m {n-1+k \choose k} = {m+n \choose n}
$$

Count the number of $(m+n)$-paths from $(0,0)$ to $(0,0)$ using steps $(1,0)$ and $(0,1)$ on $\mbb{Z}_m \times \mbb{Z}_n$. Let $k$ be the row of the first return to column $0$.STOP

\begin{verbatim}
\usepackage{youngtab}
\young(1,5,3)
\end{verbatim}STOP

$X_1, \ldots, X_n$\\

$$E(\max_i X_i) =$$\\
$$\int_0 ^\infty E(\max_i X_i|\sum_i X_i = t) \frac{d}{dt}P(\sum_i X_i\leq t) dt =$$\\
$$\int_0 ^\infty t E(\max_i X_i|\sum_i X_i=1) \frac{d}{dt}P(\sum_i X_i\leq t) dt =$$\\
$$E(\max_i X_i | \sum_i X_i = 1) E(\sum_i X_i)$$\\

If $P(X_i \leq t) = 1-e^{-t}$ for all $i$, then $E(\max_i X_i) = \sum_{k=1} ^n {n\choose k}\frac{(-1)^{k+1}}{k}$ and $E(\sum _i X_i) = n$\\

Furthermore the lengths $X_i$ are the lengths of a partition of $[0, \sum_i X_i]$ into $n$ parts chosen uniformly at random.

Thus $E(\max_i X_i) = \frac{1}{n} \sum _{k=1} ^n {n\choose k}\frac{(-1)^{k+1}}{k}$ for the lengths $X_i$ of a uniformly random chosen partition of $[0,1]$ into $n$ parts.

Todo: simplify $\sum_{k=1} ^n {n\choose k}\frac{(-1)^{k+1}}{k}$ 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. 

What is the communication complexity of finding the median of $A\cup B$ when Alice knows $A$ and Bob knows $B$? It is $O(\log(n))$ when $A,B\subseteq \{1,\dots,n\}$.

For each $n$, find $n$ lattice vectors $v_i=(x_i,y_i)$, $x_i,y_i$ integers, such that the sums $\sum_{i\in S} v_i$ are all distinct but as compact(undefined) as possible.

Percolation Voronoi (not Voronoi percolation) [after thinking about this: it seems not much can be said.] STOP\\

Minimizing the sum of squares of the errors is more natural than minimizing the sum of the absolute values of the errors if the errors are Gaussian (as physical errors are). This is because minimizing the sum of squares is a maximum likelihood estimator. (todo: remove opinion)
STOP

The relation $(s_is_j)^{m_{ij}} = e$ makes sense for reflections in arbitrary hyperplanes, not just linear ones (björner-brenti notation).
STOP

%Given is a cup of hot water, and a teabag attached to a string. My observation is that when holding the teabag by the string allowing it to rotate freely, it will rotate in different directions depending on whether the teabag is below or above the water line. This would be caused by the following statements which I think are true \\
%* the only significant force acting on the teabag is gravity\\
%* that force is negated when going from above to below the water, since the heavy stuff in the teabag floats\\
%* the function mapping gravity force to turning force is linear\\
%. STOP

A path in $PR^2$ can be thought of as a continuous movement of a pair of antipodal points on $S^2$. Restricting attention to one of these points (this is possible, even though the pair being unordered, since the two points will never equal each other) gives a path in $S^2$, which either ends in its starting point or the antipode of the starting point. (And a path in $S^2$ obviously gives a path in $PR^2$.) This is the use of the concept of 'covering space'.\\
STOP

Knowledge of $\lambda_1 ^k + \dots + \lambda_n ^k$ for $1\leq k \leq n$ as an ordered sequence is equivalent to knowing $\lambda_1, \dots, \lambda_n$ as a multiset. If $A$ is the adjacency matrix of a graph $G$, the $k$:th term in the sequence is the number of closed paths of length $k$ in $G$ (with marked 'starting' point). STOP

(non-un-)Knot around three points a,b,c such that removing any of a, b or c gives an un-knot: (aba'b')c(aba'b')'c' = aba'b'cbab'a'c', where ' is inverse of an operation and 'X' is the operation of going clockwise around point X. (Strictly these are computations in the fundamental group of the triply punctured plane, which happens to be a free three-group on generators a,b,c.)

Let $\sigma : [n] \to [n]$. The number of graphs $G$ on $[n]$ such that $xy\in E(G)$ iff $\sigma(x)\sigma(y) \notin E(G)$ is $2^{\dots}$. Replacing 'graph' by '$r$-coloring of the edges of $K_n$' replaces '2' by $r$.

%Writing $[n]_q = 1 + q + q^2 + \dots + q^{n-1}$ is the next (argument) natural generalization from generalizing $5! = 5 \cdot 4 \cdot 3 \cdot 2$ to $5! = 5\cdot 4\cdot 3\cdot 2\cdot 1$.

{\bf proposition: if $Z_1,\dots, Z_n$ are exponentially distributed with same means, then $(X_1,\dots,X_n)$, where $X_i = \frac{Z_i}{Z_1+\dots+Z_n}$, is uniformly distributed (wrt. lebesgue) on $\{(z_1,\dots,z_n): z_i \geq 0$ for all $i$, $z_1+\dots+z_n = 1\}$.}
proof. If $W$ is a random variable, then $p_W$ is its density function.

$p_X(x) = \int_0 ^\infty p_Z(tx)dt = \int _0 ^\infty e^{-(x_1+\dots+x_n)t}dt = \int_0 ^\infty e^{-t} dt$, which does not depend on $x$.

The following fact is implicit in many places, for example in the [theorem involving the word "abacus"]. However I needed some effort to state and prove it. fact:  If $\lambda$ is a partition, then if the hook number $h$ occurs in $\lambda$, then all its positive integer divisors occur as hook numbers in $\lambda$. The only crucial observation is the following: the hook number $h_{ij}$ counts the number of boxes in the unique 'snake' $S_{ij}$ with one end in row $i$ and the other in column $j$. This snake is removable, in the sense that $\lambda\backslash S_{ij}$ is a partiton diagram.

{\bf proposition}: Suppose $d|h_{11}$. Then there are $i,j$ such that $h_{ij} = d$. proof. Split the snake $S_{11}$ into $h_{11}/d$ consecutive subsnakes $S_1, \dots, S_r$, each of size $d$, where $r = h_{11}/d$. It is sufficient to show that at least one of the $S_i$ are removable, since then that snake will correspond to a hook, whose size will of course be $d$. Mark each end of each snake '+' or '-' in the following way. The left-bottom end of a snake is marked '+' iff removing the remainder(including the end itself) of the row containing the end leaves a partition. Similarly the top-right end is marked '+' if removing the remainder of the column containing that end leaves a partition. So a snake is removable if both ends (which may coincide (but only in a trivial case)) are marked '+'. We need to show that there is such a snake among the $S_i$.
Let us think of the snakes as going from left to right (they are in fact going left - down to right-up) and call the marks on each snake the left or right mark.

The following two claims easily imply that some snake has a double +.
\begin{itemize}\item If $S_L$, $S_R$ (left, right) are two consecutive snakes, then there is a + to the right in $S_L$ iff there is a '-' to the left in $S_R$.
\item $S_1$ has a '+' to the left, and $S_r$ has a '+' to the right.\end{itemize}
Example:$\young(9754+,864-S,6-S+,S+,+)$. Here $S$ denotes the middle part of each size-three snake. The hooknumber 'behind' the leftmost S is $3$.\\

Other implicit fact: Write $\lambda\to_t\lambda\p$ if $\lambda\p$ is a partition and can be obtained from the partition $\lambda$ by removing some snake of size $t$. For any partition $\lambda$ and positive integer $t$, any sequence $\lambda\to_t \lambda_1\to_t \dots \to_t \lambda_0$ of maximal length ends in the same partition $\lambda_0$.
By induction it is suffices to prove that for any partition $\lambda$ together with $\lambda_1,\lambda_2$ such that $\lambda\to_t \lambda_1$ and $\lambda\to_t \lambda_2$, there is a partition $\lambda_3$ such that $\lambda_1\to_t\lambda_3$ and $\lambda_2\to_t\lambda_3$. This is not hard to prove but it is unexpected that it is possible to prove.

In fact, if $a_t(n)$ is the number of partitions of size $n$ that can occur as $\lambda_0$'s ({\it $t$-core partitions}), then
$$
\sum_{n=0} ^\infty a_t(n)x^n = \prod_{n=1}^\infty \frac{(1-x^{tn})^t}{1-x^n}.
$$

{\bf entropy}
Suppose $X$ is a random variable such that $P[X = i] = \frac{1}{n}$ for $i\in\{1,\dots,n\}$. Then the entropy, or 'information content', $H(X)$, is defined to be $\sum_i p_i\log(\frac{1}{p_i})$. 

Why?

Of course one can think about $H(\cdot)$ as simply being a nice way to associate a number to random variables, satisfying useful properties. I will think about this informally and try to motivate calling $H(X)$ the information content of $X$. We will choose the base $2$ for the logarithm above and then the unit of $H(X)$ will be {\it bits}.

We need to informally motivate the following, after which the formula will follow from manipulations.

{\bf 1. If $X,Y$ are independent, then $H(X,Y) = H(X) + H(Y)$. Here $(X,Y)$ is the random variable taking values $(x,y)$ where $P[x,y] = P[X=x]P[Y=y]$.}

If an experiment has information content $12.34$ bits, then it is reasonable that $10$ experiments have information content $123.4$ bits.

{\bf 2. If $X$ is equal to the variable $Y$ with probability $p$ and to $Z$ with probability $(1-p)$, then $H(X) = pH(Y)+(1-p)H(Z)$}

Again thinking of $X,Y,Z$ as 'experiments', I think the following description is convincing.

Suppose we perform many $X$-experiments, in total $N$. This can be done by first selecting randomly for each experiment whether it should be a $Y$ or a $Z$. If $N$ is large, the number of $Y$:s is close to $pN$. So counting the bits in two ways we have $NH(X) = pNH(Y)+(1-p)NH(Z)$.

{\bf 3. The information content of a fair coin is $1$ bit.}

I'm fine with this statement.

{\bf The formula follows.}

Let $Y$ be a number in $[n]$ chosen uniformly at random, and $(S_i)$ be a partition of $[n]$ into $r$ subsets. We define $X$ to be the unique $x$ such that $Y \in S_x$. Let $n_i = \#S_i$. Since $Y$ is uniform, $H(Y) = \log n$ (at least for power-of-two $n$). The random variable $Y$ is described by the independent(not really) pair of random variables $(X,Z)$ where $Z$ is $Z_i$ with probability $n_i/n$, and $Z_i$ is uniform in $S_i$. Hence $\log n = H(Y) = H(X) + H(Z) = H(X) + \sum_i \frac{n_i}{n}H(Z_i) = H(X) + \sum_i \frac{n_i}{n}\log n_i$, which gives $H(X) = \sum_i \frac{n_i}{n}\log(\frac{n}{n_i})$.

I'm not totally happy about these motivations. My objective was to understand how something can have a non-integer number of 'bits' as 'information content'. It is answered above only as 'it's the result of taking averages of integer numbers of bits'. Also, not all random variables can be described as $X$ above.

{\bf mass below waist}

How does one measure the weight of the part below the waist of a person without killing it?

{\bf attempt 1}: Identify the mass distribution of the person with the function $f : \mbb{R}\to\mbb{R^+}$ with compact support. Here $f(h)dh$ is the mass of the crosscut of height $dh$ at height $h$ above the waist ($h$ may be negative). We are looking for $\int_{-\infty}^0 f(h)dh$ given some statistics $S(f)$ which can be measured without too much mess.

For $a\in\mbb{R}$ let $S_a(f) = \int_{-\infty} ^\infty hf(h+a)dh$. This is the angular momentum when balancing horizontally on a fixed point at $h = a$. Unfortunately, this just tells us the values $A=\int_{-\infty}^\infty hf(h)dh$ and $B = \int_{-\infty}^\infty f(h)dh$, and it is easy to construct two $f$:s with same values for $A$ and $B$ but different $\int_{-\infty}^0f(h)dh$.

{\bf attempt 2}: Measure the force $F$ needed to keep the waist at the water line in a pool. This just tells us what the difference of the total mass and the volume below the waist is (multiplying these by fixed physical constants so that the units are right). Clearly both these numbers are easily determined (independently of each other) in different ways. At least, we can find the mass under the waist if we somehow knew the mean density there. 

{\bf attempt 3}: Let $p_1, \dots, p_n$ be points in space with masses $m_1,\dots,m_n$.
The gravitational field at a point $q$ is $F(q):= \sum_i m_i/d(p_i,q)$.
For $n$ generic $q_i$, the vectors $v_i := (1/d(p_1,q_i), 1/d(p_2,q_i),\dots, 1/d(p_n, q_i))$ are linearly independent(i think) so knowing $F(q_i)$ suffices to determine the $m_i$.
This shows that the problem can be solved using only measurements of acceleration, but it is not very practical (or, more vaguely but more correct: using classical mechanics (no need for eg. thermodynamics)).


{\bf Putting inequalities into context}

The Cauchy-Schwarz inequality is
$$
\sum x_i y_i \leq \left(\sum_i x_i^2 \right)^{1/2} \left(\sum_i y_i^2 \right)^{1/2}.
$$
By rescaling, it suffices to prove it for unit vectors. For unit vectors it follows from the AM-GM inequality: $x_iy_i \leq (x_i^2+y_i^2)/2 \Rightarrow \sum x_i y_i \leq 1$ if $\sum_i x_i^2 + \sum_i y_i ^2 = 1$.


Now, fix nonnegative integers $m,n$ and consider an array $x_{ij}$ ($1\leq i\leq m, 1\leq j \leq n$) of positive reals together with a sequence $p_1,\dots,p_m$ of positive reals with sum $1$. By the $(m,\mbf{p})$-H\"older inequality we mean the statement that for any values $x_{ij}$, we have
$$
\sum_i \prod_j x_{ij} \leq \prod_j \left( \sum_i x_{ij}^{1/p_j} \right)^{p_j}.
$$

Thus, the $(2, 1/2, 1/2)$-H\"older inequality is the Cauchy-Schwarz inequality, and the $(2, 1/p, 1/q)$-H\"older inequality is the classical H\"older inequality.

Note that by applying Cauchy-Schwarz over and over again to $\sum_i \prod_j x_{ij}$, we obtain $(m, \mbf{p})$-H\"older inequalities for all $m,\mbf{p}$ such that the $p_i$ are powers of two (with negative integer exponents).

Now, choose $k$ and let $p = \sum_{j \leq k} p_j$, $q = \sum_{j > k} p_j$, $x_{ij} = u_i^{p_j/p}$ for $j \leq k$ and $x_{ij} = v_i ^{p_j / q}$ for $j > k$. Simplify. This shows that $(m,\mbf{p})$-H\"older implies $(2, p, q)$-H\"older.

The previous two steps and the RHS being continuous in $\mbf{p}$ imply $(2,p,q)$-H\"older for {\it all} $p,q\geq 0$ such that $p+q=1$. Finally $(2,p,q)$-H\"older implies $(2,\mbf{p})$-H\"older for all $\mbf{p}$ by induction.
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