Does anybody know about a "meaning" behind the Euler Formula, what does it really say about the primes?

I know that it is in equation to the zeta function and also how it is derived, but cannot find really anywhere a clear meaning of this function with respect to properties of prime numbers. Does it relate to a certain probability or density?

Recall that the formula is $$\zeta(s) = \prod_{p}^{\infty} \frac{1}{1-p^{-s}}.$$

Thank you for your help.

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    $\begingroup$ Euler formula is a way to encapsulate the Fundamental theorem of arithmetic in a single function. $\endgroup$
    – Siméon
    Apr 9, 2013 at 8:54
  • $\begingroup$ @Ju'x +1 for that. That's exactly what it is. Couldn't have said it better myself. $\endgroup$ Apr 9, 2013 at 8:55
  • 1
    $\begingroup$ easily said "ecapsulate": it holds because the fundamental theorem fo arithmetics hold, however, is it really only another way to express the fundamental theorem? $\endgroup$ Apr 9, 2013 at 9:21
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    $\begingroup$ A possible meaning to this question in terms of understanding it in more than one way is this Mathematica program starting below the words: "Taking the inverse of the logarithm operators matrices, we have the classical Riemann-Euler identity:" which is the second program at this link: tech.groups.yahoo.com/group/Active_Mathematica/message/3317 $\endgroup$ Aug 11, 2013 at 13:05
  • $\begingroup$ thanks Mats. Honestly I am a fan of your number theory programs. $\endgroup$ Aug 11, 2013 at 15:30

1 Answer 1


It is not clear to me what kind of "meaning" you are looking for. However, you specify probabilty, and there is indeed a probabilistic application of the Euler product of $\zeta(s)$ for integers $s>1$.

Heuristically, if $p$ and $q$ are distinct primes, then for a nice uniform probability distribution on the integers (even though it is not possible for a countable set to admit a uniform distribution), we should (or want to) expect to have that "being divisible by $p$" and "being divisible by $q$" are independent events. Now, $n$ integers are coprime if and only if for all primes $p$, they are not all divisible by the number $p$. The probability they are all divisible by $p$ is $1/p^n$, so the probability that $n$ "randomly chosen" integers are coprime is $\prod_p(1-p^{-n})=\zeta(n)^{-1}$.

BCLC edit: A simple way to obtain $\prod_{p\in\mathbb{P}}\frac{1}{1-p^{-s}}=\sum_{n=1}^{\infty}\frac{1}{n^s}$

This can be proven more formally too, under suitable interpretation. We need to specify what we're talking about: look at the probability $n$ integers randomly selected from $\{1,\cdots,N\}$ with uniform probability are coprime, and take the limit as $N\to\infty$. Another way of interpreting the product is as the probability that a randomly selected integer is $n$th-power-free, i.e. it is not divisible by a perfect $n$th power that is not a unit (the units are $\pm1$). In particular, the probability two integers are coprime is the probability an integer is sqaurefree is $6/\pi^2$.

This is not the way I like to think of the Euler product, though. The EP is a way of "factoring" a Dirichlet series, which will itself somehow encode arithmetic information. I think of DS as number-theoretic (even group-theoretic, more on that shortly) analogues of generating functions, which are nifty analytic devices used in combinatorics to succinctly encode an infinitude of counting-type information into a single analytic object, making combinatorics amenable to the techniques of analysis. Indeed, symbolic manipulations often mirror, or track, manipulations that could be done on the level of structured sets, but with less sophistication as a prerequisite.

A generalization (among many, many such) of the Riemann zeta function is the Dedekind zeta function - these things are attached to number fields, which are fields containing $\bf Q$ that are finite-dimensional as $\bf Q$-vector spaces. These zeta functions also admit Euler product factorizations. The Riemann zeta function indeed encodes the fundamental theorem of arithmetic, but in general, number fields (specifically, their rings of integers) do not admit unique factorization. This is why ideals were invented, and in fact the EP of the Dedekind zeta function encodes unique factorization of ideals instead of numbers or integers.

By "twisting" such zeta functions against characters (see Dirichlet and Hecke $L$-functions), we obtain another interesting way of generalizing $\zeta$. These also admit Euler products. They allow a marriage between number theory and Fourier analysis (which is itself to be distinguished from the spectral properties of the zeroes of these functions, a whole other topic). Indeed, each "Euler factor" of $\zeta(s)$ admits a "local" interpretation: $(1-p^{-s})^{-1}$ is the multiplicative Fourier (aka Mellin) transform of the gaussian, and the gaussian is the fixed point of the additive Fourier transform, over the $p$-adics ${\bf Q}_p$. This high-powered stuff is the substance of Tate's thesis.

Zeta functions can now be attached to a large array of different things - fields, graphs, varieties, and even groups (not quite as well-known to pure number-theorists as the others I've listed, I don't think) - and yield useful information about the objects they are attached to. In all cases, the factorization of a zeta function feels like a way of X-raying the internal structure of some object, like a number field, and peeling apart the layers so that it becomes more transparent just how they combine and overlap in the big picture.

  • $\begingroup$ Thanks. This is exactly where my question was going by "meaning". I like both interpretations and believe actually the first is a sub or consequence of the second. I dare combine the first qpproach with the beautiful sentence: Indeed, each "Euler factor" of ζ(s) admits a "local" interpretation: (1−p−s)−1 is the multiplicative Fourier (aka Mellin) transform of the gaussian, and the gaussian is the fixed point of the additive Fourier transform, over the p-adics Qp. This high-powered stuff is the substance of Tate's thesis. $\endgroup$ Apr 9, 2013 at 10:51
  • $\begingroup$ anon, one question: what would happen if we would take in your first interpretation (that means in the powerfree interpretation) not all infinite integer numbers but just a certain range of them? $\endgroup$ Apr 9, 2013 at 14:52
  • $\begingroup$ @al-Hwarizmi Not sure what you're asking. (Also, "infinite integer numbers" doesn't mean anything to me.) The probability an integer in $\{1,\cdots,N\}$ is $n$th-power-free should be about $\sum_{1<m\le N^{1/n}}\frac{\mu(m)}{N}\lfloor N/m^n\rfloor$, which tends to $\zeta(n)^{-1}$ as $n\to\infty$. Note the similarity: the DS representation of $\zeta(n)^{-1}$ is $\sum_{m\ge1}\mu(m)/m^n$, which can be shown easily with the EP factorization. Do these facts clarify what you're wondering about? $\endgroup$
    – anon
    Apr 9, 2013 at 15:28
  • $\begingroup$ You say in your first formula "about" and I think that is the exact answer I was doubtful about; I guess only when n→∞ then a fluctuation/error term diminishes and the first formula tend to the second. Thanks $\endgroup$ Apr 13, 2013 at 8:01

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