Saturday, September 11, 2021

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 6th 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 1st term of successive n-gonal series keep adding 0 to the next term (starting with 1).

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


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

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

 

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

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

 

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

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

 

Now the terms that we keep successively adding on to generate the kth 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 5th term of each polygonal series, keep adding 10 to the 5th term (starting with 5).

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

 

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

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

 

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

So the 0th 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,

(ak/0! + ak + 1/1! + ak + 2/2! + …)/e = ak+ 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 ak = 15, ak + 1 = 7

 

Then

(ak/0! + ak + 1/1! + ak + 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

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