Given a bowl with $n$ noodles. Suppose we pick two free ends of noodles at random, and join them together. Repeat this until there are no more free ends. What is the expected number of loops of noodles that we get?
Saturday, December 25, 2021
Probability of a broken stick in three pieces form a triangle.
Consider the unit interval $[0,1]$. Pick two points $A,B$ uniformly and independently in $[0,1]$. This gives three segments, $a, b, c$, formed by chopping the unit interval at $A,B$. What is the probability that the segments $a,b,c$ can form a triangle?
Chebyshev polynomials and commuting chains.
The Chebyshev polynomials is a family of polynomials $T_n(x) \in {\bf Z}[x]$
such that $$ \cos(nx) = T_n(\cos(x)) $$ with ${\text deg} T_n = n$, for each
positive integer $n$. In particular we have $$ T_1(x) = x\\ T_2(x) = 2x^2 - 2\\
T_3(x) = 4x^3 -3x \\ \vdots $$ Now an observation is that this Chebyshev family of polynomials ${\mathscr T}=\{T_n\}$ forms a commuting family of polynomials under function composition. Namely
$T_n\circ T_m = T_m \circ T_n$. In fact we have $T_n\circ T_m = T_{nm}$.
There
are other commuting families. Say $\{x + 1, x+ 2, x+3\}$, or ${\mathscr P}=\{x, x^2, x^3, x^4, \ldots\}$. Now the families $\mathscr T$ and $\mathscr P$ are called chains, where there is exactly one polynomial in each family per positive degree. As it turns out, $\mathscr T$ and $\mathscr P$ are essentially the only two commuting chains, up to similarity.
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