2
$\begingroup$

The prime number theorem accords with the following equation for the first Chebychev function that:

$$\lim_{x\rightarrow\infty}\frac{\vartheta(x)}{x}=1 \qquad (1)$$

According to Muñoz García, E. and Pérez Marco, R. "The Product Over All Primes is $4\pi^2$":

$$\prod_{p\in \Bbb P}p=4\pi^2 \qquad (2)$$ hence

$$\sum_{p\in \Bbb P}\log p=\log (4\pi^2)=2\log(2\pi) \qquad (3)$$

Recall

$$\vartheta=\sum_{p\le x}_{p\in \Bbb P} \log p \qquad (4)$$

How can both $(1)$ and $(3)$ be true and do not contradict?

Note: the second Chebychev function $\psi(x)$ could be similarly challenged.

$\endgroup$
  • 3
    $\begingroup$ Note that clearly $2$ and $3$ diverge, these are regularized products/sums, they don't represent the value of the sum/product in the principle sense. The zeta function has numerous representations valid for different domains, as a similar example we know that $$\frac{1}{1-x}=1+x+x^2+x^3..., \text{ for |x|<1}$$ And it would make sense to say that $$-1=\frac{1}{1-2}$$ But not to say $$-1=1+2+2^2+2^3+2^4...$$ But to be honest, I know very little about this kind of thing, though I think some keywords would be "Analytic continuation" and "Zeta function regularization". $\endgroup$ – Ethan Jun 14 '13 at 20:20
6
$\begingroup$

There is nothing contradictory about the two facts. Perhaps at face value it would seem like

$$\sum \log p=\log(4\pi^2)\implies \vartheta\to \log(4\pi^2),$$

which contradicts $\vartheta\sim x\to\infty$ (PNT), but the product-of-primes formula does not actually say that the sum $\sum \log p$ converges to $\log(4\pi^2)$. If you interpreted it this way, the first thing that would seem weird, before even comparing it to PNT, is that $\log p\to\infty$ so the sum is obviously divergent, contradicting the supposed fact that the sum coverges to $\log(4\pi^2)$ (a finite value).

One needs the correct way to interpret the identity "$\prod p=4\pi^2$," since it clearly is not true in the usual calculus sense of limits. In general, there are methods of reinterpreting divergent sums so that they become actual finite values - the keywords here are "regularization" & "summability."

Regularization applies to many types of expressions where infinity or divergent values appear, for example in functional integrals in quantum physics. Often it is applied to sums, and it is done using a phenomena called analytic continuation.

The general set-up is that an expression defines a function but only in a certain region of (say) the complex plane - for example $1+z+z^2+\cdots$ only converges when $|z|<1$ - and identify another function defined on a larger domain which agrees with every value of the original function on the original domain - in our example, that would be the rational function $1/(1-z)$, which is defined for all $z\in{\bf C}\setminus\{1\}$ and equals the infinite geometric series wherever the latter is defined.

One cool family of functions whose analytic continuations are interesting are given by Dirichlet series, ones of the form $L(s)=\sum_{n\ge1}a_nn^{-s}$, and in particular the Riemann zeta function $\zeta(s)$ given by $\sum_{n\ge1}n^{-s}$ for ${\rm Re}(s)>1$ (notice when $s=1$ it is the Harmonic series, which diverges), which has functional equation $\pi^{-(1-s)/2}\Gamma(\frac{1-s}{2})\zeta(1-s)=\pi^{-s/2}\Gamma(\frac{s}{2})\zeta(s)$ allowing $\zeta$ to be defined for all values $s\in{\bf C}\setminus\{1\}$, with a simple pole at $s=1$ with residue $1$.

(The $\pi^{-s/2}\Gamma(\frac{s}{2})$ factor is somewhat mysterious at first - it can be seen as the Euler factor at the so-called prime "$p=\infty$"; this derives from an elegant adelic formulation given by Tate's thesis.)

The actual details of the mathemagical regularization process to obtain $\prod p=4\pi^2$ is given in the cited article. To summarize, Möbius inversion yields

$$\zeta(s)=\sum_{n=1}^\infty\frac{1}{n^s}\iff \sum_{p\rm\,prime}\frac{1}{p^s}=\sum_{n=1}^\infty \frac{\mu(n)}{n}\log\zeta(ns)$$

Differentiate both sides of the second equality and "evaluate at $s=0$" to obtain

$$\sum_{p\rm\,prime}\frac{-\log p}{p^0} \quad ``="\quad \underbrace{\sum_{n=1}^\infty \mu(n)}_{\displaystyle\left.\frac{1}{\zeta(s)}\right|_{s=0}}\frac{\zeta'(0)}{\zeta(0)} \quad ``="\quad \frac{\zeta'(0)}{\zeta(0)^2}=-2\log(2\pi)$$

and hence $\prod p=4\pi^2$ upon exponentiating both sides of the "equality."

Ultimately, the identity is not true in the sense of limits, it is true in the sense of regularized sums and products. The regularization process is to identify the divergent expression as a more general expression evaluated at a point outside its domain, analytically continue that general expression instead and then evaluate at the given point so an actual value is obtained. In general, the value will depend on the choice of "more general expression."

As it happens, slick shortcuts in the regularization process can be obtained by maneuvering in ways that are both technically invalid but conceptually careful and clever, as above. Leonhard Euler is known for these sorts of maneuvers in obtaining finite values for divergent sums. As it happens, in some contexts (where the topology and therefore convergence is very different), the manipulations are actually perfectly technically valid - for example $1-2+4-8+\cdots=1/3$ in the $2$-adics.

The fact that the phenomena of analytic continuation exists and can be harnessed to regularize divergent sums and products seems almost mystical. In fact, contributing even further to the mystery, regularization provides empirically correct numerical values in physics sometimes (for instance, see the Casimir effect). And, according to my understanding, in theoretical physics the mathematical justification for why (zeta-)regularization works out "correctly" is still unknown.

$\endgroup$
  • $\begingroup$ thanks aon. great description. there remains however a sense of mystery and what the intuition behind "regularised product of primes" is. $\endgroup$ – al-Hwarizmi Jun 15 '13 at 8:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.