These have both Type 1 (quantitative) and Type 2 (qualitative) aspects.

Now, the Type 1 fractions can be represented as 5 equidistant points between 0 and (including) 1 on the standard number line.

By contrast the Type 2 fractions can be represented as 5 equidistant points between 0 and (including 1) on the standard unit circle (i.e. with radius 1 unit).

However if we were to stretch out the circumference on which these points lie, the length (in real linear terms) = 2π.

Therefore to convert the distance represented by the Type 2 circular "units" in the standard linear manner, one divides by 2π. This is very relevant for example in interpreting the formula for frequency of occurrence of the Zeta 1 (Riemann) zeros, which we will encounter soon!

So far we have looked on the finite version of the Zeta 2 function, i.e.

1 + s

_{2}

^{1 }+ s

_{2}

^{2 }+ .......+ s

_{2}

^{t }

^{– 1}, especially where t is prime.

The zeros of this function then represent the solutions for,

1 + s

_{2}

^{1 }+ s

_{2}

^{2 }+ .......+ s

_{2}

^{t }

^{– 1 }= 0 which are the "non-trivial" t roots of 1 (where t is prime).

The question then arises as to what value might attach to the infinite function, i.e.

ζ

_{2}(s

_{2}) = 1 + s

_{2}

^{1 }+ s

_{2}

^{2 }+ s

_{2}

^{3 }+ s

_{2}

^{4 }+….

Now the simplest example occurs where we use the "non-trivial" root of 1 in the finite case (where t = 2)

So s

_{2 }= – 1.

Now substituting this value in the infinite Zeta 2 function, we obtain,

1 – 1 + 1 – 1 + 1 – 1....

If we accept that the function has an even number of terms then,

ζ

_{2}(– 1) = 0.

However if the function has an odd number of terms then,

ζ

_{2}(– 1) = 1.

And as an equal probability attaches to the last term being either even or odd, the expected value of the function would be given as the mean average of the two values i.e. (0 + 1)/2 = 1/2.

It then is interesting to surmise as to the expected value of the infinite function for the other "non-trivial" roots of 1 (where t is prime).

For example the two "non-trivial" roots of 1 in the finite case (where t = 3) = – 1/2 + .866 i and

– 1/2 – .866 i respectively.

When we substitute the first value (i.e. – 1/2 + .866 i) in the infinite equation, 3 possible outcomes are possible

When the sequence has 3n + 1 terms (for n = 1, 2, 3,...) its value = 1

When the sequence has 3n + 2 terms, its value = 1/2 + .866 i.

When the sequence has 3n + 3 terms, its value = 0

Then, when we substitute the second value (i.e. – 1/2 – .866 i) in the infinite equation, 3 possible outcomes are again possible.

When the sequence has 3n + 1 terms (for n = 1, 2, 3,...) its value = 1

When the sequence has 3n + 2 terms, its value = 1/2 – .866 i.

When the sequence has 3n + 3 terms, its value = 0

Then to obtain the expected value, we now add up all 6 outcomes, i.e. t(t – 1), and divide by 6 to obtain (1 + 1/2 + .866 i + 0 + 1 + 1/2 – .866 i + 0)/6 = 3/6 = 1/2.

So the same expected value arises as with the "non-trivial" root of 1 for t = 2.

In fact, a very important finding can be stated, that applies for all "non-trivial" roots of 1 is that the expected value = 1/2 in all cases.

So where the values of s

_{2}are given by the "non-trivial" roots of 1 (where t is initially prime) - which represent the Zeta 2 zeros when the function is finite - the value of the corresponding Zeta infinite function = 1/2.

Thus, ζ

_{2}(s

_{2}) = 1 + s

_{2}

^{1 }+ s

_{2}

^{2 }+ s

_{2}

^{3 }+ s

_{2}

^{4 }+…. = 1/2.

Indeed this is - literally - the root reason why it is then postulated that all the Zeta 1 (Riemann) zeros lie on an imaginary line drawn through 1/2.

## No comments:

## Post a Comment