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.