### Riemann Products III

It is known that the value of $\zeta (s)$ at $s = 0$ is $\zeta (0) = - 1/2$. On the other hand, recall that the Everything Seminar showed us that

$1 - 1 + 1 - 1 + 1 - 1 + 1 - \cdots = \frac{1}{1 + 1} = \frac{1}{2}$ and

$1 + 1 + 1 + 1 + 1 + 1 + 1 + \cdots = - \frac{1}{2}$

so that the product for $1/ \zeta(s)$ tells us that

$1/ \zeta(0) = 1 - 1[1 + 1 + 1 + \cdots] $

$+ 1[1(1 + 1 + 1 + \cdots) + 1(1 + 1 + 1 + \cdots) + 1(1 + 1 + 1 + \cdots) + \cdots] - \cdots$

$= 1 + \frac{1}{2} + \frac{1}{4} + \cdots = 2$,

which is almost the right answer! What does a minus sign matter, anyway? We can probably fix that. Similarly, the Pauli product expression indicates that

$\frac{1}{\zeta (-1)} = 1 - 2 - 3 - 5 + 6 - 7 + 10 + \cdots = -12$

somehow! At least thinking of complex numbers as infinite sums is much nicer than messing around with standard analysis.

$1 - 1 + 1 - 1 + 1 - 1 + 1 - \cdots = \frac{1}{1 + 1} = \frac{1}{2}$ and

$1 + 1 + 1 + 1 + 1 + 1 + 1 + \cdots = - \frac{1}{2}$

so that the product for $1/ \zeta(s)$ tells us that

$1/ \zeta(0) = 1 - 1[1 + 1 + 1 + \cdots] $

$+ 1[1(1 + 1 + 1 + \cdots) + 1(1 + 1 + 1 + \cdots) + 1(1 + 1 + 1 + \cdots) + \cdots] - \cdots$

$= 1 + \frac{1}{2} + \frac{1}{4} + \cdots = 2$,

which is almost the right answer! What does a minus sign matter, anyway? We can probably fix that. Similarly, the Pauli product expression indicates that

$\frac{1}{\zeta (-1)} = 1 - 2 - 3 - 5 + 6 - 7 + 10 + \cdots = -12$

somehow! At least thinking of complex numbers as infinite sums is much nicer than messing around with standard analysis.

## 0 Comments:

Post a Comment

<< Home