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Balancing a disc

Balancing a circle

Let the triple $(\theta _{i},d_{i},w_{i})$ be the angle, distance and weight of a particle from the centre of the disc.

Let P be some particles on a unit disk such that a given diameter axis $d_{1}$ means that the sum of perpendiculars from P to $d_{1}$ is zero. Let $d_{2}$ be a distinct other axis with the same property.

Prove that all $d$ have this property.

Consider two axis orthogonal to each other and a set of weights with the above property (which can be achieved by calculating the position of the $n^{th}$ weight after the first $n-1$ weights have been placed).

Note that the perpendiculars to a new rotated orthogonal pair of axes from a given weight all lie on the circumference of the circle centred halfway between the origin and the weight, radius half the full distance from O to W.

This applies to each weight, and so there are a set of circles each with a common point on the circumference of the origin and the new axes intersect with these circles at the new distances.

From trigonometric identities, we have:

${\begin{aligned}\sin(\alpha +\beta )&=\sin \alpha \cos \beta +\cos \alpha \sin \beta \\\cos(\alpha +\beta )&=\cos \alpha \cos \beta -\sin \alpha \sin \beta \end{aligned}}$

We know:

${\begin{aligned}\sum w_{i}\sin \alpha _{i}&=0\\\sum w_{i}\cos \alpha _{i}&=0\end{aligned}}$

If $\alpha _{i}$ is the original angle, and $\beta$ is the rotated amount, then we have:

${\begin{aligned}w_{i}\sin(\alpha _{i}+\beta )&=w_{i}\sin \alpha _{i}\cos \beta +w_{i}\cos \alpha _{i}\sin \beta \\w_{i}\cos(\alpha _{i}+\beta )&=w_{i}\cos \alpha _{i}\cos \beta -w_{i}\sin \alpha _{i}\sin \beta \end{aligned}}$

and summing:

${\begin{aligned}\sum w_{i}\sin(\alpha _{i}+\beta )&=\sum w_{i}\sin \alpha _{i}\cos \beta +\sum w_{i}\cos \alpha _{i}\sin \beta \\\sum w_{i}\cos(\alpha _{i}+\beta )&=\sum w_{i}\cos \alpha _{i}\cos \beta -\sum w_{i}\sin \alpha _{i}\sin \beta \end{aligned}}$

${\begin{aligned}\sum w_{i}\sin(\alpha _{i}+\beta )&=\cos \beta \sum w_{i}\sin \alpha _{i}+\sin \beta \sum w_{i}\cos \alpha _{i}=0+0=0\\\sum w_{i}\cos(\alpha _{i}+\beta )&=\cos \beta \sum w_{i}\cos \alpha _{i}-\sin \beta \sum w_{i}\sin \alpha _{i}=0-0=0\end{aligned}}$

The resultant weighted vector at the origin is therefore zero, and this is invariant over axis rotation.

Or, the lines midway between two are balanced, and this process continues ad infinitum.

Convex function proofs

Consider $f(x)=x^{2}$, $x\leq y,0\leq t\leq 1$. Then:

$f(tx+(1-t)y)-tf(x)-(1-t)f(y)$

$=t^{2}x^{2}+2t(1-t)xy+(1-t)^{2}y^{2}-tx^{2}-(1-t)y^{2}$

$=t(t-1)x^{2}+2t(1-t)xy+(1-t)(-t)y^{2}$

$=-t(1-t)\left(x^{2}-2xy+y^{2}\right)$

$=-t(1-t)(x-y)^{2}$

$\leq 0$

---

Consider $f(x)=\log(x)$, $0<x\leq y,0\leq t\leq 1$. Then:

$f(tx+(1-t)y)-tf(x)-(1-t)f(y)$

$=\log(tx+(1-t)y)-(t\log(x)+(1-t)\log(y))$

$=\log(tx+(1-t)y)-\log(x^{t}y^{1-t})$

$=\log({\frac {tx+(1-t)y}{x^{t}y^{1-t}}})$

So need to prove:

${\frac {tx+(1-t)y}{x^{t}y^{1-t}}}\geq 1$

or

$tx+(1-t)y\geq x^{t}y^{1-t}$

or, dividing by $y$:

$t\left({\frac {x}{y}}-1\right)+1\geq \left({\frac {x}{y}}\right)^{t}$

So:

$t\left({\frac {x}{y}}-1\right)+1-\left({\frac {x}{y}}\right)^{t}$

$=\left({\frac {x}{y}}-1\right)\left(t-{\frac {\left({\frac {x}{y}}\right)^{t}-1}{{\frac {x}{y}}-1}}\right)$

$=\left({\frac {x}{y}}-1\right)\left(t-(1+{\frac {x}{y}}+\dots +\left({\frac {x}{y}}\right)^{t-1})\right)$

$\geq 0$

Proof of Euler's Theorem

Euler's Theorem

We have

$a^{p-1}\equiv 1{\pmod {p}}$

Therefore

$(a^{p-1})^{p}=(1+kp)^{p}\equiv 1{\pmod {p^{2}}}$

Also, if $p,q$ are primes, then

$(a^{p-1})^{q-1}=a^{(p-1)(q-1)}\equiv 1{\pmod {q}}$

$(a^{q-1})^{p-1}=a^{(p-1)(q-1)}\equiv 1{\pmod {p}}$

These are the same, and, by the Chinese Remainder Theorem

$a^{(p-1)(q-1)}\equiv 1{\pmod {pq}}$

Catalan q-polynomials

The Catalan q-polynomials $C_{k}(q)$ count the number of blocks present in the diagram under the diagonal height k, and start

• $C_{0}(q)=1$
• $C_{1}(q)=1$
• $C_{2}(q)=1+q$
• $C_{3}(q)=1+q+2q^{2}+q^{3}$
• $C_{4}(q)=1+q+2q^{2}+3q^{3}+3q^{4}+3q^{5}+q^{6}$

The sum of the coefficients of $C_{k}(q)$ give $C_{k}$.

To generate the next level, we add a horizontal and vertical step. The horizontal step is always placed at the origin, and the vertical step can be placed anywhere off the bounding diagonal, and is the first time the path touches the diagonal.

The first path from the convolution is inserted along the (k-1)-th diagonal created by the two endpoints of the new steps, and the second is placed along the k-th diagonal.

With Dyck words, where the new template is $X(k)Y(n-k)$, where $(k)$ is a Dyck word of length k. The X step is always first, and the Y step comes after a k-length Dyck word.

The new polynomial is therefore the heights of the previous polynomials, plus the rectangle created by the insertion.

$C_{n}(q)=\sum _{i=0}^{n-1}q^{i(n-i)}C_{i}(q)C_{n-i}(q)$

For example, $C_{4}(q)=q^{0}(1)(1+q+2q^{2}+q^{3})+q^{3}(1)(1+q)+q^{4}(q+1)(1)+q^{3}(1+q+2q^{2}+q^{3})(1)$.

Chinese Remainder Theorem

If p and q are prime, then the solution to $a\mod p,b\mod q$ is given by

${\frac {aq-bp}{q-p}}=a-{\frac {(b+a)p}{q-p}}=b-{\frac {(b-a)q}{q-p}}$

as the last two expressions are $a\mod p$ and $b\mod q$ respectively.

For example, $3\mod 7,6\mod {11}$ gives ${\frac {33-42}{11-7}}={\frac {-9}{4}}$, and $17\times 4=68\equiv -9\mod 77$

For a practical solution when p and q are coprime, create an array from 0 to p-1. Populate a[kp%q]=kp. Read a[1].

Cubic polynomial

Let

$f(x)=ax^{3}+bx^{2}+cx+d$

The local minima and maxima are given by the zeroes of the differential of $f(x)$.

$f'(x)=3ax^{2}+2bx+c$

$f'(x)=0$

i.e., by the quadratic formula, when

$x={\frac {-2b\pm {\sqrt {4b^{2}-12ac}}}{6a}}={\frac {-b\pm {\sqrt {b^{2}-3ac}}}{3a}}$

Let the two roots be $r_{1}<r_{2}$.

Let w.l.o.g. $f(r_{1})$ is the local maxima and $f(r_{2})$ is the local minima.

$f(r_{1})$ and :$f(r_{2})$ have several terms equal, so

$f(r_{1})-f(r_{2})={\frac {-2a}{27a^{3}}}(3b^{2}{\sqrt {b^{2}-3ac}}+{\sqrt {b^{2}-3ac}}^{3})+{\frac {-2b}{9a^{2}}}(2b{\sqrt {b^{2}-3ac}})^{2}+{\frac {-2c}{3a}}({\sqrt {b^{2}-3ac}})$

which is essentially a quadratic.

We require

$f(r_{1})>0>f(r_{2})$

and if X is the common terms of each side, then this becomes

$f(r_{1})-X>-X>f(r_{2})-X$

Round Robin

6

• 12 34 56
• 13 26 45
• 14 25 36
• 15 23 46
• 16 24 35

8

• 12 34 58 67
• 13 24 57 68
• 14 25 36 78
• 15 26 37 48
• 16 27 38 45
• 17 28 35 46
• 18 23 47 56

GF for cubes

${\frac {1+4x+x^{2}}{(1-x)^{4}}}=1+8x+27x^{2}+64x^{3}+125x^{4}+\dots$

https://www.wolframalpha.com/input?i=%281%2B4x%2Bx%5E2%29%2F%281-x%29%5E4

The general relation is given by the Eulerian numbers.

Binomials

Prove

${\binom {m}{j}}{\binom {n}{k}}\leq {\binom {m+n}{j+k}}$

$={\frac {m!n!}{j!k!(m-j)!(n-k)!}}\leq {\frac {(m+n)!}{(j+k)!(m+n-j-k)!}}$

Consider

${\frac {(m+n)!}{m!n!}}{\frac {j!k!}{(j+k)!}}{\frac {(m-j)!(n-k)!}{(m+n-j-k)!}}$

where

${\frac {(a+b)!}{a!b!}}={\binom {a+b}{a}}={\frac {(a+1)\cdots (a+b)}{b!}}=\prod _{i=1}^{b}\left(1+{\frac {a}{i}}\right)$

so

${\frac {\prod _{i=1}^{m}\left(1+{\frac {n}{i}}\right)}{\prod _{i=1}^{j}\left(1+{\frac {k}{i}}\right)\prod _{i=1}^{m-j}\left(1+{\frac {n-k}{i}}\right)}}$

and

${\frac {\binom {m+n}{m}}{{\binom {j+k}{j}}{\binom {m+n-j-k}{m-j}}}}$

The meaning of logic

TRUTH

0000 (0) FALSE

0110 (6) XOR

1001 (9) NXOR

1111 (F) TRUE

IDEMPOTENT

0001 (1) AND

0011 (3) A

0101 (5) B

0111 (7) OR

INJECTIVE

0100 (4) LT

1100 (C) NB

1101 (D) LTE

1110 (E) NAND

SURJECTIVE

0010 (2) GT

1000 (8) NOR

1011 (B) GTE

1010 (A) NA

Partial sums of binomials

$\sum _{i=0}^{k}{\binom {n-k-1+i}{i}}2^{n-k-i}$

$n=6,k=3:1+6+15+20=42$

${\binom {2}{0}}2^{3}+{\binom {3}{1}}2^{2}+{\binom {4}{2}}2+{\binom {5}{3}}=8+12+12+10=4$2.

Trichotomy

If $x\leq y$ and $y\leq x$ then $x=y$.

Proof: We have $y-x=a,x-y=b$, where both $a$ and $b$ are non-negative.

So $0=a+b$, and therefore $a=b=0$.

Pythagorean and 2 squares

Let $(a,b,c)$ be a Pythagorean triple, with a even.

Then both $c-a$ and $c+a$ are square numbers with roots r and s.

Then $c=\left({\frac {r-s}{2}}\right)^{2}+\left({\frac {r+s}{2}}\right)^{2}$

e.g. (48, 55, 73) ->73-48=25 (5) and 73+48=121 (11)

therefore 73=3^2+8^2

GF for Catalan numbers

Catalan number

Start with the GF for the central binomial coefficient.

${\frac {1}{\sqrt {1-4x}}}=\sum _{k=0}^{\infty }{\binom {2k}{k}}x^{k}$

Integrating and setting the constant to ${\frac {1}{2}}$ from the case $x=0$ yields

${\frac {1-{\sqrt {1-4x}}}{2}}=\sum _{k=0}^{\infty }{\frac {1}{k+1}}{\binom {2k}{k}}x^{k+1}$

Divide by x to get

${\frac {1-{\sqrt {1-4x}}}{2x}}=\sum _{k=0}^{\infty }C_{k}x^{k}$

Primitive roots

Primitive root modulo n

The order of an element is the smallest k where $a^{k}\equiv 1{\pmod {n}}$.

k divides $\phi (n)$. Testing a candidate against each possible k returns negative if and only if the candidate is a primitive root.

a/k	1	2	3	4	5	6
1	1	1	1	1	1	1
2	2	4	1	2	4	1
3	3	2	6	4	5	1
4	4	2	1	4	2	1
5	5	4	6	2	3	1
6	6	1	6	1	6	1

Fermat's theorem on sums of two squares

(5) can be proved by Euler's criterion. It also tells us that if $p=4k+1$,

$p|\prod _{i=1}^{4k}\left(i^{2k}+(i-1)^{2k}\right)$

which is the sum of two squares by (1). (2), (3) and (4) can then be used to find it.

Finding primitive roots for primes

For each prime p of p-1, remove k^p from 1,..,p-1 as k runs over the same range. Only primitive roots remain.

Monty Hall problem

Monty Hall problem

• If Monty reveals a goat *before* the contestant chooses a door, the odds are 50:50.

• In the original version, once the host reveals a goat, the contestant knows with 100% certainty that the other card is opposite to the one they have.

Plane partition GF

Plane partition

The GF is

$\prod _{i=1}^{\infty }{\frac {1}{(1-x^{i})^{i}}}$

Consider instead the much simpler GF of

${\frac {1}{1-x}}\prod _{i=2}^{\infty }{\frac {1}{(1-x^{i})^{2}}}$

The n-th term is the number of partitions of n into at least 0 parts + partitions of n-1 into at least 1 part + partitions of n-2 into at least 2 parts + etc...

Proof

The GF for at least k parts is

$LP_{k}=x^{k}\prod _{i=1}^{k}{\frac {1}{(1-x^{i})}}\prod _{i=1}^{\infty }{\frac {1}{(1-x^{i})}}$

and

$\sum _{k=0}^{\infty }LP_{k}=\prod _{i=1}^{\infty }{\frac {1}{(1-x^{i})^{2}}}$

and we have counted the single x twice.

Coordinates of circumcentre

Let $\triangle ABC$ be coordinates $(x_{1},y_{1}),(x_{2},y_{2}),(x_{3},y_{3})$. Then the midpoints of AB and AC are

$\left({\frac {x_{1}+x_{2}}{2}},{\frac {y_{1}+y_{2}}{2}}\right)$

$\left({\frac {x_{1}+x_{3}}{2}},{\frac {y_{1}+y_{3}}{2}}\right)$

The perpendiculars are

${\binom {y_{2}-y_{1}}{x_{1}-x_{2}}}$

${\binom {y_{3}-y_{1}}{x_{1}-x_{3}}}$

and the equations of the perpendiculars through the midpoints are

${\binom {x}{y}}={\binom {\frac {x_{1}+x_{2}}{2}}{\frac {y_{1}+y_{2}}{2}}}+s{\binom {y_{2}-y_{1}}{x_{1}-x_{2}}}$

${\binom {x}{y}}={\binom {\frac {x_{1}+x_{3}}{2}}{\frac {y_{1}+y_{3}}{2}}}+t{\binom {y_{3}-y_{1}}{x_{1}-x_{3}}}$

with equality when

${\frac {x_{1}+x_{2}}{2}}+s(y_{2}-y_{1})={\frac {x_{1}+x_{3}}{2}}+t(y_{3}-y_{1})$

${\frac {y_{1}+y_{2}}{2}}+s(x_{1}-x_{2})={\frac {y_{1}+y_{3}}{2}}+t(x_{1}-x_{3})$

so

$(x_{2}-x_{3})+2s(y_{2}-y_{1})=2t(y_{3}-y_{1})$

$(y_{2}-y_{3})+2s(x_{1}-x_{2})=2t(x_{1}-x_{3})$

and then

$(x_{2}-x_{3})(x_{1}-x_{2})+2s(x_{1}-x_{2})(y_{2}-y_{1})=2t(x_{1}-x_{2})(y_{3}-y_{1})$

$(y_{2}-y_{3})(y_{2}-y_{1})+2s(x_{1}-x_{2})(y_{2}-y_{1})=2t(x_{1}-x_{3})(y_{2}-y_{1})$

so

$(x_{2}-x_{3})(x_{1}-x_{2})-(y_{2}-y_{3})(y_{2}-y_{1})=2t(x_{1}y_{3}-x_{2}y_{3}+x_{2}y_{1}-x_{1}y_{2}+x_{3}y_{2}-x_{3}y_{1})$

zeta log functions

Namely,

$\prod _{p\in prime}1-\log \left(1-{\frac {1}{p^{s}}}\right)$

$e^{\sum \limits _{p\in prime}{\frac {1}{p^{s}}}}$

Subset complements

Let $B_{i}$ be a collection of sets such that $B_{i}\subset A$ for all $i$.

Let $C=A\cap \bigcap _{i}B_{i}'$.

Then set $C$ is a single component with a discontinuous boundary.

Every laminated map to the sphere contains at least one laminar similar to set $C$.

Hamiltonian cycles

Hamiltonian cycles

Generate the permutations of all vertices. Remove all those with non-adjacent vertices.