Polygonal Numbers and e (2)

The general formula for the nth term of an n-gonal number series is given by,

{(n  – 2)k² – (n – 4)k}/2

So therefore if for example one wishes to calculate the 6^th term of the octagonal number series, 

then n = 8 and k = 6,

Thus {(n  – 2)k² – (n – 4)k}/2

= {(6 * 6²) – (4.6)}/2 = (216 – 24)/2 = 192/2 = 96

An easy empirical way of obtaining all terms of polygonal series is given in the following manner

Let us start with the 2-gonal natural number series

1, 2, 3, 4, 5, 6, …

Then to get the 1^st term of successive n-gonal series keep adding 0 to the next term (starting with 1).

So the 1^st term of all n-gonal series = 1

To get the 2^nd term, keep adding 1 to the next term (starting with 2).

So the 2^nd term of the 3-gonal (triangular) series is 3, the 2^nd term of the 4-gonal (square) series is 4, the 2^nd term of the 5-gonal (pentagonal) series is 5 and so on.

 

To get the 3^rd term, keep adding 3 to the 3^rd term (starting with 3)

So the 3^rd term of the 3-gonal series is 6, the 3^rd term of the 4-gonal is 9, the 3^rd term of the 5-gonal, 12 and so on.

 

Then to get the 4^th term, keep adding 6 to the 4^th term (starting with 4).

So the 4^th term of the 3-gonal series is 10, the 4^th term of the 4-gonal is 16, the 4^th term of the 5-gonal, 22 and so on.

 

Now the terms that we keep successively adding on to generate the k^th term of each sequence are the successive terms of the 3-gonal (triangular) series i.e.

0, 1, 3, 6, 10, 15, …

 

To get the 5^th term of each polygonal series, keep adding 10 to the 5^th term (starting with 5).

Therefore the 5^th term of the 3-gonal series is 15, the 5^th term of the 4-gonal series 25, the 5^th term of the 5-gonal, 35 and so on.

 

To obtain the 6^th term keep adding 15 to the 6^th term (starting with 6).

Therefore the 6^th term of the 3-gonal series is 21, the 6^th term of the 4-gonal is 36, the 6^th term of the 5-gonal, 51 and so on.

 

In all the polygon series so far the k^th term has been given a positive value. However k equally can take on the value 0 and negative values – 1, – 2, – 3, …

So the 0^th term of all polygonal series = 0.

 

Then if take the pentagonal series to illustrate (where n = 5) we can derive successive terms where k =  – 1, – 2, – 3, …

So when k = – 1, {(n  – 2)k² – (n – 4)k}/2

=  (3 * (– 1)² – (– 1}/2 = 2

 

When k = – 2, {(n  – 2)k² – (n – 4)k}/2

= {(3 * (– 2)² – (– 2}/2 = 7

 

When k = – 3, {(n  – 2)k² – (n – 4)k}/2

= {(3 * (– 3)² – (– 3}/2 = 12

 

In this way we generate a new shadow pentagonal series

2, 7, 15, 26, 40, 57, 77, 100, …

 

And the same general formula applies as given in the previous entry,

(a_k/0! + a_{k + 1}/1! + a_{k + 2}/2! + …)/e = a_k+ 1 + n/2 – 1

 

Therefore in this case when k = 1,

(a₁/0! + a₂/1! + a₃/2! + …)/e = a₂ + n/2 – 1

So (2/0! + 7/1! + 15/2! + …)/e = 7 + 2.5 – 1 = 8.5

 

Then the generalised pentagonal number series, with numbers written in ascending order, containing the original sequence (for positive values of k and the shadow sequence (for negative values) is

1, 2, 5, 7, 12, 15, 22, 26, 35, 40, 51, 57, 70, 92, 100, ..,

Euler was able to use this sequence to great effect is establishing a link to the divisors of numbers and to partitions of numbers.

 

However for the formula to e to work these two sequences must be joined separately with sequences unfolding in opposite directions (with 0 in the middle common to both).

i.e.  … 15, 7, 2, 0, 1, 5, 12, 22, …

So for example with respect to this new conjoined sequence if a_k = 15, a_{k + 1} = 7

 

Then

(a_k/0! + a_{k + 1}/1! + a_{k + 2}/2! + …)/e =

= 15/0! + 7/1! + 2/2! + 0/3! + 1/4! + 5/5! + 12/6! + 22/7! + … = 7 + 2.5 – 1 = 8.5

Polygonal Numbers and e (1)

A polygonal number is a number that can be expressed as dots in the shape of a regular polygonal. So we for example the well known triangular numbers,

1, 3, 6, 10, 15, …

For simplicity we can express this series as 3-gonal (where n = 3).

 

Then we have the square numbers

1, 4, 9, 16, 25, …,

which we can express as 4-gonal (where n = 4)

 

And we have the pentagonal numbers

1, 5, 12, 22, 35, …,

which we can express as 5-gonal (where n = 5)

 

In fact the earlier series,

1, 2, 3, 4, 5, …,

can be expressed as 2-gonal (where n = 2)

And we can have continue without limit to 6-gonal 7-gonal, 8-gonal … n-gonal series.

For example the 8-gonal series (where n = 8) is that of the well known octagonal numbers,

1, 8, 21, 40, 65, …

 

Then in general for all polygonal series,

(a₁/0! + a₂/1! + a₃/2! + …)/e = a_{2 +}  n/2 – 1

where a_1, the 1^st term = 1.

 

So, for example with the simplest 2-gonal series

(1/0! + 2/1! + 3/2! + 4/3! + 5/4! + …)/e   = 2 + 1 – 1 = 2

 

And then with the 3-gonal (Triangular) series

(1/0! + 3/1! + 6/2! + 10/3! + 15/4! + …)/e   = 3 + 1.5 – 1 = 3.5

 

Then more generally

(a_k/0! + a_{k + 1}/1! + a_{k + 2}/2! + …)/e = a_{k+ 1} + n/2 – 1

 

So with the 3-gonal (Triangular) series,     

1, 3, 6, 10, 15, …, where n = 3,

if k = 4, then a_{k + 1} = 15 (i.e. the 5^th term in the series).

Therefore,

10/0! + 15/1! + 21/2! + 28/3! + …)/e  = 15 + 1.5 – 1 = 15.5

 

For the 8-gonal (Octagonal)  1, 8, 21, 40, 65, 96, …   i.e. n = 8

If k = 1, then a_{k + 1} = 8

Therefore,

1/0! + 8/1! + 21/2! + 40/3! + …)/e  = 8 + 4 – 1 = 11

 

Then if k = 3, a_{k + 1} = 40 (i.e. the 4^th term)

Therefore,

21/0! + 40/1! + 65/2! + 96/3! + …)/e  = 40 + 4 – 1 = 43

 

What is remarkable from all of this is that e can therefore be computed from any polygonal i.e. n-gonal series starting with the 1^st term, or indeed the k^th term (where k = 1, 2, 3, …) using the formula   

e = (a_k/0! + a_{k + 1}/1! + a_{k + 2}/2! + …)/(a_{k+ 1} + n/2 – 1)

 

For example when n = 24 we get what are called the icositetragonal numbers,

1, 24, 69, 136, 225, 336, 469, 624, 801, 1000, …

So if, say,  k = 3, then

e = (69/0! + 136/1! + 225/2! + 336/3! + 469/4! + 624/5! + 801/6! + 1000/7!…)/(136 + 12 – 1)

= (69/0! + 136/1! + 225/2! + 336/3! + 469/4! + 624/5! + 801/6! + 1000/7!…)/147

= 2.718044…

So using just the 1st 8 terms, we already can compute the value of e accurate to 3 decimal places.  

Generalised Formulae for e

e = 1 + 1/1! + 1/2! + 1/3! + 1/4! + …,

is a special case of

e = {1 + (1 + k)/1! + (1 + 2k)/2! + (1 + 3k)/3! + (1+ 4k)/4! + …}/(1 + k),

where k = 0.

So, when k = 1,

e = {1 + 2/1! + 3/2! + 4/3! + 5/4! + …}/2

And when for example k = 10,

e = {1 + 11/1! + 21/2! + 31/3! + 41/4! + …}/11

This relationship holds for all real values of k (including non-integers) except – 1.

 

The case of k = – 1 is fascinating as it suggests that,

e ={(1 + 0/1! – 1/2! – 2/3! – 3/4! – …)}/0

And the limiting value of the expression inside the brackets as the number of terms

n ⁓ ∞ = 0.

So in this case the suggested limiting value of 0/0 = e?

 

Then,

e^x = x⁰/0! + x¹/1! + x²/2! + x³/3! + x⁴/4! + …

So when x = 1,

e = 1 + 1/1! + 1/2! + 1/3! + 1/4! + …

 

And when x = 2,

e² = 2⁰/0! + 2¹/1! + 2²/2! + 2³/3! + 2⁴/4! + …

 

Therefore, combining the two formulae,

e = 1 + 1/1! + 1/2! + 1/3! + 1/4! + …,

is a special case of,

e^x(1 + 2k) = x⁰/0! + x¹(1 + k)/1! + x²(1 + 2k)/2! + x³(1 + 3k)/3! + x⁴(1 + 4k)/4! + …

for all x where x = 1 and k = 0.

Then when k = 0,

e ^x = x⁰/0! + x¹/1! + x²/2! + x³/3! + x⁴/4! + …

And when also x = 1,

e = 1⁰ + 1¹/1! + 1²/2! + 1³/3! + 1⁴/4! + …

However, if for example k = 2 and x = 2,

Then 5e² = 1⁰/0! + (3*2¹)/1! + (5*2²)/2! + (7*2³)/3! + (9*2⁴)//4! + …

e ² = {1⁰/0! + (3*2¹)/1! + (5*2²)/2! + (7*2³)/3! + (9*2⁴)//4! + …}/5

so that,

e = [{1⁰/0! + (3*2¹)/1! + (5*2²)/2! + (7*2³)/3! + (9*2⁴)//4! + …}/5]^1/2