## Thursday, March 28, 2013

### Two Complementary Zeta Functions (7)

We can also connect Euler's famous product formula to the Zeta 2 Function.

Indeed this can provide a new perspective on this formula (which I have not seen developed elsewhere).

For example when s = 2 in the Zeta 1 expression,

ζ1(2) = (1/1)^2 + (1/2)^2 + (1/3)^2 + (1/4)^2 + .......

= 1 + 1/4 + 1/9 + 1/16 + .... = 4/3 * 9/8 * 25/24 * 49/48 *...

So the product terms on the right hand side involve the expression p^2/(p^2 - 1) where p ranges over all the prime numbers!

However this same expression can be shown to be intimately related to the Zeta 2 Function.

Now in illustrating this it will perhaps be easier initially to start with the case corresponding in the Zeta 1 where s = 1.

Here according to Euler's formula

ζ1(1) = 1 + 1/2 + 1/3 + 1/4 = 2/1 * 3/2 * 5/4 * 7/6 * ...

So the summation series on the LHS represents the well known harmonic series while the terms in the product series on the RHS conform to p/(p - 1) where again the value of p ranges over all the prime numbers! However because Euler's formula holds only for convergent series (and these diverge) we have an exception in this case. In fact in truth where s = 1, the R.H.S. conforms to p/(p + 1),

so that

ζ1(1) = 1 + 1/2 + 1/3 + 1/4 ~ 3/2 * 4/3 * 6/5 * 8/7 i.e. (1 + 1/2) * (1 + 1/3) * (1 + 1/5) * (1 + 1/7) *....

However to illustrate our procedures we will illustrate initially by proceeding as if the Euler formula is true for the case s = 1!

As we have already seen the value of the infinite Zeta 2 Function (where s = 1/2)i.e. ζ2(1/2)

= 1 + 1/2 + 1/4 + 1/8 +..... = 2

Therefore the 1st term in the harmonic series = ζ2(1/2) - 1

Likewise the 2nd term in the harmonic series = ζ2(1/3) - 1, the 3rd term ζ2(1/4) - 1, the 4th term ζ2(1/5) - 1 etc.

Now the corresponding terms in the related product formula can be derived in a consistently simple manner with respect to the Zeta 2 expression.

So in each case where s represents the reciprocal of a prime value, we simply divide the value for the Function (representing the sum of an infinite series of terms) by the corresponding first term of the series.

So when s = 1/2 we divide 2 (the sum of the series) by 1 (the 1st term) to obtain 2/1 i.e. the 1st term in the product series

Then when s = 1/3 we divide 3/2 (the sum for the series by a) to obtain 3/2.

Then when s = 1/5 we divide 5/4 (the sum for the series by 1) to obtain 5/4.

Finally to illustrate when s = 1/7 we divide 7/6 (the sum of the series by 1) to obtain 7/6 and so on.

Now because the 1st term is always 1, the process seems somewhat trivial (and as we have seen does not actually apply in this case).

However the procedure assumes much more relevance when we deal with the product values corresponding to higher integer values of s (with respect to the Zeta 1).

So once again

ζ1(2) = 1 + 1/4 + 1/9 + 1/16 + .... = 4/3 * 9/8 * 25/24 * 49/48 *...

Now with respect to the Zeta 2 the additive terms on the LHS correspond now

to {ζ2(1/2) - 1}^2, {ζ2(1/3) - 1}^2, {ζ2(1/4) - 1}^2,{ζ2(1/5) - 1}^2 and so on!

Now this time because each terms is raised to the power of 2, to obtain the corresponding terms in the product formula we divide the sum of the Function (for each value in question) by the corresponding sum of the first two terms in the Function.

So once again where s = 1/2 the value of the Function = 2.

The sum of the corresponding 1st two terms of the Function = 1 + 1/2 = 3/2.

Then when we divide 2 by 3/2 (= 2 * 2/3) we get 4/3 i.e. the 1st term in the product formula!

The other terms in the product formula can be obtained in like manner

So for example when s = 1/3 the sum of the first two terms = 1 + 1/3 = 4/3.

When we divide the sum of the Function 3/2 by 4/3 we obtain 3/2 * 3/4 = 9/8 i.e. the 2nd value in the product formula.

The next relevant value (representing the reciprocal of a prime number) = 1/5.

Thus sum of the first two terms = 1 + 1/5 = 6/5 and the sum of the Function 5/4.

Then dividing 5/4 by 6/5 we get 5/4 * 5/6 = 25/24 i.e. the 3rd term in the product formula.

Finally to illustrate, the next relevant value for the Zeta 2 Function is 1/7

The value of the Function = 7/6 and the sum of first two terms = 1 + 1/7 = 8/7.

So 7/6 divided by 8/7 = 7/6 * 7/8 = 49/48 i.e. the next value in the product formula.

Now this process can be extended indefinitely for higher values of s.

For example when s = 3 (with respect to the Zeta 1 Function), the corresponding terms in additive series of terms in terms of the Zeta 2 will be

ζ2(1/2) - 1}^3, {ζ2(1/3) - 1}^3, {ζ2(1/4) - 1}^3,{ζ2(1/5) - 1}^3 and so on!

Then to get the corresponding terms in the product formula we divide the value of the Zeta 2 Function (ranging again over the reciprocals of all prime numbered values)by the sum of the first 3 terms of the Function in each case.

In this way both the additive and product parts of the Euler Formula can be intimately demonstrated in terms of the Zeta 2 Function (thus indicating its complementary nature to the Zeta 1).

In fact in many ways a more natural fit exists (especially for the multiplication part) in terms of the Zeta 2 expression.

Better notational representation and further illustrations of approach can be found under Section 2 at Two Zeta Functions.