Thursday, May 3, 2018

Pentagonal Number Theorem (2)

In yesterday’s entry, I mentioned in connection with the Pentagonal Number Theorem a related Euler formula i.e.

(1 – x)(1 – x2)(1 – x3)(1 – x4) …      =  1 – {x/(1 – x)} + x3/{(1 – x)(1 – x2)}
– x6/{(1 – x)(1 – x2)( 1 – x3)} + …

I then decided to use this in testing this expression by letting x = 1/2, to come up with this interesting result i.e.

1/2 * 3/4 * 7/8 * 15/16  = 1/(1 * 3) – 1(1 * 3 * 7) + 1/(1 * 3 * 7 * 15) + … = .288788…

In fact fully expressed the RHS,

 = 1 – 1/(1)  + 1/(1 * 3) – 1(1 * 3 * 7) + 1/(1 * 3 * 7 * 15) + …

I then realised that this provided just one especially interesting application of a general formula that can be written as follows

(1 – x)(1 – x2)(1 – x3)(1 – x4) … = 1 – 1/{(1/x – 1)} + 1/{(1/x – 1)(1/x2 – 1)} –
1/{(1/x – 1)(1/x2 – 1)(1/x3 – 1)} + 1/{(1/x – 1)(1/x2 – 1)(1/x3 – 1)(1/x4 – 1)} – …

And this formula will converge where x < 1.

So for example, where x = 1/3 we obtain,

2/3 * 8/9 * 26/27 * 80/81 * …  = 1 – 1/(2) + 1/(2 * 8) – 1/(2 * 8 * 26) +
1/(2 * 8 * 26 * 80) – … = .56012 …

I also soon discovered the closely related general formula that can be written as follows

(1 + x)(1 + x2)(1 + x3)(1 + x4) … = 1 + 1/{(1/x – 1)} + 1/{(1/x – 1)(1/x2 – 1)} +
1/{(1/x – 1)(1/x2 – 1)(1/x3 – 1)} + 1/{(1/x – 1)(1/x2 – 1)(1/x3 – 1)(1/x4 – 1)} + …

So once again, illustrating where x = 1/2, we obtain,

3/2 * 5/4 * 9/8 * 17/16 * …   = 1 + 1/(1) + 1/(1 * 3 * 7) + 1/(1 * 3 * 7 * 15) + …

Therefore once again, when for each term on the LHS we decrease the numerator by 2 we obtain on the RHS the alternating version of the same terms with

1/2 * 3/4 * 7/8 * 15/16  = 1/(1 * 3) – 1/(1 * 3 * 7) + 1/(1 * 3 * 7 * 15) + …

There is a fascinating connection here with e and  1/e respectively.


In other work on this blog I have shown how we can associate a unique infinite sequence of terms with every regular polynomial equation.

Perhaps the best known sequence is the Fibonacci which arises from the equation
x2 – x – 1.
Then starting with 0, 1 we multiply the coefficient of the x term (i.e. – 1) by – 1 then multiply by the 2nd of the starting digits (i.e. 1 and then likewise multiply the coefficient of the 3rd term (i.e. – 1) again by – 1 before multiplying the result by the 1st of the starting digits (i.e. 0). Then combining the 2 answers we get 1 + 0 = 1, which is then the next digit in the Fibonacci sequence.
And by concentrating on the two most recent digits generated we can continue to generate subsequent terms in the sequence.

However the same general procedure can in principle be applied to all polynomial equations (though it becomes more cumbersome for higher degree equations).

So the Fibonacci sequence is 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, …

And the significance of the terms in the sequence is that they can be used dividing the
(n + 1)th by the nth to approximate the real roots for the associated equation x2 – x – 1 = 0.

So for example 233/144 = 1.618055 already provides a good approximation of the positive real root (i.e. phi = 1.618033…).

Now if we consider (x – 1)( x – 1) we obtain the polynomial equation x2 – 2x + 1
And using the same procedure for generating subsequent terms (as for the Fibonacci sequence) we obtain 0, 1, 2, 3, 4, 5, …

In other words, the unique digit sequence associated with (x – 1)(x – 1) = (x – 1)2 
is the natural number sequence.

And of course we can use this sequence corresponding to x2 – 2x + 1 = 0, to approximate the positive real root (i.e. 1) of the equation.

So as the nth term = the (n + 1)th/n th term = (n + 1)/n ~ 1..

Now the well known expression for e is as follows

e = 1 + 1/(1) + 1/(1 * 2) + 1/(1 * 2 * 3) + 1/(1 * 2 * 3 * 4) + …

1/e = 1 –  1/(1) + 1/(1 * 2) + 1/(1 * 2 * 3) –  1/(1 * 2 * 3 * 4) + …

So the factorial expressions in the denominator entail the successive products of the natural numbers.

Let us now look at the unique digit sequence associated with (x – 1)(x – 2) = x2 – 3x + 2
which is 0, 1, 3, 7, 15, 31, …

And once again the (n + 1)th/n th term e.g. 31/15, will approximate the principle real valued root of the associated equation i.e. x2 – 3x + 2 = 0, = 2.

If we look again at the two related expressions for 3/2 * 5/4 * 9/8 * 17/16 * …   and 1/2 * 3/4 * 7/8 * 15/16 , i.e.

1 + 1/(1) + 1(1 * 3 * 7) + 1/(1 * 3 * 7 * 15) + …  and

1 – 1/ (1) + 1/(1 * 3) – 1(1 * 3 * 7) + 1/(1 * 3 * 7 * 15) + …,

we can see how they readily match the corresponding expressions for e and 1/e respectively.
Whereas in the expressions for e and (i/e) the successive products of  the unique digit sequence associated with (x – 1)(x – 1) are used, in the latter case in the corresponding expressions for 3/2 * 5/4 * 9/8 * 17/16 * …   and 1/2 * 3/4 * 7/8 * 15/16,  the successive products of the unique digit sequence associated with (x – 1)(x – 2) are now used.

So in our two related formulae,

(1 + x)(1 + x2)(1 + x3)(1 + x4) … = 1 + 1/{(1/x – 1)} + 1/{(1/x – 1)(1/x2 – 1)} +
1/{(1/x – 1)(1/x2 – 1)(1/x3 – 1)} + 1/{(1/x – 1)(1/x2 – 1)(1/x3 – 1)(1/x4 – 1)} + …

and

(1 – x)(1 – x2)(1 – x3)(1 – x4) … = 1 – 1/{(1/x – 1)} + 1/{(1/x – 1)(1/x2 – 1)} –
1/{(1/x – 1)(1/x2 – 1)(1/x3 – 1)} + 1/{(1/x – 1)(1/x2 – 1)(1/x3 – 1)(1/x4 – 1)} – …,

1/x = 2.

And the numbers in the product denominator sequence for the RHS of the expression correspond here with the unique numbers associated with (x – 1)(x – 2).

And where 1/x = n (as integer), the numbers in the product denominator sequence for the RHS of the expression correspond to (x – 1)(x – n). Strictly each number in the product sequence relates to the corresponding unique numbers - associated with (x – 1)(x – n) - multiplied by 1/x –  1. So, as we have seen, when 1/x = 2 the unique number sequence associated with (x – 1)(x – 2) directly applies!

For example when 1/x = 3, we now use the unique digits associated with (x – 1)(x – 3) = x2 – 4x + 3, which are,

0, 1, 4, 13, 40, 121, …

Therefore the appropriate denominator sequence applying entails each of these terms * 2
i.e. 2, 8, 26, 80, 242, …

And as we have already seen when 1/x = 3,

2/3 * 8/9 * 26/27 * 80/81 * …  = 1 – 1/(2) + 1/(2 * 8) – 1/(2 * 8 * 26) +
1/(2 * 8 * 26 * 80) – … = .56012 …

So likewise,

4/3 * 10/9 * 28/27 * 82/81 * …  = 1 + 1/(2) + 1/(2 * 8) + 1/(2 * 8 * 26) +
1/(2 * 8 * 26 * 80) + … = 1.56493 …

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