We might write this

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

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:

PDF article:

birthday-puzzle-connection.pdf

In an earlier article, I showed how to prove that , 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.

Throughout, and are real functions

and is a real variable.

This says, roughly,

More formally, it says:

If either

or

and also

then

Various indeterminates may be transformed to a or form as shown in the table.

form | formula | conditions | transform | transform |
---|---|---|---|---|

— | ||||

— | ||||

or |
||||

Once transformed to a or form we may apply L’Hôpital’s rule, i.e. differentiate both the numerator and denominator with respect to the limit variable, as shown in the table.

(The conditions of the previous table must hold, but have been omitted here.)

form | formula | apply L’Hôpital for | apply L’Hôpital for |
---|---|---|---|

or |
|||

In my previous post, I showed why

that is, why

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

i.e.

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

The *power mean* of a pair of values is defined by

When this is just

For contrast, we look at some other cases of power means. The analyses in these cases are very simple.

because

because

because

In this case we could manipulate the expression further:

To show that

is not so straightforward.

The definition gives

which is roughly

Thus, we have an indeterminate (or meaningless) form. We have some work to do to find a way to evaluate our limit.

L’Hôpital’s rule states that if

and

and

then

This allows a value to be given to an expression that would otherwise be indeterminate.

Unfortunately, the indeterminate form in L’Hôpital is not the one we are trying to evaluate, i.e. .

This situation can be rescued, as we can transform a power into a quotient (or rather, a function of a quotient)

So

We have that

and

Thus (assuming and , and using L’Hôpital)

But

and

So

which allows us to produce a variation of L’Hôpital’s rule for our indeterminate form:

If

and

and

then

We can apply this to our formula:

where

and

We have

and

Now,

so

and so

So, finally we have our result

For more information, see the Wikipedia article

https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule

For a proof of L’Hôpital’s rule, see for example

Mathematical Analysis

a fundamental and straightforward approach

by David S. G. Stirling

Ellis Horwood Ltd. / Halsted Press / John Wiley & Sons

or other good book on analysis, calculus, or mathematics for engineers or scientists.

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

that is, why

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.

The *power mean* is defined by

When this is just

For contrast, we look at some other cases of power means. The analyses in these cases are very simple.

because

because

because

To show that

is not so straightforward.

The definition gives

which is roughly

Thus, we have an indeterminate (or meaningless) form. We have some work to do to find a way to evaluate our limit.

L’Hôpital’s rule states that if

and

and

then

It is interesting to note that all three preconditions are required. Generally,

thus L’Hôpital’s rule can only be used in the case where it would be useful: to give meaning to the indeterminate form

This allows a value to be given to an expression that would otherwise be indeterminate.

[If the limit

is still the indeterminate form , we are free to try to apply L’Hôpital recursively, and consider further derivations

until (and if) the quotient becomes determinate. However, we will not need to do that here.]

[L’Hôpital also covers the case . Similar comments apply. But we’re only interested in the case here.]

Unfortunately, the indeterminate form in L’Hôpital is not the one we are trying to evaluate, i.e. .

This situation can be rescued, as we can transform a power into a quotient (or rather, a function of a quotient)

So

We have that

and

Thus (assuming and , and using L’Hôpital)

But

and

So

which allows us to produce a variation of L’Hôpital’s rule for our indeterminate form:

If

and

and

then

We can apply this to our formula:

where

and

We have

and

Now,

so

and so

So, finally we have our result

For more information, see the Wikipedia article

https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule

For a proof of L’Hôpital’s rule, see for example

Mathematical Analysis

a fundamental and straightforward approach

by David S. G. Stirling

Ellis Horwood Ltd. / Halsted Press / John Wiley \& Sons

or other good book on analysis, calculus, or mathematics for engineers or scientists.

]]>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 ; for 3-dimensional, which is the regular tetrahedron, the measure (capacity or volume) is ; 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, and . The value is also rational in dimensions 17 and 24.

The sequence of dimensions for such rational measures begins

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

The article includes some tables of calculated values.

= the regular -dimensional hypertetrahedron with edge length

= the measure of

= = the regular -dimensional hypertetrahedron of unit edge

= = the measure of

= orthogonal distance between a -facet and the opposing -facet in

= the value of a particular integral

= = a constant relating the measures of hypertetrahedra

The full article is attached:

hypertetrahedral-slices (v1.0) [initial version]

hypertetrahedral-slices (v1.1) [abstract added; otherwise unchanged]

The simple series

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

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

Sometimes, we also denote this series by or by .

Such series may be expanded in closed form as polynomials.

Given the expression

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

Consider, for example,

When (i.e. one million), we have

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

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

which can quickly be evaluated as

Our example’s coefficients are

As we consider different values for , 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.

The methods covered include

- Binomial Methods
- Binomial Transform Methods
- Convolution Method
- Counting Method
- Differences Methods
- Disturbance Method
- Expand-and-Contract Method
- Factorial Powers Methods
- Faulhaber and Bernoulli’s Formula
- Finite Calculus Method
- Generating Function Methods
- Hypertetrahedral Methods
- Integration Methods
- Interpolation Methods
- Iterative Method
- Matrix Methods
- Perturbation Method
- Relationship Method
- Repertoire Method
- Telescoping Methods

calculus, differentiation, derivation, integration,

binomial coefficients, Stirling numbers, Eulerian numbers, Bernoulli numbers,

power series, exponential generating function, EGF, ordinary generating function, OGF,

error, polynomial

The full article is attached: methods-for-sums-of-powers (version 1.0).

]]>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?

Here, by an n-dimensional hypersphere, I mean the boundary (‘surface’) of an n-dimensional ball.

Beware that, in the mathematical literature, the term ‘n-sphere’ is used to refer to the hypersphere having _intrinsic_ dimension n.

So an ‘n-sphere’ would be the surface of an ‘(n+1)-ball’; thus

a disk or 2-ball has as boundary a circle or 1-sphere,

a ball or 3-ball has as boundary a sphere or 2-sphere,

a 4-ball has as boundary a hyperphere or 3-sphere, and so on….

But in this article, I use the term ‘n-dimensional hypersphere’ (and ‘n-dimensional hyperball’) in a naïve way, with the n referring to an _extrinsic_ notion of dimension, or ‘natural’ dimension of euclidean space in which to embed these shapes, so, here, an n-dimensional hyperball ‘spans’ a region of n-space, and has a hyperspherical boundary that partitions the n-space into two regions: the ‘inside’ and the ‘outside’.

Before now, I have considered (and written about, elsewhere) the size of the spherical gap in a cubical-lattice arrangement of spheres.

A cubical arrangement of eight unit spheres has a gap big enough for a smaller sphere of radius √3-1 ≈ 0.732051⁻ < 1.

Going down to 2D, a square arrangement of four unit circles has a gap big enough for a smaller circle of radius √2-1 ≈ 0.414214⁻ < 1.

But going up to 4D, a tesseract of sixteen unit hyperspheres has a gap big enough for a hypersphere of radius √4-1 = 1, i.e. an equal hypersphere.

In higher dimensions the gap is bigger still.

Generally, the radius for the gap in n-dimensional space is √n-1, and that is unbounded as n increases.

In the process of calculating the gap, the hyperspheres can be placed at the easy and obvious coordinates

〈±1, ±1, …, ±1〉,

and the gap will be centred at the origin

〈0, 0, …, 0〉.

Recently, on the Today Programme on BBC Radio 4, there was a puzzle about the size of the gap within four balls placed in a triangular pyramid, or tetrahedron, so that each ball touches every other.

Puzzle for Today #261 on Wednesday 04 July.

In the very disappointing answer, only the raw answer was given, not the solution.

That was particularly a shame, as there is a particularly neat solution.

First of all, the naïve solution.

Imagine placing the first three balls in a triangle on a plane surface.

These positions may be calculated using some trigonometry.

A triange, centred at the origin, and with side of length 2, has vertices that are a distance 2√3/3 from the origin.

Thus the balls could be placed with centres at

〈cos θ, sin θ, 0〉 × 2√3/3

for θ three angles τ/3 (120°) apart, say

θ = −τ/12, −5τ/12 and +τ/4

giving coordinates of their centres

〈1, −√3/3, 0〉;

〈−1, −√3/3, 0〉;

〈0, 2√3/3, 0〉.

The fourth may be placed ‘atop’ these, with centre at

〈0, 0, 2√6/3〉.

Each pair of these points is exactly 2 apart, thus the unit balls touch.

These may be averaged to find the midpoint

〈0, 0, √6/6〉.

Optionally, if you wish for the gap to be centred at the origin,

then the whole system may be ‘dropped’ by 1/4 of the height 2√6/3 of the underlying tetrahedron, that is by √6/6.

The coordinates are then

〈1, −√3/3, −√6/6〉;

〈−1, −√3/3, −√6/6〉;

〈0, 2√3/3, −√6/6〉;

〈0, 0, √6/2〉.

Again, each pair of these points has distance exactly 2 apart.

Now, each point is exactly √6/2 from the origin.

Well, with or without the ‘drop’, that’s all a bit awkward, with all the diagonals and surds.

An alternative, much neater, solution places the four balls at the more symmetrical positions:

〈+t, +t, +t〉;

〈+t, −t, −t〉;

〈−t, +t, −t〉;

〈−t, −t, +t〉

where t = √2/2, and their centre is

〈0, 0, 0〉.

(Four non-adjacent vertices of a cube form a regular tetrahedron.)

The diagonal distances are then

2 on a face

√6 on a body diagonal

The radius is then

r = √6/2−1 = (√6−2)/2.

Alternatively, place four balls of radius R=√2 at four corners of side 2, and then scale the result.

Calculate remaining size r of small ball

by placing the centres at four corners of the cube:

〈 +1, +1, +1 〉

〈 +1, −1, −1 〉

〈 −1, +1, −1 〉

〈 −1, −1, +1 〉

The diagonal distances are then

2·√2 on a face

2·√3 on a body diagonal

The radii are then

R = √2

r = √3−√2

so

r/R = (√3−√2)/√2 = (√6-2)/2 ≈ 0.225⁻ = 22.5⁻%

Either way, that was much easier to calculate with.

That leads to the generalisation to n-dimensional space. So the question is: is there a similarly simple way to generalise this to find the size of the gap between (n+1) mutually-touching n-dimensional unit hyperspheres in n-dimensional space?

Yes!

But the method is a little different to that described above.

It’s more convenient to think in terms of n (n−1)-dimensional unit hyperspheres in (n−1)-dimensional space. Start by going up a dimension, and placing n n-dimensional hyperspheres in n-dimensional space as follows:

Place hypersphere k at a position √2 from the origin along the k-th axis:

#1 at 〈√2, 0, …, 0〉.

#2 at 〈0, √2, 0, …, 0〉.

…

#k at 〈0, …, 0, √2, 0, …, 0〉.

(k-th position)

…

#n at 〈0, …, 0, √2〉.

It is straightforward to confirm that any pair of these have distance 2 apart, and thus they ‘touch’ (or ‘kiss’).

The points are all on the axes, so this might be referred to as an ‘axial’ configuration.

The centre is

〈d, d, d, …, d〉

where d = √2 / n

The points and the centre all lie in the same diagonal ‘slice’, thus this might also be referred to as the ‘diagonal’ configuration. Any (rigid) rotation of this configuration of points will still be confined to an (n−1)-dimensional sub-space, thus any rotated variant might be referred to as facet-like. (In n-dimensional geometry, an (n−1)-dimensional element of an n-dimensional polytope is called an ‘(n−1)-face’ or ‘facet’.)

Now, take the (n−1)-dimensional ‘slice’ defined by the condition that the sum of the coordinates is equal to √2. That is, you slice through the hyperspace, and through the hyperspheres. And that’s it: you now have n (n−1)-dimensional mutually-touching unit hyperspheres in an (n−1)-dimensional (sub-)space.

As concrete examples, with the axial placements,

For the circles in 2D, we begin by placing spheres in 3D at

#1 at 〈√2, 0, 0〉;

#2 at 〈0, √2, 0〉;

#3 at 〈0, 0, √2〉

and the centre is at

〈√2/3, √2/3, √2/3〉.

For the spheres in 3D, we begin by placing hyperspheres in 4D at

#1 at 〈√2, 0, 0, 0〉;

#2 at 〈0, √2, 0, 0〉;

#3 at 〈0, 0, √2, 0〉;

#4 at 〈0, 0, 0, √2〉

and the centre is at

〈√2/4, √2/4, √2/4, √2/4〉.

It is simple to check that the sum of the coordinates of each of these points is √2.

(In linear algebra terms, the points are not linearly independent.)

The gap is easily calculated as

√(2(n−1)/n) − 1

So for n as the number of hyperspheres (in (n−1)-space):

√(2(n−1)/n) − 1

Or for n as the number of dimensions (in which there are (n+1) hyperspheres):

√(2n/(n+1)) − 1

Looking at particular cases, we have:

2 segments in 1D: gap = √(2·1/2) − 1 = 0

3 circles in 2D: gap = √(2·2/3) − 1 = 2√3/3 − 1 ≈ 0.1547001⁻

4 spheres in 3D: gap = √(2·3/4) − 1 = √6/2 − 1 ≈ 0.224745⁻

Unlike the cubical lattice case, the size of the gap is now bounded by √2-1 ≈ 0.414214⁻.

To translate the plane, and the centre of the gap, to the origin, shift by −d in each coordinate to:

#1 at 〈√2−d, −d, …, −d〉.

#2 at 〈−d, √2−d, −d, …, −d〉.

…

#k at 〈−d, …, −d, √2−d, −d, …, −d〉.

(k-th position)

…

#n at 〈−d, …, −d, √2−d〉.

where (as before) d = √2 / n

The the centre is now at the origin

O = 〈0, 0, 0, …, 0〉

For this translated configuration,

you take the (n−1)-dimensional ‘slice’ defined by the condition that the sum of the coordinates is equal to 0 (zero).

It is simple to check that the sum of the coordinates of each of these points is 0.

Likewise, each of the n points is as the same distance r from the origin where

r = √(2·(1−1/n)) = √(2·(n−1)/n)

We have produced configurations on ‘diagonal’ slices.

Some work must be done to rotate such a slice to be an orthogonal sub-space.

That is not required for gap calculation, which is most-simply undertaken in the diagonal configuration.

Obviously, the transformation to the orthogonal sub-space in not unique – the positions are free to rotate within the sub-space – but we can seek such transformations.

Here, I am using row vectors, thus transformation is made via matrix post-multiplication.

The 3D, centred, axial configuration is:

#1 at 〈2√2/3, −√2/3, −√2/3〉;

#2 at 〈−√2/3, 2√2/3, −√2/3〉;

#3 at 〈−√2/3, −√2/3, 2√2/3〉

The rotation matrix is:

To check that this is a rigid rotation, ensure that any right-handed mutually orthogonal system of three unit vectors is take to another.

That may be done by considering three unit vectors along the axes.

Alternatively, consider the essentially 2-dimensional rotation matrix factors from which the above matrix was composed:

After transformation, we have:

#1 at 〈 1, -√3/3, 0〉;

#2 at 〈 0, 2√3/3, 0〉;

#3 at 〈-1, −√3/3, 0〉.

The 4D, centred, axial configuration is:

#1 at 〈3√2/4, −√2/4, −√2/4, −√2/4〉;

#2 at 〈−√2/4, 3√2/4, −√2/4, −√2/4〉;

#3 at 〈−√2/4, −√2/4, 3√2/4, −√2/4〉;

#4 at 〈−√2/4, −√2/4, −√2/4, 3√2/4〉

The rotation matrix is:

Again, to check this, ensure that any right-handed mutually orthogonal system of three unit vectors is take to another. That may be done by considering three unit vectors along each set of three of the axes.

Again, alternatively, consider the essentially 2-dimensional rotation matrix factors from which the above matrix was composed:

After transformation, we have:

#1 at 〈 √2/2, √2/2, √2/2, 0〉;

#2 at 〈−√2/2, √2/2, −√2/2, 0〉;

#3 at 〈−√2/2, −√2/2, √2/2, 0〉;

#4 at 〈 √2/2, −√2/2, −√2/2, 0〉

That leads to a further question:

Are these values ever rational (∈ℚ) ?

There are certain dimensions in which this happens.

For example:

9 hyperspheres in 8D: gap = √(2·8/9) − 1 = 1/3 = 0.333…

50 hyperspheres in 49D: gap = √(2·49/50) − 1 = 2/5 = 0.4

289 hyperspheres in 288D: gap = √(2·288/289) − 1 = 7/17

There are infinitely many such cases.

]]>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.

The associated table is

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

The full article is available as a PDF:

Platonic_Solid_Orientations_and_the_Platonic_Constants_v1-0.pdf.

There is a well-known formula for the Fibonacci numbers

where

is the golden ratio.

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

We have

and we could consider the general `primary’ root of 1 to be defined using

where

and

and

is the circle constant (so ).

Then we have

and so

Thus we arrive at the formula for :

i.e.

Similarly, we can obtain a formula for a generalisation of the Lucas numbers :

generalising

The following are well-known:

A little algebra can be used to show that:

and

so

Similarly,

If, instead, we generalise to the real-valued function :

Then we can show that:

and

so

i.e.

Similarly, we could define

and similar results follow.

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

In the main article, I show that

This formula is essentially a polynomial of rising factorial powers.

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

and

Less important, though equally simple, is that these formulae form a sequence that continues, essentially:

and so on…

The notation here is outlined below.

(There are also alternative formulae in terms of Stirling set numbers. However, the above formulae in terms of Eulerian numbers are perhaps simpler, and utilise simpler coefficients.)

The sequence of cubes begins

1, 8, 27, 64, 125, 216, 343, 512, 729, 1000,…

The sequence of sums of cubes therefore begins

1, 9, 36, 100, 225, 441, 784, 1296, 2025, 3025, …

And we can go further, and sum those!:

1, 10, 46, 146, 371, 812, 1596, 2892, 4917, 7942,…

What is the 1000th term in each of these sequences?

Calculating these by brute force, we obtain, respectively

1000 000 000,

250 500 250 000,

50 250 416 916 700

What is the 1000 000th term in each of these sequences?

Good luck with brute force.

So, we turn to the formulae:

That is,

But then we also have:

Plugging n=1000 into these formulae we obtain:

1000 000 000,

250 500 250 000,

50 250 416 916 700

Plugging n=1000 000 into these formulae we obtain:

1000 000 000 000 000 000,

250 000 500 000 250 000 000 000,

50 000 250 000 416 666 916 666 700 000

The indirect formulae above may be used to calculate the sums of powers simply. However, if a conventional polynomial (of `ordinary’ rather than rising factorial) powers is needed, then the `indirect’ formulae may be expanded to the following `direct’ formulae:

where

where

and

where

(There are also alternative formulae in terms of Stirling set numbers.)

Hypertetrahedral numbers, or permutations-with-replacement:

The binomial coefficients:

The Eulerian numbers:

The unsigned (or signless) Stirling numbers of the first kind, or Stirling cycle numbers are denoted by:

(Please see the main article for the closed formulae defining these.)

The notation

is a -iterated summation.

(See the main article for the details, and formal definition.)

]]>