Thursday, July 5, 2012

Remarkable Relationships (3)

In an earlier posting "Interesting Prime Result", I showed how the (finite) Zeta 2 equation could be used to detect whether a number is prime.

So once the Zeta 2 Equation is defined as:

s^1 + s^2 + s^3 + s^4 +.......s^n = 0

Then dividing by the trivial solution i.e. s = 0 we obtain

1 + s^1 + s^2 + ....... s^(n - 1) = 0.

Nest by letting s = 1 in the expression

y = 1 + s^1 + s^2 + ....... s^(n - 1), by a process of continued differentiation (with respect to s) we showed how to determine whether a number is prime.

However we could also seek to proceed in a complementary direction through obtaining the integral of the same simple expression i.e.

y = 1 + s^1 + s^2 + ....... s^(n - 1).

So ʃy ds = s + (s^2)/2 + (s^3)/3 + (s^4)/4 + .... + (s^n)/n.

Then setting s = 1, we obtain the first n terms in the harmonic series i.e.

1 + 1/2 + 1/3 + 1/4 +..... + 1/n.

Now as n becomes very large the sum of this series approximates very close to log n (which measures the average spread as between prime numbers in the region of n).

For example when n = 1,000,000 the sum of the first n terms = 14.384.

So this provides an approximate measurement of the average spread (or gap) as between prime numbers (in the region of 1,000,000).
Once again this approximation will steadily improve (in relative terms) as n increases.

Thus it is interesting how a simple process of differentiation on this simple (Zeta 2) expression can determine on the one hand whether a number is prime, while the corresponding process of integration can establish the nature of the general distribution among the primes!

If we just concentrate on the first terms of the simple expression we get 1.

Then if we successively integrate with respect to s we obtain s/1!, s^2/2!, s^3/3!,
s^4/4! and so on.

Therefore by adding all these terms we obtain a simple formula for e^s (containing the first n terms of the corresponding infinite expression for e^s).

Of course where s = 1, we approximate the value of e, which becomes ever more accurate as n increases.

Returning to the roots of 1 we showed again how the average value of these roots where n is prime approximates 4/π (especially with respect to the arithmetic mean).

And this approximation steadily improves as the value of n increases.

Once again it is important to bear in mind that we use a reduced linear (1-dimensional) quantitative approach in calculating such values. This means that negative values are treated as positive and imaginary values are treated as real!

However it would also be possible to calculate values using a reduced 2-dimensional quantitative approach.

This entails in effect that negative (as well as positive) values are now considered with however imaginary once again converted to real format.

Now if we attempt to obtain the sum of roots using this approach negative will exactly cancel positive values with result = 0.

However if we obtain the product of such roots a non-trivial result will emerge which in all cases (where n is odd) can be expressed as 1/{(2^(n - 1)/2} in absolute terms.

For example where n = 5, the five roots of 1 (expressed in this 2-dimensional manner) are (to 9 decimal places),

.309016994 + .951056516 = 1.260073510
- .809016994 + .587785252 = - .221231742
.309016994 - .951056516 = - .642039522
- .809016994 - .587785252 = - 1.396802246
1 + 0 = 1

So when we multiply these 5 roots we obtain - .25.
And as in this case (n - 1)/2 = (5 - 1)/2 = 2, the absolute value of .25 = 1/(2^2) conforms to the general formula.

Though the answer in this case is negative, it can also be positive (as for example where n = 9)

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