On a Connection between the Birthday Puzzles

Sunday, 5 February 2023

The first birthday puzzle asks: How many people need to be gathered together in order that there is an at-least evens chance of a birthday coincidence? The answer is just 23.

We might write this

\displaystyle B_1(365, 0.5) = 23

The second birthday puzzle asks: How many other people need to be gathered together in order that there is an at-least evens chance of a coincidence with your birthday? The answer is 253.

Notice that

\displaystyle \binom{23}{2} = 253

It turns out that this is not a coincidence.

The value of the first birthday puzzle is awkward to calculate. In the attached article, I show how the above binomial connection leads to a good closed-form approximation for the first birthday problem:

\displaystyle \widehat{B_1}(n, p) = \frac{\sqrt{8 \frac{\ln (1 - p)}{\ln \left( 1 - \frac{1}{n} \right)} + 1} + 1}{2}

PDF article:
birthday-puzzle-connection.pdf

Advertisement

Indeterminate Forms and L’Hôpital’s Rule

Wednesday, 8 September 2021

Introduction

In an earlier article, I showed how to prove that P_0=G, i.e. the zero power mean is the geometric mean, using L’Hôpital’s rule. It was also necessary to perform some transformations on a limit expression to allow L’Hôpital’s rule to be applied.

In this article, I show how this and various other common indeterminate forms may be transformed so that L’Hôpital’s rule may similarly be applied.

I also provide a table of the results of the derivations (or ‘differentiations’) for each case, as a useful reference.

Read the rest of this entry »

P₀(x,y) = G(x,y)

Saturday, 4 September 2021

Introduction

In my previous post, I showed why

\displaystyle P_0(X) = G(X)

that is, why

\displaystyle \lim_{k \to 0} \left( \frac{ \sum_{x \in X} x^k }{|X|} \right)^{\frac{1}{k}} = { \left( \, \prod_{x \in X} x \right) }^{\frac{1}{|X|}} = \sqrt[|X|]{ \prod_{x \in X} x }

Here, I repeat the exercise for the simpler case where X has just 2 members. I.e.

\displaystyle P_0(x,y) = G(x,y)

i.e.

\displaystyle \lim_{k \to 0} \left( \frac{ x^k + y^k }{2} \right)^{\frac{1}{k}} = { \left( x y \right) }^{\frac{1}{2}} = \sqrt{ x y }

Therefore this is just a special case of what was discussed in the previous post, but which may be somewhat easier to follow.
Read the rest of this entry »

P₀ = G

Friday, 3 September 2021

Introduction

Recently, I wished to understand why the zero-power mean is the geometric mean

\displaystyle P_0 = G

that is, why

\displaystyle \lim_{k \to 0} \left( \frac{ \sum_{x \in X} x^k }{|X|} \right)^{\frac{1}{k}} = { \left( \, \prod_{x \in X} x \right) }^{\frac{1}{|X|}} = \sqrt[|X|]{ \prod_{x \in X} x }

and this sent me on much more of a mathematical adventure than I expected, involving learning about evaluating limits, L’Hôpital’s rule, and about indeterminate forms and their transformations.
Read the rest of this entry »

On the Geometry of Hypertetrahedral Slices

Friday, 27 August 2021

The full article is attached as a PDF; please see the link at the end of this post.

Abstract

In this article, I investigate the (regular) hypertetrahedra and their measures, and orthogonal slices (or sections) through them.

I consider how to configure the vertices in order easily to calculate the measure.

Formulae are found for the distance between opposing facets of a hypertetrahedron, and I investigate the behaviour of these as the dimension increases.

I discuss the “shape” of slices through hypertetrahedra, showing these to be Cartesian products of hypertetrahedra of lower dimension. Then I briefly investigate slices through slices.

Formulae are found for the measure of a hypertetrahedron as an integral over (measures of) slices. From these, I find recurrence formulae, and from these, in turn, I find a closed form for the measure of a hypertetrahedron.

For example:
the measure (length) of the 1-dimensional hypertetrahedron of unit edge, which is just the unit line segment, is 1; for 2-dimensional, which is just the equilateral triangle, the measure (area) is \sqrt{3}/4; for 3-dimensional, which is the regular tetrahedron, the measure (capacity or volume) is \sqrt{2}/12; and so on.

I determine when (i.e. for which dimensions) the measure of a hypertetrahedron is rational, and show that when it is rational, it is in fact a unit fraction.

For example:
the measures of the 7- and 8-dimensional hypertetrahedra of unit edge are, respectively, 1/20160 and 1/215040. The value is also rational in dimensions 17 and 24.

The sequence of dimensions for such rational measures begins

\displaystyle 0, 1, 7, 8, 17, 24, 31, 48, 49, 71, 80, 97, \dots

Finally, I briefly investigate the behaviour of the discrete analogue of hypertetrahedra, i.e. hypertetrahedral numbers.

The article includes some tables of calculated values.
Read the rest of this entry »

A Survey of Methods for the Evaluation of Sums of Powers

Saturday, 21 August 2021

The full article (over 100 pages) is attached as a PDF; please see the link at the end of this post.

Abstract

The simple series

\displaystyle \sum_{m=1}^{n} m^k ,

where k and n are non-negative integers, is a sum of powers

\displaystyle 1^k + 2^k + 3^k + \cdots + (n-1)^k + n^k

We regard k as being fixed and relatively small, and n as being variable and potentially very large. That is, n \gg k.

Sometimes, we also denote this series by s_k(n) or by \sum n^k.

Such series may be expanded in closed form as polynomials.

Given the expression

\displaystyle \sum n^k

for some chosen k, we wish to re-express it as a polynomial

\displaystyle a_{k+1} n^{k+1} + a_k n^k + a_{k-1} n^{k-1} + \cdots + a_2 n^2 + a_1 n + a_0

Consider, for example,

\displaystyle s_5(n) = \sum_{m=1}^{n} m^5 = 1^5 + 2^5 + 3^5 + \cdots + (n-1)^5 + n^5

When n=1 000 000 (i.e. one million), we have

\displaystyle s_5(1 000 000) = 1^5 + 2^5 + 3^5 + \cdots + 999 999^5 + 1 000 000^5

This is fairly tedious to evaluate. But we can use any of various techniques to establish that

\displaystyle s_5(n) = \frac{1}{6} n^6 + \frac{1}{2} n^5 + \frac{5}{12} n^4 - \frac{1}{12} n^2

Now, rather than having 1 000 000 terms to sum, we have just 4 terms. So

\displaystyle s_5(1 000 000) = \frac{1}{6} \, 1 000 000^6 + \frac{1}{2} \, 1 000 000^5 + \frac{5}{12} \, 1 000 000^4 - \frac{1}{12} \, 1 000 000^2

which can quickly be evaluated as

\displaystyle s_5(1 000 000) = 166 667 166 667 083 333 333 333 250 000 000 000

Our example’s coefficients are

\displaystyle \langle a_j \rangle = \left< 0, 0, -\dfrac{1}{12}, 0, +\dfrac{5}{12}, +\dfrac{1}{2}, +\dfrac{1}{6} \right>

As we consider different values for k, the coefficients of these polynomials exhibit some remarkably and unexpectedly complicated behaviour. They seem rather erratic, following no obvious pattern. Likewise, the factors exhibit no obvious pattern, and indeed are not even always real.

Mathematicians have been studying these series for hundreds of years. There are many approaches; some old, and some very recent. The approaches are quite varied. In this article, I present the many approaches I have found in the literature and online.

Read the rest of this entry »

Packing for Hyperspace

Tuesday, 17 July 2018

Overview

Given 3 mutually touching unit circles (or disks) in a (2D) plane, what is the largest circle that will fit in the gap?

Given 4 mutually touching unit spheres (or balls) in (3D) space, what is the largest sphere that will fit in the gap?

And, generally, given (n+1) mutually touching unit n-dimensional hyperspheres (or hyperballs) in a n-dimensional space, what is the largest hypersphere that will fit in the gap?

Read the rest of this entry »

Platonic Solid Orientations and Constants

Tuesday, 10 July 2018

This is an article I’ve been thinking of writing for a few years.

The preparation and writing itself has taken many tens of hours over several months.

There is nothing deep in this article; it is primarily a matter of tedious calculation, arithmetic and algebra. However, this is a reference that I wanted to have available. I have not found this information elsewhere, and so I needed to write it myself.

Here I provide explicit arithmetical (closed-form) values for the coordinates of the vertices of the five platonic solids, i.e. the convex regular polyhedra, in various simple ‘standard’ orientations, including face-ward, edge-ward, and vertex-ward. Each of these orientations is illustrated.

The article consists mainly of tables and figures.

As an example of the figures, here is the orthogonal face-ward orientation of the dodecahedron.

dodeca_face_ortho

The associated table is

para-face-dodeca-table

and then there’s a separate table giving the exact values of the c_k constants, such as

c_{15} = \dfrac{1}{2} \left( 3 - \sqrt{5} \right)

The full article is available as a PDF:
Platonic_Solid_Orientations_and_the_Platonic_Constants_v1-0.pdf.

Fractional Fibonacci Numbers

Thursday, 7 June 2018

Traditional Formula

There is a well-known formula for the Fibonacci numbers

\displaystyle F_n = \dfrac{\sqrt{5}}{5} \left( \varphi^n - (-\varphi)^{-n} \right)

where

\displaystyle \varphi = \dfrac{\sqrt{5} + 1}{2} \approx 1.618^{+}

is the golden ratio.

It turns out that there is a way to find F_x for when x is not an integer, but the values are complex rather than real.

Read the rest of this entry »

“New Approach to Sums of Powers” — Headlines and Examples

Thursday, 7 June 2018

As the article on sums of powers was rather long and dense, I thought that it would be worth giving a summary of the main results separately.

I will also show the formulae in action with a worked example.

Indirect, Simple Formulae

In the main article, I show that

\displaystyle {\bigoplus_{m=1}^{n}}{}^{(t)} \; m^k = \sum_{j=0}^{k} \left< \begin{array}{c} k \\ j \\ \end{array} \right> \triangle_{k+t}(n-j)

This formula is essentially a polynomial of rising factorial powers.

Special Cases

Perhaps the most important and useful formulae from the main article are

\displaystyle n^k = \sum_{j=0}^{k} \left< \begin{array}{c} k \\ j \\ \end{array} \right> \triangle_{k}(n-j)

and

\displaystyle \sum_{m=1}^{n} m^k = \sum_{j=0}^{k} \left< \begin{array}{c} k \\ j \\ \end{array} \right> \triangle_{k+1}(n-j)

Read the rest of this entry »