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Balancing a disc[edit]

Balancing a circle

Let the triple 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 means that the sum of perpendiculars from P to is zero. Let be a distinct other axis with the same property.

Prove that all 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 weight after the first 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:

We know:

If is the original angle, and is the rotated amount, then we have:

and summing:

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[edit]

Consider , . Then:


Consider , . Then:

So need to prove:

or

or, dividing by :

So:

Catalan q-polynomials[edit]

The Catalan q-polynomials count the number of blocks present in the diagram under the diagonal height k, and start

The sum of the coefficients of give .

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 , where 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.

For example, .

Cubic polynomial[edit]

Let

The local minima and maxima are given by the zeroes of the differential of .

i.e., by the quadratic formula, when

Let the two roots be .

Let w.l.o.g. is the local maxima and is the local minima.

and : have several terms equal, so

which is essentially a quadratic.

We require

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

Round Robin[edit]

6[edit]

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

8[edit]

  • 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[edit]

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[edit]

Prove

Consider

where

so

and

The meaning of logic[edit]

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[edit]

2.

Pythagorean and 2 squares[edit]

Let be a Pythagorean triple, with a even.

Then both and are square numbers with roots r and s.

Then

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

therefore 73=3^2+8^2

GF for Catalan numbers[edit]

Catalan number

Start with the GF for the central binomial coefficient.

Integrating and setting the constant to from the case yields

Divide by x to get

Primitive roots[edit]

Primitive root modulo n

The order of an element is the smallest k where .

k divides . 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 ,

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[edit]

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.

Plane partition GF[edit]

Plane partition

The GF is

Consider instead the much simpler GF of

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[edit]

The GF for at least k parts is

and

and we have counted the single x twice.

Coordinates of circumcentre[edit]

Let be coordinates . Then the midpoints of AB and AC are

The perpendiculars are

and the equations of the perpendiculars through the midpoints are

with equality when

so

and then

so

zeta log functions[edit]

Namely,

Hamiltonian cycles[edit]

Hamiltonian cycles

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

Binary divide[edit]

Given n and d, left-shift d until it is greater than n, and then single right shift. Subtract this value from n and repeat until remainder is less than d.

Base conversion[edit]

An integer in base b can be written as

and the same integer in base d is

If , then the first equation becomes

Expand this using the binomial formula to get

and collect by like terms in d to get the second equation.